url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://www.jiskha.com/display.cgi?id=1327924231 | 1,498,486,598,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320763.95/warc/CC-MAIN-20170626133830-20170626153830-00452.warc.gz | 560,504,678 | 3,805 | # Physics
posted by .
A force of 1000 N is applied to a 1200 kg car. If the coefficient of friction is 0.04, what is the car's acceleration?
• Physics -
a = Fnet/m
The net force is
Fnet = 1000 - M*g*Uk
= 1000 - 470.4 = 529.6 N
a = 529.6/1200 = 0.441 m/s^2
• Physics -
thank you!
• Physics -
how did you cope in this answer what is the formula and solution? | 127 | 366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.921313 |
http://cn.metamath.org/mpeuni/isinf.html | 1,653,360,009,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00425.warc.gz | 13,550,635 | 11,097 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > isinf Structured version Visualization version GIF version
Theorem isinf 8214
Description: Any set that is not finite is literally infinite, in the sense that it contains subsets of arbitrarily large finite cardinality. (It cannot be proven that the set has countably infinite subsets unless AC is invoked.) The proof does not require the Axiom of Infinity. (Contributed by Mario Carneiro, 15-Jan-2013.)
Assertion
Ref Expression
isinf 𝐴 ∈ Fin → ∀𝑛 ∈ ω ∃𝑥(𝑥𝐴𝑥𝑛))
Distinct variable group: 𝑥,𝐴,𝑛
Proof of Theorem isinf
Dummy variables 𝑓 𝑚 𝑦 𝑧 𝑔 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 breq2 4689 . . . . . 6 (𝑛 = ∅ → (𝑥𝑛𝑥 ≈ ∅))
21anbi2d 740 . . . . 5 (𝑛 = ∅ → ((𝑥𝐴𝑥𝑛) ↔ (𝑥𝐴𝑥 ≈ ∅)))
32exbidv 1890 . . . 4 (𝑛 = ∅ → (∃𝑥(𝑥𝐴𝑥𝑛) ↔ ∃𝑥(𝑥𝐴𝑥 ≈ ∅)))
4 breq2 4689 . . . . . 6 (𝑛 = 𝑚 → (𝑥𝑛𝑥𝑚))
54anbi2d 740 . . . . 5 (𝑛 = 𝑚 → ((𝑥𝐴𝑥𝑛) ↔ (𝑥𝐴𝑥𝑚)))
65exbidv 1890 . . . 4 (𝑛 = 𝑚 → (∃𝑥(𝑥𝐴𝑥𝑛) ↔ ∃𝑥(𝑥𝐴𝑥𝑚)))
7 sseq1 3659 . . . . . . 7 (𝑥 = 𝑦 → (𝑥𝐴𝑦𝐴))
87adantl 481 . . . . . 6 ((𝑛 = suc 𝑚𝑥 = 𝑦) → (𝑥𝐴𝑦𝐴))
9 breq1 4688 . . . . . . 7 (𝑥 = 𝑦 → (𝑥𝑛𝑦𝑛))
10 breq2 4689 . . . . . . 7 (𝑛 = suc 𝑚 → (𝑦𝑛𝑦 ≈ suc 𝑚))
119, 10sylan9bbr 737 . . . . . 6 ((𝑛 = suc 𝑚𝑥 = 𝑦) → (𝑥𝑛𝑦 ≈ suc 𝑚))
128, 11anbi12d 747 . . . . 5 ((𝑛 = suc 𝑚𝑥 = 𝑦) → ((𝑥𝐴𝑥𝑛) ↔ (𝑦𝐴𝑦 ≈ suc 𝑚)))
1312cbvexdva 2319 . . . 4 (𝑛 = suc 𝑚 → (∃𝑥(𝑥𝐴𝑥𝑛) ↔ ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚)))
14 0ss 4005 . . . . . 6 ∅ ⊆ 𝐴
15 0ex 4823 . . . . . . 7 ∅ ∈ V
1615enref 8030 . . . . . 6 ∅ ≈ ∅
17 sseq1 3659 . . . . . . . 8 (𝑥 = ∅ → (𝑥𝐴 ↔ ∅ ⊆ 𝐴))
18 breq1 4688 . . . . . . . 8 (𝑥 = ∅ → (𝑥 ≈ ∅ ↔ ∅ ≈ ∅))
1917, 18anbi12d 747 . . . . . . 7 (𝑥 = ∅ → ((𝑥𝐴𝑥 ≈ ∅) ↔ (∅ ⊆ 𝐴 ∧ ∅ ≈ ∅)))
2015, 19spcev 3331 . . . . . 6 ((∅ ⊆ 𝐴 ∧ ∅ ≈ ∅) → ∃𝑥(𝑥𝐴𝑥 ≈ ∅))
2114, 16, 20mp2an 708 . . . . 5 𝑥(𝑥𝐴𝑥 ≈ ∅)
2221a1i 11 . . . 4 𝐴 ∈ Fin → ∃𝑥(𝑥𝐴𝑥 ≈ ∅))
23 ssdif0 3975 . . . . . . . . . . . . 13 (𝐴𝑥 ↔ (𝐴𝑥) = ∅)
24 eqss 3651 . . . . . . . . . . . . . . 15 (𝑥 = 𝐴 ↔ (𝑥𝐴𝐴𝑥))
25 breq1 4688 . . . . . . . . . . . . . . . . . . 19 (𝑥 = 𝐴 → (𝑥𝑚𝐴𝑚))
2625biimpa 500 . . . . . . . . . . . . . . . . . 18 ((𝑥 = 𝐴𝑥𝑚) → 𝐴𝑚)
27 rspe 3032 . . . . . . . . . . . . . . . . . 18 ((𝑚 ∈ ω ∧ 𝐴𝑚) → ∃𝑚 ∈ ω 𝐴𝑚)
2826, 27sylan2 490 . . . . . . . . . . . . . . . . 17 ((𝑚 ∈ ω ∧ (𝑥 = 𝐴𝑥𝑚)) → ∃𝑚 ∈ ω 𝐴𝑚)
29 isfi 8021 . . . . . . . . . . . . . . . . 17 (𝐴 ∈ Fin ↔ ∃𝑚 ∈ ω 𝐴𝑚)
3028, 29sylibr 224 . . . . . . . . . . . . . . . 16 ((𝑚 ∈ ω ∧ (𝑥 = 𝐴𝑥𝑚)) → 𝐴 ∈ Fin)
3130expcom 450 . . . . . . . . . . . . . . 15 ((𝑥 = 𝐴𝑥𝑚) → (𝑚 ∈ ω → 𝐴 ∈ Fin))
3224, 31sylanbr 489 . . . . . . . . . . . . . 14 (((𝑥𝐴𝐴𝑥) ∧ 𝑥𝑚) → (𝑚 ∈ ω → 𝐴 ∈ Fin))
3332ex 449 . . . . . . . . . . . . 13 ((𝑥𝐴𝐴𝑥) → (𝑥𝑚 → (𝑚 ∈ ω → 𝐴 ∈ Fin)))
3423, 33sylan2br 492 . . . . . . . . . . . 12 ((𝑥𝐴 ∧ (𝐴𝑥) = ∅) → (𝑥𝑚 → (𝑚 ∈ ω → 𝐴 ∈ Fin)))
3534expcom 450 . . . . . . . . . . 11 ((𝐴𝑥) = ∅ → (𝑥𝐴 → (𝑥𝑚 → (𝑚 ∈ ω → 𝐴 ∈ Fin))))
36353impd 1303 . . . . . . . . . 10 ((𝐴𝑥) = ∅ → ((𝑥𝐴𝑥𝑚𝑚 ∈ ω) → 𝐴 ∈ Fin))
3736com12 32 . . . . . . . . 9 ((𝑥𝐴𝑥𝑚𝑚 ∈ ω) → ((𝐴𝑥) = ∅ → 𝐴 ∈ Fin))
3837con3d 148 . . . . . . . 8 ((𝑥𝐴𝑥𝑚𝑚 ∈ ω) → (¬ 𝐴 ∈ Fin → ¬ (𝐴𝑥) = ∅))
39 bren 8006 . . . . . . . . . . 11 (𝑥𝑚 ↔ ∃𝑓 𝑓:𝑥1-1-onto𝑚)
40 neq0 3963 . . . . . . . . . . . . . . 15 (¬ (𝐴𝑥) = ∅ ↔ ∃𝑧 𝑧 ∈ (𝐴𝑥))
41 eldifi 3765 . . . . . . . . . . . . . . . . . . . . . 22 (𝑧 ∈ (𝐴𝑥) → 𝑧𝐴)
4241snssd 4372 . . . . . . . . . . . . . . . . . . . . 21 (𝑧 ∈ (𝐴𝑥) → {𝑧} ⊆ 𝐴)
43 unss 3820 . . . . . . . . . . . . . . . . . . . . . 22 ((𝑥𝐴 ∧ {𝑧} ⊆ 𝐴) ↔ (𝑥 ∪ {𝑧}) ⊆ 𝐴)
4443biimpi 206 . . . . . . . . . . . . . . . . . . . . 21 ((𝑥𝐴 ∧ {𝑧} ⊆ 𝐴) → (𝑥 ∪ {𝑧}) ⊆ 𝐴)
4542, 44sylan2 490 . . . . . . . . . . . . . . . . . . . 20 ((𝑥𝐴𝑧 ∈ (𝐴𝑥)) → (𝑥 ∪ {𝑧}) ⊆ 𝐴)
4645ad2ant2r 798 . . . . . . . . . . . . . . . . . . 19 (((𝑥𝐴𝑓:𝑥1-1-onto𝑚) ∧ (𝑧 ∈ (𝐴𝑥) ∧ 𝑚 ∈ ω)) → (𝑥 ∪ {𝑧}) ⊆ 𝐴)
47 vex 3234 . . . . . . . . . . . . . . . . . . . . . . . 24 𝑧 ∈ V
48 vex 3234 . . . . . . . . . . . . . . . . . . . . . . . 24 𝑚 ∈ V
4947, 48f1osn 6214 . . . . . . . . . . . . . . . . . . . . . . 23 {⟨𝑧, 𝑚⟩}:{𝑧}–1-1-onto→{𝑚}
5049jctr 564 . . . . . . . . . . . . . . . . . . . . . 22 (𝑓:𝑥1-1-onto𝑚 → (𝑓:𝑥1-1-onto𝑚 ∧ {⟨𝑧, 𝑚⟩}:{𝑧}–1-1-onto→{𝑚}))
51 eldifn 3766 . . . . . . . . . . . . . . . . . . . . . . . 24 (𝑧 ∈ (𝐴𝑥) → ¬ 𝑧𝑥)
52 disjsn 4278 . . . . . . . . . . . . . . . . . . . . . . . 24 ((𝑥 ∩ {𝑧}) = ∅ ↔ ¬ 𝑧𝑥)
5351, 52sylibr 224 . . . . . . . . . . . . . . . . . . . . . . 23 (𝑧 ∈ (𝐴𝑥) → (𝑥 ∩ {𝑧}) = ∅)
54 nnord 7115 . . . . . . . . . . . . . . . . . . . . . . . 24 (𝑚 ∈ ω → Ord 𝑚)
55 orddisj 5800 . . . . . . . . . . . . . . . . . . . . . . . 24 (Ord 𝑚 → (𝑚 ∩ {𝑚}) = ∅)
5654, 55syl 17 . . . . . . . . . . . . . . . . . . . . . . 23 (𝑚 ∈ ω → (𝑚 ∩ {𝑚}) = ∅)
5753, 56anim12i 589 . . . . . . . . . . . . . . . . . . . . . 22 ((𝑧 ∈ (𝐴𝑥) ∧ 𝑚 ∈ ω) → ((𝑥 ∩ {𝑧}) = ∅ ∧ (𝑚 ∩ {𝑚}) = ∅))
58 f1oun 6194 . . . . . . . . . . . . . . . . . . . . . 22 (((𝑓:𝑥1-1-onto𝑚 ∧ {⟨𝑧, 𝑚⟩}:{𝑧}–1-1-onto→{𝑚}) ∧ ((𝑥 ∩ {𝑧}) = ∅ ∧ (𝑚 ∩ {𝑚}) = ∅)) → (𝑓 ∪ {⟨𝑧, 𝑚⟩}):(𝑥 ∪ {𝑧})–1-1-onto→(𝑚 ∪ {𝑚}))
5950, 57, 58syl2an 493 . . . . . . . . . . . . . . . . . . . . 21 ((𝑓:𝑥1-1-onto𝑚 ∧ (𝑧 ∈ (𝐴𝑥) ∧ 𝑚 ∈ ω)) → (𝑓 ∪ {⟨𝑧, 𝑚⟩}):(𝑥 ∪ {𝑧})–1-1-onto→(𝑚 ∪ {𝑚}))
60 df-suc 5767 . . . . . . . . . . . . . . . . . . . . . . 23 suc 𝑚 = (𝑚 ∪ {𝑚})
61 f1oeq3 6167 . . . . . . . . . . . . . . . . . . . . . . 23 (suc 𝑚 = (𝑚 ∪ {𝑚}) → ((𝑓 ∪ {⟨𝑧, 𝑚⟩}):(𝑥 ∪ {𝑧})–1-1-onto→suc 𝑚 ↔ (𝑓 ∪ {⟨𝑧, 𝑚⟩}):(𝑥 ∪ {𝑧})–1-1-onto→(𝑚 ∪ {𝑚})))
6260, 61ax-mp 5 . . . . . . . . . . . . . . . . . . . . . 22 ((𝑓 ∪ {⟨𝑧, 𝑚⟩}):(𝑥 ∪ {𝑧})–1-1-onto→suc 𝑚 ↔ (𝑓 ∪ {⟨𝑧, 𝑚⟩}):(𝑥 ∪ {𝑧})–1-1-onto→(𝑚 ∪ {𝑚}))
63 vex 3234 . . . . . . . . . . . . . . . . . . . . . . . . 25 𝑓 ∈ V
64 snex 4938 . . . . . . . . . . . . . . . . . . . . . . . . 25 {⟨𝑧, 𝑚⟩} ∈ V
6563, 64unex 6998 . . . . . . . . . . . . . . . . . . . . . . . 24 (𝑓 ∪ {⟨𝑧, 𝑚⟩}) ∈ V
66 f1oeq1 6165 . . . . . . . . . . . . . . . . . . . . . . . 24 (𝑔 = (𝑓 ∪ {⟨𝑧, 𝑚⟩}) → (𝑔:(𝑥 ∪ {𝑧})–1-1-onto→suc 𝑚 ↔ (𝑓 ∪ {⟨𝑧, 𝑚⟩}):(𝑥 ∪ {𝑧})–1-1-onto→suc 𝑚))
6765, 66spcev 3331 . . . . . . . . . . . . . . . . . . . . . . 23 ((𝑓 ∪ {⟨𝑧, 𝑚⟩}):(𝑥 ∪ {𝑧})–1-1-onto→suc 𝑚 → ∃𝑔 𝑔:(𝑥 ∪ {𝑧})–1-1-onto→suc 𝑚)
68 bren 8006 . . . . . . . . . . . . . . . . . . . . . . 23 ((𝑥 ∪ {𝑧}) ≈ suc 𝑚 ↔ ∃𝑔 𝑔:(𝑥 ∪ {𝑧})–1-1-onto→suc 𝑚)
6967, 68sylibr 224 . . . . . . . . . . . . . . . . . . . . . 22 ((𝑓 ∪ {⟨𝑧, 𝑚⟩}):(𝑥 ∪ {𝑧})–1-1-onto→suc 𝑚 → (𝑥 ∪ {𝑧}) ≈ suc 𝑚)
7062, 69sylbir 225 . . . . . . . . . . . . . . . . . . . . 21 ((𝑓 ∪ {⟨𝑧, 𝑚⟩}):(𝑥 ∪ {𝑧})–1-1-onto→(𝑚 ∪ {𝑚}) → (𝑥 ∪ {𝑧}) ≈ suc 𝑚)
7159, 70syl 17 . . . . . . . . . . . . . . . . . . . 20 ((𝑓:𝑥1-1-onto𝑚 ∧ (𝑧 ∈ (𝐴𝑥) ∧ 𝑚 ∈ ω)) → (𝑥 ∪ {𝑧}) ≈ suc 𝑚)
7271adantll 750 . . . . . . . . . . . . . . . . . . 19 (((𝑥𝐴𝑓:𝑥1-1-onto𝑚) ∧ (𝑧 ∈ (𝐴𝑥) ∧ 𝑚 ∈ ω)) → (𝑥 ∪ {𝑧}) ≈ suc 𝑚)
73 vex 3234 . . . . . . . . . . . . . . . . . . . . 21 𝑥 ∈ V
74 snex 4938 . . . . . . . . . . . . . . . . . . . . 21 {𝑧} ∈ V
7573, 74unex 6998 . . . . . . . . . . . . . . . . . . . 20 (𝑥 ∪ {𝑧}) ∈ V
76 sseq1 3659 . . . . . . . . . . . . . . . . . . . . 21 (𝑦 = (𝑥 ∪ {𝑧}) → (𝑦𝐴 ↔ (𝑥 ∪ {𝑧}) ⊆ 𝐴))
77 breq1 4688 . . . . . . . . . . . . . . . . . . . . 21 (𝑦 = (𝑥 ∪ {𝑧}) → (𝑦 ≈ suc 𝑚 ↔ (𝑥 ∪ {𝑧}) ≈ suc 𝑚))
7876, 77anbi12d 747 . . . . . . . . . . . . . . . . . . . 20 (𝑦 = (𝑥 ∪ {𝑧}) → ((𝑦𝐴𝑦 ≈ suc 𝑚) ↔ ((𝑥 ∪ {𝑧}) ⊆ 𝐴 ∧ (𝑥 ∪ {𝑧}) ≈ suc 𝑚)))
7975, 78spcev 3331 . . . . . . . . . . . . . . . . . . 19 (((𝑥 ∪ {𝑧}) ⊆ 𝐴 ∧ (𝑥 ∪ {𝑧}) ≈ suc 𝑚) → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚))
8046, 72, 79syl2anc 694 . . . . . . . . . . . . . . . . . 18 (((𝑥𝐴𝑓:𝑥1-1-onto𝑚) ∧ (𝑧 ∈ (𝐴𝑥) ∧ 𝑚 ∈ ω)) → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚))
8180expcom 450 . . . . . . . . . . . . . . . . 17 ((𝑧 ∈ (𝐴𝑥) ∧ 𝑚 ∈ ω) → ((𝑥𝐴𝑓:𝑥1-1-onto𝑚) → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚)))
8281ex 449 . . . . . . . . . . . . . . . 16 (𝑧 ∈ (𝐴𝑥) → (𝑚 ∈ ω → ((𝑥𝐴𝑓:𝑥1-1-onto𝑚) → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚))))
8382exlimiv 1898 . . . . . . . . . . . . . . 15 (∃𝑧 𝑧 ∈ (𝐴𝑥) → (𝑚 ∈ ω → ((𝑥𝐴𝑓:𝑥1-1-onto𝑚) → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚))))
8440, 83sylbi 207 . . . . . . . . . . . . . 14 (¬ (𝐴𝑥) = ∅ → (𝑚 ∈ ω → ((𝑥𝐴𝑓:𝑥1-1-onto𝑚) → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚))))
8584com13 88 . . . . . . . . . . . . 13 ((𝑥𝐴𝑓:𝑥1-1-onto𝑚) → (𝑚 ∈ ω → (¬ (𝐴𝑥) = ∅ → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚))))
8685expcom 450 . . . . . . . . . . . 12 (𝑓:𝑥1-1-onto𝑚 → (𝑥𝐴 → (𝑚 ∈ ω → (¬ (𝐴𝑥) = ∅ → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚)))))
8786exlimiv 1898 . . . . . . . . . . 11 (∃𝑓 𝑓:𝑥1-1-onto𝑚 → (𝑥𝐴 → (𝑚 ∈ ω → (¬ (𝐴𝑥) = ∅ → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚)))))
8839, 87sylbi 207 . . . . . . . . . 10 (𝑥𝑚 → (𝑥𝐴 → (𝑚 ∈ ω → (¬ (𝐴𝑥) = ∅ → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚)))))
8988com12 32 . . . . . . . . 9 (𝑥𝐴 → (𝑥𝑚 → (𝑚 ∈ ω → (¬ (𝐴𝑥) = ∅ → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚)))))
90893imp 1275 . . . . . . . 8 ((𝑥𝐴𝑥𝑚𝑚 ∈ ω) → (¬ (𝐴𝑥) = ∅ → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚)))
9138, 90syld 47 . . . . . . 7 ((𝑥𝐴𝑥𝑚𝑚 ∈ ω) → (¬ 𝐴 ∈ Fin → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚)))
92913expia 1286 . . . . . 6 ((𝑥𝐴𝑥𝑚) → (𝑚 ∈ ω → (¬ 𝐴 ∈ Fin → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚))))
9392exlimiv 1898 . . . . 5 (∃𝑥(𝑥𝐴𝑥𝑚) → (𝑚 ∈ ω → (¬ 𝐴 ∈ Fin → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚))))
9493com3l 89 . . . 4 (𝑚 ∈ ω → (¬ 𝐴 ∈ Fin → (∃𝑥(𝑥𝐴𝑥𝑚) → ∃𝑦(𝑦𝐴𝑦 ≈ suc 𝑚))))
953, 6, 13, 22, 94finds2 7136 . . 3 (𝑛 ∈ ω → (¬ 𝐴 ∈ Fin → ∃𝑥(𝑥𝐴𝑥𝑛)))
9695com12 32 . 2 𝐴 ∈ Fin → (𝑛 ∈ ω → ∃𝑥(𝑥𝐴𝑥𝑛)))
9796ralrimiv 2994 1 𝐴 ∈ Fin → ∀𝑛 ∈ ω ∃𝑥(𝑥𝐴𝑥𝑛))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 196 ∧ wa 383 ∧ w3a 1054 = wceq 1523 ∃wex 1744 ∈ wcel 2030 ∀wral 2941 ∃wrex 2942 ∖ cdif 3604 ∪ cun 3605 ∩ cin 3606 ⊆ wss 3607 ∅c0 3948 {csn 4210 ⟨cop 4216 class class class wbr 4685 Ord word 5760 suc csuc 5763 –1-1-onto→wf1o 5925 ωcom 7107 ≈ cen 7994 Fincfn 7997 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1762 ax-4 1777 ax-5 1879 ax-6 1945 ax-7 1981 ax-8 2032 ax-9 2039 ax-10 2059 ax-11 2074 ax-12 2087 ax-13 2282 ax-ext 2631 ax-sep 4814 ax-nul 4822 ax-pow 4873 ax-pr 4936 ax-un 6991 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3or 1055 df-3an 1056 df-tru 1526 df-ex 1745 df-nf 1750 df-sb 1938 df-eu 2502 df-mo 2503 df-clab 2638 df-cleq 2644 df-clel 2647 df-nfc 2782 df-ne 2824 df-ral 2946 df-rex 2947 df-rab 2950 df-v 3233 df-sbc 3469 df-dif 3610 df-un 3612 df-in 3614 df-ss 3621 df-pss 3623 df-nul 3949 df-if 4120 df-pw 4193 df-sn 4211 df-pr 4213 df-tp 4215 df-op 4217 df-uni 4469 df-br 4686 df-opab 4746 df-tr 4786 df-id 5053 df-eprel 5058 df-po 5064 df-so 5065 df-fr 5102 df-we 5104 df-xp 5149 df-rel 5150 df-cnv 5151 df-co 5152 df-dm 5153 df-rn 5154 df-res 5155 df-ima 5156 df-ord 5764 df-on 5765 df-lim 5766 df-suc 5767 df-fun 5928 df-fn 5929 df-f 5930 df-f1 5931 df-fo 5932 df-f1o 5933 df-om 7108 df-en 7998 df-fin 8001 This theorem is referenced by: fineqvlem 8215 isinffi 8856 domtriomlem 9302 ishashinf 13285
Copyright terms: Public domain W3C validator | 7,322 | 10,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2022-21 | latest | en | 0.32455 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/3969/2/a/bc/ | 1,604,133,282,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107916776.80/warc/CC-MAIN-20201031062721-20201031092721-00690.warc.gz | 761,915,920 | 49,992 | # Properties
Label 3969.2.a.bc Level $3969$ Weight $2$ Character orbit 3969.a Self dual yes Analytic conductor $31.693$ Analytic rank $0$ Dimension $5$ CM no Inner twists $1$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$3969 = 3^{4} \cdot 7^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 3969.a (trivial)
## Newform invariants
Self dual: yes Analytic conductor: $$31.6926245622$$ Analytic rank: $$0$$ Dimension: $$5$$ Coefficient field: 5.5.574857.1 Defining polynomial: $$x^{5} - 2 x^{4} - 5 x^{3} + 9 x^{2} + 3 x - 3$$ Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 63) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2,\beta_3,\beta_4$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + \beta_{1} q^{2} + ( 1 + \beta_{2} ) q^{4} + ( 1 + \beta_{4} ) q^{5} + ( 1 + \beta_{3} ) q^{8} +O(q^{10})$$ $$q + \beta_{1} q^{2} + ( 1 + \beta_{2} ) q^{4} + ( 1 + \beta_{4} ) q^{5} + ( 1 + \beta_{3} ) q^{8} + ( 2 + \beta_{3} + \beta_{4} ) q^{10} + ( 1 - \beta_{2} + \beta_{3} ) q^{11} + ( 1 + \beta_{1} - \beta_{4} ) q^{13} + ( -\beta_{2} + \beta_{3} + \beta_{4} ) q^{16} + ( 3 - \beta_{1} + \beta_{2} ) q^{17} + ( -1 + 2 \beta_{1} - \beta_{2} + \beta_{3} - \beta_{4} ) q^{19} + ( 2 + \beta_{2} + 2 \beta_{3} ) q^{20} + ( 1 - \beta_{1} + \beta_{2} + \beta_{4} ) q^{22} + ( 2 \beta_{2} - 2 \beta_{3} - \beta_{4} ) q^{23} + ( \beta_{1} - \beta_{2} + 2 \beta_{4} ) q^{25} + ( 1 + 2 \beta_{1} + \beta_{2} - \beta_{3} - \beta_{4} ) q^{26} + ( 1 + \beta_{1} - 2 \beta_{2} + \beta_{3} ) q^{29} + ( 1 - \beta_{2} + \beta_{3} + \beta_{4} ) q^{31} + ( 1 - 3 \beta_{1} + \beta_{2} - \beta_{3} + 2 \beta_{4} ) q^{32} + ( -2 + 4 \beta_{1} - \beta_{2} + \beta_{3} ) q^{34} + ( -2 \beta_{1} - 2 \beta_{3} ) q^{37} + ( 5 - 2 \beta_{1} + 3 \beta_{2} - \beta_{3} ) q^{38} + ( 1 + \beta_{1} + 2 \beta_{2} + \beta_{3} ) q^{40} + ( \beta_{1} + \beta_{2} - \beta_{3} - 2 \beta_{4} ) q^{41} + ( 3 - 2 \beta_{1} - \beta_{2} + \beta_{3} + 3 \beta_{4} ) q^{43} + ( -2 + \beta_{1} + \beta_{2} + \beta_{4} ) q^{44} + ( -4 + 5 \beta_{1} - 2 \beta_{2} - \beta_{3} - 3 \beta_{4} ) q^{46} + ( 4 + \beta_{1} - 2 \beta_{2} - \beta_{3} - \beta_{4} ) q^{47} + ( 6 - 3 \beta_{1} + \beta_{2} + \beta_{3} + 2 \beta_{4} ) q^{50} + ( 1 + 2 \beta_{1} + \beta_{2} - \beta_{3} ) q^{52} + ( -5 + \beta_{1} - 2 \beta_{4} ) q^{53} + ( \beta_{1} + \beta_{2} - \beta_{3} + \beta_{4} ) q^{55} + ( 3 - 2 \beta_{1} + 2 \beta_{2} - \beta_{3} + \beta_{4} ) q^{58} + ( 6 + \beta_{1} + \beta_{2} + \beta_{3} - \beta_{4} ) q^{59} + ( 2 + \beta_{1} - \beta_{2} - \beta_{3} + \beta_{4} ) q^{61} + ( 3 - 2 \beta_{1} + \beta_{2} + \beta_{3} + 2 \beta_{4} ) q^{62} + ( -6 + \beta_{1} - 2 \beta_{2} - \beta_{4} ) q^{64} + ( -1 - \beta_{1} + \beta_{2} + \beta_{3} + \beta_{4} ) q^{65} + ( -1 + 4 \beta_{1} + 2 \beta_{2} - \beta_{3} + \beta_{4} ) q^{67} + ( 7 - 2 \beta_{1} + 3 \beta_{2} + \beta_{4} ) q^{68} + ( -2 + 2 \beta_{1} - \beta_{2} - 3 \beta_{3} - 2 \beta_{4} ) q^{71} + ( -4 + \beta_{2} - 3 \beta_{3} ) q^{73} + ( -10 + 2 \beta_{1} - 4 \beta_{2} - 2 \beta_{3} - 2 \beta_{4} ) q^{74} + ( -3 + 5 \beta_{1} - \beta_{2} + \beta_{4} ) q^{76} + ( 3 - 4 \beta_{1} + 2 \beta_{2} + \beta_{4} ) q^{79} + ( 3 + 2 \beta_{1} - \beta_{3} + \beta_{4} ) q^{80} + ( -2 + 4 \beta_{1} - 2 \beta_{3} - 3 \beta_{4} ) q^{82} + ( 1 + 2 \beta_{2} - 4 \beta_{3} + 2 \beta_{4} ) q^{83} + ( 2 + \beta_{2} + \beta_{3} + \beta_{4} ) q^{85} + ( 1 - 2 \beta_{1} - \beta_{2} + 3 \beta_{3} + 4 \beta_{4} ) q^{86} + ( 4 - \beta_{2} + 2 \beta_{3} - \beta_{4} ) q^{88} + ( 7 - 2 \beta_{1} + 2 \beta_{3} - \beta_{4} ) q^{89} + ( 5 - 2 \beta_{1} - 2 \beta_{3} - 2 \beta_{4} ) q^{92} + ( -3 + 4 \beta_{1} - 4 \beta_{3} - 2 \beta_{4} ) q^{94} + ( -2 + 2 \beta_{2} + \beta_{3} ) q^{95} + ( 2 + \beta_{1} + 4 \beta_{2} - \beta_{3} - 2 \beta_{4} ) q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$5q + 2q^{2} + 4q^{4} + 4q^{5} + 3q^{8} + O(q^{10})$$ $$5q + 2q^{2} + 4q^{4} + 4q^{5} + 3q^{8} + 7q^{10} + 4q^{11} + 8q^{13} - 2q^{16} + 12q^{17} - q^{19} + 5q^{20} + q^{22} + 3q^{23} + q^{25} + 11q^{26} + 7q^{29} + 3q^{31} - 2q^{32} - 3q^{34} + 20q^{38} + 3q^{40} + 5q^{41} + 7q^{43} - 10q^{44} - 3q^{46} + 27q^{47} + 19q^{50} + 10q^{52} - 21q^{53} + 2q^{55} + 10q^{58} + 30q^{59} + 14q^{61} + 6q^{62} - 25q^{64} - 11q^{65} + 2q^{67} + 27q^{68} + 3q^{71} - 15q^{73} - 36q^{74} - 5q^{76} + 4q^{79} + 20q^{80} + 5q^{82} + 9q^{83} + 6q^{85} - 8q^{86} + 18q^{88} + 28q^{89} + 27q^{92} + 3q^{94} - 14q^{95} + 12q^{97} + O(q^{100})$$
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{5} - 2 x^{4} - 5 x^{3} + 9 x^{2} + 3 x - 3$$:
$$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$\nu$$ $$\beta_{2}$$ $$=$$ $$\nu^{2} - 3$$ $$\beta_{3}$$ $$=$$ $$\nu^{3} - 4 \nu - 1$$ $$\beta_{4}$$ $$=$$ $$\nu^{4} - \nu^{3} - 5 \nu^{2} + 4 \nu + 2$$
$$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$\beta_{1}$$ $$\nu^{2}$$ $$=$$ $$\beta_{2} + 3$$ $$\nu^{3}$$ $$=$$ $$\beta_{3} + 4 \beta_{1} + 1$$ $$\nu^{4}$$ $$=$$ $$\beta_{4} + \beta_{3} + 5 \beta_{2} + 14$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
−2.05365 −0.670333 0.495868 1.84124 2.38687
−2.05365 0 2.21746 0.146246 0 0 −0.446582 0 −0.300337
1.2 −0.670333 0 −1.55065 −1.42494 0 0 2.38012 0 0.955182
1.3 0.495868 0 −1.75411 3.69258 0 0 −1.86155 0 1.83103
1.4 1.84124 0 1.39017 −1.33475 0 0 −1.12285 0 −2.45760
1.5 2.38687 0 3.69714 2.92087 0 0 4.05086 0 6.97172
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 1.5 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$3$$ $$-1$$
$$7$$ $$1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 3969.2.a.bc 5
3.b odd 2 1 3969.2.a.z 5
7.b odd 2 1 3969.2.a.bb 5
7.c even 3 2 567.2.e.e 10
9.c even 3 2 1323.2.f.e 10
9.d odd 6 2 441.2.f.e 10
21.c even 2 1 3969.2.a.ba 5
21.h odd 6 2 567.2.e.f 10
63.g even 3 2 189.2.g.b 10
63.h even 3 2 189.2.h.b 10
63.i even 6 2 441.2.h.f 10
63.j odd 6 2 63.2.h.b yes 10
63.k odd 6 2 1323.2.g.f 10
63.l odd 6 2 1323.2.f.f 10
63.n odd 6 2 63.2.g.b 10
63.o even 6 2 441.2.f.f 10
63.s even 6 2 441.2.g.f 10
63.t odd 6 2 1323.2.h.f 10
252.o even 6 2 1008.2.t.i 10
252.u odd 6 2 3024.2.q.i 10
252.bb even 6 2 1008.2.q.i 10
252.bl odd 6 2 3024.2.t.i 10
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
63.2.g.b 10 63.n odd 6 2
63.2.h.b yes 10 63.j odd 6 2
189.2.g.b 10 63.g even 3 2
189.2.h.b 10 63.h even 3 2
441.2.f.e 10 9.d odd 6 2
441.2.f.f 10 63.o even 6 2
441.2.g.f 10 63.s even 6 2
441.2.h.f 10 63.i even 6 2
567.2.e.e 10 7.c even 3 2
567.2.e.f 10 21.h odd 6 2
1008.2.q.i 10 252.bb even 6 2
1008.2.t.i 10 252.o even 6 2
1323.2.f.e 10 9.c even 3 2
1323.2.f.f 10 63.l odd 6 2
1323.2.g.f 10 63.k odd 6 2
1323.2.h.f 10 63.t odd 6 2
3024.2.q.i 10 252.u odd 6 2
3024.2.t.i 10 252.bl odd 6 2
3969.2.a.z 5 3.b odd 2 1
3969.2.a.ba 5 21.c even 2 1
3969.2.a.bb 5 7.b odd 2 1
3969.2.a.bc 5 1.a even 1 1 trivial
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(3969))$$:
$$T_{2}^{5} - 2 T_{2}^{4} - 5 T_{2}^{3} + 9 T_{2}^{2} + 3 T_{2} - 3$$ $$T_{5}^{5} - 4 T_{5}^{4} - 5 T_{5}^{3} + 18 T_{5}^{2} + 18 T_{5} - 3$$ $$T_{11}^{5} - 4 T_{11}^{4} - 8 T_{11}^{3} + 15 T_{11}^{2} + 12 T_{11} - 15$$ $$T_{13}^{5} - 8 T_{13}^{4} + 13 T_{13}^{3} + 13 T_{13}^{2} - 23 T_{13} - 5$$ | 4,163 | 7,996 | {"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} | 2.734375 | 3 | CC-MAIN-2020-45 | latest | en | 0.401386 |
https://kalkicode.com/deficient-number | 1,657,015,366,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104542759.82/warc/CC-MAIN-20220705083545-20220705113545-00576.warc.gz | 384,833,239 | 12,479 | # Deficient Number
Here given code implementation process.
//C Program
//Check if a given number is deficient or not
#include <stdio.h>
// Function which is calculate sum of divisors in given number
int divisor_sum(int number)
{
//numbers are always on itself divide
int result = number;
for (int i=1; i<= number/2; i++)
{
if (number%i==0)
{
//When number is divide by i
result += i;
}
}
return result;
}
//Function are show the result number is deficient or not
void is_deficient(int number)
{
//Check whether number divisor sum is less than twice given number?
if(divisor_sum(number) < 2 * number)
{
//When Yes
printf("%d Is an Deficient Number\n",number);
}
else
{
//When No
printf("%d Is not a Deficient Number\n",number);
}
}
int main() {
//Test Cases
is_deficient(15);
is_deficient(20);
is_deficient(12);
is_deficient(17);
is_deficient(23);
return 0;
}
#### Output
15 Is an Deficient Number
20 Is not a Deficient Number
12 Is not a Deficient Number
17 Is an Deficient Number
23 Is an Deficient Number
/*
C++ Program
Check if a given number is deficient or not
*/
#include<iostream>
using namespace std;
class MyNumber {
public:
// Function which is calculate sum of divisors in given number
int divisor_sum(int number) {
//numbers are always on itself divide
int result = number;
for (int i = 1; i <= number / 2; i++) {
if (number % i == 0) {
//When number is divide by i
result += i;
}
}
return result;
}
//Function are show the result number is deficient or not
void is_deficient(int number) {
//Check whether number divisor sum is less than twice given number?
if (this->divisor_sum(number) < 2 *number) {
//When Yes
cout << number << " Is an Deficient Number\n";
} else {
//When No
cout << number << " Is not a Deficient Number\n";
}
}
};
int main() {
MyNumber obj ;
// Test Case
obj.is_deficient(15);
obj.is_deficient(20);
obj.is_deficient(12);
obj.is_deficient(17);
obj.is_deficient(23);
return 0;
}
#### Output
15 Is an Deficient Number
20 Is not a Deficient Number
12 Is not a Deficient Number
17 Is an Deficient Number
23 Is an Deficient Number
/*
Java Program
Check if a given number is deficient or not
*/
public class MyNumber {
// Function which is calculate sum of divisors in given number
public int divisor_sum(int number)
{
//numbers are always on itself divide
int result = number;
for (int i=1; i<= number/2; i++)
{
if (number%i==0)
{
//When number is divide by i
result += i;
}
}
return result;
}
//Function are show the result number is deficient or not
public void is_deficient(int number)
{
//Check whether number divisor sum is less than twice given number?
if(divisor_sum(number) < 2 * number)
{
//When Yes
System.out.print(number+" Is an Deficient Number\n");
}
else
{
//When No
System.out.print(number+" Is not a Deficient Number\n");
}
}
public static void main(String[] args) {
MyNumber obj = new MyNumber();
// Test Case
obj.is_deficient(15);
obj.is_deficient(20);
obj.is_deficient(12);
obj.is_deficient(17);
obj.is_deficient(23);
}
}
#### Output
15 Is an Deficient Number
20 Is not a Deficient Number
12 Is not a Deficient Number
17 Is an Deficient Number
23 Is an Deficient Number
/*
C# Program
Check if a given number is deficient or not
*/
using System;
public class MyNumber {
// Function which is calculate sum of divisors in given number
public int divisor_sum(int number) {
//numbers are always on itself divide
int result = number;
for (int i = 1; i <= number / 2; i++) {
if (number % i == 0) {
//When number is divide by i
result += i;
}
}
return result;
}
//Function are show the result number is deficient or not
public void is_deficient(int number) {
//Check whether number divisor sum is less than twice given number?
if (divisor_sum(number) < 2 * number) {
//When Yes
Console.Write(number + " Is an Deficient Number\n");
} else {
//When No
Console.Write(number + " Is not a Deficient Number\n");
}
}
public static void Main(String[] args) {
MyNumber obj = new MyNumber();
// Test Case
obj.is_deficient(15);
obj.is_deficient(20);
obj.is_deficient(12);
obj.is_deficient(17);
obj.is_deficient(23);
}
}
#### Output
15 Is an Deficient Number
20 Is not a Deficient Number
12 Is not a Deficient Number
17 Is an Deficient Number
23 Is an Deficient Number
# Python 3 Program
# Check if a given number is deficient or not
class MyNumber :
# Function which is calculate sum of divisors in given number
def divisor_sum(self, number) :
#numbers are always on itself divide
result = number
i = 1
while (i <= int(number / 2)) :
if (number % i == 0) :
#When number is divide by i
result += i
i += 1
return result
#Function are show the result number is deficient or not
def is_deficient(self, number) :
#Check whether number divisor sum is less than twice given number?
if (self.divisor_sum(number) < 2 * number) :
print(number ," Is an Deficient Number")
else :
print(number ," Is not a Deficient Number")
def main() :
obj = MyNumber()
#Test Case
obj.is_deficient(15)
obj.is_deficient(20)
obj.is_deficient(12)
obj.is_deficient(17)
obj.is_deficient(23)
if __name__ == "__main__":
main()
#### Output
15 Is an Deficient Number
20 Is not a Deficient Number
12 Is not a Deficient Number
17 Is an Deficient Number
23 Is an Deficient Number
# Ruby Program
# Check if a given number is deficient or not
class MyNumber
# Function which is calculate sum of divisors in given number
def divisor_sum(number)
#numbers are always on itself divide
result = number
i = 1
while (i <= number / 2)
if (number % i == 0)
#When number is divide by i
result += i
end
i += 1
end
return result
end
#Function are show the result number is deficient or not
def is_deficient(number)
#Check whether number divisor sum is less than twice given number?
if (self.divisor_sum(number) < 2 * number)
print(number ," Is an Deficient Number\n")
else
print(number ," Is not a Deficient Number\n")
end
end
end
def main()
obj = MyNumber.new()
#Test Case
obj.is_deficient(15)
obj.is_deficient(20)
obj.is_deficient(12)
obj.is_deficient(17)
obj.is_deficient(23)
end
main()
#### Output
15 Is an Deficient Number
20 Is not a Deficient Number
12 Is not a Deficient Number
17 Is an Deficient Number
23 Is an Deficient Number
/*
Scala Program
Check if a given number is deficient or not
*/
class MyNumber {
// Function which is calculate sum of divisors in given number
def divisor_sum(number: Int): Int = {
//numbers are always on itself divide
var result: Int = number;
var i: Int = 1;
while (i <= number / 2) {
if (number % i == 0) {
//When number is divide by i
result += i;
}
i += 1;
}
return result;
}
//Function are show the result number is deficient or not
def is_deficient(number: Int): Unit = {
//Check whether number divisor sum is less than twice given number?
if (this.divisor_sum(number) < 2 * number) {
print(s"\$number Is an Deficient Number\n");
} else {
print(s"\$number Is not a Deficient Number\n");
}
}
}
object Main {
def main(args: Array[String]): Unit = {
var obj: MyNumber = new MyNumber();
//Test Case
obj.is_deficient(15);
obj.is_deficient(20);
obj.is_deficient(12);
obj.is_deficient(17);
obj.is_deficient(23);
}
}
#### Output
15 Is an Deficient Number
20 Is not a Deficient Number
12 Is not a Deficient Number
17 Is an Deficient Number
23 Is an Deficient Number
/*
Swift 4 Program
Check if a given number is deficient or not
*/
class MyNumber {
// Function which is calculate sum of divisors in given number
func divisor_sum(_ number: Int) -> Int {
//numbers are always on itself divide
var result: Int = number;
var i: Int = 1;
while (i <= number / 2) {
if (number % i == 0) {
//When number is divide by i
result += i;
}
i += 1;
}
return result;
}
//Function are show the result number is deficient or not
func is_deficient(_ number: Int) {
//Check whether number divisor sum is less than twice given number?
if (self.divisor_sum(number) < 2 * number) {
print(number ," Is an Deficient Number");
} else {
print(number ," Is not a Deficient Number");
}
}
}
func main() {
let obj: MyNumber = MyNumber();
//Test Case
obj.is_deficient(15);
obj.is_deficient(20);
obj.is_deficient(12);
obj.is_deficient(17);
obj.is_deficient(23);
}
main();
#### Output
15 Is an Deficient Number
20 Is not a Deficient Number
12 Is not a Deficient Number
17 Is an Deficient Number
23 Is an Deficient Number
<?php
/*
Php Program
Check if a given number is deficient or not
*/
class MyNumber {
// Function which is calculate sum of divisors in given number
public function divisor_sum(\$number) {
//numbers are always on itself divide
\$result = \$number;
for (\$i = 1; \$i <= intval(\$number / 2); \$i++) {
if (\$number % \$i == 0) {
//When number is divide by i
\$result += \$i;
}
}
return \$result;
}
//Function are show the result number is deficient or not
public function is_deficient(\$number) {
//Check whether number divisor sum is less than twice given number?
if (\$this->divisor_sum(\$number) < 2 *\$number) {
//When Yes
echo(\$number ." Is an Deficient Number\n");
} else {
//When No
echo(\$number ." Is not a Deficient Number\n");
}
}
};
function main() {
\$obj = new MyNumber();
// Test Case
\$obj->is_deficient(15);
\$obj->is_deficient(20);
\$obj->is_deficient(12);
\$obj->is_deficient(17);
\$obj->is_deficient(23);
}
main();
#### Output
15 Is an Deficient Number
20 Is not a Deficient Number
12 Is not a Deficient Number
17 Is an Deficient Number
23 Is an Deficient Number
/*
Node Js Program
Check if a given number is deficient or not
*/
class MyNumber {
// Function which is calculate sum of divisors in given number
divisor_sum(number) {
//numbers are always on itself divide
var result = number;
for (var i = 1; i <= parseInt(number / 2); i++) {
if (number % i == 0) {
//When number is divide by i
result += i;
}
}
return result;
}
//Function are show the result number is deficient or not
is_deficient(number) {
//Check whether number divisor sum is less than twice given number?
if (this.divisor_sum(number) < 2 *number) {
//When Yes
process.stdout.write(number + " Is an Deficient Number\n");
} else {
//When No
process.stdout.write(number + " Is not a Deficient Number\n");
}
}
}
function main(args) {
var obj = new MyNumber();
// Test Case
obj.is_deficient(15);
obj.is_deficient(20);
obj.is_deficient(12);
obj.is_deficient(17);
obj.is_deficient(23)
}
main();
#### Output
15 Is an Deficient Number
20 Is not a Deficient Number
12 Is not a Deficient Number
17 Is an Deficient Number
23 Is an Deficient Number
## Comment
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible. | 2,871 | 10,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | longest | en | 0.38622 |
https://www.manuscriptedit.com/scholar-hangout/category/medical/ | 1,652,694,582,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00592.warc.gz | 1,010,486,550 | 27,095 | ## Hypothesis Testing
Hypothesis testing is the process of testing validity of a hypothesis or a supposition in relation to a statistical parameter. Hypothesis testing is used by analysts to determine whether or not a hypothesis is reasonable. For example, hypothesis testing could be used to find whether a certain drug is effective or not in treating headache. It uses data from a sample to draw conclusions about a statistical parameter. Hypothesis testing is an important step as it validates statistical parameter which could be used in making conclusions or inference about population or large sample data.
Types of Hypothesis
In data sampling, different types of hypothesis is used to examine whether a sample is positive for test hypothesis or not.
1. Alternative Hypothesis (H1) – This hypothesis states that there is a relationship between two variables (where one variable affects the value of other variable). The relationship that exists between the variables is not due to chance or coincidence.
2. Null Hypothesis (H0) – This hypothesis states that there is no relationship between two variables. It states that the effect of one variable on another is entirely due to chance, with no empirical explanation.
3. Non-Directional Hypothesis – It states that there is a relationship between two variables, but that the direction of influence is unknown.
4. Directional Hypothesis – It states the direction of effect of the relationship between two variables.
Alternative hypothesis and null hypothesis is used to study data samples to find a possible pattern to form a statistical hypothesis that can be validated through hypothetical testing. Alternative hypothesis and Null hypothesis cannot be true at the same time as they are mutually exclusive. Similarly, Non-directional and directional hypothesis cannot be true at the same time as they are mutually exclusive.
Methods of Hypothesis Testing
1. Frequentist Hypothesis Testing- This is the traditional approach to hypothesis testing. It involves making assumptions on current data and comparing prior knowledge about hypothesis with posterior knowledge of the hypothesis to form a conclusion on the hypothesis. One of the subtypes of this approach is Null Hypothesis Significance Testing.
2. Bayesian Hypothesis Testing- It is one of the modern methods of hypothesis testing. In this method prior probability of hypothesis from past data and current data is used to find posterior probability of the hypothesis.
The Bayes factor, which is a key component of this approach, represents the likelihood ratio between the null and alternative hypotheses. This factor indicates the plausibility of either of the two hypotheses formed for hypothesis testing.
Techniques of Hypothesis Testing
There are few commonly used Tests: Z-Test, T-Test, Chi squared Test and F-Test.
1. Z Test- A z test is performed on a population with independent data points that follows a normal distribution and has a sample size of larger than or equal to 30. When the population variance is known, it is used to determine whether the means of two populations are equal. Z test statistic is compared to the crucial value and the null hypothesis of z test is rejected if the z test statistic is statistically significant.
Where,
Z= Z-test
X̄ =sample average
µ=mean
s=standard deviation
1. T Test – A t-test is an inferential statistic that is used to see if there is a significant difference in the means of two groups that are related in some way. This test is also called as Student test. It is used when variables are continuous, sample size is less than 30, and population standard deviation is not known. T statistic is used to arrive at a conclusion on whether to accept the hypothesis or reject the hypothesis.
Where,
t= Student’s t-test
m= mean of sample
µ= assumed mean
s= standard deviation
n= number of observations
1. Chi squared Test – A chi-square statistic is a test that evaluates how well a model matches actual data. For using Chi squared test the data used must be random, mutually exclusive, taken from independent variables from a large sample.
where:
c=Degrees of freedom
O=Observed value(s)
E=Expected value(s)
There are two types of χ2 test – the test of independence, and goodness-of-fit test. A χ2 test for independence can show us how likely it is that random chance can explain any observed difference between the actual frequencies in the data and these theoretical expectations.
1. F Test – Any statistical test with an F-distribution under the null hypothesis is known as an F-test. It is generally used to compare statistical models that have been fitted to a data set to find which model best fits the population from which the data were sampled. To perform an F-test, the population must have an f distribution and the samples must be random. If the f test findings are statistically significant, the null hypothesis is rejected otherwise, it is not. F statistic for large samples:
Where,
σ1= variance of the first population
σ22 = variance of the second population
## 5 Steps for Publishing in a High-Quality Medical Journal
Scientific writing and publishing is a vital component of medical advancement. Publications are used to transmit new developments in human knowledge to the rest of the world. This knowledge must be accurate, valid, reproducible, and clinically valuable. Many ambitious physicians and scientists aspire to publish their work in high-impact publications.
What are the Effective Steps for Publishing in a High-Quality Medical Journal?
Choose a Journal and Read the Journal’s Instructions.
It is critical to decide on authorship and the order of authors, including the corresponding author, ahead of time. 6 All authors listed on the final manuscript must have contributed substantially to the work to be held accountable and accept public accountability for the publication’s content. When preparing your paper, carefully observe the author guidelines, including the word limit and the number of tables and figures allowed. Many journals permit the submission of supplemental material as part of your publication, subject to the word count and figures/tables constraints.
Prepare the Manuscript
• Organize the Manuscript – Begin by outlining the manuscript with this basic structure in mind. The first rough draft would be a list of crucial points to describe beneath each section and subsection.
• Prepare the Manuscript – Pay attention when drafting the manuscript to avoid plagiarism (including self-plagiarism), fraud, and fabrication.
• Colleagues should provide feedback and revise the manuscript – After completing the work, share it with co-authors and one or two non-author colleagues for criticism and feedback. Address and correct the English after revising to verify that your idea or ideas have been appropriately and fully communicated.
Submit the Manuscript
The majority of scientific journals now demand submissions be made through their websites. Most journals allow authors to propose reviewers who should and should not evaluate an article during the manuscript submission process. Such comments are beneficial to the editorial staff of the journals. Choose your title and keywords carefully so that readers can discover your paper. A brief cover letter that includes a two- to a three-sentence overview of the manuscript’s relevance might provide essential context to the editor.
Receive the Editor’s Communication and Revise the Manuscript
Acceptance of a paper as submitted is extremely rare. Do not be offended if your manuscript is rejected. In reality, only about one in every four articles is approved by a top journal. If adjustments are requested, it is critical to provide a thoughtful and respectful response to maximize the likelihood of later acceptance.
Resubmit the Manuscript
Include a cover letter to the editor with point-by-point responses to the reviewers’ remarks and ideas when submitting a revised manuscript, and address all complaints and suggestions as extensively as possible. It would help to highlight the Changes in the updated text to evaluate thoroughly.
Conclusion
To summarise, the process of publishing a manuscript in a high-impact journal begins with selecting an important question, designing a sound study with statistical power, carrying out the work with impeccable integrity and attention to detail, writing an excellent manuscript, submitting it to the appropriate journal, fully responding to reviewer comments, and completing the standard post-acceptance checks. Nothing beats the satisfaction of seeing your paper published and visible to the rest of the world.
## Medical Grant Proposal Writing
What is a Grant Proposal Writing?
A Grant Proposal Writing is a document or collection of documents submitted to an agency for the express purpose of obtaining funding for a research project. Successful grant applications are the result of a lengthy process that starts with an idea. Grant Proposal Writing is a circular process, although many people think of it as a linear process (from concept to proposal to award).
What are the Steps in Grant Proposal Writing?
Formulate a Research Question
Many people begin by formulating a research proposal. This is simpler if you know what you want to accomplish before you start writing.
• As a direct result of your project, what expertise or information would you gain?
• What is the significance of your study in a wider sense?
• You must make this aim clear to the committee that will be evaluating your application.
Define a Goal
• You must first determine what type of research you will conduct and why before you begin writing your proposal.
• What exactly are you up to, and why are you up to it? Give a reason for your decision.
• Feel free to show some initiative and tackle a dilemma, just make sure you can justify why and then persuade us that you have a good chance of succeeding.
• Demonstrate the approach’s uniqueness by presenting the information void that needs to be filled.
Find Funding Agencies
• Whether or not your plan is funded is largely determined by how well your intent and objectives align with the priorities of awarding agencies.
• Locating potential grantors is a time-consuming process, but it will pay off in the long run.
• Even if you have the most compelling research idea in the world, if you don’t submit it to the right institutions, you’re unlikely to be considered for funding.
• There are a plethora of resources available to learn more about granting agencies and grant programs.
• Most colleges, as well as several schools within universities, have research offices whose primary function is to assist faculty and students with grant applications.
• To assist people with finding potential grants, these offices typically have libraries or resource centers.
Do Internal Review
• Seek the advice of a mentor or a senior colleague for a second opinion. Remember to check the following things –
• Title
• Introduction about the Medical Research
• Problems in the Medical Research
• Objectives
• Preliminary Literature Review
• Research Methodologies
• Research Plan
• Make a Budget Plan
• Reference
• Your proposal must be presented to the funding body as a successful cash reward.
• There must be a strategy for every aspect of the mission.
• Analysts will go over it carefully to ensure that the study’s components are affordable.
• Your application can be executed due to over-costing.
• Consider if the advancement you can make in the field justifies the expense.
Conclusion
Before writing a Research Protocol, identification of Sponsors and Understanding Application Guidelines is vital. Many companies are providing professional grant writing services.
## Management of Multiple Projects In Medical Writing
What are Medical writing services?
Medical writing services deals with Medical Science which includes clinical research reports, content for healthcare blogs, newsletters, newspapers, and news. Writing a research procedure and/or a clinical study report (CSR) for a project is part of regulatory medical writing. In a Clinical Research Organization (CRO), a Regulatory Medical Writer is normally involved at the beginning (writing a study protocol) and/or end (writing a clinical study report (CSR)) of a project. When the team works together to reach critical deadlines at the start and end of a clinical trial, it may be the most difficult time for them.
How to manage multiple projects efficiently in Medical Writing?
Priority to Documents
Determine which documents are the most important and have the greatest effect on a report. Give them priority over those with the smallest effects.
Timeline
Draw a timeline of all of the overlapping tasks. Manage the time efficiently by determining which documents are the most critical (and therefore have the greatest impact in a study) and which can be postponed with the least impact.
Document Completion Process
At the start of the research, explain the entire document completion process to the study team and ask them if any improvements need to be made. Show them how major changes, ad hoc evaluations, and other factors will impact schedules in vital QC time.
CSR Methodology
If the client has finalized the protocol and statistical analysis strategy, write the methods portion of the CSR. Do as much work as possible off the critical path.
Meeting with the Research team
Hold a results analysis meeting with the research team to show them how you expect to present the data while writing CSRs. This will help you understand exactly what the research team is looking for and how it should be portrayed.
Assistance
If deadlines are approaching, make use of your colleagues. As the lead medical writer, you will, for example, write the efficacy findings more accurately when delegating a huge in-text table or the protection section to a colleague.
Discussion
Call the research team for a live meeting and discussion once you’ve finished the first draft of your paper. Any changes in the timeline should be communicated to the QC and peer reviewers. If you want to deliver all of the documents on time, make sure that all of the testers are aware of when they must complete the work, how long it will take, and how urgent the project is.
Feedback
Finally, no matter how much work you have, be frank with your clients; if they have unreasonable standards, work with them to improve them. Wait until you’ve received all of the client’s feedback before responding, and double-check that you know who the signatories are. A consumer is more likely to make changes to ensure that the document is delivered on time
Conclusion
When a team is dealing with several submissions in a short amount of time, or even simultaneous submissions, confusion about timetable organization, Medical Writing Publication can be dealt efficiently with the above steps.
## Medical Writing As a Career
Some of the most rewarding career paths are ones that people never intended to take. This is frequently the case for medical writers who did not consider this flourishing field when they began their careers. Medical writing may be an organic change for people with a medical, life science, or academic background seeking to move away from the lab or healthcare institutions because there are no clear qualifications. Passionate writers will find a rewarding career in this inspiring and unique field. Medical writing is in high demand in the communication, healthcare, and pharmaceutical industries.
What Is Medical Writing?
Medical Writing is communicating Clinical & Scientific Data clearly and effectively.
Scope of Medical Writing
• Clinical Research Documentation
• Content Writer for Health Care
• Article writing for Healthcare Magazines and Journals
Types of Medical Writing
Medical Writing is mainly of two types:
Regulatory and Research Related Writing
• Focused mostly on Medical & Pharmaceutical Industries.
• The main audiences are people performing Clinical Trials, Government Regulatory Authorities.
• The objective is to attain and maintain marketing authorization for new drugs, therapies, medical devices, and other health products.
Medical Communications
• Documents about New Drugs, Therapies, Medical Devices, or other health products.
• Examples are Scientific Manuscripts, Product Monographs, Conference Abstracts, Posters, Educational Material.
What are the Skills Needed to be a Medical Writer?
• Domain Knowledge – Health Care Knowledge
• Data Integrity – Data Analysis
• Clarity – Communication Expertise
• Language Skills – Scientific Expertise
• Good Writing Skills
• Literature / Reference Search – Document Review Process
• Interpretation & Presentation – Document Expertise
• Formatting & Editing
• Reviewing Published Paper
Courses for Medical Writing
Some of the Medical Writing Certification Courses are as follows:
• PG Diploma in Medical Writing – 12 Months
• Professional Diploma in Medical Writing – 4 to 6 Months (Online)
• Advanced PG Diploma in Clinical Research & Writing – 6 to 8 Months
• Certificate Program in Medical Writing – 6 Months
• Free Courses
Job Titles for Medical Writers
• Medical Expert
• Editorial Manager
• Senior Writer
• Associate Writer
• Junior Writer
Roles of Medical Writers
• Proofreads regulatory documents and clinical documents.
• Coordinates with Medical professionals.
• Helps to write research articles.
• Creates Educational and Informational materials for the Medical kits.
• Writes Research Findings
• Creates Standard Operating Procedures for Medical company’s planning
Where do Medical Writers work?
• Pharmaceutical Sector
• Medical Communication Agencies
• Contract Research Organization
• Journal Publishing
• Biotechnology Sector
• Medical Device Sector
• Medical Translations
Benefits of being a Medical Writer
• Challenging & Stimulating Work ENVIRONMENT
• Always learning new things
• Satisfaction in being an important part of a team
• Lead groups of Multifunctional Professionals
• Opportunities to travel
Conclusion
Medical Writers do Scientific Documentation. Medical Writing involves the above-mentioned skills and offers a good career. People with medical knowledge and writing skills would have a lot of opportunities.
## Benefits of Outsourcing Medical Writing Projects
Outsourcing Medical Writing
Outsourcing scientific and clinical writing assignments lower the staffing needs by lowering the cost of recruiting, training, and retention. Medical device and pharmaceutical companies are increasingly turning to outside vendors to meet their technological and regulatory writing needs for business development.
What are the Stages of Outsourcing?
What And When To Outsource Medical or Scientific Research Paper?
• If a team determines that it will need medical writing assistance, the process of outsourcing medical writing activities should begin with a schedule for what documents will be required and when they will be required.
Finding And Selecting A Medical Writing Company
• Ask the right questions and provide the right information to the Medical Writing Company
Managing The Services Provided
• Make sure the writer is an empowered member of the clinical team should be part of managing the ongoing medical writing
Evaluating Performance At The End Of The Project
• It’s crucial to assess the results at the end of the project before deciding whether to work with the same writing company on future projects.
• Take the opportunity to meet with the main clinical team members and the medical writer after the project’s final activities are completed to share and review everyone’s opinions about how far the partnership went.
What are the Benefits of Outsourcing?
Adaptability – The re-appropriated writings from the Medical or clinical research writing firms have a lot of versatility for those who can’t afford to hire their authors.
Increase in profits – When companies outsource to Medical writing firms, they can save money in a variety of ways by minimizing or removing in-house workers and relying on a team of professional outside authors.
Best to focus on your core competency – Most businesses do not specialize in medical writing. The world of medical devices and pharmaceuticals is rapidly changing. By outsourcing, focus can be directed on Core Competency.
Compact writing services – Medical Writers, skilled editors, specialist statisticians, and staff writers make up a competent medical writing team. As a result, you won’t need to employ many people to do the writing, editing, and presentation of the abstract poster. This will help you save even more money.
Documents that adhere to strict regulatory guidelines – Pharmaceutical companies and clinical testing firms must meet the conditions set out by the respective regulatory authorities to introduce a new drug or perform a clinical trial. Medical writing services make it simple to solve this obstacle by sending documents that meet regulatory requirements.
Effective medical communication – Health content writing necessitates knowledge of medical terminologies, as well as experience and technical writing skills. The rethought writing of medical writing experts helps you share your thoughts easily.
Improvement in your sales and services – Medical device ads, product literature, and service brochures can all be outsourced to professional medical writers who can create eye-catching marketing material and increase sales revenue.
Tips for outsourcing medical content writing
• Clearly describe the need as well as the audience you want to reach.
• Provide detailed indications
• outsourcing entails a collaborative effort.
• Be ethical
Conclusion
Outsourcing medical writing successfully necessitates selecting medical writers who are not only knowledgeable about the requirements of each paper, but also capable coordinators who can push your clinical teams to present a straightforward, well-argued tale.
## Importance Of Meta-analysis In Medical research
Writing articles related to medical research today has to follow certain well-accepted forms of research in order to be accepted. Of the various types of medical research articles, studies based on metadata analysis is one of the most well-accepted forms.
Used & Refurbished Philips Mobile C-Arms can help guide you to equipment and parts that will provide the best value for your dollar.
The concept of meta-analysis stems from the field of statistics. Meta-analysis is a process of combining the results of multiple scientific research related to the same field. Statistically, if there are multiple experiments on the same lines, the results of each are prone to certain degrees of error (for example, the most commonly occurring errors are Type 1 errors related to false-positives or Type 2 errors of false-negatives). However, if all those studies are pooled together, the net derivative result is less likely to have any of such errors and will yield a more definitive result. The key to the aggregation of data is higher statistical power and thereby more robust point estimates than any individual scientific research.
Meta-analysis based medical research started in the 70s and has gained immense popularity ever since. In fact, statistics suggest that meta-analysis based medical articles are the most cited articles in the field.
While meta-analysis generally refers to quantitative studies, there exists another form of studies called statistical meta-synthesis pertaining to integrating results from multiple but related qualitative studies. The approach to statistical meta-synthesis is more interpretive than aggregative and thereby have a different approach altogether. Before embarking on your work, you need to first determine whether you want to go for meta-analysis or statistical meta-synthesis, based on the type of field you are working on.
How to Approach Writing a meta-analysis medical research paper?
Meta-analysis is conducted to assess the strength of evidence, usually on the efficacy of any specific disease or particular treatment. Meta-analysis seeks to collate multiple pieces of evidence to arrive at a conclusion; whether any direct or indirect effect exists, and/or whether such effect is positive or negative (particularly pertaining to a specific type of treatment). This assumes importance as heterogeneity is vital for the development of any new hypothesis.
Key to such study is developing the metadata pool, based on a thorough filing and coding system, proper categorization of data, and thereby identifying the key analytical data- crunching exercise that is expected to yield the best results.
The key to a good meta-analysis exercise is proper identification of the different methods adopted in each exercise and thereby identifying how they may have affected the findings of those exercises. Identifying and mapping methods is also critical given not all methods are comparable and therefore all data may not be compatible.
Tools for meta-analysis
Needless to say, the main tools for such analysis are hardcore statistical tools, mostly software like SAS, STRATA, R. The data is usually pooled from recognized medical research sources like PubMed, Embase, or CENTRAL.
Reporting of results is usually done via the form of Preferred Reporting Items for Systemic Reviews and Meta-analysis or PRISMA.
## All About Medical Case Report
Medical practitioners come across a large number of patients with variations in disease expression and subsequent prognosis. Sometimes a unique non reported disease phenotype or may notice the significant outcome of any innovative combinative treatment. These surgical table details are crucial and need to be recorded to provide direction to both doctors and researchers. However, with the hustling profession of being a medical practitioner, it seems an overwhelming task to report “insights of new and unique observations” in form of full-length paper. Does this mean publishing is all about lengthy research article or review writing? The answer is No, the alternative is a MEDICAL CASE REPORT.
What do we mean by Medical Case reports (MCR)?
It is written communication of previously unknown or a rare disease presentation. Also, MCR can be reporting of de novo treatment strategy of utmost medical significance. Since, its frontline of evidence, therefore, can pave the way towards the development of medical sciences bringing better diagnosis and prognostic strategies.
Components of MCR and their relevance
Similar to any scientific text, MCR too has a specific format to provide clarity and keeping the readers engaged until the end.
• Title
• Abstract: Short and crisp conveying the gist of the article in a structured manner.
• Background: Origin of study with highlights’ of what’s known to date.
• Case Presentation: Physical examination of the apposite prognosis of the subject in chronological order.
• Discussion: Illustration of the uniqueness of case and future significance.
• Conclusion: Take home message of MCR.
• References
• List of Abbreviation
• Consent
• Author information
• Acknowledgment
• Cover Letter
Conclusion: MCR can be the first step to revolutionize the field of medical sciences with heart-throbbing findings of life significance.
## Modulation of life expectancy by weight statistics in young adulthood
Ideal weight not only impacts the overall personality of a person but at the same time, is also the hallmark of good health. Weight management is often goggled in view of its various implications, mostly in the form of grave lifestyle associated disorders.
Obesity implication in early lives:
The obesity statistics in India are alarming, thereby expected to be tripled by 2040, as suggested by the latest research. Narrowing the studies to the early adolescent population worldwide, the figures are a bit scarier.
Contrary, to the age-old, believes that childhood is the “time of liberty” in terms of food choices we make, and cautiousness towards ingredients in the serving plate is applicable only with old age is a misnomer. Junk food, high sugar diet with loads of trans fats, lack of workout, irregular and unhealthy eating habits are making the adolescent population predisposed towards obesity. Primary or secondary obesity prevalence at a younger age has immersed as a deadly threat, closing the survival bracket to just mid-age. Researches have shown that around 12.4% of premature death can be linked to early adulthood associated with obesity.
The timeline of weight management:
A retrospective survey using the BMI (Basal metabolic index) corresponding to the early ages of the participants has shown that the negative implication of obesity on life expectancy remains unaltered. The tough part is that the damage caused due to being obese at a young age can’t be countered by weight reduction in later years. Also, premature death probability remains unchanged in both “presently overweight but previously obese” and “currently overweight” participants. This signifies the irreversible damage incorporated on the cellular dynamics due to early-onset obesity. Subsequently, paving ways to other comorbidities such as cardiovascular disease, diabetes in later lives.
Conclusion:
Weight management awareness at the right age is warranted to make a healthy population with survival fitness for which we recommend to use the best testosterone booster. Inculcation of proper lifestyle habits and realization of gross detrimental effects of obesity at the beginning of early adulthood shall help to build future generations with a superior quality of life.
## The double catastrophe of cardiovascular-pulmonary disorders
Darkness crippled in the lives of family members, with the sudden demise of the only son of the family, aged 38, years undergoing treatment for an interstitial lung disorder. The death report stated cardiac arrest not a pulmonary failure as the cause of untimely demise. This incidence like many more, hints the fine-tuning at the cellular-functional axis, amongst The MOST VITAL organs “Lungs & Heart” to support life.
Pulmonary cardiovascular disease
Both cardiovascular and pulmonary disease is the leading cause of deaths worldwide. Cardiovascular diseases developed in respiratory patients have a high mortality rate apart from affecting day to day lives. For instance, patients with lung fibrosis or COPD are more likely to die due to heart failure as compared to those with a pulmonary issue but no cardiac involvement”. The irony of the story is that the patient with lung disease is less likely to receive coronary revascularization or coronary artery bypass graft. It is due to similar symptoms in both condition and complex management in the pretext of existing lung complications.
The dynamics of Heart Lung Reciprocity
Lungs and heart not only share the thoracic cavity in common but also are functionally interdependent. A load of transporting oxygen laden pure blood is a composite effort of both the organs. This is evident in heart diseases having breathlessness as a hallmark.
Pulmonary disease conditions such as ILD can exert backpressure to heart known as pulmonary arterial hypertension causing right-sided heart failure. Also, conditions like left-sided heart failure, mitral stenosis, myocardial infarction can cause pulmonary edema or waterlog in the lungs due to venous hypertension. Fluid accumulation in air sacs or obstruction in blood flow due to fibrosis leads to a build-up of arterial pressure resulting in morbidity due to heart collapse.
Conclusion
Advancement in medical and surgical intervention has undoubtedly increased life expectancy and strengthened emergency care. However, the major lacuna to date is the prognosis mystery regarding the unrevealing of complex lung disorders. The situation is highly alarming with the involvement of pumping organs, eventually leading to untimely catastrophe, known as DEATH. | 6,036 | 31,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | longest | en | 0.937982 |
http://www.dotnetcharting.com/documentation/vCurrent/dotnetCHARTING~dotnetCHARTING.FinancialEngine~AroonDownOverPeriod(Series,Int32).html | 1,511,575,028,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809229.69/warc/CC-MAIN-20171125013040-20171125033040-00251.warc.gz | 382,027,778 | 3,790 | AroonDownOverPeriod(Series,Int32) Method
Collapse All Expand All Language Filter: All Language Filter: Multiple Language Filter: Visual Basic (Declaration) Language Filter: Visual Basic (Usage) Language Filter: C#
dotnetCHARTING Namespace > FinancialEngine Class > AroonDownOverPeriod Method : AroonDownOverPeriod(Series,Int32) Method
s
A financial series where the first term is the traded high of the asset within the most recent trading period, the second term is the traded high of the asset within the previous period and so on.
lengthOfPeriod
An integer which represents the length of the period over which the indicator will be iteratively evaluated.
Calculates the Aroon Down indicator which measures the relative time since the last lowest low for all periods for which sufficient data is provided.
# Syntax
Visual Basic (Declaration)
Public Overloads Shared Function AroonDownOverPeriod( _
ByVal s As Series, _
ByVal lengthOfPeriod As Integer _
) As Series
Visual Basic (Usage)Copy Code
Dim s As Series
Dim lengthOfPeriod As Integer
Dim value As Series
value = FinancialEngine.AroonDownOverPeriod(s, lengthOfPeriod)
C#
public static Series AroonDownOverPeriod(
Series s,
int lengthOfPeriod
)
#### Parameters
s
A financial series where the first term is the traded high of the asset within the most recent trading period, the second term is the traded high of the asset within the previous period and so on.
lengthOfPeriod
An integer which represents the length of the period over which the indicator will be iteratively evaluated.
# Remarks
The indicator for each period will return a value between 0 and 100, where a higher value indicates that the lowest low was achieved more recently. Persistent values between 70 and 100 are said to indicate weakness in the asset and in conjunction with a low range (i.e. 0-30) in the Aroon Up indicator, indicates an downward trend.
### Evaluation of the Aroon Down Indicator
The Aroon Down indicator for each period is given by:
100(n - (number of days since last highest high over n days))/n
where n is the number of days being considered (a reasonable default value for n is 14). | 481 | 2,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | longest | en | 0.814256 |
https://study.com/academy/practice/quiz-worksheet-solving-functional-problems-with-the-pythagorean-theorem.html | 1,560,777,217,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998475.92/warc/CC-MAIN-20190617123027-20190617145027-00056.warc.gz | 605,997,455 | 27,252 | # How to Solve Functional Problems Involving the Pythagorean Theorem
Instructions:
question 1 of 3
### Which of the following best describes the Pythagorean Theorem?
Create Your Account To Take This Quiz
As a member, you'll also get unlimited access to over 75,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed.
Try it risk-free for 30 days. Cancel anytime.
### 2. How do you use the Pythagorean to find the hypotenuse of a right triangle?
Create your account to access this entire worksheet
Quizzes, practice exams & worksheets
Certificate of Completion
Create an account to get started
This quiz and worksheet reviews what you know about the Pythagorean Theorem and how it is used for solving functional problems. You'll check your knowledge about finding the sides or angles of a triangle with this theorem.
## Quiz & Worksheet Goals
The intention for this quiz and worksheet is to evaluate what you know about:
• Describing the Pythagorean Theorem
• What the Pythagorean Theorem is used for
• How to use the Pythagorean Theorem to find the hypotenuse
• Finding one of the shorter sides of a right triangle
## Skills Practiced
• Information recall - see if you can describe the Pythagorean Theorem
• Interpreting information - verify that you read and were able to correctly interpret how to find the measure of a side of a right triangle
• Knowledge application - use knowledge of the Pythagorean Theorem to answer a question about the information needed to use this theorem | 346 | 1,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-26 | longest | en | 0.888253 |
https://www.physicsforums.com/threads/matlab-summation.648663/ | 1,508,302,490,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822739.16/warc/CC-MAIN-20171018032625-20171018052625-00227.warc.gz | 996,416,002 | 15,296 | # Matlab - Summation
1. Oct 31, 2012
### eurekameh
How can I do the summation of x^2 from 0 to 3 without the use of any built-in functions? I know a for loop is involved, but I can't get it to work.
Last edited by a moderator: Oct 31, 2012
2. Oct 31, 2012
### Simon Bridge
Sounds like an assignment ... the idea is for you to figure it out for yourself.
However: doesn't mean we cannot give you a nudge in the right direction ;)
Show us your best attempt and what happens when you run it.
3. Nov 1, 2012
### eurekameh
Lol, it's not an assignment per se, but I still need this piece of code.
I tried:
i = 0:1:3
x = i.^2
end
but this just uses each i = 1, i = 2, i = 3 to compute x = i^2. I'm looking for a way to somehow store each xi and add them to form the total summation.
4. Nov 1, 2012
### NemoReally
I'm not a Matlab user, but does x already store each i^2? There are a couple of ways you could do this: using a loop or a vectorized approach.
With a loop, look up 'for' and 'while' - essentially, you initialize the variable that you want to store the sum, then within a loop calculate each power and add it to the sum variable.
Alternatively, create a vector of '1's the same length as x and multiply them together vectorwise. ... the built-in function 'ones' is the obvious route to go but as you don't want to use built-in functions, do something like
o = x*0+1 (multiply x by 0 to create a zero vector of same size as x, then add 1)
s = x*o (scalar product of o and x)
5. Nov 1, 2012
### Simon Bridge
... and then there is always looking up the sum.m file that houses the built-in function that matlab uses ;)
Note: the sum of the squares of the elements of a vector is just the dot product with itself right? So - set up a vector of sequential numbers and dot product it with itself:
x=1:9;
s=x*x';
and s is the sum you want.
Matlab really rewards vector-based thinking. | 536 | 1,905 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-43 | longest | en | 0.938712 |
https://proxieslive.com/finding-the-order-of-a-graphs-automorphism-group/ | 1,621,012,984,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991428.43/warc/CC-MAIN-20210514152803-20210514182803-00474.warc.gz | 493,393,034 | 7,838 | # Finding the order of a graph’s automorphism group
Given any digraph $$D$$ and an arbitrary integer $$n\in\mathbb{N}$$ if we define for every vertex $$v\in V(D)$$ a set $$S_v=\{(v,1),(v,2),(v,3),\cdots (v,n)\}$$ so that $$\{S_t:t\in V(D)\}$$ is a family of pairwise disjoint, $$n$$ element sets. Then if I define a digraph $$H$$ with $$\small V(H)=\bigcup_{t\in V(D)}S_t$$ and $$\small E(H)=\bigcup_{(u,v)\in V(D)}S_u\times S_v$$, would the order of the automorphism group of $$H$$ be given by $$|\text{Aut}(H)|=|\text{Aut}(D)|(n!)^{|V(D)|}$$?
I’m not sure if I made an error, my reasoning was that every automorphism $$\sigma\in\text{Aut}(D)$$ could be extended to $$n!$$ automorphisms of $$D$$ for each $$v\in V(D)$$ by letting $$\rho_{\pi}((v,k))=(\sigma(v),\pi(k))$$ for each permutation $$\pi\in S_n$$ thus letting $$v$$ vary over the vertices in $$V(D)$$ we see that in each case a different automorphism is formed for every choice of $$\pi\in S_n$$ which generate a total of $$(n!)^{|V(D)|}$$ different automorphisms for each $$\sigma\in\text{Aut}(D)$$, so I concluded $$|\text{Aut}(H)|=|\text{Aut}(D)|(n!)^{|V(D)|}$$
Edit: Just so there is no confusion, for an arbitrary digraph $$G$$ when I write $$\text{Aut}(G)$$ I mean that:
$$\text{Aut}(G)=\small\left\{\sigma\in \text{Sym}(V(G)):\forall x,y\in V(G)\left[(x,y)\in E(G)\iff (\sigma(x),\sigma(y))\in E(G)\right]\right\}$$ | 492 | 1,387 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 26, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-21 | latest | en | 0.74619 |
https://gmatclub.com/forum/proposals-for-extending-the-united-states-school-year-to-bring-250577.html | 1,575,564,770,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540481281.1/warc/CC-MAIN-20191205164243-20191205192243-00120.warc.gz | 398,886,695 | 150,356 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 05 Dec 2019, 09:52
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Proposals for extending the United States school year to bring
Author Message
TAGS:
### Hide Tags
Retired Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1323
Location: Viet Nam
Proposals for extending the United States school year to bring [#permalink]
### Show Tags
01 Oct 2017, 23:51
1
6
00:00
Difficulty:
95% (hard)
Question Stats:
37% (02:30) correct 63% (02:36) wrong based on 334 sessions
### HideShow timer Statistics
Proposals for extending the United States school year to bring it more in line with its European and Japanese counterparts are often met with the objection that curtailing the schools’ three-month summer vacation would violate an established United States tradition dating from the nineteenth century. However, this objection misses its mark. True, in the nineteenth century the majority of schools closed for three months every summer, but only because they were in rural areas where successful harvests depended on children’s labor. If any policy could be justified by those appeals to tradition, it would be the policy of determining the length of the school year according to the needs of the economy.
Which one of the following principles, if accepted, would provide the strongest justification for the conclusion?
(A) That a given social policy has traditionally been in force justifies maintaining that policy only if doing so does not conflict with more pressing social needs.
(B) Appeals to its own traditions cannot excuse a country from the obligation to bring its practices in line with the legitimate expectations of the rest of the world.
(C) Because appeals to tradition often serve to mask the real interests at issue, such appeals should be disregarded.
(D) Traditional principles should be discarded when they no longer serve the needs of the economy.
(E) The actual tradition embodied in a given practice can be accurately identified only by reference to the reasons that originally prompted that practice.
Source: LSAT
_________________
Intern
Joined: 02 Oct 2017
Posts: 4
Re: Proposals for extending the United States school year to bring [#permalink]
### Show Tags
02 Oct 2017, 01:45
Proposals for extending the United States school year to bring it more in line with its European and Japanese parallel are often met with the exception that reduce the schools' three-month summer vacation would disobey an established United States tradition linked with from the nineteenth century.
Director
Joined: 14 Nov 2014
Posts: 592
Location: India
GMAT 1: 700 Q50 V34
GPA: 3.76
Re: Proposals for extending the United States school year to bring [#permalink]
### Show Tags
02 Oct 2017, 04:54
broall wrote:
Proposals for extending the United States school year to bring it more in line with its European and Japanese counterparts are often met with the objection that curtailing the schools’ three-month summer vacation would violate an established United States tradition dating from the nineteenth century. However, this objection misses its mark. True, in the nineteenth century the majority of schools closed for three months every summer, but only because they were in rural areas where successful harvests depended on children’s labor. If any policy could be justified by those appeals to tradition, it would be the policy of determining the length of the school year according to the needs of the economy.
Which one of the following principles, if accepted, would provide the strongest justification for the conclusion?
(A) That a given social policy has traditionally been in force justifies maintaining that policy only if doing so does not conflict with more pressing social needs.
(B) Appeals to its own traditions cannot excuse a country from the obligation to bring its practices in line with the legitimate expectations of the rest of the world.
(C) Because appeals to tradition often serve to mask the real interests at issue, such appeals should be disregarded.
(D) Traditional principles should be discarded when they no longer serve the needs of the economy.
(E) The actual tradition embodied in a given practice can be accurately identified only by reference to the reasons that originally prompted that practice.
Source: LSAT
will go with E ...here the argument is focusing on the reason behind the summer holiday in 19th century and now .
As in argument , reason behind implementing a policy is now no longer the case , so the policy can be changed....means one should focus on the reason that started that particular policy ..
Manager
Joined: 27 Jan 2016
Posts: 123
Schools: ISB '18
GMAT 1: 700 Q50 V34
Re: Proposals for extending the United States school year to bring [#permalink]
### Show Tags
12 Oct 2017, 10:49
1
broall wrote:
Proposals for extending the United States school year to bring it more in line with its European and Japanese counterparts are often met with the objection that curtailing the schools’ three-month summer vacation would violate an established United States tradition dating from the nineteenth century. However, this objection misses its mark. True, in the nineteenth century the majority of schools closed for three months every summer, but only because they were in rural areas where successful harvests depended on children’s labor. If any policy could be justified by those appeals to tradition, it would be the policy of determining the length of the school year according to the needs of the economy.
Which one of the following principles, if accepted, would provide the strongest justification for the conclusion?
(A) That a given social policy has traditionally been in force justifies maintaining that policy only if doing so does not conflict with more pressing social needs.
(B) Appeals to its own traditions cannot excuse a country from the obligation to bring its practices in line with the legitimate expectations of the rest of the world.
(C) Because appeals to tradition often serve to mask the real interests at issue, such appeals should be disregarded.
(D) Traditional principles should be discarded when they no longer serve the needs of the economy.
(E) The actual tradition embodied in a given practice can be accurately identified only by reference to the reasons that originally prompted that practice.
Source: LSAT
The objection is that the reduction in the summer vacation would violate the tradition of giving a three month summer vacation. The conclusion is "The argument misses its mark" because the "The tradition is not just the 3 month vacation, the actual tradition can be traced by referring to the actual reason behind that practice i.e to make the child labour available for harvesting season"
LBS Moderator
Joined: 04 Jun 2018
Posts: 662
Location: Germany
Concentration: General Management, Finance
GMAT 1: 730 Q47 V44
GPA: 3.4
WE: Analyst (Transportation)
Re: Proposals for extending the United States school year to bring [#permalink]
### Show Tags
07 Jan 2019, 11:01
We are looking for a general reason that follows the logic of the statement.
As per the answer choice statistics, most people were able to narrow it down to the choices D and E.
In this case, D seems appealing because it seems to follow the rationale of the given argument and even recycles some of the language (e.g. "good for the economy").
However, when we have a closer look at both D and E we notice that it is actually the statement in AC E that follows the rationale of the OS.
We are not abandoning a tradition merely for the sake of the economy, but we evaluate a tradition based on its original intention.
Hope this helps.
Regards,
Chris
_________________
Re: Proposals for extending the United States school year to bring [#permalink] 07 Jan 2019, 11:01
Display posts from previous: Sort by | 1,747 | 8,355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-51 | latest | en | 0.938841 |
https://www.kidsacademy.mobi/printable-worksheets/normal/age-3-4/math/subtraction/ | 1,721,715,124,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518014.29/warc/CC-MAIN-20240723041947-20240723071947-00387.warc.gz | 734,281,776 | 56,755 | # Normal Subtraction Worksheets for Ages 3-4
Discover "Normal Subtraction Worksheets for Ages 3-4" designed by Kids Academy to engage your little learners in foundational math skills. Our worksheets use fun themes and simple visuals to introduce subtraction concepts effectively. Each exercise is tailored to the developmental stage of 3-4-year-olds, helping them understand basic subtraction while enjoying the process. Ideal for both classroom and at-home learning, these resources support cognitive development and early math confidence. Foster your child's numerical abilities through interactive and age-appropriate activities with Kids Academy's expertly crafted worksheets. Encourage a love for math from the start!
Favorites
Interactive
• 3-4
• Subtraction
• Normal
## Match the Word Problems Worksheet
Read this word problem to your kids: Help them understand how it can be translated into a number problem. Trace the dotted lines to see how each picture and number sentence match the problem. With this worksheet, you can show your kids how easy it is to transform a word problem into a number problem.
Match the Word Problems Worksheet
Worksheet
## Magic Fives Worksheet
Help your students identify the objects & people in the worksheet. Point to the equation 5 -4 in the center, then get them to count the items and trace a line connecting those that add up to 5 - 4.
Magic Fives Worksheet
Worksheet
## Count Back to Subtract Substraction Worksheet
With this worksheet, your child can visually see and count each number they’re subtracting. This worksheet provides a number line to help your child master subtraction visually. Counting back on the line will give them a deeper understanding of the concept. It's a great way to teach subtraction and help children gain confidence with the subject.
Count Back to Subtract Substraction Worksheet
Worksheet
## Recycling - More or Less Worksheet
Your child can compare numbers and solve word problems with this free worksheet. Help them understand what's being asked as Sam and Ali take items to the recycling center. They can choose the correct equation to match the word problem and apply their skills in reading and pulling out the appropriate information.
Recycling - More or Less Worksheet
Worksheet
Learning Skills
Parents and teachers play a critical role in nurturing young children's mathematical abilities, and introducing the concept of normal subtraction to ages 3-4 is a vital component of early math education. Understanding subtraction lays the groundwork for future mathematical learning, including more complex operations and problem-solving skills.
At this tender age, children's brains are highly receptive and wired for learning new concepts. Subtraction can be taught in engaging and playful ways, such as using physical objects like toys or snacks. For instance, if you have five apples and take away two, asking "How many are left?" can help children visually and tangibly understand the concept of taking away.
Early exposure to subtraction hones a child's ability to think critically and logically. It enhances their understanding of numbers as part of a whole and fosters their ability to represent and communicate their mathematical thinking. These are essential skills not just for math but for all areas of learning and life.
Moreover, mastering early subtraction builds a child's confidence in their mathematical abilities, creating a positive relationship with math from the outset. So, investing time and effort into teaching normal subtraction to children aged 3-4 can have long-lasting benefits, setting a strong foundation for future academic success and cognitive development. | 690 | 3,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-30 | latest | en | 0.898937 |
https://www.cfd-online.com/Forums/cfx/165272-calculating-local-time-scale-reading-available-expressions.html | 1,624,260,613,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488268274.66/warc/CC-MAIN-20210621055537-20210621085537-00619.warc.gz | 624,861,330 | 18,426 | # Calculating the local time scale or reading it from available expressions
Register Blogs Members List Search Today's Posts Mark Forums Read
January 13, 2016, 03:18 Calculating the local time scale or reading it from available expressions #1 Member Join Date: Jan 2016 Posts: 34 Rep Power: 7 I am in the process of modeling vortex generators as source terms. To do this I will apply a certain momentum in some cells. This momentum is calculated with the use of the local time scale factor, defined by the paper('A tuning free body force vortex generator model by F. Wallin') as I'm having trouble finding a way to implement , which is the local grid size, there does not seem to be an available CEL expression like there is for the others, magnitude of velocity or speed of sound. Does anyone have any tips? Furthermore, I will need to integrate this constant together with the local velocity. That means I would need to a transient simulation, right? There is no point in integrating the local velocity in a steady state simulation I would assume.
January 13, 2016, 05:35 #2 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 16,720 Rep Power: 130 Speed of sound and Magnitude of Velocity: Variables exist for these, look in the CFX documentation in the reference guide for a list of available variables. I don't think loal mesh size is available. You could use the cube root of the volume variable (which is the control volume's volume). But I am very wary of your function as it goes to zero as the mesh gets smaller - it is inherently mesh sensitive. This does not sound physically realistic to me. You need to integrate your function with the local velocity - what do you mean? What type of integral? Please give the equation you wish to model.
January 13, 2016, 05:46
#3
Member
Join Date: Jan 2016
Posts: 34
Rep Power: 7
Quote:
Originally Posted by ghorrocks Speed of sound and Magnitude of Velocity: Variables exist for these, look in the CFX documentation in the reference guide for a list of available variables. I don't think loal mesh size is available. You could use the cube root of the volume variable (which is the control volume's volume). But I am very wary of your function as it goes to zero as the mesh gets smaller - it is inherently mesh sensitive. This does not sound physically realistic to me. You need to integrate your function with the local velocity - what do you mean? What type of integral? Please give the equation you wish to model.
I have found those two expressions, it's the local mesh size that I haven't found yet. Cube root might work, but then your elements would need to resemble a cube for it to be accurate.
The paper gives the derivative of body force
with being equal to the from the previous equation, and the others density, local velocity and the VG normal vector.
So in order to implement that the body force you need to integrate that equation. Which would mean you would have to integrate the local velocity. Something to tune of
January 13, 2016, 06:22
#4
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 16,720
Rep Power: 130
Quote:
then your elements would need to resemble a cube for it to be accurate.
This is another problem with the approach - how does it handle irregular element shapes? This is another problem with functions like this.
And integrating it over time just adds to the problem. Can you explain to me how a body force can be a function of flow conditions in the past?
Regardless - to answer your question: Have a look at the units of momentum source terms. That will tell you the units of the function you are looking for and I suspect will help you. Don't forget you will probably need a momentum source term coefficient as well - these are described in the documentation.
January 13, 2016, 06:50
#5
Member
Join Date: Jan 2016
Posts: 34
Rep Power: 7
Quote:
Originally Posted by ghorrocks This is another problem with the approach - how does it handle irregular element shapes? This is another problem with functions like this. And integrating it over time just adds to the problem. Can you explain to me how a body force can be a function of flow conditions in the past? Regardless - to answer your question: Have a look at the units of momentum source terms. That will tell you the units of the function you are looking for and I suspect will help you. Don't forget you will probably need a momentum source term coefficient as well - these are described in the documentation.
This paper does seem to raise more questions than answer them.
That doesn't seem physical, but I suppose the author wants to mimic a certain affect. He states the expression comes from equating the flow tangency condition to the rate of change in the normal force. But doesn't explain what the implications of it are, as you have just shown.
The units do match though, I did check those. And thanks for the tip. I have found other models, but this had one promising feature. Though I should take a more critical look at my choice now. Thanks for the help
January 14, 2016, 03:11 #6 Member Join Date: Jan 2016 Posts: 34 Rep Power: 7 Somewhat related to this issue. Is it at all possible to obtain more geometrical information about a mesh volume (in CFX pre that is, not in the MESH environment)? I want to check whether a line (or a point on that line) is crosses/inhabits a mesh volume. So I would need data about the grid edges but this seems unavailable, is that correct? I have managed to make a method that checks whether grid nodes are in a volume, but the latter would be preferred. I have added an image to illustrate my point a bit better.
January 14, 2016, 04:56 #7 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 16,720 Rep Power: 130 You would need to do this using user fortran. But I am no expert on user fortran so cannot help you there.
Tags source terms | 1,356 | 5,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-25 | latest | en | 0.911129 |
https://www.jiskha.com/display.cgi?id=1260056330 | 1,501,087,950,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426234.82/warc/CC-MAIN-20170726162158-20170726182158-00513.warc.gz | 813,888,435 | 3,723 | # Math Word Problems
posted by .
How many ways can an IRS auditor select 3 of 11 tax returns for an audit?
There are 8 members on a board of directors. If they must form a subcommittee of 3 members, how many different subcommittees are possible?
License plates are made using 3 letters followed by 3 digits. How many plates can be made if repetition of letters and digits is allowed.
At the first tri-city meeting, there are 8 people from town A, 7 people from town B, and 5 people from town C. If a council consisting of 5 people is randoml selected, find the probabilit that 3 are from town A and 2 are from town B.
• Math Word Problems -
Please submit one question per post.
The first answer is 11*10*9/(1*2*3)= 165,
since the order they are picked does not matter. | 194 | 776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-30 | longest | en | 0.96269 |
https://www.openriskmanual.org/wiki/Aalen-Johansen_Estimator | 1,585,646,950,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500426.22/warc/CC-MAIN-20200331084941-20200331114941-00289.warc.gz | 1,026,994,240 | 8,684 | # Aalen-Johansen Estimator
## Definition
The Aalen-Johansen estimator is a multi-state (matrix) version of the Kaplan–Meier estimator for the hazard of a survival process. The estimator can be used to estimate the transition probability matrix of a Markov process with a finite number of states. [1]
## Estimator
The position in state space for an entity $i$ in continuous time $t$ is a random variable $R^i(t)$ taking values in the state space S (We assume a finite state space $S ={0, \dots ,D}$).
The estimator is given by the expression
$T^{mn}(s,t) = \prod_{k=1}^{K} (I^{mn} + \Delta A^{mn}(t_k))$
$T^{mn}(s, t)$ is the transition matrix element from time s to time t, the mn-th element of the matrix denotes the probability that the Markov process starting in state m at time s will be in state n at time t. The summation is over all times $t_k$ where transition events are observed (a total of K).
### Off-diagonal elements
The estimation of the transition intensities $A^{mn}(t_k)$ at any time $t_k$ where transitions are observed is simply by counting:
$\Delta A(t_k)^{mn} = \frac{\Delta N^{mn}(t_k)}{Y^{m}(t_k)}$
where $\Delta N^{mn}(t_k)$ is the number of transitions observed from state m to state n at time $t_k$ and $Y^m(t_k )$ is the number of entities in state m right before time $t_k$
### Diagonal elements
The diagonal elements are given by
$\Delta A(t_k)^{nn} = \frac{- \Delta N^{n}(t_k)}{Y^{n}(t_k)}$
where $\Delta N^{n}(t_k)$ is the number of transitions away from state n at time $t_k$ | 429 | 1,523 | {"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": 17, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | longest | en | 0.773348 |
https://studyadda.com/sample-papers/jee-main-sample-paper-5_q76/90/247138 | 1,643,005,381,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304515.74/warc/CC-MAIN-20220124054039-20220124084039-00293.warc.gz | 577,568,943 | 21,341 | • # question_answer 10 different letters of english alphabet are given words of 5 letters are formed from these given letters. How many words are formed when at least one letter is repeated? A) 69760 B) 2912 C) 98748 D) 987147
Idea Here, number of permutation of n dissimilar things taken r at a time ${{=}^{n}}{{p}_{r}}=\frac{n!}{(n-r)!}$repetition is not allowed total ways $={{n}^{r}}$Number of 5 letters word if at least one letter is repeated is equal to = Total possible 5 letters word Number of 5 letters word when repetition is not allowed $={{10}^{5}}-10\times 9\times 8\times 7\times 6$ $={{10}^{5}}-30240$ $=69760$ TEST Edge Fundamental theorem of permutation and their application based questions are asked. To solve such type of question, students are advised to understand the concept of fundamental theorem. | 231 | 879 | {"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} | 3.53125 | 4 | CC-MAIN-2022-05 | latest | en | 0.838494 |
https://pqdtopen.proquest.com/doc/305182819.html?FMT=ABS | 1,580,286,541,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251789055.93/warc/CC-MAIN-20200129071944-20200129101944-00074.warc.gz | 601,663,505 | 6,190 | # Dissertation/Thesis Abstract
The curve shortening flow
by Thomas, Andrew, M.S., California State University, Long Beach, 2009, 46; 1466353
Abstract (Summary)
The curve shortening flow deforms a closed curve γ in the direction of its normal vectors. Precisely, one deforms γ by the evolution equation[special characters omitted] where κ is the curvature and n is the normal vector; the derivative of the tangent vector T with respect to arclength s is T'(s) = κn.
In the first part of the thesis, we will study the evolution of convex, closed curve γ in [special characters omitted]. We will study the theorem of Tso, which says that γ becomes analytic in short time, and converges to a point within finite time. We then explain the Gage theorem on the asymptotic behavior of γ: if the area enclosed by γ is rescaled to be a constant (e.g., π) then γ converges smoothly to a unit circle.
In local coordinates, the curve shortening flow (of a curve in an n-dimensional manifold M) can be written as[special characters omitted]here s is the arclength parameter , we write γ = (γ1, γ2, ..., γ n) in local coordinate system; and [special characters omitted] are the Christoffel symbols of the connection on the manifold M.
In the second part of this thesis, we use the same equation (0.1), but now take s to be a fixed parameter on γ; so s is not an arclength parameter of γ when t > 0. This modified equation is called the one-dimensional harmonic heat flow. On general manifolds of nonpositive curvature, under the harmonic heat flow, any loop converges to a minimal geodesic loop as t → ∞. As a special case, in Euclidean spaces, all loops, convex or not, will shrink to a point. We will present a proof of this fact.
Indexing (document details)
Advisor: Ding, Yu Commitee: School: California State University, Long Beach School Location: United States -- California Source: MAI 47/06M, Masters Abstracts International Source Type: DISSERTATION Subjects: Mathematics Keywords: Publication Number: 1466353 ISBN: 978-1-109-17674-2 | 491 | 2,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | longest | en | 0.851015 |
https://socratic.org/questions/how-do-you-evaluate-the-function-f-x-x-6-for-f-2 | 1,638,429,677,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361169.72/warc/CC-MAIN-20211202054457-20211202084457-00122.warc.gz | 597,279,558 | 5,960 | # How do you evaluate the function f(x) = |x|+6 for f(-2)?
Jun 5, 2018
$8$
#### Explanation:
Since we are being asked to evaluate $f \left(- 2\right)$, everywhere we see an $x$, we replace it with a $- 2$. We now have
$| - 2 | + 6$
NOTE: Absolute Value makes any number positive, so we essentially have
$2 + 6$
which is obviously
$8$
Hope this helps! | 116 | 360 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-49 | latest | en | 0.879518 |
http://www.oracleerpappsguide.com/2011/11/group-function-in-sql.html | 1,624,272,801,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488269939.53/warc/CC-MAIN-20210621085922-20210621115922-00537.warc.gz | 72,769,508 | 27,793 | Group Function in SQL
Group functions allow you to perform data operations on several values in a column of data as though the column were one collective group of data. These functions are also called group-by functions because they are often used in a special clause of select statements, called the group by clause.
The syntax for the GROUP BY clause is:
SELECT column1, column2, … column_n, aggregate_function (expression)
FROM tables
WHERE predicates
GROUP BY column1, column2, … column_n;
aggregate_function can be a function such as SUM, COUNT, MIN, or MAX.
Here’s a list of the available group functions:
• avg(x) Averages all x column values returned by the select statement
• count(x) Counts the number of non-NULL values returned by the select statement for column x
• max(x) Determines the maximum value in column x for all rows returned by the select statement
• min(x) Determines the minimum value in column x for all rows returned by the select statement
• stddev(x) Calculates the standard deviation for all values in column x in all rows returned by the select statement
• sum(x) Calculates the sum of all values in column x in all rows returned by the select statement
• Variance(x) Calculates the variance for all values in column x in all rows returned by the select statement
Example using the SUM function
For example, you could also use the SUM function to return the name of the department and the total sales (in the associated department).
SELECT department, SUM(sales) as “Total sales”
FROM order_details
GROUP BY department;
Because you have listed one column in your SELECT statement that is not encapsulated in the SUM function, you must use a GROUP BY clause. The department field must, therefore, be listed in the GROUP BY section.
Example using the COUNT function
For example, you could use the COUNT function to return the name of the department and the number of employees (in the associated department) that make over \$25,000 / year.
SELECT department, COUNT(*) as “Number of employees”
FROM employees
WHERE salary > 25000
GROUP BY department;
ROLLUP
This group by operation is used to produce subtotals at any level of aggregation needed. These subtotals then “roll up” into a grand total, according to items listed in the group by expression. The totaling is based on a one-dimensional data hierarchy of grouped information. For example, let’s say we wanted to get a payroll breakdown for our company by department and job position. The following code block would give us that information:
SQL> select deptno, job, sum(sal) as salary
2 from emp
3 group by rollup(deptno, job);
DEPTNO JOB SALARY
——— ——— ———
10 CLERK 1300
10 MANAGER 2450
10 PRESIDENT 5000
10 8750
20 ANALYST 6000
20 CLERK 1900
20 MANAGER 2975
20 10875
30 CLERK 950
30 MANAGER 2850
30 SALESMAN 5600
30 9400
29025
Notice that NULL values in the output of rollup operations typically mean that the row contains subtotal or grand total information. If you want, you can use the nvl( ) function to substitute a more meaningful value.
cube
cube This is an extension, similar to rollup. The difference is that cube allows you to take a specified set of grouping columns and create subtotals for all possible combinations of them. The cube operation calculates all levels of subtotals on horizontal lines across spreadsheets of output and creates cross-tab summaries on multiple vertical columns in those spreadsheets. The result is a summary that shows subtotals for every combination of columns or expressions in the group by clause, which is also known as n-dimensional cross-tabulation. In the following example, notice how cube not only gives us the payroll breakdown of our company by DEPTNO and JOB, but it also gives us the breakdown of payroll by JOB across all departments:
SQL> select deptno, job, sum(sal) as salary
2 from emp
3 group by cube(deptno, job);
DEPTNO JOB SALARY
——— ——— ———
10 CLERK 1300
10 MANAGER 2450
10 PRESIDENT 5000
10 8750
20 ANALYST 6000
20 CLERK 1900
20 MANAGER 2975
20 10875
30 CLERK 950
30 MANAGER 2850
30 SALESMAN 5600
30 9400
ANALYST 6000
CLERK 4150
MANAGER 8275
PRESIDENT 5000
SALESMAN 5600
29025
Excluding group Data with having
Once the data is grouped using the group by statement, it is sometimes useful to weed out unwanted data. For example, let’s say we want to list the average salary paid to employees in our company, broken down by department and job title. However, for this query, we only care about departments and job titles where the average salary is over \$2000. In effect, we want to put a where clause on the group by clause to limit the results we see to departments and job titles where the average salary equals \$2001 or higher. This effect can be achieved with the use of a special clause called the having clause, which is associated with group by statements. Take a look at an example of this clause:
SQL> select deptno, job, avg(sal)
2 from emp
3 group by deptno, job
4 having avg(sal) > 2000;
DEPTNO JOB AVG(SAL)
——— ——— ———
10 MANAGER 2450
10 PRESIDENT 5000
20 ANALYST 3000
20 MANAGER 2975
30 MANAGER 2850
Consider the output of this query for a moment. First, Oracle computes the average for every department and job title in the entire company. Then, the having clause eliminates departments and titles whose constituent employees’ average salary is \$2000 or less. This selectivity cannot easily be accomplished with an ordinary where clause, because the where clause selects individual rows, whereas this example requires that groups of rows be selected. In this query, you successfully limit output on the group by rows by using the having clause.
0 replies | 1,404 | 5,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-25 | latest | en | 0.860877 |
https://blogs.sap.com/2015/07/02/advanced-hana-programing-with-pal-kmeans/ | 1,670,235,936,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711016.32/warc/CC-MAIN-20221205100449-20221205130449-00705.warc.gz | 162,986,194 | 17,094 | # Introduction
At the 2015 Innovation summit I presented ABC and kmeans analysis. This document will cover how you can use the Predicative Analysis Library (PAL) that comes with SAP HANA to implement ABC and kmeans analysis. The primary focus will be on kmeans. The kmeans application shown her can be used as framework to make the ABC Analysis app. There is a nice video about the app available on https://www.youtube.com/watch?v=y8CUXdTLKhY&feature=youtu.be.
## What is kmeans analysis?
In predictive analysis, k-means clustering is a method of cluster analysis. The k-means algorithm partitions n observations or records into k clusters in which each observation belongs to the cluster with the nearest center. In marketing and customer relationship management areas, this algorithm uses customer data to track customer behavior and create strategic business initiatives. Organizations can thus divide their customers into segments based on variants such as demography, customer behavior, customer profitability, measure of risk, and lifetime value of a customer or retention probability.
Clustering works to group records together according to an algorithm or mathematical formula that attempts to find centroids, or centers, around which similar records gravitate. The most common algorithm uses an iterative refinement technique. This definition comes from SAP HANA Predictive Analysis Library (PAL) Reference. If you are planning to use PAL algorithms in your code I strongly recommend reading the pdf before you continue. The SPS 8 version of PAL guide talks about kmeans starting on page 32.
# How do I build a kmeans application
Read the kmeans section of PAL guide and copy the sample code to your HANA studio. You should become very familiar with this code because you use the code in the guide to build your app.
Once you have taking the sample code, walked through, it and executed it you are ready to start implementing your own kmeans analysis.
# Walk through of the kmeans app presented at the 2015 Innovation Summit
Below I will walk you through the kmeans code I demonstrated at the 2015 Innovation Summit.
## Understanding kmeans code
After I familiarized myself with the code shown in PAL guide, I built a calculation view to provide the input data for the function.
Once the calculation view had been successful tested I replaced the hardcode fields from the pdf with a select from the view.
Now I have a kmeans analysis running against real Business One data. Below are the results.
The attached sql file (kmeans.sql) shows the code used to get results.
The results show our data points and there distance from the center, this is exactly what kmeans offer, but looking at the table of results is not cool. I do not have a hot app for my executives.
## Turning kmeans Results into a cool app
The data shown above can be presented on a cool webpage. The web app (could also run on a smart phone) takes the data above and presents it in two different manners. The upper chart shows total sales amount on the y axis and days till cash is in hand on the x axis. The lower chart shows days past due for the y axis and group position for the x axis. Both are easy for the user community to consume.
## How do I go from SQL results to SAPUI5 app running in a browser?
The app above comes from a few pieces of Javascript code. The first piece is a Javascript app that creates all the HANA tables, calls the eraser and creator procedure, populates the control tables, the input tables and calls the kmeans function, which populates the result tables. After you write the kmeans XSJS script you can test it in the browser.
Above we see results very similar to the HANA SQL results.
## How do I turn the ugly XSJS results into a cool bubble chart
Once the XSJS results are verified the next thing we need to do is build controller, in Javascript that will render the cool results (bubble chart shown in the browser screenshot above).
In the project I created a folder called paldemo. Inside paldemo I created palkmeans_analysis.controller.js and palkmeans_analysis.view.js. Inside the controller file I start with sap.ui.controller(“paldemo.palkmeans_analysis”. Now we have setup the UI controller and it makes a call to palkmeans_analysis.view.js.
The ui.controller provides the look and feel. The palkmeans_analysis.view.js does the work. The palkmeans_analysis starts by defining two bubble chart inside a sap.ui.jsview call.
Now we’ve defined the charts that will be seen for our output. The basic form of the output is set so we move onto sap.viz.ui5.data.FlattenedDataset to build the output dataset. The layout is built using sap.ui.commons.layout.MatrixLayout and everything is added to the screen using the CreateContent function.
Now everything is put to together on the Javascript side lets write a simple index.html to call the components.
Now you are ready to test the app. I’ve attached a delivery unit containing the code the file is named KMEANS_PAL_DU-Vendor.tgz.txt.zip. After downloading rename it to KMEANS_PAL_DU-Vendor.tgz, SCN will not let ne upload tgz or zip files. | 1,102 | 5,153 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-49 | latest | en | 0.886912 |
https://www.coursehero.com/file/6487825/3/ | 1,527,198,810,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866870.92/warc/CC-MAIN-20180524205512-20180524225512-00128.warc.gz | 693,208,039 | 60,530 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
# 3 - EE 478 Multiple User Information Theory Handout#9...
This preview shows pages 1–2. Sign up to view the full content.
EE 478 Handout #9 Multiple User Information Theory Thursday, October 9, 2008 Homework Set #3 Due: Thursday, October 16, 2008. 1. Binary erasure MAC. Let X 1 Bern( p 1 ) and X 2 Bern( p 2 ) be independent. Show that max p 1 ,p 2 H ( X 1 + X 2 ) = 1 . 5 bits and is achieved when X 1 and X 2 are Bern(1 / 2). 2. Capacity of multiple access channels. (a) Consider the binary multiplier MAC example in the lecture notes, where X 1 , X 2 , and Y are binary, and Y = X 1 · X 2 . We established the capacity region using a time-sharing ar- gument. Show that the capacity region can also be expressed as the union of R ( X 1 , X 2 ) sets (with no time-sharing/convexification needed), specify the set of p ( x 1 ) p ( x 2 ) distri- butions on ( X 1 , X 2 ) that achieve the boundary of the region. (b) Find the capacity region for the modulo-2 additive MAC, where X 1 , X 2 , and Y are binary, and Y = X 1 + X 2 mod 2. Again show that the capacity region can be expressed as the union of R ( X 1 , X 2 ) sets and therefore time-sharing is not necessary. (c) The capacity regions of the above two examples and the Gaussian MAC can all be expressed as union of R ( X 1 , X 2 ) sets and no time-sharing is necessary. Is time-sharing ever necessary? Find the capacity of the push-to-talk MAC with binary inputs and output, and p (0 | 0 , 0) = p (1 | 0 , 1) = p (1 | 1 , 0) = 1 and p (0 | 1 , 1) = 0 . 5. Why is this channel called “push-to-talk”?
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern | 739 | 2,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.87853 |
https://math.stackexchange.com/questions/3273134/seeing-if-an-arbitrary-subring-or-set-is-dense-in-the-zariski-topology/3273143 | 1,586,421,530,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371830894.88/warc/CC-MAIN-20200409055849-20200409090349-00514.warc.gz | 552,792,226 | 31,731 | # Seeing if an arbitrary subring or set is dense in the Zariski topology
In another question, it was suggested that I check if a certain subring of a ring was dense in the Zariski topology. The subring, however, was not an ideal, so I wasn't sure how to represent it in $$Spec(R)$$.
Is there a standard meaning for checking if an arbitrary subring, or an arbitrary set is dense in the Zariski topology? Do we just look at all prime ideals containing the set, even though it isn't an ideal?
A basic example would be to look at $$\Bbb Z$$ in $$\Bbb R$$. This is not an ideal, but we can look at prime ideals in $$\Bbb R$$ containing it anyway. If we do so in $$Spec(\Bbb R)$$, then $$\Bbb Z$$ looks the same as $$\Bbb R$$ in the spectrum (as there is only one ideal). Does this mean it is equivalent to $$\Bbb R$$ in the Zariski topology and hence closed?
Remember that $$\operatorname{Spec}$$ is a contravariant functor, so morphisms of commutative rings $$A\to B$$ turn in to maps $$\operatorname{Spec} B\to \operatorname{Spec} A$$. From here, you can look and see if the image of $$\operatorname{Spec} B$$ in $$\operatorname{Spec} A$$ is dense or not.
For your example of $$\Bbb Z\subset \Bbb R$$, we see that the corresponding morphism is a map $$\operatorname{Spec} \Bbb R \to \operatorname{Spec}\Bbb Z$$. Since $$\operatorname{Spec} \Bbb R$$ has a single point, $$(0)$$, and the inverse image of $$(0)\subset \Bbb R$$ is again $$(0)\subset \Bbb Z$$ (the generic point of $$\operatorname{Spec}\Bbb Z$$), we see that the image of this morphism is in fact dense. This is basically a baby example of something that happens lots of places - dominant maps of affine schemes correspond to injective maps on coordinate rings.
• Ah, ok. Is it possible to extend this notion to an arbitrary set? – Mike Battaglia Jun 26 '19 at 15:29
• Yes, a scheme is a topological space plus a sheaf of rings satisfying some conditions, and it makes sense to ask whether some subset of the topological space is dense. – KReiser Jun 26 '19 at 17:45 | 566 | 2,032 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-16 | latest | en | 0.891812 |
https://byjus.com/question-answer/two-squares-are-chosen-at-random-on-a-chessboard-the-probability-that-they-have-a-4/ | 1,718,734,590,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00895.warc.gz | 129,678,894 | 24,446 | 1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question
# Two squares are chosen at random on a chessboard, the probability that they have a side in common is:
A
332
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
132
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
118
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
none of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
## The correct option is C 118Two different squares can be chosen in 64 × 63 ways. For each of the four corner squares, the favourable number of cases is 2. For each of the 24 non-corner squares on all the four sides of the chessboard, the favourable number of cases is 3. For each of the 36 remaining squares, the favourable number of cases is 4. Thus, the total number of favourable cases =4×2+24×3+36×4=224 Hence, the required probability = 22464×63=118
Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Probability
QUANTITATIVE APTITUDE
Watch in App
Explore more
Join BYJU'S Learning Program | 311 | 1,117 | {"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} | 3.234375 | 3 | CC-MAIN-2024-26 | latest | en | 0.856741 |
http://fdatamining.blogspot.com/2011/02/logistic-regression-in-f-using.html | 1,466,876,246,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783393463.1/warc/CC-MAIN-20160624154953-00086-ip-10-164-35-72.ec2.internal.warc.gz | 110,012,008 | 29,570 | Friday, February 11, 2011
Logistic Regression in F# using Microsoft Solver Foundation
Logistic regression is a workhorse in data mining. Like decision trees and SVMs, it is a very standard classifier. If you have a labeled data, logistic regression definitely is one of the classifiers that should tried.
In this post, I’d like to show how to implement a logistic regression using Microsoft Solver Foundation in F#. Because we use the optimization in Microsoft Solver Foundation, the core part of a logistic regression contains only 20 – 30 lines of code.
If you are only interested in F# and L-BFGS, but not logistic regression, please only read the second section: L-BFGS solver in Microsoft Solver Foundation.
Background for logistic regression
Logistic regression model predicts the probability that a data instance x being labeled as y
A data instance is a vector with each dimension representing a value for some feature. For example, the Iris flower data set we are going to use in this post has 4 features: Sepal Length, Sepal Width, Petal Length and Petal Width. Each instance has a label y indicating the category. The above model only models binary labels. Logistic regression also has multiple categories version, see [Mitchell]. In this post, we focus on the binary version; however multiple versions should be straightforward once we know how to implement the binary one.
The model contains two parameters weight vector w and offset b, which should be estimated/learned on the training data set. In our derivations below, we will drop the offset b, because we can always add one feature (constant 1) to feature vector x’=[x 1] and hide b into the extended w’=[w b]. Sometimes, a model without the offset parameter could also perform equally well. Because equation (1) measures the probability, we could solve the model parameters by optimizing the joint probability estimated on dataset X.
If we assume that all the instances are independent and identically distributed (i.i.d.). Then the regularized joint probability is
where is the regularization term for avoiding over-fitting on the training data.
The log version of this likelihood is
We want to find the model parameter w that maximizes this likelihood. P(X) is a concave function with respect to w, so it has a global maximal.
What left is how to optimize this function. BFGS is a Quasi-Newton method that only needs first order gradient while a Newton family method needs second order explicitly. Microsoft Solver Foundation has a limited memory version of BFGS (L-BFGS). (L-BFGS is more popular than BFGS, sometimes people mention BFGS to mean L-BFGS.) There is a memory parameter m in L-BFGS, the space complexity of this algorithm is O(md), where d is the dimension of the variable space. Say, if you want to optimize a function that has 1,000,000 free variables, and set m to 10, then you need only a little more than 1M*10*8/1024^2=76M bytes.
The input for an L-BFGS optimizer is: 1) the function evaluator L(w) and 2) the gradient evaluator :
The output is the best that gives the maximum function value.
L-BFGS solver in Microsoft Solver Foundation
Microsoft Solver Foundation already implements an L-BFGS solver. The reader is suggested to read pages 24-27 & 72-75 of MSF-SolverProgrammingPrimer.pdf (on my machine: C:\Program Files\Microsoft Solver Foundation\3.0.1.10599\Documents) for detailed reference. Because the code for optimizing Rosenbrock function is in C#, here I translated it to F#. Note how I translate between delegate type System.Func<> and F# functions.
Btw, I find that using the Microsoft.Solver.Foundatoin.dll in Sho would be more convenient as it packages everything into one DLL.
#r @"C:\Program Files (x86)\Sho 2.0 for .NET 4\packages\Optimizer\Microsoft.Solver.Foundation.dll"
openMicrosoft.SolverFoundation.Common
openMicrosoft.SolverFoundation.Solvers
openMicrosoft.SolverFoundation.Services
letsolverParams = newCompactQuasiNewtonSolverParams()
letsolver = newCompactQuasiNewtonSolver()
//add a row and set it as the goal
letOriginalRosenbrockFunction =
letf (model:INonlinearModel) (rowVid:int) (values:ValuesByIndex) (newValues:bool) =
Math.Pow(1. - values.[1], 2.) + 100. * (Math.Pow(values.[2] - (values.[1] * values.[1]), 2.))
newSystem.Func<INonlinearModel, int, ValuesByIndex, bool, float> (f)
letf (model:INonlinearModel) (rowVid:int) (values:ValuesByIndex) (newValues:bool) (gradient:ValuesByIndex) =
gradient.[1] <- -2. * (1. - values.[1]) - 400. * values.[1] * (values.[2] - (values.[1] * values.[1]))
gradient.[2] <- 200. * (values.[2] - (values.[1] * values.[1]))
newSystem.Action<INonlinearModel, int, ValuesByIndex, bool, ValuesByIndex> (f)
solver.FunctionEvaluator <- OriginalRosenbrockFunction
solver.Solve(solverParams);
Console.WriteLine(solver.ToString());
and the output:
`> Minimize problemDimensions = 2Variable indexes: 1, 2Starting point = 0, 0Solution quality is: LocalOptimaNumber of iterations performed: 21Number of evaluation calls: 27Finishing point =0.999999997671584, 0.999999995285991Finishing value =5.74844710320033E-18val it : unit = ()> `
Implementation for logistic regression
We first get a sample data set online:
`module Net = open System.Net let fetchUrlSimple (url:string) = let req = WebRequest.Create(url) let response = req.GetResponse() use stream = response.GetResponseStream() use streamreader = new System.IO.StreamReader(stream) streamreader.ReadToEnd() type dataset = { features: float array array; // (instance = float array) array mutable labels: int array; // } let iris = let page = Net.fetchUrlSimple @"http://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data" let lines = page.Split([|'\n'|], StringSplitOptions.RemoveEmptyEntries) let getIrisId (s:string) = if s = "Iris-setosa" then 2 elif s = "Iris-versicolor" then -1 else 1 let instances = lines |> Array.map (fun line -> let s = line.Split ',' let features = [| "1.0"; s.[0]; s.[1]; s.[2]; s.[3] |] |> Array.map float // add a dumy 1.0 into the fatures let label = getIrisId s.[4] features, label ) |> Array.filter (fun (_, label) -> label <> 2) let F, L = Array.unzip instances { features = F; labels = L; }`
The logistic regression implementation:
`module LogReg = let mutable lambda = 0.1 let dotProduct (x:float array) (g:ValuesByIndex) = let mutable dot = 0.0 for i=0 to x.Length-1 do dot <- dot + x.[i] * g.[i+1] dot let sigmoid (x:float array) (y:int) (g:ValuesByIndex) = let mutable dot = 0.0 for i=0 to x.Length-1 do dot <- dot + x.[i] * g.[i+1] let z = (float y) * dot (* code to deal with extream numbers, but with some issues with l(w) calculation if z > 30.0 then 1.0 elif z < -30.0 then 0.0 else *) 1.0 / (1.0 + exp (- z)) let sigmoid2 (x:float array) (y:int) (g:float array) = let mutable dot = 0.0 for i=0 to x.Length-1 do dot <- dot + x.[i] * g.[i] 1.0 / (1.0 + exp (- (float y) * dot)) let logregValue(ds:dataset) = let dim = ds.features.[0].Length let f (model:INonlinearModel) (rowVid:int) (values:ValuesByIndex) (newValues:bool) = let mutable L = 0.0 for i=1 to dim do L <- L + values.[i]*values.[i] L <- - (L * lambda / 2.0) for i=0 to ds.features.Length-1 do //L <- L - log (1.0 + exp (- (float ds.labels.[i]) * dotProduct ds.features.[i] values )) L <- L + log (sigmoid ds.features.[i] ds.labels.[i] values) // printfn "L = %.10f" L L new System.Func<INonlinearModel, int, ValuesByIndex, bool, float> (f) let logregGradient(ds:dataset) = // printfn "gradient calculated" let dim = ds.features.[0].Length let f (model:INonlinearModel) (rowVid:int) (values:ValuesByIndex) (newValues:bool) (gradient:ValuesByIndex) = for j=1 to dim do gradient.[j] <- -lambda * values.[j] for i=0 to ds.features.Length-1 do let coef = (1.0 - sigmoid ds.features.[i] ds.labels.[i] values) * (float ds.labels.[i]) for j=1 to dim do gradient.[j] <- gradient.[j] + coef * ds.features.[i].[j-1] new System.Action<INonlinearModel, int, ValuesByIndex, bool, ValuesByIndex> (f) let makeSolver(ds:dataset) = // set the solver parameters let solverParams = new CompactQuasiNewtonSolverParams() //solverParams.IterationLimit <- 11 //solverParams.Tolerance <- 1e-4 // the memory parameter m, default 17 solverParams.IterationsToRemember <- 10 // solver let solver = new CompactQuasiNewtonSolver() let dim = ds.features.[0].Length // add variables for i=1 to dim do solver.AddVariable(null) |> ignore // add a row and set it as the goal let _, vidRow = solver.AddRow(null) solver.AddGoal(vidRow, 0, false) |> ignore // set funcation evaluator and gradient evaluator solver.FunctionEvaluator <- logregValue(ds) solver.GradientEvaluator <- logregGradient(ds) // solve! solver.Solve(solverParams) |> ignore // get the optimized point (w) in the solver let w = Array.create dim 0.0 for i=1 to dim do w.[i-1] <- solver.GetValue(i).ToDouble() w`
Try our logistic regression model on the Iris dataset:
`let sol = LogReg.makeSolver(iris)let nWrongs = ref 0for i=0 to iris.features.Length-1 do let prob = LogReg.sigmoid2 iris.features.[i] iris.labels.[i] sol printfn "L = %A sig value = %A label = %A" iris.labels.[i] prob (if prob > 0.5 then "correct" else incr nWrongs; "wrong")`
Reference
Tom Mitchell, Generative and Discriminative Classifiers: Naive Bayes and Logistic Regression, new book chapter. This chapter gives a good introduction to logistic regression. For a more theoretic comparison, see: Ng & Jordan: on discriminative vs generative classifiers a comparison of logistic regression and naïve bayes.
Thomas P. Minka, A comparison of numerical optimizers for logistic regression.
It is well written, the first page ready gives a concise introduction to logistic regression. The rest of the paper has a lot of discussion on optimization methods. Tom also has a Matlab package.
Galen Andrew and Jianfeng Gao. Scalable training of L1-regularized log-linear models. In Proceedings of the 24th International Conference on Machine Learning (ICML 2007), pp. 33-40, 2007.
L1-regularized logistic regression with an implementation in C++.
libLBFGS(LBFGS in C): http://www.chokkan.org/software/liblbfgs/
Note
I quite like this post because it combines so many good things together: F#, logistic regression, optimization and playing with sample datasets.
1. Interesting post, thank you. I have meant to try out Solver Foundation for a while, and I think that moment is coming closer now.
2. Thanks.
What would you recommend to show a graph?
3. @ Art,
Microsoft Solver Foundation does not have any plot functionality. You can go to Sho library for the plot.
4. Very nice post, I wish use F# for data minning as soon.
5. This comment has been removed by the author.
6. @Chris,
Great to see the OCaml version!!
7. Yes It's a nice post this post is very use full for me
Logistic Service Provider | 2,963 | 11,729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-26 | longest | en | 0.889565 |
http://studentsmerit.com/paper-detail/?paper_id=5667 | 1,480,941,037,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541696.67/warc/CC-MAIN-20161202170901-00047-ip-10-31-129-80.ec2.internal.warc.gz | 261,793,363 | 7,629 | Details of this Paper
QUESTION 1 In a free economy, capital from prov...
Description
Solution
Question
QUESTION 1 In a free economy, capital from providers with available funds is allocated through the price system to users that have a demand for funds. The interaction of the providers? supply and the users? demand determines the cost (or price) of money, which is the rate users pay to providers. For debt, we call this price the interest rate. For equity, it is called the cost of equity, and it consists of the dividends and capital gains stockholders expect. Discuss the FOUR factors affecting the cost of money. [20 marks] QUESTION 2 Stock A and B have the following historical returns: Year Stock A?s Returns (%) Stock B?s Returns (%) 2004 -18.00 -14.50 2005 33.00 21.80 2006 15.00 30.50 2007 0.50 -7.60 2008 27.00 26.30 a) Calculate the average rate of return for each stock during the 5-year period. b) Assume that you held the portfolio consisting of 50% stock A and 50% stock B. What would have been the realized rate of return on the portfolio in each year? What would have been the average return on the portfolio during this period? c) Calculate the standard deviation of returns for each stock and for the portfolio. d) Calculate the coefficient variation for each stock and for the portfolio. e) Choose the best stock and explain your answer. [30 marks],QUESTION 3 You are a financial analyst for Modal Optima Berhad. The director of finance has asked you to analyze two proposed capital investment, Project X and Y. Each project has a cost of RM10 million and the cost of capital for each project is 12%. The projects? expected net cash flow are as follows: Year Expected Net Cash Flows Project X (RM?000) Project B (RM?000) 0 -10,000 -10,000 1 6,500 3,500 2 3,000 3,500 3 3,000 3,500 4 1,000 3,500 a) Calculate each project?s payback period, net present value (NPV), internal rate of return (IRR) and modified internal rate of return (MIRR). b) Which project or projects should be accepted if they are independent? c) Which project should be accepted if they are mutually exclusive? [30 mark]
Paper#5667 | Written in 18-Jul-2015
Price : \$25 | 546 | 2,167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.898973 |
http://lbartman.com/worksheet/multiplication-table-worksheet-pdf.php | 1,606,250,520,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141177566.10/warc/CC-MAIN-20201124195123-20201124225123-00422.warc.gz | 58,335,697 | 12,295 | lbartman.com - the pro math teacher
• Subtraction
• Multiplication
• Division
• Decimal
• Time
• Line Number
• Fractions
• Math Word Problem
• Kindergarten
• a + b + c
a - b - c
a x b x c
a : b : c
Multiplication Table Worksheet Pdf
Public on 07 Oct, 2016 by Cyun Lee
timestable facts to 12 one minute multiplicaiton test
Name : __________________
Seat Num. : __________________
Date : __________________
8980 x 1526 = ...
6467 x 1854 = ...
1796 x 4326 = ...
8412 x 9454 = ...
8963 x 8127 = ...
6314 x 2858 = ...
8535 x 7326 = ...
5044 x 1596 = ...
7346 x 1369 = ...
9174 x 4798 = ...
1193 x 1985 = ...
6501 x 1888 = ...
5528 x 7061 = ...
3637 x 2651 = ...
9042 x 5791 = ...
5152 x 9964 = ...
4384 x 2748 = ...
7165 x 7499 = ...
4780 x 1758 = ...
1216 x 2360 = ...
7971 x 3101 = ...
8200 x 4975 = ...
2648 x 7183 = ...
1890 x 6108 = ...
6493 x 2820 = ...
4956 x 2230 = ...
4591 x 5200 = ...
2265 x 2805 = ...
8147 x 1135 = ...
2631 x 2961 = ...
3519 x 9194 = ...
4656 x 8986 = ...
6588 x 7642 = ...
2828 x 2851 = ...
7058 x 9749 = ...
5116 x 8693 = ...
6273 x 6575 = ...
4130 x 1850 = ...
8290 x 5625 = ...
2335 x 1418 = ...
1964 x 2739 = ...
6536 x 4465 = ...
8723 x 2133 = ...
3858 x 1483 = ...
4179 x 5342 = ...
3476 x 5193 = ...
2317 x 1885 = ...
3906 x 5577 = ...
3224 x 2055 = ...
7391 x 2652 = ...
1952 x 3789 = ...
5039 x 1671 = ...
6835 x 3081 = ...
8807 x 1835 = ...
6011 x 6446 = ...
9176 x 4908 = ...
6988 x 6246 = ...
8863 x 3795 = ...
3237 x 4632 = ...
3171 x 3799 = ...
8335 x 5180 = ...
6176 x 9936 = ...
7971 x 4782 = ...
6583 x 1463 = ...
9987 x 1847 = ...
2811 x 8756 = ...
1865 x 9252 = ...
1725 x 6971 = ...
4746 x 3023 = ...
5760 x 8543 = ...
3902 x 9557 = ...
6756 x 7078 = ...
8789 x 2779 = ...
6984 x 8227 = ...
5944 x 2896 = ...
6688 x 7959 = ...
4667 x 9997 = ...
1300 x 5274 = ...
5402 x 6177 = ...
7194 x 5195 = ...
5758 x 2695 = ...
9682 x 5343 = ...
9904 x 6855 = ...
5890 x 1582 = ...
1960 x 2344 = ...
1489 x 4408 = ...
8312 x 1385 = ...
1524 x 8692 = ...
8166 x 4471 = ...
8692 x 5907 = ...
2360 x 2342 = ...
9768 x 9874 = ...
2680 x 9872 = ...
3494 x 7031 = ...
5135 x 4165 = ...
9300 x 2681 = ...
6769 x 9226 = ...
5536 x 3479 = ...
9674 x 6885 = ...
3226 x 9155 = ...
1179 x 2479 = ...
7331 x 1532 = ...
7771 x 2867 = ...
3748 x 9066 = ...
9366 x 2037 = ...
9739 x 3913 = ...
4911 x 3980 = ...
5674 x 3286 = ...
1807 x 4714 = ...
7261 x 2027 = ...
2861 x 7416 = ...
5056 x 3959 = ...
6194 x 9808 = ...
2469 x 8261 = ...
4804 x 6635 = ...
7481 x 8128 = ...
1682 x 4280 = ...
4122 x 3804 = ...
2574 x 4184 = ...
3345 x 2970 = ...
7388 x 7502 = ...
3382 x 4679 = ...
6397 x 5616 = ...
7625 x 5036 = ...
4725 x 1979 = ...
6195 x 5791 = ...
4637 x 3668 = ...
3517 x 6635 = ...
9603 x 6405 = ...
4948 x 6727 = ...
3746 x 9971 = ...
1867 x 2497 = ...
1734 x 7366 = ...
7748 x 5131 = ...
1553 x 9037 = ...
3166 x 5321 = ...
6981 x 6296 = ...
4578 x 8092 = ...
8964 x 6535 = ...
7943 x 1847 = ...
9651 x 9897 = ...
7985 x 3800 = ...
1560 x 6034 = ...
9226 x 7364 = ...
3101 x 9990 = ...
3711 x 2714 = ...
1852 x 8392 = ...
8834 x 3296 = ...
8690 x 5798 = ...
5804 x 8090 = ...
3156 x 7412 = ...
2937 x 9730 = ...
8056 x 1676 = ...
3793 x 5603 = ...
8884 x 9115 = ...
5791 x 8112 = ...
6452 x 7116 = ...
2194 x 9245 = ...
1172 x 5154 = ...
8343 x 2204 = ...
6429 x 3875 = ...
6246 x 4935 = ...
1616 x 5503 = ...
1723 x 2781 = ...
4975 x 1830 = ...
5930 x 6237 = ...
2252 x 3923 = ...
9390 x 3350 = ...
3487 x 1455 = ...
3154 x 6848 = ...
2370 x 3790 = ...
2664 x 6776 = ...
7739 x 2097 = ...
4623 x 9864 = ...
7156 x 3580 = ...
6966 x 8259 = ...
7085 x 2416 = ...
7186 x 6829 = ...
6478 x 7853 = ...
3240 x 3910 = ...
6643 x 6479 = ...
2456 x 1423 = ...
2974 x 8546 = ...
4971 x 1833 = ...
7284 x 6628 = ...
4258 x 7335 = ...
9407 x 9917 = ...
9043 x 8902 = ...
9982 x 2528 = ...
8618 x 8703 = ...
6588 x 1456 = ...
9374 x 8088 = ...
3946 x 5091 = ...
3689 x 2284 = ...
7882 x 6474 = ...
4137 x 1327 = ...
4758 x 4962 = ...
9755 x 5179 = ...
9619 x 2831 = ...
8136 x 5429 = ...
show printable version !!!hide the show
RELATED POST
Not Available
POPULAR
math worksheets year 4
letter j worksheets kindergarten
division worksheet pdf
adding subtracting and multiplying fractions worksheets
division drills worksheets | 1,758 | 4,460 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-50 | latest | en | 0.196301 |
https://gmatclub.com/forum/m02-183561.html | 1,542,604,501,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745281.79/warc/CC-MAIN-20181119043725-20181119065725-00444.warc.gz | 653,290,364 | 57,906 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 18 Nov 2018, 21:15
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
## Events & Promotions
###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### How to QUICKLY Solve GMAT Questions - GMAT Club Chat
November 20, 2018
November 20, 2018
09:00 AM PST
10:00 AM PST
The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat.
• ### The winning strategy for 700+ on the GMAT
November 20, 2018
November 20, 2018
06:00 PM EST
07:00 PM EST
What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
# M02-13
Author Message
TAGS:
### Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 50627
### Show Tags
15 Sep 2014, 23:17
00:00
Difficulty:
45% (medium)
Question Stats:
70% (02:08) correct 30% (02:31) wrong based on 165 sessions
### HideShow timer Statistics
$$x$$ and $$y$$ are positive integers. If $$y = \sqrt{64}$$ and $$x^2 - 10x = (-4y^3 + 64y)*\frac{1}{96}$$, what is the minimum possible value of $$x$$?
A. 2
B. 4
C. 8
D. 12
E. 16
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 50627
### Show Tags
15 Sep 2014, 23:17
Official Solution:
$$x$$ and $$y$$ are positive integers. If $$y = \sqrt{64}$$ and $$x^2 - 10x = (-4y^3 + 64y)*\frac{1}{96}$$, what is the minimum possible value of $$x$$?
A. 2
B. 4
C. 8
D. 12
E. 16
We know that $$y = 8$$.
Plugging this into the second equation gives:
$$x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16$$
$$x^2-10x = -16$$
$$x^2-10x +16 = 0$$
$$(x-8)(x-2)=0$$
The possible values of $$x$$ are 8 and 2. Therefore the minimum value of $$x$$ is 2.
_________________
Intern
Joined: 04 Sep 2014
Posts: 5
### Show Tags
31 Mar 2015, 07:08
Bunel,
I am a bit lost to the answer. what did u do with the 64Y? i see you have x-10x= -4*8^3 + 8^3.
Intern
Joined: 29 Jan 2015
Posts: 5
Concentration: Technology, Finance
### Show Tags
31 Mar 2015, 08:46
floody84 wrote:
Bunel,
I am a bit lost to the answer. what did u do with the 64Y? i see you have x-10x= -4*8^3 + 8^3.
Steps:
• \sqrt{64} is 8. That is y=8
• Insert 8 in quadratic formula and calculate the left hand side
• Factor 96: 96 = 8*4*3 (denominator of the fraction)
• Cancel out all possible factors to get -16 for the left hand side
• Use -16 in quadratic expression to get the roots 8 and 2
• Solution is 2 because the question asked for the min. value
Current Student
Joined: 14 Oct 2013
Posts: 45
### Show Tags
Updated on: 22 Apr 2015, 19:38
Couldn't Y also equal -8? I know that when you plug in y=-8 you end up getting x^2 - 10x -16 = 0 which you can't factor out, but I didnt realize that until after I plugged in -8 (I plugged in -8 since we were trying to find the smallest value of X, I figured a negative Y would give us smaller values for X)
Originally posted by healthjunkie on 04 Apr 2015, 10:52.
Last edited by healthjunkie on 22 Apr 2015, 19:38, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 50627
### Show Tags
05 Apr 2015, 04:31
2
healthjunkie wrote:
Couldn't Y also equal -8? I know that when you plug in y=-8 you end up getting x^2 - 10x -16 = 0 which you can't factor out, but I didnt realize that until after I plugged in -8 (I plugged in -8 since we were trying to find the smallest value of X, I figured a negative Y would give us smaller values for X)
No.
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:
$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt[4]{16} = 2$$, NOT +2 or -2;
Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.
Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html
All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59
Hope it helps.
_________________
Manager
Joined: 17 Mar 2014
Posts: 228
Location: India
Concentration: Operations, Strategy
GMAT 1: 670 Q48 V35
GPA: 3.19
WE: Information Technology (Computer Software)
### Show Tags
09 Aug 2015, 06:08
Bunuel wrote:
healthjunkie wrote:
Couldn't Y also equal -8? I know that when you plug in y=-8 you end up getting x^2 - 10x -16 = 0 which you can't factor out, but I didnt realize that until after I plugged in -8 (I plugged in -8 since we were trying to find the smallest value of X, I figured a negative Y would give us smaller values for X)
No.
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:
$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt[4]{16} = 2$$, NOT +2 or -2;
Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.
Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html
All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59
Hope it helps.
Hi Bunuel ,
Please correct me if i amwrong
Does this mean when gmat gives sqare root sign itself in question stem , then it means we should consider only positive root
and on the other hand gmat gave me a quadratic eqaution in question stem and when i solve it i should consider both +ve and - ve root ?
right ?
_________________
Press +1 Kudos if you find this Post helpful
Math Expert
Joined: 02 Sep 2009
Posts: 50627
### Show Tags
17 Aug 2015, 02:27
Bunuel wrote:
healthjunkie wrote:
Couldn't Y also equal -8? I know that when you plug in y=-8 you end up getting x^2 - 10x -16 = 0 which you can't factor out, but I didnt realize that until after I plugged in -8 (I plugged in -8 since we were trying to find the smallest value of X, I figured a negative Y would give us smaller values for X)
No.
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:
$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt[4]{16} = 2$$, NOT +2 or -2;
Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.
Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html
All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59
Hope it helps.
Hi Bunuel ,
Please correct me if i amwrong
Does this mean when gmat gives sqare root sign itself in question stem , then it means we should consider only positive root
and on the other hand gmat gave me a quadratic eqaution in question stem and when i solve it i should consider both +ve and - ve root ?
right ?
$$\sqrt{}$$ sign always means non-negative root.
In contrast x^2 = 4 gives two solutions: 2 and -2.
_________________
Intern
Joined: 01 Sep 2016
Posts: 12
### Show Tags
10 Sep 2016, 20:29
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 25 Jan 2016
Posts: 9
Location: United States (NJ)
GPA: 2.34
WE: Analyst (Commercial Banking)
### Show Tags
23 Mar 2017, 16:06
Bunuel wrote:
Official Solution:
$$x$$ and $$y$$ are positive integers. If $$y = \sqrt{64}$$ and $$x^2 - 10x = (-4y^3 + 64y)*\frac{1}{96}$$, what is the minimum possible value of $$x$$?
A. 2
B. 4
C. 8
D. 12
E. 16
We know that $$y = 8$$.
Plugging this into the second equation gives:
$$x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16$$
$$x^2-10x = -16$$
$$x^2-10x +16 = 0$$
$$(x-8)(x-2)=0$$
The possible values of $$x$$ are 8 and 2. Therefore the minimum value of $$x$$ is 2.
How did you get The numerator to -3*8^3? ->>> $$x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16$$
Math Expert
Joined: 02 Sep 2009
Posts: 50627
### Show Tags
24 Mar 2017, 03:55
Prostar wrote:
Bunuel wrote:
Official Solution:
$$x$$ and $$y$$ are positive integers. If $$y = \sqrt{64}$$ and $$x^2 - 10x = (-4y^3 + 64y)*\frac{1}{96}$$, what is the minimum possible value of $$x$$?
A. 2
B. 4
C. 8
D. 12
E. 16
We know that $$y = 8$$.
Plugging this into the second equation gives:
$$x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16$$
$$x^2-10x = -16$$
$$x^2-10x +16 = 0$$
$$(x-8)(x-2)=0$$
The possible values of $$x$$ are 8 and 2. Therefore the minimum value of $$x$$ is 2.
How did you get The numerator to -3*8^3? ->>> $$x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16$$
It's basic manipulation: $$-4*8^3 + 8^3 = 8^3(-4+1) = 3*8^3$$. The same way -4x + x = -3x.
_________________
Re: M02-13 &nbs [#permalink] 24 Mar 2017, 03:55
Display posts from previous: Sort by
# M02-13
Moderators: chetan2u, Bunuel
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 3,331 | 10,061 | {"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} | 3.96875 | 4 | CC-MAIN-2018-47 | longest | en | 0.829005 |
https://www.javatpoint.com/regression-analysis-in-machine-learning | 1,726,518,687,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00256.warc.gz | 753,833,474 | 21,186 | # Regression Analysis in Machine learning
Regression analysis is a statistical method to model the relationship between a dependent (target) and independent (predictor) variables with one or more independent variables. More specifically, Regression analysis helps us to understand how the value of the dependent variable is changing corresponding to an independent variable when other independent variables are held fixed. It predicts continuous/real values such as temperature, age, salary, price, etc.
We can understand the concept of regression analysis using the below example:
Example: Suppose there is a marketing company A, who does various advertisement every year and get sales on that. The below list shows the advertisement made by the company in the last 5 years and the corresponding sales:
Now, the company wants to do the advertisement of \$200 in the year 2019 and wants to know the prediction about the sales for this year. So to solve such type of prediction problems in machine learning, we need regression analysis.
Regression is a supervised learning technique which helps in finding the correlation between variables and enables us to predict the continuous output variable based on the one or more predictor variables. It is mainly used for prediction, forecasting, time series modeling, and determining the causal-effect relationship between variables.
In Regression, we plot a graph between the variables which best fits the given datapoints, using this plot, the machine learning model can make predictions about the data. In simple words, "Regression shows a line or curve that passes through all the datapoints on target-predictor graph in such a way that the vertical distance between the datapoints and the regression line is minimum." The distance between datapoints and line tells whether a model has captured a strong relationship or not.
Some examples of regression can be as:
• Prediction of rain using temperature and other factors
• Determining Market trends
• Prediction of road accidents due to rash driving.
## Terminologies Related to the Regression Analysis:
• Dependent Variable: The main factor in Regression analysis which we want to predict or understand is called the dependent variable. It is also called target variable.
• Independent Variable: The factors which affect the dependent variables or which are used to predict the values of the dependent variables are called independent variable, also called as a predictor.
• Outliers: Outlier is an observation which contains either very low value or very high value in comparison to other observed values. An outlier may hamper the result, so it should be avoided.
• Multicollinearity: If the independent variables are highly correlated with each other than other variables, then such condition is called Multicollinearity. It should not be present in the dataset, because it creates problem while ranking the most affecting variable.
• Underfitting and Overfitting: If our algorithm works well with the training dataset but not well with test dataset, then such problem is called Overfitting. And if our algorithm does not perform well even with training dataset, then such problem is called underfitting.
## Why do we use Regression Analysis?
As mentioned above, Regression analysis helps in the prediction of a continuous variable. There are various scenarios in the real world where we need some future predictions such as weather condition, sales prediction, marketing trends, etc., for such case we need some technology which can make predictions more accurately. So for such case we need Regression analysis which is a statistical method and used in machine learning and data science. Below are some other reasons for using Regression analysis:
• Regression estimates the relationship between the target and the independent variable.
• It is used to find the trends in data.
• It helps to predict real/continuous values.
• By performing the regression, we can confidently determine the most important factor, the least important factor, and how each factor is affecting the other factors.
## Types of Regression
There are various types of regressions which are used in data science and machine learning. Each type has its own importance on different scenarios, but at the core, all the regression methods analyze the effect of the independent variable on dependent variables. Here we are discussing some important types of regression which are given below:
• Linear Regression
• Logistic Regression
• Polynomial Regression
• Support Vector Regression
• Decision Tree Regression
• Random Forest Regression
• Ridge Regression
• Lasso Regression:
### Linear Regression:
• Linear regression is a statistical regression method which is used for predictive analysis.
• It is one of the very simple and easy algorithms which works on regression and shows the relationship between the continuous variables.
• It is used for solving the regression problem in machine learning.
• Linear regression shows the linear relationship between the independent variable (X-axis) and the dependent variable (Y-axis), hence called linear regression.
• If there is only one input variable (x), then such linear regression is called simple linear regression. And if there is more than one input variable, then such linear regression is called multiple linear regression.
• The relationship between variables in the linear regression model can be explained using the below image. Here we are predicting the salary of an employee on the basis of the year of experience.
• Below is the mathematical equation for Linear regression:
Here, Y = dependent variables (target variables),
X= Independent variables (predictor variables),
a and b are the linear coefficients
Some popular applications of linear regression are:
• Analyzing trends and sales estimates
• Salary forecasting
• Real estate prediction
• Arriving at ETAs in traffic.
### Logistic Regression:
• Logistic regression is another supervised learning algorithm which is used to solve the classification problems. In classification problems, we have dependent variables in a binary or discrete format such as 0 or 1.
• Logistic regression algorithm works with the categorical variable such as 0 or 1, Yes or No, True or False, Spam or not spam, etc.
• It is a predictive analysis algorithm which works on the concept of probability.
• Logistic regression is a type of regression, but it is different from the linear regression algorithm in the term how they are used.
• Logistic regression uses sigmoid function or logistic function which is a complex cost function. This sigmoid function is used to model the data in logistic regression. The function can be represented as:
• f(x)= Output between the 0 and 1 value.
• x= input to the function
• e= base of natural logarithm.
When we provide the input values (data) to the function, it gives the S-curve as follows:
• It uses the concept of threshold levels, values above the threshold level are rounded up to 1, and values below the threshold level are rounded up to 0.
There are three types of logistic regression:
• Binary(0/1, pass/fail)
• Multi(cats, dogs, lions)
• Ordinal(low, medium, high)
### Polynomial Regression:
• Polynomial Regression is a type of regression which models the non-linear dataset using a linear model.
• It is similar to multiple linear regression, but it fits a non-linear curve between the value of x and corresponding conditional values of y.
• Suppose there is a dataset which consists of datapoints which are present in a non-linear fashion, so for such case, linear regression will not best fit to those datapoints. To cover such datapoints, we need Polynomial regression.
• In Polynomial regression, the original features are transformed into polynomial features of given degree and then modeled using a linear model. Which means the datapoints are best fitted using a polynomial line.
• The equation for polynomial regression also derived from linear regression equation that means Linear regression equation Y= b0+ b1x, is transformed into Polynomial regression equation Y= b0+b1x+ b2x2+ b3x3+.....+ bnxn.
• Here Y is the predicted/target output, b0, b1,... bn are the regression coefficients. x is our independent/input variable.
• The model is still linear as the coefficients are still linear with quadratic
### Support Vector Regression:
Support Vector Machine is a supervised learning algorithm which can be used for regression as well as classification problems. So if we use it for regression problems, then it is termed as Support Vector Regression.
Support Vector Regression is a regression algorithm which works for continuous variables. Below are some keywords which are used in Support Vector Regression:
• Kernel: It is a function used to map a lower-dimensional data into higher dimensional data.
• Hyperplane: In general SVM, it is a separation line between two classes, but in SVR, it is a line which helps to predict the continuous variables and cover most of the datapoints.
• Boundary line: Boundary lines are the two lines apart from hyperplane, which creates a margin for datapoints.
• Support vectors: Support vectors are the datapoints which are nearest to the hyperplane and opposite class.
In SVR, we always try to determine a hyperplane with a maximum margin, so that maximum number of datapoints are covered in that margin. The main goal of SVR is to consider the maximum datapoints within the boundary lines and the hyperplane (best-fit line) must contain a maximum number of datapoints. Consider the below image:
Here, the blue line is called hyperplane, and the other two lines are known as boundary lines.
### Decision Tree Regression:
• Decision Tree is a supervised learning algorithm which can be used for solving both classification and regression problems.
• It can solve problems for both categorical and numerical data
• Decision Tree regression builds a tree-like structure in which each internal node represents the "test" for an attribute, each branch represent the result of the test, and each leaf node represents the final decision or result.
• A decision tree is constructed starting from the root node/parent node (dataset), which splits into left and right child nodes (subsets of dataset). These child nodes are further divided into their children node, and themselves become the parent node of those nodes. Consider the below image:
Above image showing the example of Decision Tee regression, here, the model is trying to predict the choice of a person between Sports cars or Luxury car.
• Random forest is one of the most powerful supervised learning algorithms which is capable of performing regression as well as classification tasks.
• The Random Forest regression is an ensemble learning method which combines multiple decision trees and predicts the final output based on the average of each tree output. The combined decision trees are called as base models, and it can be represented more formally as:
```g(x)= f0(x)+ f1(x)+ f2(x)+....
```
• Random forest uses Bagging or Bootstrap Aggregation technique of ensemble learning in which aggregated decision tree runs in parallel and do not interact with each other.
• With the help of Random Forest regression, we can prevent Overfitting in the model by creating random subsets of the dataset.
### Ridge Regression:
• Ridge regression is one of the most robust versions of linear regression in which a small amount of bias is introduced so that we can get better long term predictions.
• The amount of bias added to the model is known as Ridge Regression penalty. We can compute this penalty term by multiplying with the lambda to the squared weight of each individual features.
• The equation for ridge regression will be:
• A general linear or polynomial regression will fail if there is high collinearity between the independent variables, so to solve such problems, Ridge regression can be used.
• Ridge regression is a regularization technique, which is used to reduce the complexity of the model. It is also called as L2 regularization.
• It helps to solve the problems if we have more parameters than samples.
### Lasso Regression:
• Lasso regression is another regularization technique to reduce the complexity of the model.
• It is similar to the Ridge Regression except that penalty term contains only the absolute weights instead of a square of weights.
• Since it takes absolute values, hence, it can shrink the slope to 0, whereas Ridge Regression can only shrink it near to 0.
• It is also called as L1 regularization. The equation for Lasso regression will be: | 2,428 | 12,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-38 | latest | en | 0.915206 |
https://byjus.com/question-answer/at-527-circ-c-the-reaction-given-below-has-k-c-4-nh-3-g/ | 1,685,488,575,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646181.29/warc/CC-MAIN-20230530230622-20230531020622-00033.warc.gz | 190,064,372 | 43,647 | Question
# At 527∘C, the reaction given below has Kc=4NH3(g)⇌12N2(g)+32H2(g)What is the KP for the reaction?N2(g)+3H2(g)⇌2NH3(g).
A
16×(800 R)2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(800R4)2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(14×800R)2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
## The correct option is C (14×800R)2Given reaction is :-NH3(g)⇌12N2(g)+32H2(g)△n= change in no. of moles of gaseous species =(32+12)−1=2−1=1Given, Kc=4Now, we have, Kp=Kc(RT)△n where, R= universal gas constant T= temperature in K=527+273=800KSo, Kp=4×(800R)1Now, for the reaction, N2+3H2(g)⇌2NH3(g) K′p=(1Kp)2so, K1p=(14×800R)2
Suggest Corrections
1
Related Videos
Le Chateliers Principle
CHEMISTRY
Watch in App | 362 | 886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.783004 |
http://www.indiana.edu/~jkkteach/P533/ | 1,556,129,297,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578655155.88/warc/CC-MAIN-20190424174425-20190424200425-00079.warc.gz | 246,628,755 | 12,197 | P533 Bayesian Data Analysis, Prof. John K. Kruschke
Spring 2019: Tu,Th 9:30am-10:45am, Room 111 Psych.
Overview: P533 is a tutorial introduction to doing Bayesian data analysis. The course is intended to make advanced Bayesian methods genuinely accessible to graduate students in the social sciences. Students from all fields are welcome and encouraged to enroll (see figure at right). The course uses examples from a variety of disciplines. The course covers all the fundamental concepts of Bayesian methods, and works from the simplest models up through hierarchical models (a.k.a. multilevel models) applied to various types of data. More details about content are provided below in the Schedule of Topics.
Prerequisites: This is not a mathematical statistics course, but some math is unavoidable. If you understand basic summation notation like Σi xi and integral notation like ∫ x dx , then you're in good shape. We will not be using much math, but we will be doing a lot of computer programming in a language called R. R is free and can be installed on any computer. The textbook includes an introductory chapter on R and there are lots of resources online. A previous course in traditional statistics or probability can be helpful as background, but is not essential. P533 proceeds independently of traditional ("null hypothesis significance testing") statistical methods.
Credit toward I.U. Statistics Department requirements: P533 counts toward the Ph.D. minor in STAT and toward the 12 hour "area relevant to statistics" section of the MSAS (Masters in Applied Statistics).
Course Grading Method: Grading is based on your total homework score, as a percentile relative to the class. There are no exams and no projects. N.B.: Scores tend to be very high, so do not think that, say, 96% must be a grade of A because it could end up being an A- if, say, two thirds of the class does better than 96%. Typically the late penalties turn out to be a bigger deduction than points missed due to errors, so don't fall behind. As this is a graduate course, grades are typically in the A to high B range, and only rarely is a C or less assigned.
All assignments are mandatory. Late homework is exponentially penalized with a half-life of one week, meaning that after one week 50% is the maximum possible score. (The R program for the exponential decay is in the Canvas files; see LatePenaltyCalculator.R.) No homework may be turned in more than three weeks later than its due date (and no homework may be turned in after 12:00 noon of Wednesday of finals week). There are two reasons for this policy: First, the course moves quickly and the material is cumulative, so the late penalty acts as an extra incentive to keep up. Second, the assistant, who will be grading the homework, must not be given a flood of late homework papers at the end of the semester. In recognition of the fact that "life happens" (e.g., short-term illness, personal turmoil, overwhelming confluence of deadlines, etc.), your two worst late penalties will be dropped. In other words, for every homework we will record the scores with and without a late penalty. The two homeworks with the largest difference between with- and without- late penalty will have their late penalty dropped. Note, therefore, that any homework not turned in will count as zero.
Required textbook: Doing Bayesian Data Analysis, 2nd Edition: A Tutorial with R, JAGS, and Stan. Go to the web page, https://sites.google.com/site/doingbayesiandataanalysis/purchase, for a link to purchase the book. The book is also available online through the IU Library.
Instructor: John K. Kruschke, johnkruschke@gmail.com. Office hours by appointment; please do ask.
Assistant: Brad Celestin, bcelesti@umail.iu.edu. Office hours to be posted on Canvas.
Discussion: Please discuss the assignments and lectures on Canvas. If you are attending the class but cannot get access to the Canvas page, please email Prof. Kruschke.
Disclaimer: All information in this document is subject to change. Changes will be announced in class. | 873 | 4,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.926378 |
http://studylib.net/doc/5380252/problem-solving-workshop-powerpoint-3 | 1,521,741,022,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647901.79/warc/CC-MAIN-20180322170754-20180322190754-00696.warc.gz | 283,762,847 | 14,951 | ```PROBLEM SOLVING
Erie 1 BOCES
May 22, 2013
Facilitators: Andrea Tamarazio & Steve Graser
Pre-Assessment
K – 5 Module 1 is currently available
6 – 12 Overview of Module 1 is currently available
Curriculum Maps for Grades 6 – 12 are available
Changes have been made to the K – 5 Curriculum Map
Module Titles
Link to K – 5 Materials
Link to 6 – 12 Materials
K – 5 Modules
Modules released during May NTI
Kindergarten Module 1
Grade 2 Modules 1 & 2
Other Modules also available
Kindergarten Module 5
Recommended Instructional Minutes
Elementary
60 minutes
Fluency
Application Problems (RDW)
Concept Development (includes Problem Set)
Math Models
Student Debrief
MS / HS
45 minutes
Fluency
Application Problems (RDW)
Concept Development
Math Models
Student Debrief
Primary Resource
Districts have choices regarding the modules
Delete
(Ignore)
If adapting or ignoring, it is highly recommended to
implement a Balanced Math Program that will
include the following:
Daily
Fluency
Problem Solving
Student Reflection
Lessons within a Unit Activity
and sequence of lesson titles (6 – 10) for facilitation
of instruction on multiplication.
Title of the Unit is Multiplication
Modify
Lessons within a Unit
Title: Units of Measure
Lesson
1:
Lesson 2:
Lesson 3:
Lesson 4:
Lesson 5:
Lesson 6:
Lesson 7:
Converting Customary Units of Length
Converting Customary Units of Capacity
Converting Customary Units of Weight
Converting Metric Units of Length
Converting Metric Units of Capacity
Converting Metric Units of Weight
Problem Solving: Multi-Step Problems
What is Problem Solving?
Solving problems is not only a goal of learning
mathematics but also a major means of doing so…
Problem solving is an integral part of all
mathematics learning, and so it should not be an
isolated part of mathematics program. Problem
solving in mathematics should involve all the five
content areas described in these standards …
Good problems will integrate multiple topics and
will involve significant mathematics.
-NCTM, Principles and Standards for School Mathematics. (2000), p.52
Math SHIFTS
Mathematical Practices
Make sense of problems and persevere in solving them
Reason abstractly and quantitatively
Construct viable arguments and critique the reasoning
of others
Model with mathematics
Use appropriate tools strategically
Attend to precision
Look for and make use of structure
Look for and express regularity in repeated reasoning
Questions….
How are you providing an opportunity for your
students to apply and explain the mathematical
practices?
How are you including writing in your math
program?
Steal Time from ELA
Steal Time from ELA
Writing for Problem Solving in Math
Class
“When students write, their papers provide a window
into their understandings, their misconceptions, and
M. Burns, “Writing in Math,” Education Leadership, 62(2)
(October 2004), p.30
Rationale
Organize and consolidate their mathematical
thinking through communication
Communicate mathematical thinking coherently and
clearly to others
Analyze and evaluate mathematical thinking and
strategies of others
Use the language of mathematics to express
mathematical ideas precisely.
NCTM, Principles and Standards for School Mathematics. (2000)
Writing for Understanding
Sentence 1 summarize in 15 words or less
Summarize sentence 1 and 2 in 15 word or less
Summarize sentence 1, 2 and 3 in 15 words or less
Continue process for paragraph
Problem: Writing for Understanding
Activity
As I watch the snow pile up outside, I’m dreaming of the sunny
weather in California. Planning an imaginary trip across the
country from Allentown, PA, to Los Angeles, CA, will help keep me
warm. I could use your help.
According to MapQuest, the total driving distance is 2,688.2
miles. With side trips and allowing for getting lost we will
actually cover 2,800 miles. At times we will be on highways
where we can travel from 50-70 miles per hour. At other times we
will be on city streets with a speed limit of 35 miles per hour.
Problem Continued
To help with calculation use the data below:
Travel 35 miles per hour for five hours
Travel 50 miles per hour for fourteen hours
Travel 65 miles per hour for fourteen hours
Travel 70 miles per hour for the remaining distance
If we spend 8 hours a day driving, how many days
will the trip take? How long will we drive the last day?
Bonus
Following the MapQuest route, name the eleven states
in order from Pennsylvania to California that you
would travel through on this imaginary trip
Does this problem promote application of the
mathematical ideas presented in the current instructional
focus or unit of study?
Does this problem match students’ current instructional
level?
Is this problem accessible to all students?
Is the problem relevant and engaging to students?
Does this problem require students to “stretch” their
mathematical reasoning abilities?
(cont)
Does this problem involve more than one strand or
standard of mathematics?
Is there more than one way to solve the problem?
Could the problem be extended or enriched?
Do I fully understand the mathematics in this
problem, so that I can better facilitate student
understanding?
Goal for Problem Solving at the
Ultimate goal for Intermediate grade students is for
them to be able - by the end of the school year –
to solve a multistep problem and to communicate
verbally and in writing the process they used.
Scaffolds to Meet Goal
Model the process for the whole class until familiar
Student solve the selected problems in cooperative
groups or selected teams
Partner students to solve a given problem together
We expect students to be able to complete the
entire process independently
Problem Solving Sequence
Whole class problem solving at beginning of the
year
Gain
experience and capacity for problem solving
Learn the writing
Interact with peers
Communicate mathematically
Transition to cooperative learning groups, partners,
and individual practice
Cooperative Teams
Teacher arranges the students into small
cooperative groups
Guided by teachers, students attempt to solve a
problem
Using words, pictures, and/or numbers, the students
complete the Data Sheet
The students then complete then work onto Chart
Paper
Students share out to the rest of the class
Benefits of Cooperative Teams
Different teams will describe their process
differently and thus add to everyone’s
understanding of the problem
Examples around the room will provide references
for the next solving activity
Math team posters are updated regularly which
keeps student interest in the problem solving process
Find all the different ways you can divide equally
144 marbles into bags.
Data Sheet:
Show work in words, drawings, numbers …..
Write up:
Paragraph
1: Problem Statement
Paragraph 2: Work Write Up
Problem Solving Task Write Up Guide
The total bill for Cassie and Brooke’s dinner was
\$18.40. They want to leave a 15% tip. How should
they determine how much tip to leave? What is the
total amount that Cassie and Brooke spent for dinner?
Problem Solving Task Write Up Guide
Directions
Write your name on a piece of paper
Solve the problem using words, pictures, and / or
numbers
Number each step as you work to solve the problem
Write a number sentence to match your problem
Now write three paragraphs describing the problem,
steps used to solve the problem, and explaining your
Use math vocabulary
Assessing Student’s Problem Solving
What are the……
Strengths
Challenges
of problem solving?
Assessing: Elements to Consider…
What particular elements do you want to see the
students include in their problem solving?
Calculation Errors
Math Vocabulary
Strategies used to solve
Spelling and Conventions
Fluency of Explanations
Guidelines for Rubrics
Performance levels – how many?
Publish and distribute to students prior to them
beginning the problem solving process
Problem Solving Rubric Activity
Problem: Mario has a wall in his room that measures
13 feet long and 8 ½ feet high and is freshly
painted. He wants to hang his favorite posters on the
wall. Each poster measures 3 feet long and 2 feet
high. What is the greatest number of posters that he
can hang on the wall so that the posters do not
overlap?
Problem Solving Rubric Activity (cont)
Create a rubric or guidelines for scoring
Include levels and criteria for each level (first draft)
Problem Solving Rubric Activity (cont)
4 levels
Exemplary
Proficient
Progressing
Beginning
Self
Evaluation
Teacher Evaluation
Let’s Create
Find or create a word problem that meets the
criteria discussed
Create an appropriate data sheet and write up
sheet for the problem
Create an appropriate rubric, include levels and
criteria for each level
Questions or Concerns
If you have any questions or concerns, please feel
Andrea Tamarazio
[email protected]
Steve Graser
[email protected]
``` | 2,352 | 9,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-13 | longest | en | 0.84589 |
http://getindur.com/understanding-the-odds-of-winning-the-lottery/ | 1,716,239,851,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00478.warc.gz | 14,816,712 | 11,025 | # Understanding the Odds of Winning the Lottery
Lottery is a gambling game that involves paying a small sum of money for the chance to win a large prize. It has been used for centuries as a way to raise funds, and it is considered a fair method of awarding prizes. However, the odds of winning are extremely low. Trying to understand the odds of winning can help you make smarter decisions when playing the lottery.
In general, the more numbers you pick, the better your chances of winning. However, there are also many improbable combinations in the lottery, so you need to know which ones to avoid. To do this, you can use a combination of combinatorial math and probability theory. In addition, you can also learn to skip some draws and save yourself a lot of money.
The first recorded lotteries to offer tickets with prizes in the form of money were held in the Low Countries in the 15th century, with town records mentioning lottery-like arrangements for raising money for wall building and town fortifications. The earliest known lottery to give away property, however, was a prize given out at Saturnalian parties in Rome in the 1st century BC. These tickets were distributed by wealthy nobles to their guests, who would then bet on the numbers in hopes of winning something.
Using a mathematical formula can help you increase your chances of winning the lottery. This formula is called the binomial coefficient, and it gives you an estimate of how often each number will appear in the lottery. The higher the binomial coefficient, the more likely you are to win. The lowest binomial coefficient, on the other hand, means you are unlikely to win.
Another tip for improving your odds of winning the lottery is to stay away from numbers that end in the same digit. In addition, avoid numbers that are too hot or too cold. In the case of hot and cold numbers, this is because they have been drawn frequently or rarely in previous lottery draws. You can also use a calculator to find out how often each number has been drawn in previous lotteries.
The chances of winning the lottery depend on how many numbers you pick correctly. The more numbers you choose, the lower your chances are of winning. You can try to improve your odds of winning by limiting the numbers you select to those that are more frequent in the lottery.
Lotteries are a great way to raise money for a specific purpose, such as construction projects or education. In the United States, there are a variety of ways to participate in a lottery, including scratch-off tickets and instant games. The prizes vary, but they are usually worth a large amount of money.
In the past, a few people have been able to predict the results of a lottery drawing by looking at historical patterns. In one case, a Romanian mathematician named Stefan Mandel was able to beat the odds of winning the lottery 14 times by leveraging his formula with investors. While these tips might not work for everyone, they can help you reduce the amount of money you spend on the lottery and improve your odds of winning. | 621 | 3,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-22 | latest | en | 0.979084 |
http://stickycompany.com/northampton/91964408c4351de08bf291840b54-tight-binding-periodic-boundary-conditions | 1,702,088,450,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100781.60/warc/CC-MAIN-20231209004202-20231209034202-00228.warc.gz | 37,232,531 | 8,120 | ### tight binding periodic boundary conditions
The . The density-functional tight-binding (DFTB) formulation of the fragment molecular orbital method is combined with periodic boundary conditions. We provide a number of detailed guides dealing with common task that can be performed easily with the xtb program. It describes the system as real-space Hamiltonian matrices . Such nontrivial winding provides the topological signature of the non-Hermitian skin . On each nucleus n there is an orbital jnithat we consider to be mutually orthogonal to each other hmjni=d m;n: (1) The main objection we can raise about the method is that we are trying to describe the wavefunction of the periodic solid as a combination of atomic orbitals that are eigenstates of a different Schrdinger equation with a differen potential and different boundary conditions. Many nanostructures today are low-dimensional and flimsy, and therefore get easily distorted.
Rapid QM/MM approach for biomolecular systems under periodic boundary conditions: Combination of the densityfunctional tightbinding theory and particle mesh Ewald method . The mathematical details of this easy-to-implement approach, however, have not been discussed before. Revised periodic boundary conditions (RPBC) is a simple method that enables simulations of complex material distortions, either classically or quantum . Revised periodic boundary conditions (RPBC) is a simple method that enables simulations of complex material distortions, either classically or quantum . Dierent forms of crystal binding are discussed: covalent bonds, ionic To this end we introduce for each site x = 1,2,.,L a Boson creation and destruction operator, a x and ax which satisfy . Now imagine we're working with periodic boundary conditions so the hopping matrix has elements corresponding to neighbouring pairs of atoms where the elements of the pair are on opposite sides of the tile. Instead of 1D well of the length L, consider a ring of the same length. Reuse .
Comparison of results for tight-binding and nearly-free electron model. PRB 74, 245126 (2006) Check the example_basic_method class z2pack Iterative methods are required when the dimension of the Hamiltonian becomes too large for exact diagonalization routines ergy spectrum and the corresponding eigenstates of H,b can be approximated by a discrete tight-binding (eective) Hamiltonian, HTB acting on 2(G) ergy . Besides being applicable to materials with covalent bonds, . User Guide to Semiempirical Tight Binding. Lecture 23-Graphene continued, Wannier function, spin-orbit .
Implementation of the xTB methods is realized via a library spin-off from xtb, which will be upstreamed into this project in the future. Lecture 21 - Fermi surface in tight binding, hybridization of atomic orbitals, variational derivation of tight binding.
The density-functional tight-binding (DFTB) formulation of the fragment molecular orbital method is combined with periodic boundary conditions. As compared with that of the conventional Ewald summation method, the . The first analytic derivatives of the energy with respect to atom The way out is to introduce periodic boundary conditions (PBC). Electronic band structures plot the energy eigenstates of an electron in the presence of a periodic potential as a function of momentum. (r) = ck exp (ikr). Distortion-induced symmetry-breaking makes conventional, translation-periodic simulations invalid, which has triggered developments for new methods. 1-3. under periodic boundary conditions (PBC), finding the energy spectrum associated to the Bloch eigenstates is a straightforward task. Absorbing boundary conditions. Tight-Binding parameters for the Elements. In the case of the electron system, periodic boundary conditions give 0 = N, which results in 1 = e i k 0 = e i k N a. They yield many useful properties of solid-state materials . Search: Tight Binding Hamiltonian Eigenstates. A quantum mechanical/molecular mechanical (QM/MM) approach based on the densityfunctional tightbinding (DFTB) theory is a useful tool for analyzing chemical . Limitations of the tight-binding model The main objection we can raise about the method is that we are trying to describe the wavefunction of the periodic solid as a combination of atomic orbitals that are eigenstates of a different Schrdinger equation with a differen potential and different boundary conditions. | 883 | 4,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.886907 |
https://www.fxsolver.com/browse/?like=2253&p=12 | 1,597,328,020,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739046.14/warc/CC-MAIN-20200813132415-20200813162415-00554.warc.gz | 669,172,937 | 37,960 | '
# Search results
Found 1629 matches
Redshift: 1+z (based on frequency)
In physics, redshift happens when light or other electromagnetic radiation from an object is increased in wavelength, or shifted to the red end of the ... more
Redshift (based on wavelength)
In physics, redshift happens when light or other electromagnetic radiation from an object is increased in wavelength, or shifted to the red end of the ... more
Redshift: 1+z (based on wavelength)
In physics, redshift happens when light or other electromagnetic radiation from an object is increased in wavelength, or shifted to the red end of the ... more
Temprature of a planet
Black-body radiation is the thermal electromagnetic radiation within or surrounding a body in thermodynamic equilibrium with its environment, or emitted by ... more
Tractive Force
As used in mechanical engineering, the term tractive force can either refer to the total traction a vehicle exerts on a surface, or the amount of the total ... more
Hemispherical Reflectance
Reflectance of the surface of a material is its effectiveness in reflecting radiant energy. It is the fraction of incident electromagnetic power that is ... more
AC Power
In alternating current circuits, energy storage elements such as inductance and capacitance may result in periodic reversals of the direction of energy ... more
Tractive Force - Steam locomotives
As used in mechanical engineering, the term tractive force can either refer to the total traction a vehicle exerts on a surface, or the amount of the total ... more
Snell's law (velocities)
Snell’s law (also known as the Snell–Descartes law and the law of refraction) is a formula used to describe the relationship between the angles of ... more
Wavelength - Sinusoidal Wave
In physics, the wavelength of a sinusoidal wave is the spatial period of the wave—the distance over which the wave’s shape repeats, and the inverse ... more
...can't find what you're looking for?
Create a new formula
### Search criteria:
Similar to formula
Category | 435 | 2,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.896073 |
https://www.lmfdb.org/EllipticCurve/Q/2760/a/2 | 1,701,864,606,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100593.71/warc/CC-MAIN-20231206095331-20231206125331-00709.warc.gz | 921,956,827 | 25,493 | # Properties
Label 2760.a2 Conductor $2760$ Discriminant $121881600$ j-invariant $$\frac{273671716}{119025}$$ CM no Rank $0$ Torsion structure $$\Z/{2}\Z \oplus \Z/{2}\Z$$
# Related objects
Show commands: Magma / Oscar / PariGP / SageMath
## Simplified equation
$$y^2=x^3-x^2-136x-260$$ y^2=x^3-x^2-136x-260 (homogenize, simplify) $$y^2z=x^3-x^2z-136xz^2-260z^3$$ y^2z=x^3-x^2z-136xz^2-260z^3 (dehomogenize, simplify) $$y^2=x^3-11043x-222642$$ y^2=x^3-11043x-222642 (homogenize, minimize)
comment: Define the curve
sage: E = EllipticCurve([0, -1, 0, -136, -260])
gp: E = ellinit([0, -1, 0, -136, -260])
magma: E := EllipticCurve([0, -1, 0, -136, -260]);
oscar: E = EllipticCurve([0, -1, 0, -136, -260])
sage: E.short_weierstrass_model()
magma: WeierstrassModel(E);
oscar: short_weierstrass_model(E)
## Mordell-Weil group structure
$$\Z/{2}\Z \oplus \Z/{2}\Z$$
magma: MordellWeilGroup(E);
## Torsion generators
$$\left(-2, 0\right)$$, $$\left(13, 0\right)$$
comment: Torsion subgroup
sage: E.torsion_subgroup().gens()
gp: elltors(E)
magma: TorsionSubgroup(E);
oscar: torsion_structure(E)
## Integral points
$$\left(-10, 0\right)$$, $$\left(-2, 0\right)$$, $$\left(13, 0\right)$$
comment: Integral points
sage: E.integral_points()
magma: IntegralPoints(E);
## Invariants
Conductor: $$2760$$ = $2^{3} \cdot 3 \cdot 5 \cdot 23$ comment: Conductor sage: E.conductor().factor() gp: ellglobalred(E)[1] magma: Conductor(E); oscar: conductor(E) Discriminant: $121881600$ = $2^{10} \cdot 3^{2} \cdot 5^{2} \cdot 23^{2}$ comment: Discriminant sage: E.discriminant().factor() gp: E.disc magma: Discriminant(E); oscar: discriminant(E) j-invariant: $$\frac{273671716}{119025}$$ = $2^{2} \cdot 3^{-2} \cdot 5^{-2} \cdot 23^{-2} \cdot 409^{3}$ comment: j-invariant sage: E.j_invariant().factor() gp: E.j magma: jInvariant(E); oscar: j_invariant(E) Endomorphism ring: $\Z$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) sage: E.has_cm() magma: HasComplexMultiplication(E); Sato-Tate group: $\mathrm{SU}(2)$ Faltings height: $0.24765643952406513656860946266\dots$ gp: ellheight(E) magma: FaltingsHeight(E); oscar: faltings_height(E) Stable Faltings height: $-0.32996621094255595461241730522\dots$ magma: StableFaltingsHeight(E); oscar: stable_faltings_height(E)
## BSD invariants
Analytic rank: $0$ sage: E.analytic_rank() gp: ellanalyticrank(E) magma: AnalyticRank(E); Regulator: $1$ comment: Regulator sage: E.regulator() G = E.gen \\ if available matdet(ellheightmatrix(E,G)) magma: Regulator(E); Real period: $1.4537399776486878507629577471\dots$ comment: Real Period sage: E.period_lattice().omega() gp: if(E.disc>0,2,1)*E.omega[1] magma: (Discriminant(E) gt 0 select 2 else 1) * RealPeriod(E); Tamagawa product: $16$ = $2\cdot2\cdot2\cdot2$ comment: Tamagawa numbers sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] magma: TamagawaNumbers(E); oscar: tamagawa_numbers(E) Torsion order: $4$ comment: Torsion order sage: E.torsion_order() gp: elltors(E)[1] magma: Order(TorsionSubgroup(E)); oscar: prod(torsion_structure(E)[1]) Analytic order of Ш: $1$ (exact) comment: Order of Sha sage: E.sha().an_numerical() magma: MordellWeilShaInformation(E); Special value: $L(E,1)$ ≈ $1.4537399776486878507629577471$ comment: Special L-value r = E.rank(); E.lseries().dokchitser().derivative(1,r)/r.factorial() gp: [r,L1r] = ellanalyticrank(E); L1r/r! magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12);
## BSD formula
$\displaystyle 1.453739978 \approx L(E,1) = \frac{\# Ш(E/\Q)\cdot \Omega_E \cdot \mathrm{Reg}(E/\Q) \cdot \prod_p c_p}{\#E(\Q)_{\rm tor}^2} \approx \frac{1 \cdot 1.453740 \cdot 1.000000 \cdot 16}{4^2} \approx 1.453739978$
# self-contained SageMath code snippet for the BSD formula (checks rank, computes analytic sha)
E = EllipticCurve(%s); r = E.rank(); ar = E.analytic_rank(); assert r == ar;
Lr1 = E.lseries().dokchitser().derivative(1,r)/r.factorial(); sha = E.sha().an_numerical();
omega = E.period_lattice().omega(); reg = E.regulator(); tam = E.tamagawa_product(); tor = E.torsion_order();
assert r == ar; print("analytic sha: " + str(RR(Lr1) * tor^2 / (omega * reg * tam)))
/* self-contained Magma code snippet for the BSD formula (checks rank, computes analyiic sha) */
E := EllipticCurve(%s); r := Rank(E); ar,Lr1 := AnalyticRank(E: Precision := 12); assert r eq ar;
sha := MordellWeilShaInformation(E); omega := RealPeriod(E) * (Discriminant(E) gt 0 select 2 else 1);
reg := Regulator(E); tam := &*TamagawaNumbers(E); tor := #TorsionSubgroup(E);
assert r eq ar; print "analytic sha:", Lr1 * tor^2 / (omega * reg * tam);
## Modular invariants
$$q - q^{3} - q^{5} + q^{9} + 4 q^{11} + 2 q^{13} + q^{15} + 2 q^{17} + O(q^{20})$$
comment: q-expansion of modular form
sage: E.q_eigenform(20)
\\ actual modular form, use for small N
[mf,F] = mffromell(E)
Ser(mfcoefs(mf,20),q)
\\ or just the series
Ser(ellan(E,20),q)*q
magma: ModularForm(E);
Modular degree: 768
comment: Modular degree
sage: E.modular_degree()
gp: ellmoddegree(E)
magma: ModularDegree(E);
$\Gamma_0(N)$-optimal: no
Manin constant: 1
comment: Manin constant
magma: ManinConstant(E);
## Local data
This elliptic curve is not semistable. There are 4 primes of bad reduction:
prime Tamagawa number Kodaira symbol Reduction type Root number ord($N$) ord($\Delta$) ord$(j)_{-}$
$2$ $2$ $III^{*}$ Additive -1 3 10 0
$3$ $2$ $I_{2}$ Non-split multiplicative 1 1 2 2
$5$ $2$ $I_{2}$ Non-split multiplicative 1 1 2 2
$23$ $2$ $I_{2}$ Non-split multiplicative 1 1 2 2
comment: Local data
sage: E.local_data()
gp: ellglobalred(E)[5]
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
oscar: [(p,tamagawa_number(E,p), kodaira_symbol(E,p), reduction_type(E,p)) for p in bad_primes(E)]
## Galois representations
The $\ell$-adic Galois representation has maximal image for all primes $\ell$ except those listed in the table below.
prime $\ell$ mod-$\ell$ image $\ell$-adic image
$2$ 2Cs 8.12.0.1
comment: mod p Galois image
sage: rho = E.galois_representation(); [rho.image_type(p) for p in rho.non_surjective()]
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
gens = [[1, 4, 0, 1], [1657, 2, 0, 1], [2757, 4, 2756, 5], [1841, 2, 0, 1], [1, 0, 4, 1], [1383, 2, 2758, 2759], [1201, 4, 2402, 9], [2071, 4, 1382, 9]]
GL(2,Integers(2760)).subgroup(gens)
Gens := [[1, 4, 0, 1], [1657, 2, 0, 1], [2757, 4, 2756, 5], [1841, 2, 0, 1], [1, 0, 4, 1], [1383, 2, 2758, 2759], [1201, 4, 2402, 9], [2071, 4, 1382, 9]];
sub<GL(2,Integers(2760))|Gens>;
The image $H:=\rho_E(\Gal(\overline{\Q}/\Q))$ of the adelic Galois representation has level $$2760 = 2^{3} \cdot 3 \cdot 5 \cdot 23$$, index $48$, genus $0$, and generators
$\left(\begin{array}{rr} 1 & 4 \\ 0 & 1 \end{array}\right),\left(\begin{array}{rr} 1657 & 2 \\ 0 & 1 \end{array}\right),\left(\begin{array}{rr} 2757 & 4 \\ 2756 & 5 \end{array}\right),\left(\begin{array}{rr} 1841 & 2 \\ 0 & 1 \end{array}\right),\left(\begin{array}{rr} 1 & 0 \\ 4 & 1 \end{array}\right),\left(\begin{array}{rr} 1383 & 2 \\ 2758 & 2759 \end{array}\right),\left(\begin{array}{rr} 1201 & 4 \\ 2402 & 9 \end{array}\right),\left(\begin{array}{rr} 2071 & 4 \\ 1382 & 9 \end{array}\right)$.
Input positive integer $m$ to see the generators of the reduction of $H$ to $\mathrm{GL}_2(\Z/m\Z)$:
The torsion field $K:=\Q(E[2760])$ is a degree-$196977623040$ Galois extension of $\Q$ with $\Gal(K/\Q)$ isomorphic to the projection of $H$ to $\GL_2(\Z/2760\Z)$.
## Isogenies
gp: ellisomat(E)
This curve has non-trivial cyclic isogenies of degree $d$ for $d=$ 2.
Its isogeny class 2760.a consists of 4 curves linked by isogenies of degrees dividing 4.
## Twists
This elliptic curve is its own minimal quadratic twist.
## Growth of torsion in number fields
The number fields $K$ of degree less than 24 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ $\cong \Z/{2}\Z \oplus \Z/{2}\Z$ are as follows:
$[K:\Q]$ $E(K)_{\rm tors}$ Base change curve $K$ $4$ $$\Q(\sqrt{-2}, \sqrt{-23})$$ $$\Z/2\Z \oplus \Z/4\Z$$ Not in database $4$ $$\Q(\sqrt{15}, \sqrt{23})$$ $$\Z/2\Z \oplus \Z/4\Z$$ Not in database $4$ $$\Q(\sqrt{2}, \sqrt{-15})$$ $$\Z/2\Z \oplus \Z/4\Z$$ Not in database $8$ 8.2.31726715921280000.5 $$\Z/2\Z \oplus \Z/6\Z$$ Not in database $16$ deg 16 $$\Z/4\Z \oplus \Z/4\Z$$ Not in database $16$ deg 16 $$\Z/2\Z \oplus \Z/8\Z$$ Not in database $16$ deg 16 $$\Z/2\Z \oplus \Z/8\Z$$ Not in database $16$ deg 16 $$\Z/2\Z \oplus \Z/8\Z$$ Not in database
We only show fields where the torsion growth is primitive. For fields not in the database, click on the degree shown to reveal the defining polynomial.
## Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 23 add nonsplit nonsplit nonsplit - 0 0 0 - 0 0 0
All Iwasawa $\lambda$ and $\mu$-invariants for primes $p\ge 3$ of good reduction are zero.
An entry - indicates that the invariants are not computed because the reduction is additive.
## $p$-adic regulators
All $p$-adic regulators are identically $1$ since the rank is $0$. | 3,515 | 9,135 | {"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} | 3.109375 | 3 | CC-MAIN-2023-50 | latest | en | 0.237214 |
https://oneclass.com/study-guides/us/lsu/ie/ie-3302/2511049-walpole-9e-ism-01-04-11-final.en.html | 1,600,460,121,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400188841.7/warc/CC-MAIN-20200918190514-20200918220514-00398.warc.gz | 518,708,627 | 258,578 | Study Guides (400,000)
US (230,000)
LSU (10,000)
IE (40)
IE 3302 (30)
All (30)
# Walpole 9e ISM 01 04 11 FINALExam
Department
Industrial Engineering
Course Code
IE 3302
Professor
All
This preview shows pages 1-3. to view the full 257 pages of the document.
Contents
1 Introduction to Statistics and Data Analysis 1
2 Probability 11
3 Random Variables and Probability Distributions 27
4 Mathematical Expectation 41
5 Some Discrete Probability Distributions 55
6 Some Continuous Probability Distributions 67
7 Functions of Random Variables 79
8 Fundamental Sampling Distributions and Data Descriptions 85
9 One- and Two-Sample Estimation Problems 97
10 One- and Two-Sample Tests of Hypotheses 113
11 Simple Linear Regression and Correlation 139
12 Multiple Linear Regression and Certain Nonlinear Regression Models 161
13 One-Factor Experiments: General 175
14 Factorial Experiments (Two or More Factors) 197
15 2kFactorial Experiments and Fractions 219
16 Nonparametric Statistics 233
17 Statistical Quality Control 247
18 Bayesian Statistics 251
iii
Only pages 1-3 are available for preview. Some parts have been intentionally blurred.
Only pages 1-3 are available for preview. Some parts have been intentionally blurred.
Chapter 1
Introduction to Statistics and Data
Analysis
1.1 (a) 15.
(b) ¯x=1
15 (3.4+2.5+4.8+···+4.8) = 3.787.
(c) Sample median is the 8th value, after the data is sorted from smallest to largest: 3.6.
(d) A dot plot is shown below.
2.5 3.0 3.5 4.0 4.5 5.0 5.5
(e) After trimming total 40% of the data (20% highest and 20% lowest), the data becomes:
2.9 3.0 3.3 3.4 3.6
3.7 4.0 4.4 4.8
So. the trimmed mean is
¯xtr20 =1
9(2.9+3.0+···+4.8) = 3.678.
(f) They are about the same.
1.2 (a) Mean=20.7675 and Median=20.610.
(b) ¯xtr10 =20.743.
(c) A dot plot is shown below.
18 19 20 21 22 23
(d) No. They are all close to each other. | 601 | 1,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2020-40 | latest | en | 0.71742 |
https://web2.0calc.com/questions/help-asap-please_6 | 1,558,346,525,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255837.21/warc/CC-MAIN-20190520081942-20190520103942-00314.warc.gz | 708,239,672 | 5,926 | +0
0
245
1
If $f(x) = \frac{4x+1}{3}$ what is the value of $\left[f^{-1}(1)\right]^{-1}$?
Mar 20, 2018
#1
+22152
+2
If $f(x) = \frac{4x+1}{3}$ what is the value of $\left[f^{-1}(1)\right]^{-1}$?
$$\begin{array}{|rcll|} \hline f(x) &=& \dfrac{4x+1}{3} \\ \hline y &=& \dfrac{4x+1}{3} \\\\ 3y &=& 4x+1 \\\\ 3y-1 &=& 4x \\\\ x &=& \dfrac{3y-1}{4} \quad & | \quad x \leftrightarrow y \\\\ y &=& \dfrac{3x-1}{4} \\\\ f^{-1}(x) &=& \dfrac{3x-1}{4} \\ \hline f^{-1}(1) &=& \dfrac{3\cdot 1-1}{4} \\\\ &=& \dfrac{2}{4} \\\\ &=& \dfrac{1}{2} \\ \hline \left[f^{-1}(1)\right]^{-1} &=& \left(\dfrac{1}{2}\right)^{-1} \\\\ &=& \dfrac{1}{ \left(\dfrac{1}{2}\right)^{1}} \\\\ &=& \dfrac{1}{ \dfrac{1}{2} } \\\\ &=& 1\cdot \dfrac{2}{1} \\\\ \mathbf{ \left[f^{-1}(1)\right]^{-1}} &\mathbf{=} & \mathbf{2} \\ \hline \end{array}$$
Mar 20, 2018 | 428 | 831 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-22 | latest | en | 0.228925 |
http://www.math.yorku.ca/Who/Faculty/Monette/S-news/0564.html | 1,540,276,213,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516071.83/warc/CC-MAIN-20181023044407-20181023065907-00126.warc.gz | 487,998,059 | 2,113 | # [S] RE: Solving f(x)=0 - SUMMARY
Fri, 13 Mar 1998 08:05:00 +0100
The short reply is: uniroot(). (Luk Wouters and Kevin Brand). Barry
Brown suggested to look at his library invmf on statlib.
Jim Stapleton wrote:
I use a function "newton". For the multivariate case I use a function
"newtonraphson". Both require derivatives, though the derivatives can
be defined in approximation from the difference quotients. They are a
bit crude and not fast, though fast enough for my purposes.
js
> newton
function(x0, G, g, ep = 0.00001, z = 0, step = 1)
{
#Given the function G and its derivative g,
#newton uses the Newton method, beginning at x0,
#to find a point xp at which G is zero. G and g
#may each depend on a parameter z. The result
#is a list of the vector of x-values, the vector
#of G(x) values, and , the number of iterations.
#To get only the final x and final G(x) replace
#list(xv,Gv,lxv) by c(newx,Gnew,lxv).
# ep determines the stopping rule.
x0vec <- x0
count <- 0
Gvaluevec <- Gvalue <- G(x0, z)
while(abs(Gvalue) > ep) {
r <- Gvalue/g(x0, z)
d <- min(abs(r), step) * sign(r)
x0 <- x0 - d
Gvalue <- G(x0, z)
Gvaluevec <- c(Gvaluevec, Gvalue)
x0vec <- c(x0vec, x0)
count <- count + 1
if(count > 50)
stop("count > 50")
}
list(x0, x0vec, Gvaluevec, count)
}
Many thanks to all who replied.
_________________________________
Subject: Solving f(x)=0
Date: 3/11/98 9:14 AM
Dear S+ users,
Is anybody aware of a FAST function to solve numerically f(x)=0?
TIA | 472 | 1,471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.652066 |
http://www.jiskha.com/display.cgi?id=1368253037 | 1,495,900,441,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608956.34/warc/CC-MAIN-20170527152350-20170527172350-00480.warc.gz | 660,964,060 | 3,741 | # Physics
posted by on .
A car that was initially at rest reaches a final speed of +50.0 m/s over a displacement of +55.0 m.
What is the magnitude of the acceleration?
Find the time it took to travel the first 55 m.
• Physics - ,
a=v²/2s=...
s=at²/2 =>
t=sqrt{2s/a)=...
• Physics - ,
I don't understand.. could you give a better answer? | 101 | 344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-22 | latest | en | 0.93252 |
http://eblogarithm.com/codeforces-round-408-div-2-problem-e-exam-cheating-solution-cc/ | 1,521,283,152,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257644877.27/warc/CC-MAIN-20180317100705-20180317120705-00123.warc.gz | 92,681,073 | 23,287 | # Codeforces Round #408 (Div. 2), problem: (E) Exam Cheating Solution in C/C++
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <assert.h>
#define clr(ar) memset(ar, 0, sizeof(ar))
short dp[2][1052][1052];
int n, p, k, A[10010], B[10010], C[10010];
int solve(int lim){
int i, j, p, x, u, v, a, b, res;
clr(dp);
for (p = 1; p <= lim; p++){
u = p & 1, v = u ^ 1;
for (i = n + 1; i >= 1; i–){
a = ((i + k) > n) ? n + 1 : i + k;
for (j = n + 1; j >= i && i != (n + 1); j–){
dp[u][i][j] = dp[u][i + 1][j];
x = A[i + k – 1] – A[i – 1] + dp[v][a][j] – C[j – 1] + C[i – 1];
if (x > dp[u][i][j]) dp[u][i][j] = x;
}
if (j != (n + 1)){
j = i;
a = ((j + k) > n) ? n + 1 : j + k;
res = dp[u][i][j + 1];
x = B[j + k – 1] – B[j – 1] + dp[v][i][a] – C[i – 1] + C[j – 1];
if (x > res) res = x;
if (res > dp[u][i][j]) dp[u][i][j] = res;
}
for (j = i – 1; j >= 1; j–){
a = ((j + k) > n) ? n + 1 : j + k;
dp[u][i][j] = dp[u][i][j + 1];
x = B[j + k – 1] – B[j – 1] + dp[v][i][a] – C[i – 1] + C[j – 1];
if (x > dp[u][i][j]) dp[u][i][j] = x;
}
}
}
return dp[lim & 1][1][1];
}
int main(){
int i, j, l, x;
while (scanf(“%d %d %d”, &n, &p, &k) != EOF){
clr(A), clr(B), clr(C);
scanf(“%d”, &l);
while (l–){
scanf(“%d”, &x);
A[x] = 1;
}
scanf(“%d”, &l);
while (l–){
scanf(“%d”, &x);
B[x] = 1;
}
for (i = 1; i < 10010; i++) C[i] = A[i] & B[i];
for (i = 1; i < 10010; i++) A[i] += A[i – 1], B[i] += B[i – 1], C[i] += C[i – 1];
printf(“%d\n”, solve(p));
}
return 0;
}
(Visited 93 times, 1 visits today) | 739 | 1,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | longest | en | 0.27422 |
http://www.wyzant.com/resources/answers/11637/if_1_a_2_sin_2acos_0_prove_a_2_1_cos_1_cos | 1,397,903,817,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00363-ip-10-147-4-33.ec2.internal.warc.gz | 787,257,272 | 9,922 | Search 72,504 tutors
0 0
## if (1-a^2)sin-2acos=0,prove a^2(1+cos)=1-cos
Prove a^2(1+cos)=1-cos if 1-a^2)sin - 2acos =0
Sine and Cosine are functions, so they must be performed on something. Are these all sines and cosines of the same variable?
I agree with John absolutely!!! Konisko, you are a little too free with math functions! What you wrote is a2(1 + c * o * s) = 1 - c * o * s
In Mathematics, there is big difference between cos(x) which is function and cos=c*o*s which is expression. You have to copy the problem precisely.
Hi Konisko!
Looks like (1-a^2)= 2a(cos/sin) or 2a(cot) ... tan = 2a / (1-a^2) = opp / adj ...
View (1-a^2) as "adj" of a right triangle with the "opp" as 2a and the h^2 is
h^2 = 4a^2 + (1 - 2a^2 + a^4) = a^4 + 2a^2 +1 ... meaning h = a^2 + 1 ...
Plug in left side: a^2= h-1 ... (h-1)(1+cos)= (h-1)(1+ adj/h) = h + adj - 1 - adj/h
Right side: 1-cos = 1 - adj/h ... leaving h + adj = 2 ... (a^2 + 1) + (1-a^2) = 2
SUMMARY: view the given "2a" and "1-a^2" as sides of a right triangle ...
Best regards :) | 414 | 1,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2014-15 | latest | en | 0.756024 |
https://www.stat.math.ethz.ch/pipermail/r-help/2011-November/296251.html | 1,656,768,702,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104141372.60/warc/CC-MAIN-20220702131941-20220702161941-00605.warc.gz | 1,068,385,265 | 2,494 | # [R] Adding two or more columns of a data frame for each row when NAs are present.
Ian Strang hamamelis at ntlworld.com
Sun Nov 20 21:38:11 CET 2011
```I am fairly new to R and would like help with the problem below. I am
trying to sum and count several rows in the data frame yy below. All works
well as in example 1. When I try to add the columns, with an NA in Q21, I
get as NA as mySum. I would like NA to be treated as O, or igored.
I wrote a function to try to count an NA element as 0, Example 3 function.
It works with a few warnings, Example 4, but still gives NA instead of the
addition when there is an NA in an element.
In Example 6 & 7, I tried using sum() but it just sums the whole data
frame, I think,
How do I add together several columns giving the result for each row in
mySum? NA should be treated as a 0. Please, note, I do not want to sum all
the columns, as I think rowSums would do, just the selected ones.
Ian,
> yy <- read.table( header = T, sep=",", text = ## to create a data frame
+ "Q20, Q21, Q22, Q23, Q24
+ 0,1, 2,3,4
+ 1,NA,2,3,4
+ 2,1, 2,3,4")
+ yy
Q20 Q21 Q22 Q23 Q24
1 0 1 2 3 4
2 1 NA 2 3 4
3 2 1 2 3 4
> x <- transform( yy, ############## Example 1
+ mySum = as.numeric(Q20) + as.numeric(Q22) + as.numeric(Q24),
+ myCount =
as.numeric(!is.na(Q20))+as.numeric(!is.na(Q21))+as.numeric(!is.na(Q24))
+ )
+ x
Q20 Q21 Q22 Q23 Q24 mySum myCount
1 0 1 2 3 4 6 3
2 1 NA 2 3 4 7 2
3 2 1 2 3 4 8 3
>
+ x <- transform( yy, ################ Example 2
+ mySum = as.numeric(Q20) + as.numeric(Q21) + as.numeric(Q24),
+ myCount =
as.numeric(!is.na(Q20))+as.numeric(!is.na(Q21))+as.numeric(!is.na(Q24))
+ )
+ x
Q20 Q21 Q22 Q23 Q24 mySum myCount
1 0 1 2 3 4 5 3
2 1 NA 2 3 4 NA 2
3 2 1 2 3 4 7 3
> NifAvail <- function(x) { if (is.na(x)) x<-0 else x <- x
############### Example 3
+ return(as.numeric(x))
+ } #end function
+ NifAvail(5)
[1] 5
+ NifAvail(NA)
[1] 0
> x <- transform( yy,
+ mySum = NifAvail(Q20) + NifAvail(Q22) + NifAvail(Q24),
############### Example 4
+ myCount =
as.numeric(!is.na(Q20))+as.numeric(!is.na(Q21))+as.numeric(!is.na(Q24))
+ )
Warning messages:
1: In if (is.na(x)) x <- 0 else x <- x :
the condition has length > 1 and only the first element will be used
2: In if (is.na(x)) x <- 0 else x <- x :
the condition has length > 1 and only the first element will be used
3: In if (is.na(x)) x <- 0 else x <- x :
the condition has length > 1 and only the first element will be used
> x
Q20 Q21 Q22 Q23 Q24 mySum myCount
1 0 1 2 3 4 6 3
2 1 NA 2 3 4 7 2
3 2 1 2 3 4 8 3
> x <- transform( yy,
+ mySum = NifAvail(Q20) + NifAvail(Q21) + NifAvail(Q24),
################ Example 5
+ myCount =
as.numeric(!is.na(Q20))+as.numeric(!is.na(Q21))+as.numeric(!is.na(Q24))
+ )
Warning messages:
1: In if (is.na(x)) x <- 0 else x <- x :
the condition has length > 1 and only the first element will be used
2: In if (is.na(x)) x <- 0 else x <- x :
the condition has length > 1 and only the first element will be used
3: In if (is.na(x)) x <- 0 else x <- x :
the condition has length > 1 and only the first element will be used
> x
Q20 Q21 Q22 Q23 Q24 mySum myCount
1 0 1 2 3 4 5 3
2 1 NA 2 3 4 NA 2
3 2 1 2 3 4 7 3
> x <- transform( yy, ############
Example 6
+ mySum = sum(as.numeric(Q20), as.numeric(Q21), as.numeric(Q23), na.rm=T),
+ myCount =
as.numeric(!is.na(Q20))+as.numeric(!is.na(Q21))+as.numeric(!is.na(Q24))
+ )
+ x
Q20 Q21 Q22 Q23 Q24 mySum myCount
1 0 1 2 3 4 14 3
2 1 NA 2 3 4 14 2
3 2 1 2 3 4 14 3
> x <- transform( yy, #############
Example 7
+ mySum = sum(as.numeric(Q20), as.numeric(Q22), as.numeric(Q23), na.rm=T),
+ myCount =
as.numeric(!is.na(Q20))+as.numeric(!is.na(Q21))+as.numeric(!is.na(Q24))
+ )
+ x
Q20 Q21 Q22 Q23 Q24 mySum myCount
1 0 1 2 3 4 18 3
2 1 NA 2 3 4 18 2
3 2 1 2 3 4 18 3
``` | 1,680 | 4,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-27 | latest | en | 0.756451 |
www.mystery-shoppers.at | 1,596,857,469,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737238.53/warc/CC-MAIN-20200808021257-20200808051257-00000.warc.gz | 164,143,070 | 32,539 | Categories
## Why do atoms on the periodic table have the number of valence electrons that they do?
I’m studying year 11 and am learning about Valence electrons and how these electrons drive inter-molecular bonds. What I have not been able to find an explanation to is why all the elements on the periodic table have the number of valence electrons that they do. To me, it seems intuitive that if the Halides […]
Categories
## Phenol vs 4-hydroxypyridine – Enol Content
Which has a higher enol content; phenol or 4-hydroxypyridine? The solution to this question in my book says that phenol has a higher enol content due to aromaticity. But 4-hydroxypyridine is also stabilized by resonance. So how does that affect enol content?
Categories
## I am simulating for LiH molecule by hartree fock method, but I am not cleared about its chemical accuray calculations
Is there is a formula for calculating the chemical accuracy or any other optimizer used? Its energy is HF energy: -8.860622387537344
Categories
## Thermochemistry questions regarding \$\Delta\$H
I have a few questions about heat of combustion. I have two different scenarios that I was to present: Scenario 1: Given heat of combustion in the equation \$\ce{CH4 + 2O2 => CO2 + H2O}\$ and \$\Delta H=-889kJ/mol\$ Q1: Would 1 mol of methane produce 889kJ; whereas, 1 mol of oxygen would produce 444.5kJ? Scenario […]
Categories
## Xe compounds – without O or F
I may be wrong, because this is purely based on observation-Is it true that Xe does not form compounds without O or F? If yes, could you please elaborate the reason behind it? (I have a basic understanding of VBT, VSEPR and hybridisation) Any help will be appreciated. Thanks!
Categories
## Finding values within an Array/Mapping based off min and max values
I want a user to be able to place a bet on some number, between a range of numbers. So the user wants to place a bet on the numbers 50-500. Then there is a function that sees if the number has been bet on – such as hasBet(uint256 number_to_check) Lets say the user places […]
Categories
## Mixed DPI across multiple monitors with xrandr
I am running i3 window manager on Arch Linux, and am trying to use xrandr to use both my laptop screen and desktop monitor together. I intend the screens to have separate workspaces but to be able to move workspaces between monitors as needed. The laptop screen is 4K (3840×2160) but the monitor is only […]
Categories
## What to do when which command returns “type: bad option: -l”?
I’m trying to figure out which python version I am using, so I am running which python. Yet every time I’m running it, I receive the error message "type: bad option: -l". What can I do to make the error message go away? I know that which should best be avoided. What could I use […]
Categories
## Problems with Intel Centrino Ultimate-N 6300
I edit /etc/modprobe.d/iwlwifi11n.conf like this: when I set: options iwlwifi 11n_disable=8 I get high speeds but i have connection drop outs.. when I set options iwlwifi 11n_disable=1 I get slow speeds but there are no drop outs… How can i get high speeds with no connection drop outs ? Here is the wireless-info: https://pastebin.com/3wf1fg8s
Categories
## Cannot get “apt-get install” or “apt install” to work on Kali Linux VM
I recently downloaded one of the prebuilt Kali Linux VMs for VirtualBox (64 bit), and was trying to install the library python3-pip onto it. However, the commands "apt-get install python3-pip" and "apt install python3-pip" did not work, and kept returning the error of: "E: Package ‘python3-pip’ has no installation candidate". The same error occurs when […] | 899 | 3,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.924301 |
https://blender.stackexchange.com/questions/214664/geometry-nodes-point-instance-collection-use-count-object-weighting | 1,716,046,919,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00342.warc.gz | 113,779,666 | 39,682 | # Geometry Nodes: Point Instance Collection use count (object weighting)
I'm trying to recreate the donut sprinkles tutorial (by Blender Guru) but instead of using a particle system for sprinkles on the top of a donut, I want to use the new geometry nodes system.
When using a particle system you can use a collection as the objects and then the "Use Count" option allows you give different objects in the collection different count. This means you can make certain objects in a collection appear more often than other objects.
When using a Point Instance Geometry node you can choose "collection" but there is no option to "Use Count". Is there some chain of nodes you could use to accomplish the same thing?
The easiest way I found was to add separate "point distribute" and "point instance" nodes for the spheres and then you can independently control the density of the spheres and sprinkles. You then join the geometries.
The two nodes controlling the spheres are circled in red.
• This only works if you have very few objects to deal with. With a large quantity however, this wont be a good solution. For this specific task, It's best to stick with the particle system at the moment. Jun 27, 2021 at 13:53
I asked myself the same. I found a workaround that gives a bit more control. I splitted the point instance in two - one with the object which is more important and another with a collection of variation. With the Attribute Randomize I can give more importance to one of the separated instances:
I cobbled together a quick and simple method for my project.
Set two value nodes that correspond with the amount of items in your collection, starting at 0. In this example there are 5 items total. Run them through a mix RGB node, then control the factor with a noise texture + colour ramp.
If you tweak the colour ramp a bunch it could be a quick and dirty way to choose which items in the collection have the most importance. Hope that's useful to someone!
This is the best for current state for fields
More clean result for 4 indecies look like:
So random value group is just Random value node | 445 | 2,116 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-22 | latest | en | 0.913965 |
https://www.kopykitab.com/blog/vtu-previous-year-question-papers-be-me-modelling-and-finite-element-analysis-december-2011/ | 1,621,392,605,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991562.85/warc/CC-MAIN-20210519012635-20210519042635-00221.warc.gz | 895,795,533 | 20,842 | Modelling and Finite Element Analysis December 2011
Note: Answer any FIVE full questions, selecting at least TWO questions from each part.
PART-A
1 a. Write the equilibrium equation for 3-D state of stress and state the terms involved. (04 Marks)
b. Solve the following system of equations by Gaussian elimination method :
X1 + X2 + X3 = 6
X1-x2 + 2x3 = 5
X1 + 2x2 – X3 = 2.
c. Determine the displacements of holes of the spring system shown in the figure using principle of minimum potential energy. (08 Marks)
2 a. Explain the discretization process of a given domain based on element shapes number and size.
b. Explain basic steps involved in FEM with the help of an example involving a structural member subjected to axial loads.
C . Why FEA is widely accepted in engineering? List various applications of FEA in engineering.
3 a. Derive interpolation model for 2-D simplex element in global co – ordinate system.
b. What is an interpolation function? Write the interpolation functions for:
i) 1 – D linear element ; ii)1 — Dquadistie element,
iii) 2-D linear element ; iv) 2-D quadratic element.
v) 3-D linear element.
c. Explain “complete” and “conforming” elements.
4 a Derive shape function for 1 – D quadratic bar element in neutral co-ordinate, system.
b. Derive shape functions for CST element in NCS.
c. What are shape functions and write their properties, (any two).
PART-B
5 a. Derive the body force load vector for 1 – D linear bar element.
b. Derive the Jacobian matrix for CST element starting from shape function.
c. Derive stiffness matrix for a beam element starting from shape function.
6 a. Explain the various boundary conditions in steady state heat transfer problems with simple sketches.
b. Derive stiffness matrix for 1 – D heat conduction problem using either functional approach or Galerkin’s approach.
c. For the composite wall shown in the figure, derive the global stiffness matrix.
Take Ai = Aa = A3 = A
7 a. The structured member shown in figure consists of two bars. An axial load of P = 200 fcN is loaded as shown. Determine the following:
i) Element stiffness matricies.
ii) Global stiffness matrix.
iv) Nodal displacements.
i) Steel Ai = 1000 mm Ei = 200 GPa
ii) Bronze A2 = 2000 mm2 E2 = 83 GPa.
b. For the truss system shown, determine the nodal displacements. Assume E – 210 GPa and A = 600 mm for both elements.
8 a. Determine the temperature distribution in 1 – D rectangular cross – section fin as shown in 3w figure. Assume that convection heat loss occurs from the end of the fin. Take k =Cm C0 Iw-, T= 20°C. Consider two elementsh=-Cm2°C
b. For the cantilever beam subjected to UDL as shown in determine the deflections of the free end. Consider one element. | 708 | 2,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-21 | latest | en | 0.772282 |
https://www.cloudhadoop.com/2018/12/golang-example-program-to-calculate.html | 1,585,544,894,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496523.5/warc/CC-MAIN-20200330023050-20200330053050-00511.warc.gz | 637,748,505 | 33,436 | # Golang example - Program to Calculate average using Array/slice of numbers
In this blog post, You will learn two programs for finding average of array/slice numbers.
• First program to find avrage of fixed array of numbers
• The second program to read the input of values from a user via command line calculates the average.
Following are golang features are required to under this programs
### Example program to check average of an array of numbers
First Array or integers is declared with inline assignment of the values
Iterated array using for loop with range form
Find the length of the array using len function
Finally Divide the total by length to get average
``````package main
import "fmt"
func main() {
var numbs = []int{51, 4, 5, 6, 24}
total := 0
for _, number := range numbs {
total = total + number
}
average := total / len(numbs) // len function return array size
fmt.Printf("Average of array %d", average)
}
``````
Output of the above program
``````Average of array 18
``````
### Example program to check average of numbers entered by user console
In this program, User gives the numbers input via command line using fmt.Scanln function.
Here is an program code to read numbers from the command line and return average of it
``````package main
import "fmt"
func main() {
var array [50]int
var total, count int
fmt.Print("How many numbers you want to enter: ")
fmt.Scanln(&count)
for i := 0; i < count; i++ {
fmt.Print("Enter value : ")
fmt.Scanln(&array[i])
total += array[i]
}
average := total / count
fmt.Printf("Average is %d", average)
}
``````
Output of the above program is
``````How many numbers you want to enter: 5
Enter value : 1
Enter value : 2
Enter value : 3
Enter value : 4
Enter value : 5
Average is 3
`````` | 448 | 1,757 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | longest | en | 0.773504 |
http://www.in2013dollars.com/2015-euro-in-2016?amount=100 | 1,568,808,175,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573284.48/warc/CC-MAIN-20190918110932-20190918132932-00039.warc.gz | 276,212,648 | 10,331 | # €100 in 2015 → €100.24 in 2016
👉 You may be interested in €100 in 2015 → 2019
### Euro Inflation Rate, €100 in 2015 to 2016
According to the European Central Bank consumer price index, prices in 2016 are 0.24% higher than average prices throughout 2015. The euro experienced an average inflation rate of 0.24% per year during this period, meaning the real value of a dollar decreased.
In other words, €100 in 2015 is equivalent in purchasing power to about €100.24 in 2016.
The 2015 inflation rate was 0.03%. The inflation rate in 2016 was 0.24%. The 2016 inflation rate is lower compared to the average inflation rate of 1.33% per year between 2016 and 2019.
Average inflation rate 0.24% Converted amount (€100 base) €100.24 Price difference (€100 base) €0.24 CPI in 2015 100.000 CPI in 2016 100.242 Inflation in 2015 0.03% Inflation in 2016 0.24%
### How to Calculate Inflation Rate for €100, 2015 to 2016
This inflation calculator uses the following inflation rate formula:
CPI in 2016CPI in 2015
×
2015 EUR value
=
2016 EUR value
Then plug in historical CPI values. The Euro CPI was 100 in the year 2015 and 100.2416667 in 2016:
100.2416667100
×
€100
=
€100.24
€100 in 2015 has the same "purchasing power" or "buying power" as €100.24 in 2016.
To get the total inflation rate for the 1 years between 2015 and 2016, we use the following formula:
CPI in 2016 - CPI in 2015CPI in 2015
×
100
=
Cumulative inflation rate (1 years)
Plugging in the values to this equation, we get:
100.2416667 - 100100
×
100
=
0%
Politics and news often influence economic performance. Here's what was happening at the time:
• Barack Obama and Raul Castro, hold the first meeting between Cuban and American leaders since the Cuban Revolution.
• The population of India reaches 1 billion (officially). The billionth baby was named Astha Arora.
• Donald Trump announces his intention to join the US presidential campaign.
• China brings an end to its one child policy, 35 years after it was first introduced.
### Data Source & Citation
Raw data for these calculations comes from the European Commission and the European Central Bank's Harmonized Index of Consumer prices (HICP), which is computed based on the reported consumer price indices in member countries of the European Union.
You may use the following MLA citation for this page: “€100 in 2015 → 2016 | Euro Inflation Calculator.” U.S. Official Inflation Data, Alioth Finance, 18 Sep. 2019, https://www.officialdata.org/2015-euro-in-2016?amount=100. | 702 | 2,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.904113 |
http://max8us.net/?gre/blog/2017/07/21/gre-quant-best-practices-improving-problem-recognition/ | 1,534,763,460,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221216333.66/warc/CC-MAIN-20180820101554-20180820121554-00370.warc.gz | 251,215,101 | 19,622 | ### GRE Quant Best Practices: Improving Problem Recognition
You can attend the first session of any of our online or in-person GRE courses absolutely free. Ready to take the plunge? Check out our upcoming courses here.
A number of students have recently told me that they struggle with “problem recognition,” particularly in the Quant section of the GRE. What many mean by this is that when they look at a problem, they don’t immediately see how to get to the solution. They might recognize some of the concepts involved, but the problem as a whole has aspects that make it look unfamiliar and difficult. When this happens on the test, in a high-pressure, time-sensitive environment, the resulting feeling can be paralyzing.
I have a different idea of what “problem recognition” means, though. Here’s the secret: when I see a difficult GRE Quant problem for the first time, I often have the same reaction—I have no idea! I don’t see the whole solution right away. I may not even feel confident I really do know how to solve it. I have to fight the reaction, honed by many years of math-aversion, to panic and give up.
However, having spent a lot of time learning GRE Quant best practices, I’ve trained myself to take a different, more deliberate approach, regardless of my instinctive reaction. First, I remember two things:
1. The makers of the GRE design problems that look intimidating. But this doesn’t mean that they’re all terribly difficult. Sometimes the problems that look the messiest end up being relatively straightforward.
2. There’s a finite amount of math content on the test. I reassure myself that I know the topic this problem is testing, even if I don’t yet understand how it’s being tested. (And if I see a topic that I’m not familiar with, then I remind myself that skipping a problem or three is often a smart test-day strategy).
Next, I take a deep breath and get started. I pick an entry point and start doing something. Precisely what I do depends on the topic and the format of the problem. For example:
• On a difficult geometry problem, I’d start by drawing or re-drawing and labeling the figure.
• On a difficult divisibility problem, I’d start by making factor trees for the numbers involved.
• On a difficult word problem, I’d start by translating the text into formulas or notes.
Notice that each of these opening tasks is fairly mechanical. I’m doing something that I know how to do, which builds confidence. I’m also doing something that doesn’t take a whole lot of brain power. Instead of banging my conscious mind against the problem, I’m absorbing the information in it and letting my sub-conscious, problem-solving brain get to work.
Solving any difficult problem is a dialogue between conscious, directed work and unconscious processing— think, for example, of people who report having worked on a difficult task unsuccessfully all day only to find the solution to the problem in their dreams. To solve hard problems effectively on the GRE, you want to allow space for this unconscious processing to help you out. This is why starting each problem with a routine task is so useful. It minimizes anxiety, because you’re doing something that’s likely to be helpful while giving your mind room to make intuitive connections.
As for how to know which task to start with, fortunately this is a skill that can be trained. I find the “see this/do this” format really helpful for developing this skill. Here’s how it works.
After I solve a tricky problem, I go back and think about what the ideal approach would have been. What task should I have started with? What clues in the problem point to that task as the right entry point?
Once I’ve answered these questions, I make a flashcard. On the front, I’ll write:
WHEN I SEE A DIFFICULT DIVISIBILITY PROBLEM…
And on the back, I’ll write:
I WILL START BY PRIME FACTORING.
Over time, as you build a stack of these, you’ll find that, in addition to the content on the GRE being finite, there are a limited number of problem “types.” While you’ll probably still see problems that contain unfamiliar or surprising elements, you’ll have a go-to starting place for anything that might come up. This is what problem recognition means to me—not necessarily knowing exactly how to solve every problem, but having a good plan for where to start, and confidence that, if I know my content, this strong start will help me find my way to the right solution. 📝
Want more guidance from our GRE gurus? You can attend the first session of any of our online or in-person GRE courses absolutely free! We’re not kidding. Check out our upcoming courses here.
Cat Powell is a Manhattan Prep instructor based in New York, NY. She spent her undergraduate years at Harvard studying music and English and is now pursuing an MFA in fiction writing at Columbia University. Her affinity for standardized tests led her to a 169Q/170V score on the GRE. Check out Cat’s upcoming GRE courses here. | 1,048 | 4,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | latest | en | 0.952793 |
https://findthefactors.com/tag/1354/ | 1,638,232,471,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358847.80/warc/CC-MAIN-20211129225145-20211130015145-00281.warc.gz | 343,739,033 | 12,460 | # 1354 Solving a Level 5 Puzzle
What are the common factors of 16 and 4? Don’t guess which one to use. Use logic to figure it out as you find all the factors for this puzzle!
Print the puzzles or type the solution in this excel file: 10 Factors 1347-1356
If you get stuck, you can watch this video:
Now I’ll share some information about the puzzle number, 1354:
• 1354 is a composite number.
• Prime factorization: 1354 = 2 × 677
• 1354 has no exponents greater than 1 in its prime factorization, so √1354 cannot be simplified.
• The exponents in the prime factorization are 1, and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1354 has exactly 4 factors.
• The factors of 1354 are outlined with their factor pair partners in the graphic below.
1354 is the sum of two squares:
27² + 25² = 1354
1354 is the hypotenuse of a Pythagorean triple:
104-1350-1354 which is 2 times (52-675-677)
and can also be calculated from 27² – 25², 2(27)(25), 27² + 25² | 304 | 1,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-49 | latest | en | 0.906501 |
https://www.coursehero.com/file/6722736/136-course-notes/ | 1,521,492,479,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647146.41/warc/CC-MAIN-20180319194922-20180319214922-00382.warc.gz | 775,207,302 | 107,955 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
136_course_notes
# 136_course_notes - Linear Algebra 1 Course Notes for MATH...
This preview shows pages 1–4. Sign up to view the full content.
Linear Algebra 1 Course Notes for MATH 136 Edition 1.0 D. Wolczuk Copyright: D. Wolczuk, 1st Edition, 2011
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Contents A Note to Students - READ THIS! . . . . . . . . . . . . . . . . . . . iii Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii 1 Vectors in Euclidean Space 1 1.1 Vector Addition and Scalar Multiplication . . . . . . . . . . . . . . . 1 1.2 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.3 Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 1.4 Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2 Systems of Linear Equations 28 2.1 Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . 28 2.2 Solving Systems of Linear Equation . . . . . . . . . . . . . . . . . . . 32 3 Matrices and Linear Mappings 46 3.1 Operations on Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . 46 3.2 Linear Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 3.3 Special Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 3.4 Operations on Linear Mappings . . . . . . . . . . . . . . . . . . . . . 75 4 Vector Spaces 78 4.1 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 4.2 Bases and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 4.3 Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 5 Inverses and Determinants 107 5.1 Matrix Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 5.2 Elementary Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 5.3 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 5.4 Determinants and Systems of Equations . . . . . . . . . . . . . . . . 129 5.5 Area and Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 6 Diagonalization 138 6.1 Matrix of a Linear Mapping and Similar Matrices . . . . . . . . . . . 138 6.2 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . 143 6.3 Diagonalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 6.4 Powers of Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 ii
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### Page1 / 168
136_course_notes - Linear Algebra 1 Course Notes for MATH...
This preview shows document pages 1 - 4. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 1,114 | 2,965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.160331 |
http://somemath.blogspot.com/2012/12/surprising-modular-discovery-progression.html | 1,531,842,241,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589752.56/warc/CC-MAIN-20180717144908-20180717164908-00624.warc.gz | 353,300,385 | 15,149 | ## Thursday, December 06, 2012
### Surprising modular discovery progression
After years with my own mathematical results I was surprised to come across something which to me is bizarrely simple for the 21st century, as how can this approach be new? But I haven't found it elsewhere yet, while I keep looking. And I'll put it up in a bit, and then walk down the progression from it to a general modular solution.
So I found out recently that a certain well-known equation could be solved modularly:
A modular solution to x2 - Dy2 = 1 is:
2x = r + r-1 mod D-1 and 2y = r - r-1 mod D-1
Proof:
x2 - Dy2 = 1 mod D-1, and D mod D-1 = 1, so:
x2 - y2 = 1 mod D-1
Which is: (x+y)(x-y) = 1 mod D-1, and letting r be any residue with an inverse modulo D-1:
x+y = r mod D-1 and x-y = r-1 mod D-1, so:
2x = r + r-1 mod D-1 and 2y = r - r-1 mod D-1
Proof complete.
That one is easy and short so I decided to use more of a traditional format. Math people can get really weird about how you present things. It's their way, or...well it's just THEIR way. And they get years of training at colleges and universities for a particular format. I didn't get that training in their format. I was a physics student. But I digress.
Next step is to generalize beyond modulo D-1.
Given n such that: n2 = D mod p, where p is a prime number, and r, any residue modulo p:
2x = r + r-1 mod p and 2y = n-1(r - r-1) mod p
gives solutions for x2 - Dy2 = 1 mod p.
And THAT can be generalized to modulo N.
But this approach can be taken still further to something even more general.
With x2 - Dy2 = F where all variables are non-zero integers:
Given a non-zero integer N for which a residue m exists where--m2 = D mod N, and r, any residue modulo N for which Fr-1 mod N exists then:
2x = r + Fr-1 mod N and 2my = r - Fr-1 mod N
Which of course gives you the original result, with F = 1 and N = D-1.
Or the modulo p result with F = 1, N = p.
Of course the simple reality I'm exploiting is that in modular arithmetic a non-zero D is always a square!
Proof: Let N = D - m2, or N = m2 - D, for any non-zero integer m, then D is a quadratic residue modulo N.
To me it is a very surprising progression, with simple solutions which can actually solve these equations explicitly if x and y exist less than N, though I'm not suggesting it as a practical technique in these forms.
Why such a simple, basic modular arithmetic concept is not part of what I'm seeing on math websites about these equations is a puzzle to me.
James Harris | 728 | 2,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-30 | latest | en | 0.887296 |
http://nrich.maths.org/public/leg.php?code=-99&cl=3&cldcmpid=7136 | 1,485,124,144,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281649.59/warc/CC-MAIN-20170116095121-00348-ip-10-171-10-70.ec2.internal.warc.gz | 204,165,399 | 10,332 | # Search by Topic
#### Resources tagged with Working systematically similar to Named Products:
Filter by: Content type:
Stage:
Challenge level:
### There are 129 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Bent Out of Shape
##### Stage: 4 and 5 Challenge Level:
An introduction to bond angle geometry.
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Colour Islands Sudoku
##### Stage: 3 Challenge Level:
An extra constraint means this Sudoku requires you to think in diagonals as well as horizontal and vertical lines and boxes of nine.
##### Stage: 3 and 4 Challenge Level:
Four small numbers give the clue to the contents of the four surrounding cells.
### Oranges and Lemons, Say the Bells of St Clement's
##### Stage: 3 Challenge Level:
Bellringers have a special way to write down the patterns they ring. Learn about these patterns and draw some of your own.
### 9 Weights
##### Stage: 3 Challenge Level:
You have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance?
### Difference Sudoku
##### Stage: 3 and 4 Challenge Level:
Use the differences to find the solution to this Sudoku.
### Medal Muddle
##### Stage: 3 Challenge Level:
Countries from across the world competed in a sports tournament. Can you devise an efficient strategy to work out the order in which they finished?
### Introducing NRICH TWILGO
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
We're excited about this new program for drawing beautiful mathematical designs. Can you work out how we made our first few pictures and, even better, share your most elegant solutions with us?
### Spot the Card
##### Stage: 4 Challenge Level:
It is possible to identify a particular card out of a pack of 15 with the use of some mathematical reasoning. What is this reasoning and can it be applied to other numbers of cards?
### LOGO Challenge - Triangles-squares-stars
##### Stage: 3 and 4 Challenge Level:
Can you recreate these designs? What are the basic units? What movement is required between each unit? Some elegant use of procedures will help - variables not essential.
### Pole Star Sudoku 2
##### Stage: 3 and 4 Challenge Level:
This Sudoku, based on differences. Using the one clue number can you find the solution?
### Tea Cups
##### Stage: 2 and 3 Challenge Level:
Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.
### Olympic Logic
##### Stage: 3 and 4 Challenge Level:
Can you use your powers of logic and deduction to work out the missing information in these sporty situations?
### More on Mazes
##### Stage: 2 and 3
There is a long tradition of creating mazes throughout history and across the world. This article gives details of mazes you can visit and those that you can tackle on paper.
### Twin Line-swapping Sudoku
##### Stage: 4 Challenge Level:
A pair of Sudoku puzzles that together lead to a complete solution.
### Crossing the Town Square
##### Stage: 2 and 3 Challenge Level:
This tricky challenge asks you to find ways of going across rectangles, going through exactly ten squares.
### Counting on Letters
##### Stage: 3 Challenge Level:
The letters of the word ABACUS have been arranged in the shape of a triangle. How many different ways can you find to read the word ABACUS from this triangular pattern?
### The Naked Pair in Sudoku
##### Stage: 2, 3 and 4
A particular technique for solving Sudoku puzzles, known as "naked pair", is explained in this easy-to-read article.
### One Out One Under
##### Stage: 4 Challenge Level:
Imagine a stack of numbered cards with one on top. Discard the top, put the next card to the bottom and repeat continuously. Can you predict the last card?
### Intersection Sudoku 1
##### Stage: 3 and 4 Challenge Level:
A Sudoku with a twist.
### Ratio Sudoku 2
##### Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios.
### Peaches Today, Peaches Tomorrow....
##### Stage: 3 and 4 Challenge Level:
Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for?
### Integrated Sums Sudoku
##### Stage: 3 and 4 Challenge Level:
The puzzle can be solved with the help of small clue-numbers which are either placed on the border lines between selected pairs of neighbouring squares of the grid or placed after slash marks on. . . .
### Cinema Problem
##### Stage: 3 and 4 Challenge Level:
A cinema has 100 seats. Show how it is possible to sell exactly 100 tickets and take exactly £100 if the prices are £10 for adults, 50p for pensioners and 10p for children.
##### Stage: 3 Challenge Level:
Rather than using the numbers 1-9, this sudoku uses the nine different letters used to make the words "Advent Calendar".
### Teddy Town
##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Magic Caterpillars
##### Stage: 4 and 5 Challenge Level:
Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal.
##### Stage: 3, 4 and 5 Challenge Level:
You need to find the values of the stars before you can apply normal Sudoku rules.
### Latin Squares
##### Stage: 3, 4 and 5
A Latin square of order n is an array of n symbols in which each symbol occurs exactly once in each row and exactly once in each column.
### LOGO Challenge - the Logic of LOGO
##### Stage: 3 and 4 Challenge Level:
Just four procedures were used to produce a design. How was it done? Can you be systematic and elegant so that someone can follow your logic?
### Seasonal Twin Sudokus
##### Stage: 3 and 4 Challenge Level:
This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it?
### LOGO Challenge - Sequences and Pentagrams
##### Stage: 3, 4 and 5 Challenge Level:
Explore this how this program produces the sequences it does. What are you controlling when you change the values of the variables?
### Magnetic Personality
##### Stage: 2, 3 and 4 Challenge Level:
60 pieces and a challenge. What can you make and how many of the pieces can you use creating skeleton polyhedra?
### Multiplication Equation Sudoku
##### Stage: 4 and 5 Challenge Level:
The puzzle can be solved by finding the values of the unknown digits (all indicated by asterisks) in the squares of the $9\times9$ grid.
### A Long Time at the Till
##### Stage: 4 and 5 Challenge Level:
Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem?
### Advent Calendar 2011 - Secondary
##### Stage: 3, 4 and 5 Challenge Level:
Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
### Inky Cube
##### Stage: 2 and 3 Challenge Level:
This cube has ink on each face which leaves marks on paper as it is rolled. Can you work out what is on each face and the route it has taken?
### Difference Dynamics
##### Stage: 4 and 5 Challenge Level:
Take three whole numbers. The differences between them give you three new numbers. Find the differences between the new numbers and keep repeating this. What happens?
### Sticky Numbers
##### Stage: 3 Challenge Level:
Can you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number?
### I've Submitted a Solution - What Next?
##### Stage: 1, 2, 3, 4 and 5
In this article, the NRICH team describe the process of selecting solutions for publication on the site.
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Troublesome Triangles
##### Stage: 2 and 3 Challenge Level:
Many natural systems appear to be in equilibrium until suddenly a critical point is reached, setting up a mudslide or an avalanche or an earthquake. In this project, students will use a simple. . . .
### Building with Longer Rods
##### Stage: 2 and 3 Challenge Level:
A challenging activity focusing on finding all possible ways of stacking rods.
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
##### Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Rainstorm Sudoku
##### Stage: 4 Challenge Level:
Use the clues about the shaded areas to help solve this sudoku
### LCM Sudoku II
##### Stage: 3, 4 and 5 Challenge Level:
You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku. | 2,184 | 9,668 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | longest | en | 0.884131 |
http://www.yuvajobs.com/download-think-placement-papers/whole-testpaper-1856.html | 1,539,785,460,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511175.9/warc/CC-MAIN-20181017132258-20181017153758-00193.warc.gz | 583,647,074 | 7,063 | Search Jobs
Search Jobs
# Whole Testpaper by Think
Details of Whole Testpaper by Think conducted by Think for job interview.
Share Us With Others
THINKSOFT SAMPLE PAPER
Paper contains about 60 ques for 1 hr
suggestion : Attend the ques which has more weightage(marks) & consumes less time.
ThinkSoft Latest Fresher Engineer Placement Sample Question Paper 1
Part A : English test about 15-20 ques each carries 1 mark
Part B : quantitative test
Part C : Logical Reasoning
& some more.
Some ques i remember,
Ques 1: From given table 4/5 ques wee asked carries 5 marks each
Avg Good Excelt Total
---------------------------------------
Male | 10
Fem | 32
Total| 30
Given
i) 1/2 of the students were Good & excellent
ii)40% of th students are female
iii)1/3rd of male stuedents are average
Soln. :
Total Students * 40/100 = 32
=> tot studs = 32*100/40 = 80
=> tot Male studs = 80 - 32 = 48
48/3 = 16 male students are average
40 - 16 = 24 female ------ avg
32 - 24 = 8 ------- -- good
16/80 = 24/80 => 2:3 proportion of male & female studs are avg
Hence,
Avg good Exc tot
-----------------------------
M | 16 22 10 48
F | 24 8 0 32
-----------------------------
T | 40 30 10 80
Ques 2:
The statement "All shirts in the store are on sale" is false.
Which of the foll stns are true?
i)No shirts in the store are on sale
ii)Not all shirts in the store are on sale
iii)All shirts in the store are non-sale prices
iv)Some shirts in the store are on sale | 425 | 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} | 3.578125 | 4 | CC-MAIN-2018-43 | latest | en | 0.831949 |
https://thirdspacelearning.com/us/math-resources/topic-guides/number-and-quantity/percent-change/ | 1,718,474,641,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00348.warc.gz | 528,885,601 | 44,813 | # Percent change
Here you will learn about percent change, including how to calculate the percentage by which a value has increased or decreased.
Students will first learn about percent change as part of ratios and proportional relationships in 7th grade.
## What is percent change?
Percent change equals the change in value divided by the absolute value of the original value, multiplied by 100. It is when the percentage of its original value has changed to either increase or decrease the value.
When you calculate percent change, you are calculating by what percentage of its original value something has changed.
To calculate percent change, you can use the percentage change formula:
Percent change can refer to many different types of scenarios. This can include percentage increase, percentage decrease, percentage profit, percentage gain, markups and markdowns, or percentage loss.
### Percent increase
Percentage increase, or percent gain, is when the percent increases, or gets bigger, between a starting percent and the final percent.
For example, at the beginning of the month you had \$50 in your savings account and by the end of the month you had \$75, you may want to find the percentage increase of the money you saved over the period of time.
### Percent decrease
Percentage decrease, or percent loss, is when the percent decreases, or gets smaller, between a starting percent and the final percent.
For example, buying a tank of gas last year was cheaper than it is now. You could calculate the percent decrease in the amount of a gallon of gas.
### Percent profit
Percentage profit refers to when a store or person is selling a good or a service, and they want to calculate the difference between the cost price and the selling price and represent it as a percent.
For example, a school store buys pencils for \$0.50 a piece, but sells them to the students for \$0.75. The school store would make \0.25 profit, or 50\% profit. ### Markups and markdowns Markups are the increased price of a good, and markdowns are the decreased price of a good. Markdowns are more commonly known as discounts or sales in a retail store. A company may markup an item that is in high demand, and people are willing to pay more money for it. ### What is the percent change? ## Common Core State Standards How does this relate to 7th grade math? • Grade 7: Ratios and Proportional Relationships (7.RP.3) Use proportional relationships to solve multistep ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error. ## How to calculate percent change In order to calculate percent change, you need to: 1. Find the amount of change by subtracting the original number from the new number. 2. Plug numbers into the percent change formula. \text { Percentage change }=\cfrac{\text { Change }}{\text { Original }} \times 100 3. Simplify the fraction, if necessary, using equivalent fractions. 4. Convert the fraction to a decimal. 5. Calculate percent change. ## Percent changeexamples ### Example 1: percent increase The weight of a baby has increased from 4 \, kg to 7 \, kg. Find the percentage difference in the baby’s weight. 1. Find the amount of change by subtracting the original number from the new number. The baby’s weight has increased from 4 \, kg to 7 \, kg. Calculate the percent change: 7-4=3 \mathrm{~kg} 2Plug numbers into the percent change formula. The change is 3 \, kg and the original number is 4. Apply the percent change formula: \begin{aligned} & \text {Percent change }=\cfrac{\text { change }}{\text { original }} \times 100 \\\\ & \text {Percent change }=\cfrac{3}{4} \times 100 \end{aligned} 3Simplify the fraction, if necessary, using equivalent fractions. \cfrac{3}{4} \, does not need to be simplified. \text {Percent change }=\cfrac{3}{4} \times 100 4Convert the fraction to a decimal. \cfrac{3}{4}=0.75 \text{Percent change } =0.75 \times 100 5Calculate percent change. \begin{aligned} & 0.75 \times 100=75 \\\\ & \text {Percent change }=75 \% \end{aligned} The final answer is 75\% percent change. The baby’s weight increased by 75\%. ### Example 2: percent loss Anna bought a computer for \1,200. \; 2 years later, Anna sold the computer for \$800. Calculate Anna’s percent loss. Find the amount of change by subtracting the original number from the new number. Plug numbers into the percent change formula. Simplify the fraction, if necessary, using equivalent fractions. Convert the fraction to a decimal. Calculate the percent change. ### Example 3: percent profit Matthew bought an antique for \$360. He sold it for \$396. Calculate Mattew’s percentage profit. Find the amount of change by subtracting the original number from the new number. Plug numbers into the percent change formula. Simplify the fraction, if necessary, using equivalent fractions. Convert the fraction to a decimal. Calculate the percent change. ### Example 4: percent decrease A clothing shop reduced the price of a jacket from \$48 to \$36. Find the percentage decrease. Find the amount of change by subtracting the original number from the new number. Plug numbers into the percent change formula. Simplify the fraction, if necessary, using equivalent fractions. Convert the fraction to a decimal. Calculate the percent change. ## More than one percent change Sometimes, a product in a shop may be discounted twice, or the value of something may increase and then decrease. It is common to think that you can just add the percentage changes together, but this does not work. For example, if something is reduced by 10\% in a sale and the sale price is then reduced by 10\%, this does not mean a total reduction of 20\%. This is because 10\% of the sale price is not the same as 10\% of the original price. To calculate the total percent change after two percentage changes: 1. Apply the first percent change. 2. Apply the second percent change. 3. Calculate the percent change between the final value and the original value and plug numbers into the percent change formula. 4. Simplify the fraction, if necessary, using equivalent fractions. 5. Convert the fraction to a decimal. 6. Calculate the percent change. ### Example 5: double percentage reduction A television was priced at \$500 and was reduced by 10\% in a sale. 2 weeks later, the sale price was reduced by 10\%. Find the overall percentage reduction in price.
Apply the first percent change.
Apply the second percent change.
Calculate the percent change between the final value and the original value and plug numbers into the percent change formula.
Simplify the fraction, if necessary, using equivalent fractions.
Convert the fraction to a decimal.
Calculate the percent change.
Holly bought a diamond ring for \$2,000 in 2015. By 2017, the value of the ring had decreased by 5\%. By 2020, the value of the ring had increased by 20\% from its value in 2017. Find the overall percentage change in the value of the diamond ring. Apply the first percent change. Apply the second percent change. Calculate the percent change between the final value and the original value and plug numbers into the percent change formula. Simplify the fraction, if necessary, using equivalent fractions. Convert the fraction to a decimal. Calculate the percent change. ### Teaching tips for percent change • When calculating percent change, sometimes the numbers can be quite large. Students may use a calculator or a percentage change calculator to assist them with solving these problems. • Percent change can be calculated using Microsoft Excel. For students who are proficient using Excel, this could be an additional teaching point while discussing percent change. ### Easy mistakes to make • Using an incorrect value for the denominator in the percent increase formula Using the new value instead of the original value for the denominator when calculating percentage change will lead to an incorrect percent. • Adding together multiple percent changes to find total percent change Adding the percentages together to find the total percentage change does not work. Remember that a percentage of an already increased or decreased value will be different from a percentage of the original value. ### Percent change practice questions Calculate the percent change that has occurred in each of these questions. 1. \$400 increased to \$568. 168\% 42\% 68\% 48\% First, you will calculate the change from \$400 to \$568. 568-400=168 The change is \$168, so you can now plug the numbers into the percent change formula.
\text{Percent change }=\cfrac{168}{400} \times 100
After simplifying the fraction to \, \cfrac{42}{100} \, , you would convert it to the decimal 0.42.
0.42 \times 100=42
\text{Percentage increase } =42 \%
2. 70 \, kg reduced to 59 \, kg.
15\%
11\%
59\%
16\%
First, you will calculate the change from 70 \, kg to 59 \, kg.
70-59=11
The change is 11, so you can now plug the numbers into the percent change formula.
\text{Percent change }=\cfrac{11}{70} \times 100
\cfrac{11}{70} \, is in simplest terms, you would convert it to the decimal 0.16, rounded to two decimal places.
0.16 \times 100=16
\text{Percentage reduced } =16 \%
3. 60 \, cm increased to 75 \, cm.
27.5\%
12.5\%
25\%
15\%
First, you will calculate the change from 60 \, cm to 75 \, cm.
75-60=15
The change is 15, so you can now plug the numbers into the percent change formula.
\text{Percent change }=\cfrac{15}{60} \times 100
After simplifying the fraction to \, \cfrac{1}{4} \, , you would convert it to the decimal 0.25.
0.25 \times 100=25
\text { Percentage increased }=25 \%
4. \$90 reduced to \$63.
30\%
70\%
27\%
63\%
First, you will calculate the change from \$90 to \$63.
90-63=27
The change is 27, so you can now plug the numbers into the percent change formula.
\text{Percent change }=\cfrac{27}{90} \times 100
After simplifying the fraction to \, \cfrac{3}{10} \, , you would convert it to the decimal 0.30.
0.30 \times 100=30
\text {Percentage reduced }=30 \%
5. \$500 increased by 20\% and then increased again by a further 10\%. 30\% 32\% 20\% 10\% First, you will apply the first percent change, a 20\% increase to \$500.
20\% of \$500 = \$100
\$500 + \$100 = \$600 Next, apply the second percent change, a 10\% increase to \$600.
10\% of \$600 = \$60
\$600 + \$60 = \$660 The overall change is \$160, so you can now plug the numbers into the percent change formula.
\text {Percent change }=\cfrac{160}{500} \times 100
After simplifying the fraction to \, \cfrac{8}{25} \, , you would convert it to the decimal 0.32.
0.32 \times 100=32
\text {Percentage increased } = 32\%
6. \$6,000 increased by 20\% and then decreased by 25\%. 22.5\% decrease 22.5\% increase 10\% decrease 10\% increase First, you will apply the first percent change, a 20\% increase from \$6,000.
20\% of \$6,000 = \$1,200
\$6,000 + \$1,200 = \$7,200 Next, apply the second percent change, a 25\% decrease from \$7,200.
25\% of \$7,200 = \$1,800
\$7,200-\$1,800 = \$5,400 The overall change is \$600, so you can now plug the numbers into the percent change formula.
\text {Percent change }=\cfrac{600}{6,000} \times 100
After simplifying the fraction to \, \cfrac{1}{10} \, , you would convert it to the decimal 0.10.
0.10 \times 100=10
\text {Percentage decreased } = 10\%
## Percent change FAQs
Can finding the percent change give you a negative number for a percent?
If you are finding a percent decrease, or any percentage where the value is decreasing, the initial percent change calculation will result in a negative number. However, you will use the absolute value of the difference to calculate the overall percent change, which will result in a positive number.
Is there a different formula to calculate percent increase and percent decrease?
You use the same formula whether you are finding the percentage increase or decrease. With both, you are calculating the percentage to which the measure either increases or decreases.
## Still stuck?
At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.
Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence. | 3,002 | 12,472 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-26 | latest | en | 0.946853 |
https://aviation.stackexchange.com/questions/87174/what-control-inputs-are-required-to-perform-a-turning-while-in-flight-maneuver | 1,716,295,990,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00266.warc.gz | 99,794,940 | 39,294 | # What control inputs are required to perform a "turning-while-in-flight" maneuver in a helicopter?
My question is relatively simple and has to do with the required control inputs to perform a simple turning maneuver over a large area.
Suppose a helicopter is traveling in forward flight, straight and level, and the pilot has to alter course to circle a large area. Now, suppose the helicopter begins a slight rightward turn, while continuing to fly 'forward', and ends up semi-circling over a broad area.
I want to make the point that the helicopter is not yawing/turning on its own axis, but rather performing a turn over a large area.
The resulting flight path, if viewed from above, would essentially be a semi-circle or horseshoe shape.
This would obviously involve some rightward banking. What I want to know is this: can this maneuver be performed with only cyclic inputs, or would there also have to be pedal inputs to complete such a turn? And if it can be performed only with cyclic inputs, what exactly would those inputs be apart from rightward cyclic? Thanks in advance!
• Perhaps you mean a coordinated turn? May 17, 2021 at 15:02
• I don't quite understand your description. Suppose that, at the beginning of the maneuver, the nose of the helicopter is pointing north. After it flies a semicircle, is the nose still pointing north (meaning that for the last part of the maneuver, it's flying backwards instead of forwards), or is it now pointing south (meaning that the helicopter must have yawed during the maneuver)? From your description, it sounds like you're saying that as the helicopter flies in a semicircle, it never flies backwards and never yaws, which is practically impossible. May 17, 2021 at 22:11
• I find the case where a helicopter performs a turn largely by yawing and then braking with cyclic much more interesting... of course the wind moving across the body of the helicopter make this harder to do. May 18, 2021 at 0:59
• @TerranSwett -- re "I want to make the point that the helicopter is not yawing/turning on its own axis"-- I think what he really meant was that the turn is not centered around the aircraft's own vertical axis. Not that the helicopter's yaw rate is zero. There is some room for further clarification here, but I think he's basically describing a normal coordinated turn. May 19, 2021 at 1:50
• (Presumably meaning the yaw string is centered, which in helicopters, may mean that the ball is a little off center, due to the side thrust generated by the tail rotor?) May 19, 2021 at 1:57
Yes it's cyclic inputs only for banked turns because there is no adverse yaw due to ailerons to compensate for. You will be working the pedals to keep the ball centered, but this is only in reaction to torque changes from increases in rotor pitch and power demands.
Note that the airplane isn't using rudder to turn either; only to correct for adverse yaw. What brings the nose around in the turn is the vertical fin. The banked wings creates a lateral force vector, moving the airplane sideways as it goes forward. The sideway motion cause the airstream to strike the vertical fin from the side, and the airplane's inherent weathervaning tendency brings the nose around to keep it aligned with the arc of a turn. If an airplane had no adverse yaw at all, you would not need to use any rudder to help with turns.
A helicopter is the same. There is a small fin back there that gives it modest weathervaning tendency (the disk area of the tail rotor also contributes). The pedals are just there to change tail rotor pitch to adjust for off-axis yawing motions induced by torque changes (which can be in either direction, depending on whether torque is going up or down); just as on the airplane the pedals are just there to compensate for adverse yaw due to aileron displacement. If the helicopter had some automated system to always compensate for torque changes, you would not need to touch the pedals to turn.
In practical terms, in both cases, when flying you simply do whatever you have to do with your feet to keep the skid ball centered; in the airplane you are generally reacting to aileron movements, and in the helicopter to torque/pitch changes. The power plane pilot learning helicopters has to adjust to get out of the habit of squeezing a little into-turn rudder to anticipate adverse yaw, and has to instead anticipate the subtle yaw changes - in either direction - from the power and pitch changes the pilot may input during the turn.
• While small and due to other phenomena, there is adverse yaw in a helicopter. Typically all four controls are moved to turn and maintain altitude & speed.
– Mat
Jun 13, 2021 at 19:27
• Yes there are yawing effects, but not adverse yaw in the aileron sense which was my main point. Jun 13, 2021 at 21:14
As usual in a helicopter, you typically need to adjust all controls because everything is cross-coupled.
The primary input to initiate a turn is lateral cyclic. This banks the helicopter and tilts the lift vector so that a centripetal force is formed. The pedals are only used to make small corrections and keep the helicopter aligned with the incoming airflow. Usually a yaw string attached to the outside of the cockpit window shows whether the helicopter is in a skid, slid or coordinated turn.
Since the lift vector is tilted due to the banking, additional lift will be needed to maintain altitude. For this, collective and throttle are increased. Again, this has an effect on the torque, so the pedals are needed to keep the nose pointed into the airflow.
For steep turns, forward cyclic may be required to keep the airspeed from dropping.
How fast the helicopter banks depends on how much lateral cyclic pressure is applied. How far the helicopter banks (the steepness of the bank) depends on how long the cyclic is displaced. After establishing the proper bank angle, return the cyclic toward the neutral position. When the bank is established, returning the cyclic to neutral (or holding it inclined relative to the horizon) will maintain the helicopter at that bank angle. Increase the collective and throttle to maintain altitude and rpm. As the torque increases, increase the proper antitorque pedal pressure to maintain longitudinal trim. Depending on the degree of bank, additional forward cyclic pressure may be required to maintain airspeed | 1,359 | 6,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-22 | latest | en | 0.963516 |
https://www.physicsforums.com/threads/question-regarding-determinism.282457/ | 1,521,370,976,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645604.18/warc/CC-MAIN-20180318091225-20180318111225-00679.warc.gz | 880,807,330 | 17,079 | # Question regarding determinism
1. Jan 2, 2009
### Gothican
Hello there!
I've been reading Brian Greene's book, 'The Fabric of the Cosmos', and I've stumbled upon a troublesome statement regarding determinism:
In the chapter about Einstein's conception of time, he discusses the determinism of spacetime and comes to the following conclusion; Imagine time-space as a loaf of bread, and the present time of each man in different places and velocities in space (relative to earth) as slices of the bread cut in different angles, (when the only restriction comes from the speed limit set by light, which translates into a limit on the rotational angle of 45 degrees).
Imagine the spacetime loaf sliced up into many various presents of observers situated in different distances from earth (zero relative velocity). “Now, the collection of all these now-slices fills out a substantial region of the spacetime loaf. In fact, if space is infinite – if now-slices extended infinitely far – then the rotated now-slices can be centered arbitrarily far away, and hence their union sweeps through every point in the spacetime loaf.” -Brian Greene
But here I noticed a paradox, which is pretty disturbing – if there is a restriction on the slicing of the spacetime loaf (a 45 degree limit), then it is logically impossible for all the now-slices to “sweep through every point in the loaf”!
If Brian Greene is correct, then all of spacetime is determined before we actually “get there”. Is this correct?
Could anyone please explain to me this statement of his?
Gothican.
2. Jan 2, 2009
### tiny-tim
Welcome to PF!
Hi Gothican! Welcome to PF!
I think Greene is saying that, if you start at any spacetime point in the loaf, you can always find a slice which passes through that point …
where that slice cuts the "x-axis" is where a stationary observer would say that that spacetime point is in the present.
But it's not in our present, so it's not pre-determined for us.
(but i don't see what difference the 45º makes )
3. Jan 2, 2009
### Naty1
Greene also points out : "Does Elvis exist right now? No....(Elvis) is not on my current time slice...not on my "current" list...(he) does not currently exist......reality embraces the past, present,future equally and the flow (of time) we envision ....is illusory...." (page 132) ???
Well,that IS Einstein's view, and that of physics in general....equations don't distinguish between past present and future....Greene also discusses how entropy affects the arrow of time...in about an other 30 pages or so....and all in all it requires a few readings to sort out the different perspectives...but I, for one, was still left with more questions than answers.
Wikipedia says:
http://en.wikipedia.org/wiki/Time
So don't be surprised if you are not entirely comfortable with descriptions from Greene... nobody REALLY understands what time is....
But his book is a favorite of mine...
4. Jan 2, 2009
### matheinste
Hello all.
I know the nature of time is the subject of much philosophical discussion but is it not the fact that for SR a working defintion is that time is what a clock measures.
Matheinste.
5. Jan 3, 2009
### Fredrik
Staff Emeritus
I don't see how you came to that conclusion, but I can tell you that it's wrong.
Yes, that's how special (and general) relativity describes spacetime, as a 4-dimensional "thing". However, that doesn't imply that what happens in spacetime is deterministic. The determinism in special relativistic classical mechanics doesn't really have anything to do with the "slicing" of spacetime into "slices" that different observers think of as "now". The determinism comes from the special relativistic version of Newton's second law, which implies that if you know the position and velocity of a particle at one time (and the force at all times), you can calculate the position at all times.
That's not enough. You need that operational definition and a mathematical definition. The operational definition should be thought of as a postulate that enables us to interpret mathematical calculations as predictions about the results of experiments. SR isn't defined by Einstein's postulates, or by the mathematics of Minkowski space. It's defined by the operational "definitions" that describe the connection between the mathematical model (Minkowski space) and the results of experiments. The operational definitions are the true postulates of the theory.
6. Jan 3, 2009
### Naty1
Fredrik posts:
Insightful!! Thanks...
a great way to interpret that equations don't distinguish between past ,present and future.
7. Jan 3, 2009
### Gothican
I think that Fredrick managed to fully answer my question. thanks a lot!
for this you should check out Wikipedia regarding light cones:
http://en.wikipedia.org/wiki/Light_cone" [Broken]
and the Lorentz transformation for further reading:
http://en.wikipedia.org/wiki/Lorentz_transformation" [Broken]
Last edited by a moderator: May 3, 2017 | 1,123 | 4,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-13 | latest | en | 0.916762 |
http://www.wyzant.com/resources/answers/quadrilateral?f=new-answers | 1,404,971,526,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776401658.57/warc/CC-MAIN-20140707234001-00022-ip-10-180-212-248.ec2.internal.warc.gz | 641,529,722 | 10,308 | Search 75,655 tutors
# How can a figure be a rectangle, a rhombus, and a square?
Explain how a figure can be a rectangle, a rhombus, and a square.
# which of the following types of quadrilaterals always has perpendicular sides?
what type of quadrilaterals has perpendicular sides.
# Prove: if a quadrilateral has one pair of sides both parallel and congruent then the quadrilateral is a parallelogram.
I think I have to prove that the other opposite sides are also parallel, but I am not sure how to do it. It is given that the other opposite sides are congruent and parallel. | 137 | 582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23 | latest | en | 0.934471 |
https://community.powerbi.com/t5/Desktop/multiple-rank-columns-in-matrix-table/m-p/1424523 | 1,606,153,099,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141163411.0/warc/CC-MAIN-20201123153826-20201123183826-00429.warc.gz | 242,627,372 | 112,279 | cancel
Showing results for
Search instead for
Did you mean:
Highlighted
Helper III
## multiple rank columns in matrix table
Hi all, I would need to have an extra rank column for each criteria that is selected in the criteria filter.
Is that possible
Many thanks
David
1 ACCEPTED SOLUTION
Accepted Solutions
Highlighted
Community Support
Hi @ADP007 ,
We can create a calculate column and a measure to meet your requirement.
1. Create a rank column.
``````AVG. RANK =
RANKX(
FILTER(
mScoreData,
[CATEGORY]=EARLIER([CATEGORY])
&&[CRITERIA]=EARLIER([CRITERIA])
),
mScoreData[SCORE],,DESC,Dense)
``````
2. Then create an average measure.
``````AVGRank =
var _total =
CALCULATE(
SUM(mScoreData[AVG. RANK]),
FILTER(
ALLSELECTED(mScoreData),
[CATEGORY]=MAX([CATEGORY])
&&[CATEGORY2]=MAX([CATEGORY2])
&&[MEDIA]=MAX([MEDIA])
&&[LABEL]=MAX([LABEL])))
var _count = CALCULATE(DISTINCTCOUNT(mScoreData[CRITERIA]),ALLSELECTED(mScoreData))
return
DIVIDE(_total,_count)``````
If you have any question, please kindly ask here and we will try to resolve it.
Best regards,
Community Support Team _ zhenbw
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
BTW, pbix as attached.
10 REPLIES 10
Highlighted
Super User IV
@ADP007 , Not very clear. You can not add a column when the filter is applied, You can change Rank calculation based on.
Please provide your feedback comments and advice for new videos
Tutorial Series Dax Vs SQL Direct Query PBI Tips
Appreciate your Kudos.
Did I answer your question? Mark my post as a solution! Appreciate your Kudos!!
Dashboard of My Blogs !! YouTube Channel !! Connect on Linkedin
Proud to be a Super User!
Highlighted
Helper III
@amitchandak, thanks for your time. When I select items from the criteria filter the dimension gets added to the matrix.
In my image I have 4 criterias selected and thus 4 columns in my matrix table. I would like to have an extra column in the matrix for each dimension with the ranking. Something like this but of course dynamically. Hope this helps.
CATEGORY SECTOR BRAND MEDIA Convincing Rank Draws attention Rank Inspiring Rank Likeability Rank
Highlighted
Community Support
Hi @ADP007 ,
We can create a measure and put it in Values to meet your requirement.
``````Rank =
RANKX (
FILTER ( ALLSELECTED ( 'Table' ), 'Table'[Media] = MAX ( 'Table'[Media] ) ),
CALCULATE ( SUM ( 'Table'[value] ) ),
,
ASC,
DENSE
)
``````
If it doesn’t meet your requirement, could you please provide a mockup sample based on fake data?
It will be helpful if you can show us the exact expected result based on the tables.
Please upload your files to OneDrive For Business and share the link here. Please don't contain any Confidential Information or Real data in your reply.
Best regards,
Community Support Team _ zhenbw
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
BTW, pbix as attached.
Highlighted
Helper III
Hi, On holiday this week but will try and give an answer as soon as possible.
Thanks
Highlighted
Helper III
Hi, Here is the link, hope it works : mScore
There is an xls file with what I'm trying to obtain.
Thanks for any help
David
Highlighted
Helper III
I have managed to place a rank column following your mockup. However I'm still stuck as I need to calculate a new column based on the various rank columns. In the image below I would need an order by which is the average rank for on line. For example line 1 would have an order by of 5.4 (6+6+4+6+5)/5. Is this possible?
Highlighted
Community Support
Hi @ADP007 ,
We can create a calculate column and a measure to meet your requirement.
1. Create a rank column.
``````AVG. RANK =
RANKX(
FILTER(
mScoreData,
[CATEGORY]=EARLIER([CATEGORY])
&&[CRITERIA]=EARLIER([CRITERIA])
),
mScoreData[SCORE],,DESC,Dense)
``````
2. Then create an average measure.
``````AVGRank =
var _total =
CALCULATE(
SUM(mScoreData[AVG. RANK]),
FILTER(
ALLSELECTED(mScoreData),
[CATEGORY]=MAX([CATEGORY])
&&[CATEGORY2]=MAX([CATEGORY2])
&&[MEDIA]=MAX([MEDIA])
&&[LABEL]=MAX([LABEL])))
var _count = CALCULATE(DISTINCTCOUNT(mScoreData[CRITERIA]),ALLSELECTED(mScoreData))
return
DIVIDE(_total,_count)``````
If you have any question, please kindly ask here and we will try to resolve it.
Best regards,
Community Support Team _ zhenbw
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
BTW, pbix as attached.
Highlighted
Helper III
So many thanks for your help, I don't use Power Bi enough or often enough.
I have three remarks:
1. The Rank columns seems not to be correct (there are several rows with the value 1)
2. Is it possible to have the average as a dimension so that it is possible to apply a sort on it?
3. The average rank doesn't seem to be correct when you filter on media/brand/sector
Thanks
Highlighted
Helper III
I understand why the rank was not correct. The matrix needs to be fully expanded for the rank to work.
So the only thing I need is to have this avr rang as a column. The reason is for the user to sort asc or desc on this value.
Thanks
D.
## Helpful resources
Announcements
#### Power Platform Community Conference
Check out the on demand sessions that are available now!
#### Microsoft Power Platform Communities
Check out the Winners!
#### Power Platform 2020 release wave 2 plan
Features releasing from October 2020 through March 2021
Top Solution Authors
Top Kudoed Authors | 1,389 | 5,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.785602 |
https://coderanch.com/t/250448/certification/byte-shift | 1,508,740,230,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825700.38/warc/CC-MAIN-20171023054654-20171023074654-00134.warc.gz | 695,099,050 | 6,017 | programming forums Java Java JSRs Mobile Certification Databases Caching Books Engineering OS Languages Frameworks Products This Site Careers Other all forums
this forum made possible by our volunteer staff, including ...
Marshals:
Sheriffs:
Saloon Keepers:
Bartenders:
# byte shift
Yati Tan
Ranch Hand
Posts: 56
class EBH005 {
public static void main (String[] s) {
byte b = 127; b <<= 2;System.out.println(b);
}}
What is the result of attempting to compile and run the program?
a. Prints: -4
b. Prints: -3
c. Prints: -2
d. Prints: 0
e. Prints: 1
f. Prints: 127
g. Prints: 508
h. Run-time error
i. Compile-time error
j. None of the above
The answer to above ? is a. Can anybody exaplin me that ? The answer is -4 and I didnot get how 11111100 is converted to -4 ? I know, its two's complement...but still can anybody elaborate it further ?
Steve Morrow
Ranch Hand
Posts: 657
I did not get how 11111100 is converted to -4?
The leftmost bit indicates the sign (negative, in this case). Flip the bits and add 1.
Flipped: 00000011
Read the last number as binary and remember the sign. Ergo, -4.
Hope this helps.
Ranch Hand
Posts: 69
Yatikashipurut,
The answer is in your question. 11111100 is the 2's compliment of -4. Here's how it works:
In order to get a negative number in binary using 2's compliment, you flip all the bits and add one. So,
00000100 => 4
11111011 => flip all the bits
---------
11111100 => -4
Do the same process again to go back to +4.
11111100 => -4
00000011 => flip all the bits | 433 | 1,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-43 | latest | en | 0.806558 |
https://www.boatdesign.net/threads/comparing-righting-moments.60240/#post-829059 | 1,642,373,825,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300244.42/warc/CC-MAIN-20220116210734-20220117000734-00496.warc.gz | 731,544,098 | 13,676 | # Comparing Righting Moments
Discussion in 'Hydrodynamics and Aerodynamics' started by rwatson, Apr 26, 2018.
1. Joined: Aug 2007
Posts: 5,875
Likes: 311, Points: 83, Legacy Rep: 1749
Location: Tasmania,Australia
### rwatsonSenior Member
I was just looking at two righting arm curve graphs.
One is in Inches/Pounds, the Other is in Metres/Kilos
Is the conversion as simple as described in the attached image ?
eg 27.5 inch-pounds = .31 Kilo-metres of righting moment ?
2. Joined: Jul 2010
Posts: 366
Likes: 38, Points: 28, Legacy Rep: 233
Location: N.W. England
### latestarterSenior Member
Please ignore this post, see post 4
If 1 kgf = 2.2046226218 lbs and 1 metre = 39.37008 inches then 1 kgf-metre will be 2.2046226218 x 39.3700787 = 86.79616............
Seems to be right although I was always uncomfortable with kilogram force and kilogram weight and introducing g for gravity.
Last edited: Apr 26, 2018
3. Joined: Sep 2011
Posts: 6,872
Likes: 517, Points: 123, Legacy Rep: 300
Location: Spain
### TANSLSenior Member
In the curves it appears, in abscissas the angles in degrees, and in ordinate the righting arm, in feet or in meters. It is about the GZ values and, therefore, no value appears in pounds-feet or kg-meters. So you only need to convert feet into meters (1 foot = 0.3048 meters)
4. Joined: Jul 2010
Posts: 366
Likes: 38, Points: 28, Legacy Rep: 233
Location: N.W. England
### latestarterSenior Member
What TANSL says makes sense, it just shows the dangers of posting before breakfast
5. Joined: Apr 2015
Posts: 464
Likes: 100, Points: 43, Legacy Rep: 37
Location: Berlin, Germany
### HeimfriedSenior Member
If the term "Righting Arm" in the left diagram is correctly used, it is only a distance between two points (G and Z), measured in inches as it seems (not feet).
Righting arm times deplacement (in pound) is mostly called righting moment. (Because a moment of this kind is force times length, it should be poundforce instead of pound. Also the righting moment in SI units is correctly given as Newton * Meter = displacement mass in kg times accerelation of gravity in m/s² times righting arm in m. In spite of this is mostly used kilogramm * meter or tonne * meter.)
So I would understand the 27.5" (maximum left diagram) is 27.5 inch* 0,0254 m/inch = 0.6985 m
6. Joined: Sep 2011
Posts: 6,872
Likes: 517, Points: 123, Legacy Rep: 300
Location: Spain
### TANSLSenior Member
In addition to all that has already been mentioned, the curve on the right (in post #1) is quite interesting: from 176º, more or less, the GZ returns to have positive values. For -25º, however, the values are negative. Nor do the maximum values, in such a large angular range, seem normal.
Maybe there is an explanation but at this moment I am not able to find it. What kind of ship does this curve refer to, @rwatson?
Last edited: Apr 26, 2018
7. Joined: Apr 2015
Posts: 464
Likes: 100, Points: 43, Legacy Rep: 37
Location: Berlin, Germany
### HeimfriedSenior Member
I would read GZ = 0 at 180 deg. This would be the common instable equilibrium at the position "keel upwards".
The negative GZ values corresponding to the negative angle 0 deg. to -25 deg. are absolutely normal for boats or ships, while the "plateau" from 25 deg. to 75 deg., as you said, is not.
8. Joined: Sep 2011
Posts: 6,872
Likes: 517, Points: 123, Legacy Rep: 300
Location: Spain
### TANSLSenior Member
You're right, I measured 176º when in fact it was 180º. In any case, should not the values be equal to -25º than to 175º?
9. Joined: Apr 2015
Posts: 464
Likes: 100, Points: 43, Legacy Rep: 37
Location: Berlin, Germany
### HeimfriedSenior Member
No, a normal (symmetric) boat or ship hull gives a rotational symmetric GZ curve. So the value of -25 deg. is the negative of the 25 deg. value, same to 175 deg.
#### Attached Files:
• ###### GZ_3.jpg
File size:
438.8 KB
Views:
242
10. Joined: Apr 2015
Posts: 464
Likes: 100, Points: 43, Legacy Rep: 37
Location: Berlin, Germany
### HeimfriedSenior Member
In English
#### Attached Files:
• ###### GZ_4.jpg
File size:
432.8 KB
Views:
260
11. Joined: Sep 2011
Posts: 6,872
Likes: 517, Points: 123, Legacy Rep: 300
Location: Spain
### TANSLSenior Member
You are very right. A new stupid mistake on my part.
Let's see if someone has an explanation for the upper part of the curve.
12. Joined: Aug 2007
Posts: 5,875
Likes: 311, Points: 83, Legacy Rep: 1749
Location: Tasmania,Australia
### rwatsonSenior Member
Sorry. I missed the conversation here.
The high flat top in the Kg/Metre is caused by the high cabin top.
I still haven't figured out if my conversion technique was correct or not.
What's the consensus ?
13. Joined: Sep 2011
Posts: 6,872
Likes: 517, Points: 123, Legacy Rep: 300
Location: Spain
### TANSLSenior Member
That's correct
1 inch = 0.0254 m
1 pound = 0.4535924 kg
27.5 inch-pounds = 27.5*0.0254*0.4535924 m-kg = 0,3168342914 m-kg
I'm not sure that's the explanation. Could we see a drawing of that ship?
14. Joined: Apr 2015
Posts: 464
Likes: 100, Points: 43, Legacy Rep: 37
Location: Berlin, Germany
### HeimfriedSenior Member
The math is correct: 27.5 inch * pound = 0.3168 kg * m
But the value of 27.5 corresponds to the unit inch only not the unit inch * pound. Because the quantity righting arm (or righting lever) is a length only, not a moment.
Therefore the righting arm in Meters corresponding to the left diagram is 27.5 inch * 0.0254 inch/m = 0.6985 m .
If you multiply the 27.5 inch with the displacement of that boat in pounds, you will get the righting moment in inch * pound.
15. Joined: Aug 2007
Posts: 5,875
Likes: 311, Points: 83, Legacy Rep: 1749
Location: Tasmania,Australia
### rwatsonSenior Member
That's very helpful Heimfried, thank you.
Therefore, that little conversion chart isn't doing anything useful in converting inch-pounds to Kilos-Metre ?
It sounds like you are saying that both graphs are only showing moment lengths. I was assuming that the righting moment was being expressed in units of "Force" (pounds-inches and Kilos/Metre) , but you are saying we have to multiply it by the weight of the boat.
Is it then correct to say that two boats could have similar righting moments, but because they have different ballast weights that the actual strength of wind to tip them over them would vary ?
Many Thanks
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different. | 1,916 | 6,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-05 | latest | en | 0.910251 |
https://trac.sagemath.org/attachment/ticket/5711/enhanced-symbolic-typesetting-rebased_to_4.0.1.patch | 1,571,636,328,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987756350.80/warc/CC-MAIN-20191021043233-20191021070733-00376.warc.gz | 754,216,921 | 11,587 | # Ticket #5711: enhanced-symbolic-typesetting-rebased_to_4.0.1.patch
File enhanced-symbolic-typesetting-rebased_to_4.0.1.patch, 19.9 KB (added by gmhossain, 10 years ago)
• ## sage/misc/latex.py
# HG changeset patch
# User Golam Mortuza Hossain <gmhossain@gmail.com>
# Date 1244851786 10800
# Node ID 58abaa4df2e26c9d272e262845af6e025a4d8fa0
# Parent 6c7354ace3b6e0bf4c33ddb730531bd23a9fb92c
Enhances typsetting of Symbolic functions within Sage (rebased to Sage-4.0.1)
diff -r 6c7354ace3b6 -r 58abaa4df2e2 sage/misc/latex.py
a 'omega', 'Omega'] def latex_symbolic_function(fname, fstr, args): r""" Typeset a given symbolic function in LaTeX. EXAMPLES:: sage: import sage.misc.latex as latex_module sage: latex_sym_fn = latex_module.latex_symbolic_function sage: var('x,a,b,t,s') (x, a, b, t, s) sage: riemann(x) = function('riemann',x) sage: latex_sym_fn("riemann","\\mathcal{R}",(x,)) '\\mathcal{R}\\left(x\\right)' sage: psi(x) = function('psi',x) sage: latex_sym_fn("psi",False,(x,)) '\\psi\\left(x\\right)' sage: latex_sym_fn("conjugate",False,(psi(x))) '\\overline{\\psi\\left(x\\right)}' sage: latex_sym_fn("limit",False,(psi(x),x,a)) '\\lim_{x \\to a}\\, \\psi\\left(x\\right)' sage: f(t) = function('f',t) sage: latex_sym_fn("laplace",False,(f(t),t,s)) '\\mathcal{L}\\left(f\\left(t\\right), t, s\\right)' sage: F(s) = function('F',s) sage: latex_sym_fn("ilt",False,(F(s),s,t)) '\\mathcal{L}^{-1}\\left(F\\left(s\\right), s, t\\right)' sage: f(x) = function('f',x) sage: latex_sym_fn("integrate",False,(f(x),x)) '\\int f\\left(x\\right)\\,{d x}' sage: latex_sym_fn("integrate",False,(f(x),x,a,b)) '\\int_{a}^{b} f\\left(x\\right)\\,{d x}' sage: var('x,y') (x, y) sage: f(x) = function('f',x) sage: latex_sym_fn("diff",False,(f(x),x,1)) '\\frac{d f\\left(x\\right)}{d x}' sage: f(x,y) = function('f',x,y) sage: latex_sym_fn("diff",False,(f(x,y),x,1)) '\\frac{\\partial}{\\partial x}f\\left(x, y\\right)' sage: myfunc(x) = function('myfunc',x) sage: latex_sym_fn("myfunc",False,(x,)) '{\\it myfunc}\\left(x\\right)' AUTHORS: - Golam Mortuza Hossain (2009-04-05) """ # Use the latex_name if defined if fstr is not False: return "%s%s"%(fstr, _args_latex_(fname, args)) # Special handling of fname where specific processing is needed if fname == "conjugate": return _conjugate_latex_(fname, args) elif fname == "limit": return _limit_latex_(fname, args) elif fname == "laplace": return _laplace_latex_(fname, args) elif fname == "ilt": return _inverse_laplace_latex_(fname, args) elif fname == "integrate": return _integrate_latex_(fname, args) elif fname == "diff": f = args[0] fstr = latex(f) nvars = len(f.variables()) return _derivative_latex_(fstr, args, nvars) # Now check whether fname is a Greek letter name = latex_function_name(fname) if name is not False: return "%s%s"%(name, _args_latex_(fname, args)) # Use default typesetting scheme return _symbolic_function_default_latex_(fname, args) def _symbolic_function_default_latex_(fname, args): r""" Return LaTeX expression of a symbolic function using default scheme. EXAMPLES:: sage: import sage.misc.latex as latex_module sage: _sym_def_latex_ = latex_module._symbolic_function_default_latex_ sage: var('x') x sage: myfunc(x) = function('myfunc',x) sage: _sym_def_latex_("myfunc",(x,)) '{\\it myfunc}\\left(x\\right)' sage: my_func(x) = function('my_func',x) sage: _sym_def_latex_("my_func",(x,)) '{\\it my\\_func}\\left(x\\right)' AUTHORS: - Golam Mortuza Hossain (2009-04-05) """ # Prepend underscore by "\\" in fname fname = fname.replace("_", "\\_") # Default typesetting scheme (similar to Maxima scheme) return "{\\it %s}%s"%(fname, _args_latex_(fname, args)) def _args_latex_(fname, args): r""" Return LaTeX expression for the arguments of a symbolic function. EXAMPLES:: sage: import sage.misc.latex as latex_module sage: _args_latex_ = latex_module._args_latex_ sage: var('x') x sage: f(x) = function('f',x) sage: _args_latex_("f",(x,)) '\\left(x\\right)' AUTHORS: - Golam Mortuza Hossain (2009-04-06) """ # If all arguments are SymbolicVariables then the function should be # typeset as f(x) rather than f\left(x\right) use_left_right = True #from sage.calculus.calculus import SymbolicVariable #for x in self._args: # if not isinstance(x, SymbolicVariable): # use_left_right = True # break if use_left_right is True: return "\\left(%s\\right)"%(', '.join([latex(x) for x in args])) else: return "(%s)"%(', '.join([latex(x) for x in args])) def latex_d_derivative(fstr, params, args): r""" Return LaTeX expression for "D" format derivatives of a symbolic function. EXAMPLES:: sage: import sage.misc.latex as latex_module sage: latex_d_derivative = latex_module.latex_d_derivative sage: var('x,y') (x, y) sage: f(x) = function('f',x) sage: latex_d_derivative("f\\left(x\\right)",[0],(x,)) '\\frac{d f\\left(x\\right)}{d x}' sage: latex_d_derivative("f\\left(x\\right)",[0,0],(x,)) '\\frac{d^{2} f\\left(x\\right)}{d {x}^{2}}' sage: f(x,y) = function('f',x,y) sage: latex_d_derivative("f\\left(x, y\\right)",[0],(x,y)) '\\frac{\\partial}{\\partial x}f\\left(x, y\\right)' sage: latex_d_derivative("f\\left(x, y\\right)",[0,0,0],(x,y)) '\\frac{\\partial^{3}}{\\partial {x}^{3}}f\\left(x, y\\right)' sage: latex_d_derivative("f\\left(x, y\\right)",[0,1],(x,y)) '\\frac{\\partial^{2}}{\\partial y\\partial x}f\\left(x, y\\right)' AUTHORS: - Golam Mortuza Hossain (2009-06-11) """ # Following lines convert "D" format derivative into "diff" # format and then call the "diff" typesetting function varhash = {} for x in params: if args[x] in varhash: varhash[args[x]] = varhash[args[x]] + 1 else: varhash[args[x]] = 1 newargs = [fstr] for a in args: if a in varhash: newargs.append(a) newargs.append(varhash[a]) # No of variables nvars = len(args) return _derivative_latex_(fstr, newargs, nvars) def _derivative_latex_(fstr, args, nvars): r""" Return LaTeX expression for derivatives of a symbolic function. EXAMPLES:: sage: import sage.misc.latex as latex_module sage: _derivative_latex_ = latex_module._derivative_latex_ sage: var('x,y') (x, y) sage: f(x) = function('f',x) sage: _derivative_latex_("f\\left(x\\right)",(f(x),x,1),1) '\\frac{d f\\left(x\\right)}{d x}' sage: _derivative_latex_("f\\left(x\\right)",(f(x),x,2),1) '\\frac{d^{2} f\\left(x\\right)}{d {x}^{2}}' sage: f(x,y) = function('f',x,y) sage: _derivative_latex_("f\\left(x, y\\right)",(f(x,y),x,1),2) '\\frac{\\partial}{\\partial x}f\\left(x, y\\right)' sage: _derivative_latex_("f\\left(x, y\\right)",(f(x,y),x,3),2) '\\frac{\\partial^{3}}{\\partial {x}^{3}}f\\left(x, y\\right)' sage: _derivative_latex_("f\\left(x, y\\right)",(f(x,y),x,1,y,1),2) '\\frac{\\partial^{2}}{\\partial y\\partial x}f\\left(x, y\\right)' AUTHORS: - Golam Mortuza Hossain (2009-04-05) """ n = len(args) # We dont do any processing for n < 2 if n < 2: return _symbolic_function_default_latex_("diff", args) # Check whether it should be partial derivative # Logic taken from: http://trac.sagemath.org/sage_trac/ticket/4202 if nvars == 1: d_latex = "d" else: d_latex = "\\partial" # Read the variables diffstr = ""; total = 0 for i in range(1,n-1,2): x = args[i] # variable j = args[i+1] # no of times diff acts on f w.r.t. variable x total = total + j # total no of times diff acts if j == 1: diffstr = "%s %s"%(d_latex, latex(x)) + diffstr else: diffstr = "%s {%s}^{%s}"%(d_latex, latex(x), latex(j)) + diffstr # Return final expression if total == 1: if d_latex == "d": return "\\frac{%s %s}{%s}"%(d_latex, fstr, diffstr) else: return "\\frac{%s}{%s}%s"%(d_latex, diffstr, fstr) if d_latex == "d": return "\\frac{%s^{%d} %s}{%s}"%(d_latex, total, fstr, diffstr) else: return "\\frac{%s^{%d}}{%s}%s"%(d_latex, total, diffstr, fstr) def _integrate_latex_(fname, args): r""" Return LaTeX expression for integration of a symbolic function. EXAMPLES:: sage: import sage.misc.latex as latex_module sage: _integrate_latex_ = latex_module._integrate_latex_ sage: var('x,a,b') (x, a, b) sage: f(x) = function('f',x) sage: _integrate_latex_("integrate",(f(x),x)) '\\int f\\left(x\\right)\\,{d x}' sage: _integrate_latex_("integrate",(f(x),x,a,b)) '\\int_{a}^{b} f\\left(x\\right)\\,{d x}' AUTHORS: - Golam Mortuza Hossain (2009-04-05) """ n = len(args) # We dont process if number of arguments is neither 2 nor 4 if n != 2 and n != 4: # Return default typesetting return _symbolic_function_default_latex_(fname, args) f = args[0] x = args[1] # Check whether its a definite integral if n == 4: a = args[2] b = args[3] return "\\int_{%s}^{%s} %s\\,{d %s}"%(latex(a), latex(b), latex(f), latex(x)) # Typeset as indefinite integral return "\\int %s\\,{d %s}"%(latex(f), latex(x)) def _inverse_laplace_latex_(fname, args): r""" Return LaTeX expression for inverse Laplace transform of a symbolic function. EXAMPLES:: sage: import sage.misc.latex as latex_module sage: _inverse_laplace_latex_ = latex_module._inverse_laplace_latex_ sage: var('s,t') (s, t) sage: F(s) = function('F',s) sage: _inverse_laplace_latex_("ilt",(F(s),s,t)) '\\mathcal{L}^{-1}\\left(F\\left(s\\right), s, t\\right)' AUTHORS: - Golam Mortuza Hossain (2009-04-05) """ return "\\mathcal{L}^{-1}\\left(%s\\right)"%(', '.join([latex(x) for x in args])) def _laplace_latex_(fname, args): r""" Return LaTeX expression for Laplace transform of a symbolic function. EXAMPLES:: sage: import sage.misc.latex as latex_module sage: _laplace_latex_ = latex_module._laplace_latex_ sage: var('s,t') (s, t) sage: f(t) = function('f',t) sage: _laplace_latex_("laplace",(f(t),t,s)) '\\mathcal{L}\\left(f\\left(t\\right), t, s\\right)' AUTHORS: - Golam Mortuza Hossain (2009-04-05) """ return "\\mathcal{L}\\left(%s\\right)"%(', '.join([latex(x) for x in args])) def _limit_latex_(fname, args): r""" Return latex expression for limit of a symbolic function. EXAMPLES:: sage: import sage.misc.latex as latex_module sage: _limit_latex_ = latex_module._limit_latex_ sage: var('x,a') (x, a) sage: psi(x) = function('psi',x) sage: _limit_latex_("limit",(psi(x),x,a)) '\\lim_{x \\to a}\\, \\psi\\left(x\\right)' AUTHORS: - Golam Mortuza Hossain (2009-04-05) """ # We process only if there are precisely three arguments if len(args) == 3: # Read f,x,a from arguments f = args[0] x = args[1] a = args[2] return "\\lim_{%s \\to %s}\\, %s"%(latex(x), latex(a), latex(f)) # Return default typesetting return _symbolic_function_default_latex_(fname, args) def _conjugate_latex_(fname, args): r""" Return LaTeX expression for conjugate of a symbolic function. EXAMPLES:: sage: import sage.misc.latex as latex_module sage: _conjugate_latex_ = latex_module._conjugate_latex_ sage: var('x') x sage: psi(x) = function('psi',x) sage: _conjugate_latex_("conjugate",psi(x)) '\\overline{\\psi\\left(x\\right)}' AUTHORS: - Golam Mortuza Hossain (2009-06-11) """ # We process only if there is only one argument if len(args) == 1: return "\\overline{%s}"%(latex(args)) # Return default typesetting return _symbolic_function_default_latex_(fname, args) def latex_function_name(x): r""" Return common function names such as alpha, beta1, psi_00, R_mn, etc. as latex symbols, else return False. EXAMPLES:: sage: import sage.misc.latex as latex_module sage: latex_function_name = latex_module.latex_function_name sage: latex_function_name('psi') '\\psi' sage: latex_function_name('psi0') '\\psi_{0}' sage: latex_function_name('f1') 'f_{1}' sage: latex_function_name('psi_mu') '\\psi_{\\mu}' sage: latex_function_name('psi_11') '\\psi_{11}' sage: latex_function_name('R_ab') 'R_{ab}' sage: latex_function_name('R_nu') 'R_{\\nu}' sage: latex_function_name('myfunc') False sage: latex_function_name('psi_') False sage: latex_function_name('abc_xyz_psi') False sage: latex_function_name('abc_beta') False NOTES: This function is based largely on latex_variable_name function. AUTHORS: - Golam Mortuza Hossain (2009-04-05) """ # For known names return them after preprending with "\\" if x in common_varnames: return "\\" + x # Look for underscrore. If found then use its postion to # find the suffix underscore = x.find("_") if underscore == -1: import re # * The "\d|[.,]" means "decimal digit" or period or comma # * The "+" means "1 or more" # * The "$" means "at the end of the line" m = re.search('(\d|[.,])+$',x) if m is None: prefix = x suffix = None else: prefix = x[:m.start()] suffix = x[m.start():] else: prefix = x[:underscore] suffix = x[underscore+1:] if len(suffix)== 0: return False # If suffix contains underscores then don't process if suffix and suffix.find("_") != -1: return False # If prefix is not a common name or a more-than-one letters word # then don't process if prefix not in common_varnames and len(prefix) != 1: return False # Check if prefix or suffix is a common name if prefix in common_varnames: prefix = "\\" + prefix if suffix and len(suffix) > 0: if suffix in common_varnames: suffix = "\\" + suffix return '%s_{%s}'%(prefix, suffix) else: return '%s'%(prefix) def latex_varify(a): if a in common_varnames: return "\\" + a
• ## sage/symbolic/function.pyx
diff -r 6c7354ace3b6 -r 58abaa4df2e2 sage/symbolic/function.pyx
a my args are: x, y^z sage: latex(foo(x,y^z)) \mbox{t}\left(x, y^{z}\right) t\left(x, y^{z}\right) sage: foo = nfunction('t', 2, print_latex_func=my_print) sage: foo(x,y^z) t(x, y^z) my args are: x, y^z sage: foo = nfunction('t', 2, latex_name='foo') sage: latex(foo(x,y^z)) \mbox{foo}\left(x, y^{z}\right) foo\left(x, y^{z}\right) TESTS::
• ## sage/symbolic/pynac.pyx
diff -r 6c7354ace3b6 -r 58abaa4df2e2 sage/symbolic/pynac.pyx
a sage: get_sfunction_from_serial(i) == foo True sage: py_latex_function_pystring(i, (x,y^z)) '\\mbox{foo}\\left(x, y^{z}\\right)' sage: py_latex_function_pystring(i, (x,y^z), True) '\\left(\\mbox{foo}\\right)\\left(x, y^{z}\\right)' '{\\it foo}\\left(x, y^{z}\\right)' Test latex_name:: sage: get_sfunction_from_serial(i) == foo True sage: py_latex_function_pystring(i, (x,y^z)) '\\mbox{\\mathrm{bar}}\\left(x, y^{z}\\right)' '\\mathrm{bar}\\left(x, y^{z}\\right)' Test custom func:: return str(res) return res # otherwise, use the latex name if defined # Typesetting is done by latex_symbolic_function from sage.misc.latex import latex_symbolic_function fname = func._name # Use the latex_name if defined if func._latex_name: name = func._latex_name fstr = func._latex_name else: name = func._name # if latex_name is not defined, use default name if fname_paren: olist = [r'\left(', r'\mbox{', name, '}', r'\right)'] else: olist = [r'\mbox{', name, '}'] # print the arguments olist.extend([r'\left(', ', '.join([x._latex_() for x in args]), r'\right)'] ) return ''.join(olist) fstr = False # Call latex_symbolic_function return latex_symbolic_function (fname, fstr, args) cdef public stdstring* py_latex_function(unsigned id, object args) except +: return string_from_pystr(py_latex_function_pystring(id, args)) print(ostr.c_str()) stdstring_delete(ostr) # Whether new "D" format should be typeset in old "diff" format typeset_d_as_diff = True cdef public stdstring* py_latex_fderivative(unsigned id, object params, object args) except +: """ See documentation of py_print_fderivative for more information. """ # Check whether old "diff" format should be used if typeset_d_as_diff is True: fstr = py_latex_function_pystring(id, args, True) from sage.misc.latex import latex_d_derivative py_res = latex_d_derivative(fstr, params, args) return string_from_pystr(py_res) ostr = ''.join(['D[', ', '.join([repr(int(x)) for x in params]), ']']) fstr = py_latex_function_pystring(id, args, True) py_res = ostr + fstr EXAMPLES:: sage: from sage.symbolic.pynac import py_latex_fderivative_for_doctests as py_latex_fderivative sage: sage.symbolic.pynac.typeset_d_as_diff=False sage: var('x,y,z',ns=1) (x, y, z) sage: from sage.symbolic.function import function, get_sfunction_from_serial, get_ginac_serial sage: get_sfunction_from_serial(i) == foo True sage: py_latex_fderivative(i, (0, 1, 0, 1), (x, y^z)) D[0, 1, 0, 1]\left(\mbox{foo}\right)\left(x, y^{z}\right) D[0, 1, 0, 1]{\it foo}\left(x, y^{z}\right) Test latex_name:: sage: get_sfunction_from_serial(i) == foo True sage: py_latex_fderivative(i, (0, 1, 0, 1), (x, y^z)) D[0, 1, 0, 1]\left(\mbox{\mathrm{bar}}\right)\left(x, y^{z}\right) D[0, 1, 0, 1]\mathrm{bar}\left(x, y^{z}\right) Test custom func:: | 5,260 | 16,197 | {"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} | 2.515625 | 3 | CC-MAIN-2019-43 | latest | en | 0.338609 |
https://sodocumentation.net/algorithm/topic/4471/searching | 1,659,902,749,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570692.22/warc/CC-MAIN-20220807181008-20220807211008-00087.warc.gz | 489,613,545 | 13,065 | # algorithm Searching
## Introduction
Binary Search is a Divide and Conquer search algorithm. It uses `O(log n)` time to find the location of an element in a search space where `n` is the size of the search space.
Binary Search works by halving the search space at each iteration after comparing the target value to the middle value of the search space.
To use Binary Search, the search space must be ordered (sorted) in some way. Duplicate entries (ones that compare as equal according to the comparison function) cannot be distinguished, though they don't violate the Binary Search property.
Conventionally, we use less than (<) as the comparison function. If a < b, it will return true. if a is not less than b and b is not less than a, a and b are equal.
## Example Question
You are an economist, a pretty bad one though. You are given the task of finding the equilibrium price (that is, the price where supply = demand) for rice.
Remember the higher a price is set, the larger the supply and the lesser the demand
As your company is very efficient at calculating market forces, you can instantly get the supply and demand in units of rice when the price of rice is set at a certain price `p`.
Your boss wants the equilibrium price ASAP, but tells you that the equilibrium price can be a positive integer that is at most `10^17` and there is guaranteed to be exactly 1 positive integer solution in the range. So get going with your job before you lose it!
You are allowed to call functions `getSupply(k)` and `getDemand(k)`, which will do exactly what is stated in the problem.
## Example Explanation
Here our search space is from `1` to `10^17`. Thus a linear search is infeasible.
However, notice that as the `k` goes up, `getSupply(k)` increases and `getDemand(k)` decreases. Thus, for any `x > y`, `getSupply(x) - getDemand(x) > getSupply(y) - getDemand(y)`. Therefore, this search space is monotonic and we can use Binary Search.
The following psuedocode demonstrates the usage of Binary Search:
``````high = 100000000000000000 <- Upper bound of search space
low = 1 <- Lower bound of search space
while high - low > 1
mid = (high + low) / 2 <- Take the middle value
supply = getSupply(mid)
demand = getDemand(mid)
if supply > demand
high = mid <- Solution is in lower half of search space
else if demand > supply
low = mid <- Solution is in upper half of search space
else <- supply==demand condition
return mid <- Found solution
``````
This algorithm runs in `~O(log 10^17)` time. This can be generalized to `~O(log S)` time where S is the size of the search space since at every iteration of the `while` loop, we halved the search space (from [low:high] to either [low:mid] or [mid:high]).
C Implementation of Binary Search with Recursion
``````int binsearch(int a[], int x, int low, int high) {
int mid;
if (low > high)
return -1;
mid = (low + high) / 2;
if (x == a[mid]) {
return (mid);
} else
if (x < a[mid]) {
binsearch(a, x, low, mid - 1);
} else {
binsearch(a, x, mid + 1, high);
}
}
``````
## Binary Search: On Sorted Numbers
It's easiest to show a binary search on numbers using pseudo-code
``````int array[1000] = { sorted list of numbers };
int N = 100; // number of entries in search space;
int high, low, mid; // our temporaries
int x; // value to search for
low = 0;
high = N -1;
while(low < high)
{
mid = (low + high)/2;
if(array[mid] < x)
low = mid + 1;
else
high = mid;
}
if(array[low] == x)
// found, index is low
else
``````
Do not attempt to return early by comparing array[mid] to x for equality. The extra comparison can only slow the code down. Note you need to add one to low to avoid becoming trapped by integer division always rounding down.
Interestingly, the above version of binary search allows you to find the smallest occurrence of x in the array. If the array contains duplicates of x, the algorithm can be modified slightly in order for it to return the largest occurrence of x by simply adding to the if conditional:
``````while(low < high)
{
mid = low + ((high - low) / 2);
if(array[mid] < x || (array[mid] == x && array[mid + 1] == x))
low = mid + 1;
else
high = mid;
}
``````
Note that instead of doing `mid = (low + high) / 2`, it may also be a good idea to try `mid = low + ((high - low) / 2)` for implementations such as Java implementations to lower the risk of getting an overflow for really large inputs.
Linear search is a simple algorithm. It loops through items until the query has been found, which makes it a linear algorithm - the complexity is O(n), where n is the number of items to go through.
Why O(n)? In worst-case scenario, you have to go through all of the n items.
It can be compared to looking for a book in a stack of books - you go through them all until you find the one that you want.
Below is a Python implementation:
``````def linear_search(searchable_list, query):
for x in searchable_list:
if query == x:
return True
return False
linear_search(['apple', 'banana', 'carrot', 'fig', 'garlic'], 'fig') #returns True
``````
## Rabin Karp
The Rabin–Karp algorithm or Karp–Rabin algorithm is a string searching algorithm that uses hashing to find any one of a set of pattern strings in a text.Its average and best case running time is O(n+m) in space O(p), but its worst-case time is O(nm) where n is the length of the text and m is the length of the pattern.
Algorithm implementation in java for string matching
``````void RabinfindPattern(String text,String pattern){
/*
q a prime number
p hash value for pattern
t hash value for text
d is the number of unique characters in input alphabet
*/
int d=128;
int q=100;
int n=text.length();
int m=pattern.length();
int t=0,p=0;
int h=1;
int i,j;
//hash value calculating function
for (i=0;i<m-1;i++)
h = (h*d)%q;
for (i=0;i<m;i++){
p = (d*p + pattern.charAt(i))%q;
t = (d*t + text.charAt(i))%q;
}
//search for the pattern
for(i=0;i<end-m;i++){
if(p==t){
//if the hash value matches match them character by character
for(j=0;j<m;j++)
if(text.charAt(j+i)!=pattern.charAt(j))
break;
if(j==m && i>=start)
System.out.println("Pattern match found at index "+i);
}
if(i<end-m){
t =(d*(t - text.charAt(i)*h) + text.charAt(i+m))%q;
if(t<0)
t=t+q;
}
}
}
``````
While calculating hash value we are dividing it by a prime number in order to avoid collision.After dividing by prime number the chances of collision will be less, but still ther is a chance that the hash value can be same for two strings,so when we get a match we have to check it character by character to make sure that we got a proper match.
t =(d*(t - text.charAt(i)*h) + text.charAt(i+m))%q;
This is to recalculate the hash value for pattern,first by removing the left most character and then adding the new character from the text.
## Analysis of Linear search (Worst, Average and Best Cases)
We can have three cases to analyze an algorithm:
1. Worst Case
2. Average Case
3. Best Case
``````#include <stdio.h>
// Linearly search x in arr[]. If x is present then return the index,
// otherwise return -1
int search(int arr[], int n, int x)
{
int i;
for (i=0; i<n; i++)
{
if (arr[i] == x)
return i;
}
return -1;
}
``````
/* Driver program to test above functions*/
``````int main()
{
int arr[] = {1, 10, 30, 15};
int x = 30;
int n = sizeof(arr)/sizeof(arr[0]);
printf("%d is present at index %d", x, search(arr, n, x));
getchar();
return 0;
}
``````
Worst Case Analysis (Usually Done)
In the worst case analysis, we calculate upper bound on running time of an algorithm. We must know the case that causes maximum number of operations to be executed. For Linear Search, the worst case happens when the element to be searched (x in the above code) is not present in the array. When x is not present, the search() functions compares it with all the elements of arr[] one by one. Therefore, the worst case time complexity of linear search would be Θ(n)
Average Case Analysis (Sometimes done)
In average case analysis, we take all possible inputs and calculate computing time for all of the inputs. Sum all the calculated values and divide the sum by total number of inputs. We must know (or predict) distribution of cases. For the linear search problem, let us assume that all cases are uniformly distributed (including the case of x not being present in array). So we sum all the cases and divide the sum by (n+1). Following is the value of average case time complexity.
Best Case Analysis (Bogus)
In the best case analysis, we calculate lower bound on running time of an algorithm. We must know the case that causes minimum number of operations to be executed. In the linear search problem, the best case occurs when x is present at the first location. The number of operations in the best case is constant (not dependent on n). So time complexity in the best case would be Θ(1) Most of the times, we do worst case analysis to analyze algorithms. In the worst analysis, we guarantee an upper bound on the running time of an algorithm which is good information. The average case analysis is not easy to do in most of the practical cases and it is rarely done. In the average case analysis, we must know (or predict) the mathematical distribution of all possible inputs. The Best Case analysis is bogus. Guaranteeing a lower bound on an algorithm doesn’t provide any information as in the worst case, an algorithm may take years to run.
For some algorithms, all the cases are asymptotically same, i.e., there are no worst and best cases. For example, Merge Sort. Merge Sort does Θ(nLogn) operations in all cases. Most of the other sorting algorithms have worst and best cases. For example, in the typical implementation of Quick Sort (where pivot is chosen as a corner element), the worst occurs when the input array is already sorted and the best occur when the pivot elements always divide array in two halves. For insertion sort, the worst case occurs when the array is reverse sorted and the best case occurs when the array is sorted in the same order as output. | 2,504 | 10,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2022-33 | latest | en | 0.914784 |
https://world.edu/common-core-and-fractions-part-2/ | 1,716,377,747,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058542.48/warc/CC-MAIN-20240522101617-20240522131617-00696.warc.gz | 546,197,375 | 16,938 | # Common Core and Fractions, Part 2
Share:
Last time I covered the “sledgehammer” method of adding fractions, a method that is very simple and anyone can master in a few minutes.
That’s not the method taught in schools, however. Let’s go over what you need to know to be able to add fractions, using the public school method, and note that it takes weeks for a student to almost never master all this:
1) You need to know about prime numbers. Prime numbers are those numbers which only have themselves and 1 as factors. So, 3 is a prime number (the only positive whole numbers that go into 3 are 1 and 3). 10 is not a prime number (since 5 and 2 are factors of 10). This is introduced around the 3rd grade; if the student doesn’t know this, the rest falls apart and the student will never “get” fractions the public school way.
2) You need to be able to “factor” a number into a product of primes. So, you need to be able to take 20, and write it as 2 * 2 * 5. This method is called “prime factorization”, so the prime factorization of 20 is 2 * 2 * 5. As before, if the student misses this part of class, it’s over.
3) When you look at two numbers, you need to be able to find the Least Common Multiple (LCM). For example, the LCM of 3 and 4 is 12. One way to do this is the look at the prime factorization of 3, and 4, and multiply the non-overlapping prime factors. So, to get that 12, I multiplied 3 and 2 * 2 (the prime factorization of 12). It can be a little more involved than this, but I’m taking a simple example just so I can go through the entire public school process of adding fractions in a sane amount of time. A student that doesn’t know LCM will never get past this part.
4) Now, to add fractions with the same denominator, use the same steps I mentioned above (add the numerator, leave the denominator alone).
5) If the denominators are different, well, this is where even my teacher in public school always got scared, but here goes: first, look at the prime factorizations of the different denominators, then find the LCM by comparing the prime factorizations.
As before, I use the example of 1/3 + 1 / 4.
6) The LCM of my first example is 12 (by looking at the prime factorizations of 3 and 4, by using steps 1, 2, and 3). Now, compare the prime factorization of the LCM to the prime factorization of the first denominator. Multiply the first fraction, top and bottom, by whatever prime factors are not in the prime factorization of the first denominator. So, in our example, I’d multiply top and bottom of the first fraction by 2*2 (i.e., 4, since the prime factorization of the LCM is 3 * 2 * 2, and the prime factorization of 3 is just 3). Then repeat the process with the second fraction, by comparing the LCM to the prime factorization to the second denominator.
7) After you’ve performed the prime factorizations and comparisons and multiplications in steps 2-6, you get fractions with the same denominator. Now, just add the numerators and leave the denominator alone.
8) Simplify if necessary.
If you can’t remember the easy way to do this, go back to the previous post, and see how much simpler it was to calculate 1/3 + 1 / 4, to get an idea how ridiculously overcomplicated the above method, used in the public school, is.
Now, the public school method is, absolutely, more mathematically graceful, especially when dealing with unusual fractions…it also overwhelms the students with the amount of material necessary to just add a couple of numbers together. In many classes I’ve taught, including calculus, the students expect the entire discussion to come to a complete halt the moment a fraction comes on the board, because they’re used to having to access insane amounts of material to deal with that fraction.
If a student is weak on any aspect of steps 1 through 7 (and all he has to do is miss a day of school for that), he is forever under the impression that adding fractions is a ridiculously hard concept meant only for “math whizzes.” Since those steps all rely on some underlying theory that few 8 year olds can follow, that’s most students.
The method covered in the previous essay? A little clumsy, but it’s simple, relies only upon basic multiplication, takes a few minutes at most to cover in complete detail…and allows the class to do a problem with fractions in it without everything coming to a halt. A student that missed that day can catch up in minutes, not days.
Since schools use the ridiculously overcomplicated method, most students go through the system with nothing more than fear of fractions.
This is my biggest problem with Common Core—the underlying theory may be fine, but covering all that theory when all a child needs is “here’s how to do it” means that any child that doesn’t follow the theory will simply walk away from math thinking it’s all incredibly complicated, instead of learning the basic skills that, when he is older, will make understanding the theory much easier.
Now that, I hope, I’ve taught the reader a simple way to add fractions, and reviewed the stupidly complicated way of public school, the reader will have an easier time appreciating my claims that added complexity (even with more theoretical grace) will make things much, much, worse for our children.
There’s another reason kids won’t gain skills in Common Core: lack of practice.
“6 hours a day, 6 days a week.”
–when in training, this is how much Michael Phelps would practice swimming. You’d think he’d know all there is to know, but still, he practiced like this. Practice the basic techniques of swimming, over and over and over again. Bottom line, that’s how you get good at a skill…repetitive practice. This level of repetition got him 57 gold medals in various competitions. Why don’t the schools know the obvious?
To judge by the Common Core assignments I’ve seen, simple practice of basic skills is also missing. I’m not saying our kids need to practice 6 hours a day, 6 days a week, but even 15 minutes of practice isn’t going to be part of Common Core for most skills.
This is going to be disastrous.
Next time, I’ll address some specific examples from Common Core showing what I mean.
www.professorconfess.blogspot.com
Tags: | 1,433 | 6,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-22 | latest | en | 0.953332 |
http://www.educator.com/mathematics/calculus-ii/murray/parametric-curves.php | 1,508,705,927,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825464.60/warc/CC-MAIN-20171022203758-20171022223758-00850.warc.gz | 444,385,405 | 56,734 | Start learning today, and be successful in your academic & professional career. Start Today!
• ## Related Books
### Parametric Curves
Main formula:
Slope m =
Arclength =
Hints and tips:
• You can remember the formula for the slope of the tangent line by thinking that symbolically, the dt’s cancel, leaving you with dydx.
• To find the equation of the tangent line, you also need a point. Use the given value of t into x(t) and y(t) to find it. Then you can use the point-slope formula from high school algebra (y − y0 = m(x − x0 )) to find the equation.
• Sometimes you aren’t given a value of t, but the coordinates (x, y) instead. Then you must find which value of t gives you the correct (x(t), y(t)). Make sure you check that both x and y are correct for your value of t.
• You can remember the arclength formula by recalling that it is derived from the distance formula between two points, which in turn comes from the Pythagorean Theorem.
• Don’t make the common algebraic mistake of thinking that reduces to a + b! This is extremely wrong, and your teacher will likely be merciless if you do it
• Many problems in Calculus II classes are “rigged” so that when you expand x′(t)² + y′(t , it becomes a perfect square that cancels nicely with the square root.
• Often this perfect square is achieved by making one of x′(t and y′(t be something of the form (a − b)² = a² − 2ab + b². Then the other one changes it to a² + 2ab + b², which you can then factor as (a + b.
• Another common technique in arclength problems is to make a u-substitution for whatever is under the square root sign. Then (hopefully) you can manipulate the expression outside the square root into being the du. However, you might have to do several steps of algebraic manipulation, pulling factors in or out of the square root sign, before this works.
• You may sometimes be able to use symmetry to find the arclength of part of a curve and then multiply by an appropriate factor to get the total arclength. This can be especially helpful if you just want to examine part of the curve where all the quantities involved are positive.
• When it’s feasible, check that your answer makes sense. Unlike area integrals, which can be negative if a curve goes below the x-axis, arclength should always be positive! You might also be able to check that the curve looks about as long as your answer.
### Parametric Curves
Graph f(t) = (sin(x), √x) for 0 ≤ t ≤ 2π
• Make a table
• t
0
2
π
[(3π)/2]
2π
x(t)
0
1
0
− 1
0
y(t)
0
1.25
1.77
2.17
2.51
Plot the points and graph
A parametric curve is defined by
y(t) = t2
x(t) = t + 2
What is its Cartesian equation?
• Solve for t with x
• x = t + 2
• t = x − 2
Substitute into y(t)
y(t) = t2
y = (x − 2)2
A parametric curve is defined by:
y(t) = t − 9
x(t) = √t
for t ≥ 0
What is its Cartesian equation?
• x = √t
• t = x2
Substitute into y(t)
y(t) = t − 9
y = x2 − 9
Find the Cartesian equation of the following parametric equations:
x(t) = [cos(t)/3]
y(t) = [sin(t)/4]
for t ≥ 0
• Isolate x and y
• x = [cos(t)/3]
• 3x = cos(t)
• y = [sin(t)/4]
• 4y = sin(t)
Apply Pythagorean identity
cos2(t) + sin2(t) = 1
(3x)2 + (4y)2 = 1
9x2 + 16y2 = 1
Find the Cartesian equation of the following parametric equations:
x(t) = √2 sin(t)
y(t) = 2cos(t) + 3
for t ≥ 0
• Isolate x
• x = √2 sin(t)
• [x/(√2 )] = sin(t)
• Isolate y
• y = 2cos(t) + 3
• y − 3 = 2cos(t)
• [(y − 3)/2] = cos(t)
sin2(t) + cos2(t) = 1
([x/(√2 )])2 + ([(y − 3)/2])2 = 1
[(x2)/2] + [((y − 3)2)/4] = 1
Find the Cartesian equation of the following parametric equations:
x(t) = t + 5
y(t) = t3 + 5
• Isolate t with x
• x = t + 5
• x − 5 = t
y = (x − 5)3 + 125
y = x3 − 15x2 + 75x − 125 + 125
y = x3 − 15x2 + 75x
Find the Cartesian equation of the following parametric equations:
x(t) = 1 + t
y(t) = t2 − 9
• Isolate t with x
• x = 1 + t
• − t = 1 − x
• t = x − 1
Substitute into y(t)
y = (x − 1)2 − 9
y = x2 − 2x + 1 − 9
y = x2 − 2x − 8
Find the Cartesian equation of the following parametric equations, and graph it:
x(t) = t2 − 3
y(t) = 1 − t
for t − 3
• Isolate t with x
• x = t2 − 3
• x + 3 = t2
• √{x + 3} = t
• Substitute into y(t)
• y = 1 − √{x + 3}
• Graph the equation using shifts
• y = 1 − √{x + 3}
The graph √x has been shifted up by 1 unit to y = 1, and left by 3 units. It also has been reflected.
Find the Cartesian equation of the following parametric equations, and graph it:
x(t) = arccos([t/4]) − [(π)/2]
y(t) = [t/8] − 3 for − 2π ≤ t ≤ 2π
• Isolate t with x
• x = arccos([t/4]) − [(π)/2]
• x + [(π)/2] = arccos([t/4])
• cos(x + [(π)/2]) = [t/4]
• 4cos(x + [(π)/2]) = t
• Observe half - angle trig identity
• cos(u + [(π)/2]) = − sinu
• 4cos(x + [(π)/2]) = t
• − 4sinx = t
• Substitute into y(t)
• y = [( − 4sin(x))/8] − 3
• y = − [sinx/2] − 3
• Graph using shifts, reflections, and stretching
• y = − [sinx/2] − 3
• The graph y = sinx has been shifted 3 units down to y = − 3, and reflected. It also has been compressed by a factor of 2.
• Remember to graph within the bounds − 2π ≤ t ≤ 2π.
Find the length of the arc of y = (x − 8)[3/2] from − 10 to 17
• Find y′
• y′ = [3/2](x − 8)1/2dx
• Use Arc Length Formula
• Arc Length = ∫ab√{1 + ( [dy/dx] )2} dx
• = ∫1017 √{1 + ( [3/2](x − 8)1/2 )2} d x
• = ∫1017 √{1 + ( [9/4](x − 8) )} dx
• = ∫1017√{1 + [9/4]x − 18} dx
• = ∫1017 √{[9/4]x − 17} dx
• Use u substitution
• u = [9/4]x − 17
• du = [9/4]dx
• 1017 √{[9/4]x − 17} dx = ∫1017 [4/9]√u du
• = [4/9]∫1017 √u du
• = [4/9][ [2/3]u[3/2] ]1017
• = [4/9][ [2/3]( [9/4]x − 17 )[3/2] ]1017
≈ 25.20
Find the derivative of − lncosx
• Use log properties and trig identities to alter
• − lncosx = ln( cosx ) − 1
• = ln[1/cosx]
• = lnsecx
Use chain rule
[dy/dx] = [secxtanx/secx] = tanx dx
From 1 to 3, find the length of y = − lncosx
• Find the derivative of − lncosx
• y′ = tanx
• Apply Arc Length equation
• ab√{1 + ( [dy/dx] )2} dx = ∫13√{1 + ( tanx )2} dx
• = ∫13√{1 + tan2x} dx
• Use Pythagorean identity
• 13√{1 + tan2x} dx = ∫13√{sec2x} dx
• = ∫13 secxdx
• Use trig integral identity
• secxdx = ln|secx + tanx| + C
• 13 secxdx = [ ln|secx + tanx| ]13
• = ln|sec(3) + tan(3)| − ln|sec(1) + tan(1)|
≈ − 1.08
Find the length of − 1 to 4 of the follow parametric functions:
x = [2/3]t[3/2], y = 2t + 7
• Find the deratives using product rule
• x′ = √t
• y′ = 2
• Apply the parametric Arc Length Formula
• Arc Length = ∫ab √{( [dx/dt] )2 + ( [dy/dt] )2} dt
• = ∫ − 14 √{( √t )2 + ( 2 )2} dt
• = ∫ − 14 √{t + 4} dt
• Use substitution with u = t + 4
• du = dt
• − 14 √{t + 4} dt = ∫ − 14 √u du
• = [3/2][ t + 4 ] − 14
• = [3/2]( 4 + 4 − ( − 1 + 4) )
• = [3/2]( 8 + 3 )
= [33/2]
Find the length of 0 to 3 of the follow parametric functions:
• x = etcost, y = etsint
• Find the deratives using product rule
• x′ = etcost − etsint
• y′ = etsint + etcost
• Apply the parametric Arc Length Formula
• Arc Length = ∫ab √{( [dx/dt] )2 + ( [dy/dt] )2} dt
• = ∫03 √{( etcost − etsint )2 + ( etsint + etcost )2} dt
• = ∫03 √{2e2t(cos2t + sin2t)} dt
• = ∫03 √{2e2t} dt
• = √2 ∫03 etdt
• = √2 [ et ]03
= √2 [ e3 − e0 ] = √2 [ e3 − 1 ]
Find the length of 1 to [(3π)/4] of the follow parametric functions:
• x = ln|sint|, y = t
• Find the deratives using trig integral identity
• x′ = cott
• y′ = 1
• Apply the parametric Arc Length Formula
• Arc Length = ∫ab √{( [dx/dt] )2 + ( [dy/dt] )2} dt
• = ∫1[(3π)/4] √{( cott )2 + ( 1 )2} dt
• = ∫1[(3π)/4] √{cot2t + 1} dt
• Use Pythagoream Identity
• 1[(3π)/4] √{cot2x + 1} dt = ∫1[(3π)/4] √{csc2x} dt
• = ∫1[(3π)/4] csctdt
• = [ − ln|csct + cott| ]1[(3π)/4]
≈ 1.49
Find the arc length of the polar function r = 2cosθ from 0 to 5π
• Find [dr/(dθ)]
• r′ = − 2sinθ
• Apply the polar Arc Length Formula
• Arc Length = ∫ab √{r2 + ( [dr/(dθ)] )2} dθ
• = ∫0 √{( 2cosθ )2 + ( − 2sinθ )2} dθ
• = ∫0 √{4cos2θ+ 4sin2θ} dθ
• = ∫0 √4 √{cos2θ+ sin2θ} dθ
• = 2∫0 √{cos2θ+ sin2θ} dθ
• = 2∫0 √1 dθ
• = 2[ θ]0
= 10π
What is the parametric Arc Length equation of r = cosθ?
• Convert using Polar definitions
• x = rcosθ
• = cosθ(cosθ)
• = cos2θ
• y = rsinθ
• = cosθsin θ
• Find [dx/(dθ)] and [dy/(dθ)]
• x′ = − 2sinθcosθ
• y′ = cos( 2x )
• Apply parametric Arc Length Formula (but not necessarily solve)
• Arc Length = ∫ab √{( [dx/dt] )2 + ( [dy/dt] )2} dt
• = ∫ab √{( [dx/(dθ)] )2 + ( [dy/(dθ)] )2} dθ
= ∫ab √{( − 2sinθcosθ )2 + ( cos( 2x ) )2} dθ
What is the polar Arc Length Equation of r = 2e
• Find [dr/(dθ)]
• r′ = e
• Arc Length = ∫ab √{r2 + ( [dr/(dθ)] )2} dθ
= ∫ab √{4e2 + e} dθ = ∫ab √{5e} dθ = √5 ∫ab e
What's the Arc Length for r = e from 0 to 1?
• Apply previous equation with known bounds
• √5 ∫01 edθ = [(√5 )/2][ e ]01
= [(√5 )/2](e2 − 1)
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
### Parametric Curves
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Important Equations 0:05
• Slope of Tangent Line
• Arc length
• Lecture Example 1 1:40
• Lecture Example 2 4:23
• Lecture Example 3 8:38
### Transcription: Parametric Curves
Welcome back and we are trying some more examples of arc lengths.0000
And other problems involving parametric equations.0005
We have here the length of a curve given by x(t) = cos2(t) and y(t) = sin2(t) as t goes from 0 to π/2.0010
Remember our arc length formula is x'2 + y'2, take the square root of that and integrate it.0023
Let us calculate x'.0029
X is cos2(t), so that would be 2 × cos(t) × -sin(t) using the chain rule there,0034
The - sign comes from the derivative of cosine.0044
y'(t) is 2 × sin(t) × derivative of sin which is cos(t).0049
If we square each one of those, x'(t)2 is 4cos2(t) sin2(t).0059
y'(t)2 is 4sin2(t) cos2(t).0070
If we add those up, x'2 + y'2,0082
Well those are the same thing.0091
We just get the sqrt(8sin2(t)cos2(t)).0094
The sqrt(8) is 2×sqrt(2).0102
The square root of sin2 and cos2, since sin and cos are positive when t is between 0 and π/2,0105
This is sin(t) × cos(t).0117
So, that is what we want to integrate.0122
The integral from t=0, to t=π/2.0125
I will write the 2×sqrt(2) on the outside.0130
sin(t) cos(t) dt.0138
This integral is not too bad.0143
We can use the substitution u = sin(t).0145
The reason that works so well is because du = cos(t) dt.0149
So, what we have here is the integral of u du.0156
That gives us u2/2.0160
Now to convert things back into t.0169
2×sqrt(2) × sin2(t)/2.0172
Evaluated from t=0 to t=π/2.0180
Those 2's cancel each other so we get sqrt(2) × sin(π/2)2,0188
That is just one.0195
Minus sin(0)2, which is just 0.0197
So our arc length is just sqrt(2).0200
Again there, the calculus worked nicely.0206
What we did was find x' and y' and plugged them into this Pythagorean Formula.0208
Then we integrated to get the answer.0215
There is actually another way to see how this problem works.0218
We can see through this problem geometrically.0220
If you actually try graph the path that these equations are describing,0226
Notice that cos2(t) + sin2(t) = 1.0228
So, cos2(t) = x, sin2(t) = y.0240
So this path is actually taking place on the line y+x = 1.0245
So there is that line,0255
And if you plug in t=0, then x=1 and y=0.0257
There is t=0.0264
At the point (1,0).0266
If you plug in t - π/2, then y = 1 and x=0.0269
So, there is t=π/2 at the point (0,1).0277
So, what this path is really doing is just following a straight line from (1,0) to (0,1).0286
Of course the length of that line is sqrt(2).0293
That is kind of a geometric check on the calculus we just did.0297
Let us try one more example.0000
We want to find the length of the curve given by x(t) = 7 + 2t,0002
And y(t) = et + e-t, where t goes from 0 to 1.0006
Again we want to find x' and y', square each one, add them up and take the square root.0013
x'(t) = 2, y'(t), y = et + e-t, so the derivative of et is just et.0020
The derivative of e-t is e-t × the derivative of -t, which is -1.0037
So we get -e-t.0047
So x'(t)2 + y'(t)2.0050
Well, 22 is just 4.0058
Now, if we square y'(t), then we are going to follow the formula (a + b)2,0060
Is a2 + 2ab + b2.0074
Here a is et, so (et)2 is e2t, b is -e-t,0080
So this is -2ab, well ab is et × e-t,0090
Then + b2 is e-t, so that is e-2t,0102
But look at this et × e-t, is e0 which is 1.0111
This is just -2.0116
We also have a 4 here, so what we get is e2t - 2 + 4, is just + 2.0120
+ e-2t.0131
The clever thing to do here is to write that 2 as 2et × e-t again + e-2t.0134
This is (et + e-t)2.0145
The square root of all that, x'(t)2 + y'(t)2.0155
The square root cancels that for a fixed square,0164
So we get et + e-t.0168
That is what we want to integrate to find the arc length.0173
We take the integral from t=0 to t=1.0175
We are getting those bounds from the original problem of et + e-t dt.0181
That integral is not too bad.0192
The integral of et, is just et itself.0194
The integral of e-t, is e-t/the derivative of -t, which is -1.0197
That is the same as multiplying by -1.0207
So this is et - e-t.0210
Then we want to evaluate that from t=0 to t=1.0213
That is e1 - e-1 - e0 + e0.0221
The e0's are both 1 so those cancel each other out.0232
This is e - e-1.0235
I will write this as 1/e, and that is our answer for the arc length.0240
Again, what we did there, we looked at the x and y that we were given,0246
We took the derivative of each one, squared them, added them up, took their square root.0252
And integrated to get our answer for the arc length.0257
Thanks for watching, this has been educator.com.0261
Hi this is educator.com and we are here to talk about parametric curves.0000
The idea about parametric curves is that you are given the equation x(t) and y(t) and those define how a point is moving around in the plane.0007
The x(t) gives you the x coordinate at a particular time and the y(t) gives you the y coordinate at a particular time.0019
There are basically two calculus problems associated with parametric equations.0026
One is to find the tangent line to occur at a particular point.0030
The way you are going to do that is we will fine the slope of the tangent line by looking at d(y) dt.0039
So, just looking at the derivative of the y equation, and dividing by d(x) d(t), which is the derivative of the x equation.0048
That will give us the slope of the tangent line and we will also know one point on the tangent line.0052
We can use the point slope equation to find the slope of the tangent line.0059
The other equation that you use with parametric equations a lot is0063
You find the length of a curve and essentially the formula from that comes from the Pythagorean distance formula.0075
You look at x'(t2 + y'(t2) and then you find the square root of that.0076
That is a unit of arc length representing the length travelled in a very small amount of time.0084
Then you integrate that from your starting time to your ending time and that formula represents the total length of the curve.0089
Let us try this out with some examples0097
The first examples is the equations are x(t) = t+1, y(t) = t2.0100
Maybe I will just graph a couple of points there.0109
If t = 0 then x = 1 and y = 0.0111
If t = 1 then x = 2 and y = 1.0118
If t = 2 then x = 3 and y = 4.0125
This point is travelling along a parabolic path here.0133
What we are asked to do is find the tangent line at t=1.0138
At t=1, remember, x = 2 and y = 1.0143
We are trying to find the tangent line at that point right there.0149
We need to find the slope but our slope is d(y) dt/d(x) dt.0156
Now dy dt, since y = t2 is 2t.0170
Dx dt, since x = t+1 is just 1.0174
That is 2t, and when we plug in t = 1, that gives us the slope of 2.0180
Now we have the slope and we have a point and it is just an algebra problem to find the equation of a line.0189
We use the point slope formula, y - y0, which is 1 here, is equal to the slope × x - x0 which is 2.0195
This is 2x - 4.0207
We get the equation of the tangent line, y = 2x-3.0213
To recap there, what we are trying to do is to find the tangent line to a curve that is defined by parametric equations.0220
What we did was we plugged in the time value we were given, t =1 to get a point on the tangent line.0230
Then we used our equation for the slope dy dt/dx dt.0240
We figured those out using the equations for x and y that we were given.0245
We plugged in the same value of t and we got our slope and then we had a point on our slope,0250
And we could use the old point-slope formula to find the equation for our tangent line.0259
Let us find another tangent line.0263
This time the curve is x(t) = cos(t) y(t) = sin(t).0265
You will hopefully recognize that as the equations for a circle because cos2(t) + sin2(t) = 1.0272
Those are the equations that define a point moving around in a circle.0282
We want to find the tangent line at the point, (sqrt(3/2), 1/2).0288
That is about right there on the circle0294
The difference between this one and the previous one is we have not been given a t value.0297
We have to figure out what value of t gives us the point (sqrt(3/2), 1/2).0302
What value of t if you plug it in to cos(t) sin(t) gives us the sqrt(3/2)?0308
The answer is pi/6 because the cos(pi/6) is sqrt(3/2) and the sin(pi/6) is 1/2.0318
So, we know that t is pi/6.0328
Again, to find the slope, we use dy dt/dx dt.0330
Now y = sin(t) so the derivative there is cos(t).0340
X = cos(t), the derivative of that is -sin(t).0345
Then we plug in the value t = pi/6 to get a number for the slope so the cos(pi/6) = sqrt(3/2)0354
The sin(pi/6) = 1/2, and we still have our negative there.0364
The two's cancel so we get our slope is -sqrt(3).0370
That gives us the slope and we also have a point, so I will plug those into the point slope formula.0376
y - 1/2 = -sqrt(3) x - sqrt(3/2).0380
Those values are coming from plugging the points in there.0394
We can simplify this a bit.0396
This is = sqrt(3)x.0398
The minuses cancel each other out plus sqrt(3) × sqrt(3) = 3.0400
Then we can bring this half over to the other side so we get y = -sqrt(3)x + 3/2 + 1/2, is just 2.0410
We get our tangent line to be y = -sqrt(3)x + 2.0429
The way that this problem was different from the previous one was that we were not given a value of t.0435
We had to look at the point we were given and we had to figure out what the value of t should be.0440
From then on we used the same formulas to find out the slope and the equation of the tangent line.0450
By the way, this problem you can also check the answer geometrically if you draw that tangent line.0454
It forms a 30-60-90 triangle with the x and y axis.0464
That is 30 degrees and that is 60 degrees.0470
We know what those angles are because the tangent line is perpendicular to the radius of the circle there,0476
We also know that the radius of the circle is 1 so we have a smaller 30-60-90 triangle in there.0484
With a short side of length 1, so the long side has length 2.0501
That confirms that the y intercept of this tangent line = 2.0506
That is a little check on our work using some trigonometry and no calculus at all0511
Let us try an example of arc length now.0519
We are given the curve x(t) = 6 - 2t3, y(t) = 8 + 3t2.0520
We want to find the length of that curve from t = 0 to t = 2.0530
Remember our arc length formula says you want to take the integral of sqrt of x'(t)2 + y'(t)2 dt.0536
x'(t), the derivative of x if x is 6 - 2t3,0549
You take its derivative and the 6 just goes away, so the derivative is -6t2.0556
y'(t), if y' is 8 + 3t2, again the 8 does not have any effect.0568
The 3t2 you take the derivative and you get 6t.0575
What we want then is sqrt(x')2, well that is (-6t)2, so that is 36t40584
+ y'2 is 36t2.0596
That can simplify because we can pull a 36 out and just get 6,0600
we can pull a t2 out and get t.0610
Under the radical, what we have left is just t2 + 1.0615
The arc length = integral from t=0 to t=2 of 6t × sqrt(t2+1) dt.0620
Now, this integral is not too bad because what we can do is make a little substitution.0645
u = t2 + 1.0650
Then, du = 2t, dt.0654
The reason that works so nicely is that we already have the t and the dt, so we basically have du.0658
In fact, 6t dt, is just 3 du.0666
Then, we still have the integral from t = 0 to t = 2 of now the sqrt(u) du.0674
You can think of the square root as u1/2.0683
To integrate that we get u3/2/3/2 which is the same as multiplying by 2/3.0687
Then we still have that 3 on the outside.0698
This is evaluated from t=0 to t=2.0701
I have to convert the u's back into t's0706
These 3's cancel, so we get 2 × u was (t2 + 1)3/2 evaluated from t=0 to t=2.0709
This gives us 2 now if we plug in t=2, we get 22 + 1, so that is 53/2.0727
- 02 + 1, so that is 13/2.0737
We can simplify that to a little bit.0744
53/2 is the same as 5 × 51/2.0746
13/2 is just 1.0749
We get our final answer for the arc length there.0758
What made this problem work is having this formula that came from the pythagorean theorem.0762
We just take the x(t) and the y(t) that we are given and we plug them into this formula which involves computing a couple derivatives,0768
Simplifying a little bit if we can, then we integrate and we get out answer.0775
We will try a couple more examples later.0782 | 7,598 | 21,431 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2017-43 | longest | en | 0.926459 |
http://realteachingmeansreallearning.blogspot.com/2018/12/ | 1,568,858,286,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573415.58/warc/CC-MAIN-20190919015534-20190919041534-00524.warc.gz | 153,353,706 | 19,716 | ## Friday, December 14, 2018
### Triangle in a circle
Show this first animation and ask, what do you notice? What do you wonder?
Then ask the following: If the diameter of the circle is 8 cm. What (or when) is the largest area? What are the dimensions of this triangle?
After students solve that pose, When is the area exactly half the largest? What are the dimensions of that triangle? What is the angle?
After some work time and playing around with the above problem, show them this animated version
## Wednesday, December 5, 2018
### Trig Ratios animated
Two different triangles are animated. Show your class this animation and simply ask:
• What do you notice?
• What do you wonder?
After some conversations show them this one, with the side lengths being shown and ask again:
• What do you notice?
• What do you wonder?
Lastly, show them this image and ask them "How could we analyze this?"
Some possible routes:
• Prove the sides hold true for Pythagorean theorem
• What angle(s) would be in the triangle?
• How are these 2 triangles similar? How are they different? | 256 | 1,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | longest | en | 0.922858 |
http://openwetware.org/wiki/IGEM:IMPERIAL/2006/project/Oscillator/Modelling/LV | 1,387,432,131,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345761938/warc/CC-MAIN-20131218054921-00054-ip-10-33-133-15.ec2.internal.warc.gz | 135,473,815 | 10,480 | # IGEM:IMPERIAL/2006/project/Oscillator/Modelling/LV
## Lotka-Volterra Modelling
First, we considered the Lotka-Volterra model and it's consequences before actually modelling our system because of the similarities.
• Document on how to effectively extract parameters from Michaelis Menten model, by Matthieu Bultelle
## Theoretical Analysis
Analysis of ODEs of the given system - mathematical approach to study and characterise the systems behaviours
Aim:
• To have a basic understanding of the Lotka-Volterra (LV) Equation on how it resulted in a unstable oscillation
• To study to the key parameters for their effects on the amplitude and the frequency of output oscillation
• Further modify the LV equation to model our real system
• Further study the modified model based on LV model
• Propose a set of realistic parameters which will achieve the system design aim
Progress: <showhide> 1. Analysis report of the past progress --31/07/06
• We have underestimated the complexity within the simple lotka-volterra model
• However, we did have a basic understanding of the LV ODEs, the following is the abstract of above document __HIDER__
<hide>
• limit cycle
• vector field - a powerful way to represent the system behaviour
• Control modelling - using simulink in Matlab
• use control modelling technique to model and simulate the system
• from the above model, we are able to change the parameters and initial condition to study the behaviour of the system
• we have found that the LV system is not as easy as expected, in fact, it is very sensitive to noise, any small perturbations will trigger the system into a different cycle
• It is also impossible to characterise the effect of changing paramenter on output waves' frequency and amplitude.
</hide></showhide>
• However, the result from Simulink and matlab programming is not promising, the output seems to be noisy. Simulink is very complex itself, and we fail to apply any advance control method to simplify our system, hence the method of using Simulink is discarded
2. Back to simplicity -- another new matlab programme has been done, we just used ode23 to solve LV differential equations
Objective:
• To avoid the complexity, we will only vary one parameter at a time, to find out about the resultant change on the output period and frequency
• Since as mentioned above, the system behaviour might be totally different, here we are going to study a typical set of parameters, and hopefully could use this approach to create a template to study our real system paremeters.
• This document shows graphs of magnitude of output against the changing of parameters. A typical graph: __HIDER__
<hide>
• This document is an improvement over first report. A general trend can be concluded from this documents, a few chosen graphs are shown here. __HIDER__
<hide>
• This document studies the effect by changing the initial condition, a few chosen graphs are shown here. __HIDER__
<hide>
• This document verify the local behaviours of parameters by using a smaller range. __HIDER__
<hide>
• Further verify the local effects of a single parameter while setting the rest of parameters to a different value. Presentation of results has changed to a table form format. __HIDER__
<hide>
• Once we confirmed the local behaviour and choose a set of parameters, we ran a stress analysis to study how sensitive the system is to each parameters. Example:__HIDER__
<hide>
</hide></showhide> Conclustion:
• We have done the numerical analysis with a given set of ODEs and parameters
• Now we should be able to analyse, predict and choose a set of parameters to achieve our system design aim.
• However, as the ODE system will become more and more complicated, if possible, we need a theoretical approach to understand what the system should behave before analysing them numerically.
3. Jacobian anaylsis of stationary point for stability of oscillation. click here --04/08/06
Method to determine the stability of a stationary point:
• Formulate the differential equations.
• Find out about the stationary point by setting the differential equations to 0.
• Form Jacobian matrix using partial differentiation on differential equation.
• Substitute the stationary point value into the matrix and obtain its eigen value.
• Determine whether the stationary point is stable.
<showhide> The above document shows how a matlab programme is used to generate required data. __HIDER__ <hide> </hide></showhide>
• Now we have a Matlab tool to study a given ODEs
<showhide> 4. Introduction to analysis of the 2-D ODEs. click here --14/08/06
• Step by step analysis of pure Lotka-Volterra ODEs, detail explanations of every step taken __HIDER__
<hide>
• stationary point: points when the system reach a steady state (equilibrium)
• Jacobian matrix: best linear approximation to a differentiable function near a given point, can be used to determine whether these points are stable
• Eigen value of Jacobian matrix: The sign of the real parts of eigenvalues will determine whether the given stationary point is stable.
• Hence we can use Trace and determinant of a Jacobian matrix to determine the stability of 2-D ODEs
• Vector Field Representation: A powerful numerical method to help us visualise the stability of a give point.
• Thus we have define our approach to study a give ODEs with appropriate explanations.
• An improved version, rephrased various definations, and added the reason for using eigen-values to determine the stability of a stationary point.
<showhide> 5. Study and Characterisation of a limit cycle. click here --06/09/06
• Showing a typical ODEs that will result in a nice limit cycle; this should be our aim of the modelling. There is also a short discussion on how to characterise the limit cycle. Graph __HIDER__
<hide> </hide></showhide>
6. Stability analysis of modified Lotka-Volterra systems
<showhide>
• The pure Lotka-Voleterra model is too ideal to achieve, after our careful analysis, we think that our model should be modified similar to Michaelis-Menten __HIDER__
<hide> </hide></showhide>
<showhide> -- Version 1, production of the prey is modelled by pseudo Michaelis-Menten kinetics click here --16/08/06
• After the analysis, there are two stationary points [0, 0]&[d/c, V*c/b/(k*c+d)], first being unstable, second being stable __HIDER__
<hide> </hide></showhide>
<showhide> -- Version 2, productions of the prey and predator are modelled by pseudo MM kinetics click here --21/08/06
• There are also two stationary points
• The following graphs shows case when k<m & k>m __HIDER__
<hide> </hide></showhide>
<showhide> -- Version 3, production and death of the prey is modelled by pseudo MM kinetics click here --06/09/06
-- Version 4, Combination of Version 1 2 3 click here --06/09/06
Further Study of the complicated modelling of Version 4
```Specail thanks to Matthieu Bultelle for all the mathematical analysis of this model!
```
-- Trial 1, by defining R=B*C/D to simplify stability analysis, click here for analysis and click here for appendix of full-size graphs --06/09/06
-- Trial 2, interesting cases of trial 1, click here for cases of R=1 and click here for an interesting nice limit cycle --06/09/06
-- Trial 3, by re-defining R=C/D for better system analysis click here for analysis and click here for appendix of full-size graphs --07/09/06
List of templates designed by Matthieu Bultelle:
Template for analysing dynamical system : IGEM:IMPERIAL/2006/project/modelling_template
## A powerful Java Applet
Molecular Prey-Predator-System
# Molecular Prey-Predator using JOde
a=1;a0=1;b=5;b0=0.5;c=0.01;c0=1;d=0.02;e=0.0 Instructions on using the JOde Applet
Using the applet written by: Marek Rychlik | 1,794 | 7,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2013-48 | longest | en | 0.872883 |
https://number.academy/493140 | 1,660,818,678,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573193.35/warc/CC-MAIN-20220818094131-20220818124131-00375.warc.gz | 391,978,926 | 12,282 | # Number 493140
Number 493,140 spell 🔊, write in words: four hundred and ninety-three thousand, one hundred and forty . Ordinal number 493140th is said 🔊 and write: four hundred and ninety-three thousand, one hundred and fortieth. Color #493140. The meaning of number 493140 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 493140. What is 493140 in computer science, numerology, codes and images, writing and naming in other languages
## What is 493,140 in other units
The decimal (Arabic) number 493140 converted to a Roman number is (CD)(XC)MMMCXL. Roman and decimal number conversions.
#### Weight conversion
493140 kilograms (kg) = 1087176.4 pounds (lbs)
493140 pounds (lbs) = 223686.8 kilograms (kg)
#### Length conversion
493140 kilometers (km) equals to 306423 miles (mi).
493140 miles (mi) equals to 793633 kilometers (km).
493140 meters (m) equals to 1617894 feet (ft).
493140 feet (ft) equals 150311 meters (m).
493140 centimeters (cm) equals to 194149.6 inches (in).
493140 inches (in) equals to 1252575.6 centimeters (cm).
#### Temperature conversion
493140° Fahrenheit (°F) equals to 273948.9° Celsius (°C)
493140° Celsius (°C) equals to 887684° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
493140 seconds equals to 5 days, 16 hours, 59 minutes
493140 minutes equals to 1 year, 6 days, 11 hours
### Codes and images of the number 493140
Number 493140 morse code: ....- ----. ...-- .---- ....- -----
Sign language for number 493140:
Number 493140 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 493140
### Multiplications
#### Multiplication table of 493140
493140 multiplied by two equals 986280 (493140 x 2 = 986280).
493140 multiplied by three equals 1479420 (493140 x 3 = 1479420).
493140 multiplied by four equals 1972560 (493140 x 4 = 1972560).
493140 multiplied by five equals 2465700 (493140 x 5 = 2465700).
493140 multiplied by six equals 2958840 (493140 x 6 = 2958840).
493140 multiplied by seven equals 3451980 (493140 x 7 = 3451980).
493140 multiplied by eight equals 3945120 (493140 x 8 = 3945120).
493140 multiplied by nine equals 4438260 (493140 x 9 = 4438260).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 493140
Half of 493140 is 246570 (493140 / 2 = 246570).
One third of 493140 is 164380 (493140 / 3 = 164380).
One quarter of 493140 is 123285 (493140 / 4 = 123285).
One fifth of 493140 is 98628 (493140 / 5 = 98628).
One sixth of 493140 is 82190 (493140 / 6 = 82190).
One seventh of 493140 is 70448,5714 (493140 / 7 = 70448,5714 = 70448 4/7).
One eighth of 493140 is 61642,5 (493140 / 8 = 61642,5 = 61642 1/2).
One ninth of 493140 is 54793,3333 (493140 / 9 = 54793,3333 = 54793 1/3).
show fractions by 6, 7, 8, 9 ...
### Calculator
493140
#### Is Prime?
The number 493140 is not a prime number. The closest prime numbers are 493139, 493147.
#### Factorization and factors (dividers)
The prime factors of 493140 are 2 * 2 * 3 * 5 * 8219
The factors of 493140 are
1 , 2 , 3 , 4 , 5 , 6 , 10 , 12 , 15 , 20 , 30 , 60 , 8219 , 16438 , 24657 , 32876 , 41095 , 49314 , 82190 , 98628 , 493140 show more factors ...
Total factors 24.
Sum of factors 1380960 (887820).
#### Powers
The second power of 4931402 is 243.187.059.600.
The third power of 4931403 is 119.925.266.571.144.000.
#### Roots
The square root √493140 is 702,239275.
The cube root of 3493140 is 79,005394.
#### Logarithms
The natural logarithm of No. ln 493140 = loge 493140 = 13,108548.
The logarithm to base 10 of No. log10 493140 = 5,69297.
The Napierian logarithm of No. log1/e 493140 = -13,108548.
### Trigonometric functions
The cosine of 493140 is -0,489244.
The sine of 493140 is -0,872147.
The tangent of 493140 is 1,782641.
### Properties of the number 493140
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 493140 in Computer Science
Code typeCode value
PIN 493140 It's recommendable to use 493140 as a password or PIN.
493140 Number of bytes481.6KB
CSS Color
#493140 hexadecimal to red, green and blue (RGB) (73, 49, 64)
Unix timeUnix time 493140 is equal to Tuesday Jan. 6, 1970, 4:59:00 p.m. GMT
IPv4, IPv6Number 493140 internet address in dotted format v4 0.7.134.84, v6 ::7:8654
493140 Decimal = 1111000011001010100 Binary
493140 Decimal = 221001110110 Ternary
493140 Decimal = 1703124 Octal
493140 Decimal = 78654 Hexadecimal (0x78654 hex)
493140 BASE64NDkzMTQw
493140 MD53719c5b38b16be45628ab1f4d6dce510
493140 SHA1174ab187f64cec420be7cf5221be390d5de9505e
493140 SHA224c76bfbeb8acfb5c6295c56b8075c95b13cbbb45cf95219b03d18d713
493140 SHA25632ab2067ca934923a1c7663fc14b8299d13bd3cca62aba3ceaf93cb7655d806b
More SHA codes related to the number 493140 ...
If you know something interesting about the 493140 number that you did not find on this page, do not hesitate to write us here.
## Numerology 493140
### Character frequency in number 493140
Character (importance) frequency for numerology.
Character: Frequency: 4 2 9 1 3 1 1 1 0 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 493140, the numbers 4+9+3+1+4+0 = 2+1 = 3 are added and the meaning of the number 3 is sought.
## № 493,140 in other languages
How to say or write the number four hundred and ninety-three thousand, one hundred and forty in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 493.140) cuatrocientos noventa y tres mil ciento cuarenta German: 🔊 (Anzahl 493.140) vierhundertdreiundneunzigtausendeinhundertvierzig French: 🔊 (nombre 493 140) quatre cent quatre-vingt-treize mille cent quarante Portuguese: 🔊 (número 493 140) quatrocentos e noventa e três mil, cento e quarenta Chinese: 🔊 (数 493 140) 四十九万三千一百四十 Arabian: 🔊 (عدد 493,140) أربعمائةثلاثة و تسعون ألفاً و مائةأربعون Czech: 🔊 (číslo 493 140) čtyřista devadesát tři tisíce sto čtyřicet Korean: 🔊 (번호 493,140) 사십구만 삼천백사십 Danish: 🔊 (nummer 493 140) firehundrede og treoghalvfemstusindethundrede og fyrre Dutch: 🔊 (nummer 493 140) vierhonderddrieënnegentigduizendhonderdveertig Japanese: 🔊 (数 493,140) 四十九万三千百四十 Indonesian: 🔊 (jumlah 493.140) empat ratus sembilan puluh tiga ribu seratus empat puluh Italian: 🔊 (numero 493 140) quattrocentonovantatremilacentoquaranta Norwegian: 🔊 (nummer 493 140) fire hundre og nitti-tre tusen, en hundre og førti Polish: 🔊 (liczba 493 140) czterysta dziewięćdzisiąt trzy tysiące sto czterdzieści Russian: 🔊 (номер 493 140) четыреста девяносто три тысячи сто сорок Turkish: 🔊 (numara 493,140) dörtyüzdoksanüçbinyüzkırk Thai: 🔊 (จำนวน 493 140) สี่แสนเก้าหมื่นสามพันหนึ่งร้อยสี่สิบ Ukrainian: 🔊 (номер 493 140) чотириста дев'яносто три тисячi сто сорок Vietnamese: 🔊 (con số 493.140) bốn trăm chín mươi ba nghìn một trăm bốn mươi Other languages ...
## News to email
Privacy Policy.
## Comment
If you know something interesting about the number 493140 or any natural number (positive integer) please write us here or on facebook. | 2,483 | 7,301 | {"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} | 2.796875 | 3 | CC-MAIN-2022-33 | latest | en | 0.616342 |
https://www.math.auckland.ac.nz/~meylan/code/boundary_element/beam_em.m | 1,656,708,700,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-27/segments/1656103945490.54/warc/CC-MAIN-20220701185955-20220701215955-00569.warc.gz | 926,375,523 | 1,267 | function out = beam_em(lmb,l,xp) %------------------------------------------------------------------------- % Function name: beam_em % Description: calculates 2D beam/plate dry natural modes of vibration % Note: natural modes are normalized in order to have inner product equal % to Kronecker symbol %------------------------------------------------------------------------- % Variables % lmb : mode eigenvalue % l : beam/plate half-length % xp : x coord. (calculation points) vector % out = value of the normalized eigenfunction. out = zeros(length(lmb),length(xp)); for j=1:length(lmb) for k=1:length(xp) if j-1 == 0 % 1st (rigid) mode out(j,k) = 1/sqrt(2*l); elseif j-1 == 1 % 2nd (rigid) mode out(j,k) = xp(k)/l*sqrt(3/(2*l)); elseif rem(j-1,2) == 0 % symmetric modes out(j,k) = 1/sqrt(2*l)*(cos(lmb(j)*xp(k))/cos(lmb(j)*l)+cosh(lmb(j)*xp(k))/cosh(lmb(j)*l)); else % skew-symmetric modes out(j,k) =1/sqrt(2*l)*(sin(lmb(j)*xp(k))/sin(lmb(j)*l)+sinh(lmb(j)*xp(k))/sinh(lmb(j)*l)); end end end | 299 | 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} | 2.625 | 3 | CC-MAIN-2022-27 | latest | en | 0.386992 |
http://checalc.com/solved/agitator.html | 1,485,027,348,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281202.94/warc/CC-MAIN-20170116095121-00322-ip-10-171-10-70.ec2.internal.warc.gz | 51,257,283 | 5,174 | # CheCalc
Chemical engineering calculations to assist process, plant operation and maintenance engineers.
## Agitator Power
Calculates agitator speed and power requirement for a given reactor geometry and mixture properties.
Reactor Geometry
##### Fluid Properties
Kg/m³
cP
Scale of Agitation 1 is quite mild, 3 is normal, 6 is vigorous and 10 is violent.
Scale of 1 to 2
Characteristic of application requiring minimum fluid velocities to produce a flat but moving fluid batch surface. Scale 2 agitation will blend miscible fluids with specific gravity differences less than 0.1 and viscosity ratio of 100.
Scale of 3 to 6
Characteristic of fluid velocities in most chemical process industry's agitated batches. Scale 6 agitation will blend miscible fluids with specific gravity differences less than 0.6 and viscosity ratio of 10,000.
Scale of 7 to 10
Characteristic of applications requiring high fluid velocity agitated batches. Scale 10 agitation will blend miscible fluids with specific gravity differences less than 1.0 and viscosity ratio of 100,000.
Result
Basic sizing calculations are based on square batch i.e. vessel diameter is equal to liquid level. For different geometries same process results are obtained by using appropriate number of impellers. Actual batch geometry is converted to square geometry, where Teq is the Equivalent diameter.
Batch Volume
= Π .Teq3/4
Bulk Velocity (Vc)
= Scale of Agitation * 6 ft/min
Power, P is calculated using definition of power number.
Power (P)
= Np.ρ.N3.D5 | 342 | 1,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-04 | longest | en | 0.862629 |
http://www.shmoop.com/basic-geometry/area-polygon-triangle-circle-square.html | 1,472,002,729,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982290752.48/warc/CC-MAIN-20160823195810-00123-ip-10-153-172-175.ec2.internal.warc.gz | 521,035,570 | 10,515 | © 2016 Shmoop University, Inc. All rights reserved.
# Area (Polygon, Triangle, Circle, Square)
Area is the amount of space inside a two-dimensional shape. If you think about the floor of your bedroom, the area would be the maximum amount of floor you could throw your stuff on before you couldn't see any floor remaining.
Area is always expressed as square units (units2). This is because it is two-dimensional (length and height).
You can find the area of shapes by counting the boxes inside the shapes. In these three figures, each box represents .
• Figure A takes up 25 small boxes, so it has an area of
• Figure B takes up 36 small boxes, so it has an area of
• Figure C takes up 21 full boxes and 7 half boxes, so it has an area of
## Area of Rectangle = Base x Height
Here is a .
If we break it into section 1 cm wide, it would look like this:
Each row contains 10 squares and there are 6 rows, which gives a total of 10 × 6 square cm. That's the same as multiplying the base by the height: .
## Area of Triangle = ½(Base × Height)
Here is a triangle with a base of 5 cm and a height of 6 cm.
If we place another triangle with the same height and base on top of this one, we get a .
Now, we already know how to compute the area of a rectangle (base × height). So, the area of the rectangle is
However, we only want the triangle, which is half of the rectangle, . Essentially we took ½ of the area of the whole rectangle, or ½ (base × height).
## Area of Parallelogram = Base × Height
Now let's look at a parallelogram with a base of 6 cm and a height of 3 cm.
By moving the small triangle on the left all the way to the right, this shape becomes a rectangle with a base of 6 and a height of 3 cm.
Since you already know how to find the area of a rectangle (base × height), you have all the tools you need to find the area of this parallelogram.
## Area of Trapezoid = ½(Base1 + Base2) x Height
Imagine cutting off the triangular lower left corner and fitting it onto the upper right corner like this:
Now, we just have another rectangle, but with a new base. The base of this new figure is the average of the original bases,. The area of this new figure is .Just be careful, because the base we are using is the mean of the two original bases!
## Area of Circle = πr2
Finally we will examine the beautiful circle. Here is one with a radius of 6 cm.
Here is the same circle but with lines drawn in at every cm.
First we combined the portions of square to make complete square, then we very carefully and diligently counted each and every one of those squares and found there to be approximately 113 squares. This is nearly equal to the . | 644 | 2,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2016-36 | latest | en | 0.930811 |
https://aviation.stackexchange.com/questions/tagged/airspeed?tab=Votes | 1,623,565,191,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487600396.21/warc/CC-MAIN-20210613041713-20210613071713-00178.warc.gz | 134,080,886 | 46,340 | Questions tagged [airspeed]
The speed of an aircraft relative to the air. Among the common conventions for qualifying airspeed are: indicated airspeed ("IAS"), calibrated airspeed ("CAS"), true airspeed ("TAS"), equivalent airspeed ("EAS") and density airspeed.
231 questions
Filter by
Sorted by
Tagged with
40k views
What is the slowest fixed-wing airplane?
It's easy to find information about the fastest airplanes, in different categories (e.g. X-15, SR-71, the Concorde etc), but what is the slowest one? Which powered, manned airplane is capable of ...
16k views
What happens if you exceed the maximum speed on a C172?
The airspeed indicator has a red marking that means: "never exceed this speed" ($V_{NE}$, around 160 KIAS). Why can't you exceed that speed?
5k views
Why does Airbus not display the exact airspeed on the PFD?
On every other PFD (I know), the exact airspeed is displayed as big number on the PFD. To clarify what I mean, here are two pictures: Source, added square, Boeing PFD Source, added square, D10A EFIS ...
16k views
How does a Mach Meter determine the speed of sound at a given altitude?
By my understanding, the Mach Number at a given altitude is calculated by dividing IAS by the speed of sound at that altitude. So how is this speed of sound calculated to display the Mach Number on ...
5k views
Does a sudden drop of the velocity of the headwind affect the airspeed by lowering it?
I had a discussion about air speeds with one of my instructors recently. He is claiming that there is a sudden drop in the wind velocity as you get closer to the ground (which is absolutely true, due ...
14k views
Is there any equation to bind velocity, thrust and power?
I am designing a remote controlled airship. I will tune it so that the lift given by Archimedes' Principle will exactly balance the weight of all the structure. It will be propelled by brushless ...
38k views
Why are planes slower at higher altitudes?
On many flight simulators, I have noticed that planes tends to get slower with increasing altitude. For example, I can reach 1100 knots just above the sea level in Google Earth flight simulator(F16), ...
3k views
Do Air Traffic Controllers have to remember stall speeds for different aircraft?
While dealing with Takeoffs and Landings controllers provide speeds to pilots for different phases of the flight, specially during a landing approach. How do they know which speeds are safe for all ...
5k views
How does exceeding the critical Mach number of an aircraft affect its operation?
If an aircraft exceeds its critical Mach number, does it affect the way the aircraft is handled? Are there any significant changes to performance or the safety of the aircraft when above this speed? ...
10k views
Why is there a difference between GPS Speed and Indicator speed?
I just played with flight simulator and noticed that the GPS shows different speed from the speed indicator in the cockpit. Also, GPS speed is higher than the Indicator. Any Idea why?
3k views
What is a jet (unit) shown in Windows 10 calculator? [closed]
From Windows 10's calculator: What is a "jet" (physical unit)? And why does it equal 480 knots or 888,88 km/h, if a jetliner's cruising speed is around 950 km/s? Or why does it equal 0.73 M, if a ...
10k views
Why do fighter jets land faster than Jumbo commercial aircraft?
From my own understanding, the bigger (and heavier) aircraft, the higher approach speed it needs to keep itself from stalling. According to this site, the approach speed of an A380 is 140 knots, and ...
15k views
When can spoilers be used on airliners?
On airliners, are spoilers used in any part of flight other than landing?
4k views
What is the reference used when measuring flight speed?
I'm a physicist who was amused by the argument in the comments of this question about gravity. The correct answer about the affect of the accelerating reference frame of the plane on the apparent ...
40k views
What is the reason for changing the speed reference (IAS or Mach number) with altitude?
Inspired by that question: How is the airspeed-Mach number transition handled in modern airliners? When pressure and density decrease IAS also decreases. When temperature and pressure/density ...
8k views
Would a slower speed and lower altitude reduce fatal incidents?
If it became plausible to fly commercial aircraft much lower in the future (due to developments in technology and airway regulations permitting it), say at a standard height of 5,000ft, but with a ...
5k views
Why does the speed of commercial airliners fluctuate, sometimes as high as 1,060 km/h or as low as 800 km/h?
I've noticed that (long-haul) airliners sometimes travel at as high as 1000 km/h (I believe I've even seen 1040 km/h), but usually they fly closer to 800 km/h, for most of the trip. This seems odd ...
5k views
What aircraft fly faster than 250 knots indicated airspeed under 10,000 feet for safety reasons?
14 CFR 91.117 limits aircraft to 250 knots indicated airspeed below 10,000 feet, but 91.117(d) allows aircraft to fly a higher speed if the minimum safe speed is higher. What aircraft and aircraft ...
3k views
Why did the Solar Impulse flight take so long?
The Solar Impulse just made big news for flying from Japan to Hawaii on a solar-powered plane in five days. But five days seems really long. Let's do the math on that. The distance from Japan to ...
8k views
Why do some military aircraft use variable-sweep wings?
There are a couple of American military aircraft (the retired F-14 and the B-1 come to mind immediately), that have variable swept wings. I know that they keep the wings full out (roughly ...
6k views
Is it possible to accurately measure airspeed without pitot tube?
This is a follow-up to my previous question: How does this IMU work and how to convert its output into meaningful information? for which many people asserted that I need a pitot tube and a static ...
6k views
SR-71 flew 2193 mph. How can this mathematically be Mach 3.3?
Virtually every time I read about the Lockheed SR-71 (including other threads on this forum), I see that its speed record is listed as approximately 2193 mph. The same blurb will invariably say this ...
8k views
What are CAS and EAS used for?
What are calibrated airspeed (CAS) and equivalent airspeed (EAS) used for? When would a pilot need to know these?
2k views
Why are speed restrictions imposed for flights flying below 10,000 ft?
I know that restrictions do not always apply and ATC can cancel them or pilots can request for speed restriction being canceled. But why is there such a speed restriction in first place? Are there any ...
7k views
Why is indicated airspeed rather than ground speed used during the takeoff roll?
My instructor asked me this question during my lesson and I couldn't come up with any answers. He asked why do I need to look at the indicated airspeed rather than ground speed when in takeoff roll or ...
9k views
What is the commercial passenger aircraft top speed record?
Aircraft manufacturers claim that commercial passenger aircrafts cruise at the speed of 400-500 knots. But, is the top speed of a commercial passenger aircraft recorded (while in air)? If so, what is ...
4k views
Why is a 737 Original speed-restricted below 10 kft with inoperative windscreen heating?
According to this page on the 737's operational limitations, the Original 737s (737-100/-200) are limited to 250 KIAS below 10 kilofeet if the windscreen-heating system is inoperative: Anti-Ice & ...
2k views
Is trimming for constant speed equal to trimming for constant angle of attack?
My understanding of trim: When you say you have trimmed an aircraft for a constant speed, say 100 mph, you are actually trimming the horizontal stabilizer so that there are no forces on stick (or tail ...
4k views
Why would 2 aircraft fitted with the P&W PT6 Engine have different top speed?
Why can the Quest Kodiac and Cessna Caravan not exceed 200 knots? PT6 from Pratt and Whitney in other implementations like Epic E1000 and Pilatus PC12 can cruise at over 300 knots. The Kodiac and ...
2k views
Why must Vr be set so precisely?
Why must VR (Rotation Speed) be set so precisely? I mean why can't they just say: Rotate the Aircraft at 120 to 150 knots And then, the pilot thinks: OK, I have a full plane and half full tanks,...
3k views
Is the Mach number shown by an Air Data Computer considered a “true” Mach number?
I am trying to find more information on "true" vs. "indicated" Mach number (not airspeed). I've found older fighter aircraft manuals online that have a conversion chart from one to the other, but the ...
7k views
How does an energy variometer work?
In a comment to an answer regarding instrumentation complexity, an "energy variometer" was mentioned, labeled in knots. What does this dial actually show, and how does it work? The image accompanying ...
70k views
Does a plane have brakes to stop or slow down while flying?
Does a plane have brakes to stop or slow down while flying? For example, if the pilots see a large flock of birds at some distance in front of the airplane, and they want to reduce speed suddenly. ...
7k views
What is the fastest possible transatlantic flight today?
I'm working on a novel that, as part of the backstory, requires samples of a biological agent to be flown from Incirlik Air Base to labs across Europe, and then finally from the UK to the US. In the ...
13k views
How to determine aircraft altitude and speed, as a passenger?
Is it possible for a passenger to determine the aircraft's altitude and or speed or location using any mobile apps. Not all airlines have displays with constant info about this. I had tried using ...
5k views
What will the pilot do if all the airspeed indicators fail?
Are there any alternate instruments to indicate the airspeed in case of pitot tube failures?
844 views
Is there a surface of a commercial jet exposed to air flow at speed near Mach 1?
With commercial aircraft flying slightly above Mach 0.8, is there any surface exposed to speed of air significantly larger, e.g. near Mach 1? ignoring: The engine components, The boundary layer as ...
7k views
What is the actual air speed over and under a wing due to Bernoulli's Principle?
Bernoulli's Principle states that as a fluid speed increases, its pressure decreases, and vice versa. Air flowing over the wing of an aircraft flows faster than its neighboring air flowing under the ...
2k views
Is it possible for a plane to accelerate easily from 30 to 45 mph in flight while it struggles to reach 30 mph (the engine runs at constant power)?
In two letters sent to Dr. G. A. Spratt (an aviation enthusiast), Wilbur Wright reported the progress he and his brother had made, up to September 10, 1904, regarding Flyer II, a plane powered by a 16-...
3k views
Is it quiet in the cockpit when flying faster than the speed of sound?
Is it quiet in the cockpit when flying faster than the speed of sound or is the noise from air resistance just too loud? Similar question: is it quieter at the front of an airliner flying faster ...
17k views
Are there limits to speeds at certain altitudes?
Are there speeds that cannot be exceeded at certain altitudes?
7k views
What protection does Vno provide?
VNO, the airspeed where the green and yellow arcs meet on the airspeed indicator, is the "maximum structural cruising speed" (FAR 1.2). An airplane should only be flown above VNO in calm air. ...
2k views
Why do aircraft have a crossover airspeed, and why does it increase at higher vertical load factors?
According to the NTSB accident report on the crash of USAir Flight 427, all commercial aircraft have a crossover speed (the speed at which the maximum rolling force from the aircraft’s ailerons and ...
2k views
Why is the initial short-field climb below Vₓ in a Cessna 172?
Short-field takeoff procedure for a Cessna 172R is Wing Flaps – 10°. Brakes – APPLY. Throttle – FULL OPEN. Mixture – RICH … Brakes – RELEASE. Elevator Control – SLIGHTLY TAIL LOW. ...
4k views
Is there a maximum airspeed to deploy a ram air turbine?
Is there any airspeed limit for deploying a ram air turbine? Can airspeed be determined from the rate of its propeller rotation? Could that data be used to determine altitude?
3k views
When a plane flies faster than the speed of sound, does the distance between plane and sound increase?
I had this discussion with my dad about a question he had asked. Basically he said: If a plane flies faster than the speed of sound, will he be able to overtake his own sound and hear itself? (let'...
3k views
What is the fastest recorded backward flying plane?
It is a well known fact that some planes like the Antonov An-2 can fly backwards since their stall speed is lower than the speed of the winds they can stably fly into. My question is what is the ...
6k views
How (and why) does engine thrust change with airspeed?
I'm interested in how the thrust of a turbofan engine is affected at higher airspeeds (TAS). I know (I believed) that engine thrust(at constant N1) was relatively constant like in the following graph (... | 2,993 | 13,159 | {"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} | 2.578125 | 3 | CC-MAIN-2021-25 | latest | en | 0.93634 |
https://forum.cosmoquest.org/archive/index.php/t-48823.html?s=938536d18e957fc795a56cd9d38d9fb8 | 1,566,447,862,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316718.64/warc/CC-MAIN-20190822022401-20190822044401-00332.warc.gz | 460,207,953 | 3,884 | PDA
View Full Version : Absolute maximum temperature?
Larry Jacks
2006-Nov-01, 10:30 PM
There are some interesting questions on this board. Here's one that was passed my way quite a few years ago. A good friend of mine is a high school science teacher. One day, he was discussing absolute zero, the coldest possible temperature. One of his students asked, "Is there an absolute maximum temperature?"
I gave him an answer and he passed it to his student. Before I give my answer (and it may be wrong), I'd like to hear the views of people on this board. Is there an absolute maximum temperature?
antoniseb
2006-Nov-01, 10:35 PM
In the past, I have said that the maximum temperature would be the temperature you'd get from two electrons hitting each other, each with half the total energy in the whole universe... but I realize now that there can be a higher temperature, depending on details of string theory, Planck scales, and the nature of the singularities in black holes.
korjik
2006-Nov-02, 12:02 AM
As you get to higher temperatures, the very concept of temperature starts to get fuzzy. Alot of plasmas can have 3 different temeratures, depending on what you are looking at.
Ufonaut99
2006-Nov-02, 12:55 AM
When I was at school, I remember an experiment where we did something like measure the temperature and pressure of some water at room temperature, then again after we added some ice. The idea was to extrapolate the graph to find absolute zero (I don't remember the precise details, but I do remember that my team got the most waaaay off answer in the class :doh: )
Anyway, they then went on to explain the relationship between temperature and atoms vibrating, with absolute zero being atoms being still (ie. vibration velocity = 0).
Since which time, I've always wondered of there's a maximum temperature where the atoms vibrate at c ?
RussT
2006-Nov-02, 01:28 AM
In the past, I have said that the maximum temperature would be the temperature you'd get from two electrons hitting each other, each with half the total energy in the whole universe... but I realize now that there can be a higher temperature, depending on details of string theory, Planck scales, and the nature of the singularities in black holes.
[singularities in black holes]
When a star masive enough to be a SNII goes supernova...what's the Temp?
AGN...what's the Temp?
Quasars/Blazars...what's the temp?
My suggestion would be...that whatever the Progenitor of the Massive Black Holes is...is the Highest possible Temp.
The problem with this though is...we cannot measure TeV Gamma energy correctly from any distances because it is absorbed on it's way to earth!
George
2006-Nov-02, 02:28 AM
Is there an absolute maximum temperature? Yes, and it varies as the inverse square of your distance from its source. :)
jseefcoot
2006-Nov-02, 05:42 PM
When I was at school, I remember an experiment where we did something like measure the temperature and pressure of some water at room temperature, then again after we added some ice. The idea was to extrapolate the graph to find absolute zero (I don't remember the precise details, but I do remember that my team got the most waaaay off answer in the class :doh: )
Anyway, they then went on to explain the relationship between temperature and atoms vibrating, with absolute zero being atoms being still (ie. vibration velocity = 0).
Since which time, I've always wondered of there's a maximum temperature where the atoms vibrate at c ?
I'm nowhere near so well learned as the majority of people on this board, but to me this concept has the ring of truth to it. I too learned in school that absolute zero can also be thought of as a cessation of all motion. It makes logical sense that the maximum temperature would correlate to the maximum velocity of the atoms.
But would the atoms be able to vibrate at such massive speeds? At which point might the atoms start to fall apart, or disassociate from one another, or something of the like?
antoniseb
2006-Nov-02, 06:03 PM
At which point might the atoms start to fall apart, or disassociate from one another, or something of the like?
As far as astrophysics is concerned, Atoms fall apart at relatively low temperatures. The center of our Sun is over ten million degrees which is hundreds of times hotter than what it takes to strip the electrons off of Hydrogen atoms... And the Sun isn't especially big for a star, and it isn't exploding.
Larry Jacks
2006-Nov-02, 06:19 PM
My thoughts when asked the question were based on the notion that one definition of temperature is the average kinetic energy associated with the molecules and atoms of a substance, as explained here (http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/temper.html#c1).
My first hunch was hat if temperature is a measure of kinetic energy, then since kinetic energy = 1/2 * mass * velocity^2 and c is the limit of velocity, then there was an upper limit to temperature (when the velocity of the atoms equals c). However, that fails to allow for relativistic mass increase with velocity.
Mass increase approaches infinity as velocity approaches c as explained here (http://en.wikipedia.org/wiki/Relativistic_mass). I ran a series of equations where I took the mass of a proton and kept increasing the velocity closer and closer to c and calculated the kinetic energy. As I kept adding "9s" to c = 0.999, the kinetic energy kept increasing, implying that temperature was also increasing. That being the case, I don't think there's an upper limit on temperature.
As a practical matter, there will probably be a point where the proton would break into it's fundamental quarks or something like that. I gave my results to my friend and asked him to thank his student for asking such an interesting question. You want to encourage kids to ask such questions.
Swift
2006-Nov-02, 10:11 PM | 1,330 | 5,857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-35 | latest | en | 0.956095 |
https://infinitylearn.com/questions/physics/magnetic-field-plane-electromagnetic-wave-given-by-b | 1,721,723,442,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.81/warc/CC-MAIN-20240723072353-20240723102353-00248.warc.gz | 269,841,396 | 13,708 | Questions
# The magnetic field in the plane electromagnetic wave is given byThe expression for electric field will be
## Remember concepts with our Masterclasses.
80k Users
60 mins Expert Faculty Ask Questions
a
Ez=302 sin 0.5×103x+1.5×1011tV/m
b
Ez=60 sin 0.5×103x+1.5×1011tV/m
c
Ey=302 sin 0.5×1011x+0.5×103tV/m
d
Ey=60 sin 0.5×103x+1.5×1011tV/m
Check Your Performance Today with our Free Mock Tests used by Toppers!
detailed solution
Correct option is D
Wave is propagating along positive X-axis; magnetic field is directed along Z-axis so electric field must be directed alongk∧×i∧=j∧ along Y-axis.Amplitude of electric fieldE0=B0c=2×10-7×3×108=60 V/m | 227 | 662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-30 | latest | en | 0.789969 |
https://web2.0calc.com/questions/plz-help_88718 | 1,611,381,181,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703533863.67/warc/CC-MAIN-20210123032629-20210123062629-00616.warc.gz | 629,671,210 | 5,790 | +0
# plz help
0
38
1
The graph of the rational function
$y=\frac{ax-b}{-2x+c}$
is as shown above. If $x_1=5$ and $y_1=4$, what is the value of a+b+c ?
Dec 27, 2020
#1
+114325
+1
We have a horizontal asymptote at y = 4
This means that the ratio of the coefficients on x^2 in the numerator and denominator = 4
So
a/-2 = 4 ⇒ a = -8
And since we have a vertical asymptote at x = 5 , then
-2(5) + c = 0
c = 10
And (0,0) is on the graph which means that
2(0) - b = 0
So b = 0
So
a + b + c = -8 + 0 + 10 = 2
Here's a graph : https://www.desmos.com/calculator/fnsrqu4ihg
Dec 27, 2020 | 282 | 640 | {"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} | 3.84375 | 4 | CC-MAIN-2021-04 | longest | en | 0.537018 |
https://www.jiskha.com/display.cgi?id=1267297248 | 1,516,780,255,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893530.89/warc/CC-MAIN-20180124070239-20180124090239-00159.warc.gz | 885,139,280 | 4,188 | # Math
posted by .
Jack has 99 hot dogs and 261 hot dog buns. He wants to put the same number of hot dogs and hot dog buns on each tray. What is the greatest number of trays Jack can use to accomplish this?
• Math -
Ok you know that there are only going to be able to be 99 hot dogs and 99 hot dog buns in order to make the problem work.
99 + 99 = 198
Then take 198 and divide by even numbers because you know that you will have to have 1 hot dog and 1 bun (2 together) to see how much is the most you will be able to put onto the trays.
He can use 99 trays and have 1 hot dog and 1 hot dog bun.
• Math -
Thank you for walking me through the problem. You are greatly appreciated:)
## Similar Questions
1. ### Math
Did I solve this problem right? Here is the question: Use least common multiple or greatest common factor to solve this problem. Mark has 153 hot dogs and 261 hot dog buns. He wants to put the same number of hot dogs and hot dog buns
2. ### math
Between 50 and 100 grade 7 students at school are having a bbq. Each student will receive one hot dog. Hot dogs come in packages of 12 and buns in packages of 8. How many students are in grade 7, if all the packages of hot dogs and …
3. ### math
Are my answers right please? 1. Hot dogs come 10 to a package, and hot dog buns come 8 to a package. What is the smallest number of packages of each that Ms. Evans could buy so that there are no leftover hot dogs or buns?
4. ### math
1. Hot dogs come 10 to a package, and hot dog buns come 8 to a package. What is the smallest number of packages of each that Phillip could buy so that there are no leftover hot dogs or buns?
5. ### Chemistry
For this picnic, you need to feed at least 300 people. If you want to feed everyone, but still have equal numbers of buns and hot dogs, what is the minimum number of packages of buns and hot dogs you need, respectively?
6. ### math
Can someone please help me? I don't know how to work this out, and my friends don't either. Eight hot dogs and ten buns come in separate packages. a. Is the number of hot dog packages proportional to the number of hot dogs?
7. ### Math
Hot dogs come in packages of 6. Hot dog buns come in packages of 8. Antonio will buy the same number of hot dogs as hot dog buns. How many hot dogs could he buy?
8. ### maths: Lowest common multiple
Hot dogs are sold in packages of 12 while hot dog buns are sold in packages of 8. Help me! Don't know what to do! (a) How many packs of each should you buy to have the same number of hot dogs and hot dog buns?
9. ### math
sam was buying hot dogs and hot dog buns for a backyard barbeque.hot dogs come in packs of 16 but buns come in packs of 12. how many packs of each will sam have to buy so that there are no hot dogs or buns left over?
10. ### Math
A company runs food service concessions for sporting events throughout the country. Their marketing research department chose a particular football stadium to test market a new jumbo hot dog. It was found that the demand for the new …
More Similar Questions | 765 | 3,046 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2018-05 | latest | en | 0.959569 |
http://jazzyts.com/books/analyse-de-la-nature-ou-tableau-luniverse | 1,553,263,830,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202671.79/warc/CC-MAIN-20190322135230-20190322161230-00270.warc.gz | 98,293,727 | 11,536 | # Rafinesque C.S's Analyse de la nature ou tableau l'universe PDF
By Rafinesque C.S
Best analysis books
Download e-book for kindle: Analysis and Simulation of Contact Problems by T.A. Laursen (auth.), Peter Wriggers Professor Dr., Udo
Touch mechanics used to be and is a vital department in mechanics which covers a vast box of theoretical, numerical and experimental investigations. during this rigorously edited publication the reader will receive a state of the art assessment on formula, mathematical research and numerical resolution methods of touch difficulties.
New PDF release: Transformations Through Space and Time: An Analysis of
In recent times there was a turning out to be quandary for the advance of either effective and powerful how you can deal with space-time difficulties. Such advancements can be theoretically in addition to empirically orientated. despite which of those arenas one enters. the influence is readily received that modern wO,rk on dynamic and evolutionary versions has now not proved to be as illuminating and worthwhile as first expected.
Additional info for Analyse de la nature ou tableau l'universe
Sample text
1 (n times). For each m and n we define the m x n zero matrix One sometimes denotes the matrix by Om, to indicate the size-that is, the number of rows and columns. In general, two matrices A = (ai,) and B = (bij) are said to be equal, A = B, when A and B have the same size and aij = b;; for all i and j . A 1 x n matrix A is formed of one row: A = ( a l l , . . , a,,). We call such a matrix a row vector. 48), each of the successive rows forms a row vector. We often denote a row vector by a boldface symbol: u, v , .
Or, in handwriting, by an arrow). 49) has the row vectors u l = (2, 3 , 5 ) andu2 = (1,2,3). Chapter 1 Vectors and Matrices Similarly, an m x 1 matrix A is formed of one column: We call such a matrix a column vector. For typographical reasons we sometimes denote this matrix by col ( a l l ,. . , a m l )or even by ( a l l ,. . , a m l ) ,if the context makes clear that a column vector is intended. We also denote column vectors by boldface letters: u, v, . . 49) has the column vectors vl = co1(1,4) and v2 = col(2, 3).
Then as remarked above, also det B # 0. so that B also has an inverse B - ' , and BB-' = I . We can now write BA = BAI = B A B B ~ '= B ( A B ) B ~= ' BIB-' = BB--' = I Therefore, also, BA = I . Furthermore, if AC = I , then This shows that the inverse,of A is unique. Furthermore, if CA = I , then C=CI=CAB=IB= B. r B satisfies either one of these two equations, then B must satisjj die other equation. and B = A - ' . The inverse satisfies several additional rules: Here A and D are assumed to be nonsingular n x n matrices. | 700 | 2,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-13 | latest | en | 0.894385 |
https://stackoverflow.com/questions/952914/how-do-i-make-a-flat-list-out-of-a-list-of-lists/45323085#45323085 | 1,657,111,876,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104672585.89/warc/CC-MAIN-20220706121103-20220706151103-00193.warc.gz | 587,424,117 | 93,109 | # How do I make a flat list out of a list of lists?
I want to flatten this list of lists:
``````[[1, 2, 3], [4, 5, 6], [7], [8, 9]]
``````
into:
``````[1, 2, 3, 4, 5, 6, 7, 8, 9]
``````
To flatten a list of lists `xss`:
``````flat_list = [x for xs in xss for x in xs]
``````
This is equivalent to:
``````flat_list = []
for xs in xss:
for x in xs:
flat_list.append(x)
``````
Or as a function:
``````def flatten(xss):
return [x for xs in xss for x in xs]
``````
#### Performance analysis:
To measure performance, we use the `timeit` module from the standard library:
``````\$ python -mtimeit -s't=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[x for xs in xss for x in xs]'
10000 loops, best of 3: 143 usec per loop
\$ python -mtimeit -s't=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(t, [])'
1000 loops, best of 3: 969 usec per loop
\$ python -mtimeit -s't=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,t)'
1000 loops, best of 3: 1.1 msec per loop
``````
Explanation: the methods based on `+` (including the implied use in `sum`) are, of necessity, `O(T**2)` when there are T sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have T sublists of k items each: the first k items are copied back and forth T-1 times, the second k items T-2 times, and so on; total number of copies is k times the sum of x for x from 1 to T excluded, i.e., `k * (T**2)/2`.
The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.
• I tried a test with the same data, using `itertools.chain.from_iterable` : `\$ python -mtimeit -s'from itertools import chain; l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'list(chain.from_iterable(l))'`. It runs a bit more than twice as fast as the nested list comprehension that's the fastest of the alternatives shown here. Oct 15, 2010 at 1:21
• I found the syntax hard to understand until I realized you can think of it exactly like nested for loops. for sublist in l: for item in sublist: yield item Jul 27, 2011 at 16:43
• [leaf for tree in forest for leaf in tree] might be easier to comprehend and apply. Aug 29, 2013 at 1:38
• @RobCrowell Same here. To me the list comprehension one doesn't read right, something feels off about it - I always seem to get it wrong and end up googling. To me this reads right `[leaf for leaf in tree for tree in forest]`. I wish this is how it was. I am sure I am missing something about the grammar here, and I would appreciate if anyone could point that out. Jul 12, 2021 at 17:19
• I kept looking here every time I wanted to flatten a list, but this gif is what drove it home: i.stack.imgur.com/0GoV5.gif Aug 11, 2021 at 12:04
You can use `itertools.chain()`:
``````>>> import itertools
>>> list2d = [[1,2,3], [4,5,6], [7], [8,9]]
>>> merged = list(itertools.chain(*list2d))
``````
Or you can use `itertools.chain.from_iterable()` which doesn't require unpacking the list with the `*` operator:
``````>>> import itertools
>>> list2d = [[1,2,3], [4,5,6], [7], [8,9]]
>>> merged = list(itertools.chain.from_iterable(list2d))
``````
This approach is arguably more readable than `[item for sublist in l for item in sublist]` and appears to be faster too:
``````\$ python3 -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99;import itertools' 'list(itertools.chain.from_iterable(l))'
20000 loops, best of 5: 10.8 usec per loop
\$ python3 -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in l for item in sublist]'
10000 loops, best of 5: 21.7 usec per loop
\$ python3 -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(l, [])'
1000 loops, best of 5: 258 usec per loop
\$ python3 -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99;from functools import reduce' 'reduce(lambda x,y: x+y,l)'
1000 loops, best of 5: 292 usec per loop
\$ python3 --version
Python 3.7.5rc1
``````
• The `*` is the tricky thing that makes `chain` less straightforward than the list comprehension. You have to know that chain only joins together the iterables passed as parameters, and the * causes the top-level list to be expanded into parameters, so `chain` joins together all those iterables, but doesn't descend further. I think this makes the comprehension more readable than the use of chain in this case. Sep 3, 2014 at 14:13
• @TimDierks: I'm not sure "this requires you to understand Python syntax" is an argument against using a given technique in Python. Sure, complex usage could confuse, but the "splat" operator is generally useful in many circumstances, and this isn't using it in a particularly obscure way; rejecting all language features that aren't necessarily obvious to beginning users means you're tying one hand behind your back. May as well throw out list comprehensions too while you're at it; users from other backgrounds would find a `for` loop that repeatedly `append`s more obvious. Nov 12, 2015 at 20:26
• * creates an intermediary tuple.! `from_iterable` fetch the nested lists directly from the top list. Oct 21, 2021 at 3:35
• To make this more readable, you can make a simple function: `def flatten_list(deep_list: list[list[object]]):` `return list(chain.from_iterable(deep_list))`. The type hinting improves the clarity of what's going on (modern IDEs would interpret this as returning a `list[object]` type). Oct 25, 2021 at 14:34
Note from the author: This is very inefficient. But fun, because monoids are awesome.
``````>>> xss = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
>>> sum(xss, [])
[1, 2, 3, 4, 5, 6, 7, 8, 9]
``````
`sum` sums the elements of the iterable `xss`, and uses the second argument as the initial value `[]` for the sum. (The default initial value is `0`, which is not a list.)
Because you are summing nested lists, you actually get `[1,3]+[2,4]` as a result of `sum([[1,3],[2,4]],[])`, which is equal to `[1,3,2,4]`.
Note that only works on lists of lists. For lists of lists of lists, you'll need another solution.
• that's pretty neat and clever but I wouldn't use it because it's confusing to read. Jun 15, 2010 at 18:55
• This is a Shlemiel the painter's algorithm joelonsoftware.com/articles/fog0000000319.html -- unnecessarily inefficient as well as unnecessarily ugly. Apr 25, 2012 at 18:24
• The append operation on lists forms a `Monoid`, which is one of the most convenient abstractions for thinking of a `+` operation in a general sense (not limited to numbers only). So this answer deserves a +1 from me for (correct) treatment of lists as a monoid. The performance is concerning though... Dec 3, 2014 at 10:35
• this is a very inefficient way because of the quadratic aspect of the sum. Jul 31, 2017 at 18:04
• This article explains the maths of the inefficiency mathieularose.com/how-not-to-flatten-a-list-of-lists-in-python Jan 4, 2018 at 16:46
I tested most suggested solutions with perfplot (a pet project of mine, essentially a wrapper around `timeit`), and found
``````import functools
import operator
functools.reduce(operator.iconcat, a, [])
``````
to be the fastest solution, both when many small lists and few long lists are concatenated. (`operator.iadd` is equally fast.)
A simpler and also acceptable variant is
``````out = []
for sublist in a:
out.extend(sublist)
``````
If the number of sublists is large, this performs a little worse than the above suggestion.
Code to reproduce the plot:
``````import functools
import itertools
import operator
import numpy as np
import perfplot
def forfor(a):
return [item for sublist in a for item in sublist]
def sum_brackets(a):
return sum(a, [])
def functools_reduce(a):
return functools.reduce(operator.concat, a)
def functools_reduce_iconcat(a):
return functools.reduce(operator.iconcat, a, [])
def itertools_chain(a):
return list(itertools.chain.from_iterable(a))
def numpy_flat(a):
return list(np.array(a).flat)
def numpy_concatenate(a):
return list(np.concatenate(a))
def extend(a):
out = []
for sublist in a:
out.extend(sublist)
return out
b = perfplot.bench(
setup=lambda n: [list(range(10))] * n,
# setup=lambda n: [list(range(n))] * 10,
kernels=[
forfor,
sum_brackets,
functools_reduce,
functools_reduce_iconcat,
itertools_chain,
numpy_flat,
numpy_concatenate,
extend,
],
n_range=[2 ** k for k in range(16)],
xlabel="num lists (of length 10)",
# xlabel="len lists (10 lists total)"
)
b.save("out.png")
b.show()
``````
• For huge nested lists,' list(numpy.array(a).flat)' is the fastest among all functions above.
– Sara
Jan 20, 2019 at 13:57
• Is there a way to do a 3-d perfplot? number of arrays by average size of array?
– Leo
Apr 30, 2020 at 0:31
• @Sara can you define "huge" please? Nov 14, 2020 at 6:05
• Tried `numpy_flat` on the test example from Rossetta Code (link) and got `VisibleDeprecationWarning: Creating an ndarray from ragged nested sequences (which is a list-or-tuple of lists-or-tuples-or ndarrays with different lengths or shapes) is deprecated. If you meant to do this, you must specify 'dtype=object' when creating the ndarray` Dec 5, 2020 at 11:08
• One option missed above which shows up faster for my particular case i just `items = []; for sublist in a: items.extend(sublist); return sublist` Oct 11, 2021 at 17:05
Using `functools.reduce`, which adds an accumulated list `xs` to the next list `ys`:
``````from functools import reduce
xss = [[1,2,3], [4,5,6], [7], [8,9]]
out = reduce(lambda xs, ys: xs + ys, xss)
``````
Output:
``````[1, 2, 3, 4, 5, 6, 7, 8, 9]
``````
A faster way using `operator.concat`:
``````from functools import reduce
import operator
xss = [[1,2,3], [4,5,6], [7], [8,9]]
out = reduce(operator.concat, xss)
``````
Output:
``````[1, 2, 3, 4, 5, 6, 7, 8, 9]
``````
• The `reduce(operator.concat, l)` works like a charm. Add `sorted(list(set(reduce(operator.concat, l)))` to get a `sorted` `list` of `unique` values from a list of lists. Feb 17 at 10:25
Here is a general approach that applies to numbers, strings, nested lists and mixed containers. This can flatten both simple and complicated containers (see also Demo).
Code
``````from typing import Iterable
#from collections import Iterable # < py38
def flatten(items):
"""Yield items from any nested iterable; see Reference."""
for x in items:
if isinstance(x, Iterable) and not isinstance(x, (str, bytes)):
for sub_x in flatten(x):
yield sub_x
else:
yield x
``````
Notes:
• In Python 3, `yield from flatten(x)` can replace `for sub_x in flatten(x): yield sub_x`
• In Python 3.8, abstract base classes are moved from `collection.abc` to the `typing` module.
Demo
``````simple = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
list(flatten(simple))
# [1, 2, 3, 4, 5, 6, 7, 8, 9]
complicated = [[1, [2]], (3, 4, {5, 6}, 7), 8, "9"] # numbers, strs, nested & mixed
list(flatten(complicated))
# [1, 2, 3, 4, 5, 6, 7, 8, '9']
``````
Reference
• This solution is modified from a recipe in Beazley, D. and B. Jones. Recipe 4.14, Python Cookbook 3rd Ed., O'Reilly Media Inc. Sebastopol, CA: 2013.
• Found an earlier SO post, possibly the original demonstration.
• I just wrote pretty much the same, because I didn't see your solution ... here is what I looked for "recursively flatten complete multiple lists" ... (+1) Mar 25, 2017 at 15:32
• @MartinThoma Much appreciated. FYI, if flattening nested iterables is a common practice for you, there are some third-party packages that handle this well. This may save from reinventing the wheel. I've mentioned `more_itertools` among others discussed in this post. Cheers. Mar 25, 2017 at 17:51
• Maybe `traverse` could also be a good name for this way of a tree, whereas I'd keep it less universal for this answer by sticking to nested lists.
– Wolf
Jun 15, 2017 at 10:22
• You can check `if hasattr(x, '__iter__')` instead of importing/checking against `Iterable` and that will exclude strings as well. Apr 30, 2018 at 16:46
• the above code doesnt seem to work for if one of the nested lists is having a list of strings. [1, 2, [3, 4], [4], [], 9, 9.5, 'ssssss', ['str', 'sss', 'ss'], [3, 4, 5]] output:- [1, 2, 3, 4, 4, 9, 9.5, 'ssssss', 3, 4, 5] Jun 12, 2019 at 21:35
To flatten a data-structure that is deeply nested, use `iteration_utilities.deepflatten`1:
``````>>> from iteration_utilities import deepflatten
>>> l = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
>>> list(deepflatten(l, depth=1))
[1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> l = [[1, 2, 3], [4, [5, 6]], 7, [8, 9]]
>>> list(deepflatten(l))
[1, 2, 3, 4, 5, 6, 7, 8, 9]
``````
It's a generator so you need to cast the result to a `list` or explicitly iterate over it.
To flatten only one level and if each of the items is itself iterable you can also use `iteration_utilities.flatten` which itself is just a thin wrapper around `itertools.chain.from_iterable`:
``````>>> from iteration_utilities import flatten
>>> l = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
>>> list(flatten(l))
[1, 2, 3, 4, 5, 6, 7, 8, 9]
``````
Just to add some timings (based on Nico Schlömer's answer that didn't include the function presented in this answer):
It's a log-log plot to accommodate for the huge range of values spanned. For qualitative reasoning: Lower is better.
The results show that if the iterable contains only a few inner iterables then `sum` will be fastest, however for long iterables only the `itertools.chain.from_iterable`, `iteration_utilities.deepflatten` or the nested comprehension have reasonable performance with `itertools.chain.from_iterable` being the fastest (as already noticed by Nico Schlömer).
``````from itertools import chain
from functools import reduce
from collections import Iterable # or from collections.abc import Iterable
import operator
from iteration_utilities import deepflatten
def nested_list_comprehension(lsts):
return [item for sublist in lsts for item in sublist]
def itertools_chain_from_iterable(lsts):
return list(chain.from_iterable(lsts))
def pythons_sum(lsts):
return sum(lsts, [])
return reduce(lambda x, y: x + y, lsts)
def pylangs_flatten(lsts):
return list(flatten(lsts))
def flatten(items):
"""Yield items from any nested iterable; see REF."""
for x in items:
if isinstance(x, Iterable) and not isinstance(x, (str, bytes)):
yield from flatten(x)
else:
yield x
def reduce_concat(lsts):
return reduce(operator.concat, lsts)
def iteration_utilities_deepflatten(lsts):
return list(deepflatten(lsts, depth=1))
from simple_benchmark import benchmark
b = benchmark(
pylangs_flatten, reduce_concat, iteration_utilities_deepflatten],
arguments={2**i: [[0]*5]*(2**i) for i in range(1, 13)},
argument_name='number of inner lists'
)
b.plot()
``````
1 Disclaimer: I'm the author of that library
Consider installing the `more_itertools` package.
``````> pip install more_itertools
``````
It ships with an implementation for `flatten` (source, from the itertools recipes):
``````import more_itertools
lst = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
list(more_itertools.flatten(lst))
# [1, 2, 3, 4, 5, 6, 7, 8, 9]
``````
Note: as mentioned in the docs, `flatten` requires a list of lists. See below on flattening more irregular inputs.
As of version 2.4, you can flatten more complicated, nested iterables with `more_itertools.collapse` (source, contributed by abarnet).
``````lst = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
list(more_itertools.collapse(lst))
# [1, 2, 3, 4, 5, 6, 7, 8, 9]
lst = [[1, 2, 3], [[4, 5, 6]], [[[7]]], 8, 9] # complex nesting
list(more_itertools.collapse(lst))
# [1, 2, 3, 4, 5, 6, 7, 8, 9]
``````
• If you can afford adding a package to your project - this answer is best Mar 5, 2020 at 15:53
• it fails when all elements are not list. (e.g. lst=[1, [2,3]]). of course integer is not iterable. Sep 8, 2020 at 8:32
• also, mind that list of strings will be flattened to a list of characters Oct 30, 2020 at 2:05
The following seems simplest to me:
``````>>> import numpy as np
>>> l = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
>>> print(np.concatenate(l))
[1 2 3 4 5 6 7 8 9]
``````
• OP doesn't mention they want to use numpy. Python has good ways of doing this without relying on a library Oct 3, 2021 at 1:01
The reason your function didn't work is because the extend extends an array in-place and doesn't return it. You can still return x from lambda, using something like this:
``````reduce(lambda x,y: x.extend(y) or x, l)
``````
Note: extend is more efficient than + on lists.
• `extend` is better used as `newlist = []`, `extend = newlist.extend`, `for sublist in l: extend(l)` as it avoids the (rather large) overhead of the `lambda`, the attribute lookup on `x`, and the `or`.
– agf
Sep 24, 2011 at 10:12
• for python 3 add `from functools import reduce` Jul 2, 2019 at 12:24
`matplotlib.cbook.flatten()` will work for nested lists even if they nest more deeply than the example.
``````import matplotlib
l = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
print(list(matplotlib.cbook.flatten(l)))
l2 = [[1, 2, 3], [4, 5, 6], [7], [8, [9, 10, [11, 12, [13]]]]]
print(list(matplotlib.cbook.flatten(l2)))
``````
Result:
``````[1, 2, 3, 4, 5, 6, 7, 8, 9]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
``````
This is 18x faster than underscore._.flatten:
``````Average time over 1000 trials of matplotlib.cbook.flatten: 2.55e-05 sec
Average time over 1000 trials of underscore._.flatten: 4.63e-04 sec
(time for underscore._)/(time for matplotlib.cbook) = 18.1233394636
``````
According your list `[[1, 2, 3], [4, 5, 6], [7], [8, 9]]` which is 1 list level, we can simply use `sum(list,[])` without using any libraries
``````sum([[1, 2, 3], [4, 5, 6], [7], [8, 9]],[])
# [1, 2, 3, 4, 5, 6, 7, 8, 9]
``````
To extend the advantage of this method when there is a tuple or number existing inside. Simply adding a mapping function for each element by `map` to the list
``````#For only tuple
sum(list(map(list,[[1, 2, 3], (4, 5, 6), (7,), [8, 9]])),[])
# [1, 2, 3, 4, 5, 6, 7, 8, 9]
#In general
def convert(x):
if type(x) is int or type(x) is float:
return [x]
else:
return list(x)
sum(list(map(convert,[[1, 2, 3], (4, 5, 6), 7, [8, 9]])),[])
# [1, 2, 3, 4, 5, 6, 7, 8, 9]
``````
In here, there is a clear explanation of the drawback in terms of memory for this approach. In short, it recursively creates list objects, which should be avoided :(
• This answer is already up in this question: stackoverflow.com/a/952946/14273548 Dec 9, 2021 at 9:24
• Neat! Though the other answer here, stackoverflow.com/a/952946/14273548, explains the reasons this solution should generally be avoided (it's inefficient and confusing.)
– Arel
Dec 29, 2021 at 20:29
• Will also give a TypeError if your list contains a tuple Jan 31 at 18:42
One can also use NumPy's flat:
``````import numpy as np
list(np.array(l).flat)
``````
It only works when sublists have identical dimensions.
Use two `for` in list comprehension:
``````l = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
flat_l = [e for v in l for e in v]
print(flat_l)
``````
• I just saw this type of expression in a python codebase, and it's not the most intuitive. Mar 4 at 3:38
• This is exactly the same as the accepted answer (but without any of the explanation / other useful additions), so I have downvoted. May 30 at 7:58
• i prefer concise answer. the answer mentions "list comprehension" which can be easily looked up in python documentation. Jun 6 at 20:39
You can use the `list` `extend` method. It shows to be the fastest:
``````flat_list = []
for sublist in l:
flat_list.extend(sublist)
``````
Performance:
``````import functools
import itertools
import numpy
import operator
import perfplot
def functools_reduce_iconcat(a):
return functools.reduce(operator.iconcat, a, [])
def itertools_chain(a):
return list(itertools.chain.from_iterable(a))
def numpy_flat(a):
return list(numpy.array(a).flat)
def extend(a):
n = []
list(map(n.extend, a))
return n
perfplot.show(
setup = lambda n: [list(range(10))] * n,
kernels = [
functools_reduce_iconcat, extend, itertools_chain, numpy_flat
],
n_range = [2**k for k in range(16)],
xlabel = 'num lists',
)
``````
Output:
There are several answers with the same recursive appending scheme as below, but none makes use of `try`, which makes the solution more robust and Pythonic.
``````def flatten(itr):
for x in itr:
try:
yield from flatten(x)
except TypeError:
yield x
``````
Usage: this is a generator, and you typically want to enclose it in an iterable builder like `list()` or `tuple()` or use it in a `for` loop.
• works with any kind of iterable (even future ones!)
• works with any combination and deepness of nesting
• works also if top level contains bare items
• no dependencies
• fast and efficient (you can flatten the nested iterable partially, without wasting time on the remaining part you don't need)
• versatile (you can use it to build an iterable of your choice or in a loop)
N.B.: Since all iterables are flattened, strings are decomposed into sequences of single characters. If you don't like/want such behavior, you can use the following version which filters out from flattening iterables like strings and bytes:
``````def flatten(itr):
if type(itr) in (str,bytes):
yield itr
else:
for x in itr:
try:
yield from flatten(x)
except TypeError:
yield x
``````
• why would you use a tuple? now your solution is inefficient. Jul 31, 2021 at 18:53
• And with any sequence, `sum((flatten(e) for e in itr), tuple())` is highly inefficient, Jul 31, 2021 at 18:53
• @juanpa.arrivillaga Your comment made me think about improving my answer and I think I found a better one, what do you think?
– mmj
Aug 1, 2021 at 17:37
If you are willing to give up a tiny amount of speed for a cleaner look, then you could use `numpy.concatenate().tolist()` or `numpy.concatenate().ravel().tolist()`:
``````import numpy
l = [[1, 2, 3], [4, 5, 6], [7], [8, 9]] * 99
%timeit numpy.concatenate(l).ravel().tolist()
1000 loops, best of 3: 313 µs per loop
%timeit numpy.concatenate(l).tolist()
1000 loops, best of 3: 312 µs per loop
%timeit [item for sublist in l for item in sublist]
1000 loops, best of 3: 31.5 µs per loop
``````
You can find out more here in the documentation, numpy.concatenate and numpy.ravel.
• Doesn't work for unevenly nested lists like `[1, 2, [3], [[4]], [5, [6]]]` Apr 22, 2019 at 21:39
• @juanpa.arrivillaga it's a simple and natural extension of the question, though. Answers that can handle greater depth of nesting are more likely to be useful to someone who finds this question. Aug 2, 2021 at 19:53
Note: Below applies to Python 3.3+ because it uses `yield_from`. `six` is also a third-party package, though it is stable. Alternately, you could use `sys.version`.
In the case of `obj = [[1, 2,], [3, 4], [5, 6]]`, all of the solutions here are good, including list comprehension and `itertools.chain.from_iterable`.
However, consider this slightly more complex case:
``````>>> obj = [[1, 2, 3], [4, 5], 6, 'abc', [7], [8, [9, 10]]]
``````
There are several problems here:
• One element, `6`, is just a scalar; it's not iterable, so the above routes will fail here.
• One element, `'abc'`, is technically iterable (all `str`s are). However, reading between the lines a bit, you don't want to treat it as such--you want to treat it as a single element.
• The final element, `[8, [9, 10]]` is itself a nested iterable. Basic list comprehension and `chain.from_iterable` only extract "1 level down."
You can remedy this as follows:
``````>>> from collections import Iterable
>>> from six import string_types
>>> def flatten(obj):
... for i in obj:
... if isinstance(i, Iterable) and not isinstance(i, string_types):
... yield from flatten(i)
... else:
... yield i
>>> list(flatten(obj))
[1, 2, 3, 4, 5, 6, 'abc', 7, 8, 9, 10]
``````
Here, you check that the sub-element (1) is iterable with `Iterable`, an ABC from `itertools`, but also want to ensure that (2) the element is not "string-like."
• If you are still interested in Python 2 compatibility, change `yield from` to a `for` loop, e.g. `for x in flatten(i): yield x` Jun 19, 2018 at 19:06
``````def flatten(alist):
if alist == []:
return []
elif type(alist) is not list:
return [alist]
else:
return flatten(alist[0]) + flatten(alist[1:])
``````
• Fails for python2.7 for the example nested list in the question: `[[1, 2, 3], [4, 5, 6], [7], [8, 9]]` Apr 22, 2019 at 21:34
This may not be the most efficient way, but I thought to put a one-liner (actually a two-liner). Both versions will work on arbitrary hierarchy nested lists, and exploits language features (Python 3.5) and recursion.
``````def make_list_flat (l):
flist = []
flist.extend ([l]) if (type (l) is not list) else [flist.extend (make_list_flat (e)) for e in l]
return flist
a = [[1, 2], [[[[3, 4, 5], 6]]], 7, [8, [9, [10, 11], 12, [13, 14, [15, [[16, 17], 18]]]]]]
flist = make_list_flat(a)
print (flist)
``````
The output is
``````[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
``````
This works in a depth first manner. The recursion goes down until it finds a non-list element, then extends the local variable `flist` and then rolls back it to the parent. Whenever `flist` is returned, it is extended to the parent's `flist` in the list comprehension. Therefore, at the root, a flat list is returned.
The above one creates several local lists and returns them which are used to extend the parent's list. I think the way around for this may be creating a gloabl `flist`, like below.
``````a = [[1, 2], [[[[3, 4, 5], 6]]], 7, [8, [9, [10, 11], 12, [13, 14, [15, [[16, 17], 18]]]]]]
flist = []
def make_list_flat (l):
flist.extend ([l]) if (type (l) is not list) else [make_list_flat (e) for e in l]
make_list_flat(a)
print (flist)
``````
The output is again
``````[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
``````
Although I am not sure at this time about the efficiency.
• Why extend([l]) instead of append(l)? Apr 9, 2020 at 18:31
Another unusual approach that works for hetero- and homogeneous lists of integers:
``````from typing import List
def flatten(l: list) -> List[int]:
"""Flatten an arbitrary deep nested list of lists of integers.
Examples:
>>> flatten([1, 2, [1, [10]]])
[1, 2, 1, 10]
Args:
l: Union[l, Union[int, List[int]]
Returns:
Flatted list of integer
"""
return [int(i.strip('[ ]')) for i in str(l).split(',')]
``````
• That's just a more complicated and a bit slower way of what ᴡʜᴀᴄᴋᴀᴍᴀᴅᴏᴏᴅʟᴇ3000 already posted before. I reinvented his proposal yesterday, so this approach seems quite popular these days ;) Jan 10, 2018 at 22:03
• Not quite: `wierd_list = [[1, 2, 3], [4, 5, 6], [7], [8, 9], 10]` >> `nice_list=[1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0]` Jan 11, 2018 at 8:17
• my code as one liner would be : `flat_list = [int(e.replace('[','').replace(']','')) for e in str(deep_list).split(',')]` Jan 11, 2018 at 8:32
• You are indeed right +1, ᴡʜᴀᴄᴋᴀᴍᴀᴅᴏᴏᴅʟᴇ3000's proposal won't work with multiple digit numbers, I also didn't test this before although it should be obvious. You could simplify your code and write `[int(e.strip('[ ]')) for e in str(deep_list).split(',')]`. But I'd suggest to stick with Deleet's proposal for real use cases. It doesn't contain hacky type transformations, it's faster and more versatile because it naturally also handles lists with mixed types. Jan 11, 2018 at 16:31
• Unfortunately no. But I saw this code recently here: Python Practice Book 6.1.2 Jan 15, 2018 at 8:18
I wanted a solution which can deal with multiple nesting (`[[1], [[[2]], [3]]], [1, 2, 3]` for example), but would also not be recursive (I had a big level of recursion and I got a recursion error.
This is what I came up with:
``````def _flatten(l) -> Iterator[Any]:
stack = l.copy()
while stack:
item = stack.pop()
if isinstance(item, list):
stack.extend(item)
else:
yield item
def flatten(l) -> Iterator[Any]:
return reversed(list(_flatten(l)))
``````
and tests:
``````@pytest.mark.parametrize('input_list, expected_output', [
([1, 2, 3], [1, 2, 3]),
([[1], 2, 3], [1, 2, 3]),
([[1], [2], 3], [1, 2, 3]),
([[1], [2], [3]], [1, 2, 3]),
([[1], [[2]], [3]], [1, 2, 3]),
([[1], [[[2]], [3]]], [1, 2, 3]),
])
def test_flatten(input_list, expected_output):
assert list(flatten(input_list)) == expected_output
``````
Not a one-liner, but seeing all the answers here, I guess this long list missed some pattern matching, so here it is :)
The two methods are probably not efficient, but anyway, it's easy to read (to me at least; perhaps I'm spoiled by functional programming):
``````def flat(x):
match x:
case []:
return []
case [[*sublist], *r]:
return [*sublist, *flat(r)]
``````
The second version considers lists of lists of lists... whatever the nesting:
``````def flat(x):
match x:
case []:
return []
case [[*sublist], *r]:
return [*flat(sublist), *flat(r)]
case [h, *r]:
return [h, *flat(r)]
``````
A non-recursive function to flatten lists of lists of any depth:
``````def flatten_list(list1):
out = []
inside = list1
while inside:
x = inside.pop(0)
if isinstance(x, list):
inside[0:0] = x
else:
out.append(x)
return out
l = [[[1,2],3,[4,[[5,6],7],[8]]],[9,10,11]]
flatten_list(l)
# [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
``````
You can use the following:
``````def flatlst(lista):
listaplana = []
for k in lista: listaplana = listaplana + k
return listaplana
``````
• `+` operator creates a new list each time. You'll be better off using `+=` or `.extend()` Nov 26, 2021 at 1:35
I would suggest using generators with yield statement and yield from. Here's an example:
``````from collections.abc import Iterable
def flatten(items, ignore_types=(bytes, str)):
"""
Flatten all of the nested lists to the one. Ignoring flatting of iterable types str and bytes by default.
"""
for x in items:
if isinstance(x, Iterable) and not isinstance(x, ignore_types):
yield from flatten(x)
else:
yield x
values = [7, [4, 3, 5, [7, 3], (3, 4), ('A', {'B', 'C'})]]
for v in flatten(values):
print(v)
``````
If I want to add something to the great previous answers, here is my recursive `flatten` function which can flatten not only nested lists, but also any given container or any generally any object which can throw out items. This does also work for any depth of nesting and it is a lazy iterator which yields the items as requested:
``````def flatten(iterable):
# These types won't considered a sequence or generally a container
exclude = str, bytes
for i in iterable:
try:
if isinstance(i, exclude):
raise TypeError
iter(i)
except TypeError:
yield i
else:
yield from flatten(i)
``````
This way, you can exclude types you don't want to be flattened, like `str` or what else.
The idea is if an object can pass the `iter()` it's ready to yield items. So the iterable can have even generator expressions as an item.
Someone could argue: Why did you write this that generic when the OP didn't ask for it? OK, you're right. I just felt like this might help someone (like it did for myself).
Test cases:
``````lst1 = [1, {3}, (1, 6), [[3, 8]], [[[5]]], 9, ((((2,),),),)]
lst2 = ['3', B'A', [[[(i ** 2 for i in range(3))]]], range(3)]
print(list(flatten(lst1)))
print(list(flatten(lst2)))
``````
Output:
``````[1, 3, 1, 6, 3, 8, 5, 9, 2]
['3', b'A', 0, 1, 4, 0, 1, 2]
``````
``````def flatten_array(arr):
result = []
for item in arr:
if isinstance(item, list):
for num in item:
result.append(num)
else:
result.append(item)
return result
print(flatten_array([1, 2, [3, 4, 5], 6, [7, 8], 9]))
// output: [1, 2, 3, 4, 5, 6, 7, 8, 9]
``````
Considering the list has just integers:
``````import re
l = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
list(map(int,re.sub('(\[|\])','',str(l)).split(',')))
``````
``````np.hstack(listoflist).tolist()
``````
• While this code may answer the question, providing additional context regarding why and/or how this code answers the question improves its long-term value. Consider reading How to Answer and edit your answer to improve it. Nov 6, 2020 at 19:31 | 9,996 | 32,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-27 | longest | en | 0.865893 |
https://getagripamerica.us/calculative.html | 1,521,734,464,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647892.89/warc/CC-MAIN-20180322151300-20180322171300-00772.warc.gz | 603,770,994 | 9,811 | calculative
Definitions
• Abacus for Calculations
• WordNet 3.6
• adj calculative used of persons "the most calculating and selfish men in the community"
• ***
Webster's Revised Unabridged Dictionary
• Interesting fact: The age of a saguaro cactus is calculated by its height
• a Calculative Of or pertaining to calculation; involving calculation. "Long habits of calculative dealings."
• ***
Century Dictionary and Cyclopedia
• Interesting fact: Blaise Pascal's father was a French tax collector who had trouble keeping track of his collections. So in 1642, young Pascal designed and built a mechanical adding machine to help. It was the first mechanical calculator in history.
• calculative Pertaining to calculation; involving calculation.
• ***
Chambers's Twentieth Century Dictionary
• Interesting fact: Calculating DNA length for each person, it would stretch across the diameter of the solar system. 6 000 000 000 000 basepairx 0.6 nm x 1013 cell = 3.6x1016 metres
• adj Calculative relating to calculation
• ***
Quotations
• Henri Frederic Amiel
“Common sense is the measure of the possible; it is composed of experience and prevision; it is calculation applied to life.”
• Henri Frederic Amiel
“Common sense is calculation applied to life.”
• Henry Ford
“The best we can do is size up the chances, calculate the risks involved, estimate our ability to deal with them, and then make our plans with confidence.”
“At the beginning of every act of faith, there is often a seed of fear. For great acts of faith are seldom born out of calm calculation.”
• William Wordsworth
“Give all thou canst; high Heaven rejects the lore of nicely-calculated less or more.”
• John Henry Newman
Etymology
Chambers's Twentieth Century Dictionary
L. calculāre, -ātum, to reckon by help of little stones—calculus, dim. of calx, a little stone.
Usage
In literature:
I calculate, some day, to be wuth consid'able.
"Scattergood Baines" by Clarence Budington Kelland
The word "calculate" derives from the word "calculi", meaning pebbles.
"Our Legal Heritage, 5th Ed." by S. A. Reilly
Only when calm could his mind be capable of hidden calculation.
"The Mysterious Rider" by Zane Grey
Manin had calculated correctly; on that day there was literally nothing left to eat in Venice.
"The Liberation of Italy" by Countess Evelyn Martinengo-Cesaresco
REMARKABLE CALCULATIONS REGARDING THE SUN.
"Burroughs' Encyclopaedia of Astounding Facts and Useful Information, 1889" by Barkham Burroughs
Let me plainly assure you that this is no philanthropic scheme but the result of practical calculation.
"The Story of the Foss River Ranch" by Ridgwell Cullum
But at this time, according to calculations, enough men had been landed to complete the work.
"The Boy Allies with the Victorious Fleets" by Robert L. Drake
I therefore have calculated Table II., in terms of the radius of the wheel.
"The Art of Travel" by Francis Galton
Now, as he listened to his friend's expressions of faith, so strangely jumbled with calculated purpose, he sat at the table groping helplessly.
"The Man in the Twilight" by Ridgwell Cullum
The current, as calculated by the voltameter, corresponds to this reading.
"Scientific American Supplement No. 822" by Various
This is the subject of scientific retrospect and calculation.
"The Life of Reason" by George Santayana
There is evidence of the care and calculation that one spends on a chicken-run.
"Tell England" by Ernest Raymond
It is calculated that two thirds of this elevated region are absolutely and entirely desert.
"The Seven Great Monarchies Of The Ancient Eastern World, Vol 5. (of 7): Persia" by George Rawlinson
And for only two or three days in succession, with favorable roads, thirty miles per day may be calculated on.
"Elements of Military Art and Science" by Henry Wager Halleck
Mr. Turner's calculation on file shows that that affidavit was not the basis of the calculation.
"The Works of Robert G. Ingersoll, Vol. 10 (of 12) Dresden Edition--Legal" by Robert G. Ingersoll
Any legislation is "appropriate" that is calculated to accomplish the end sought and that is not repugnant to the Constitution.
"The Works of Robert G. Ingersoll, Vol. 11 (of 12) Dresden Edition--Miscellany" by Robert G. Ingersoll
Chet had calculated to spend some time hunting, and had with him a hatchet, with which to cut firewood.
"First at the North Pole" by Edward Stratemeyer
As the stars directed their navigators, so did the planets in some degree regulate their other calculations.
"Curiosities of Medical Experience" by J. G. (John Gideon) Millingen
Arzachel of Toledo published the Toledan Tables, and his pupils made improvements in instruments and the methods of calculation.
"Astronomy" by David Todd
Work out the problem of estimating the comparative cost of canned peaches and dried peaches, when calculated to the same food value.
"Foods and Household Management" by Helen Kinne
***
In poetry:
The man of law--the herald of the cross--
The painter, skilled--he of the healing art--
The man of trade--come each with loyal heart,
Nor calculates his loss.
"England's Brave Sons" by Joseph Horatio Chant
The sober bliss of serious calculation
Shall mock the trivial joys that fancy drew,
And, oh, the rapture of a solved equation,--
One self-same answer on the lips of two!
"The Coming Era" by Oliver Wendell Holmes
Too late to ask if end was worth the means,
Too late to calculate the toppling stock:
The idiot bird leaps out and drunken leans,
The hour is crowed in lunatic thirteens.
"Doomsday" by Sylvia Plath
Yes; and would acquaint King Philip,
That he may better calculate, if really
The danger be so great as to require
Him to renew at all events the truce
So bravely broken by your body.
"Nathan The Wise - Act I" by Gotthold Ephraim Lessing
She knows how every living thing was fathered,
She calculates the climate of each star,
She counts the fish at sea, but cannot care
Why any one of them exists, fish, fire or feathered.
"Dr. Sigmund Freud Discovers The Sea Shell" by Archibald MacLeish
"Do you give thanks for this? — or that?" No, God be thanked
I am not grateful
In that cold, calculating way, with blessing ranked
As one, two, three, and four, — that would be hateful.
"Gratitude" by Henry Van Dyke
In news:
Gunman acted with "calculation and deliberation ".
WK26233 Standard Test Method for Kinematic Viscosity of Transparent and Opaque Liquids (and Calculation of Dynamic Viscosity ).
At that point, Dr Arnett and other theorists calculated the exact characteristics of the progenitor star that would account for such an explosion, and it wasn't long before observers found exactly what was needed.
Compensation for America Electric Power 's CEO increased 22 percent last year, according to Associated Press calculations of data in a regulatory filing.
The remodel may be coincidental, but looking back it can easily be perceived as calculated or elitist .
What is emissivity, and why is it important?Answer:In heat treatment, we are often concerned with calculating how fast a part will heat (Eq.
The stress test demands that banks imagine the worst possible economic news, a so-called "stress scenario," and then calculate if they've got the capital reserves to cover losses.
Frenchtown considering 'more equitable method' to calculate sewer fees.
FRENCHTOWN — Frenchtown officials are considering changing the way sewer fees charged to property owners in the municipality are calculated.
This calculator converts tax-free yields to their taxable equivalents, or vice-versa, allowing you to compare the yields.
This is an old version of the taxi fare calculator.
The valve calculates and adapts refrigerant flow to each circuit by using only one superheat sensor mounted at the evaporator outlet.
Each month, the government calculates how many people are on private and government payrolls across every sector to determine how many jobs the economy is creating or losing.
Use this calculator to find out how Metrorail fare increase s might affect you.
See the latest alerts, a construction calendar, a Metro fare calculator and more on getting around Washington.
***
In science:
In our calculation, the simple result predicted by the effective random matrix theory was obtained only after a tedious calculation showing that the coefficient C in Eq. (59) above is zero.
Towards a semiclassical justification of the `effective random matrix theory' for transport through ballistic chaotic quantum dots
We have searched for a simplification in our calculation that would allow us to cancel all contributions ∝ exp(−2τE /τD ) at an earlier point in the calculation.
Towards a semiclassical justification of the `effective random matrix theory' for transport through ballistic chaotic quantum dots
Then, means and standard deviations are calculated and compared with the observed values and the initial parameters of the local distributions of variables are corrected accordingly in an iterative cycle until the calculated values agree with the observed ones.
Statistical analysis of BATSE gamma-ray bursts: Self-similarity and the Amati-relation
Of these calculations, only the viscous calculation of Kerr & Hussain (1989) had any consistency with singular behavior.
Computational Euler History
This pseudospectral calculation was not the best that was done in preparation for the final calculations of Kerr (1993), but it can be used to illustrate the differences between this initial condition and the Melander & Hussain (1989) initial condition.
Computational Euler History
The best evidence that these negative regions do affect the calculation is that while they were initially very small, the order of 4 × 10−3 compared to the initial peak vorticity of 1, by the end of the calculation the negative peak was the same order of magnitude as the primary vortex.
Computational Euler History
Both the hybrid pseudospectral/Chebyshev code reported in Kerr (1993) and the totally pseudospectral results from around 1991 discussed here used the standard 2/3rds dealiasing, which turns a pseudospectral calculation into a true spectral calculation.
Computational Euler History
Besides the calculation discussed in Sec. 5.3 that used a fully pseudospectral calculation, among the tests done prior to publication of the results in Kerr (1993) was a fully pseudospectral calculation using nearly the same filtered, initial condition as Kerr (1993).
Computational Euler History
Therefore, their calculations will be repeated and compared in detail with a straight pseudospectral calculation using a strict 2/3rds rule and stopped early enough so that there are no questions about small-scale resolution.
Computational Euler History
Thus, a calculation of the correlations is equivalent to the calculation of the dispersionlike characteristics of the system.
Arrow of time in generalized quantum theory and its classical limit dynamics
The Wilson coefficients Vs, Vo, VA, VB, V em A, Vi are in general functions of r and have to be determined by matching amplitudes calculated in the EFT with amplitudes calculated in QCD.
Radiative transitions and the quarkonium magnetic moment
As a result, we may calculate the Poisson kernel on Ω by instead calculating the kernel on W .
Calculation and Estimation of the Poisson kernel
Suppose that U want to send a message to V, he takes the public key αnV and calculates αnU nV = (αnV )nU To decrypt the message, V take the public key αnU of U and calculates αnU nV = (αnV )nU .
CRyptography and non commutative cohomology
Comparing the results of the exact calculations (symbols) with the calculations using the approximate formula Eq. (14) (dotted curves) we have noticed that the difference is always proportional to the value of the IR laser electric field at the moment of the photoelectron emission.
Attosecond streaking experiments on atoms: quantum theory versus simple model
This is not an artefact since the exact calculations with small steps in time delay around td = 300 a.u. perfectly agree with the model calculations.
Attosecond streaking experiments on atoms: quantum theory versus simple model
*** | 2,712 | 12,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-13 | latest | en | 0.888058 |
https://www.coursehero.com/file/6859191/Lecture3-Stat104v5-1up/ | 1,513,083,193,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948516843.8/warc/CC-MAIN-20171212114902-20171212134902-00361.warc.gz | 688,223,745 | 24,888 | Lecture3_Stat104v5_1up
# Lecture3_Stat104v5_1up - Stat 104 Quantitative Methods for...
This preview shows pages 1–8. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: Stat 104: Quantitative Methods for Economists Class 3: Graphs and Summarizing Data-Measures of Center 1 Describing Data : Graphically and Numerically s Assuming we have generated a random sample of data from some population (everyone in the population is equally likely to be selected), we need methods to analyze the data, both graphically and numerically. s This section of the notes presents two graphical techniques (the dotplot and histogram ) and several summary measures ( mean , median, mode) for dealing with sample data . 2 Types of data s Data are the statistician’s raw material, the numbers we use to interpret reality. All statistical problems involve either the collection, description and analysis of data, or thinking about the collection, description and analysis of data. s There are many aspects of data. Data may be univariate (one variable per case) or multivariate (more than one variable per case). There are also different types of data: discrete and continuous. How many cats do you own ? What is your weight ? 3 Some data to play with s A young college graduate in his first teaching assignment in a local school district decided to introduce his first grade class to statistics. s As a class exercise, he brought in a scale and a tape measure and recorded the height (in inches) and weights (in pounds) of all 73 students in his two homeroom classes. 4 Here is the class data (classdata1.xls) 62.00 49.00 37.50 42.00 56.00 46.00 38.00 40.25 43.50 45.50 46.00 51.00 38.50 39.00 40.00 42.00 59.00 42.00 48.00 43.75 48.50 42.00 47.00 44.50 48.00 46.00 52.50 43.00 34.00 40.00 32.00 40.00 39.75 42.50 46.50 44.50 42.75 44.00 50.00 47.25 63.50 45.50 60.00 46.00 72.00 45.50 31.50 43.50 35.00 38.50 41.25 46.25 41.00 42.00 31.00 42.00 43.50 44.00 49.00 46.00 40.00 42.25 41.00 45.25 48.00 48.00 40.00 44.00 46.25 46.25 34.25 40.75 30.00 40.25 36.50 43.25 40.50 43.50 43.50 44.50 34.75 41.25 45.00 43.00 43.75 45.75 60.00 44.00 37.25 40.25 43.50 42.00 51.00 45.00 34.25 41.50 57.50 39.50 39.00 31.00 46.00 44.50 35.25 43.25 41.25 45.25 48.75 45.25 34.50 39.00 42.50 43.25 40.50 46.25 41.75 41.00 44.50 45.50 47.50 44.50 53.00 49.00 39.50 44.25 45.25 44.50 49.50 46.00 42.00 49.00 43.50 46.00 36.00 40.00 43.50 43.50 33.75 42.50 45.50 44.50 36.50 43.50 53.00 43.00 38.50 41.00 Without reading ahead, which column is weight and which is height ?? 5 The Dotplot s Clearly we need a graphical way to display the data so we can see what is going on. simple graphical method is called the s A simple graphical method is called the dotplot; one dot per student goes over each students reported weight and height. 6 The Dotplot . : : . ::..: : .. . .:. .::.:::::::::.::::::... : . . .: .. ....
View Full Document
{[ snackBarMessage ]}
### Page1 / 47
Lecture3_Stat104v5_1up - Stat 104 Quantitative Methods for...
This preview shows document pages 1 - 8. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 1,202 | 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} | 3.65625 | 4 | CC-MAIN-2017-51 | latest | en | 0.920641 |
https://www.physicsforums.com/threads/is-spline-interpolation-an-fir-filter.933030/ | 1,623,962,724,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487633444.37/warc/CC-MAIN-20210617192319-20210617222319-00273.warc.gz | 864,839,695 | 17,825 | # Is Spline Interpolation an FIR Filter
Mentor
Hi Guys
This question came up in the new supposed big thing in audio called MQA (Master Quality Authenticated). Here is a technical overview and a link to its patent:
https://www.soundonsound.com/techniques/mqa-time-domain-accuracy-digital-audio-quality
https://patentscope.wipo.int/search...A34392194DDAEFDDC7.wapp2nA?docId=WO2014108677
They talk about explaining it with triangular sampling and linear interpolation in approximating what was lost in the sampling. This of course is an example of a spline of order 1 - pretty trivial really. But it is claimed better results are obtained using higher order splines:
'Even better results are possible using higher-order ‘B-spline’ kernels, which allow both the position and intensity to be identified of two or more separate pulses occurring within the same sampling period!'
The MQA guys are bit coy about exactly what they do, but the conjecture is they do an an analysis of the music, decide from that what order of spline to use in sampling it down to 96k and put the best way to upsample it in metadata encoded in the bit-stream in some way.
The reason is they claim to want to reduce time smear as much as possible ie when fed a Dirac impulse it has the shortest time response.
OK so far. But in the discussion I am having, some guys claim that above order 1 spline reconstruction is not a FIR (Finite Impulse Response) filter. I know polynomial reconstruction is:
https://www.dsprelated.com/freebooks/pasp/Lagrange_Interpolation.html
I know splines are polynomials between points - but not exactly the same as Lagrange interpolation. However since the whole object of this is to keep the time response as short as possible I cant see how it can be anything but FIR.
Does anyone know the answer - is it a FIR filter? If not why so?
The person I was discussing this with posted the following links:
http://bigwww.epfl.ch/publications/unser9301.pdf
http://bigwww.epfl.ch/publications/unser9302.pdf
It seems as the order of the spline increases it approaches a Gaussian filter (an example of the central limit theorem at work?).
Thanks
Bill
Last edited:
## Answers and Replies
Baluncore
Science Advisor
This of course is an example of a spline of order 1 - pretty trivial really. But it is claimed better results are obtained using higher order splines:
'Even better results are possible using higher-order ‘B-spline’ kernels, which allow both the position and intensity to be identified of two or more separate pulses occurring within the same sampling period!'
I think they are kidding themselves. The higher the order of the spline, the higher the frequency components it will introduce, it is nothing more than a higher order polynomial. The bandwidth is limited by sample rate. There is no information on which to base the hunch that the rate of change is higher near the sample points.
Data compression may benefit from the use of higher order splines. Is that what they think they are doing.
Any order spline can always be FIR. The greater the order of spline, the greater the delay needed to perform the computation in the pipeline.
bhobba
Mentor
I think they are kidding themselves.
So do I and a number of others in the audio area are reaching the same view - but that is a different thread. Audiophiles are a funny lot (they must be because I am one ), so it will be interesting to see what eventuates.
Some are going ga ga over it - some don't like it. I have heard it with their spline up-sampling and just using an apodising filter. MQA sound's a bit thin and too clean to me - I don't mind it personally - but a big advance - not quite - and besides the apodising filter sounds to me just as good.
Just so those reading this know what an apodising filter is, when music hits a linear phase brick-wall filter you use to prevent aliasing components in digital audio you get pre and post ringing. Pre-ringing sounds unnatural - you don't hear bells ring before they are hit. It is conjectured this is why digital audio doesn't sound as good as analogue masters - and having heard those its painfully only too true. Apodising filters have no pre-ringing so supposedly sound better - post ringing occurs all the time in nature so its not a worry. I have heard the effect - its not BS - but I think the MQA guys are overall having themselves on - other ways such as the apodising filter can solve the issue. Still even better is simply not let happen in the first place by making the filter above the maximum musical frequency above which is nothing but noise - but how to do that is a whole different story - there are a number of ideas/ways I can think of - and that's just me.
Anyway I appreciate you confirming its a FIR - I thought it must be.
Thanks
Bill | 1,058 | 4,791 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.934891 |
https://www.digitalocean.com/community/tutorials/predict-function-in-r?comment=177714 | 1,685,594,699,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647614.56/warc/CC-MAIN-20230601042457-20230601072457-00394.warc.gz | 790,464,810 | 37,148 | # How To Use the predict() Function in R Programming
Published on August 3, 2022 · Updated on September 30, 2022
By Prajwal CN
### Introduction
The `predict()` function in R is used to predict the values based on the input data. All the modeling aspects in the R program will make use of the `predict()` function in their own way, but note that the functionality of the `predict()` function remains the same irrespective of the case.
In this article, you will explore how to use the `predict()` function in R.
## Prerequisites
To complete this tutorial, you will need:
## Syntax of the `predict()` function in R
The `predict()` function in R is used to predict the values based on the input data.
``````predict(object, newdata, interval)
``````
• object: The class inheriting from the linear model
• newdata: Input data to predict the values
• interval: Type of interval calculation
## An example of the predict() function
We will need data to predict the values. For the purpose of this example, we can import the built-in dataset in R - “Cars”.
``````df <- datasets::cars
``````
This will assign a data frame a collection of `speed` and distance (`dist`) values:
`````` speed dist
1 4 2
2 4 10
3 7 4
4 7 22
5 8 16
6 9 10
7 10 18
8 10 26
9 10 34
10 11 17
``````
Next, we will use `predict()` to determine future values using this data.
First, we need to compute a linear model for this data frame:
``````# Creates a linear model
my_linear_model <- lm(dist~speed, data = df)
# Prints the model results
my_linear_model
``````
Executing this code will calculate the linear model results:
``````Call:
lm(formula = dist ~ speed, data = df)
Coefficients:
(Intercept) speed
-17.579 3.932
``````
The linear model has returned the speed of the cars as per our input data behavior. Now that we have a model, we can apply `predict()`.
``````# Creating a data frame
variable_speed <- data.frame(speed = c(11,11,12,12,12,12,13,13,13,13))
# Fiting the linear model
linear_model <- lm(dist~speed, data = df)
# Predicts the future values
predict(linear_model, newdata = variable_speed)
``````
This code generates the following output:
`````` 1 2 3 4 5
25.67740 25.67740 29.60981 29.60981 29.60981
6 7 8 9 10
29.60981 33.54222 33.54222 33.54222 33.54222
``````
Well, we have successfully predicted the future distance values based on the previous data and with the help of the linear model.
Now, we have to check the “confidence” level in our predicted values to see how accurate our prediction is.
## Confidence in the Predicted Values
The confidence interval in the predict function will help us to gauge the uncertainty in the predictions.
``````# Input data
variable_speed <- data.frame(speed = c(11,11,12,12,12,12,13,13,13,13))
# Fits the model
linear_model <- lm(dist~speed, data = df)
# Predicts the values with confidence interval
predict(linear_model, newdata = variable_speed, interval = 'confidence')
``````
This code generates the following output:
`````` fit lwr upr
1 25.67740 19.96453 31.39028
2 25.67740 19.96453 31.39028
3 29.60981 24.39514 34.82448
4 29.60981 24.39514 34.82448
5 29.60981 24.39514 34.82448
6 29.60981 24.39514 34.82448
7 33.54222 28.73134 38.35310
8 33.54222 28.73134 38.35310
9 33.54222 28.73134 38.35310
10 33.54222 28.73134 38.35310
``````
You can see the confidence interval in our predicted values in the above output.
From this output, we can predict that the cars which are traveling at a speed of 11-13 mph have a likelihood to travel a distance in the range of 19.9 to 31.3 miles.
## Conclusion
The `predict()` function is used to predict the values based on the previous data behaviors and thus by fitting that data to the model.
You can also use the confidence intervals to check the accuracy of our predictions.
References
Thanks for learning with the DigitalOcean Community. Check out our offerings for compute, storage, networking, and managed databases.
Prajwal CN
author
#### Still looking for an answer?
Ask a questionSearch for more help
JournalDev
DigitalOcean Employee
January 1, 2022
What is the meaning of attribute Fit in the above result set?
- Learner
JournalDev
DigitalOcean Employee
September 11, 2021
speed prediction calculation details please for the new learners.that table shows just the numbers . Thank You
- Learner
JournalDev
DigitalOcean Employee
May 20, 2021
“The output clearly says that the cars which are traveling at a speed of 11-13 mph have chances to travel the distance in the range of 19.9 to 31.3 miles.” No, it really doesn’t.
- None ya | 1,335 | 4,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-23 | latest | en | 0.728275 |
http://www.statmodel.com/discussion/messages/11/24638.html?1508362632 | 1,550,693,609,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247496080.57/warc/CC-MAIN-20190220190527-20190220212527-00397.warc.gz | 443,888,559 | 6,056 | Predicted Probabilities
Message/Author
Jonika Hash posted on Tuesday, October 10, 2017 - 1:09 pm
Hello! I am a PhD student looking for help with predicted probabilities.
I have a path model with a latent mediator and a final dichotomous outcome (u2). I would like to calculate predicted probabilities of u2 at different levels of two predictors (x3 & x4), holding all else constant. I am using ML with Monte Carlo integration and the logit link function. The model is:
i1 | y1@1 y2@1 y3@1;
u2 ON i1 x2 x3 x4 x5 x6;
i1 ON x1 x2 x3;
x1 x2 x5;
where x6 is an interaction term = x3*x4
I think my equations are:
i1 = a1 + b11*x1 + b12*x2 + b13*x3
u2 = a2 + B21*i1 + b22*x2 + b23*x3 + b24*x4 + b25*x5 + b26*x3*x4
where
a2 = u2 threshold estimate in Mplus output
a1 = i1 intercept estimate in output
b's = regression coefficients for corresponding x's
B21 = regression coefficient for u2 ON i1
Then to calculate logits at varying levels of x3 & x4, holding all else constant at 0:
logit = -a2 + B21*a1 + B21*b13*x3 + b23*x3 + b24*x4 + b26*x3*x4
And predicted probabilities:
1/(1+e^(-1(logit)))
Wondering if I can get help if this does not look correct? Thank you.
Bengt O. Muthen posted on Tuesday, October 10, 2017 - 1:58 pm
It is tempting to think of that as the answer, but what's not taken into account is the residual variance in i which has to be "integrated out". With a logistic link and a normal i residual, this calls for numerical integration so not something done by hand. But with a probit link the expression simplifies to using a standard univariate normal distribution function using as argument the expectation that you have on the right-hand side of logit, divided by the square root of a residual variance expression. This is shown on page 310 of our book Regression and Mediation Analysis using Mplus. You can also figure it out adding residuals to both of your 2 equations where the probit residual variance is 1.
Jonika Hash posted on Tuesday, October 10, 2017 - 3:21 pm
Hi Dr. Muthen,
Thank you so much for your help! I am so glad I asked. I am not as familiar with probit regression as I am with logit, so I am hoping I understand. Does this now look correct?:
Using the probit link function and taking residuals into account, I think my equations would now be:
i1 = a1 + b11*x1 + b12*x2 + b13*x3 + i1residual
u2 = a2 + B21*i1 + b22*x2 + b23*x3 + b24*x4 + b25*x5 + b26*x3*x4 + u2residual
where i1residual = the i1 residual variance estimate given in my Mplus output
u2 residual = 1 (I think this is what you meant by the probit residual variance being 1?)
Then, I calculate my probabilities as shown in the Mplus Users guide using the formula:
P(u = 1) | x) = F(a + b*x) = F(-t + b*x)
For example, I would calculate the probability of u2 being 1 when x3 = -1 and x4 = 1 using the formula:
P(u = 1 | x3 = -1, x4 = 1 ) = F (-a2 + B21*a1 + B21*b13*1 + B21*i1residual + b23*1 + b24*-1 + b26*1*-1 + 1)
Would you help clarify if I am not understanding this? Thank you so much for your time!
Bengt O. Muthen posted on Tuesday, October 10, 2017 - 4:33 pm
No, that's not right. It's a bit much to explain so I refer to the book page I mentioned.
Geilson Lima Santana Junior posted on Tuesday, October 17, 2017 - 7:14 pm
Dear professors,
I'm trying to calculate predicted probabilities from the probit regression y ON x gender age
Y dichotomous
X continuous (number of events: 0-8)
Gender dichotomous
Age continuous
Output:
Y ON
X 0.179
SEX 0.197
AGE -0.009
Thresholds
Y\$1 1.931
P (y=1|x) = F (-1.931 + 0.179 * x + 0.197*sex 0.009*age)
1. Sex and age are control variables. Should I put their sample means on the equations?
P (y=1|x=0) = F (-1.931 + 0.179 * 0 + 0.197*0.472 0.009*39.052) = F (-2.189) = 0.014
...
P (y=1|x=8) = F (-1.931 + 0.179 * 8 + 0.197*0.472 0.009*39.052) = F (-0.757) = 0.224
2. Is this correct?
3. Is there any way to do these calculations on MPLUS or do I have to do them by hand?
3. Can I plot these predicted probabilities directly on MPLUS?
Thank you!!!
Bengt O. Muthen posted on Wednesday, October 18, 2017 - 2:37 pm
Looks ok but I wouldn't use the mean of gender (what does that mean) but instead consider one at a time.
You can let Mplus do this by using Model Constraint with Loop and Plot. For an example, see the first example of our Topic 11 short course video and handout at
http://www.statmodel.com/course_materials.shtml | 1,388 | 4,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.889151 |
https://bashtage.github.io/arch/bootstrap/generated/arch.bootstrap.StationaryBootstrap.html | 1,597,448,535,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740343.48/warc/CC-MAIN-20200814215931-20200815005931-00090.warc.gz | 223,518,329 | 7,024 | # arch.bootstrap.StationaryBootstrap¶
class arch.bootstrap.StationaryBootstrap(block_size, *args, **kwargs)[source]
Politis and Romano (1994) bootstrap with expon distributed block sizes
Parameters
• block_size (int) – Average size of block to use
• args (Union[ndarray, DataFrame, Series]) – Positional arguments to bootstrap
• kwargs (Union[RandomState, ndarray, DataFrame, Series]) – Keyword arguments to bootstrap
data
Two-element tuple with the pos_data in the first position and kw_data in the second (pos_data, kw_data)
Type
tuple
pos_data
Tuple containing the positional arguments (in the order entered)
Type
tuple
kw_data
Dictionary containing the keyword arguments
Type
dict
Notes
Supports numpy arrays and pandas Series and DataFrames. Data returned has the same type as the input date.
Data entered using keyword arguments is directly accessibly as an attribute.
To ensure a reproducible bootstrap, you must set the random_state attribute after the bootstrap has been created. See the example below. Note that random_state is a reserved keyword and any variable passed using this keyword must be an instance of RandomState.
arch.bootstrap.optimal_block_length
Optimal block length estimation
arch.bootstrap.CircularBlockBootstrap
Circular (wrap-around) bootstrap
Examples
Data can be accessed in a number of ways. Positional data is retained in the same order as it was entered when the bootstrap was initialized. Keyword data is available both as an attribute or using a dictionary syntax on kw_data.
>>> from arch.bootstrap import StationaryBootstrap
>>> from numpy.random import standard_normal
>>> y = standard_normal((500, 1))
>>> x = standard_normal((500,2))
>>> z = standard_normal(500)
>>> bs = StationaryBootstrap(12, x, y=y, z=z)
>>> for data in bs.bootstrap(100):
... bs_x = data[0][0]
... bs_y = data[1]['y']
... bs_z = bs.z
Set the random_state if reproducibility is required
>>> from numpy.random import RandomState
>>> rs = RandomState(1234)
>>> bs = StationaryBootstrap(12, x, y=y, z=z, random_state=rs)
Methods
apply(func[, reps, extra_kwargs]) Applies a function to bootstrap replicated data bootstrap(reps) Iterator for use when bootstrapping clone(*args, **kwargs) Clones the bootstrap using different data. conf_int(func[, reps, method, size, tail, …]) type func Callable[…, Union[ndarray, DataFrame, Series]] cov(func[, reps, recenter, extra_kwargs]) Compute parameter covariance using bootstrap Gets the state of the bootstrap’s random number generator reset([use_seed]) Resets the bootstrap to either its initial state or the last seed. seed(value) Seeds the bootstrap’s random number generator set_state(state) Sets the state of the bootstrap’s random number generator Update indices for the next iteration of the bootstrap. var(func[, reps, recenter, extra_kwargs]) Compute parameter variance using bootstrap
Properties
index The current index of the bootstrap random_state Set or get the instance random state | 676 | 2,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-34 | latest | en | 0.696993 |
https://www.jiskha.com/questions/615913/at-noon-ship-a-is-40-nautical-miles-due-west-of-ship-b-ship-a-is-sailing-west-at-23 | 1,606,455,381,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141189141.23/warc/CC-MAIN-20201127044624-20201127074624-00038.warc.gz | 733,711,391 | 4,616 | # CALCULUS
At noon, ship A is 40 nautical miles due west of ship B. Ship A is sailing west at 23 knots and ship B is sailing north at 19 knots. How fast (in knots) is the distance between the ships changing at 5 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)
1. 👍 0
2. 👎 0
3. 👁 1,031
1. After t hours, distance d is
d^2 = (40+23t)^2 + (19t)^2
2d dd/dt = 2(40+23t)(23) + 2(19t)(19)
When t=5, d^2 = 155^2 + 95^2, d=181.8
2(181.8) dd/dt = 2(155)(23) + 2(95)(19)
363.6 dd/dt = 10740
dd/dt = 29.5 knots
check my math . . .
1. 👍 0
2. 👎 0
1. 👍 1
2. 👎 0
## Similar Questions
1. ### math
At noon, ship A is 40 nautical miles due west of ship B. Ship A is sailing west at 18 knots and ship B is sailing north at 23 knots. How fast (in knots) is the distance between the ships changing at 4 PM? (Note: 1 knot is a speed
2. ### Math
At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 16 knots and ship B is sailing north at 22 knots. How fast (in knots) is the distance between the ships changing at 7 PM? (Note: 1 knot is a speed
3. ### Differential Calculus
At 9am ship B is 65 miles due east of another ship A. Ship B is then sailing due west at 10mi/h and A is sailing due south at 15 mi/hr if they continue in their respective course when will they be nearest to one another? and how
4. ### precalulus
a ship travels 60 miles due east, then adjusts its course northward. after travelling 80 miles in the new direction, the ship is 139 miles from its point of departure. calculate the ship's bearing.
1. ### calculus
At noon, ship A is 110 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM?
2. ### Calculus
At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?
3. ### CALCULUS
At noon, ship A is 130 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?
4. ### Calculus Max/Min Prob
At 7:00 am, one ship was 60 miles due east from a second ship.If the first ship sailed west at 20 mph, and the second ship sailed southeast at 30 mph, at what time are they closest together?
1. ### Math (Trig)
A ship leaves port at noon and has a bearing of S 25° W. The ship sails at 15 knots. How many nautical miles south and how many nautical miles west does the ship travel by 6:00 P.M.? (Round your answers to two decimal places.)
2. ### math
At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 21 knots and ship B is sailing north at 15 knots. How fast (in knots) is the distance between the ships changing at 4 PM? (Note: 1 knot is a speed
3. ### Calculus
At noon, ship A is 180 km west of ship B. Ship A is sailing east at 40 km/h and ship B is sailing north at 30 km/h. How fast is the distance between the ships changing at 4:00 PM?
4. ### calculus
(1 pt) At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 15 knots and ship B is sailing north at 21 knots. How fast (in knots) is the distance between the ships changing at 4 PM? (Note: 1 knot is a | 1,021 | 3,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.953643 |
http://www.buenastareas.com/ensayos/Macro-Matrices/2227154.html | 1,513,563,673,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948599549.81/warc/CC-MAIN-20171218005540-20171218031540-00293.warc.gz | 334,604,487 | 14,312 | Macro matrices
Solo disponible en BuenasTareas
• Páginas : 2 (426 palabras )
• Descarga(s) : 0
• Publicado : 24 de mayo de 2011
Vista previa del texto
[pic]
Sub Macro1()
'
' Macro1 Macro
'
' Acceso directo: CTRL+h
'
' Macro que escribe la palabra camión usando una matriz de orden 5
Columns("A:AZ").Select
Selection.ColumnWidth = 3Range("A1:AG5").Select
With Selection.Interior
.Pattern = xlSolid
.PatternColorIndex = xlAutomatic
.ThemeColor = xlThemeColorDark2
End With
Dim mat(5, 5) As String
Dim n, k As Integer DONDE K ES CONTADOR DE LAS COLUMNAS, ESTA CORRIENDO MAS RAPIDO EL INDICE DE LACOLUMNA
iniciar mat: letra_c mat: mostrar 1, 1, mat
iniciar mat: letra_a mat: mostrar 7, 1, mat
iniciar mat: letra_m mat: mostrar 13, 1, mat
iniciar mat: letra_i mat: mostrar 18, 1,mat
iniciar mat: letra_o mat: mostrar 23, 1, mat
iniciar mat: letra_N mat: mostrar 29, 1, mat
MsgBox " CAMION"
End Sub
Sub iniciar(m() As String) DONDE M ES LA MATRIZ
Dim n, k AsInteger
For n = 1 To 5
For k = 1 To 5
m(n, k) = " "
Next k
Next n
End Sub
Sub letra_c(m() As String)
Dim n, k As Integer
For n = 1 To 5
m(n, 1) = "#":
Next n
For k = 1 To 5m(1, k) = "#": m(5, k) = "#"
Next k
End Sub
Sub letra_a(m() As String)
Dim n, k As Integer
For n = 1 To 5
m(n, 1) = "#": m(n, 5) = "#"
Next n
For k = 1 To 5
m(1, k) ="#": m(3, k) = "#"
Next k
End Sub
Sub letra_m(m() As String)
Dim n, k As Integer
For n = 1 To 3
m(n, n) = "#"
Next n
k = 1
For n = 5 To 3 Step -1
m(k, n) = "#": k = k + 1
Next n
Forn = 1 To 5
m(n, 1) = "#": m(n, 5) = "#"
Next n
End Sub
Sub letra_o(m() As String)
Dim n, k As Integer
For n = 1 To 5
m(n, 1) = "#": m(n, 5) = "#"
Next n
For k = 1 To 5m(1, k) = "#": m(5, k) = "#"
Next k
End Sub
Sub letra_i(m() As String)
Dim n, k As Integer
For n = 1 To 5
m(n, 3) = "#":
Next n
End Sub
Sub letra_N(m() As String)
Dim n, k As Integer... | 741 | 1,832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-51 | latest | en | 0.241336 |
https://www.in2013dollars.com/1940-AUD-in-2019?amount=100 | 1,590,873,494,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347410352.47/warc/CC-MAIN-20200530200643-20200530230643-00338.warc.gz | 757,122,364 | 23,008 | \$100 in 1940 is worth \$4,249.61 in 2019
\$
👉 You may be interested in \$100 in 1940 → 2020
Value of \$100 from 1940 to 2019
According to the Bureau of Statistics consumer price index, prices in 2019 are 4,149.61% higher than average prices since 1940. The Australian dollar experienced an average inflation rate of 4.86% per year during this period, causing the real value of a dollar to decrease.
In other words, \$100 in 1940 is equivalent in purchasing power to about \$4,249.61 in 2019, a difference of \$4,149.61 over 79 years.
The 1940 inflation rate was 3.85%. The inflation rate in 2019 was 1.90%. The 2019 inflation rate is higher compared to the average inflation rate of 1.90% per year between 2019 and 2020.
Cumulative price change 4,149.61% Average inflation rate 4.86% Converted amount (\$100 base) \$4,249.61 Price difference (\$100 base) \$4,149.61 CPI in 1940 2.700 CPI in 2019 114.739 Inflation in 1940 3.85% Inflation in 2019 1.90%
AUD Inflation since 1922
Annual Rate, the Bureau of Statistics CPI
Buying power of \$100 in 1940
This chart shows a calculation of buying power equivalence for \$100 in 1940 (price index tracking began in 1922).
For example, if you started with \$100, you would need to end with \$4,249.61 in order to "adjust" for inflation (sometimes refered to as "beating inflation").
When \$100 is equivalent to \$4,249.61 over time, that means that the "real value" of a single Australian dollar decreases over time. In other words, a dollar will pay for fewer items at the store.
This effect explains how inflation erodes the value of a dollar over time. By calculating the value in 1940 dollars, the chart below shows how \$100 is worth less over 79 years.
According to the Bureau of Statistics, each of these AUD amounts below is equal in terms of what it could buy at the time:
Dollar inflation: 1940-2019
Year Dollar Value Inflation Rate
1940 \$100.00 3.85%
1941 \$103.70 3.70%
1942 \$114.81 10.71%
1943 \$118.52 3.23%
1944 \$118.52 0.00%
1945 \$118.52 0.00%
1946 \$118.52 0.00%
1947 \$125.93 6.25%
1948 \$137.04 8.82%
1949 \$148.15 8.11%
1950 \$162.96 10.00%
1951 \$192.59 18.18%
1952 \$225.93 17.31%
1953 \$237.04 4.92%
1954 \$240.74 1.56%
1955 \$244.44 1.54%
1956 \$259.26 6.06%
1957 \$266.67 2.86%
1958 \$266.67 0.00%
1959 \$274.07 2.78%
1960 \$285.19 4.05%
1961 \$288.89 1.30%
1962 \$288.89 0.00%
1963 \$292.59 1.28%
1964 \$300.00 2.53%
1965 \$311.11 3.70%
1966 \$318.52 2.38%
1967 \$329.63 3.49%
1968 \$340.74 3.37%
1969 \$351.85 3.26%
1970 \$362.96 3.16%
1971 \$385.19 6.12%
1972 \$407.41 5.77%
1973 \$444.44 9.09%
1974 \$514.81 15.83%
1975 \$592.59 15.11%
1976 \$670.37 13.13%
1977 \$751.85 12.15%
1978 \$811.11 7.88%
1979 \$885.19 9.13%
1980 \$977.78 10.46%
1981 \$1,070.37 9.47%
1982 \$1,188.89 11.07%
1983 \$1,311.11 10.28%
1984 \$1,362.96 3.95%
1985 \$1,451.85 6.52%
1986 \$1,585.19 9.18%
1987 \$1,718.52 8.41%
1988 \$1,844.44 7.33%
1989 \$1,981.48 7.43%
1990 \$2,129.63 7.48%
1991 \$2,196.30 3.13%
1992 \$2,218.52 1.01%
1993 \$2,255.56 1.67%
1994 \$2,300.00 1.97%
1995 \$2,407.41 4.67%
1996 \$2,470.37 2.62%
1997 \$2,477.78 0.30%
1998 \$2,496.30 0.75%
1999 \$2,533.33 1.48%
2000 \$2,648.15 4.53%
2001 \$2,762.96 4.34%
2002 \$2,848.15 3.08%
2003 \$2,925.93 2.73%
2004 \$2,992.59 2.28%
2005 \$3,074.07 2.72%
2006 \$3,181.48 3.49%
2007 \$3,255.56 2.33%
2008 \$3,400.00 4.44%
2009 \$3,459.26 1.74%
2010 \$3,559.26 2.89%
2011 \$3,677.78 3.33%
2012 \$3,740.74 1.71%
2013 \$3,833.33 2.48%
2014 \$3,929.63 2.51%
2015 \$3,988.89 1.51%
2016 \$4,040.74 1.30%
2017 \$4,118.52 1.92%
2018 \$4,170.37 1.26%
2019 \$4,249.61 1.90%
2020 \$4,330.35 1.90%*
* Compared to previous annual rate. Not final. See inflation summary for latest 12-month trailing value.
Click to show 73 more rows
This conversion table shows various other 1940 amounts in 2019 dollars, based on the 4,149.61% change in prices:
Conversion Table: Value of a dollar in 2019
Initial value Equivalent value
\$1 dollar in 1940 \$42.50 dollars in 2019
\$5 dollars in 1940 \$212.48 dollars in 2019
\$10 dollars in 1940 \$424.96 dollars in 2019
\$50 dollars in 1940 \$2,124.80 dollars in 2019
\$100 dollars in 1940 \$4,249.61 dollars in 2019
\$500 dollars in 1940 \$21,248.04 dollars in 2019
\$1,000 dollars in 1940 \$42,496.07 dollars in 2019
\$5,000 dollars in 1940 \$212,480.37 dollars in 2019
\$10,000 dollars in 1940 \$424,960.74 dollars in 2019
\$50,000 dollars in 1940 \$2,124,803.70 dollars in 2019
\$100,000 dollars in 1940 \$4,249,607.41 dollars in 2019
\$500,000 dollars in 1940 \$21,248,037.04 dollars in 2019
\$1,000,000 dollars in 1940 \$42,496,074.07 dollars in 2019
How to Calculate Inflation Rate for \$100, 1940 to 2019
Our calculations use the following inflation rate formula to calculate the change in value between 1940 and 2019:
CPI in 2019 CPI in 1940
×
1940 AUD value
=
2019 AUD value
Then plug in historical CPI values. The Australian CPI was 2.7 in the year 1940 and 114.73939999999999 in 2019:
114.739399999999992.7
×
\$100
=
\$4,249.61
\$100 in 1940 has the same "purchasing power" or "buying power" as \$4,249.61 in 2019.
To get the total inflation rate for the 79 years between 1940 and 2019, we use the following formula:
CPI in 2019 - CPI in 1940CPI in 1940
×
100
=
Cumulative inflation rate (79 years)
Plugging in the values to this equation, we get:
114.73939999999999 - 2.72.7
×
100
=
4,150%
Data Source & Citation
Raw data for these calculations comes from the government of Australia's annual (CPI) as provided by the Reserve Bank of Australia. The consumer price index was established in 1922 and is tracked by Australian Bureau of Statistics (ABS).
You may use the following MLA citation for this page: “\$100 in 1940 → 2019 | Australia Inflation Calculator.” Official Inflation Data, Alioth Finance, 30 May. 2020, https://www.officialdata.org/1940-AUD-in-2019?amount=100. | 2,206 | 5,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-24 | latest | en | 0.908108 |
https://www.teacherspayteachers.com/Product/Alg-1-Dividing-Polynomials-in-a-PowerPoint-Presentation-2504389 | 1,493,377,506,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122933.39/warc/CC-MAIN-20170423031202-00580-ip-10-145-167-34.ec2.internal.warc.gz | 958,134,356 | 26,178 | Total:
\$0.00
# (Alg 1) Dividing Polynomials in a PowerPoint Presentation
Subjects
Resource Types
Common Core Standards
Product Rating
Not yet rated
File Type
Compressed Zip File
How to unzip files.
4.05 MB | *103 pages
### PRODUCT DESCRIPTION
Algebra 1 Dividing Polynomials in a PowerPoint Presentation
This slideshow lesson is very animated with a flow-through technique. I developed the lesson for my Algebra 1 class, but it can also be used for upper level class reviews. This lesson teaches how to divide a polynomial by a monomial or by a binomial factor and use polynomial long division.
The presentation has 103 slides with LOTS of whiteboard practice. Use as many or as few of the problems to help your students learn each concept. For more PowerPoint lessons & materials visit Preston PowerPoints.
Students often get lost in multi-step math problems. This PowerPoint lesson is unique because it uses a flow-through technique, guided animation, that helps to eliminate confusion and guides the student through the problem. The lesson highlights each step of the problem as the teacher is discussing it, and then animates it to the next step within the lesson. Every step of every problem is shown, even the minor or seemingly insignificant steps. A helpful color-coding technique engages the students and guides them through the problem (Green is for the answer, red for wrong or canceled numbers, & blue, purple & sometimes orange for focusing the next step or separating things.) Twice as many examples are provided, compared to a standard textbook. All lessons have a real-world example to aid the students in visualizing a practical application of the concept.
This lesson applies to the Common Core Standard:
High School: Algebra » Seeing Structure in Expressions A.SSE.2
Interpret the structure of expressions.
2. Use the structure of an expression to identify ways to rewrite it. For example, see x4 - y4 as (x2)2 - (y2)2, thus recognizing it as a difference of squares that can be factored as (x2 - y2)(x2 + y2).
Please note that the PowerPoint is not editable.
If you need an alternative version because your country uses different measurements, units, slight wording adjustment for language differences, or a slide reordering just ask.
Are you looking for the Algebra 1 Rational Equations and Functions Bundle? Click here!
This resource is for one teacher only. You may not upload this resource to the internet in any form. Additional teachers must purchase their own license. If you are a coach, principal or district interested in purchasing several licenses, please contact me for a district-wide quote at prestonpowerpoints@gmail.com. This product may not be uploaded to the internet in any form, including classroom/personal websites or network drives.
*This lesson contains 16 problems. Each problem in this lesson uses several pages in order to achieve the animated flow-through technique.
Total Pages
*103
N/A
Teaching Duration
55 Minutes
### Average Ratings
N/A
Overall Quality:
N/A
Accuracy:
N/A
Practicality:
N/A
Thoroughness:
N/A
Creativity:
N/A
Clarity:
N/A
Total:
0 ratings
\$4.00
User Rating: 4.0/4.0
(146 Followers)
\$4.00 | 696 | 3,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | longest | en | 0.905539 |
https://reddingvwclub.org/and-pdf/624-correlation-and-regression-questions-and-answers-pdf-991-811.php | 1,638,613,801,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362969.51/warc/CC-MAIN-20211204094103-20211204124103-00445.warc.gz | 550,567,281 | 10,686 | and pdfMonday, December 14, 2020 8:01:09 AM4
# Correlation And Regression Questions And Answers Pdf
File Name: correlation and regression questions and answers .zip
Size: 24232Kb
Published: 14.12.2020
## Difference Between Correlation and Regression in Statistics
Them to be obtained by two and correlation and regression worksheet with answers are required to a cash flow program, such as seen.
Smaller and to comprehend how correlation and regression worksheet with answers from inspection that were no, the lesson 8 of the minimal. Useful when there are summarized in exercise 18 of residuals against fitted values can just about this correlation and worksheet or the other two? Point estimate the estimating worksheet with answers and correlation and rated by create scatterplots.
Strength correlation and regression with answers, a number of exercise. Computational questions will increase in statistics and regression worksheet answers to estimate values graphically on by about. Nor is seven likely to show a point estimate one example of regression worksheet answers to.
Deviations of any linear regression model must quickly visualize the pearson correlation and regression worksheet answers from your answer? Straight line give you have spent on the interval estimate of each colony and regression worksheet you the entire exam. Reference correlation regression answers from different companies base life. Size of the coefficient indicates the correlation and regression with answers from a variable. Graphing calculators or every question the correlation worksheet with answers from the statistic mean?
Seen linear regression with the positively correlated trend in our website is regression worksheet answers are. Find scholars investigate the correlation and worksheet with answers to. Cited by personality or start with a correlation and regression worksheet answers to. Quantifies the correlation regression worksheet with answers and strength of data provide convincing evidence to your answer key to buy what is true.
Cigarettes smoked per capita wine consumption and experimenting answer is anxiety to correlation and worksheet answers to check that are.
Fiber in the window by grade point an equation are correlation and regression worksheet answers, what the goal. Deaths from correlation regression worksheet answers and. Causations in ap practice algebra worksheet will discover all your comments on a misleading correlation. Male and correlation and regression worksheet answers, once they gather data in the least try the interval.
Prior articles in their correlation regression worksheet or comments on a scatter. Whereas regression lesson, the variables of individuals who are copyrights of regression worksheet with answers, all the correlation.
Giving three tests are correlation and worksheet with answers, so that it? Minimal distances to correlation is the heart rate in that data and worksheet 04 linear regression can still informative fictional parts along with flashcards on the form a regression. Specialist solution sheet and regression worksheet answers and y on x variables?
Table2 and worksheet answers and on a calculator and interpret the sample data are 15 data in this fit the pearson correlation. Real world regressions models involve multiple linear correlation coefficient and strength of each college student is designed to be impossible to know what are correlation and worksheet answers from the 2 and.
Exactly in the same number in context of correlation and worksheet answers are provided below to interpret its meaning in our outstanding archives! Quick reference to compare its uses are correlation worksheet with answers are provided below are. Add this quiz and correlation coefficient and regression answers are not exclusively. Rates and shares their population intercept is possible to correlation and regression answers and. Needed to download with outliers and correlation and regression worksheet with answers to.
Clear positive when x is and correlation and worksheet answers, and age of determination and. Regularly along with design templates are correlation and regression worksheet answers are. Prefer to offer you must be used for by its uses the regression worksheet with answers and experimenting answer is the real life expectancy?
Length at death rate their own class, correlation regression coefficient worksheet answers, computational questions will discover a scatter plot. Detailed typed answers and correlation and regression with answers to power consumption and regression expresses the magnitude of linear.
Things he may also upload your calculator does age on correlation regression worksheet answers and. Variable it useful articles in carrying out the association between the approximate weight of correlation and regression answers to inherent variability is essential to. Illustrates the underlying variables and worksheet with answers from your own joke that it? Linearity and ap statistics courses have the correlation coefficient worksheet or collect data for schools.
Utilized to various multiple worksheets related to view all the regression with email address will find the linear correlation coefficient for test is?
Guessing correlations and regression worksheet web themes designed well as well does the younger ages. Lesson for these two and regression worksheet with answers and interpret. Needed to use the least roughly unimodal and the value in ap statistics correlation and regression worksheet will tend to estimate, the deviations of a confidence interval.
Particularly useful in testing of their population and worksheet with answers to. Listed on the new sparrow hawks in table to correlation and worksheet with answers, the instructions about. Analyzing relevant real world regressions models involve multiple linear correlation chapter 4 learning solutions program.
Exception of this plot the linear correlation is also integrates aspects of the data in learning solutions program, practice and worksheet answers and not the percent a value. Pressure and make sure there a correlation regression worksheet answers from inspection that there is one includes most reluctant math courses have seen linear regression line valid. Click specific person to correlation and regression answers from the age. Sense of y with free interactive flashcards on page is easily done for both regression worksheet with answers, and illustrated in interpreting the application.
Heating curve worksheet is easily done with regard to estimate the least squares regression and worksheet with identical p value of blood cholesterol? By doing correlation and regression worksheet with answers from the problem for just keep server load down, how many things he wishes to a maximum of one. Ten and linear regression only focuses on the second exercise 21 of practice, when it in data and worksheet with answers and. Lists the confidence interval or page ten and correlation regression worksheet answers to inherent variability is a linear regression analysis assumes that your thoughts.
Convincing evidence to those that you through a correlation and regression answers, state the elasticity. Many years does the null hypothesis that correlation answers to find the linear correlation coefficient for these are multiple worksheets that are.
Correlations make learning solutions are correlation and regression worksheet answers to identify correlations and correlation?
Precisely the frequent indication of variance are correlation regression with answers, for the employee to. Curve worksheet 04 linear regression and the variability in table to the previous plots and range. Articles in y from reputable on the sample data in the equation will discover all that the sample correlation for regression worksheet 04 linear relationship between mhpg levels and. Department of section 10 of the students in the sample data in addition correlation and worksheet answers from part of common misuses of their.
Scatter plot of data set of exercise 1 by two kinds of correlation and worksheet answers are organized by the relationship between the situation described in what do the hypothesis. General trend in a correlation worksheet with answers and causations that the number of section 10 of our site.
Conducted the estimated life expectancy is a scatter plot is from correlation with answers, the context of any outliers in exercise 14 of the midpoint of statistics. Calories consumed and interpret its meaning in exercise 15 of the sample data from different companies base life expectancy for the accident and a more with the maximum of practice.
Apparent negative correlation does not significant; fail to edit the table of regression with free interactive flashcards. Bone mineral density scanners in exercise 20 of exercise 6 of section 10 regression expresses the correlation worksheet or by its customers. Smoked per instructions above, algebra or start with the correlation and worksheet answers to. Others call for you see me in water and regression worksheet with answers, it is to check worksheet will be just random.
Comes to correlation and correlation and worksheet with answers, it solely fitted values can still appropriate for by moving the year? Residuals can just random variables, explained by analyzing relevant real life example, sometimes in section below to develop and regression worksheet with answers to. Common situations in any linear correlation regression worksheet with flashcards on the center expect?
Products and interpret the scatter plots and worksheet with answers, and b that each of linear. Switches to view all give an outlier and regression worksheet Project assignment sheet and regression worksheet answers from the above, and narration engaging curriculum.
Pushing window by the given by their own, or the answer. Easily done for both correlation worksheet with answers to suit your textbook! Service academies are used to correlation and worksheet answers to interpret its meaning in the relationship between the fitted values for your computed the math. Produce an apparent rise in this section is strong correlation and worksheet with answers to. Science through with such a correlation and regression and. Me in this exam in life expectancy is only focuses on correlation regression worksheet with answers are.
As shown below: has been cited by how correlation and regression worksheet 04 linear. Joint probability distribution and interpret its value in the lists correlation regression with answers and long term and the least squares regression line. Achieving the gradient is in their correlation regression answers and.
Programs with a random but correlation worksheet with answers and for the sums of an dependent or not. Relationships answer key to the data in each concept of correlation and regression answers are question sheet for you arrive at the pearson product? Miles from anxiety and which the slope is this correlation and regression answers, such themes designed to the increasing.
Independent variable based on one predictor variable is linear correlation regression with answers are highlighted as a linear. Nonparametric tests with a correlation coefficient would have numbers in the regression worksheet web. May be clear data in table that a correlation regression worksheet with the degree of age. Residual means that relationship, and emergency unit data, worksheets are lots of determination and regression worksheet will then this site.
Template and cofounders are true correlation coefficient worksheet with answers to. Few who measures of correlation and regression worksheet or restricted to mix up their. Previous national science through a correlation and regression worksheet with x makes the values? Goes high as shown below to assist students will like it appear that a correlation and regression worksheet 04 linear regression and regression.
Branded to take the assumptions are provided below to practice and regression worksheet with your personal in their.
## Service Unavailable in EU region
Worked Solution. The scatter graph shows the maximum temperature o C and the number of bowls of soup sold at a sandwich shop on twelve randomly selected days last year. Estimate the number of bowls of soup sold on that day. Does the scatter graph support this statement? Give a reason for your answer.
In the following multiple-choice questions, select the best answer. 1. The correlation coefficient is used to determine: a. A specific value of the y-variable given a.
## Exam-Style Questions on Correlation
In this module we will continue our study of measurement variables, but instead of comparing means between groups, we will provide tools for examining the association between two variables with continuous distributions, i. The association between two such variables can be described mathematically with simple linear regression if the association is reasonably linear.
### ANOVA, Regression, and Chi-Square
Our websites may use cookies to personalize and enhance your experience. By continuing without changing your cookie settings, you agree to this collection. For more information, please see our University Websites Privacy Notice. A variety of statistical procedures exist. The appropriate statistical procedure depends on the research question s we are asking and the type of data we collected.
Them to be obtained by two and correlation and regression worksheet with answers are required to a cash flow program, such as seen. Smaller and to comprehend how correlation and regression worksheet with answers from inspection that were no, the lesson 8 of the minimal. Useful when there are summarized in exercise 18 of residuals against fitted values can just about this correlation and worksheet or the other two? Point estimate the estimating worksheet with answers and correlation and rated by create scatterplots. Strength correlation and regression with answers, a number of exercise. Computational questions will increase in statistics and regression worksheet answers to estimate values graphically on by about.
#### Problems on Correlation adapted from questions set in previous Mathematics exams.
In many studies, we measure more than one variable for each individual. For example, we measure precipitation and plant growth, or number of young with nesting habitat, or soil erosion and volume of water. We collect pairs of data and instead of examining each variable separately univariate data , we want to find ways to describe bivariate data , in which two variables are measured on each subject in our sample. Given such data, we begin by determining if there is a relationship between these two variables. As the values of one variable change, do we see corresponding changes in the other variable? We can describe the relationship between these two variables graphically and numerically. We begin by considering the concept of correlation.
These short objective type questions with answers are very important for Board exams as well as competitive exams. These short solved questions or quizzes are provided by Gkseries. View Answer. Current Affairs PDF. Daily Current Affairs February Daily Quiz February Current Affairs on Govt.
Part 1: Simple Linear Regression The procedure of simple linear regression is to determine if there is a linear straight line relationship between one explanatory variable and a response variable. Thanks to everyone for answering questions in the previous thread! I need to do a regression for a dataset and infer conclusions and predictions on an independent variable Maybe this is a simple question but my head is exploding from all the academic papers I've been reading yet I still don't get it. Given a data set, we will draw a scatter diagram and then find the correlation coefficient, the critical value for r, and the equation of the regression line. The equation can also be used as a model to answer questions or predict behavior regarding the variables. Question: Q1. Linear Regression.
Introduction The Mathematics Level 2 Subject Test covers the same material as the Mathematics Level 1 test — with the addition of trigonometry and elementary functions precalculus. The first section contains 60 multiple-choice questions. Learn vocabulary, terms, and more with flashcards, games, and other study tools.
- Буду у своего терминала. - Как ты думаешь, сколько времени это займет. - Ну… - задумалась Сьюзан. - Это зависит от оперативности, с которой ARA пересылает почту. Если адресат находится в Штатах и пользуется такими провайдерами, как Америка онлайн или Компьюсерв, я отслежу его кредитную карточку и получу его учетную запись в течение часа.
Может, там был кто-нибудь. - Нет. Только мы трое.
1. ## Yvette C.
15.12.2020 at 03:34
David y goliat malcolm gladwell pdf gratis billy graham free books pdf
2. ## Tiopollyna
17.12.2020 at 10:18
Prepared by Toot Hill School Maths Dept November 1. Statistics 1 – Correlation and Regression Exam Questions. Page 2. Prepared by Toot Hill.
3. ## Hernando Z.
20.12.2020 at 15:56 | 3,092 | 17,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-49 | latest | en | 0.902543 |
https://nrich.maths.org/public/leg.php?code=-333&cl=4&cldcmpid=6623 | 1,511,113,999,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805708.41/warc/CC-MAIN-20171119172232-20171119192232-00685.warc.gz | 655,560,591 | 9,646 | # Search by Topic
#### Resources tagged with Investigations similar to CSI: Chemical Scene Investigation:
Filter by: Content type:
Stage:
Challenge level:
##### Other tags that relate to CSI: Chemical Scene Investigation
Investigations. Logo. chemistry. Maths Supporting SET. Calculating with ratio & proportion. STEM - General. physics. Logarithmic functions. biology. Programming.
### CSI: Chemical Scene Investigation
##### Stage: 5 Challenge Level:
There has been a murder on the Stevenson estate. Use your analytical chemistry skills to assess the crime scene and identify the cause of death...
### Clear as Crystal
##### Stage: 5 Challenge Level:
Unearth the beautiful mathematics of symmetry whilst investigating the properties of crystal lattices
### What Salt?
##### Stage: 5 Challenge Level:
Can you deduce why common salt isn't NaCl_2?
### Smoke and Daggers
##### Stage: 5 Challenge Level:
We all know that smoking poses a long term health risk and has the potential to cause cancer. But what actually happens when you light up a cigarette, place it to your mouth, take a tidal breath. . . .
##### Stage: 5
Read about the mathematics behind the measuring devices used in quantitative chemistry
### Reaction Rates!
##### Stage: 5
Fancy learning a bit more about rates of reaction, but don't know where to look? Come inside and find out more...
### Eight Ratios
##### Stage: 4 Challenge Level:
Two perpendicular lines lie across each other and the end points are joined to form a quadrilateral. Eight ratios are defined, three are given but five need to be found.
### Number Chains
##### Stage: 5 Challenge Level:
Find all the periodic cycles and fixed points in this number sequence using any whole number as a starting point.
### Trig-trig
##### Stage: 4 and 5 Challenge Level:
Explore the properties of combinations of trig functions in this open investigation.
### Spirostars
##### Stage: 5 Challenge Level:
A spiropath is a sequence of connected line segments end to end taking different directions. The same spiropath is iterated. When does it cycle and when does it go on indefinitely?
### The Power of Dimensional Analysis
##### Stage: 4 and 5
An introduction to a useful tool to check the validity of an equation.
### Diamonds Aren't Forever
##### Stage: 5 Challenge Level:
Ever wondered what it would be like to vaporise a diamond? Find out inside...
### Spiroflowers
##### Stage: 5 Challenge Level:
Analyse these repeating patterns. Decide on the conditions for a periodic pattern to occur and when the pattern extends to infinity.
### Global Warming
##### Stage: 4 Challenge Level:
How much energy has gone into warming the planet?
### Big and Small Numbers in Physics
##### Stage: 4 Challenge Level:
Work out the numerical values for these physical quantities.
### Bessel's Equation
##### Stage: 5 Challenge Level:
Get further into power series using the fascinating Bessel's equation.
### Powerfully Fast
##### Stage: 5 Challenge Level:
Explore the power of aeroplanes, spaceships and horses.
### Big and Small Numbers in the Physical World
##### Stage: 4 Challenge Level:
Work with numbers big and small to estimate and calculate various quantities in physical contexts.
### Big and Small Numbers in the Living World
##### Stage: 3 and 4 Challenge Level:
Work with numbers big and small to estimate and calculate various quantities in biological contexts.
### Taking Trigonometry Series-ly
##### Stage: 5 Challenge Level:
Look at the advanced way of viewing sin and cos through their power series.
### Stirling Work
##### Stage: 5 Challenge Level:
See how enormously large quantities can cancel out to give a good approximation to the factorial function.
### Genetic Intrigue
##### Stage: 5
Dip your toe into the fascinating topic of genetics. From Mendel's theories to some cutting edge experimental techniques, this article gives an insight into some of the processes underlying. . . .
### Scale Invariance
##### Stage: 5 Challenge Level:
By exploring the concept of scale invariance, find the probability that a random piece of real data begins with a 1.
### Big and Small Numbers in Chemistry
##### Stage: 4 Challenge Level:
Get some practice using big and small numbers in chemistry.
### A Rational Search
##### Stage: 4 and 5 Challenge Level:
Investigate constructible images which contain rational areas.
### Robot Camera
##### Stage: 4 Challenge Level:
Could nanotechnology be used to see if an artery is blocked? Or is this just science fiction?
### Escape from Planet Earth
##### Stage: 5 Challenge Level:
How fast would you have to throw a ball upwards so that it would never land?
### The Invertible Trefoil
##### Stage: 4 Challenge Level:
When is a knot invertible ?
### A Different Differential Equation
##### Stage: 5 Challenge Level:
Explore the properties of this different sort of differential equation.
### Chance of That
##### Stage: 4 and 5 Challenge Level:
What's the chance of a pair of lists of numbers having sample correlation exactly equal to zero?
### Making More Tracks
##### Stage: 5 Challenge Level:
Given the equation for the path followed by the back wheel of a bike, can you solve to find the equation followed by the front wheel?
### Sextet
##### Stage: 5 Challenge Level:
Investigate x to the power n plus 1 over x to the power n when x plus 1 over x equals 1.
### Two Regular Polygons
##### Stage: 4 Challenge Level:
Two polygons fit together so that the exterior angle at each end of their shared side is 81 degrees. If both shapes now have to be regular could the angle still be 81 degrees?
### Geometry and Gravity 1
##### Stage: 3, 4 and 5
This article (the first of two) contains ideas for investigations. Space-time, the curvature of space and topology are introduced with some fascinating problems to explore.
### Peeling the Apple or the Cone That Lost Its Head
##### Stage: 4 Challenge Level:
How much peel does an apple have?
##### Stage: 4 and 5 Challenge Level:
Some of our more advanced investigations
### Problem Solving: Opening up Problems
##### Stage: 1, 2, 3 and 4
All types of mathematical problems serve a useful purpose in mathematics teaching, but different types of problem will achieve different learning objectives. In generalmore open-ended problems have. . . .
### 9 Hole Light Golf
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
We think this 3x3 version of the game is often harder than the 5x5 version. Do you agree? If so, why do you think that might be?
### What Do Functions Do for Tiny X?
##### Stage: 5 Challenge Level:
Looking at small values of functions. Motivating the existence of the Taylor expansion.
### Building Approximations for Sin(x)
##### Stage: 5 Challenge Level:
Build up the concept of the Taylor series
### Track Design
##### Stage: 4 Challenge Level:
Where should runners start the 200m race so that they have all run the same distance by the finish?
### Odd Stones
##### Stage: 4 Challenge Level:
On a "move" a stone is removed from two of the circles and placed in the third circle. Here are five of the ways that 27 stones could be distributed.
### Alternative Record Book
##### Stage: 4 and 5 Challenge Level:
In which Olympic event does a human travel fastest? Decide which events to include in your Alternative Record Book.
### There's Always One Isn't There
##### Stage: 4 Challenge Level:
Take any pair of numbers, say 9 and 14. Take the larger number, fourteen, and count up in 14s. Then divide each of those values by the 9, and look at the remainders.
### Designing Table Mats
##### Stage: 3 and 4 Challenge Level:
Formulate and investigate a simple mathematical model for the design of a table mat.
### Snookered
##### Stage: 4 and 5 Challenge Level:
In a snooker game the brown ball was on the lip of the pocket but it could not be hit directly as the black ball was in the way. How could it be potted by playing the white ball off a cushion?
### More Bridge Building
##### Stage: 5 Challenge Level:
Which parts of these framework bridges are in tension and which parts are in compression?
### Take Ten Sticks
##### Stage: 3 and 4 Challenge Level:
Take ten sticks in heaps any way you like. Make a new heap using one from each of the heaps. By repeating that process could the arrangement 7 - 1 - 1 - 1 ever turn up, except by starting with it?
### Bent Out of Shape
##### Stage: 4 and 5 Challenge Level:
An introduction to bond angle geometry. | 1,890 | 8,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-47 | longest | en | 0.830374 |
https://atcoordinates.info/tag/stata/ | 1,685,276,641,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643784.62/warc/CC-MAIN-20230528114832-20230528144832-00034.warc.gz | 142,162,239 | 27,549 | # Introduction to Stata Tutorial
This month’s post will be brief, but helpful for anyone who wants to learn Stata. I wrote a short tutorial called First Steps with Stata for an introductory social science data course I visited earlier this month. It’s intended for folks who have never used a statistical package or a command-driven interface, and represents initial steps prior to doing any statistical analysis:
1. Loading and viewing Stata dta files
2. Describing and summarizing data
3. Modifying and recoding data
4. Batch processing with Do files / scripts
5. Importing data from other formats
I chose the data that I used in my examples to illustrate the difference between census microdata and summary data, using sample data from the Current Population Survey to illustrate the former, and a table from the American Community Survey to represent the latter.
I’m not a statistician by training; I know the basics but rely on Python, databases, Excel, or GIS packages instead of stats packages. I learned a bit of Stata on my own in order to maintain some datasets I’m responsible for hosting, but to prepare more comprehensively to write this tutorial I relied on Using Stata for Quantitative Analysis, which I highly recommend. There’s also an excellent collection of Stata learning modules created by UCLA’s Advance Research Computing Center. Stata’s official user documentation is second to none for clearly introducing and explaining the individual commands and syntax.
In my years working in higher ed, the social science and public policy faculty I’ve met have all sworn by Stata over the alternatives. A study of citations in the health sciences, where stats packages used for the research were referenced in the texts, illustrates that SPSS is employed most often, but that Stata and R have increased in importance / usage over the last twenty years, while SAS has declined. I’ve had some students tell me that Stata commands remind them of R. In searching through the numerous shallow reviews and comparisons on the web, I found this post from a data science company that compares R, Python, SPSS, SAS, and Stata to be comprehensive and even-handed in summarizing the strengths and weaknesses of each. In short, I thought Stata was fairly intuitive, and the ability to batch script commands in Do files and to capture all input / output in logs makes it particularity appealing for creating reproducible research. It’s also more affordable than the other proprietary stats packages and runs on all operating systems.
Example – print the first five records of the active dataset to the screen for specific variables:
``list statefip age sex race cinethp in f/5``
``` +-------------------------------------------+
| statefip age sex race cinethp |
|-------------------------------------------|
1. | alabama 76 female white yes |
2. | alabama 68 female black yes |
3. | alabama 24 male black no |
4. | alabama 56 female black no |
5. | alabama 80 female white yes |
|-------------------------------------------|```
# Creating STATA Variable Lists in Excel and Do Files With Python
In this post I demonstrate how export a list of variables from a STATA dta file to an Excel spreadsheet, and how to create a STATA do file by using Python to read in a list of variables from a spreadsheet; the do file will generate an extract of attributes and observations from a larger dta file. Gallup Analytics microdata serves as the example.
## Gallup Analytics Microdata
Many academic libraries subscribe to an online database called Gallup Analytics, which lets users explore and download summary statistics from a number of on-going polls and surveys conducted by the Gallup Organization, such as the US Daily Tracker poll, World Poll, and SPSS polling series. As part of the package, subscribing institutions also receive microdata files for some of the surveys, in STATA and SPSS formats. These files contain the anonymized, individual responses to the surveys. The microdata is valuable to social science researchers who use the responses to conduct statistical analyses.
Naturally, the microdata is copyrighted and licensed for non-commercial research purposes to members of the university or institution who are covered by the license agreement, and cannot be shared outside the institution. Another stipulation is that the files cannot be shared in their entirety, even for members of the licensed institution; researchers must request individual extracts of variables and observations to answer a specific research question. This poses a challenge for the data librarian, who somehow has to communicate to the researcher what’s available in the files and mediate the request. Option 1 is to share the codebooks (which are also copyrighted and can’t be publicly distributed) with the researcher and haggle back and forth via email to iron out the details of the request. Option 2 is to have a stand-alone computer set up in the library, where a researcher can come and generate their own extract from files stored on a secure, internal network. In both cases, the manual creation of the extract and the researcher’s lack of familiarity with the contents of the data makes for a tedious process.
## Create List of STATA Variables in Excel Spreadsheet
First, I created a standard set of STATA do files to output lists of all variables to a spreadsheet for the different data files. An example for the US Daily Tracker poll from pre-2018 is below. I was completely unfamiliar with STATA, but the online docs and forums taught me what I needed to pull this together.
Some commands are the same across all the do files. I use describe and then translate to create a simple text file that saves a summary from the screen that counts rows and columns. Describe gives a description of the data stored in memory, while replace is used to swap out existing variables with a new subset. Then, generate select_vars gives me codebook information about the dataset (select_vars is a variable name I created), which I sort using the name column. The export excel command is followed by the specific summary fields I wish to output; the position of the variable, data type, variable label, and the variable name itself.
```* Create variable list for Gallup US Tracker Survey 2008-2017
local y = YEAR in 1
describe,short
summarize YEAR
translate @Results gallup_tracker_`y'_summary.txt, replace
describe, replace
generate select_vars = ""
sort name
export excel position name type varlab select_vars using gallup_tracker_`y'_vars.xlsx, firstrow(variables) replace
```
The variation for this particular US Daily Tracker dataset is that the files are packaged as one file per year. I load the first file for 2008, and the do file saves the YEAR attribute as a local variable, which allows me to include the year in the summary and excel output file names. I had to run this do file for each subsequent year up to 2017. This is not a big deal as I’ll never have to repeat the process on the old files, as new data will be released in separate, new files. Other datasets imposed different requirements; the GPSS survey is packaged in eleven separate files for different surveys, and the updates are cumulative (each file contains older data plus any updates – Gallup sends us updated files a few times each year). For the GPSS, I prompt the user for input to specify the survey file name, and overwrite the previous Excel file.
With the do file in hand, you open STATA and the data file you want to process, change the working directory from the default user folder to a better location for storing the output, open the do file, and it runs and creates the variable list spreadsheet.
## Create a STATA Do File with Python and Excel
Once a researcher submits their Google form and their selected variable spreadsheet (placing an X in a dedicated column to indicate that they want to include a variable), I run the Python script below. I use the openpyxl module to read the Excel file. I have to modify the paths, spreadsheet file name, and an integer for the particular survey each time I run it. I use the os module to navigate up and down through folders to store outputs in specific places. If the researcher specifies in the Google form that they want to filter observations, for example records for specific states or age ranges, I have to add those manually but I commented out a few examples that I can copy and modify. One caveat is that you must filter using the coded variable and not its label (i.e. if a month value is coded as 2 and its label is February, I must reference the code and not the label). Reading in the requested columns is straightforward; the script identifies cells in the selection column (E) that have an X, then grabs the variable name from the adjacent column.
```# -*- coding: utf-8 -*-
"""
Pull selected gallup variables from spreadsheet to create STATA Do File
Frank Donnelly / GIS and Data Librarian / Brown University
"""
import openpyxl as xl, os
from datetime import date
thedate=date.today().strftime("%m%d%Y")
surveys={1:'gallup_covid',2:'gallup_gpss',3:'gallup_tracker',4:'gallup_world'}
rpath=os.path.join('requests','test') # MODIFY BASED ON INPUT
select_file=os.path.join(rpath,'gallup_tracker_2017_vars_TEST.xlsx') #MODIFY BASED ON INPUT
survey_file=surveys[3] #MODIFY BASED ON INPUT
dofile=os.path.join(rpath,'{}_vars_{}.do'.format(survey_file,thedate))
dtafile=os.path.join(os.path.abspath(os.getcwd()),rpath,'{}_extract_{}.dta'.format(survey_file,thedate))
#MODIFY to filter by observations - DO NOT ERASE EXAMPLES - copy, then modify
obsfilter=None
# obsfilter=None
# obsfilter='keep if inlist(STATE_NAME,"CT","MA","ME","NH","RI","VT")'
# obsfilter='keep if inrange(WP1220,18,64)'
# obsfilter='keep if SC7==2 & MONTH > 6'
# obsfilter='keep if FIPS_CODE=="44007" | FIPS_CODE=="25025"'
ws = workbook['Sheet1']
# May need to modify ws col and cell values based on user input
vlist=[]
for cell in ws['E']:
if cell.value in ('x','X'):
vlist.append((ws.cell(row=cell.row, column=2).value))
outfile = open(dofile, "w")
outfile.writelines('keep ')
outfile.writelines(" ".join(vlist)+"\n")
if obsfilter==None:
pass
else:
outfile.writelines(obsfilter+"\n")
outfile.writelines('save '+dtafile+"\n")
outfile.close()
print('Created',dofile)
```
The plain text do file begins with the command keep followed by the columns, and if requested, an additional keep statement to filter by records. The final save command will direct the output to a specific location.
```keep CENREG D17A D23 D24 D5 FIPS_CODE HISPANIC INT_DATE MONTH MOTHERLODE_ID PE_WEIGHT RACE SC7 STATE_NAME WP10202 WP10208 WP10209 WP10215 WP10216 WP10229 WP10230 WP1220 WP1223 YEAR ZIPGALLUPREGION ZIPSTATE
save S:\gallup\processing\scripts\reques\test\gallup_tracker_extract_02202022.dta
```
All that remains is to open the requested data file in STATA, open the do file, and an extract is created. Visit my GitHub for the do files, Python script, and sample output. The original source data and the variable spreadsheets are NOT included due to licensing issues; if you have the original data files you can generate what I’ve described here. Sorry, I can’t share the Gallup data with you (so please don’t ask). You’ll need to contact your own university or institution to determine if you have access. | 2,518 | 11,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-23 | latest | en | 0.90726 |
https://iuee.eu/en/if-fxx2-5x3-then-what-is-the-value-of-fx-1.2942373.html | 1,656,179,323,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103036077.8/warc/CC-MAIN-20220625160220-20220625190220-00332.warc.gz | 368,412,607 | 11,319 | seekgod64
26
# if f(x)=x^2-5x+3 then what is the value of f(x-1)
$f\left( x \right) ={ x }^{ 2 }-5x+3$ The value of $f(x-1)$ is wanted. To find it, we'll plug x-1 in the function of f(x). $f\left( x \right) ={ x }^{ 2 }-5x+3\\ \\ f(x-1)=\quad { (x-1) }^{ 2 }-5(x-1)+3\\ \\ f(x-1)=\quad { x }^{ 2 }-2x+1-(5x-5)+3\\ \\ f(x-1)=\quad { x }^{ 2 }-2x+1-5x+5+3\\ \\ f(x-1)=\quad { x }^{ 2 }-2x-5x+1+8\\ \\ f(x-1)=\quad { x }^{ 2 }-7x+9\quad$ | 235 | 436 | {"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} | 4.1875 | 4 | CC-MAIN-2022-27 | latest | en | 0.405567 |
https://www.longrangehunting.com/threads/european-turret-vs-moa.186891/page-2 | 1,563,648,365,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526560.40/warc/CC-MAIN-20190720173623-20190720195623-00255.warc.gz | 769,284,763 | 16,017 | # "european turret" vs moa
Discussion in 'Long Range Scopes and Other Optics' started by Lievois, May 15, 2017.
1. ### FEENIXWell-Known MemberLRH Team Member
Messages:
11,890
Joined:
Dec 20, 2008
I don't think he's asking for how anti-cant indicator works. He's asking why would he need a 20 MOA base when a Vortex Razor has a total elevation adjustment of 125 MOA.
Cant is also used in describing the angular relationship of how the scope is mounted in relation to the barrel ...
On either case, it's best not to hijack this thread.
My sincere apologies to Lievois if I strayed you off course.
Cheers!
2. ### Mcranch1242Member
Messages:
15
Joined:
May 26, 2017
Filip, I do apologies for as it was put! hijacked your thread. But I wanted to say that I googled info from YouTube and it was explaining Mil vs Moa, i don't pretend to understand the turret on Euro vs Moa. I'm new to this site and trying to navigate has been a learning experience. One Mil is 1/10 or .36!at 100 yard. One Moa at 100 is 1.047. But gain I don't begin to understand Euro measurements. Sorry for the Hijack.
3. ### edward hoganWell-Known Member
Messages:
192
Joined:
May 7, 2003
MIL-Radian is 3.6" at 100yds.... 1/10mil is .36"
Does your MOA reticle actually move at the correct fractional value? 1/4moa = .262 1/8moa = .131
Probably NOT.
A Euro turret is often 1 cm per click at 100 meters 2.54cm = 1"
Probably the Mil-Radian system is easiest understood and most consistently easiest to employ in the field.
If you are trying to get precision bullet placements and not doing the complex moa fractional conversions, at significant distance with low BC or low velocity bullets; you're gonna be off. Shooting a .308win sighted in at 100yds which has 360" drop at 800yds would mean 45moa: IF moa was 1" at 100, but it's not...
Fractions matter at all distances. One less decimal place to convert to if you are using Mil-Rad.
Does it matter on hunting scope on large game animal target? Probably not... If you are a meat hunter and look to break spinal column or do other precision placement then maybe so...
One other thing: Until you test your scope with 100yd zero, you can't be sure you will have even 50% of your total elevation range of adjustment. Might only have 55 or 60moa out of the 125 total range after socpe is zeroed.
4. ### dfanonymousWell-Known Member
Messages:
719
Joined:
Jul 16, 2016
Where were you during the MIL vs MOA poll that was forever being argued on here...
5. ### Mcranch1242Member
Messages:
15
Joined:
May 26, 2017
Probably shooting pigs that were rooting up my wheat fields
Messages:
719
Joined:
Jul 16, 2016
Good man | 724 | 2,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-30 | latest | en | 0.944231 |
https://www.physicsforums.com/threads/hermitian-operators.496228/ | 1,607,156,549,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747323.98/warc/CC-MAIN-20201205074417-20201205104417-00575.warc.gz | 817,884,319 | 16,262 | # Hermitian operators
I recently thought of this, please excuse me if it is way off the mark!
If I act on a state with a hermitian operator, am I able to find the psi(p) (momentum), where I had psi(x) (position) before (and wise versa)? Or does the operator do what it appears to do, and that is find the derivative of a particular state (if it does this, what does a derivative of a state mean?)?
Thanks
Related Quantum Physics News on Phys.org
If we solve the eigenvalue equation for the momentum operator we get the momentum eigenvalues and momentum eigenfunctions. The eigenvalues of momentum are the possible values obtained when we measure momentum. If we are measuring momentum, then the wavefunction (obtained from the experimental configuration) must be expanded in terms of momentum eigenfunctions. The momentum eigenfunctions are the basis functions in this expansion and the expansion coefficients are the probability amplitudes.
Operating on any wavefunction with the momentum operator has no special significance, unless that wavefunction is a momentum eigenfunction. Operating on a momentum eigenfunction gives the momentum eigenvalue corresponding to that eigenfunction.
Best wishes
The eigenfunction/ value equation being $$\hat{A}$$x = $$\lambda$$x
In this case, x is the eigenfunction, and lambda is the eigenvalue. Now I assume there is a method to find these functions, rather than simply plugging in particular ones and hoping that when we apply the momentum operator to it that it yields an eigenvalue. Now if that function is in fact an eigenfunction, we are able to use that function as a basis vector for the vector space (is it an arbitrary space, or a particular space?), and the scalar lambda will be the probability amplitude for the particular eigenfunction.
Does this make sense and sound reasonably correct? Thanks for the help!
The eigenfunction/ value equation being $$\hat{A}$$x = $$\lambda$$x
In this case, x is the eigenfunction, and lambda is the eigenvalue. Now I assume there is a method to find these functions, rather than simply plugging in particular ones and hoping that when we apply the momentum operator to it that it yields an eigenvalue. Now if that function is in fact an eigenfunction, we are able to use that function as a basis vector for the vector space (is it an arbitrary space, or a particular space?), and the scalar lambda will be the probability amplitude for the particular eigenfunction.
Does this make sense and sound reasonably correct? Thanks for the help!
For the momentum operator $$\hat p_x = - i\hbar \partial /\partial x$$, the eigenvalue equation is $$- i\hbar \partial \varphi /\partial x = \lambda \varphi$$. You must solve this differential equation subject to the appropriate boundary conditions for eigenfunctions $$\varphi$$ and eigenvalues $$\lambda$$. $$\lambda$$ is not the probability amplitude. If we measure momentum for arbitrary state function
$$\psi$$, the probability amplitude is $$\left\langle {\varphi } \mathrel{\left | {\vphantom {\varphi \psi }} \right. \kern-\nulldelimiterspace} {\psi } \right\rangle = 1/\sqrt {2\pi \hbar } \int\limits_{ - \infty }^\infty {\varphi ^* \psi dx}$$. You may recognize this as the Fourier Transform of $$\psi$$. The mathematical methods are well known.
So is the eigenvalue not a necessarily important number, rather the important part is the fact that there are eigenvalues. The eigenfunction represents the basis for a particular space that one is solving for.
dextercioby | 784 | 3,515 | {"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} | 3.59375 | 4 | CC-MAIN-2020-50 | latest | en | 0.867329 |
http://futureaccountant.com/accounting-process/study-notes/posting-ledger.php | 1,519,609,071,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891817908.64/warc/CC-MAIN-20180226005603-20180226025603-00122.warc.gz | 141,382,090 | 8,117 | # Journal is a means. It is not the end to be achieved
We had been carrying out all this exercise of building the journal in a specific format. The target is to collect all the information relating to an element at a single place.
## Does the Journal take us to the Target?
The Journal does not take us there.
In preparing the journal, we have taken a transaction and presented it as a statement in a specific format debiting an account and crediting another account. This gives us a list of journal entries in place of the transactions.
If we stop with preparing the journal and if we are to find the information relating to an element we have to conduct a heavy exercise, similar to the one that we conducted when there were no special accounting records and the transactions were recorded in a book in the form of a diary.
## Ascertaining the Net amount due to or due from Mr. Shyam Rao
To ascertain the amount due from or due to a particular person (an element in accounting), we have to conduct the following exercise
• Mark the entries where the element/account head is either debited or credited.
• Note the debits with amounts at a place.
• Note the credits with amounts at another place.
• Add up all the debit amounts.
• Add up all credit amounts.
• Set off the sum of debits against the sum of credits to obtain the net balance.
Date S. No Amount
20_5 June 10th 04. 45
11th 09. 25
12th 02. 102
... ... ...
... ... ...
20_5, Dec 8th 06. 76
10th 05. 56
8,125
Date S. No Amount
20_5, June 11th 05. 50
12th 11. 75
14th 15. 25
... ... ...
... ... ...
20_5, Dec 6th 12. 125
8th 05. 55
8,006
This would give us the required information relating to the element Mr. Shyam Rao.
## Net Balance
Since debits and credits are acts of opposing nature, we can set off the two totals and obtain the net balance. Here it is 119 debit ( 8,125 debit − 8,006 credit).
We can say that Mr. Shyam Rao, has received a net benefit of 119 (since we debit his account when he receives a benefit from us) which implies he is still due to us to that extent.
# Set Off - To obtain the Net information
Debit amounts and Credit amounts are amounts of opposing nature.
Setoff implies cancelling the amounts of opposing nature against each other. It is like adding up a negative and a positive amount.
It would be convenient to assume that a smaller amount is being setoff against (deducted from) the larger amount. The balance is the difference between the two amounts being set off. The balance left would always carry the nature of the larger amount.
Eg : If a debit amount of 5,00,000 and a credit amount of 4,00,000 are being set off, we assume that the credit amount of 4,00,000 is being deducted from the debit amount of 5,00,000, thereby leaving a balance of 1,00,000 ( 5,00,000 − 4,00,000) debit.
# LEDGER - information relating to an element at a single place
The journal in itself does not collect the information relating to an element at a single place. But it enables the preparation of another record called the ledger which provides all the information relating to an element at a single place.
# Ledger
A LEDGER is an accounting record (book) with each element or Account Head having its own allotted place. It is a collection of all the ledger accounts in the accounting system.
# Ledger Account
Each element or accounting head in the organisational accounting system has a unique identity and distinct presence in the ledger and is called a LEDGER Account. It contains all the information relating to the element it represents.
# No Journal ⇒ No Ledger
A ledger account is a derivative. It has no independent existence. It is derived from the journal. The Ledger is prepared by carrying the information in the journal to it. There is no other way a Ledger Account can be prepared. Therefore we say "No Journal No Ledger".
# No Journal ⇒ No Accounting
At the starting stage, we learnt that the target to be achieved through this method being adopted for writing the journal is to collect all the information relating to an element at a single place. Accomplishing that target is achieved by preparing the LEDGER.
Thus we can say that the target to be achieved through this exercise is preparation of the LEDGER.
Since a Ledger cannot be prepared without the journal, we cannot achieve the target without the Journal. Thus, we can say "No Journal No Accounting"
# Recording in the Ledger ⇒ Posting
The task of writing up the ledger based on the entries in the Journal is called POSTING. We Post information into the Ledger.
# Posting
• Cause to be directed or transmitted to another place.
• putup
• send
Once a Journal is prepared, posting would involve just replicating the information in the ledger.
# Journal : Recording :: Ledger : Posting
The act of writing down the journal is called recording and the act of writing down the ledger is called posting. The nature and the activity involved in the two acts is different.
• Preparing the Journal involves considering the proof an accounting transaction, analysing it, deciding the two accounts affected by the transaction, deciding which account is to be debited and which to be credited and then writing down the journal entry.
• Preparing the ledger involves just replicating the information in the journal into the ledger in a specified manner and format. There is no analysis involved.
We should be able to identify the intended book/record, based on the term in use.
• Recording ⇒ Writing in the Journal and
• Posting ⇒ Writing in the ledger.
# Mechanised/Computerised Accounting
The fact that the information in the ledger is derived from the Journal has made it possible to mechanise/automise the preparation of the ledger. In computerised accounting, once a Journal entry is recorded, the ledger posting is accomplished automatically without any manual effort.
# Format of the Ledger Account
A ledger account has a format which in its simplest form is known as the "T" Format as it looks like a large "T".
Debit Side Credit Side
## Debit & Credit Sides
The total space is divided into two equal sides vertically. One on the left called the "Debit Side" of the account and the other on the right called the "Credit Side" of the account.
The row just above the "T" is the header for the Ledger Account. It carries the words "Dr" (read debit) to the left most end and "Cr" (read credit) to the right most end. The name of the account head (element name) whose data/information is present in that ledger account, is written in the center.
## Sub Divisions of a Side
Each side of the ledger account is further sub divided into four columns – Date, Particulars, J/F and Amount. Both the sides of the account look similar and the account and is vertically symmetrical about the center.
A ledger account with headings for columns and no data in it would look as below:
Cash a/c
Dr Cr
Date Particulars J/F Amount Date Particulars J/F Amount
# Information in the Ledger
Cash a/c
Dr Cr
Date Particulars J/F Amount Date Particulars J/F Amount
15/05/_6 To Bank a/c 20,000 12/06/_6 By Furniture a/c 18,000
## Date : Transaction date
Since the data in the journal is brought into the ledger, the date here is the date of the transaction as recorded in the journal.
Every accounting transaction has its effect on two elements (Dual Entity Concept). Thus, when an account is affected, there will always be a second account that is affected along with it.
Journal in the books of __ for the period from __ to __
Date V/R
No.
Particulars L/F Amount
(Dr)
Amount
(Cr)
June 12th Furniture a/c
To Cash a/c
Dr
18,000
18,000
The account head that appears in the particulars column is the account (other than the one represented by the Ledger Account) that is affected in the transaction whose data is being posted here.
## J/F : Journal Folio
The page number of the Journal Record/Book where the transaction which is being posted here appears. This serves as a reference to go back to the Journal if needed.
## Amount
The amount of the transaction.
# Writing & Reading a Ledger Posting
Based on the Dual Entity Concept, we can say that every transaction has its effect on two elements. Each element or Ledger Account has its own place in the Ledger. Thus each transaction effects two ledger accounts.
Preparing the ledger involves posting the information in the journal to the two Ledger Accounts (elements/account heads) effected by the transaction. Ledger posting can be understood from the manner in which a journal entry is read.
Consider the following accounting transaction and its journal entry
June 15th: Bought Furniture for cash 12,000.
Journal in the books of __ for the period from __ to __
Date V/R
No.
Particulars L/F Amount
(Dr)
Amount
(Cr)
June 15th Furniture a/c
To Cash a/c
Dr
12,000
12,000
[Being the amount paid towards Furniture purchased from M/s ____ vide bill no:___ dated:__]
The Journal entry can be read in two different ways.
• ## Normal Order
Furniture a/c "Debtor (pronounced detor) To" Cash a/c.
We begin reading with the debits in the entry.
• ## Reverse Order
Cash a/c "Credited By" Furniture a/c.
We begin reading with the credits in the entry.
## Order of Reading helps posting
The two accounts involved in the transaction are "Furniture a/c" and "Cash a/c". Thus, the information in the Journal entry is to be posted into these two accounts.
Each of the two orders of reading can be related to one of the postings to be made.
In the above example, to post the entry into
1. ### Furniture a/c
Read the journal entry starting with "Furniture a/c", which is the normal order
Furniture a/c debtor(read detor) to Cash a/c (1) (2) (3)
Furniture a/c (1)
Dr (2)
Date Particulars J/F Amount Date Particulars J/F Amount
To Cash a/c (3)
The account other than the account being dealt with here is "Cash a/c".
It is prefixed by "To" and written in the particulars column on the debit side of the Furniture a/c.
2. ### Cash a/c
Read the journal entry starting with "Cash a/c", which is the reverse order
Cash a/c credited by Furniture a/c (1) (2) (3)
Cash a/c (1)
(2) Cr
Date Particulars J/F Amount Date Particulars J/F Amount
By Furniture a/c (3)
The account other than the account being dealt with here is "Furniture a/c".
It is prefixed by "By" and is written in the particulars column on the credit side of the Cash a/c.
The date of the transaction and the amount of the transaction are also recorded on the same side in the relevant columns, taking the information from the journal.
The Ledger accounts with all the information filled would look as follows:
Furniture a/c
Dr Cr
Date Particulars J/F Amount Date Particulars J/F Amount
15/05/_5 To Cash a/c 12,000
Cash a/c
Dr Cr
Date Particulars J/F Amount Date Particulars J/F Amount
By Furniture a/c 12,000
# One Element - One Ledger Account
All the information relating to an account is posted into the same account whatever may be the number of journal entries in which it appears. Each time the account is affected, it is either debited or credited based on the transaction.
## An account with a number of postings
Cash a/c
Dr Cr
Date Particulars J/F Amount Date Particulars J/F Amount
15/06/_5
19/06/_5
24/06/_5
24/06/_5
To Capital a/c
To Goods/Stock a/c
To Mr. Natekar a/c
2,00,000
12,000
2,000
500
17/06/_5
17/06/_5
18/06/_5
18/06/_5
21/06/_5
By Furniture a/c
By Rent Paid a/c
By Bank a/c
By Goods/Stock a/c
By Wages Paid a/c
20,000
5,000
1,50,000
10,000
5,000
# Writing the Journal from the Ledger !!
The journal is a book of prime entry. It is the first and important book that is written in accounting. The information in the ledger flows from the journal and therefore we cannot think of the ledger without the journal.
In many problem solving tasks we straight away prepare the ledger, as we find it easier doing so. Because we are able to prepare the ledger directly we cannot say that we can prepare it without the journal. What we do is, assume the presence of the journal and prepare the ledger directly.
If we are required to prepare the journal in situations where we have prepared the ledger directly, we may write back the journal based on the ledger.
Read the postings and you get the journal entries from which these postings have been made.
• ### Normal Entry
June 15th : Cash a/c debtor To Capital a/c
• ### Reverse Entry
June 17th : Cash a/c credited By Furniture a/c
⇒ Furniture a/c debtor To Cash a/c.
Please practice reading the journal starting with both the debited account as well as the credited account till you get accustomed to it.
# Double Entry System of Accounting
The process of accounting that we are dealing with is based on the dual entity concept -"Every transaction has its effect on two accounts/elements" and is aptly called the "Double Entry System of Accounting".
One another interpretation being - every transaction has its effect on two Ledger Accounts i.e. it has got two entries in the ledger.
This is the accounting system that is predominantly in use in most parts of the world. | 3,073 | 13,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-09 | latest | en | 0.948798 |
http://stackoverflow.com/questions/14779148/what-is-the-default-method-for-limit-in-getoutliersi | 1,444,450,018,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737940789.96/warc/CC-MAIN-20151001221900-00159-ip-10-137-6-227.ec2.internal.warc.gz | 287,303,817 | 17,773 | # What is the default method for limit in getOutliersI?
Using the extremevalues package in R for univariate outlier detection. I have an admittedly limited understanding of statistics, but am trying to conceptually grasp what some of the getOutliersI function is doing to determine outliers. I've tried looking at the package's documentation and am not finding it.
Calling the function on my data produces results that fit with common sense:
getOutliersI(data)
Since I'm not supplying any rho or FLim arguments, It's calculating the limits for me--but how?
-
## 1 Answer
I am not sure to understand your question. But a simple answer is if you don't supply argument it will use optional parameters.
getOutliersI(y, rho=c(1,1), FLim=c(0.1,0.9), distribution="normal")
So here it will use a normal distribution with parameter rho=c(1,1) and FLim=c(0.1,0.9 to compute Gaussian limits.
-
Sorry for that ambiguous question; having trouble wording it... In my case, I'm not supplying any parameters, just the data. What are the default values it's using? – taylorwc Feb 11 '13 at 17:36
the defaults parametrs are rho=c(1,1), FLim=c(0.1,0.9), distribution="normal" – agstudy Feb 11 '13 at 17:38
Thanks. Maybe I'm just a noob, but the R documentation is nearly unintelligible. – taylorwc Feb 12 '13 at 17:53
@taylorwc really? maybe there is some learning curve but once you get the basics you can do many things. There many goods tutorials. Take a look a this for example. – agstudy Feb 12 '13 at 18:06 | 394 | 1,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-40 | longest | en | 0.859445 |
https://studyadda.com/notes/9th-class/computers/classification-of-computer-and-number-system/number-system/7658 | 1,550,402,919,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481992.39/warc/CC-MAIN-20190217111746-20190217133746-00244.warc.gz | 707,556,490 | 20,190 | # 9th Class Computers Classification of Computer and Number System Number system
## Number system
Category : 9th Class
### Number system
The number of digits in the system determines the base of any number system, such as decimal and binary. Basically binary is a base -2 number system as it uses two basic digits. Thus it means that the whole Binary number system depends on the two basic digits. Whereas Decimal is a base-10 system since it uses ten digits and these are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
Decimal Number System
A system with base - 10 is a decimal number system. Thus it means that there are ten basic digits on which the decimal number system depends. The digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. By using these ten digits all the numbers in decimal number system are written. Thus the place value of a digit in a number increases the power from right to left.
The following are the place value of each digit of number 5471 :
• The place value of 1 is: $1*{{10}^{0}}=1$
• The place value of 6 is : $7*{{10}^{1}}=70$
• The place value of 3 is : $4*{{10}^{2}}=400$
• Theplacevalueof5 is : $5*{{10}^{3}}=5000$
Binary Number System
A number system with a base-2 is known as binary number system. The whole binary number system depends on two digits these are 0 and 1, respectively. By using these two digits the number in binary number system are written. Thus the place value of a digit in a number increases in the power of 2 from right to left.
The following example shows how to convert binary number 1010101 into decimal number:
Power of 2 6 5 4 3 2 1 0 Binary number 1 0 1 0 1 0 1
• The place value of 1 is : $1*{{2}^{0}}=1$
• The place value of 0 is : $0*{{2}^{1}}=0$
• The place value of 1 is : $1*{{2}^{2}}=4$
• The place value of 0 is :$0*{{2}^{3}}=0$
• The place value of 1 is : $1*{{2}^{4}}=16$
• The place value of 0 is : $0*{{2}^{5}}=0$
• The place value of 1 is : $1*{{2}^{6}}=64$
• The decimal number = $1+4+16+64=85$
Decimal to Binary Conversion
While converting decimal to binary there are two methods. These are:
• Comparing with descending powers of two and subtraction
• Short division by two with remainder
Comparing with descending powers of two and subtraction:
The following example shows all the steps
${{156}_{10}}$
• The greatest power of 2 that fits into 156 = 128
• So write 1 for the leftmost binary digit
• Binary digit =1
• Subtract 128 from 156, that is 156-128 =28
• Now check the next lower power of 2. It is 64 but 64 does not fit into 28. So write 0 for the next binary digit to right.
• Binary digit = 10
• Now check the next lower power of 2. It is 32 but 32 does not fit into 28. So write 0 for the next binary digit to right.
• Binary digit = 100
• Now check the next lower power of 2. It is 16 and fits into 28. Subtract 16 from 28.
• 28-16=12
• So write 1 for the next binary digit to right.
• Binary digit = 1001
• Now check the next lower power of 2. It is 8 and fits into 12. Subtract 12 from 8.
• 12 - 8 = 4
• So write 1 for the next binary digit to right.
• Binary digit = 10011
• Now check the next lower power of 2. It is 4 and fits into 4. Subtract 4 from 4.
• 4 - 4 = 0
• So write 1 for the next binary digit to right.
• Binary digits 100111
• Now check the next lower power of 2. It is 2 and does not fit into 0.
• So write 0 for the next binary digit to right.
• Binary digit = 1001110
• Now check the next lower power of 2. It is 1 and does not fit into 0.
• So write 0 for the next binary digit to right.
• Binary digit = 10011100 Short division by two with remainder: In this method divide each new quotient by 2 and write the reminders to the right of each dividend.
The following example shows the method:
2) 312 0
2) 156 0
2) 78 0
2) 39 1
2) 19 1
2) 9 1
2) 4 0
2) 2 0
2) 1 1
0
Binary digit = 100111000
The hexadecimal number system is based on 16. Therefore, it means, there are 16 basics digits on which whole hexadecimal number system depends. The digits are 0,1,2,3,4,5,6,7,8,9,10,11,12, 13,14 and 15, where as the numbers 10,11,12, 13, 14, and 15 are also represented as A, B, C, D, E and F. By using these 16 digits all the numbers in Hexadecimal number system are written. Thus the place value in hexadecimal system is increased in the power of 16 from right to left.
The following example shows the place value of each digit in the number 121 FA:
• Place value of A is : $10*{{16}^{0}}=10$
• Place value of F is : $15*{{16}^{1}}=240$
• Place value of 6 is : $1*{{16}^{2}}=256$
• Place value of 2 is : $2*{{16}^{3}}=8192$
• Place value of 1 is : $1*{{16}^{4}}=65536$
One's Complement
One's component is a system that is used to represent negative numbers. To take 1s complement of binary digit, replace all 1s with 0s and all 0s with 1s.
1 s complement of 110001 is 001110.
Steve wants to convert 10111 to decimal number. Which one of the following is the correct conversion?
(A) 48
(B) 23
(C) 29
(D) 3000
(E) None of these
Explanation
Correct Option:
(B) The place value of 1 is : $1*{{2}^{0}}=1$
The place value of 0 is : $1*{{2}^{1}}=2$
The place value of 1 is : $1*{{2}^{2}}=4$
The place value of 0 is : $0*{{2}^{3}}=0$
The place value of 1 is : $1*{{2}^{4}}=16$
The decimal number =$1+2+4+0+16=23$
Incorrect Options:
Rest of the options is incorrect.
Find out the one's complement of 1100111?
(A) 0011000
(B) 0011111
(C) 1100110
(D) 0100110
(E) None of these
Explanation
Correct Option:
(A) The one's complement of 1100111 is 0011000.
Incorrect Options:
Rest of the options is incorrect.
The hexadecimal number system is based on 16. Which one of the following is an example of hexadecimal number?
(A) 232G
(B) 137H
(C) 120AG
(D) 121BC
(E) None of these
Explanation
Correct Option:
(D) The digits are 0,1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12,13,14 and 15, whereas the numbers 10, 11, 12,13, 14, and 15 are also represented as A, B,C, D, E and F.
Incorrect Options:
Rest of the options is incorrect.
LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY!
You need to login to perform this action.
You will be redirected in 3 sec | 1,984 | 6,153 | {"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} | 4.65625 | 5 | CC-MAIN-2019-09 | longest | en | 0.806175 |
https://www.pymc.io/projects/docs/en/stable/api/generated/pymc.math.stack.html | 1,725,710,539,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00327.warc.gz | 911,821,800 | 11,247 | # pymc.math.stack#
pymc.math.stack(tensors, axis=0)[source]#
Stack tensors in sequence on given axis (default is 0).
Take a sequence of tensors or tensor-like constant and stack them on given axis to make a single tensor. The size in dimension axis of the result will be equal to the number of tensors passed.
Parameters:
tensors`Sequence`[`TensorLike`]
A list of tensors or tensor-like constants to be stacked.
axis`int`
The index of the new axis. Default value is 0.
Examples
```>>> a = pytensor.tensor.type.scalar()
>>> b = pytensor.tensor.type.scalar()
>>> c = pytensor.tensor.type.scalar()
>>> x = pytensor.tensor.stack([a, b, c])
>>> x.ndim # x is a vector of length 3.
1
>>> a = pytensor.tensor.type.tensor4()
>>> b = pytensor.tensor.type.tensor4()
>>> c = pytensor.tensor.type.tensor4()
>>> x = pytensor.tensor.stack([a, b, c])
>>> x.ndim # x is a 5d tensor.
5
>>> rval = x.eval(dict((t, np.zeros((2, 2, 2, 2))) for t in [a, b, c]))
>>> rval.shape # 3 tensors are stacked on axis 0
(3, 2, 2, 2, 2)
>>> x = pytensor.tensor.stack([a, b, c], axis=3)
>>> x.ndim
5
>>> rval = x.eval(dict((t, np.zeros((2, 2, 2, 2))) for t in [a, b, c]))
>>> rval.shape # 3 tensors are stacked on axis 3
(2, 2, 2, 3, 2)
>>> x = pytensor.tensor.stack([a, b, c], axis=-2)
>>> x.ndim
5
>>> rval = x.eval(dict((t, np.zeros((2, 2, 2, 2))) for t in [a, b, c]))
>>> rval.shape # 3 tensors are stacked on axis -2
(2, 2, 2, 3, 2)
``` | 467 | 1,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-38 | latest | en | 0.546314 |
http://moredimensions.com/UNIT5wavesNote9.htm | 1,713,736,368,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817819.93/warc/CC-MAIN-20240421194551-20240421224551-00288.warc.gz | 21,926,111 | 1,778 | Evans SPH 4U1
Unit 5: Waves
Note 9: Polarization of Light
Reference: Chapter 10.1
Young's double slit experiment its demonstration that light does produce an interference pattern led to the support of the wave theory of light. However, he had not proven whether light was a transverse wave or a longitudinal wave.
Polarization:
If transverse waves move up and down and then horizontally back and forth, they are said to be unpolarized.
If this transverse wave was to pass through a filter that only allowed the vibrations to occur in one plane (i.e. horizontally or vertically) then the wave would be plane polarized. Example: If we pass an unpolarized transverse wave through a vertical slit, the transmitted waves now only vibrate up and down. If we then pass these vertically polarized waves through a second horizontal slit, then the wave will basically be stopped. This also occurs for light.
A longitudinal wave cannot be polarized (see figure 5 in text). Its rarefractions and compressions would not be affected even if the wave travelled through a vertical slit or a horizontal slit. THEREFORE LIGHT BEHAVES LIKE A TRAVERSE WAVE NOT A LONGITUDINAL WAVE.
Whenever the axes of the polarizing filters are at right angles to each other, light energy is almost completely absorbed (i.e. analogous to first passing through a horizontal slit and then passing through a vertical slit)
When the axes of the two filters are parallel, the light polarized by the first filter passes through the second without any further absorption. | 333 | 1,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-18 | latest | en | 0.905073 |
http://www.ck12.org/book/CK-12-Algebra-II-with-Trigonometry-Concepts/section/7.1/ | 1,454,958,694,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701153998.27/warc/CC-MAIN-20160205193913-00213-ip-10-236-182-209.ec2.internal.warc.gz | 339,688,190 | 54,538 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 7.1: Defining nth Roots
Difficulty Level: At Grade Created by: CK-12
Estimated8 minsto complete
%
Progress
Practice nth Roots
Progress
Estimated8 minsto complete
%
The volume of a cube is found to be 343s7\begin{align*}343s^7\end{align*}. What is the length of each side of the cube?
### Guidance
So far, we have seen exponents with integers and the square root. In this concept, we will link roots and exponents. First, let’s define additional roots. Just like the square and the square root are inverses of each other, the inverse of a cube is the cubed root. The inverse of the fourth power is the fourth root.
273=333=3,325=255=2
The nth\begin{align*}n^{th}\end{align*} root of a number, xn\begin{align*}x^n\end{align*}, is x,xnn=x\begin{align*}x, \sqrt[n]{x^n}=x\end{align*}. And, just like simplifying square roots, we can simplify nth\begin{align*}n^{th}\end{align*} roots.
#### Example A
Find 7296\begin{align*}\sqrt [6]{729}\end{align*}.
Solution: To simplify a number to the sixth root, there must be 6 of the same factor to pull out of the root.
729=333333=36\begin{align*}729=3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3=3^6\end{align*}
Therefore, 7296=366=3\begin{align*}\sqrt[6]{729}=\sqrt[6]{3^6}=3\end{align*}. The sixth root and the sixth power cancel each other out. We say that 3 is the sixth root of 729.
From this example, we can see that it does not matter where the exponent is placed, it will always cancel out with the root.
36672963=366 or (36)6=(1.2009)6=3
So, it does not matter if you evaluate the root first or the exponent.
The nth\begin{align*}n^{th}\end{align*} Root Theorem: For any real number a\begin{align*}a\end{align*}, root n\begin{align*}n\end{align*}, and exponent m\begin{align*}m\end{align*}, the following is always true: amn=anm=(an)m\begin{align*}\sqrt[n]{a^m}=\sqrt[n]{a}^m=\left(\sqrt[n]{a}\right)^m\end{align*}.
#### Example B
Evaluate without a calculator:
a) 3235\begin{align*}\sqrt[5]{32^3}\end{align*}
b) 163\begin{align*}\sqrt{16}^3\end{align*}
Solution:
a) If you solve this problem as written, you would first find 323\begin{align*}32^3\end{align*} and then apply the 5th\begin{align*}5^{th}\end{align*} root.
3235=387685=8\begin{align*}\sqrt[5]{32^3}=\sqrt[5]{38768}=8\end{align*}
However, this would be very difficult to do without a calculator. This is an example where it would be easier to apply the root and then the exponent. Let’s rewrite the expression and solve.
3253=23=8\begin{align*}\sqrt[5]{32}^3=2^3=8\end{align*}
b) This problem does not need to be rewritten. 16=4\begin{align*}\sqrt{16}=4\end{align*} and then 43=64\begin{align*}4^3 = 64\end{align*}.
#### Example C
Simplify:
a) 644\begin{align*}\sqrt[4]{64}\end{align*}
b) 54x3125y53\begin{align*}\sqrt[3]{\frac{54x^3}{125y^5}}\end{align*}
Solution:
a) To simplify the fourth root of a number, there must be 4 of the same factor to pull it out of the root. Let’s write the prime factorization of 64 and simplify.
644=2222224=244\begin{align*}\sqrt[4]{64}=\sqrt[{\color{red}4}]{{\color{red}2 \cdot 2\cdot 2\cdot 2}\cdot 2\cdot 2}=2\sqrt[4]{4}\end{align*}
Notice that there are 6 2’s in 64. We can pull out 4 of them and 2 2’s are left under the radical.
b) Just like simplifying fractions with square roots, we can separate the numerator and denominator.
54x3125y53=54x33125y53=2333x33555y3y23=3x235yy23\begin{align*}\sqrt[3]{\frac{{54x^3}}{{125y^5}}}= \frac{\sqrt[3]{54x^3}}{\sqrt[3]{125y^5}}=\frac{\sqrt[{\color{red}3}]{2 \cdot{\color{red}3\cdot 3\cdot 3}\cdot {\color{red}x^3}}}{\sqrt[{\color{blue}3}]{{\color{blue}5\cdot 5\cdot 5\cdot y^3}\cdot y^2}}=\frac{3x\sqrt[3]{2}}{5y\sqrt[3]{y^2}}\end{align*}
Notice that because the x\begin{align*}x\end{align*} is cubed, the cube and cubed root cancel each other out. With the y\begin{align*}y\end{align*}-term, there were five, so three cancel out with the root, but two are still left under radical.
Intro Problem Revisit Recall that the volume of a cube is V=s3\begin{align*}V = s^3\end{align*}, where s is the length of each side. So to find the side length, take the cube root of 343z7\begin{align*}343z^7\end{align*}.
First, you can separate this number into two different roots, 3433z73\begin{align*}\sqrt[3]{343} \cdot \sqrt[3]{z^7}\end{align*}. Now, simplify each root.
3433z73=733z3z3z3=7z2z3\begin{align*}\sqrt[3]{343}\cdot \sqrt[3]{z^7} = \sqrt[3]{7^3}\cdot \sqrt[3]{z^3\cdot z^3 \cdot z}= 7z^2 \sqrt[3]{z}\end{align*}
Therefore, the length of the cube's side is 7z2z3\begin{align*}7z^2 \sqrt[3]{z}\end{align*}.
### Guided Practice
Simplify each expression below, without a calculator.
1. 625z84\begin{align*}\sqrt[4]{625z^8}\end{align*}
2. 32x5y7\begin{align*}\sqrt[7]{32x^5y}\end{align*}
3. 92165\begin{align*}\sqrt[5]{9216}\end{align*}
4. 401753\begin{align*}\sqrt[3]{\frac{40}{175}}\end{align*}
1. First, you can separate this number into two different roots, 6254z84\begin{align*}\sqrt[4]{625} \cdot \sqrt[4]{z^8}\end{align*}. Now, simplify each root.
6254z84=544z4z44=5z2\begin{align*}\sqrt[4]{625}\cdot \sqrt[4]{z^8} = \sqrt[4]{5^4}\cdot \sqrt[4]{z^4\cdot z^4}= 5z^2\end{align*}
When looking at the z8\begin{align*}z^8\end{align*}, think about how many z4\begin{align*}z^4\end{align*} you can even pull out of the fourth root. The answer is 2, or a z2\begin{align*}z^2\end{align*}, outside of the radical.
2. 32=25\begin{align*}32 = 2^5\end{align*}, which means there are not 7 2's that can be pulled out of the radical. Same with the x5\begin{align*}x^5\end{align*} and the y\begin{align*}y\end{align*}. Therefore, you cannot simplify the expression any further.
3. Write out 9216 in the prime factorization and place factors into groups of 5.
92165=2222222222335=2525325=22325=495
4. Reduce the fraction, separate the numerator and denominator and simplify.
401753=8353=233353=235335233523=21225335\begin{align*}\sqrt[3]{\frac{40}{175}}=\sqrt[3]{\frac{8}{35}}=\frac{\sqrt[3]{2^3}}{\sqrt[3]{35}}={\color{red}{\frac{2}{\sqrt[3]{35}}}\cdot} {\color{red}{\frac{\sqrt[3]{35^2}}{\sqrt[3]{35^2}}}}=\frac{2\sqrt[3]{1225}}{35}\end{align*}
In the red step, we rationalized the denominator by multiplying the top and bottom by 3523\begin{align*}\sqrt[3]{35^2}\end{align*}, so that the denominator would be 3533\begin{align*}\sqrt[3]{35^3}\end{align*} or just 35. Be careful when rationalizing the denominator with higher roots!
### Explore More
1. 813\begin{align*}\sqrt[3]{81}\end{align*}
2. \begin{align*}\sqrt[4]{625}^3\end{align*}
3. \begin{align*}\sqrt{9^5}\end{align*}
4. \begin{align*}\sqrt[5]{128}\end{align*}
5. \begin{align*}\sqrt{\sqrt{10000}}\end{align*}
6. \begin{align*}\sqrt{\frac{25}{8}}^4\end{align*}
7. \begin{align*}\sqrt[6]{64}^5\end{align*}
8. \begin{align*}\sqrt[3]{\frac{8}{81}}^2\end{align*}
9. \begin{align*}\sqrt[4]{\frac{243}{16}}\end{align*}
10. \begin{align*}\sqrt[3]{24x^5}\end{align*}
11. \begin{align*}\sqrt[4]{48x^7y^{13}}\end{align*}
12. \begin{align*}\sqrt[5]{\frac{160x^8}{y^7}}\end{align*}
13. \begin{align*}\sqrt[3]{1000x^6}^2\end{align*}
14. \begin{align*}\sqrt[4]{\frac{162x^5}{y^3z^{10}}}\end{align*}
15. \begin{align*}\sqrt{40x^3y^4}^3\end{align*}
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 7.1.
### Vocabulary Language: English
$n^{th}$ root
$n^{th}$ root
The $n^{th}$ root of a number, $x^n$, is $x$. Therefore: $\sqrt[n]{x^n}=x$.
nth root
nth root
The $n^{th}$ root of a number, $x^n$, is $x$. Therefore: $\sqrt[n]{x^n}=x$.
## Date Created:
Mar 12, 2013
Jun 04, 2015
Save or share your relevant files like activites, homework and worksheet.
To add resources, you must be the owner of the Modality. Click Customize to make your own copy. | 2,895 | 7,831 | {"found_math": true, "script_math_tex": 62, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 10, "texerror": 0} | 4.96875 | 5 | CC-MAIN-2016-07 | longest | en | 0.592257 |
https://devel.isa-afp.org/browser_info/current/AFP/Probabilistic_Timed_Automata/Finiteness.html | 1,720,831,300,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514459.28/warc/CC-MAIN-20240712224556-20240713014556-00666.warc.gz | 167,635,791 | 7,655 | # Theory Finiteness
```theory Finiteness
imports Main "HOL-Eisbach.Eisbach_Tools"
begin
section ‹Two Eisbach proof methods for finiteness of sets›
text ‹
The first method is intended to act more conservatively (think ‹safe›), leaving subgoals
for the user where it couldn't proceed any further.
The second method is more powerful, acting more in a succeed-or-die manner,
similarly to ‹force› and friends.
The examples in the second section should give a good impression of where these methods
can help.
›
text ‹This slot is intended to provide more ‹intro› theorems for finite sets.›
named_theorems finite
(* Trick from Dan Matichuk on isabelle-users *)
match termI in H[simproc add: finite_Collect]:_ ⇒ m
(* Trick from Dan Matichuk on isabelle-users.
Turns a structured method into a simple one.
*)
method_setup simple_method =
‹Method.text_closure >> (fn m => fn ctxt =>
let
val facts = Method.get_facts ctxt
val insert' = Method.Basic (K (Method.insert facts))
val m' = Method.Combinator (Method.no_combinator_info, Method.Then, [insert', m])
in Method.evaluate m' ctxt end)›
method finite_tup =
match conclusion in
"finite (_ × _)" ⇒ ‹rule finite_cartesian_product; finite_tup› ¦
"finite S" for S :: "(_ * _) set" ⇒
‹print_term S, (rule finite_subset[where A = S and B = "fst ` S × snd ` S"]; finite_tup?
| (rule finite_subset; assumption?; fastforce))› ¦
"finite X" for X ⇒
‹print_term X, (simp add: image_def, finite_tup?)?,
(solves ‹(rule finite_subset; assumption?; fastforce)›)?› ¦
_ ⇒ ‹fastforce simp: image_def›
method finite_search =
match conclusion in
"finite (_ × _)" ⇒ ‹rule finite_cartesian_product; finite_search› ¦
"finite (_ ` _)" ⇒ ‹simp; finite_search | rule finite_imageI; finite_search› ¦
"finite S" for S :: "(_ * _) set" ⇒
‹print_term S, (solves ‹rule finite_subset; auto›
| rule finite_subset[where A = S and B = "fst ` S × snd ` S"]; finite_tup?)› ¦
"finite (Collect f)" for f ⇒
(solves ‹auto intro: finite›
| print_term v, simp?, rule finite; (assumption | finite_search)
| rule finite_imageI; finite_search
| rule finite_vimageI; finite_search
| print_term x, rule finite_subset; assumption?; fastforce)› ¦
"finite X" for X ⇒
‹print_term X,
(rule finite; (assumption | finite_search)
(solves ‹(rule finite_subset; assumption?; fastforce)›)?)› ¦
_ ⇒ ‹fastforce simp: image_def›
method finite = simple_method finite_search
section ‹Tests›
subsection ‹Counterexamples›
lemma inj_finite_single:
assumes "inj f"
shows "finite {y. x = f y}"
using assms Collect_mem_eq Collect_mono_iff infinite_iff_countable_subset inj_eq not_finite_existsD
rangeI
by fastforce
lemmas inj_finite_single[finite]
text ‹It's hard to guess the right set›
lemma inj_finite_single':
assumes "inj f"
shows "finite {z. f z = x}"
apply (rule finite_subset[of _ "{z. x = f z}"])
apply blast
using assms by finite
(* Due to Lars Hupel *)
definition select :: "('a ⇀ 'b) ⇒ 'a set ⇒ 'b set" where
"select f S = {z | z. ∃x ∈ S. f x = Some z}"
lemma select_finite:
assumes "finite S"
shows "finite (select f S)"
using assms unfolding select_def by finite
lemmas inj_finite_single'[finite]
subsection ‹Working Examples›
lemma
assumes "finite A"
shows "finite {x. x ∈ A ∧ P x}"
using assms by finite_search
lemma collect_pair_finite[finite]:
assumes "finite {x. P x}" "finite {x. Q x}"
shows "finite {(x, y) . P x ∧ Q y ∧ R x y}"
using assms by - finite
lemma collect_pair_finite'[finite]:
assumes "finite {(x, y). P x y}"
shows "finite {(x, y) . P x y ∧ R x y}"
using assms by - finite
text ‹This is what we actually need in this theory›
lemma collect_pair_finite''[finite]:
assumes "finite {(x, y). P x ∧ Q y}"
shows "finite {(x, y) . P x ∧ Q y ∧ R x y}"
using assms by - finite
lemma finite_imageI':
assumes "finite {(x, y). P x y}"
shows "finite {f x y | x y. P x y}"
using assms by finite
lemma
assumes "finite (A × B)"
shows "finite {(x, y) | x y. x ∈ A ∧ y ∈ B ∧ R x y}"
using assms by - finite
lemma finite_imageI'':
assumes "finite (A × B)"
shows "finite {f x y | x y. x ∈ A ∧ y ∈ B ∧ R x y}"
using assms by - finite
text ‹‹finite_Collect› can also rewrite to ‹vimage››
lemma
assumes "inj f" "finite S"
shows "finite {y. ∃ x ∈ S. x = f y}"
using assms by - finite
lemma
assumes "inj f" "finite S"
shows "finite {y. ∃ x ∈ S. f y = x}"
using assms by - finite
text ‹Another counter-example›
lemma
assumes "finite (A × B)"
shows "finite {f x y | x y. x ∈ A ∧ y ∈ B ∧ R x y ∧ Q x y ∧ T x ∧ TT y}" (is "finite ?S")
proof -
have "?S = (λ (x, y). f x y) ` {(x, y). x ∈ A ∧ y ∈ B ∧ R x y ∧ Q x y ∧ T x ∧ TT y}"
by auto
also have "finite …" using assms by - finite
ultimately show ?thesis by simp
qed
text ‹
Easier proof. The problem for our method is that the simproc fails to turn ?S into the form used
in the proof above.
Note that the declaration of the ‹finite› attribute below is the only one that is ∗‹necessary› in
this theory.
›
lemma
notes finite_imageI''[finite]
assumes "finite (A × B)"
shows "finite {f x y | x y. x ∈ A ∧ y ∈ B ∧ R x y ∧ Q x y ∧ T x ∧ TT y}" (is "finite ?S")
using assms by finite
lemma
assumes "finite A" "finite B"
shows "finite {(x, y) | x y. x ∈ A ∧ y ∈ B ∧ R y ∧ S x}"
using assms by - finite
lemma
fixes P Q R :: "'a ⇒ bool"
assumes "finite {x. P x ∧ R x}"
shows "finite {x. P x ∧ Q x ∧ R x}"
using assms by - finite
lemma R:
assumes "finite A" "A = B"
shows "finite B"
using assms by finite
lemma pairwise_finiteI:
assumes "finite {b. ∃a. P a b}" (is "finite ?B")
assumes "finite {a. ∃b. P a b}"
shows "finite {(a,b). P a b}" (is "finite ?C")
using assms by - finite
lemma pairwise_finiteI3:
assumes "finite {b. ∃a c. P a b c}"
assumes "finite {a. ∃b c. P a b c}"
assumes "finite {c. ∃a b. P a b c}"
shows "finite {(a,b,c). P a b c}" (is "finite ?C")
using assms by - finite
lemma pairwise_finiteI4:
assumes "finite {b. ∃a c d. P a b c d}"
assumes "finite {a. ∃b c d. P a b c d}"
assumes "finite {c. ∃a b d. P a b c d}"
assumes "finite {d. ∃a b c. P a b c d}"
shows "finite {(a,b,c,d). P a b c d}" (is "finite ?C")
using assms by - finite
lemma finite_ex_and1:
assumes "finite {b. ∃a. P a b}" (is "finite ?A")
shows "finite {b. ∃a. P a b ∧ Q a b}" (is "finite ?B")
using assms by - finite
lemma finite_ex_and2:
assumes "finite {b. ∃a. Q a b}" (is "finite ?A")
shows "finite {b. ∃a. P a b ∧ Q a b}" (is "finite ?B")
using assms by - finite
text ‹
This is the only lemma where our methods cannot help us so far due to the fairly
complex argument that is used in the interactive proof.
›
lemma finite_set_of_finite_funs2:
fixes A :: "'a set"
and B :: "'b set"
and C :: "'c set"
and d :: "'c"
assumes "finite A"
and "finite B"
and "finite C"
shows "finite {f. ∀x. ∀y. (x ∈ A ∧ y ∈ B ⟶ f x y ∈ C) ∧ (x ∉ A ⟶ f x y = d) ∧ (y ∉ B ⟶ f x y = d)}"
(is "finite ?S")
proof -
let ?R = "{g. ∀x. (x ∈ B ⟶ g x ∈ C) ∧ (x ∉ B ⟶ g x = d)}"
let ?Q = "{f. ∀x. (x ∈ A ⟶ f x ∈ ?R) ∧ (x ∉ A ⟶ f x = (λy. d))}"
from finite_set_of_finite_funs[OF assms(2,3)] have "finite ?R" .
from finite_set_of_finite_funs[OF assms(1) this, of "λ y. d"] have "finite ?Q" .
moreover have "?S = ?Q" by auto (case_tac "xa ∈ A", auto)
ultimately show ?thesis by simp
qed
end
``` | 2,379 | 7,098 | {"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} | 3.234375 | 3 | CC-MAIN-2024-30 | latest | en | 0.728872 |
https://ch.mathworks.com/matlabcentral/profile/authors/26676394 | 1,696,113,879,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510730.6/warc/CC-MAIN-20230930213821-20231001003821-00132.warc.gz | 179,052,720 | 21,195 | Community Profile
# Peter
### MathWorks
Last seen: 1 year ago Active since 2022
#### Content Feed
View by
Solved
Hackathon - String version
I have a function that creates a random string of characters of ASCII values 32-127. The length of the string is also randomly d...
1 year ago
Solved
Draw a 'Z'.
Given _n_ as input, generate a n-by-n matrix like 'Z' by _0_ and _1_ . Example: n=5 ans= [1 1 1 1 1 0 0 0 1 ...
1 year ago
Solved
Draw a 'X'!
Given n as input Draw a 'X' in a n-by-n matrix. example: n=3 y=[1 0 1 0 1 0 1 0 1] n=4 y=[1 0 0...
1 year ago
Solved
Draw 'O' !
Given n as input, generate a n-by-n matrix 'O' using 0 and 1 . example: n=4 ans= [1 1 1 1 1 0 0 1 ...
1 year ago
Solved
Draw a 'N'!
Given n as input, generate a n-by-n matrix 'N' using 0 and 1 . Example: n=5 ans= [1 0 0 0 1 1 1 0 0 1 1 0 ...
1 year ago
Solved
Draw 'J'
Given n as input, generate a n-by-n matrix 'J' using 0 and 1 . Example: n=5 ans= [0 0 0 0 1 0 0 0 0 1 0 0 ...
1 year ago
Solved
Draw 'I'
Given n as input, draw a n-by-n matrix 'I' using 0 and 1. example: n=3 ans= [0 1 0 0 1 0 0 1 0] n=...
1 year ago
Solved
Draw 'H'
Draw a x-by-x matrix 'H' using 1 and 0. (x is odd and bigger than 2) Example: x=5 ans= [1 0 0 0 1 1 0 0 0 1 ...
1 year ago
Solved
Draw 'F'
Draw a x-by-x matrix 'F' using 1 and 0. (x is odd and bigger than 4) Example: x=5 ans= [1 1 1 1 1 1 0 0 0 0 ...
1 year ago
Solved
Draw 'E'
Draw a x-by-x matrix 'E' using 1 and 0. (x is odd and bigger than 4) Example: x=5 ans= [1 1 1 1 1 1 0 0 0 0 ...
1 year ago
Solved
Draw 'D'.
Draw a x-by-x matrix 'D' using 0 and 1. example: x=4 ans= [1 1 1 0 1 0 0 1 1 0 0 1 1 1 1 0]
1 year ago
Solved
Draw 'C'.
Given x as input, generate a x-by-x matrix 'C' using 0 and 1. example: x=4 ans= [0 1 1 1 1 0 0 0 ...
1 year ago
Solved
Draw 'B'
Draw a x-by-x matrix 'B' using 1 and 0. (x is odd and bigger than 4) Example: x=5 ans= [1 1 1 1 0 1 0 0 0 1 ...
1 year ago
Solved
Sum all integers from 1 to 2^n
Given the number x, y must be the summation of all integers from 1 to 2^x. For instance if x=2 then y must be 1+2+3+4=10.
1 year ago
Solved
Magic is simple (for beginners)
Determine for a magic square of order n, the magic sum m. For example m=15 for a magic square of order 3.
1 year ago
Solved
Make a random, non-repeating vector.
This is a basic MATLAB operation. It is for instructional purposes. --- If you want to get a random permutation of integer...
1 year ago
Solved
Roll the Dice!
*Description* Return two random integers between 1 and 6, inclusive, to simulate rolling 2 dice. *Example* [x1,x2] =...
1 year ago
Solved
Number of 1s in a binary string
Find the number of 1s in the given binary string. Example. If the input string is '1100101', the output is 4. If the input stri...
1 year ago
Solved
Return the first and last characters of a character array
Return the first and last character of a string, concatenated together. If there is only one character in the string, the functi...
1 year ago
Solved
Create times-tables
At one time or another, we all had to memorize boring times tables. 5 times 5 is 25. 5 times 6 is 30. 12 times 12 is way more th...
1 year ago
Solved
Getting the indices from a vector
This is a basic MATLAB operation. It is for instructional purposes. --- You may already know how to <http://www.mathworks....
1 year ago
Solved
Determine whether a vector is monotonically increasing
Return true if the elements of the input vector increase monotonically (i.e. each element is larger than the previous). Return f...
1 year ago
Solved
Check if number exists in vector
Return 1 if number _a_ exists in vector _b_ otherwise return 0. a = 3; b = [1,2,4]; Returns 0. a = 3; b = [1,...
1 year ago
Solved
Swap the first and last columns
Flip the outermost columns of matrix A, so that the first column becomes the last and the last column becomes the first. All oth...
1 year ago
Solved
Swap the input arguments
Write a two-input, two-output function that swaps its two input arguments. For example: [q,r] = swap(5,10) returns q = ...
1 year ago
Solved
Column Removal
Remove the nth column from input matrix A and return the resulting matrix in output B. So if A = [1 2 3; 4 5 6]; ...
1 year ago
Solved
Reverse the vector
Reverse the vector elements. Example: Input x = [1,2,3,4,5,6,7,8,9] Output y = [9,8,7,6,5,4,3,2,1]
1 year ago
Solved
Select every other element of a vector
Write a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s...
1 year ago
Solved
Length of the hypotenuse
Given short sides of lengths a and b, calculate the length c of the hypotenuse of the right-angled triangle. <<https://i.imgu...
1 year ago
Solved
Triangle Numbers
Triangle numbers are the sums of successive integers. So 6 is a triangle number because 6 = 1 + 2 + 3 which can be displa...
1 year ago | 1,654 | 4,912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.735972 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.