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# 2_1_2_2 - Sections 2.1 and 2.2 Limit Instructor Ms Hoa...
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Sections 2.1 and 2.2: Limit Instructor: Ms. Hoa Nguyen What is Calculus? Calculus is the mathematics of motion and change. Why is Limit Theory the basis of Calculus? The major concepts of Calculus such as continuity, derivative and definite integral are all defined by limit. How do limits arise in the tangent problem? The tangent problem: Given a point P on a curve C , find an equation of the tangent line to the curve C at the point P . Steps : 1) Draw a secant line PQ where Q is a nearby point. 2) The slope m of the tangent line at P is the limit of the slopes of the secant line PQ as Q approaches P ( slope = rise over run ). Remark : The slope m of the tangent line to the curve y = f ( x ) at the point P ( a, f ( a )) is: m = lim x a f ( x ) - f ( a ) x - a Let h = x - a , then m = lim h 0 f ( a + h ) - f ( a ) h 3) The equation of the tangent line at the point P is an equation of the line passing through the point P and having slope m : y - y P = m ( x - x P ) ( point-slope form of the equation of a line ).
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https://www.molympiad.net/2019/05/mathematical-danube-competition-juniors-2016.html | 1,723,351,552,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640975657.64/warc/CC-MAIN-20240811044203-20240811074203-00322.warc.gz | 686,504,178 | 160,128 | ## $hide=mobile$type=ticker$c=12$cols=3$l=0$sr=random$b=0 # ĐẶT MUA TẠP CHÍ / PURCHASE JOURNALS 1. Let$S=x_{1} x_{2}+x_{3} x_{4}+\ldots+x_{2015} x_{2016}$, where$x_{1}, x_{2} \ldots, x_{2006} \in \{\sqrt{3}-\sqrt{2}, \sqrt{3}+\sqrt{2}\} .$Is the equality$S=2016$possible? 2. Determine the poxitive integers$n>1$such that, for any divisar$d$of$n$, the numbers$d^{2} d+1$and$d^{2}-d+1$are prime. 3. Let$A B C$' be a triangle with$A B < A C$,$I$its incenter, and$M$the midpoint of the side$B C$. If$I A=I M$, determine the smallest possible value of the angle$AIM$. 4. A unit square is removed from the corner of the$n \times n$grid where$n \geq 2$. Prowe that the remainder can be covered by copies of the$L$-shapes consisting of$3$or$5$unit squares depicted in the figure, Every aquare must be covered once and the$L$-shapes must not go over the bounds of the grid. ##$hide=mobile$type=ticker$c=36$cols=2$l=0$sr=random$b=0
Name
Abel,5,Albania,2,AMM,2,Amsterdam,4,An Giang,45,Andrew Wiles,1,Anh,2,APMO,21,Austria (Áo),1,Ba Lan,1,Bà Rịa Vũng Tàu,77,Bắc Bộ,2,Bắc Giang,62,Bắc Kạn,4,Bạc Liêu,18,Bắc Ninh,53,Bắc Trung Bộ,3,Bài Toán Hay,5,Balkan,41,Baltic Way,32,BAMO,1,Bất Đẳng Thức,69,Bến Tre,72,Benelux,16,Bình Định,65,Bình Dương,38,Bình Phước,52,Bình Thuận,42,Birch,1,BMO,41,Booklet,12,Bosnia Herzegovina,3,BoxMath,3,Brazil,2,British,16,Bùi Đắc Hiên,1,Bùi Thị Thiện Mỹ,1,Bùi Văn Tuyên,1,Bùi Xuân Diệu,1,Bulgaria,6,Buôn Ma Thuột,2,BxMO,15,Cà Mau,22,Cần Thơ,27,Canada,40,Cao Bằng,12,Cao Quang Minh,1,Câu Chuyện Toán Học,43,Caucasus,3,CGMO,11,China - Trung Quốc,25,Chọn Đội Tuyển,515,Chu Tuấn Anh,1,Chuyên Đề,125,Chuyên SPHCM,7,Chuyên SPHN,30,Chuyên Trần Hưng Đạo,3,Collection,8,College Mathematic,1,Concours,1,Cono Sur,1,Contest,675,Correspondence,1,Cosmin Poahata,1,Crux,2,Czech-Polish-Slovak,28,Đà Nẵng,50,Đa Thức,2,Đại Số,20,Đắk Lắk,76,Đắk Nông,15,Danube,7,Đào Thái Hiệp,1,ĐBSCL,2,Đề Thi,1,Đề Thi HSG,2249,Đề Thi JMO,1,DHBB,30,Điện Biên,15,Định Lý,1,Định Lý Beaty,1,Đỗ Hữu Đức Thịnh,1,Do Thái,3,Doãn Quang Tiến,5,Đoàn Quỳnh,1,Đoàn Văn Trung,1,Đồng Nai,64,Đồng Tháp,63,Du Hiền Vinh,1,Đức,1,Dương Quỳnh Châu,1,Dương Tú,1,Duyên Hải Bắc Bộ,30,E-Book,31,EGMO,30,ELMO,19,EMC,11,Epsilon,1,Estonian,5,Euler,1,Evan Chen,1,Fermat,3,Finland,4,Forum Of Geometry,2,Furstenberg,1,G. 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2011-2012,44,HSG 12 2012-2013,58,HSG 12 2013-2014,53,HSG 12 2014-2015,44,HSG 12 2015-2016,37,HSG 12 2016-2017,46,HSG 12 2017-2018,55,HSG 12 2018-2019,43,HSG 12 2019-2020,43,HSG 12 2020-2021,52,HSG 12 2021-2022,35,HSG 12 2022-2023,42,HSG 12 2023-2024,23,HSG 12 2023-2041,1,HSG 12 An Giang,8,HSG 12 Bà Rịa Vũng Tàu,13,HSG 12 Bắc Giang,18,HSG 12 Bạc Liêu,3,HSG 12 Bắc Ninh,13,HSG 12 Bến Tre,19,HSG 12 Bình Định,17,HSG 12 Bình Dương,8,HSG 12 Bình Phước,9,HSG 12 Bình Thuận,8,HSG 12 Cà Mau,7,HSG 12 Cần Thơ,7,HSG 12 Cao Bằng,5,HSG 12 Chuyên SPHN,11,HSG 12 Đà Nẵng,3,HSG 12 Đắk Lắk,21,HSG 12 Đắk Nông,1,HSG 12 Điện Biên,3,HSG 12 Đồng Nai,20,HSG 12 Đồng Tháp,18,HSG 12 Gia Lai,14,HSG 12 Hà Nam,5,HSG 12 Hà Nội,17,HSG 12 Hà Tĩnh,16,HSG 12 Hải Dương,16,HSG 12 Hải Phòng,20,HSG 12 Hậu Giang,4,HSG 12 Hòa Bình,10,HSG 12 Hưng Yên,10,HSG 12 Khánh Hòa,4,HSG 12 KHTN,26,HSG 12 Kiên Giang,12,HSG 12 Kon Tum,3,HSG 12 Lai Châu,4,HSG 12 Lâm Đồng,11,HSG 12 Lạng Sơn,8,HSG 12 Lào Cai,17,HSG 12 Long An,18,HSG 12 Nam Định,7,HSG 12 Nghệ An,13,HSG 12 Ninh Bình,12,HSG 12 Ninh Thuận,7,HSG 12 Phú Thọ,18,HSG 12 Phú Yên,13,HSG 12 Quảng Bình,14,HSG 12 Quảng Nam,11,HSG 12 Quảng Ngãi,6,HSG 12 Quảng Ninh,20,HSG 12 Quảng Trị,10,HSG 12 Sóc Trăng,4,HSG 12 Sơn La,5,HSG 12 Tây Ninh,6,HSG 12 Thái Bình,11,HSG 12 Thái Nguyên,13,HSG 12 Thanh Hóa,17,HSG 12 Thừa Thiên Huế,19,HSG 12 Tiền Giang,3,HSG 12 TPHCM,13,HSG 12 Tuyên Quang,3,HSG 12 Vĩnh Long,7,HSG 12 Vĩnh Phúc,20,HSG 12 Yên Bái,6,HSG 9,573,HSG 9 2009-2010,1,HSG 9 2010-2011,21,HSG 9 2011-2012,42,HSG 9 2012-2013,41,HSG 9 2013-2014,35,HSG 9 2014-2015,41,HSG 9 2015-2016,38,HSG 9 2016-2017,42,HSG 9 2017-2018,45,HSG 9 2018-2019,41,HSG 9 2019-2020,18,HSG 9 2020-2021,50,HSG 9 2021-2022,53,HSG 9 2022-2023,55,HSG 9 2023-2024,15,HSG 9 An Giang,9,HSG 9 Bà Rịa Vũng Tàu,8,HSG 9 Bắc Giang,14,HSG 9 Bắc Kạn,1,HSG 9 Bạc Liêu,1,HSG 9 Bắc Ninh,12,HSG 9 Bến Tre,9,HSG 9 Bình Định,11,HSG 9 Bình Dương,7,HSG 9 Bình Phước,13,HSG 9 Bình Thuận,5,HSG 9 Cà Mau,2,HSG 9 Cần Thơ,4,HSG 9 Cao 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UTF-8 | 7,009 | 15,112 | {"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} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.430066 |
https://jp.mathworks.com/matlabcentral/answers/885029-i-m-trying-to-convert-a-1-dimension-lat-lon-data-to-2d-data-without-interpolating-without-griddata?s_tid=srchtitle | 1,632,615,645,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057787.63/warc/CC-MAIN-20210925232725-20210926022725-00628.warc.gz | 380,830,676 | 26,060 | # I'm trying to convert a 1 dimension lat,lon,data to 2d data without interpolating (without griddata command). Is this possible?
5 ビュー (過去 30 日間)
DIPENDRA007 2021 年 7 月 24 日
I have a one dimensional latitude,longitude, data. Essentially all are in one dimension. However I want to create a 2D plot with the data.
So, what I want is the data should be in 2D format without using the griddate which uses interpolation. So, in 2D format of data there should be original value of 1D data present at every original latitude, longitude point and other points it will be NaN or zero.
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### 採用された回答
Chunru 2021 年 7 月 24 日
Updated solution:
dg = 1;
longrid = (-180:dg:180);
latgrid = (-90:dg:90);
z = nan(length(latgrid), length(longrid));
for ilat=1:length(latgrid)
for ilon=1:length(longrid)
ii = T1.LATITUDE-latgrid(ilat)>=-dg/2 & T1.LATITUDE-latgrid(ilat)<dg/2 & ...
T1.LONGITUDE-longrid(ilon)>=-dg/2 & T1.LONGITUDE-longrid(ilon)<dg/2;
z(ilat, ilon) = mean(T1.DATA(ii));
end
end
figure
imagesc(longrid, latgrid, z);
axis xy
xlabel('lon');
ylabel('lat');
colorbar
figure;
cmap = turbo(1024); cmap = cmap(257:768,:);
ax = axesm('MapProjection','apianus','MapLatLimit',[-90 90], 'MapLonLimit', [-180 180]);
[lon, lat] = meshgrid(longrid, latgrid);
geoshow(ax, lat, lon, z, cmap , 'DisplayType', 'image')
Your description of the problem is not detailed. Is this what you want?
n = 100;
lat = rand(n, 1);
lon = rand(n, 1);
z = rand(n, 1);
figure
stem3(lon, lat, z);
xlabel('lon'); ylabel('lat'); zlabel('z');
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Translated by | 525 | 1,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.433428 |
https://www.jiskha.com/display.cgi?id=1227734431 | 1,511,569,089,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809160.77/warc/CC-MAIN-20171124234011-20171125014011-00503.warc.gz | 810,184,828 | 3,443 | math
posted by .
lula wants to round 76.24491 to the nearest hundreth. how do you do it?
• math -
76.24491 = 76.24
Remember that after the decimal point are:
tenths
hundredths
thousandths
ten-thousandths
You look at the number after the hundredths place. If it's 4 or less, you drop it. If it's 5 or more, you round up to the next hundredth.
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More Similar Questions | 490 | 1,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-47 | latest | en | 0.886991 |
https://younesse.net/assets/Slides/Deep-Signalling-Pathway-Slides/Final_Presentation.html | 1,553,518,183,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203947.59/warc/CC-MAIN-20190325112917-20190325134917-00220.warc.gz | 964,263,457 | 21,486 | # Why does evolution favor deep signalling pathways?
## Example: the MAPK/ERK pathway
Animation courtsey of Ribosome Studio Youtube channel
### Why does evolution favor deep signalling pathways?
Natural selection: Deeper stories $⟹$ Improving biological fitness
BUT counter-intuitive
• energetically
• probabilistically: Indeed: if
\begin{align*} & 𝔼(\max_i \depth(P_i)) \\ & ≤ 𝔼\Big(\ln\Big(\underbrace{\sum\limits_{ i=0 }^n \exp(\depth(P_i))}_{\; ≝ \; α_n}\Big)\Big) \\ & ≤ \ln \underbrace{𝔼(α_n)}_{\rlap{\substack{= \; 𝔼(α_{n-1}) + 𝔼(\exp(\depth(P_n))) \\ = \, (1 + \frac e i) \, 𝔼(α_{n-1})}}}\\ & = \ln \Big(\prod\limits_{ i=1 }^n \Big(\underbrace{1 + \frac e i}_{≤\, \exp (\frac e i)}\Big) \Big) = e H_n \sim \boxed{e \ln n} \end{align*}
Possible explanations:
Greater depth:
$$\, \\ \\ \\ \begin{cases} \text{ enhances stability } \\ \text{ deletions make pathways non-functional more often than additions do } \end{cases}$$
# II. Theoretical model
Does refinement increase depth?
## Genetic algorithm
An heuristic optimisation that loop on two steps:
• MUTATION: map ancesters to $n$ descendent with random mutations
• SELECTION: filter descendent on their fitness (ERK quantity)
Model biais: biological clock assumption, spontaneous discrete generation, etc …
## III. Refinements
$$\MEK(x_{pp}), \MAPK(x_{p}) ⟶_{τ_b} \MEK(x_{pp}^{\color{DeepPink}{1}}), \MAPK(x_{p}^{\color{DeepPink}{1}})\\ \color{DarkCyan}{\MEK(x_{pp}^1), \MAPK(x_{p}^1) ⟶_{τ_u} \MEK(x_{pp}), \MAPK(x_{p})}\\ \MEK(x_{pp}^1), \MAPK(x_{p}^1) ⟶_{τ_p} \MEK(x_{pp}^1), \color{OrangeRed}{\MAPK(x_{pp}^1)}\\$$
• $\color{DarkCyan}{τ_u}$ too big: unbind before phosphorylation (too liquid)
• $\color{DarkCyan}{τ_u}$ too small: never unbind (too sticky)
## Category of site graphs: $\SGph$
Objects $(V, λ, σ, μ)$:
• $V$: set of nodes
• $λ: V ⟶ 𝒜$: name assignment
• $σ: V ⟶ 𝒫(𝒮)$: site assignment
• $μ: \underbrace{\text{ matching}}_{\rlap{\text{irreflexive symmetric binary relation}}}$ over $\sum\limits_{ v ∈ V } σ(v)$
Morphism $f: (V, λ, σ, μ) ⟶ (V', λ', σ', μ')$:
Name/site/edge-preserving and edge reflecting monomorphism $f: V ⟶ V'$
Signature $Σ: 𝒜 ⟶ 𝒮$:
• $x ≤ Σ \quad ⟺ \quad σ_x(V_x) ⊆ Σ(λ_x(V_x))$
• $Σ ≤ x \quad ⟺ \quad Σ(λ_x(V_x)) ⊆ σ_x(V_x)$
### Epimorphisms
Forgetful functor to the category of graphs:
$$U: \SGph ⟶ \Gph$$
Epimorphisms from $x$ to $y$:
$$[x, y]^e \; ≝ \; \big\lbrace h ∈ \underbrace{\Hom[\SGph]{x, y}}_{\text{denoted by } [x,y]} \; \mid \; h \text{ is an epi (i.e. right-cancellable)} \big\rbrace$$
Lemma:
$$h ∈ \overbrace{\Hom[\SGph]{x, y}}^{\text{denoted by } [x,y]} \text{ is an epi } \\ \; ⟺ \; ∀ c_y ⊆ y \text{ connected component}, \, h^{-1}(c_y) ≠ ∅$$
### Epi-mono factorisation
Factorisation:
$f ∈ [s, x]$ is said to be factored by $t$ if $f = γ ϕ$ for $ϕ ∈ [s, t]^e$ and $γ ∈ [t, x]$
Every iso $α ∈ [t, t']$ conjugates the factorisations $ϕ, γ$ and $ϕ', γ'$:
⟹ Equivalence relation:
$$ϕ, γ \; ≃_{tt'} \; ϕ', γ'$$
The group $[t, t]$ acts freely on $[s, t]^e × [t, x]$ (as we have epis and monos), so by Burnside’s lemma:
$$\vert \underbrace{[s, t]^e × [t, x] / [t, t]}_{\text{denoted by } [s, t]^e ×_{[t, t]} [t, x]} \vert = \frac 1 {\vert [t,t] \vert} \sum\limits_{ (ϕ, γ) ∈ [s, t]^e × [t, x]} \underbrace{\vert Stab_{[t, t]}((ϕ, γ)) \vert}_{= 1} \\ = \vert [s, t]^e × [t, x] \vert/\vert [t,t] \vert$$
### Object refinement
A refinement $Σ(s)$ of $s ≤ Σ$ under $Σ$:
is a set comprised of one element per isomorphism class in $\lbrace t : Σ \; \mid \; [s, t]^e ≠ ∅ \rbrace$
Theorem: if $s ≤ Σ$ and $x : Σ$:
$$[s,x] \; ≅ \; \sum\limits_{ t ∈ Σ(s) } [s, t]^e ×_{[t,t]} [t, x]$$
### Rule refinement
Well defined atomic action (in a rewriting rule):
Labelled transition:
$$x \quad ⟶_f^R \quad f(α) \cdot x$$
where
• the rule $R \; ≝ \; \underbrace{s}_{\text{object}}, \underbrace{α}_{\text{action}}, \underbrace{τ}_{\text{rate}}$, whose activity $a(x, r) \; ≝ \; τ \vert [s, x] \vert$
• $f ∈ [s, x]$
If $θ$ is an iso: $θ(r) \; ≝ \; θ(s), θ(α), τ$ satisfies:
$$x \, ⟶_f^R \, f(α) \cdot x \qquad ⟺ \qquad x \, ⟶_{fθ^{-1}}^{θ(R)} \, fθ^{-1}(θ(α)) \cdot x$$
Rule refinement:
If $s ≤ Σ$, the refinement of the rule $R \; ≝ \; s, α, τ$ under $Σ$ is:
$$Σ(s, α, τ) \; ≝ \; (t, ϕ(α), τ)_{t ∈ Σ(s), ϕ ∈ [s,t]^e/[t,t]}$$
# IV. Methods
Our naive code implementation was split between:
• KaSimir: Genetic algorithm implementation, using KaSim to compute fitness
• KaStorama: wrapper around KaStor to retreive causal story depth from KaSim traces
• KaRapuce: Slitghly change rates of refined rules as mutation event
https://github.com/yvan-sraka/KaSimir
Demo time!
# Bibliography
• P. Boutillier, J. Feret, J. Krivine, and W. Fontana, “The Kappa Language
and Kappa Tools,” p. 52.
• “Signaling Pathways,” Tocris Bioscience. https://www.tocris.com/signaling-pathways.
• V. Danos, J. Feret, W. Fontana, R. Harmer, and J. Krivine, “Rule-Based
Modelling, Symmetries, Refinements,” in Formal Methods in Systems
Biology
, vol. 5054, J. Fisher, Ed. Berlin, Heidelberg: Springer Berlin
Heidelberg, 2008, pp. 103–122.
• E. Murphy, V. Danos, J. Féret, J. Krivine, and R. Harmer, “Rule-Based
Modeling and Model Refinement,” in Elements of Computational Systems
Biology
, H. M. Lodhi and S. H. Muggleton, Eds. Hoboken, NJ, USA: John
Wiley & Sons, Inc., 2010, pp. 83–114.
• “List of signalling pathways,” Wikipedia. 30-Nov-2016. | 2,061 | 5,361 | {"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} | 2.609375 | 3 | CC-MAIN-2019-13 | longest | en | 0.524598 |
http://au.metamath.org/mpegif/stoweidlem20.html | 1,513,212,495,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948532873.43/warc/CC-MAIN-20171214000736-20171214020736-00500.warc.gz | 24,348,259 | 36,321 | Mathbox for Glauco Siliprandi < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > stoweidlem20 Unicode version
Theorem stoweidlem20 27872
Description: If a set A of real functions from a common domain T is closed under the sum of two functions, then it is closed under the sum of a finite number of functions, indexed by G. (Contributed by Glauco Siliprandi, 20-Apr-2017.)
Hypotheses
Ref Expression
stoweidlem20.1
stoweidlem20.2
stoweidlem20.3
stoweidlem20.4
stoweidlem20.5
stoweidlem20.6
Assertion
Ref Expression
stoweidlem20
Distinct variable groups: ,,,, ,, ,,,, ,,, ,,
Allowed substitution hints: () (,) (,,,) (,)
Proof of Theorem stoweidlem20
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 stoweidlem20.2 . 2
2 id 19 . . . 4
3 stoweidlem20.3 . . . . . 6
4 nnre 9769 . . . . . 6
53, 4syl 15 . . . . 5
6 leid 8932 . . . . 5
75, 6syl 15 . . . 4
82, 7jca 518 . . 3
9 eleq1 2356 . . . . . . 7
10 breq1 4042 . . . . . . . . 9
1110anbi2d 684 . . . . . . . 8
12 oveq2 5882 . . . . . . . . . . 11
1312sumeq1d 12190 . . . . . . . . . 10
1413mpteq2dv 4123 . . . . . . . . 9
1514eleq1d 2362 . . . . . . . 8
1611, 15imbi12d 311 . . . . . . 7
179, 16imbi12d 311 . . . . . 6
18 breq1 4042 . . . . . . . . . 10
1918anbi2d 684 . . . . . . . . 9
20 oveq2 5882 . . . . . . . . . . . 12
2120sumeq1d 12190 . . . . . . . . . . 11
2221mpteq2dv 4123 . . . . . . . . . 10
2322eleq1d 2362 . . . . . . . . 9
2419, 23imbi12d 311 . . . . . . . 8
25 breq1 4042 . . . . . . . . . 10
2625anbi2d 684 . . . . . . . . 9
27 oveq2 5882 . . . . . . . . . . . 12
2827sumeq1d 12190 . . . . . . . . . . 11
2928mpteq2dv 4123 . . . . . . . . . 10
3029eleq1d 2362 . . . . . . . . 9
3126, 30imbi12d 311 . . . . . . . 8
32 breq1 4042 . . . . . . . . . 10
3332anbi2d 684 . . . . . . . . 9
34 oveq2 5882 . . . . . . . . . . . 12
3534sumeq1d 12190 . . . . . . . . . . 11
3635mpteq2dv 4123 . . . . . . . . . 10
3736eleq1d 2362 . . . . . . . . 9
3833, 37imbi12d 311 . . . . . . . 8
39 breq1 4042 . . . . . . . . . 10
4039anbi2d 684 . . . . . . . . 9
41 oveq2 5882 . . . . . . . . . . . 12
4241sumeq1d 12190 . . . . . . . . . . 11
4342mpteq2dv 4123 . . . . . . . . . 10
4443eleq1d 2362 . . . . . . . . 9
4540, 44imbi12d 311 . . . . . . . 8
46 stoweidlem20.1 . . . . . . . . . . . 12
47 1z 10069 . . . . . . . . . . . . . . 15
4847a1i 10 . . . . . . . . . . . . . 14
49 stoweidlem20.4 . . . . . . . . . . . . . . . . . . . . . 22
50 nnuz 10279 . . . . . . . . . . . . . . . . . . . . . . . 24
513, 50syl6eleq 2386 . . . . . . . . . . . . . . . . . . . . . . 23
52 eluzfz1 10819 . . . . . . . . . . . . . . . . . . . . . . 23
5351, 52syl 15 . . . . . . . . . . . . . . . . . . . . . 22
5449, 53jca 518 . . . . . . . . . . . . . . . . . . . . 21
55 ffvelrn 5679 . . . . . . . . . . . . . . . . . . . . 21
5654, 55syl 15 . . . . . . . . . . . . . . . . . . . 20
572, 56jca 518 . . . . . . . . . . . . . . . . . . 19
58 eleq1 2356 . . . . . . . . . . . . . . . . . . . . . . 23
5958anbi2d 684 . . . . . . . . . . . . . . . . . . . . . 22
60 feq1 5391 . . . . . . . . . . . . . . . . . . . . . 22
6159, 60imbi12d 311 . . . . . . . . . . . . . . . . . . . . 21
62 stoweidlem20.6 . . . . . . . . . . . . . . . . . . . . . 22
6362a1i 10 . . . . . . . . . . . . . . . . . . . . 21
6461, 63vtoclga 2862 . . . . . . . . . . . . . . . . . . . 20
6556, 64syl 15 . . . . . . . . . . . . . . . . . . 19
6657, 65mpd 14 . . . . . . . . . . . . . . . . . 18
6766adantr 451 . . . . . . . . . . . . . . . . 17
68 simpr 447 . . . . . . . . . . . . . . . . 17
6967, 68jca 518 . . . . . . . . . . . . . . . 16
70 ffvelrn 5679 . . . . . . . . . . . . . . . 16
7169, 70syl 15 . . . . . . . . . . . . . . 15
72 recn 8843 . . . . . . . . . . . . . . 15
7371, 72syl 15 . . . . . . . . . . . . . 14
7448, 73jca 518 . . . . . . . . . . . . 13
75 fveq2 5541 . . . . . . . . . . . . . . 15
7675fveq1d 5543 . . . . . . . . . . . . . 14
7776fsum1 12230 . . . . . . . . . . . . 13
7874, 77syl 15 . . . . . . . . . . . 12
7946, 78mpteq2da 4121 . . . . . . . . . . 11
80 ffn 5405 . . . . . . . . . . . . 13
8166, 80syl 15 . . . . . . . . . . . 12
82 dffn5 5584 . . . . . . . . . . . 12
8381, 82sylib 188 . . . . . . . . . . 11
8479, 83eqtr4d 2331 . . . . . . . . . 10
8584, 56eqeltrd 2370 . . . . . . . . 9
8685adantr 451 . . . . . . . 8
87 simprl 732 . . . . . . . . . . . . 13
88 simpll 730 . . . . . . . . . . . . 13
89 simprr 733 . . . . . . . . . . . . 13
9087, 88, 893jca 1132 . . . . . . . . . . . 12
91 simp1 955 . . . . . . . . . . . . . . 15
92 nnre 9769 . . . . . . . . . . . . . . . . . . 19
93923ad2ant2 977 . . . . . . . . . . . . . . . . . 18
9493lep1d 9704 . . . . . . . . . . . . . . . . 17
95 simp3 957 . . . . . . . . . . . . . . . . 17
9694, 95jca 518 . . . . . . . . . . . . . . . 16
97 1re 8853 . . . . . . . . . . . . . . . . . . . 20
9897a1i 10 . . . . . . . . . . . . . . . . . . 19
9993, 98readdcld 8878 . . . . . . . . . . . . . . . . . 18
10033ad2ant1 976 . . . . . . . . . . . . . . . . . . 19
101100, 4syl 15 . . . . . . . . . . . . . . . . . 18
10293, 99, 1013jca 1132 . . . . . . . . . . . . . . . . 17
103 letr 8930 . . . . . . . . . . . . . . . . 17
104102, 103syl 15 . . . . . . . . . . . . . . . 16
10596, 104mpd 14 . . . . . . . . . . . . . . 15
10691, 105jca 518 . . . . . . . . . . . . . 14
10790, 106syl 15 . . . . . . . . . . . . 13
108 simplr 731 . . . . . . . . . . . . 13
109107, 108mpd 14 . . . . . . . . . . . 12
11090, 109jca 518 . . . . . . . . . . 11
111 nfv 1609 . . . . . . . . . . . . . . 15
112 nfv 1609 . . . . . . . . . . . . . . 15
11346, 111, 112nf3an 1786 . . . . . . . . . . . . . 14
114 simpl2 959 . . . . . . . . . . . . . . . . 17
11550eleq2i 2360 . . . . . . . . . . . . . . . . 17
116114, 115sylib 188 . . . . . . . . . . . . . . . 16
117 simpll1 994 . . . . . . . . . . . . . . . . . 18
11847a1i 10 . . . . . . . . . . . . . . . . . . . . 21
119 nnz 10061 . . . . . . . . . . . . . . . . . . . . . . . . 25
1203, 119syl 15 . . . . . . . . . . . . . . . . . . . . . . . 24
1211203ad2ant1 976 . . . . . . . . . . . . . . . . . . . . . . 23
122121adantr 451 . . . . . . . . . . . . . . . . . . . . . 22
123122adantr 451 . . . . . . . . . . . . . . . . . . . . 21
124 elfzelz 10814 . . . . . . . . . . . . . . . . . . . . . 22
125124adantl 452 . . . . . . . . . . . . . . . . . . . . 21
126118, 123, 1253jca 1132 . . . . . . . . . . . . . . . . . . . 20
127 elfzle1 10815 . . . . . . . . . . . . . . . . . . . . . 22
128127adantl 452 . . . . . . . . . . . . . . . . . . . . 21
129 elfzle2 10816 . . . . . . . . . . . . . . . . . . . . . . . 24
130129adantl 452 . . . . . . . . . . . . . . . . . . . . . . 23
131 simpll3 996 . . . . . . . . . . . . . . . . . . . . . . 23
132130, 131jca 518 . . . . . . . . . . . . . . . . . . . . . 22
133 zre 10044 . . . . . . . . . . . . . . . . . . . . . . . . . 26
134124, 133syl 15 . . . . . . . . . . . . . . . . . . . . . . . . 25
135134adantl 452 . . . . . . . . . . . . . . . . . . . . . . . 24
13699adantr 451 . . . . . . . . . . . . . . . . . . . . . . . . 25
137136adantr 451 . . . . . . . . . . . . . . . . . . . . . . . 24
138101adantr 451 . . . . . . . . . . . . . . . . . . . . . . . . 25
139138adantr 451 . . . . . . . . . . . . . . . . . . . . . . . 24
140135, 137, 1393jca 1132 . . . . . . . . . . . . . . . . . . . . . . 23
141 letr 8930 . . . . . . . . . . . . . . . . . . . . . . 23
142140, 141syl 15 . . . . . . . . . . . . . . . . . . . . . 22
143132, 142mpd 14 . . . . . . . . . . . . . . . . . . . . 21
144128, 143jca 518 . . . . . . . . . . . . . . . . . . . 20
145126, 144jca 518 . . . . . . . . . . . . . . . . . . 19
146 elfz4 10807 . . . . . . . . . . . . . . . . . . 19
147145, 146syl 15 . . . . . . . . . . . . . . . . . 18
148 simplr 731 . . . . . . . . . . . . . . . . . 18
149117, 147, 1483jca 1132 . . . . . . . . . . . . . . . . 17
150 simp1 955 . . . . . . . . . . . . . . . . . . . . . 22
15149adantr 451 . . . . . . . . . . . . . . . . . . . . . . . . 25
152 simpr 447 . . . . . . . . . . . . . . . . . . . . . . . . 25
153151, 152jca 518 . . . . . . . . . . . . . . . . . . . . . . . 24
154 ffvelrn 5679 . . . . . . . . . . . . . . . . . . . . . . . 24
155153, 154syl 15 . . . . . . . . . . . . . . . . . . . . . . 23
1561553adant3 975 . . . . . . . . . . . . . . . . . . . . . 22
157150, 156jca 518 . . . . . . . . . . . . . . . . . . . . 21
158 eleq1 2356 . . . . . . . . . . . . . . . . . . . . . . . . . 26
159158anbi2d 684 . . . . . . . . . . . . . . . . . . . . . . . . 25
160 feq1 5391 . . . . . . . . . . . . . . . . . . . . . . . . 25
161159, 160imbi12d 311 . . . . . . . . . . . . . . . . . . . . . . . 24
162161, 63vtoclga 2862 . . . . . . . . . . . . . . . . . . . . . . 23
163155, 162syl 15 . . . . . . . . . . . . . . . . . . . . . 22
1641633adant3 975 . . . . . . . . . . . . . . . . . . . . 21
165157, 164mpd 14 . . . . . . . . . . . . . . . . . . . 20
166 simp3 957 . . . . . . . . . . . . . . . . . . . 20
167165, 166jca 518 . . . . . . . . . . . . . . . . . . 19
168 ffvelrn 5679 . . . . . . . . . . . . . . . . . . 19
169167, 168syl 15 . . . . . . . . . . . . . . . . . 18
170 recn 8843 . . . . . . . . . . . . . . . . . 18
171169, 170syl 15 . . . . . . . . . . . . . . . . 17
172149, 171syl 15 . . . . . . . . . . . . . . . 16
173 fveq2 5541 . . . . . . . . . . . . . . . . 17
174173fveq1d 5543 . . . . . . . . . . . . . . . 16
175116, 172, 174fsump1 12235 . . . . . . . . . . . . . . 15
176 simpr 447 . . . . . . . . . . . . . . . . . 18
177 fzfid 11051 . . . . . . . . . . . . . . . . . . 19
178 simpll1 994 . . . . . . . . . . . . . . . . . . . . 21
17947a1i 10 . . . . . . . . . . . . . . . . . . . . . . . 24
180122adantr 451 . . . . . . . . . . . . . . . . . . . . . . . 24
181 elfzelz 10814 . . . . . . . . . . . . . . . . . . . . . . . . 25
182181adantl 452 . . . . . . . . . . . . . . . . . . . . . . . 24
183179, 180, 1823jca 1132 . . . . . . . . . . . . . . . . . . . . . . 23
184 elfzle1 10815 . . . . . . . . . . . . . . . . . . . . . . . . 25
185184adantl 452 . . . . . . . . . . . . . . . . . . . . . . . 24
186 simpll 730 . . . . . . . . . . . . . . . . . . . . . . . . . 26
187 simpr 447 . . . . . . . . . . . . . . . . . . . . . . . . . 26
188186, 187jca 518 . . . . . . . . . . . . . . . . . . . . . . . . 25
189181, 133syl 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
190189adantl 452 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
19193adantr 451 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
192 elfzle2 10816 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
193192adantl 452 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
194190, 191, 1933jca 1132 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
195 letrp1 9614 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
196194, 195syl 15 . . . . . . . . . . . . . . . . . . . . . . . . . . 27
197 simpl3 960 . . . . . . . . . . . . . . . . . . . . . . . . . . 27
198196, 197jca 518 . . . . . . . . . . . . . . . . . . . . . . . . . 26
19999adantr 451 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
200101adantr 451 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
201190, 199, 2003jca 1132 . . . . . . . . . . . . . . . . . . . . . . . . . . 27
202201, 141syl 15 . . . . . . . . . . . . . . . . . . . . . . . . . 26
203198, 202mpd 14 . . . . . . . . . . . . . . . . . . . . . . . . 25
204188, 203syl 15 . . . . . . . . . . . . . . . . . . . . . . . 24
205185, 204jca 518 . . . . . . . . . . . . . . . . . . . . . . 23
206183, 205jca 518 . . . . . . . . . . . . . . . . . . . . . 22
207206, 146syl 15 . . . . . . . . . . . . . . . . . . . . 21
208 simplr 731 . . . . . . . . . . . . . . . . . . . . 21
209178, 207, 2083jca 1132 . . . . . . . . . . . . . . . . . . . 20
210209, 169syl 15 . . . . . . . . . . . . . . . . . . 19
211177, 210fsumrecl 12223 . . . . . . . . . . . . . . . . . 18
212176, 211jca 518 . . . . . . . . . . . . . . . . 17
213 eqid 2296 . . . . . . . . . . . . . . . . . 18
214213fvmpt2 5624 . . . . . . . . . . . . . . . . 17
215212, 214syl 15 . . . . . . . . . . . . . . . 16
216215oveq1d 5889 . . . . . . . . . . . . . . 15
217175, 216eqtr4d 2331 . . . . . . . . . . . . . 14
218113, 217mpteq2da 4121 . . . . . . . . . . . . 13
219218adantr 451 . . . . . . . . . . . 12
220 simpl 443 . . . . . . . . . . . . . . . . . . . . 21
22147a1i 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
222 peano2nn 9774 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
223 nnz 10061 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
224222, 223syl 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2252243ad2ant2 977 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
226221, 121, 2253jca 1132 . . . . . . . . . . . . . . . . . . . . . . . . . . 27
227 nnge1 9788 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
228222, 227syl 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2292283ad2ant2 977 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
230229, 95jca 518 . . . . . . . . . . . . . . . . . . . . . . . . . . 27
231226, 230jca 518 . . . . . . . . . . . . . . . . . . . . . . . . . 26
232 elfz4 10807 . . . . . . . . . . . . . . . . . . . . . . . . . 26
233231, 232syl 15 . . . . . . . . . . . . . . . . . . . . . . . . 25
23491, 233jca 518 . . . . . . . . . . . . . . . . . . . . . . . 24
235 simpl 443 . . . . . . . . . . . . . . . . . . . . . . . . . 26
23649adantr 451 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
237 simpr 447 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
238236, 237jca 518 . . . . . . . . . . . . . . . . . . . . . . . . . . 27
239 ffvelrn 5679 . . . . . . . . . . . . . . . . . . . . . . . . . . 27
240238, 239syl 15 . . . . . . . . . . . . . . . . . . . . . . . . . 26
241235, 240jca 518 . . . . . . . . . . . . . . . . . . . . . . . . 25
242241simprd 449 . . . . . . . . . . . . . . . . . . . . . . . 24
243234, 242syl 15 . . . . . . . . . . . . . . . . . . . . . . 23
24491, 243jca 518 . . . . . . . . . . . . . . . . . . . . . 22
245 simpr 447 . . . . . . . . . . . . . . . . . . . . . . . 24
246 eleq1 2356 . . . . . . . . . . . . . . . . . . . . . . . . . . 27
247246anbi2d 684 . . . . . . . . . . . . . . . . . . . . . . . . . 26
248 feq1 5391 . . . . . . . . . . . . . . . . . . . . . . . . . 26
249247, 248imbi12d 311 . . . . . . . . . . . . . . . . . . . . . . . . 25
250249, 63vtoclga 2862 . . . . . . . . . . . . . . . . . . . . . . . 24
251245, 250syl 15 . . . . . . . . . . . . . . . . . . . . . . 23
252251pm2.43i 43 . . . . . . . . . . . . . . . . . . . . . 22
253244, 252syl 15 . . . . . . . . . . . . . . . . . . . . 21
254220, 253syl 15 . . . . . . . . . . . . . . . . . . . 20
255254, 176jca 518 . . . . . . . . . . . . . . . . . . 19
256 ffvelrn 5679 . . . . . . . . . . . . . . . . . . 19
257255, 256syl 15 . . . . . . . . . . . . . . . . . 18
258176, 257jca 518 . . . . . . . . . . . . . . . . 17
259 eqid 2296 . . . . . . . . . . . . . . . . . 18
260259fvmpt2 5624 . . . . . . . . . . . . . . . . 17
261258, 260syl 15 . . . . . . . . . . . . . . . 16
262261oveq2d 5890 . . . . . . . . . . . . . . 15
263113, 262mpteq2da 4121 . . . . . . . . . . . . . 14
264263adantr 451 . . . . . . . . . . . . 13
265 simpl1 958 . . . . . . . . . . . . . . 15
266 biid 227 . . . . . . . . . . . . . . . . 17
267266biimpi 186 . . . . . . . . . . . . . . . 16
268267adantl 452 . . . . . . . . . . . . . . 15
269233adantr 451 . . . . . . . . . . . . . . . . 17
270265, 269jca 518 . . . . . . . . . . . . . . . 16
271 ffn 5405 . . . . . . . . . . . . . . . . . . . 20
272252, 271syl 15 . . . . . . . . . . . . . . . . . . 19
273 dffn5 5584 . . . . . . . . . . . . . . . . . . 19
274272, 273sylib 188 . . . . . . . . . . . . . . . . . 18
275241, 274syl 15 . . . . . . . . . . . . . . . . 17
276275, 240eqeltrrd 2371 . . . . . . . . . . . . . . . 16
277270, 276syl 15 . . . . . . . . . . . . . . 15
278265, 268, 2773jca 1132 . . . . . . . . . . . . . 14
279 stoweidlem20.5 . . . . . . . . . . . . . . 15
280 nfmpt1 4125 . . . . . . . . . . . . . . 15
281 nfmpt1 4125 . . . . . . . . . . . . . . 15
282279, 280, 281stoweidlem8 27860 . . . . . . . . . . . . . 14
283278, 282syl 15 . . . . . . . . . . . . 13
284264, 283eqeltrrd 2371 . . . . . . . . . . . 12
285219, 284eqeltrd 2370 . . . . . . . . . . 11
286110, 285syl 15 . . . . . . . . . 10
287286ex 423 . . . . . . . . 9
288287ex 423 . . . . . . . 8
28924, 31, 38, 45, 86, 288nnind 9780 . . . . . . 7
290289a1i 10 . . . . . 6
29117, 290vtoclga 2862 . . . . 5
2923, 291syl 15 . . . 4
2933, 292mpd 14 . . 3
2948, 293mpd 14 . 2
2951, 294syl5eqel 2380 1
Colors of variables: wff set class Syntax hints: wi 4 wa 358 w3a 934 wnf 1534 wceq 1632 wcel 1696 class class class wbr 4039 cmpt 4093 wfn 5266 wf 5267 cfv 5271 (class class class)co 5874 cc 8751 cr 8752 c1 8754 caddc 8756 cle 8884 cn 9762 cz 10040 cuz 10246 cfz 10798 csu 12174 This theorem is referenced by: stoweidlem32 27884 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1536 ax-5 1547 ax-17 1606 ax-9 1644 ax-8 1661 ax-13 1698 ax-14 1700 ax-6 1715 ax-7 1720 ax-11 1727 ax-12 1878 ax-ext 2277 ax-rep 4147 ax-sep 4157 ax-nul 4165 ax-pow 4204 ax-pr 4230 ax-un 4528 ax-inf2 7358 ax-cnex 8809 ax-resscn 8810 ax-1cn 8811 ax-icn 8812 ax-addcl 8813 ax-addrcl 8814 ax-mulcl 8815 ax-mulrcl 8816 ax-mulcom 8817 ax-addass 8818 ax-mulass 8819 ax-distr 8820 ax-i2m1 8821 ax-1ne0 8822 ax-1rid 8823 ax-rnegex 8824 ax-rrecex 8825 ax-cnre 8826 ax-pre-lttri 8827 ax-pre-lttrn 8828 ax-pre-ltadd 8829 ax-pre-mulgt0 8830 ax-pre-sup 8831 This theorem depends on definitions: df-bi 177 df-or 359 df-an 360 df-3or 935 df-3an 936 df-tru 1310 df-ex 1532 df-nf 1535 df-sb 1639 df-eu 2160 df-mo 2161 df-clab 2283 df-cleq 2289 df-clel 2292 df-nfc 2421 df-ne 2461 df-nel 2462 df-ral 2561 df-rex 2562 df-reu 2563 df-rmo 2564 df-rab 2565 df-v 2803 df-sbc 3005 df-csb 3095 df-dif 3168 df-un 3170 df-in 3172 df-ss 3179 df-pss 3181 df-nul 3469 df-if 3579 df-pw 3640 df-sn 3659 df-pr 3660 df-tp 3661 df-op 3662 df-uni 3844 df-int 3879 df-iun 3923 df-br 4040 df-opab 4094 df-mpt 4095 df-tr 4130 df-eprel 4321 df-id 4325 df-po 4330 df-so 4331 df-fr 4368 df-se 4369 df-we 4370 df-ord 4411 df-on 4412 df-lim 4413 df-suc 4414 df-om 4673 df-xp 4711 df-rel 4712 df-cnv 4713 df-co 4714 df-dm 4715 df-rn 4716 df-res 4717 df-ima 4718 df-iota 5235 df-fun 5273 df-fn 5274 df-f 5275 df-f1 5276 df-fo 5277 df-f1o 5278 df-fv 5279 df-isom 5280 df-ov 5877 df-oprab 5878 df-mpt2 5879 df-1st 6138 df-2nd 6139 df-riota 6320 df-recs 6404 df-rdg 6439 df-1o 6495 df-oadd 6499 df-er 6676 df-en 6880 df-dom 6881 df-sdom 6882 df-fin 6883 df-sup 7210 df-oi 7241 df-card 7588 df-pnf 8885 df-mnf 8886 df-xr 8887 df-ltxr 8888 df-le 8889 df-sub 9055 df-neg 9056 df-div 9440 df-nn 9763 df-2 9820 df-3 9821 df-n0 9982 df-z 10041 df-uz 10247 df-rp 10371 df-fz 10799 df-fzo 10887 df-seq 11063 df-exp 11121 df-hash 11354 df-cj 11600 df-re 11601 df-im 11602 df-sqr 11736 df-abs 11737 df-clim 11978 df-sum 12175
Copyright terms: Public domain W3C validator | 10,156 | 19,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-51 | longest | en | 0.094686 |
https://community.esri.com/t5/arcgis-explorer-desktop-questions/normalization-percent-of-total-question/m-p/199739 | 1,675,713,101,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500357.3/warc/CC-MAIN-20230206181343-20230206211343-00540.warc.gz | 186,244,273 | 39,862 | # Normalization: Percent of Total (question)
1882
2
06-27-2014 07:43 AM
New Contributor III
Hi all,
I have a map of a county precincts, all 195. The attribute table shows the FID number, the Precinct number, a few columns such as district and shape, and most importantly it shows the number of Dems, Reps, and Unaffiliated voters.
I'm looking at making chloropleth maps which reflect the political party of the people in the precinct. It's easy enough to map the number of Dems or Reps, but I'm curious how to use the percent of total function. I'd like to show Dems are for ex: 50% of a precinct. However, I think the Normalization tab is also tabulating the area, some of the district numbers, and other irrelevant information for this task.
My question is...how would I go about using just the columns with the political parties as a percent of total?
Tags (4)
2 Replies
Occasional Contributor
Hi Elterrell
You might want to post this question under the ArcMap forum. ArcGIS Explorer Desktop doesn't have a normalization function.
Good luck
Ellen
Hi all,
I have a map of a county precincts, all 195. The attribute table shows the FID number, the Precinct number, a few columns such as district and shape, and most importantly it shows the number of Dems, Reps, and Unaffiliated voters.
I'm looking at making chloropleth maps which reflect the political party of the people in the precinct. It's easy enough to map the number of Dems or Reps, but I'm curious how to use the percent of total function. I'd like to show Dems are for ex: 50% of a precinct. However, I think the Normalization tab is also tabulating the area, some of the district numbers, and other irrelevant information for this task.
My question is...how would I go about using just the columns with the political parties as a percent of total? | 439 | 1,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.932095 |
https://phasergames.com/scaling-games-in-phaser-3/ | 1,723,093,827,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640719674.40/warc/CC-MAIN-20240808031539-20240808061539-00658.warc.gz | 368,847,493 | 30,149 | # Scaling Games in Phaser 3 with an Alignment Grid
## Scaling Games in Phaser 3 with an Alignment Grid
As game developers, one of the challenges we have faced since the invention of the smartphone is how to scale the game so it will fit on whatever device the player has. Scaling games in Phaser 3 is no exception. There are a lot of different ways programmer will achieve this. I’m going to show you the way I’ve been doing it for work and my own projects. I’m not saying it is absolutely the best way, but it works best for me. It is also an easy concept to explain because it doesn’t rely on anything built into Phaser itself, but rather simple mathematics. It will work on any platform where you can measure the widths and heights of objects and of the canvas itself. I’ll be using the Phaser 3 basic template to jump-start the project. This template has a static size on computers and a dynamic one on mobile. In other words, the canvas size matches the screen size on a phone or tablet.
This was originally a project I did for Phaser CE but I’ve made a lot of improvements since then.
## How does the Alignment Grid work?
Using a dynamic canvas size means we don’t know the size of the game when we start. The alignment grid works because instead of placing an object at a static point we place it at a percentage of the canvas size. Say for example a sprite at an x position of 200 and a y position of 150, we place it at 25% of the screen’s width and 50% of the screen’s height.
Even though we don’t know the size, we know there will always be percentages. The Alignment grid gives a visual way to do that. We start by setting up how many rows and how many columns we want to split the game into.
```var gridConfig = {
'scene': this,
'cols': 5,
'rows': 5
}```
Now let’s set up the class that uses this config, the Alignment Grid. Since this is a simple object and we will be passing the scene, we don’t need to extend it with anything.
```class AlignGrid {
constructor(config) {
}
}```
The only required parameter is the scene, everything else is optional. We will verify if the scene is present and if not, we will log an error and return. If the optional parameters are missing we will set defaults. Also, we will promote some of the config object’s properties to class level variables. this.scene = config.scene for example.
```class AlignGrid {
constructor(config) {
if (!config.scene) {
console.log("missing scene!");
return;
}
if (!config.rows) {
config.rows = 3;
}
if (!config.cols) {
config.cols = 3;
}
if (!config.width) {
config.width = game.config.width;
}
if (!config.height) {
config.height = game.config.height;
}
this.h = config.height;
this.w = config.width;
this.rows = config.rows;
this.cols = config.cols;
this.scene = config.scene;
}```
## Doing the Math!
Now here is where the magic happens. We will take the height and width passed to the AlignmentGrid class, by default the height and width of the game, and divide it by the numbers of rows and columns and put it in a cell width variable and cell height variable.
```//cw cell width is the scene width divided by the number of columns
this.cw = this.w / this.cols;
//ch cell height is the scene height divided the number of rows
this.ch = this.h / this.rows;```
## Seeing the grid!
Now we have the numbers we need to start placing by percentage. To make it easier for scaling games in Phaser 3 it is better to be able to see the grid. Let’s make a function called show() with a single parameter to alpha the lines, so we can turn the lines down in case we need to see the objects we are positioning better.
```//mostly for planning and debugging this will
//create a visual representation of the grid
show(a = 1) {
this.graphics.lineStyle(4, 0xff0000, a);
//
//
//this.graphics.beginPath();
for (var i = 0; i < this.w; i += this.cw) {
this.graphics.moveTo(i, 0);
this.graphics.lineTo(i, this.h);
}
for (var i = 0; i < this.h; i += this.ch) {
this.graphics.moveTo(0, i);
this.graphics.lineTo(this.w, i);
}
this.graphics.strokePath();
}```
Let’s see the grid in action now.
In your create function of sceneMain place this code
```var gridConfig = {
'scene': this,
'cols': 5,
'rows': 5
}
this.aGrid = new AlignGrid(gridConfig);
this.aGrid.showNumbers();```
Here is the result.
## Placing Object on the grid
Now that we have a grid we are back on familiar ground as programmers because we know how to place objects at x and y coordinates. Let’s set up a function to do that.
```//place an object in relation to the grid
placeAt(xx, yy, obj) {
//calculate the center of the cell
//by adding half of the height and width
//to the x and y of the coordinates
var x2 = this.cw * xx + this.cw / 2;
var y2 = this.ch * yy + this.ch / 2;
obj.x = x2;
obj.y = y2;
}```
## Set up a test object to place on the grid
This part should be fairly well known to you, we will simply load an image and make a sprite.
```preload() {
}
create() {
var gridConfig = {
'scene': this,
'cols': 5,
'rows': 5
}
this.aGrid = new AlignGrid(gridConfig);
this.aGrid.show();
//
//
this.face = this.add.image(0, 0, "face");
}```
Now place it on the grid by using this line of code:
`this.aGrid.placeAt(2,2,this.face);`
This will place it 2 squares over and 2 squares down, counting from 0. So this will place it in the middle of the screen
## Scaling to the grid
Notice though that the face doesn’t fit in the square. Since we are dividing the width of the game by 5 to make the grid we can resize the face by scaling it related to the game’s width too.
```//scale the face
this.face.displayWidth = game.config.width / 5;
this.face.scaleY = this.face.scaleX;```
for more on scaling in Phaser 3 please see this post.
### Making it a little easier
For the purpose of scaling games in Phaser 3, this is enough, but it was slowing me down constantly counting the rows and columns of the grids. I am usually using an 11 x 11 grid, and my productivity started to slide. So I came up with a numbering system. This function will call the show function and places numbers in each cell.
```showNumbers(a = 1) {
this.show(a);
var n = 0;
for (var i = 0; i < this.rows; i++) {
for (var j = 0; j < this.cols; j++) {
var numText = this.scene.add.text(0, 0, n, {
color: 'red'
});
numText.setOrigin(0.5, 0.5);
this.placeAt(j, i, numText);
n++;
}
}
}```
I then made a new function to place by index rather than x and y. It simply converts the index to the grid’s x and y and calls the placeAt function.
```placeAtIndex(index, obj) {
var yy = Math.floor(index / this.cols);
var xx = index - (yy * this.cols);
this.placeAt(xx, yy, obj);
}```
So now replace the placeAt(2,2,this.face) with this code:
`this.aGrid.placeAtIndex(18, this.face);`
and replace the grid.show() with:
`this.aGrid.showNumbers();`
Here is the result:
### Testing the scaling
In your browser’s tools, resize the screen for various devices. On chrome, this is done by pressing f-12 and then clicking the mobile icon.
Always refresh the page after changing devices. There are a few occasions where I’ve had to tweak placements or sizes for certain devices, but I’ve been using this method in my professional work for almost 2 years now, and I’ve found it has done 99% of the work for me. Although that was Phaser CE, because we are using mathematics this technique will work with any framework. I hope it helps you with scaling games in Phaser 3. Happy coding!!! | 1,858 | 7,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-33 | latest | en | 0.899332 |
https://www.lmfdb.org/EllipticCurve/Q/17661/f/ | 1,581,973,368,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143373.18/warc/CC-MAIN-20200217205657-20200217235657-00394.warc.gz | 836,543,969 | 43,694 | # Properties
Label 17661.f Number of curves 6 Conductor 17661 CM no Rank 1 Graph
# Related objects
Show commands for: SageMath
sage: E = EllipticCurve("17661.f1")
sage: E.isogeny_class()
## Elliptic curves in class 17661.f
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality
17661.f1 17661a5 [1, 1, 0, -659361, -206353626] [2] 100352
17661.f2 17661a3 [1, 1, 0, -41226, -3234465] [2, 2] 50176
17661.f3 17661a4 [1, 1, 0, -32816, 2260629] [2] 50176
17661.f4 17661a6 [1, 1, 0, -28611, -5235204] [2] 100352
17661.f5 17661a2 [1, 1, 0, -3381, -17640] [2, 2] 25088
17661.f6 17661a1 [1, 1, 0, 824, -1661] [2] 12544 $$\Gamma_0(N)$$-optimal
## Rank
sage: E.rank()
The elliptic curves in class 17661.f have rank $$1$$.
## Modular form 17661.2.a.f
sage: E.q_eigenform(10)
$$q + q^{2} - q^{3} - q^{4} - 2q^{5} - q^{6} - q^{7} - 3q^{8} + q^{9} - 2q^{10} - 4q^{11} + q^{12} - 2q^{13} - q^{14} + 2q^{15} - q^{16} + 6q^{17} + q^{18} - 4q^{19} + O(q^{20})$$
## Isogeny matrix
sage: E.isogeny_class().matrix()
The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering.
$$\left(\begin{array}{rrrrrr} 1 & 2 & 8 & 4 & 4 & 8 \\ 2 & 1 & 4 & 2 & 2 & 4 \\ 8 & 4 & 1 & 8 & 2 & 4 \\ 4 & 2 & 8 & 1 & 4 & 8 \\ 4 & 2 & 2 & 4 & 1 & 2 \\ 8 & 4 & 4 & 8 & 2 & 1 \end{array}\right)$$
## Isogeny graph
sage: E.isogeny_graph().plot(edge_labels=True)
The vertices are labelled with LMFDB labels. | 700 | 1,538 | {"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} | 3.109375 | 3 | CC-MAIN-2020-10 | longest | en | 0.448411 |
https://adprun.net/accounting/page/2/ | 1,713,729,615,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817819.93/warc/CC-MAIN-20240421194551-20240421224551-00873.warc.gz | 70,662,952 | 6,619 | # Accounting
## What Is Variable Costing?
It can change its entire labor force, managerial as well as line workers. In general, companies with a high proportion of variable costs relative to fixed costs are considered to be less volatile, as their profits are more dependent on the success of their sales. Businesses mustalways paytheir fixed costs regardless of how well they are doing. By contrast, variable costs only occur once there is a good or service being produced. Fixed costs are paid regardless of how much a business produces, s […]
## What Is The Difference Between Vertical Analysis And Horizontal Analysis?
The lower portion of the chart shows how each of the company’s products contributed to the company’s total sales for the year. A common size income statement is an income statement in which each line item is expressed as a percentage of the value of sales, to make analysis easier. This shows that the amount of cash at the end of 2018 is 141% of the amount it was at the end of 2014. By doing the same analysis for each item on the balance sheet and income statement, one can see how ea […]
## What Is The Difference Between Revenues And Earnings?
Because revenue is typically listed at the top of the income statement. Revenue is simply all the income your business generates before subtracting any other expenses. In other words, it is the money your company receives in exchange for goods or services. To better understand the main differences between revenue vs. profit, let’s compare the two concepts head-to-head. A U.S. corporation’s revenues are reported on the top line of its income statement, while its earnings are reported […]
## What Is Variable Cost? Learn Why Variable Costs Are Important To A Business
By contrast, variable costs are calculated using multiplication. You can plug production data into the variable cost formula to determine total cost. They are fixed because they are paid out regularly and are independent of revenue level or production volume. But, other forms of labor are dependent on these factors, according to Accounting Tools. Over a five-year horizon, all costs can become variable costs. It can change its entire labor force, managerial as well as line workers. In general, c […]
## What Is The Difference Between A Trial Balance And A Balance Sheet?
Tree NodeSelect a node from the consolidation tree by which you want to further filter trial balance results. Get All Lowest Level Nodes Displays the information for the children at the lowest level of the tree for the selected node. View activity for the specified period only, or view cumulative activity from the beginning of the fiscal year through the specified period, or view a mix of cumulative and period activity, based on account type. Learn accounting fundamentals and how to read financ […]
## What Is The Difference Between Negative Assurance And Positive Assurance?
Negative assurance relates to limited assurance engagements. Due to the lower assurance required, auditors use a negatively worded conclusion. The term audit refers to examination or investigation. An audit is a process through which independent auditors assess a subject matter.
What is example of assurance?
The definition of assurance is an affirmation and commitment. An example of an assurance is a construction firm stating that a job will be finished by the original projected date. A st […] | 648 | 3,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-18 | latest | en | 0.950596 |
https://m.taodocs.com/topdoc/97452-0-0-1.html | 1,718,655,687,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861737.17/warc/CC-MAIN-20240617184943-20240617214943-00504.warc.gz | 340,767,601 | 8,681 | ---------------------------------作者:_____________-----------------------------日期::_____________小...
(375+1034)+(966+125) (2130+783+270)+1017 99+999+9999+99999899+344 2357-183-317-357 2365-1086-214497-...
300÷125÷8=300÷(125×8)=300÷1000=0.3396-96-172-28=(396-96)-(172+28)=300-200=100125*24=125*8*3=100...
184+98695+202864-199738-301380+476+120(569+468)+(432+131)704×25256-147-53373-129+29189-(89+74)28×4...
2小学四年级数学简便计算题.doc
(181+2564)+2719378+44+114+242+222 276+228+353+219(375+1034)+(966+125)(2130+783+270)+101799+999+9999+...
158+262+138375+219+381+2255001-247-1021-232(181+2564)+2719378+44+114+242+222276+228+353+219(375+1034...
---------------------------------作者:_____________-----------------------------日期::_____________四...
---------------------------------作者:_____________-----------------------------日期::_____________小...
158+262+138375+219+381+2255001-247-1021-232(181+2564)+2719378+44+114+242+222276+228+353+219(375+1034...
158+262+138375+219+381+2255001－247－1021－232（181+2564）+2719378+44+114+242+222276+228+353+219(375...
158+262+138 375+219+381+225 5001－247－1021－232（181+2564）+2719 378+44+114+242+222 ...
Evaluation Warning: The document was created with Spire.Doc for .NET.简便计算分类练习题98 X199 58 X9...
.精选文档.158+262+138 375+219+381+225 5001－247－1021－232（181+2564）+2719 378+44+114+...
.精选文档.简便计算分类练习题第一种(300+6)x12???????????? ??25x(4+8) 32×(25+125) ...
Evaluation Warning: The document was created with Spire.Doc for .NET.简便计算分类练习题第一种(300+6)...
1 / 7简略计算练习题一、2356-（1356-721 ） 1235- （1780-1665 ） 75×27+19×2531×870+13×310 25×65+2...
-. 优选-简便计算分类... | 666 | 1,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-26 | latest | en | 0.239736 |
http://www.stata.com/statalist/archive/2002-12/msg00020.html | 1,495,845,989,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608726.4/warc/CC-MAIN-20170527001952-20170527021952-00035.warc.gz | 814,925,737 | 2,574 | st: Gravity Model ML estimation
From "Julia A Gamas Buentello" To Subject st: Gravity Model ML estimation Date Mon, 2 Dec 2002 14:32:42 -0500
```______________________________________________
Hi,
Does anybody have a subroutine that will solve the following problem?:
I have a gravity model of the form:
Tij = Ai *Oi* Bj* Dj * (d to the minus beta), where
Ai = Sumation over i of [Bj* Dj * (d to the minus beta)]
and
Bj = Summation over j of [Ai Oi * (d to the minus beta)]
The purpose is to find beta using non-linear maximum likelihood. The Ai
and Bj are balancing factors that "guarantee" that two constraints are met.
I want to estimate it using maximum likelihood. The procedure is to start
with a beta=1, then use it to iterate Ai and Bj until Ai and Bj no longer
change, then to go back and check that a constraint equation is met. The
constraint equation is:
estimated sum over i and j of Tij ln(dij) = the "real" sum over i and j of
Tij ln(dij)
Does Stata have a subroutine that already does this? I am hoping to save
some graduate student hours and not have to write up the program from
scratch.
Julia Gamas
Boston University Center for Transportation Studies
*
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 361 | 1,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-22 | latest | en | 0.852805 |
http://clay6.com/qa/27121/a-tailor-purchased-125-meter-of-cloth-to-stitch-one-dress-he-needs-2-25-met | 1,511,421,517,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806760.43/warc/CC-MAIN-20171123070158-20171123090158-00263.warc.gz | 55,597,141 | 27,372 | # A tailor purchased 125 meter of cloth. To stitch one dress he needs 2.25 meter. How many dresses can he stitch using 125 meter of cloth? Will there be any cloth left over? If so how much?
Cloth needed to stitch 1 dress = 2.25 meter
Total cloth available = 125 meter
Number of dresses that can be stitched = 125 / 2.25 = 55 dresses
Remaining cloth = 125 - (55 * 2.25) = 125 - 123.75 = 1.25 meter
edited Feb 7, 2014 by fingeazy | 133 | 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} | 3.96875 | 4 | CC-MAIN-2017-47 | longest | en | 0.857798 |
https://math.stackexchange.com/questions/1703268/how-to-solve-the-following-equation-involving-an-exponential-function/1703301 | 1,718,624,477,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861701.67/warc/CC-MAIN-20240617091230-20240617121230-00646.warc.gz | 342,484,849 | 40,773 | # How to solve the following equation involving an exponential function
How do you solve $10^x = x$? I'm not sure how to solve this algebraically. Using log functions wasn't enough.
• @JKnecht I don't think most people can try very much and actually get anywhere with this kind of problem. Commented Mar 18, 2016 at 18:23
• To be honest, I feel as though most people should try searching for the answer first, there are plenty of questions almost titled exactly "How to solve this exponential equation" Commented Mar 18, 2016 at 18:26
• @SimpleArt I think he should have shown where he got stuck when he used the "log functions". Even if his approach was wrong or only a couple of steps. Regarding your other comment i completely agree. Commented Mar 18, 2016 at 18:37
• @JKnecht I'd agree, but when I personally tried to solve this for the first time, the paper I was using went in about 20 different directions, none of which were any closer to the solution than the other. Commented Mar 18, 2016 at 18:46
## 6 Answers
The graph above should immediately tell you that there are no real solutions to the equation. If you're interested in the complex solutions, here's how we can proceed $$10^x = x$$ $$1=\frac{x}{10^x}$$ $$1=\frac{x}{e^{\ln 10^x}}$$ $$1=xe^{-x\ln 10}$$ $$-\ln 10=(-x\ln 10)e^{(-x\ln 10)}$$ Therefore $$W(-\ln 10)=-x\ln 10$$ $$x=-\frac{W(-\ln 10)}{\ln 10}$$ Where $W$ is the Lambert W function.
• Both sides of the coin. (+1) Commented Mar 19, 2016 at 23:20
Hint:
Draw the graph of $y=10^x$ and that of $y=x$, will they ever meet?
• I'd start with this as well. If you actually get any interception, then continue down the RabbitHole :) Commented Mar 20, 2016 at 14:33
Before we try to find numerical solutions, let us first see if there are any solutions in the first place. A quick sketch will show that there should be no solution. Let us prove this algebraically.
First notice that for all real $x,$ $10^{x} > 0.$ A positive is always bigger than a negative, so for $x < 0,$ $10^{x} > x.$
For $x > 0,$ we see that $10^{x}$ grows far faster than $x.$ When $x$ is $0,$ $10^{x} = 1 > 0,$ and beyond that, $10^{x}$ always grows faster. So we have that $10^{x} > x$ for all $x.$ Thus, there is $\boxed{\text{no solution}}.$
• Nice, I guess you could use derivatives if you wanted to show that it grows faster. Commented Mar 19, 2016 at 9:01
As others have noted, there are no real solutions in this case. There are complex solutions. They are $$-{\frac {{\rm W} \left(-\ln \left( 10 \right) \right)}{\ln \left( 10 \right) }}$$ where $W$ is any branch of the Lambert W function. The first few solutions in order of increasing real part are
$$- 0.1191930734 \pm 0.7505832941\,i, 0.5294805081 \pm 3.342716202\,i, 0.7877834910 \pm 6.083768254\,i, 0.9480581767 \pm 8.821952931\,i, 1.064691576 \pm 11.55730317\,i$$
• You could use W$_k$ to show each solution separately. Commented Mar 18, 2016 at 18:24
I can show you some interesting ways to find the answer.
Start with $x=\log_{10}(x)$.
Substitute this into itself to get $x=\log_{10}(\log_{10}(x))$
Repeat this infinitely: $$x=\log_{10}(\log_{10}(\log_{10}(\dots\log_{10}(x)\dots)))$$
Try plugging in a random number for the $x$ inside the logarithms and use a calculator that can calculate complex numbers with logarithms to find the result.
Plugging in different numbers may produce different answers, but all answers should work to solve $x=10^x$.
For numerical reassurance (I used google)
$$\log(2)=0.30102999566$$
$$\log(\log(2))=-0.52139022765$$
$$\log(\log(\log(2)))=-0.2828+1.3643i$$
$$\log(\log(\log(\dots\log(2)\dots)))=-0.119193073+0.750583294i$$
When I try to plug this back into $x=10^x$, I get that it works, with a very small amount of error.
Also note that:
$$10^x=10^{x\pm\log_{10}(e)2\pi in},n=0,1,2,3,\dots$$
And using that, we get $$x=10^{x+\log_{10}(e)2\pi in}\implies x=\log_{10}(x)\mp\log_{10}(e)2\pi in$$
And putting this into our substitution method:
$$x=\log_{10}(\log_{10}(\dots\log_{10}(x)\mp\log_{10}(e)2\pi in)\mp\log_{10}(e)2\pi in)\mp\log_{10}(e)2\pi in$$
For $x>0$ you have no answers beacause of the derivatives. For $x<0$ you have no answers beacause $10^x>0>x$. So there are no answers.
• This question was tagged as algebra-precalculus, so I'm not quiet sure whether the OP has any familiarity with derivatives. Commented Mar 18, 2016 at 16:51 | 1,408 | 4,376 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-26 | latest | en | 0.937206 |
https://coindataflow.com/en/pair/uma-cnc | 1,624,153,507,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487653461.74/warc/CC-MAIN-20210619233720-20210620023720-00130.warc.gz | 176,821,117 | 14,285 | \$1,53T
Total marketcap
\$3,81T
Total volume
43.32%
BTC dominance
# UMA to CNC Exchange Rate - 1 UMA in Global China Cash
0
• BTC 0.0003
• ETH 0.005
Vol [24h]
\$0
## uma to cnc converter
Exchange Pair Price 24h volume
AEX UMA/CNC \$9,6293 \$0
## UMA/CNC Exchange Rate Overview
Name Ticker Price % 24h 24h high 24h low 24h volume
UMA uma \$10,5409 -3.6993% \$10,9859 \$10,5391 \$16,55M
Global China Cash cnc \$0,1572 0.1343% \$0,1593 \$0,1539 \$100,12M
Selling 1 UMA you get 0 Global China Cash cnc.
UMA 2/4/21 had the highest price, at that time trading at its all-time high of \$41,56.
136 days have passed since then, and now the price is 25.36% of the maximum.
Based on the table data, the UMA vs CNC exchange volume is \$0.
Using the calculator/converter on this page, you can make the necessary calculations with a pair of UMA/Global China Cash.
## Q&A
### What is the current UMA to CNC exchange rate?
Right now, the UMA/CNC exchange rate is 0.
### What has been the UMA to Global China Cash trading volume in the last 24 hours?
Relying on the table data, the UMA to Global China Cash exchange volume is \$0.
### How can I calculate the amount of CNC? / How do I convert my UMA to Global China Cash?
You can calculate/convert CNC from UMA to Global China Cash converter. Also, you can select other currencies from the drop-down list.
## UMA to CNC Conversation Table
UMA CNC
0.04 UMA = 0 CNC
0.09 UMA = 0 CNC
0.1 UMA = 0 CNC
0.2 UMA = 0 CNC
0.4 UMA = 0 CNC
0.9 UMA = 0 CNC
1 UMA = 0 CNC
5 UMA = 0 CNC
9 UMA = 0 CNC
100 UMA = 0 CNC
1000 UMA = 0 CNC
10000 UMA = 0 CNC
100000 UMA = 0 CNC | 551 | 1,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-25 | latest | en | 0.84387 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-7-functions-and-graphs-7-5-formulas-applications-and-variation-7-5-exercise-set-page-488/74 | 1,545,031,964,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828448.76/warc/CC-MAIN-20181217065106-20181217091106-00553.warc.gz | 904,380,372 | 12,832 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$\color{blue}{y=\frac{3}{40}xz}$
RECALL: When $y$ varies jointly as $x$ and $z$, the variation is represented by the equation $y=kxz$ where $k$ is the constant of variation. $y$ varies jointly as $x$ and $z$ so the equation of the variation is $y=kxz$ with $k$=constant of variation. To find the value of $k$, substitute the given values into $y=kxz$ to obtain: $$y=kxz \\\frac{3}{2}=k(2)(10) \\\frac{3}{2}=20k \\\dfrac{\frac{3}{2}}{20}=\frac{20k}{20} \\\frac{3}{2(20)}=k \\\frac{3}{40}=k$$ Thus, the equation of the variation is: $\color{blue}{y=\frac{3}{40}xz}$. | 222 | 643 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 4.40625 | 4 | CC-MAIN-2018-51 | latest | en | 0.769628 |
https://www.mastguru.com/hcf-and-lcm-questions-answers/question/28 | 1,631,994,710,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056572.96/warc/CC-MAIN-20210918184640-20210918214640-00023.warc.gz | 886,989,659 | 14,205 | # HCF and LCM Questions Answers
• #### 1. Find the HCF of \begin{aligned} 2^2 \times 3^2 \times 7^2, 2 \times 3^4 \times 7 \end{aligned}
1. 128
2. 126
3. 146
4. 434
Explanation:
HCF is Highest common factor, so we need to get the common highest factors among given values. So we got
2 * 3*3 * 7
• #### 2. Find the HCF of 54, 288, 360
1. 18
2. 36
3. 54
4. 108
Explanation:
Lets solve this question by factorization method.
\begin{aligned}
18 = 2 \times 3^2, 288 = 2^5 \times 3^2, 360 = 2^3 \times 3^2 \times 5
\end{aligned}
So HCF will be minimum term present in all three, i.e.
\begin{aligned}
2 \times 3^2 = 18
\end{aligned}
• #### 3. Reduce \begin{aligned} \frac{368}{575} \end{aligned} to the lowest terms.
1. \begin{aligned} \frac{30}{25} \end{aligned}
2. \begin{aligned} \frac{28}{29} \end{aligned}
3. \begin{aligned} \frac{28}{29} \end{aligned}
4. \begin{aligned} \frac{16}{25} \end{aligned}
Explanation:
We can do it easily by in two steps
Step1: We get the HCF of 368 and 575 which is 23
Step2: Divide both by 23, we will get the answer 16/25
• #### 4. Reduce \begin{aligned} \frac{803}{876} \end{aligned} to the lowest terms.
1. \begin{aligned} \frac{11}{12} \end{aligned}
2. \begin{aligned} \frac{23}{24} \end{aligned}
3. \begin{aligned} \frac{26}{27} \end{aligned}
4. \begin{aligned} \frac{4}{7} \end{aligned}
Explanation:
HCF of 803 and 876 is 73, Divide both by 73, We get the answer 11/12
• #### 5. HCF of \begin{aligned} 2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11 \end{aligned} is
1. \begin{aligned} 2^4 \times 3^4 \times 5^3 \end{aligned}
2. \begin{aligned} 2^4 \times 3^4 \times 5^3 \times 11 \end{aligned}
3. \begin{aligned} 2^2 \times 3^2 \times 5^2 \end{aligned}
4. \begin{aligned} 2 \times 3 \times 5 \end{aligned}
Explanation:
As in HCF we will choose the minimum common factors among the given.. So answer will be third option
• #### 6. What will be the LCM of 8, 24, 36 and 54
1. 54
2. 108
3. 216
4. 432
Explanation:
LCM of 8-24-36-54 will be
2*2*2*3*3*3 = 216
• #### 7. Find the HCF of \begin{aligned} \frac{2}{3}, \frac{4}{6}, \frac{8}{27} \end{aligned}
1. \begin{aligned} \frac{2}{27} \end{aligned}
2. \begin{aligned} \frac{8}{3} \end{aligned}
3. \begin{aligned} \frac{2}{3} \end{aligned}
4. \begin{aligned} \frac{8}{27} \end{aligned}
Explanation:
Whenever we have to solve this sort of question, remember the formula.
HCF = \begin{aligned} \frac{HCF of Numerators}{LCM of Denominators} \end{aligned}
So answers will be option 1
• #### SAMARPITA DUTTA 5 years ago
HOW LCM COMES 27 WHERE 3 NUMBERS ARE 3,6 AND 27 ?? PLEASE EXPLAIN THIS...
• #### Sumit 5 years ago
Why nobody notice it. In the 7th question the lcm of denominator is 54 and not 27. 27 is not a multiple of 6
• #### Anbu Sudha 6 years ago
Very useful sir.
But one doubt LCM IS 24. ONE NUMBER IS 8 WHAT IS HCF AND OTHER NUMBER PLEASE TELL WITH FULL STEP ANSWER
#### mastguru 6 years ago replied
welcome Tanima !
#### Arun 5 years ago replied
Indeed the LCM would be 54 and so the answer would be 1/27 and not 2/27
#### i-love-samarpita-dutta 4 years ago replied
Same query, I think it is a mistake.
#### mastguru 6 years ago replied
thank you Taha !
• #### poonam rajpal 6 years ago
SIR,THIS MUCH OF PREPARTION IS ENOUGH FOR PASSING THE ENTRANCE....
• #### nadeem 7 years ago
very helpful wbsite i love it
• #### Jayesh 7 years ago
Awesome website for competitive exam preparation
Thank you sir
#### mastguru 7 years ago replied
Thank you Jayesh for your feedback..
• #### Alok Singh 7 years ago
In Q.7 , all the given options are wrong. the right answer is 1/27.
Because the H.C.F of 2,4 and 8 is 2
and L.C.M of 3,6 and 27 is 54
so it will be 2/54 = 1/27.
tnx sir....
#### mastguru 7 years ago replied
Most welcome :)
• #### Ruchi Malik Sharma 7 years ago
solution of the que no 16 not appropriate please provide us right solution of the que no 16
• #### Ruchi Malik Sharma 7 years ago
Q12 : havin wrong Ans
HCF of : 1575,1125,765
45 Ans
• #### Sri 7 years ago
very very useful
• #### Swati 7 years ago
Thanx for your great questions...It helps me a lot in my preparation
• #### neetu 8 years ago
thanx mastguru....
#### Ashish rajput 6 years ago replied
thanku mastguru side..
• #### Sahil Arora 8 years ago
Thanks mastguru.. keep it up.
• #### Shikha 8 years ago
Thanks for awesome questions answers collection.. | 1,549 | 4,400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2021-39 | latest | en | 0.651363 |
https://rpg.stackexchange.com/questions/131365/wod-in-anydice-rolling-for-minimum-number-of-successes | 1,713,622,132,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00455.warc.gz | 456,888,432 | 36,231 | # WoD in anydice: rolling for minimum number of successes
We all know how to roll dice in WoD in anydice, but I stand in front of a different yet related problem:
There are some gifts (for example eye pluck for Corax) [and other powers] that demand a certain number of successes against a difficulty. In the named example, 4 successes against the difficulty 9 are needed.
This sounded ludicrous, till I ran the numbers and found out that (by manually summing the results) 6 dice would result in a 1.34% chance to do this, 7.17 with willpower, 11.29% with a specialization only and 22.27% with both. But... that is tedious.
How to model rolling Xd10 against difficulty Y with a demand of Z successes, only displaying the percentage of "minimum and more successes" "below threshold" and "botch"? If computation time limits wouldn't bitch out about it, it would be nice to run a loop of Z to 10 dice in addition.
## How you'd like it
If you really want to do the summing directly in code, probably the easiest way would be to use a helper function like this:
function: COUNT:n minimum MIN:n {
if COUNT < 0 { result: -1 }
if COUNT < MIN { result: 0 }
result: 1
}
This just takes a number COUNT (which could be the result of any dice roll) and compares it to a target number MIN, returning -1 if the count is negative, 0 if it's non-negative but less than the target, and 1 if it's equal to or greater than the target. Plotting e.g. [[roll X d NORMAL] minimum 4] will then directly give you the probabilities you asked for.
Of course, it's also easy enough to loop this over a range of dice pool sizes, if that's what you want.
## The other way
There's no need to write any extra code for this, since the AnyDice user interface already provides the "At Least" and "At Most" modes that automatically sum the output probabilities.
For example, running the code by Jasper Flick from this answer (which, by default, uses DIFFICULTY: 7 and X: 4 dice) and clicking the "At Least" button gives the following output:
Looking at the bar labeled "4" in each graph, we can see that the probability of rolling at least 4 successes (with 4 dice against difficulty 7, in this case) is normally 2.56%, and rises to 11.86% with specialization, 15.36% with willpower and 27.94% with both.
Similarly, looking at the bar labeled "0" in the same output gives the probability of not botching (since the code treats a botched roll as -1 successes), which is 93.29% without willpower (and 100% with it). To get the probability of botching the roll, you can either subtract that from 100% (and, hopefully, get 6.71%), or just switch to "At Most" mode and look at the "-1" bar instead.
• I did switch the two around, because I did want just the solution you crammed in at the end - the other solution is not helping in the process to analyse and probably fix the problem of those ludicious gifts. It only matters if it is a botch, success or fail. All the other numbers? not helpful. Sep 12, 2018 at 15:59
• I realise where my error came from: I did up the difficulty to 9 in one test run, and only 7 in the other. Sep 12, 2018 at 16:10 | 793 | 3,122 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-18 | latest | en | 0.919323 |
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# In tossing a coin, find the chance of throwing head and tail
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03 Jun 2004, 02:18
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In tossing a coin, find the chance of throwing head and tail alternately in 3 successive trials?
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03 Jun 2004, 02:49
1/4
only two alternates HTH THT, possibilities 8; 2/8=1/4
Display posts from previous: Sort by | 373 | 1,403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-26 | longest | en | 0.878282 |
https://wikivisually.com/wiki/Map_(mathematics) | 1,534,288,542,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209650.4/warc/CC-MAIN-20180814225028-20180815005028-00184.warc.gz | 870,121,048 | 52,346 | # Map (mathematics)
In mathematics, the term mapping, sometimes shortened to map, refers to either a function, often with some sort of special structure, or a morphism in category theory, which generalizes the idea of a function. There are also a few, less common uses in logic and graph theory.
## Maps as functions
In many branches of mathematics, the term map is used to mean a function, sometimes with a specific property of particular importance to that branch. For instance, a "map" is a continuous function in topology, a linear transformation in linear algebra, etc.
Some authors, such as Serge Lang,[1] use "function" only to refer to maps in which the codomain is a set of numbers (i.e. a subset of the fields R or C) and the term mapping for more general functions.
Sets of maps of special kinds are the subjects of many important theories: see for instance Lie group, mapping class group, permutation group.
In the theory of dynamical systems, a map denotes an evolution function used to create discrete dynamical systems. See also Poincaré map.
A partial map is a partial function, and a total map is a total function. Related terms like domain, codomain, injective, continuous, etc. can be applied equally to maps and functions, with the same meaning. All these usages can be applied to "maps" as general functions or as functions with special properties.
## Maps as morphisms
In category theory, "map" is often used as a synonym for morphism or arrow, thus for something more general than a function.[2]
## Other uses
### In logic
In formal logic, the term map is sometimes used for a functional predicate, whereas a function is a model of such a predicate in set theory.
### In graph theory
An example of a map in graph theory
In graph theory, a map is a drawing of a graph on a surface without overlapping edges (an embedding). If the surface is a plane then a map is a planar graph, similar to a political map.[3]
### In computer science
In the communities surrounding programming languages that treat functions as first-class citizens, a map often refers to the binary higher-order function that takes a function f and a list [v0, v1, ..., vn] as arguments and returns [f(v0), f(v1), ..., f(vn)], where n ≥ 0. | 499 | 2,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2018-34 | longest | en | 0.928232 |
https://aviation.stackexchange.com/questions/39690/what-is-the-meaning-of-the-percentage-used-to-describe-the-thinness-of-this-glid/39710 | 1,576,427,201,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541308604.91/warc/CC-MAIN-20191215145836-20191215173836-00074.warc.gz | 278,569,723 | 31,907 | # What is the meaning of the percentage used to describe the thinness of this glider's airfoil?
In a glider review for the Schempp-Hirth Discus, I encountered the following paragraph:
The Discus 2 airfoil is thin (about 14.5 percent), incorporating studies by K.H. Horstmann and Dr. Wuerz (wing) and Luc Boermans (tail).
After researching this a bit, I've been unable to find an explanation for what "14.5 percent" was referring to.
I thought perhaps that it is the ratio (14.5/100) of the thickness of the wing to the width of the wing, but that is really just an educated guess.
The Wikipedia page on airfoils has a section entitled Thin airfoil theory, with two bullets which also might qualify:
(1) on a symmetric airfoil, the center of pressure and aerodynamic center lies exactly one quarter of the chord behind the leading edge (2) on a cambered airfoil, the aerodynamic center lies exactly one quarter of the chord behind the leading edge
Is this "one quarter" the ratio that is mentioned in the article?
I did see this post, but had trouble seeing how the 90% and 99% numbers mentioned in the answer are related.
• 14.5% is a relative thickness (thickness divided by chord) – Gypaets Jul 5 '17 at 22:22
• You could call it the 'cross-sectional aspect ratio'. – amI Jul 5 '17 at 22:51
14.5% given in the article is the (maximum) thickness to chord ratio, expressed in percentage. It is usually used to explain how thin (or thick) the airfoil is. Note that this is a characteristic of the airfoil.
From the NASA page Wing geometry definitions:
Airfoil thickness to chord ratio, image from NASA page Wing geometry definitions
... The straight line drawn from the leading to trailing edges of the airfoil is called the chord line. ... The maximum distance between the two lines is called the camber, which is a measure of the curvature of the airfoil (high camber means high curvature).
The maximum distance between the upper and lower surfaces is called the thickness. Often you will see these values divided by the chord length to produce a non-dimensional or "percent" type of number.
• Well, at least the answer was on the page "intended for college, high school, or middle school students" :-). Seriously, the linked page does a great job of putting this stuff into layman's language. Thanks for the answer! – bclarkreston Jul 6 '17 at 20:23
• @bclarkreston, yes, the black line is the chord. It is also the $x$ axis, because the profile is normalized to have the leading edge at origin and the trailing edge at $(1, 0)$. The gray line is the camber line. – Jan Hudec Jul 6 '17 at 18:58 | 638 | 2,612 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2019-51 | latest | en | 0.94772 |
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April 26th, 2018, 09:28 PM #1 Senior Member Joined: Aug 2014 From: India Posts: 343 Thanks: 1 What is the meaning of "every choice we pick"? In how many ways, can you distribute 3 pencils to 8 people? In this case the order we pick people don’t matter. If I give a pencil first to A, then to B and then to C, it’s same as giving it to C first, followed by A and B. So, if we have 3 pencils to distribute, there are 3! or 6 variations for every choice we pick. What is the meaning of "every choice we pick"?
April 27th, 2018, 05:26 AM #2 Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 First one way to look at this problem is that there are 8 choices to give the first pencil, then 7 people left without a pencil, so 7 choices to give the second pencil, which leaves 6 people without a pencil so 6 choices to give the third pencil. That would be $\displaystyle 8(7)(6)= 336$ ways which could also be written as $\displaystyle \frac{8(7)(6)(5)(4)(3)(2)(1)}{5(4)(3)(2)(1)}= \frac{8!}{5!}= \frac{8!}{(8- 3)!}$. But that would be much too large because it counts the same three people, in different orders, as different answers. That is wrong because "the order we pick people don’t matter" Label the people A, B, C, D, E, F, G, H. One "choice" would be to give the first pencil to "C", the second pencil to "F", and the third pencil to "D": "CFD" in that order. But since "the order we pick people don’t matter", "CDF", "DCF", "DFC", "FDC", and "FCD" are the same three people chosen in different orders and should NOT be counted as different answers. There are 3!= 6 different orders in which we could pick the same three people. "Every choice we pick", the three people we pick to give pencils to, could have been chosen in 3!= 6 different ways. All of those different ways was counted in $\displaystyle frac{8!}{5!}$. In order NOT to count those different orders we must divide by 3!. The number of way we could give 3 pencils to 8 people NOT counting different orders (who got the first pencil, who got the second) is $\displaystyle \frac{8!}{5!3!}= 56$.
April 27th, 2018, 05:38 AM #3 Senior Member Joined: May 2016 From: USA Posts: 1,249 Thanks: 515 They are thinking about the problem in two steps. First step is a permutation. We have 8 ways to choose the first person to get a pencil, 7 ways to choose the second person, and 6 ways to choose the third person. That is $\dfrac{8!}{(8 - 3)!} = \dfrac{8 * 7 * 6 * 5!}{5!} = 8 * 7 * 6 = 336.$ But for whatever one of those 336 choices, say we chose abc, we actually have 3! variants in terms of order of each choice of three people, namely abc, acb, bac, bca, cab, cba. But in this case we don't care about order so we go to the second step and divide our answer from the first step by 6 or 3! $56 = \dfrac{336}{6} = \dfrac{\dfrac{8!}{5!}}{3!} = \dfrac{8!}{3! * (8 - 3)!}.$ This two-step process explains the formula used for counting combinations. Last edited by JeffM1; April 27th, 2018 at 05:41 AM.
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The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it? | 2,354 | 10,112 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-34 | latest | en | 0.865822 |
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LESSON 7-1
## LESSON 7-1
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1. LESSON 7-1 Preparing an Income Statement Component Percentages
2. INCOME STATEMENT INFORMATION ON A WORK SHEET page 182 LESSON 7-1
3. HEADING OF AN INCOME STATEMENT page 182 1 2 3 1. Center the name of the company on the first line. 2. Center the name of the report on the second line. 3. Center the date of the report on the third line. LESSON 7-1
4. Single line 7 Revenue 1 Revenue amount 3 Account title 2 Expenses 4 Expense amount 6 Account titles 5 Total expenses 8 Total of expenses 9 Single line 11 Net Income 12 Record net income 13 Calculate net income 10 Double lines 14 REVENUE, EXPENSES, AND NET INCOME SECTIONS OF AN INCOME STATEMENT page 183 LESSON 7-1
5. COMPONENT PERCENTAGE ANALYSIS OF AN INCOME STATEMENT page 184 LESSON 7-1
6. Total Revenue 5 Revenue 1 Revenue amounts 3 Account titles 2 Total of revenue 4 Net Loss 6 Record Net Loss 7 INCOME STATEMENT WITH TWO SOURCES OF REVENUE AND A NET LOSS page 185 LESSON 7-1
7. 7 Computing Component Percentage page 185 Total Expenses/ Total Sales \$4466.00 / \$3610.00= 1.2372 - MOVE the decimal 123.7% Net Loss / Total Sales (\$856.00) / \$3610.00= (.2372) - MOVE the decimal (23.7%) LESSON 7-1 | 450 | 1,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.537549 |
https://forum.lazyprogrammer.me/viewtopic.php?f=2&t=102&sid=3960e4730bddc5db1f08e981c16c3c67 | 1,638,824,100,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363312.79/warc/CC-MAIN-20211206194128-20211206224128-00340.warc.gz | 331,532,802 | 6,347 | ## Numpy exercise
Anything related to deep learning, machine learning, data science, and artificial intelligence. Course topic requests can go here.
Rodri
Posts: 4
Joined: Mon Jun 15, 2020 8:33 am
### Numpy exercise
Hi to all! I would like to share my results of Numpy exercise.
I have used a similar method used in numpy lecture, but using "time.perf_counter()" instead of "datetime.now()", why? because I did not find any difference and I usually use that.
The code is shown as follows:
Code: Select all
``````import numpy as np
import time
A = np.array([[1,2,3], [4,5,6], [7,8,9]])
#A = np.ones((3,3)) * 3
B = np.ones((3,3)) * 2
X = np.zeros((3,3)) #Matrix dot solution
sum1 = 0
sum2 = 0
med1 = 0
med2 = 0
#Manual method
for loop in range(0,500):
t1 = time.perf_counter()
for k in range(0,2+1):
for i in range(0,2+1):
e = 0
for j in range(0,2+1):
e = e + (A[k,j] * B[j,i])
X[k,i] = e
t2 = time.perf_counter()
sum1 = sum1 + (t2 - t1)
med1 = sum1/50
for loop2 in range(0,500):
t1 = time.perf_counter()
A.dot(B)
t2 = time.perf_counter()`````` | 370 | 1,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.84687 |
https://web2.0calc.com/questions/algebra_25283 | 1,726,315,133,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00610.warc.gz | 570,076,912 | 5,467 | +0
# Algebra
0
1
1
+80
Find all values of \$x\$ such that \$x^2 - 5x + 4 = -x^2 - 25x - 14\$. If you find more than one value, then list your solutions, separated by commas.
Aug 29, 2024
### 1+0 Answers
#1
+1018
0
x^2 - 5x + 4 = -x^2 - 25x - 14
2x2 + 20x + 18 = 0
x2 + 10x + 9 = 0
(x + 1)(x + 9) = 0
(x + 1) = 0 ==>> x = –1
(x + 9) = 0 ==>> x = –9
x = –1, –9
.
Aug 29, 2024 | 214 | 410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-38 | latest | en | 0.39996 |
https://www.aqua-calc.com/calculate/gravel-quarter-cylinder/substance/caribsea-coma-and-blank-marine-coma-and-blank-arag-alive-coma-and-blank-special-blank-grade-blank-reef | 1,713,834,725,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00359.warc.gz | 589,486,711 | 8,551 | # Qt Cyl: CaribSea Marine Arag-Alive Special Grade Reef
## quarter cylindrical aquarium sand, gravel and substrate calculato...57
### The weight of CaribSea, Marine, Arag-Alive, Special Grade Reef, 2 inches high, in a quarter cylindrical aquarium with the radius of a base of 10 inches
gram 3 504.86 ounce 123.63 kilogram 3.5 pound 7.73
### The volume of CaribSea, Marine, Arag-Alive, Special Grade Reef, 2 inches high, in a quarter cylindrical aquarium with the radius of a base of 10 inches
centimeter³ 2 574.07 inch³ 157.08 foot³ 0.09 meter³ 0
• CaribSea, Marine, Arag-Alive, Special Grade Reef weighs 1.36 gram per cubic centimeter or 1 361.6 kilogram per cubic meter, i.e. density of caribSea, Marine, Arag-Alive, Special Grade Reef is equal to 1 361.6 kg/m³. In Imperial or US customary measurement system, the density is equal to 85 pound per cubic foot [lb/ft³], or 0.79 ounce per cubic inch [oz/inch³] .
• CaribSea, Marine, Arag-Alive, Special Grade Reef weighs 1 361.6 kg/m³ (85.00191 lb/ft³) with specific gravity of 1.3616 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
• The corner bow front aquariums are designed to fit in the corner of a room and to provide optimum viewing space. There is a quarter of a circle lies at the base of the aquarium with radius (r).
The number in square brackets, to the right of the gravel name, is the gravel specific gravity or relative density, as compared to pure water.
The calculator computes the weight (1) and volume (2) of gravel using the following formulas:
(1) m = ρ × V and (2) V = π×r²×h ⁄ 4, where ρ is the density, V — volume, m — weight (or mass), and h is the height of gravel; π is approximately equal to 3.14159265359, and r is the radius of a base of an aquarium.
See also our gravel calculators for cylindrical and rectangular shaped aquariums and ponds.
#### Foods, Nutrients and Calories
POP!, GOURMET POPCORN, INFUSED CHOCOLATE CARAMEL SEA SALT, UPC: 855332005229 contain(s) 467 calories per 100 grams (≈3.53 ounces) [ price ]
86615 foods that contain Potassium, K. List of these foods starting with the highest contents of Potassium, K and the lowest contents of Potassium, K
#### Gravels, Substances and Oils
CaribSea, Marine, Aragonite, Super Reef weighs 1 361.6 kg/m³ (85.00191 lb/ft³) with specific gravity of 1.3616 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Paving Asphalt Cement, AR-2000 weighs 1 022.12 kg/m³ (63.80887 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Walnut oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
A Petaohm is a SI-multiple (see prefix Peta) of the electric resistance unit ohm and equal to 1.0 × 10+15 Ω)
The electric potential φ(A) of a point A (in the electric field) specifies the work to be done by the force F = −Q × E in order to move the charge Q from a fixed reference point P to the point A.
lb/ft³ to long tn/fl.oz conversion table, lb/ft³ to long tn/fl.oz unit converter or convert between all units of density measurement.
#### Calculators
Calculate gravel and sand coverage in a corner bow front aquarium | 967 | 3,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-18 | latest | en | 0.771473 |
https://bernardparent.ca/viewforum.php?f=54&sid=fe460eecdeaa4e61ca82353f061a761e | 1,532,159,152,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592420.72/warc/CC-MAIN-20180721071046-20180721091046-00460.warc.gz | 609,285,793 | 6,574 | AE23837 Numerical Analysis Numerical analysis is the branch of mathematics involved with the development of efficient methods for solving problems using a computer. In this second-year undergraduate course, we will cover various numerical methods including numerical integration, numerical differentiation, root finding of linear and non-linear systems of equations, and curve fitting. For each numerical method shown in class, the student will be taught how to write a C code.
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Numerical Analysis Questions & Answers 01.03.2018 STICKY QNA 1 ... 12 , 13 , 14
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Numerical Analysis Scores 12.21.2017 STICKY
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Numerical Analysis Class Absence 11.22.2017 STICKY DISCUSSION 1 , 2
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Numerical Analysis C IDE 09.09.2017 STICKY
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Numerical Analysis Syllabus 08.30.2016 STICKY
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Numerical Analysis Final Exam Poll 12.14.2017 DISCUSSION 1 , 2
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2017 Numerical Analysis Final Exam 12.12.2017
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2017 Numerical Analysis Midterm Exam 10.20.2017
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Numerical Analysis Assignment 2 — Root Finding 10.02.2017
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Numerical Analysis Assignment 1 — IEEE Arithmetic 09.13.2017
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Numerical Analysis Assignment 8 — Numerical Differentiation 08.29.2017
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Numerical Analysis Assignment 4 — Partial Pivoting and Non-Linear Systems 08.29.2017
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Numerical Analysis Assignment 7 — Numerical Integration 08.29.2017
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Numerical Analysis Assignment 6 — Piecewise Interpolation and Splines 08.29.2017
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Numerical Analysis Assignment 5 — Curve Fitting and Interpolation 08.29.2017
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Numerical Analysis Assignment 3 — Systems of Equations 08.29.2017
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Numerical Analysis Information 12.28.2016
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2016 Numerical Analysis Midterm Exam 12.26.2016
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2016 Numerical Analysis Final Exam 12.26.2016
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Numerical Analysis Assignment 0 — Registration 08.30.2016
$\pi$ | 588 | 1,858 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-30 | latest | en | 0.7336 |
http://jean-pierre.moreau.pagesperso-orange.fr/Pascal/carpol_pas.txt | 1,606,594,497,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-50/segments/1606141195745.90/warc/CC-MAIN-20201128184858-20201128214858-00426.warc.gz | 47,940,452 | 2,136 | {******************************************************** * Calculate the coefficients of the characteristic poly-* * nomial P(l)=A-l*I of a real tridiagonal matrix A(i,j) * * ----------------------------------------------------- * * Ref.: "Algèbre, Algorithmes et programmes en Pascal * * By Jean-Louis Jardrin, DUNOD Paris, 1988" * * [BIBLI 10]. * * ----------------------------------------------------- * * SAMPLE RUN: * * (Find the caracteristic polynomial of matrix: * * 4 1 0 0 0 0 * * 2 4 1 0 0 0 * * A = 0 2 4 1 0 0 * * 0 0 2 4 1 0 * * 0 0 0 2 4 1 * * 0 0 0 0 2 4 ) * * * * Input matrix size (max 100): 6 * * Input 5 elements of subdiagonal: * * Element 1: 2 * * Element 2: 2 * * Element 3: 2 * * Element 4: 2 * * Element 5: 2 * * Input 6 elements of main diagonal: * * Element 1: 4 * * Element 2: 4 * * Element 3: 4 * * Element 4: 4 * * Element 5: 4 * * Element 6: 4 * * Input 5 elements of supradiagonal: * * Element 1: 1 * * Element 2: 1 * * Element 3: 1 * * Element 4: 1 * * Element 5: 1 * * * * Chracteristic polynomial P(l): * * Coefficient for degree 6: 1.0000000000E+000 * * Coefficient for degree 5: -2.4000000000E+001 * * Coefficient for degree 4: 2.3000000000E+002 * * Coefficient for degree 3: -1.1200000000E+003 * * Coefficient for degree 2: 2.9040000000E+003 * * Coefficient for degree 1: -3.7760000000E+003 * * Coefficient for degree 0: 1.9120000000E+003 * * * * (The characteristic polynomial of matrix A is: * * P(l)=l^6-24*l^5+230*l^4-1120*l^3+2904*l^2-3376*l * * +1912 ) * * * * English version by J-P Moreau, Paris * * (with dynamically allocated vectors). * * (www.jpmoreau.fr) * ********************************************************} Program CARPOL; Uses WinCrt; CONST NMAX = 100; NMAXP = 101; TYPE pV = ^VEC; VEC = Array[1..NMAXP] of Double; VAR k,n: Integer; B,C,D,E,P: pV; {**************************************************** * The procedure PCTR calculates the coefficients of * * the characteristic polynomial P(l)=A-l*I of a real* * square tridiagonal matrix A(i,j). * * * * Note: the roots of P(l) are the eigenvalues of * * matrix A(i,j). * * ------------------------------------------------- * * INPUTS: * * n: zize of matrix (integer) * * B: vector of codiagonal terms products * * (see main program). * * D: vector of main diagonal terms * * OUTPUT: * * P: vector of coefficients of P(l) * * * * The type pV (pointer to a real vector) must be * * declared and allocated in the calling program. * ****************************************************} Procedure PCTR(n:integer; B,D:pV; VAR P:pV); Var i,k:integer; T:pV; Begin New(T); P^[2]:=D^[1]; T^[1]:=1.0; P^[1]:=-T^[1]; For k:=2 to n do begin P^[k+1]:=D^[k]*P^[k]-B^[k-1]*T^[k-1]; for i:=k downto 3 do begin T^[i]:=P^[i]; P^[i]:=-T^[i]+D^[k]*P^[i-1]-B^[k-1]*T^[i-2] end; T^[2]:=P^[2]; P^[2]:=-T^[2]+D^[k]*P^[1]; T^[1]:=P^[1]; P^[1]:=-T^[1] end; Dispose(T) End; Procedure Read_data; Var i:Integer; Begin writeln; write(' Input matrix size (max ',NMAX,'): '); readln(n); writeln(' Input ',n-1,' elements of subdiagonal:'); for i:=1 to n-1 do begin write(' Element ',i,': '); readln(C^[i]) end; writeln(' Input ',n,' elements of main diagonal:'); for i:=1 to n do begin write(' Element ',i,': '); readln(D^[i]) end; writeln(' Input ',n-1,' elements of supradiagonal:'); for i:=1 to n-1 do begin write(' Element ',i,': '); readln(E^[i]) end End; {main program} BEGIN New(B); New(C); New(D); New(E); New(P); Read_data; {prepare B vector} for k:=1 to n-1 do B^[k]:=C^[k]*E^[k]; Dispose(C); Dispose(E); {call routine to calculate coefficients} PCTR(n,B,D,P); {print results} writeln; writeln(' Characteristic polynomial P(l):'); For k:=1 to n+1 do writeln(' Coefficient for degree ',n-k+1,': ',P^[k]); {exit section} Readkey; Dispose(B); Dispose(D); Dispose(P); DoneWinCrt END. {end of file carpol.pas} | 1,284 | 3,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-50 | longest | en | 0.509786 |
https://www.mathplanet.com/education/act/problems-21-40/34-what-is-the-larger-number | 1,695,444,555,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506479.32/warc/CC-MAIN-20230923030601-20230923060601-00302.warc.gz | 983,632,214 | 7,799 | # 34. What is the larger number?
$\begin{array}{lcl} The\;sum\;of\;two\;numbers\;is\;9\;and\\ one\;number\;is\;5\;times\;the\;other.\\ What\;is\;the\;larger\;of\;the\;two\;numbers?\\ \\ (F)\; 3\\ (G)\; 4.5\\ (H)\; 5\\ (J)\; 6\\ (K)\; 7.5\\ \end{array}$ | 133 | 253 | {"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} | 3.0625 | 3 | CC-MAIN-2023-40 | longest | en | 0.224202 |
http://math.stackexchange.com/questions/660046/find-this-limit-without-using-lhospitals-rule | 1,469,514,910,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824756.90/warc/CC-MAIN-20160723071024-00148-ip-10-185-27-174.ec2.internal.warc.gz | 156,408,949 | 17,435 | # Find this limit without using L'Hospital's rule
I have to find this limit without using l'Hôspital's rule:
$$\lim_{x\to 0} \frac{\alpha \sin \beta x - \beta \sin \alpha x}{x^2 \sin \alpha x}$$
Using L'Hôspital's rule gives:
$$\frac{\beta}{6(\alpha^2 - \beta^2)}$$ I am stuck where to begin without using the rule.
-
If you haven't learned taylor series yet (tbongers answer), I think you can use the squeeze theorem to do this one, although I suspect the solution will be far more difficult. – MHH Feb 2 '14 at 1:03
Possible duplicate of this. – user 170039 Oct 19 '14 at 17:07
Using the Taylor series
$$\sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \dots$$
the numerator is
\begin{align*} \alpha \left(\beta x - \frac{(\beta x)^3}{3!} + O(x^5)\right) - \beta \left(\alpha x - \frac{(\alpha x)^3}{3!} + O(x^5)\right) = \frac{\beta \alpha^3 - \alpha \beta^3}{6} x^3 + O(x^5) \end{align*}
Then the fraction can be written as
\begin{align*} \frac{\dfrac{\beta \alpha^3 - \alpha \beta^3}{6} x^3 + O(x^5)}{\alpha x^3 + O(x^5)} \to \frac{\beta \alpha^3 - \alpha \beta^3}{6\alpha} \end{align*}
as $x \to 0$.
-
I think that you had a mistake somewhere when you applied L'Hospital's rule and arrived to $$\frac{\beta}{6(\alpha^2 - \beta^2)}$$ as reported in your post.
To get rid of problems with $x$ in the denominator, you must apply L'Hospital's rule three times and arrive to $$\frac{\beta \alpha^3 - \alpha \beta^3}{6\alpha}$$ which, fortunately (!), matches what T. Bongers obtained using Taylor series.
- | 524 | 1,521 | {"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} | 4.34375 | 4 | CC-MAIN-2016-30 | latest | en | 0.797308 |
https://kerodon.net/tag/02SM | 1,680,349,955,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949958.54/warc/CC-MAIN-20230401094611-20230401124611-00405.warc.gz | 379,632,870 | 8,749 | # Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
### 5.7.9 Proof of the Universality Theorem
Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a cocartesian fibration of simplicial sets, and suppose that the fiber $\operatorname{\mathcal{E}}_{C} = \{ C\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}$ is essentially small for each vertex $C \in \operatorname{\mathcal{C}}$. Our goal in this section is to prove Theorem 5.7.8.3, which that the space of transport witnesses $\operatorname{TW}(\operatorname{\mathcal{E}}/\operatorname{\mathcal{C}})$ of Notation 5.7.8.1 is a contractible Kan complex. The main step is to establish the following:
Lemma 5.7.9.1. Let $U: \operatorname{\mathcal{E}}\rightarrow \Delta ^1$ be a cocartesian fibration having fibers $\operatorname{\mathcal{E}}_{0} = \{ 0\} \times _{\Delta ^1} \operatorname{\mathcal{E}}$ and $\operatorname{\mathcal{E}}_{1} = \{ 1\} \times _{\Delta ^1} \operatorname{\mathcal{E}}$. Then the restriction map
$\theta : \operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^1 ) \rightarrow \operatorname{TW}( \operatorname{\mathcal{E}}_0 \amalg \operatorname{\mathcal{E}}_{1} / \operatorname{\partial \Delta }^{1} )$
is a trivial Kan fibration of simplicial sets.
Proof. It follows from Lemma 5.7.8.10 that $\theta$ is a Kan fibration; we wish to show that it is a trivial Kan fibration. Fix a pair of small $\infty$-categories $\operatorname{\mathcal{D}}_0$ and $\operatorname{\mathcal{D}}_1$. Set $\operatorname{\mathcal{E}}'_0 = \{ \operatorname{\mathcal{D}}_0 \} \times _{\operatorname{\mathcal{QC}}} \operatorname{\mathcal{QC}}_{\operatorname{Obj}}$ and $\operatorname{\mathcal{E}}'_1 = \{ \operatorname{\mathcal{D}}_1 \} \times _{ \operatorname{\mathcal{QC}}} \operatorname{\mathcal{QC}}_{\operatorname{Obj}}$, and let $\operatorname{Equiv}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_0 )$ and $\operatorname{Equiv}( \operatorname{\mathcal{E}}_1, \operatorname{\mathcal{E}}'_1 )$ be the Kan complexes introduced in Example 5.7.8.2, so that the fiber
$\{ (\operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1) \} \times _{ \operatorname{Fun}( \operatorname{\partial \Delta }^1, \operatorname{\mathcal{QC}}) } \operatorname{TW}( \operatorname{\mathcal{E}}_0 \amalg \operatorname{\mathcal{E}}_1 / \operatorname{\partial \Delta }^1 )$
can be identified with the product $\operatorname{Equiv}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_0) \times \operatorname{Equiv}( \operatorname{\mathcal{E}}_{1}, \operatorname{\mathcal{E}}'_1 )$. Let $\operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^{1})_{\operatorname{\mathcal{D}}_0,\operatorname{\mathcal{D}}_1}$ denote the fiber product $\{ (\operatorname{\mathcal{D}}_0,\operatorname{\mathcal{D}}_1) \} \times _{ \operatorname{Fun}( \operatorname{\partial \Delta }^1, \operatorname{\mathcal{QC}}) } \operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^1)$, so that $\theta$ restricts to a Kan fibration
$\theta _{\operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1}: \operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^{1})_{\operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1} \rightarrow \operatorname{Equiv}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_0 ) \times \operatorname{Equiv}( \operatorname{\mathcal{E}}_1, \operatorname{\mathcal{E}}'_1 ).$
Note that every fiber of $\theta$ can also be viewed as a fiber of $\theta _{\operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1}$ for suitably chosen $\infty$-categories $\operatorname{\mathcal{D}}_0$ and $\operatorname{\mathcal{D}}_1$. Consequently, to show that $\theta$ is a trivial Kan fibration, it will suffice to show that each of the morphisms $\theta _{\operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1}$ is a trivial Kan fibration (Proposition 3.2.6.15), or equivalently that it is a homotopy equivalence (Corollary 3.2.7.4).
For the remainder of the proof, we will regard the $\infty$-categories $\operatorname{\mathcal{D}}_0$ and $\operatorname{\mathcal{D}}_1$ as fixed. Let $\operatorname{\mathcal{B}}^{+}$ denote the fiber product
$\operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{D}}_0 , \operatorname{\mathcal{D}}_1) \times _{ \operatorname{Fun}( \Delta ^1 \times \operatorname{\mathcal{E}}_{0}, \operatorname{\mathcal{QC}})^{\simeq } } \operatorname{Fun}( \Delta ^1 \times \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{QC}}_{\operatorname{Obj}} )^{\simeq }.$
Let $\pi ^{+}: \operatorname{\mathcal{B}}^{+} \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1)$ be given by projection onto the first factor, and let
$r_{0}^{+}: \operatorname{\mathcal{B}}^{+} \rightarrow \operatorname{Fun}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_0)^{\simeq } \quad \quad r^{+}_1: \operatorname{\mathcal{B}}^{+} \rightarrow \operatorname{Fun}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_1)^{\simeq }$
be given by restriction to the simplicial subsets $\{ 0\} \times \operatorname{\mathcal{E}}_{0}$ and $\{ 1\} \times \operatorname{\mathcal{E}}_{0}$, respectively. Combining Propositions 4.4.5.1 and 4.4.3.7, we deduce that the map
$(r^{+}_0, r^{+}_1, \pi ^{+}): \operatorname{\mathcal{B}}^{+} \rightarrow \operatorname{Fun}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_{0})^{\simeq } \times \operatorname{Fun}( \operatorname{\mathcal{E}}_{0}, \operatorname{\mathcal{E}}'_{1})^{\simeq } \times \operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1)$
is a Kan fibration. In particular, the simplicial set $\operatorname{\mathcal{B}}^{+}$ is a Kan complex.
Let $\operatorname{\mathcal{B}}$ denote the summand $\operatorname{\mathcal{B}}^{+}$ spanned by those pairs $(e, \widetilde{e})$, where $e: \operatorname{\mathcal{D}}_0 \rightarrow \operatorname{\mathcal{D}}_1$ is a functor and $\widetilde{e}: \Delta ^1 \times \operatorname{\mathcal{E}}_{0} \rightarrow \operatorname{\mathcal{QC}}_{\operatorname{Obj}}$ is a morphism fitting into a commutative diagram
$\xymatrix@R =50pt@C=50pt{ \Delta ^1 \times \operatorname{\mathcal{E}}_{0} \ar [r]^-{ \widetilde{e} } \ar [d] & \operatorname{\mathcal{QC}}_{\operatorname{Obj}} \ar [d]^{V} \\ \Delta ^1 \ar [r]^-{e} & \operatorname{\mathcal{QC}}}$
which satisfies the following pair of conditions:
• The restriction $\widetilde{e}|_{ \{ 0\} \times \operatorname{\mathcal{E}}_0 }: \operatorname{\mathcal{E}}_0 \rightarrow \operatorname{\mathcal{E}}'_{0}$ is an equivalence of $\infty$-categories.
• For each object $Z \in \operatorname{\mathcal{E}}_{0}$, the composite map
$\Delta ^1 \times \{ Z\} \hookrightarrow \Delta ^1 \times \operatorname{\mathcal{E}}_0 \xrightarrow { \widetilde{e} } \operatorname{\mathcal{QC}}_{\operatorname{Obj}}$
is a $V$-cocartesian morphism of $\operatorname{\mathcal{QC}}_{\operatorname{Obj}}$.
Condition $(i)$ ensures that $r_{0}^{+}$ restricts to a morphism of Kan complexes $r_0: \operatorname{\mathcal{B}}\rightarrow \operatorname{Equiv}( \operatorname{\mathcal{E}}_{0}, \operatorname{\mathcal{E}}'_{0} )$. Moreover, $\pi ^{+}$ and $r_{1}^{+}$ restrict to morphisms $\pi : \operatorname{\mathcal{B}}\rightarrow \operatorname{Hom}_{\operatorname{\mathcal{QC}}}(\operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1)$ and $r_{1}: \operatorname{\mathcal{B}}\rightarrow \operatorname{Fun}(\operatorname{\mathcal{E}}_{0},\operatorname{\mathcal{E}}'_1)^{\simeq }$, respectively. Since $\operatorname{\mathcal{B}}$ is a summand of $\operatorname{\mathcal{B}}^{+}$, the map
$(r_0, r_1,\pi ): \operatorname{\mathcal{B}}\rightarrow \operatorname{Equiv}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_0) \times \operatorname{Fun}( \operatorname{\mathcal{E}}_{0}, \operatorname{\mathcal{E}}'_1)^{\simeq } \times \operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1)$
is also a Kan fibration.
It follows from Theorem 5.2.1.1 that composition with $V$ induces a cocartesian fibration $V': \operatorname{Fun}( \operatorname{\mathcal{E}}_{0}, \operatorname{\mathcal{QC}}_{\operatorname{Obj}} ) \rightarrow \operatorname{Fun}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{QC}})$. Moreover, a morphism of the $\infty$-category $\operatorname{Fun}( \operatorname{\mathcal{E}}_{0}, \operatorname{\mathcal{QC}}_{\operatorname{Obj}} )$ is $V'$-cocartesian if and only if it corresponds to a morphism of simplicial sets $\widetilde{e}: \Delta ^1 \times \operatorname{\mathcal{E}}_0 \rightarrow \operatorname{\mathcal{QC}}_{\operatorname{Obj}}$ satisfying condition $(ii)$. Let $\operatorname{Fun}'( \Delta ^1 \times \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{QC}}_{\operatorname{Obj}})$ denote the full subcategory of $\operatorname{Fun}( \Delta ^1 \times \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{QC}}_{\operatorname{Obj}} )$ spanned by morphisms which satisfy this condition. Unwinding the definitions, we have a pullback square
$\xymatrix@R =50pt@C=-70pt{ & \operatorname{\mathcal{B}}\ar [dl]_{ (r_0, \pi ) } \ar [dr] & \\ \operatorname{Equiv}( \operatorname{\mathcal{E}}_{0}, \operatorname{\mathcal{E}}'_{0} ) \times \operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1) \ar [dr] & & \operatorname{Fun}'( \Delta ^1 \times \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{QC}}_{\operatorname{Obj}}) \ar [dl] \\ & \operatorname{Fun}( \{ 0\} \times \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{QC}}_{\operatorname{Obj}} ) \times _{ \operatorname{Fun}( \{ 0\} \times \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{QC}}) } \operatorname{Fun}( \Delta ^1 \times \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{QC}}), & }$
where the bottom right map is a trivial Kan fibration (Proposition 5.2.1.3). It follows that the map $(r_0, \pi ): \operatorname{\mathcal{B}}\rightarrow \operatorname{Equiv}( \operatorname{\mathcal{E}}_{0}, \operatorname{\mathcal{E}}'_{0} ) \times \operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1)$ is a trivial Kan fibration of simplicial sets.
Let $s: \operatorname{Equiv}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_{0} ) \times \operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1 ) \rightarrow \operatorname{\mathcal{B}}$ be a section of the trivial Kan fibration $(r_0, \pi )$, and let $T$ denote the composite map
$\operatorname{Equiv}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_{0} ) \times \operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1) \xrightarrow {s} \operatorname{\mathcal{B}}\xrightarrow { (r_0, r_1) } \operatorname{Equiv}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_{0} ) \times \operatorname{Fun}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_{1} )^{\simeq }.$
For every equivalence of $\infty$-categories $F: \operatorname{\mathcal{E}}_{0} \rightarrow \operatorname{\mathcal{E}}'_{0}$, we can regard $T|_{ \{ F\} \times \operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1) }$ as a morphism of Kan complexes $T_{F}: \operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1) \rightarrow \operatorname{Fun}( \operatorname{\mathcal{E}}_{0}, \operatorname{\mathcal{E}}'_{1} )^{\simeq }$. Unwinding the definitions, we can identify $T_{F}$ with the composition
$\operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1) \xrightarrow {T'} \operatorname{Fun}( \operatorname{\mathcal{E}}'_{0}, \operatorname{\mathcal{E}}'_{1} )^{\simeq } \xrightarrow { \circ F } \operatorname{Fun}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_{1} )^{\simeq },$
where $T'$ is given by parametrized covariant transport for the cocartesian fibration $V: \operatorname{\mathcal{QC}}_{\operatorname{Obj}} \rightarrow \operatorname{\mathcal{QC}}$ (Definition 5.2.8.1). It follows from Proposition 5.6.6.14 that $T'$ is a homotopy equivalence. Our assumption that $F$ is an equivalence of $\infty$-categories then guarantees that $T_{F}$ is also a homotopy equivalence. Allowing $F \in \operatorname{Equiv}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_{0} )$ to vary and applying Proposition 3.2.8.1, we conclude that $T$ is a homotopy equivalence. Since $s$ is homotopy inverse to the trivial Kan fibration $(r_0, \pi )$, it is also a homotopy equivalence. Applying the two-out-of-three property (Remark 3.1.6.7), we conclude that the map
$(r_0, r_1): \operatorname{\mathcal{B}}\rightarrow \operatorname{Equiv}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_{0}) \times \operatorname{Fun}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_{1} )^{\simeq }$
is also a homotopy equivalence. Since $(r_0, r_1)$ is also a Kan fibration, it is a trivial Kan fibration (Proposition 3.3.7.4).
Using Proposition 5.2.2.8, we can choose a functor $\lambda : \operatorname{\mathcal{E}}_0 \rightarrow \operatorname{\mathcal{E}}_{1}$ and a natural transformation $h: \Delta ^1 \times \operatorname{\mathcal{E}}_0 \rightarrow \operatorname{\mathcal{E}}$ which witnesses $\lambda$ as given by covariant transport along the nondegenerate edge of $\Delta ^1$ (in the sense of Definition 5.2.2.4). Form a pullback diagram
5.64
$$\begin{gathered}\label{equation:universality-formulation-0} \xymatrix@R =50pt@C=50pt{ \widetilde{\operatorname{\mathcal{B}}} \ar [r] \ar [d]^{ ( \widetilde{r}_0, \widetilde{r}_1)} & \operatorname{\mathcal{B}}\ar [d]^{ (r_0, r_1) } \\ \operatorname{Equiv}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_{0} ) \times \operatorname{Equiv}( \operatorname{\mathcal{E}}_{1}, \operatorname{\mathcal{E}}'_{1}) \ar [r]^-{\circ \lambda } & \operatorname{Equiv}( \operatorname{\mathcal{E}}_{0}, \operatorname{\mathcal{E}}'_{0} ) \times \operatorname{Fun}(\operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_{1} )^{\simeq }. } \end{gathered}$$
Let $\operatorname{\mathcal{M}}$ denote the pushout $( \Delta ^1 \times \operatorname{\mathcal{E}}_0) \coprod _{ ( \{ 1\} \times \operatorname{\mathcal{E}}_0 ) } \operatorname{\mathcal{E}}_{1}$, so that we can identify $\widetilde{\operatorname{\mathcal{B}}}$ with a summand of the Kan complex
$\operatorname{Hom}_{\operatorname{\mathcal{QC}}}(\operatorname{\mathcal{D}}_0,\operatorname{\mathcal{D}}_1) \times _{ \operatorname{Fun}( \operatorname{\mathcal{M}}, \operatorname{\mathcal{QC}})^{\simeq } } \operatorname{Fun}( \operatorname{\mathcal{M}}, \operatorname{\mathcal{QC}}_{\operatorname{Obj}} )^{\simeq }.$
Note that $h$ induces a categorical equivalence of simplicial sets $h^{+}: \operatorname{\mathcal{M}}\rightarrow \operatorname{\mathcal{E}}$ (Corollary 5.2.4.2). We have a commutative diagram of Kan complexes
5.65
$$\begin{gathered}\label{equation:universality-formulation} \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1) \times _{ \operatorname{Fun}( \operatorname{\mathcal{E}}, \operatorname{\mathcal{QC}})^{\simeq } } \operatorname{Fun}( \operatorname{\mathcal{E}}, \operatorname{\mathcal{QC}}_{\operatorname{Obj}} )^{\simeq } \ar [d] \ar [r] & \operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1) \ar [d] \\ \operatorname{Fun}( \operatorname{\mathcal{E}}, \operatorname{\mathcal{QC}}_{\operatorname{Obj}} )^{\simeq } \ar [d]^{\circ h^{+}} \ar [r]^-{ V \circ } & \operatorname{Fun}( \operatorname{\mathcal{E}}, \operatorname{\mathcal{QC}})^{\simeq } \ar [d]^-{ \circ h^{+}} \\ \operatorname{Fun}( \operatorname{\mathcal{M}}, \operatorname{\mathcal{QC}}_{\operatorname{Obj}} )^{\simeq } \ar [r]^-{ V \circ } & \operatorname{Fun}( \operatorname{\mathcal{M}}, \operatorname{\mathcal{QC}})^{\simeq }, } \end{gathered}$$
where the upper vertical are homotopy equivalences (since $h^{+}$ is a categorical equivalence) and the horizontal maps are Kan fibrations (Corollary 4.4.5.7). Note that the top and bottom squares of (5.65) are homotopy pullback squares (Example 3.4.1.3 and Corollary 3.4.1.5). It follows that the outer rectangle is also a homotopy pullback square (Proposition 3.4.1.11): that is, precomposition with $h^{+}$ induces a homotopy equivalence of Kan complexes
$\xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\operatorname{\mathcal{QC}}}(\operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1) \times _{ \operatorname{Fun}( \operatorname{\mathcal{E}}, \operatorname{\mathcal{QC}})^{\simeq } } \operatorname{Fun}( \operatorname{\mathcal{E}}, \operatorname{\mathcal{QC}}_{\operatorname{Obj}} )^{\simeq } \ar [d]^{ \varphi } \\ \operatorname{Hom}_{\operatorname{\mathcal{QC}}}(\operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1) \times _{ \operatorname{Fun}( \operatorname{\mathcal{M}}, \operatorname{\mathcal{QC}})^{\simeq } } \operatorname{Fun}( \operatorname{\mathcal{M}}, \operatorname{\mathcal{QC}}_{\operatorname{Obj}} )^{\simeq }. }$
Applying Remark 5.7.5.3, we see that $\operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^1)_{\operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1}$ can be identified with the inverse image of $\widetilde{\operatorname{\mathcal{B}}}$ under the homotopy equivalence $\varphi$. In particular, $\varphi$ restricts to a homotopy equivalence $\varphi _0: \operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^1 )_{ \operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1} \rightarrow \widetilde{\operatorname{\mathcal{B}}}$. Unwinding the definitions, we see that the morphism
$\theta _{\operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1}: \operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^1 )_{ \operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1} \rightarrow \operatorname{Equiv}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_{0} ) \times \operatorname{Equiv}( \operatorname{\mathcal{E}}_{1}, \operatorname{\mathcal{E}}'_{1})$
coincides with the the composition $( \widetilde{r}_0, \widetilde{r}_1 ) \circ \varphi _0$. Since $( \widetilde{r}_0, \widetilde{r}_1)$ is a pullback of the trivial Kan fibration $(r_0, r_1): \operatorname{\mathcal{B}}\rightarrow \operatorname{Equiv}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_{0} ) \times \operatorname{Fun}( \operatorname{\mathcal{E}}_0, \operatorname{\mathcal{E}}'_{1} )^{\simeq }$, it is also a trivial Kan fibration. In a particular, $( \widetilde{r}_0, \widetilde{r}_1 )$ is a homotopy equivalence, so that the composite map $\theta _{\operatorname{\mathcal{D}}_0, \operatorname{\mathcal{D}}_1} = ( \widetilde{r}_0, \widetilde{r}_1 ) \circ \varphi _0$ is also a homotopy equivalence, as desired. $\square$
Lemma 5.7.9.2. Let $\operatorname{\mathcal{E}}$ be an essentially small $\infty$-category. Then the simplicial set $\operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^0 )$ is a contractible Kan complex.
Proof. It follows from Lemma 5.7.8.7 that the simplicial set $\operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^0 )$ is a Kan complex. Since $\operatorname{\mathcal{E}}$ is essentially small, the Kan complex $\operatorname{TW}(\operatorname{\mathcal{E}}/ \Delta ^0)$ is nonempty. It will therefore suffice to show that the diagonal map
$\delta : \operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^0 ) \rightarrow \operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^0 ) \times \operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^0 )$
is a homotopy equivalence (Corollary 3.2.7.6). Unwinding the definitions, we see that $\delta$ factors as a composition
\begin{eqnarray*} \operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^0 ) & \xrightarrow {\delta '} & \operatorname{Fun}( \Delta ^1, \operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^0) ) \\ & \simeq & \operatorname{TW}( \Delta ^1 \times \operatorname{\mathcal{E}}/ \Delta ^1 ) \\ & \xrightarrow {\delta ''} & \operatorname{TW}( \operatorname{\partial \Delta }^1 \times \operatorname{\mathcal{E}}/ \operatorname{\partial \Delta }^1 ) \\ & \simeq & \operatorname{TW}( \operatorname{\mathcal{E}}/ \Delta ^0 ) \times \operatorname{TW}(\operatorname{\mathcal{E}}/ \Delta ^0 ). \end{eqnarray*}
Since the $1$-simplex $\Delta ^1$ is contractible (Example 3.2.6.3), the morphism $\delta '$ is a homotopy equivalence. It will therefore suffice to show that the restriction map $\delta ''$ is a homotopy equivalence, which follows from Lemma 5.7.9.1. $\square$
Proof of Theorem 5.7.8.3. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a cocartesian fibration of simplicial sets. Assume that, for each vertex $C \in \operatorname{\mathcal{C}}$, the $\infty$-category $\operatorname{\mathcal{E}}_{C} = \{ C\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}$ is essentially small. We wish to show that the simplicial set $\operatorname{TW}(\operatorname{\mathcal{E}}/ \operatorname{\mathcal{C}})$ is a contractible Kan complex.
For every simplicial set $\operatorname{\mathcal{C}}_0$ equipped with a morphism $\operatorname{\mathcal{C}}_0 \rightarrow \operatorname{\mathcal{C}}$, let $X( \operatorname{\mathcal{C}}_0 )$ denote the simplicial set $\operatorname{TW}( \operatorname{\mathcal{E}}_0 / \operatorname{\mathcal{C}}_0 )$, where $\operatorname{\mathcal{E}}_0$ is the fiber product $\operatorname{\mathcal{C}}_0 \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}$. Note that the simplicial set $X(\operatorname{\mathcal{C}}) = \operatorname{TW}(\operatorname{\mathcal{E}}/ \operatorname{\mathcal{C}})$ can be realized as the inverse limit of the tower
$\cdots \rightarrow X( \operatorname{sk}_{2}(\operatorname{\mathcal{C}}) ) \rightarrow X( \operatorname{sk}_{1}(\operatorname{\mathcal{C}}) ) \rightarrow X(\operatorname{sk}_{0}(\operatorname{\mathcal{C}}) ),$
where each of the transition maps is a Kan fibration (Lemma 5.7.8.10). Consequently, to show that $X(\operatorname{\mathcal{C}})$ is a contractible Kan complex, it will suffice to show that each of the simplicial sets $X( \operatorname{sk}_{k}(\operatorname{\mathcal{C}}) )$ is a contractible Kan complex. Replacing $\operatorname{\mathcal{C}}$ by $\operatorname{sk}_{k}(\operatorname{\mathcal{C}})$, we can assume that the simplicial set $\operatorname{\mathcal{C}}$ has dimension $\leq k$, for some integer $k \geq -1$.
We now proceed by induction on $k$. In the case $k=-1$, the simplicial set $\operatorname{\mathcal{C}}$ is empty and $\operatorname{TW}( \operatorname{\mathcal{E}}/ \operatorname{\mathcal{C}})$ is isomorphic to $\Delta ^0$. We may therefore assume without loss of generality that $k \geq 0$. Let $S$ be the collection of nondegenerate $k$-simplices of $\operatorname{\mathcal{C}}$, so that Proposition 1.1.3.13 supplies a pushout diagram of simplicial sets
$\xymatrix@R =50pt@C=50pt{ \underset { \sigma \in S }{\coprod } \operatorname{\partial \Delta }^{k} \ar [r] \ar [d] & \underset { \sigma \in S }{\coprod } \Delta ^{k} \ar [d] \\ \operatorname{\mathcal{C}}_0 \ar [r] & \operatorname{\mathcal{C}}, }$
where $\operatorname{\mathcal{C}}_0 = \operatorname{sk}_{k-1}(\operatorname{\mathcal{C}})$ is the $(k-1)$-skeleton of $\operatorname{\mathcal{C}}$. It follows from our inductive hypothesis that the simplicial set $X(\operatorname{\mathcal{C}}_0)$ is a contractible Kan complex. Consequently, to show that $X(\operatorname{\mathcal{C}})$ is a contractible Kan complex, it will suffice to show that the restriction map $\theta : X(\operatorname{\mathcal{C}}) \rightarrow X(\operatorname{\mathcal{C}}_0)$ is a trivial Kan fibration. Note that $\theta$ is a pullback of the restriction map
$\theta _0: X( \underset { \sigma \in S }{\coprod } \Delta ^{k} ) \rightarrow X( \underset { \sigma \in S }{\coprod } \operatorname{\partial \Delta }^{k} ).$
We will complete the proof by showing that $\theta _0$ is a trivial Kan fibration. Since $\theta _0$ is a Kan fibration (Lemma 5.7.8.10), this is equivalent to the assertion that $\theta _0$ is a homotopy equivalence (Corollary 3.2.7.4). Our inductive hypothesis guarantees that the Kan complex $X( \underset { \sigma \in S }{\coprod } \operatorname{\partial \Delta }^{k} )$ is contractible. We are therefore reduced to showing that the Kan complex $X( \underset { \sigma \in S }{\coprod } \Delta ^{k} )$ is also contractible. Since the collection of contractible Kan complexes is closed under products, we are reduced to verifying the contractibility of the simplicial set $X( \operatorname{\mathcal{C}}_0 )$ in the special case where $\operatorname{\mathcal{C}}_0 = \Delta ^{k}$ is a standard simplex of dimension $k$. We now consider several cases:
• In the case $k=0$, the desired result follows from Lemma 5.7.9.2.
• In the case $k=1$, Lemma 5.7.9.1 supplies a trivial Kan fibration $X( \Delta ^1 ) \rightarrow X( \operatorname{\partial \Delta }^1 )$. Our inductive hypothesis guarantees that the Kan complex $X( \operatorname{\partial \Delta }^1 )$ is contractible, so that $X( \Delta ^1)$ is also contractible.
• In the case $k \geq 2$, we can choose an integer $0 < i < k$. In this case, the inclusion $\Lambda ^{k}_{i} \hookrightarrow \Delta ^{k}$ is inner anodyne, so the restriction map $X( \Delta ^{k} ) \rightarrow X( \Lambda ^{k}_{i} )$ is a trivial Kan fibration (Lemma 5.7.8.10). Our inductive hypothesis guarantees that the Kan complex $X( \Lambda ^{k}_{i} )$ is contractible, so that $X( \Delta ^{k} )$ is also contractible.
$\square$ | 8,746 | 25,787 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-14 | latest | en | 0.406382 |
https://cw.fel.cvut.cz/wiki/courses/rm35koa/labs | 1,716,769,488,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00828.warc.gz | 155,514,176 | 9,020 | # Labs
The goal of the labs is to exercise the topics presented on lectures.
On some labs you will receive homework assignment, which are implementation of an algorithm or a method solving some interesting combinatorial optimization problem. In all cases, the solutions to homework assignments are submitted to Brute where they are automatically checked and evaluated. Late upload will be penalized by -1 point for each week after deadline. Completing all homeworks successfully (i.e., the output is classified as correct according to Brute) is a mandatory requirement for the assessment. Moreover, we encourage you to solve the homeworks since in the practical test you will use algorithms implemented for the homeworks.
## Plan of the Labs
Week No. Title Notes Materials
1 Introduction, Gurobi installation, LP basics Grading and rules
Gurobi installation guide
Gurobi code example (code)
LP basics (Wolfram notebook)
Introduction (colab notebook)
Transportation problem (colab notebook)
Simplex worst case notes (optional)
Wolfram simplex (optional)
LP algorithms (optional)
2 Semestral work, ILP basics Cocontest (pdf), Public instances, Public instances threshold
Introduction (colab solved)
Gurobi official interactive examples
Mathematical programming book with examples
Computable vs uncomputable functions (optional)
3 ILP 1 HW1 ILP basics (colab TODO), ILP basics (colab solved)
Transportation (colab TODO), Transportation (colab solved)
HW1 assignment (pdf), Public instances (zip)
4 ILP 2 Call center (colab TODO), Call Center (colab solved)
The Catering Problem (colab TODO), Catering (colab solved), Video, Solution
Tiles (colab TODO), Tiles (colab solved)
Dlužníček (Colab TODO), Dlužníček (colab solved), Video, Solution
5 ILP 3 HW2 HW2 assignment (pdf) Public instances, Video
Lazy Constraints: Handout (pdf), Video, Colab Notebook, Circle approximation (zip), Wolfram notebook (optional)
Peaking Power Plants (colab TODO), Video, Peaking Power Plants (colab solved), Solution (video)
6 Unexpected ILP applications Game of Fivers: Game, Video, Handout (pdf), Colab (TODO), Colab (solution)
Rubik's Cube: Video, Handout (pdf),Colab
Verification of DNNs: Video, Handout (pdf), Colab
7 Metaheuristics + consultations Metaheuristics (colab TODO) , Metaheuristics (colab solved) , Public instances threshold
8 SPT CC-O Handout (pdf), Function approximation (colab) czech_republic.txt, Content-aware image resizing, Online demo
9 Max flow HW3 Max-Flow: Handout (pdf), Video, Lazy separation notes (optional)
HW3: HW3 assignment (pdf), Initial feasible flow for FF (video), Public instances
10 Min-cost flow Min-cost flow: Handout (pdf), Object tracking (pdf), Object tracking (video)
11 DP/approximation Min-cost flow: Image reconstruction (colab)
Multicommodity flows: Pipe puzzle (colab)
DP: Handout (pdf), Optimal coin system design (colab)
12 cancelled Optional materials: Nonogram (colab TODO), Game of Life sythetize (colab solved), Game of Life (video), Game of Life (game)
Computational proof of Euler's impossible puzzle (colab)
13 Scheduling HW4 HW4 Assignment (+ Handout), Public Instances, Video
The Scheduling ZOO
14 Reserve, CP masterclass CP: Masterclass (colab), data (zip), CP overview (slides)
Circle packing: Colab (solved)
Cocontest: Results (pdf)
## Classroom computers
OS: Debian Linux 64b, select Ubuntu during booting
Login: uses credentials from Department of Computers. If you haven't use them before, setup them at https://www.felk.cvut.cz/labpass/
Development environments: CLion (C++), IntelliJ (Java), PyCharm (Python), GVim, Netbeans are installed. CLion, IntelliJ and PyCharm are installed in /opt and their license have to be activated via creating JetBrains account with faculty email.
VPN: Gurobi (Academic) licence can be obtained for free, but only from the university network; to access it remotely, you can use VPN - for more info see FEL VPN (CZ)/ FEE VPN (EN). | 953 | 3,921 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-22 | latest | en | 0.836019 |
https://www.physicsforums.com/members/krhisjun.226913/recent-content | 1,632,024,522,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056711.62/warc/CC-MAIN-20210919035453-20210919065453-00203.warc.gz | 989,441,486 | 10,873 | # Recent content by krhisjun
1. ### Thermodynamics, two equal volumes of gas at STP mix, entropy change?
As far as ive researched it is infact zero due to the fact that the molecules are indistinguishable from each other, the partition can just be reinserted and the thermodynamic state is the same so it hasnt changed in entropy.
2. ### Thermodynamics, two equal volumes of gas at STP mix, entropy change?
but how do i go about proving entropy change is zero
3. ### Thermodynamics, two equal volumes of gas at STP mix, entropy change?
Homework Statement Calculate entropy change if two equal volumes of gas (1 cm3) at standard temperature and pressure are separated from each other by a partition which is removed to create twice the volume at standard temperature and pressure Homework Equations dS = k.N. dV/V + C/T dT...
5. ### Characterize a co2 TEA class 4 laser
Homework Statement Im working on a lab experiment in my undergraduate physics degree requiring the use of a class 4 laser, i need to characterize this in terms of its power, coherence, pulse duration, wavelength and irradiance. This is a pulsed laser with rough parameters of 1-2J pulse energy...
6. ### Easy Question, Sunspots Astronomy Umbra and Penumbral sunspot areas
Thats a good link but doesnt explain the significance of the areas of the umbra and penumbra areas shamefully....
7. ### Easy Question, Sunspots Astronomy Umbra and Penumbral sunspot areas
Can anyone explain the significance of the umbra and penumbra areas?
8. ### Easy Question, Sunspots Astronomy Umbra and Penumbral sunspot areas
Ive jsut spoken to a friend that used photoshop and the lasso tool to outline the areas then using one of the options it will give a pixel count. i can understand how this will give me the area, and thus the ratio, but not how i could quantify any errors in this calculation.. Thanks for the reply
9. ### Easy Question, Sunspots Astronomy Umbra and Penumbral sunspot areas
Didnt know if this is the write palce to put this, but im stuck on this assingment and i think it must have an easy solution so if someone could point me in the right direction then it would be much appreciated. Homework Statement Using an appropriate software package, quantify the...
10. ### Calculating values of electric and amgnetic fields of laser beam
Ok, didnt realise that equation, so using those values E = \sqrt{2S\mu c} ? Therefore E = 119959.9933 V m^-1 ? H = E/c\mu ? Therefore H = 318.4160428 A m^-1 ? I tried to confirm the equations using dimensional analysis: E = V m^-1 mu= kg·m·s−2·A−2 C = m S^-1 S = J s^-1 m^-2 I cant get that...
11. ### Calculating values of electric and amgnetic fields of laser beam
for fear of asking the obvious, S being?
12. ### Calculating values of electric and amgnetic fields of laser beam
Sorry, i mean to say i dont know how i would get H B or E. i can see easily how with any of the variables allows the other for calculation but im at a loss to get any.
13. ### Calculating values of electric and amgnetic fields of laser beam
Bu then HHow do i calculate H?
14. ### Calculating values of electric and amgnetic fields of laser beam
Homework Statement A continuous wave laser beam in free space carries a power of 15w and has a circular cross section with diameter 1mm. Calculate peak values of the oscillatory electric and magnetic fields Eo and Ho repectively. Homework Equations Eox = (\mu/\epsilon)^1/2 Hoy \pir^2...
15. ### Calculate capacitance to adjust current in ac circuit
Homework Statement A Purely resistive resistive 100w, 110v lamp is to be supplied by an a.c. source of 230v r.m.s at 50Hz. what value of pure capacitance could be put in series to adjust the current flow to a suitable value? Homework Equations p=iv q=cv v=ir i know a number of other... | 926 | 3,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-39 | latest | en | 0.860138 |
https://www.mathworks.com/matlabcentral/profile/authors/8545900?detail=cody&page=3 | 1,670,011,478,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00735.warc.gz | 936,402,042 | 4,229 | Solved
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Student’s Name: Professor’s Name: Course: Date: Trigonometry Part a: sinusoidal function for number of daylight hours Amplitude is equal to: (15.3 hours days after the longest day of the year therefore: f (97) = 3.3 cos (2π / 365*97) + 12 = 11.6 Hours [...]
## Order Description:
The hours of daylight, throughout the year, in a particular town can be graphed using trigonometric functions. On June 21, the longest day of the year, there are 15.3 daylight hours. On Dec. 21, the shortest day of the year, there are 8.7 daylight hours. a) Determine the function for the number of daylight hours with respect to the number of days since Jan. 1st. (Hint: Jan. 1st is day 1) b) Determine the number of daylight hours the town will have on March 27th and October 2nd. Answer in Radians and show ALL work.
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4.5 | 567 | 2,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-06 | latest | en | 0.869355 |
https://access.openupresources.org/curricula/our-k5-math/en/grade-2/unit-8/section-a/lesson-3/student.html | 1,713,390,081,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817181.55/warc/CC-MAIN-20240417204934-20240417234934-00041.warc.gz | 67,834,505 | 6,344 | # Lesson 3 Is it Odd or Even?
• Let’s explain why the number of objects in a group is even or odd.
## Warm-up Choral Count: Skip Count by 2
Count by 2, starting at 2.
What patterns do you see?
## Activity 1 Color by Number
Han wants to color a design with the same number of yellow and blue shapes.
1. Which designs could Han choose? Show your thinking using diagrams, symbols, or other representations. Use counters if it helps.
2. Draw a circle design that would work for Han and a circle design that would not work.
3. Priya drew a design that has 6 circles, 3 triangles, and 3 squares. Would Han be able to color it the way he wants? Show your thinking using diagrams, symbols, or other representations. Use counters if it helps.
## Activity 2 Card Sort: Even or Odd
1. Sort your cards into a group that shows an even number and a group that shows an odd number. | 214 | 877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-18 | latest | en | 0.901959 |
https://www.indiabix.com/logical-reasoning/statement-and-assumption/discussion-689 | 1,723,389,711,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00562.warc.gz | 648,149,400 | 7,960 | # Logical Reasoning - Statement and Assumption - Discussion
Discussion Forum : Statement and Assumption - Section 1 (Q.No. 2)
Directions to Solve
In each question below is given a statement followed by two assumptions numbered I and II. You have to consider the statement and the following assumptions and decide which of the assumptions is implicit in the statement.
• (A) If only assumption I is implicit
• (B) If only assumption II is implicit
• (C) If either I or II is implicit
• (D) If neither I nor II is implicit
• (E) If both I and II are implicit.
2.
Statement: It is desirable to put the child in school at the age of 5 or so.
Assumptions:
1. At that age the child reaches appropriate level of development and is ready to learn.
2. The schools do not admit children after six years of age.
Only assumption I is implicit
Only assumption II is implicit
Either I or II is implicit
Neither I nor II is implicit
Both I and II are implicit
Explanation:
Since the statement talks of putting the child in school at the age of 5, it means that the child is mentally prepared for the same at this age. So, I is implicit. But nothing about admission after 6 years of age is mentioned in the statement. So, II is not implicit.
Discussion:
6 comments Page 1 of 1.
Priya said: 5 years ago
Here only 1 statement is implicit as at very early age the child reaches the level of development and able to understand things but fixation of 6 years in a general case in not possible. So, 2 do not follow.
(1)
Ram said: 8 years ago
School does not admit after 6 years of the case is a general case which doesn't need to be dealt with the statements and hence IInd is explicit, whereas 1 is implicit because this is only the age at which child is mentally prepared to join the school.
Shashanka s said: 8 years ago
At that age means which age?
At the age of 5 the child is vehemently ready to learn and being the period of discovery I support A as an implicit by guys you need to educate us how to make such conclusions.
Statement talks about desirability of putting a child in school.
Therefore it is assumed that child has reached the the age where he is ready to learn.
Negating 2nd Assumption we have "School do admit students after 6 Years of age. ".
If it be the case still one can talk of putting the child in school at the age of 5 years.
So the two are independent and not related anyway.
Therefore II is not an assumption. | 571 | 2,443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-33 | latest | en | 0.951323 |
nigawadsreeja1.medium.com | 1,627,456,676,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153531.10/warc/CC-MAIN-20210728060744-20210728090744-00307.warc.gz | 435,568,613 | 26,110 | # Cyber Crime-Confusion Matrix
## What is Cyber Crime ?
Cybercrime is an illegal action against any person using a computer, its systems, and its online or offline applications.
## Types of Cyber Attacks :
1.Brute-force attack : Trail and error approach to guess the password successfully.
2.Credential Stuffing : In this attack a cybercriminal uses stolen usernames and passwords from one organization(purchased off of the dark web)to access user accounts at another organization.
3.Phishing and Spear Phishing: It’s the practice of sending emails from a trusted-seeming source to gain personal information.
4.Malware attacks: This attack is due to a malicious software that is downloaded in your system without you being aware of its presence.
## What is Confusion Matrix ?
A confusion matrix is a tabular summary of the number of correct and incorrect predictions made by a classifier. It is used to measure the performance of a classification model. It can be used to evaluate the performance of a classification model through the calculation of performance metrics like accuracy, precision, recall, and F1-score.
For a binary classification use case ,2*2 matrix is used as shown below
From the above diagram:
1. Left side of Confusion matrix indicates the actual values of dataset.
2. Top column indicates the predicted values of the machine learning model.
3. If the actual and predicted values are same, then we can say that it is either True positive or True Negative.
4. If the actual and predicted values are different ,then we can say that it is either False Positive or False Negative.
TN (True Negative) : It is an outcome where model correctly predicts the negative class.
TP (True Positive) : It is an outcome where model correctly predicts the positive class.
FN (False Negative): it is an outcome where model incorrectly predicts the negative class. It is type 2 error.
FP (False Positive) : It is an outcome where model incorrectly predict the positive class. It is type 1 error more dangerous than type 2 error.
## Performance of confusion matrix depends on the following components:
1. Accuracy : It is the closeness of measured value to the true value.
2. Precision : The ratio of correct positive predictions to the total predicted positives.
3. Recall : The ratio of true positives to the to the total number of true positives and false negatives.
## How Confusion Matrix relate to Cybercrime ?
Let’s take an example
From above diagram there are total 165 records
Positive indicate attack has not happened and Negative indicate cyber crime has happened.
Among 165 records
TP :100 are True positive ,it means machine predicted cyber attack has not happened and it’s actually true.
TN : 50 are True negative ,it means machine predicted cyber attack has happened and it’s actually true.
FP : 10 are False positive , it means machine predicted cyber attack has not happened but it’s actually not true. This is type 1 error ,is more dangerous than type 2 error.
FN : 5 are False negative , it means machine predicted cyber attack has happened but it’s actually not true. This is type 2 error.
Accuracy : (TP + TN)/n =(100+50)/165 = 0.90
Precision : TP/(TP + FP) =100/(100 + 10) = 0.90
Recall : TP/(TP+FN) =100/(100+5) = 0.95 | 702 | 3,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.848313 |
https://blazinggames.blogspot.com/2018/03/32-variables-and-math.html | 1,675,655,059,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500303.56/warc/CC-MAIN-20230206015710-20230206045710-00117.warc.gz | 153,793,300 | 11,630 | ## Saturday, March 24, 2018
### 3.2 Variables and Math
A variable is essentially a name that refers to a bit of information, such as a name. The information that the variable can hold can be changed which is why it is called a variable. While there are different types of variables JavaScript doesn't care what information the variable contains and can change the type of information that a variable contains. This is known as dynamic typing. It is also a big source of errors which is why many languages do require that you have specific types of variables for holding specific types of information. Later versions of the ECMA-Script specification do support optional types which is a good compromise.
You set a value to a variable by using the equals sign and can do various operations on the variable. Variables can be set to numeric values, String values, and can even be a function or instance of an object. We will discuss these things in further detail in later sections and chapters. Here is a bit of sample code that shows various variables being assigned values. The optional var keyword is used to indicate the declaration of a variable, but unknown variables will be created the first time they are seen. Variables that are declared without a var statement are assumed to be global in scope meaning that everyone can see the variable. This feature of automatically creating a variable the first time it is encountered makes it easy to write code but also means that a typo can result in very hard to find bugs.
var x = 11;
y = 7;
text = "Strings need to be enclosed in quotes";
function_variable = function() { /*code here*/ };
When using variables that use numbers in JavaScript, it is possible to do mathematical operations on the variables. Math in JavaScript uses algebraic order of operations which means that numbers will be multiplied or divided before they are added or subtracted unless brackets are used to group operations. Adding and subtracting uses the plus and minus sign. Multiplication is done by using the asterisk (*) symbol and dividing is done by using the slash (/) symbol. You can also get the remainder of a division by using the percent (%) symbol.
JavaScript also supports special decrement and increment operators. The double minus signs represent the decrement operation and essentially means deduct one from the variable this operator is used on. There is also an increment operator which is represented by double plus signs. The operator can be placed before or after the variable. If placed before the variable, the operation is done immediately. If placed after the operator, the operation will only be performed after any other operations are done. For instance, the statement x=y++ would place the value of y into x and then increase y.
In most programming languages it is valid to have a statement such as x = x + 3 as the = means assignment so the statement would take the value that is currently in x and increase it by 3. Equals is represented by two or three equal signs as will be explained in the next section. Applying mathematics to the value in a variable and storing it back into itself is a very common thing to do in programming so there is a shortcut for doing these operations. By placing a mathematical operator (+, -, *, /) before the equals sign you tell flash to do that operation on the left-hand variable. The statement x *= 2 would be resolved as x = x * 2. While this doesn’t sound like that big of a time saver, if you use long variable names you will quickly come to appreciate this shortcut.
More advanced mathematical operations, such as sin and cosine, can be done by using the Math class. You can call these functions simply by using the format Math.function(). I will explain these functions as we use them, but any book focused on JavaScript programming would have them.
Strings, a term used for a block of text, also can use one particular math operator. That being the + operator. Adding strings, better known as concatenation, simply joins the strings together. Adding a number to a string concatenates the string equivalent of the number to the end of the string. This is a problem area for JavaScript as it will convert types on you so if it thinks the variable num is the string “2” then the statement num = num + 2; would result in the value of “22”. | 871 | 4,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2023-06 | latest | en | 0.928403 |
https://jeopardylabs.com/print/physics-chapter-3-review-3 | 1,631,874,322,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055632.65/warc/CC-MAIN-20210917090202-20210917120202-00596.warc.gz | 351,892,612 | 4,706 | Chapter Vocabulary
Motion With Constant Acceleration
Acceleration vs. Velocity
Accelerated Motion
Free Fallin'
100
The area under a velocity time graph is the ________.
Displacement.
100
A car with a velocity of 30 m/s is accelerated uniformly at the rate of 2.0 m/s2 for 10 s. What is its final velocity?
50 m/s
100
Can a car traveling on an interstate highway have a negative velocity and a positive acceleration at the same time? Explain.
Yes, a car's velocity is positive or negative with respect to its direction of motion from some point of reference. A positive acceleration with a negative velocity would imply the car is traveling in the negative direction and slowing down.
100
This is the slope of the line tangent to the curve of a velocity-time graph at a particular instant in time.
What is instantaneous acceleration
100
What is the acceleration due to gravity?
-9.8 m/s2
200
The term used to describe a falling object for which air has no appreciable effect on its motion.
What is free fall.
200
Find the uniform acceleration that causes a car's velocity to change from 27 m/s to 45 m/s in a 6.0-s period.
3.0 m/s2
200
True or False. Can the velocity of an object change if the acceleration is constant? Give an example or counterexample.
Yes. Dropping a book. The longer it drops the faster it goes but the acceleration is constant.
200
A ball falls freely from rest for 15.0 s. Calculate the ball's velocity after 15.0 s.
-147 m/s
200
A stone that starts at rest is in free fall for 8.0s. What is the stone's displacement during this time?
-313.6 meters
300
Acceleration describes a change in ________.
Velocity.
300
If a car is traveling 100 km/hr and comes to a stop in 3.0 min, what is the car's acceleration in m/s2?
-0.15 m/s2
300
A race car is moving at a constant velocity of 25 m/s for 15 seconds. What is the cars acceleration?
Zero. Objects traveling at a constant speed are not accelerating!
300
How far does a car travel in 30.0 s while its velocity is changing from 50.0 m/s to 80.0 m/s at a uniform rate of acceleration?
What is 1950 meters
300
You throw a beanbag in the air at 15.0 m/s. How far did it travel before reaching its maximum height?
11.47 meters
400
You did it !!
400
A car is accelerated uniformily at the rate of 0.50 m/s2 for 1.0 s. Its final velocity is 23 m/s. What is the initial velocity?
22.5 m/s
400
Does a car that is slowing down always have a negative acceleration? Why? Give an example or counterexample.
No, if the positive axis points in the direction opposite the velocity, the acceleration will be positive. If a car is traveling in the negative direction, only a positive acceleration will slow it down. A negative acceleration will cause an increase in speed.
400
If a car is traveling 100 m/s and comes to a stop in 3.0 min, what is the car's acceleration?
- 0.55 m/s2
400
You throw a ball downward from a window at a speed of 2.0 m/s. How fast will it be moving when it hits the sidewalk 2.5 meters below?
7.3 m/s
500
The _____ ______ of an object is the slope of its velocity-time graph.
Average acceleration.
500
What is the minimum length runway needed to accommodate airplanes that can accelerate uniformly at 2.7 m/s2 and must reach a ground velocity of 64 m/s before they can take off?
758 meters
500
What is acceleration? (Described in terms of velocity)
Acceleration is the rate of change of velocity. How much the velocity changed divided by how long it took the change to occur.
500
A tennis ball is dropped from 1.5 m above the ground, touches the ground for 0.008 s and rebounds to a height of 0.75 m. What is the ball's velocity when it hits the ground?
-5.42 m/s
500
If you throw a ball upwards out of a window at 2.0 m/s. How fast will it be moving when it hits the sidewalk 2.5 meters below?
7.3 m/s
Click to zoom | 971 | 3,813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2021-39 | latest | en | 0.915014 |
https://exercism.io/tracks/scala/exercises/collatz-conjecture/solutions/a68fb4ae2f3847fbb259df8d0163b95b | 1,627,207,454,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151641.83/warc/CC-MAIN-20210725080735-20210725110735-00580.warc.gz | 259,184,179 | 6,363 | # EricHype's solution
## to Collatz Conjecture in the Scala Track
Published at Jan 22 2021 · 0 comments
Instructions
Test suite
Solution
The Collatz Conjecture or 3x+1 problem can be summarized as follows:
Take any positive integer n. If n is even, divide n by 2 to get n / 2. If n is odd, multiply n by 3 and add 1 to get 3n + 1. Repeat the process indefinitely. The conjecture states that no matter which number you start with, you will always reach 1 eventually.
Given a number n, return the number of steps required to reach 1.
## Examples
Starting with n = 12, the steps would be as follows:
1. 12
2. 6
3. 3
4. 10
5. 5
6. 16
7. 8
8. 4
9. 2
10. 1
Resulting in 9 steps. So for input n = 12, the return value would be 9.
The Scala exercises assume an SBT project scheme. The exercise solution source should be placed within the exercise directory/src/main/scala. The exercise unit tests can be found within the exercise directory/src/test/scala.
To run the tests simply run the command `sbt test` in the exercise directory.
Please see the learning and installation pages if you need any help.
## Source
An unsolved problem in mathematics named after mathematician Lothar Collatz https://en.wikipedia.org/wiki/3x_%2B_1_problem
## Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
### CollatzConjectureTest.scala
``````import org.scalatest.{Matchers, FunSuite}
/** @version 1.2.0 */
class CollatzConjectureTest extends FunSuite with Matchers {
test("zero steps for one") {
CollatzConjecture.steps(1) should be (Some(0))
}
test("divide if even") {
pending
CollatzConjecture.steps(16) should be (Some(4))
}
test("even and odd steps") {
pending
CollatzConjecture.steps(12) should be (Some(9))
}
test("Large number of even and odd steps") {
pending
CollatzConjecture.steps(1000000) should be (Some(152))
}
test("zero is an error") {
pending
CollatzConjecture.steps(0) should be (None)
}
test("negative value is an error") {
pending
CollatzConjecture.steps(-15) should be (None)
}
}``````
``````object CollatzConjecture {
def steps(num: Int): Option[Int] = {
def _steps(num: Int, currentSteps: Int): Int = {
num match {
case 1 => currentSteps
case even if num % 2 == 0 => _steps(even / 2, currentSteps + 1)
case _ => _steps(3 * num + 1, currentSteps + 1)
}
}
if ( num < 1 ){
None
} else {
Some(_steps(num, 0))
}
}
}`````` | 696 | 2,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-31 | latest | en | 0.774918 |
https://edurev.in/course/quiz/attempt/3213_Test-Symmetry-2/93bcea71-96d4-4f2c-94a3-9ac9b3ea1a00 | 1,623,873,202,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626008.14/warc/CC-MAIN-20210616190205-20210616220205-00261.warc.gz | 216,190,322 | 40,383 | Courses
# Test: Symmetry - 2
## 20 Questions MCQ Test Mathematics (Maths) Class 6 | Test: Symmetry - 2
Description
This mock test of Test: Symmetry - 2 for Class 6 helps you for every Class 6 entrance exam. This contains 20 Multiple Choice Questions for Class 6 Test: Symmetry - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Symmetry - 2 quiz give you a good mix of easy questions and tough questions. Class 6 students definitely take this Test: Symmetry - 2 exercise for a better result in the exam. You can find other Test: Symmetry - 2 extra questions, long questions & short questions for Class 6 on EduRev as well by searching above.
QUESTION: 1
Solution:
QUESTION: 2
Solution:
QUESTION: 3
### What is the other name for a line of symmetry of a circle?
Solution:
A diameter divides the circle into 2 equal parts. So, it can be considered as a line of symmetry.
QUESTION: 4
Which of these letters of the English alphabet has reflectional symmetry about a vertical mirror?
Solution:
U appears the same in mirror too.
QUESTION: 5
Find the number of lines of symmetry in regular hexagon.
Solution:
QUESTION: 6
Letter ‘A’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
Solution:
QUESTION: 7
Letter ‘C’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
Solution:
QUESTION: 8
Letter ‘E’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
Solution:
QUESTION: 9
Letter ‘H’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
Solution:
QUESTION: 10
Letter ‘M’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
Solution:
QUESTION: 11
Letter ‘T’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
Solution:
QUESTION: 12
Letter ‘G’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
Solution:
QUESTION: 13
Letter ‘D’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
Solution:
QUESTION: 14
Letter ‘B’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
Solution:
QUESTION: 15
Letter ‘I’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
Solution:
QUESTION: 16
Find the number of lines of symmetry in the below figure:
Solution:
QUESTION: 17
Find the number of lines of symmetry in the below figure:
Solution:
QUESTION: 18
Find the number of lines of symmetry in the below figure:
Solution:
QUESTION: 19
Find the number of lines of symmetry in the below figure:
Solution:
QUESTION: 20
Find the number of lines of symmetry in a circle.
Solution: | 689 | 2,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.884771 |
https://www.experts-exchange.com/questions/21911222/Escape-in-eval-function-on-Single-Quotation.html | 1,503,383,521,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110471.85/warc/CC-MAIN-20170822050407-20170822070407-00699.warc.gz | 906,237,749 | 28,075 | Solved
# Escape in eval function on Single Quotation
Posted on 2006-07-07
Medium Priority
714 Views
Try these two sample scripts:
This works:
<script>
var boys = 'He said \'howdy\', I said \'hi\'.' // nesting single quotes within single quotes
</script>
This does not work:
<script>
var boys = eval("'He said \'howdy\', I said \'hi\'.'") // nesting single quotes within single quotes
</script>
Why is this? Does not eval handle escape singel character?
And what should I put instead of \' in the eval function to make it work?
Thank you,
Best regards,
Marius
0
Question by:traelnes
[X]
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LVL 2
Expert Comment
ID: 17057510
Hi Marius,
try this
<script>
var val='He said \'howdy\', I said \'hi\'.';
var boys = eval("val"); // nesting single quotes within single quotes
</script>
regards,
vikrant
0
LVL 22
Expert Comment
ID: 17057519
Hello traelnes,
because \' itself is eval again and after eval("'") you've get single quote that have to be evaled again and you get unterminated string constant.
Why you have to eval this? just concatenate...
HTH
I
0
LVL 63
Accepted Solution
Zvonko earned 2000 total points
ID: 17057693
Escape chars has to be escaped in eval():
var boys = eval("'He said \\'howdy\\', I said \\'hi\\'.'");
0
Author Comment
ID: 17057957
Thank you Zvonko for this. It works.
Just curious about what ivostoykov mean by concatenate?
0
LVL 63
Expert Comment
ID: 17058099
You are welcome.
Let us wait and see what concatenate trick comes along ;-)
0
Author Comment
ID: 17066600
No Concentate. Ok, Zvonko, you are the winner!
0
LVL 63
Expert Comment
ID: 17067556
:)
Thanks.
0
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# UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
## International General Certificate of Secondary Education
MATHEMATICS 0580/04
0581/04
Paper 4 (Extended)
May/June 2006
Electronic calculator
Geometrical instruments
Graph paper (2 sheets)
Mathematical tables (optional)
Tracing paper (optional)
Write your name, Centre number and candidate number on all the work you hand in.
Write in dark blue or black pen on both sides of the paper.
You may use a soft pencil for any diagrams or graphs.
Do not use staples, paper clips, highlighters, glue or correction fluid.
At the end of the examination, fasten all your work securely together.
The number of marks is given in brackets [ ] at the end of each question or part question.
All working must be clearly shown. It should be done on the same sheet as the rest of the answer.
Marks will be given for working which shows that you know how to solve the problem even if you get the
The total of the marks for this paper is 130.
Electronic calculators should be used.
If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to
three significant figures.
Give answers in degrees to one decimal place.
For π use either your calculator value or 3.142.
## This document consists of 7 printed pages and 1 blank page.
IB06 06_0580_04/7RP
UCLES 2006 [Turn over
2
1 (a) A train completed a journey of 850 kilometres with an average speed of 80 kilometres per hour.
Calculate, giving exact answers, the time taken for this journey in
## (ii) hours, minutes and seconds. [1]
(b) Another train took 10 hours 48 minutes to complete the same 850 km journey.
## (i) It departed at 19 20.
At what time, on the next day, did this train complete the journey? [1]
(ii) Calculate the average speed, in kilometres per hour, for the journey. [2]
(c)
25
C D
20
Speed 15 BB
(metres per second)
10
5 A
O 1 2 3 4 5 6 7 8 9 10
Time (seconds)
The solid line OABCD on the grid shows the first 10 seconds of a car journey.
(i) Describe briefly what happens to the speed of the car between B and C. [1]
(ii) Describe briefly what happens to the acceleration of the car between B and C. [1]
## (iii) Calculate the acceleration between A and B. [2]
(iv) Using the broken straight line OC, estimate the total distance travelled by the car in the
whole 10 seconds. [3]
(v) Explain briefly why, in this case, using the broken line makes the answer to part (iv) a good
estimate of the distance travelled. [1]
(vi) Calculate the average speed of the car during the 10 seconds.
## © UCLES 2006 0580/04, 0581/04 Jun 2006
3
2
NOT TO B C
SCALE
B
C
12 cm
18 cm
A O
12 cm
A
40 cm E 22 cm D
E D
Diagram 1 Diagram 2
## Diagram 1 shows a closed box. The box is a prism of length 40 cm.
The cross-section of the box is shown in Diagram 2, with all the right-angles marked.
AB is an arc of a circle, centre O, radius 12 cm.
ED = 22 cm and DC = 18 cm.
Calculate
## (a) the perimeter of the cross-section, [3]
(b) the area of the cross-section, [3]
(c) the volume of the box, [1]
(d) the total surface area of the box. [4]
## 3 Answer the whole of this question on a sheet of graph paper.
(a) Find the values of k, m and n in each of the following equations, where a > 0.
(i) a0 = k, [1]
1
(ii) am = , [1]
a
(iii) an = a3 . [1]
(b) The table shows some values of the function f(x) = 2x.
## x −2 −1 −0.5 0 0.5 1 1.5 2 3
f(x) r 0.5 0.71 s 1.41 2 2.83 4 t
## (i) Write down the values of r, s and t. [3]
(ii) Using a scale of 2 cm to represent 1 unit on each axis, draw an x-axis from −2 to 3 and a
y-axis from 0 to 10. [1]
(iii) On your grid, draw the graph of y = f(x) for −2 x 3. [4]
## (c) The function g is given by g(x) = 6 – 2x.
(i) On the same grid as part (b), draw the graph of y = g(x) for – 2 x 3. [2]
(ii) Use your graphs to solve the equation 2x = 6 – 2x. [1]
(iii) Write down the value of x for which 2x < 6 – 2x for x ∈ {positive integers}. [1]
4
North
A B
O
NOT TO
SCALE
## The diagram shows a plan for a new city.
It is to be built inside a circle of radius 5 km.
The areas where homes can be built are shaded on the diagram.
The homes must be at least 2 km from the centre of the city, O.
The homes must also be at least 0.5 km from two main roads CD and AB, which are in North-South and
West-East directions.
(a) Using 1 cm to represent 1 km, make an accurate scale drawing showing the areas for the homes.
(You do not need to shade these areas.) [4]
(b) The town hall, T, will be built so that it is equidistant from the roads OA and OC.
It will be 1 km from O and West of CD.
(i) On your scale drawing, using a straight edge and compasses only, draw the locus of points, inside
the town, which are equidistant from OA and OC. [2]
## (ii) Mark and label the point T. [1]
(c) The police station, P, will be built so that it is equidistant from T and B.
It will be 3 km from O and North of AB.
Showing all your construction lines, find and label the point P. [3]
(d) What will be the actual straight line distance between the town hall and the police station? [1]
5
## (a) Write down a general equation for x and y.
Show that when x = 5 and y = 4.8 the equation becomes x 2 y = 120 . [2]
6
13 cm NOT TO
SCALE
D C
E 6 cm
A 8 cm B
## The diagram shows a pyramid on a horizontal rectangular base ABCD.
The diagonals of ABCD meet at E.
P is vertically above E.
AB = 8 cm, BC = 6 cm and PC = 13 cm.
## (b) Calculate the volume of the pyramid.
1
[The volume of a pyramid is given by 3
× area of base × height.] [2]
## (d) M is the mid-point of AD and N is the mid-point of BC.
Calculate angle MPN. [3]
## (ii) K lies on PB so that BK = 4 cm.
Calculate the length of KC. [3]
## © UCLES 2006 0580/04, 0581/04 Jun 2006 [Turn over
6
3
7 Transformation T is translation by the vector .
2
Transformation M is reflection in the line y = x.
## (b) Find the 2 by 2 matrix M, which represents the transformation M. [2]
(c) Show that, for any value of k, the point Q (k – 2, k – 3) maps onto a point on the line y = x following
the transformation TM(Q). [3]
## (d) Find M-1, the inverse of the matrix M. [2]
0 3 0 4
(e) N is the matrix such that N + = .
1 0 0 0
8 (a)
2x + 4
x+2 x NOT TO
SCALE
x2 – 40
## The diagram shows a trapezium.
Two of its angles are 90o.
The lengths of the sides are given in terms of x.
The perimeter is 62 units.
(i) Write down a quadratic equation in x to show this information. Simplify your equation. [2]
7
(b)
2y – 1
y NOT TO
SCALE
y+2
## The diagram shows a right-angled triangle.
The lengths of the sides are given in terms of y.
## (i) Show that 2y2 – 8y – 3 = 0. [3]
(ii) Solve the equation 2y2 – 8y – 3 = 0, giving your answers to 2 decimal places. [4]
(iii) Calculate the area of the triangle. [2]
## 9 (a) The numbers 0, 1, 1, 1, 2, k, m, 6, 9, 9 are in order (k ≠ m).
Their median is 2.5 and their mean is 3.6.
## (i) Write down the mode. [1]
(ii) Find the value of k. [1]
(iii) Find the value of m. [2]
(iv) Maria chooses a number at random from the list.
The probability of choosing this number is 1 . Which number does she choose? [1]
5
## (b) 100 students are given a question to answer.
The time taken (t seconds) by each student is recorded and the results are shown in the table.
## t 0<t 20 20<t 30 30<t 35 35<t 40 40<t 50 50<t 60 60<t 80
Frequency 10 10 15 28 22 7 8
## (ii) Two students are picked at random.
What is the probability that they both took more than 50 seconds?
## Answer part (c) on a sheet of graph paper.
(c) The data in part (b) is re-grouped to give the following table.
Frequency p q 8
## (i) Write down the values of p and q. [2]
(ii) Draw an accurate histogram to show these results.
Use a scale of 1 cm to represent 5 seconds on the horizontal time axis.
Use a scale of 1 cm to 0.2 units of frequency density (so that 1 cm2 on your histogram represents
1 student). [4]
## © UCLES 2006 0580/04, 0581/04 Jun 2006
8
BLANK PAGE
Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has
been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make
amends at the earliest possible opportunity.
University of Cambridge International Examinations is part of the University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the
University of Cambridge. | 2,609 | 8,590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-43 | latest | en | 0.842841 |
http://www.java-forums.org/new-java/49958-recursive-exercises-print.html | 1,477,685,303,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988725470.56/warc/CC-MAIN-20161020183845-00198-ip-10-171-6-4.ec2.internal.warc.gz | 513,493,327 | 3,916 | # Recursive exercises.
• 10-16-2011, 09:57 PM
danthegreat
Recursive exercises.
I'm having a problem with this recursive function
I can't seem to grasp the thought process for recursion. It's difficult for me and gives me a headache. It's not how do it, its just how i dont know when it will stop...
the function below needs to return true if the parameter is a palindrome. What's wrong with the code? The output is wrong but in my head it seems correct..I've tried outputting serveral variables and nothing really leads me to source of the problem.
Code:
``` public static boolean isPalindrome(String phrase) { int length=phrase.length(); String reverse=""; if(length>0) { reverse+=phrase.charAt(length-1); System.out.print(reverse); isPalindrome(phrase.substring(0, length-1)); } if(phrase.equals(reverse)) { return true; } else { return false; } }```
phrase currently = racecar
output is
racecarfalse
it should be racecartrue
something with that if statement. Is this a problem with recursion? Is the true being replaced with falses?
Any suggestions?
• 10-16-2011, 10:20 PM
sunde887
Re: Recursive exercises.
You have good logic but you are making a mistake. The reverse string is local to the method, meaning a new one is created each time a recursive call is made. Try changing the System.out.print(...) to a System.out.println(...) to highlight the problem more. You can keep reverse variable but it has to be declared elsewhere. Or you can probably change the logic to not need a string variable.
• 10-16-2011, 10:35 PM
danthegreat
Re: Recursive exercises.
My teacher says we are not allowed to change the parameters. How do I "change" my logic to not require the use of a reverse...I've tried replacing it with a boolean, a counter, but all of those are variables so they all get replaced...
• 10-16-2011, 10:39 PM
sunde887
Re: Recursive exercises.
You can't change the parameter, but reverse is not a parameter, it is a locally declared and initialized string. The logic change method would require some extra thinking, given a word with 7 letters that IS a palindrome what is true about the first letter and the last letter? How about the second letter and the second to last letter? Repeat this logic recursively.
If you do not want to rethink the logic, what you have will work, you will just need to change where the variable reverse is declared and initialized (move it outside the method).
• 10-16-2011, 10:48 PM
danthegreat
Re: Recursive exercises.
I thought of this. However, it only checks correctly Length-1 and (Length+1)-(Length). Therefore, it won't check position 0, (Length-Length)...That's the only problem.
Code:
```public static boolean isPalindrome(String phrase) { int length=phrase.length(); if(length>0) { if(phrase.charAt(length-1)==phrase.charAt((length+1)-length)) { isPalindrome(phrase.substring(0, length-1)); } else { return false; } } return true; }```
• 10-16-2011, 10:51 PM
sunde887
Re: Recursive exercises.
Why not simply check charAt(0) to charAt(length-1), and with each recursive call shrink the string on both sides.
• 10-16-2011, 11:24 PM
danthegreat
Re: Recursive exercises.
Code:
```public static boolean isPalindrome(String phrase) { int length=phrase.length(); if(length>0) { if(phrase.charAt(0)==phrase.charAt(length-1)) { isPalindrome(phrase.substring(XXXXX), length-1)); } else { return false; } } return true; }```
Where X is placed, its unclear to me how to solve this. Cutting the string on the beginning would require to write (length+1)-(length). however, this will not check position 0. How would you be able to fufill both requests
• 10-16-2011, 11:39 PM
gcalvin
Re: Recursive exercises.
1. If the length of the String is 1 or less (1, not 0), it is a palindrome. Return true. (This solves "I don't know when it will stop.")
2. If the first character is not the same as the last character, it is not a palindrome. Return false. Otherwise...
3. If the part between the first and last characters is a palindrome, it is a palindrome. If it's not, it's not.
• 10-17-2011, 12:34 AM
sunde887
Re: Recursive exercises.
The XXXX can be simply 1, you check the 0 character to get into that condition so it has been checked already. Also, see gcalvins points.
• 10-17-2011, 01:08 AM
danthegreat
Re: Recursive exercises.
I get an error:
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: -1
at java.lang.String.substring(String.java:1937)
at howard_recursiveExercises.isPalindrome(recursiveEx ercises.java:175)
at howard_recursiveExercises.isPalindrome(recursiveEx ercises.java:175)
at howard_recursiveExercises.isPalindrome(recursiveEx ercises.java:175)
at howard_recursiveExercises.isPalindrome(recursiveEx ercises.java:175)
at howard_recursiveExercises.main(_recursiveExercises .java:12)
• 10-17-2011, 01:42 AM
danthegreat
Re: Recursive exercises.
nvm figured it out, its if length>1 sigh...
lol still don't understand how or why but....it works! | 1,353 | 6,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | latest | en | 0.788975 |
http://atlas.dr-mikes-maths.com/atlas/960/10886/4.html | 1,579,823,571,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250614086.44/warc/CC-MAIN-20200123221108-20200124010108-00377.warc.gz | 16,175,989 | 2,259 | Questions?
See the FAQ
or other info.
# Polytope of Type {4,6,10}
Atlas Canonical Name : {4,6,10}*960d
if this polytope has a name.
Group : SmallGroup(960,10886)
Rank : 4
Schlafli Type : {4,6,10}
Number of vertices, edges, etc : 4, 24, 60, 20
Order of s0s1s2s3 : 20
Order of s0s1s2s3s2s1 : 2
Special Properties :
Universal
Non-Orientable
Flat
Related Polytopes :
Facet
Vertex Figure
Dual
Facet Of :
{4,6,10,2} of size 1920
Vertex Figure Of :
{2,4,6,10} of size 1920
Quotients (Maximal Quotients in Boldface) :
2-fold quotients : {4,6,5}*480b, {2,6,10}*480e
4-fold quotients : {2,3,10}*240a, {2,6,5}*240b
8-fold quotients : {2,3,5}*120
Covers (Minimal Covers in Boldface) :
2-fold covers : {4,12,10}*1920f, {4,12,10}*1920g, {8,6,10}*1920f, {4,6,10}*1920d
Permutation Representation (GAP) :
```s0 := (4,6);;
s1 := ( 3, 4)( 5, 6)( 8, 9)(10,11);;
s2 := ( 1, 2)( 7, 8)(10,11);;
s3 := ( 8,10)( 9,11);;
poly := Group([s0,s1,s2,s3]);;
```
Finitely Presented Group Representation (GAP) :
```F := FreeGroup("s0","s1","s2","s3");;
s0 := F.1;; s1 := F.2;; s2 := F.3;; s3 := F.4;;
rels := [ s0*s0, s1*s1, s2*s2, s3*s3, s0*s2*s0*s2,
s0*s3*s0*s3, s1*s3*s1*s3, s0*s1*s0*s1*s0*s1*s0*s1,
s0*s1*s2*s1*s0*s1*s2*s1, s1*s2*s1*s2*s1*s2*s1*s2*s1*s2*s1*s2,
s1*s2*s3*s2*s1*s2*s1*s2*s3*s2*s1*s2,
s3*s2*s3*s1*s2*s3*s1*s2*s3*s1*s2*s3*s2*s1*s2 ];;
poly := F / rels;;
```
Permutation Representation (Magma) :
```s0 := Sym(11)!(4,6);
s1 := Sym(11)!( 3, 4)( 5, 6)( 8, 9)(10,11);
s2 := Sym(11)!( 1, 2)( 7, 8)(10,11);
s3 := Sym(11)!( 8,10)( 9,11);
poly := sub<Sym(11)|s0,s1,s2,s3>;
```
Finitely Presented Group Representation (Magma) :
```poly<s0,s1,s2,s3> := Group< s0,s1,s2,s3 | s0*s0, s1*s1, s2*s2,
s3*s3, s0*s2*s0*s2, s0*s3*s0*s3, s1*s3*s1*s3,
s0*s1*s0*s1*s0*s1*s0*s1, s0*s1*s2*s1*s0*s1*s2*s1,
s1*s2*s1*s2*s1*s2*s1*s2*s1*s2*s1*s2,
s1*s2*s3*s2*s1*s2*s1*s2*s3*s2*s1*s2,
s3*s2*s3*s1*s2*s3*s1*s2*s3*s1*s2*s3*s2*s1*s2 >;
```
References : None.
to this polytope | 973 | 1,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-05 | latest | en | 0.402984 |
http://convertwizard.com/100-square_meters-to-square_miles | 1,560,706,708,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998288.34/warc/CC-MAIN-20190616162745-20190616184745-00530.warc.gz | 41,394,424 | 22,787 | # 100 Square meters to Square miles (100 m2 to mile2)
Convert 100 Square meters to Square miles (m2 to mile2) with our conversion calculator and conversion tables. To convert 100 m2 to mile2 use direct conversion formula below.
100 m2 = 3.8610038610039E-5 mile2.
You also can convert 100 Square meters to other Area (popular) units.
100 SQUARE METERS
=
3.8610038610039E-5 SQUARE MILES
Direct conversion formula: 1 Square meters * 2590000 = 1 Square miles
Opposite conversion: 100 Square miles to Square meters
## Conversion table: Square meters to Square miles
SQUARE METERS SQUARE MILES
1 = 3.8610038610039E-7
2 = 7.7220077220077E-7
3 = 1.1583011583012E-6
4 = 1.5444015444015E-6
5 = 1.9305019305019E-6
7 = 2.7027027027027E-6
8 = 3.0888030888031E-6
9 = 3.4749034749035E-6
10 = 3.8610038610039E-6
SQUARE MILES SQUARE METERS
1 = 2590000
2 = 5180000
3 = 7770000
4 = 10360000
5 = 12950000
7 = 18130000
8 = 20720000
9 = 23310000
10 = 25900000
## Nearest numbers for 100 Square meters
SQUARE METERS SQUARE MILES
101.22 m2 = 3.9081081081081E-5 mile2
103.1 m2 = 3.980694980695E-5 mile2
103.9 m2 = 4.011583011583E-5 mile2
104 m2 = 4.015444015444E-5 mile2
104.66 m2 = 4.0409266409266E-5 mile2
105.9 m2 = 4.0888030888031E-5 mile2
107.6 m2 = 4.1544401544402E-5 mile2
109.2 m2 = 4.2162162162162E-5 mile2
111.62 m2 = 4.3096525096525E-5 mile2
117.5 m2 = 4.5366795366795E-5 mile2
119 m2 = 4.5945945945946E-5 mile2
126 m2 = 4.8648648648649E-5 mile2
126.9 m2 = 4.8996138996139E-5 mile2
132.1 m2 = 5.1003861003861E-5 mile2
133.6 m2 = 5.1583011583012E-5 mile2
135 m2 = 5.2123552123552E-5 mile2
135.8 m2 = 5.2432432432432E-5 mile2
135.82 m2 = 5.2440154440154E-5 mile2
140 m2 = 5.4054054054054E-5 mile2
140.9 m2 = 5.4401544401544E-5 mile2 | 762 | 1,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.537602 |
https://howlingpixel.com/i-en/Moment-generating_function | 1,531,833,652,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589726.60/warc/CC-MAIN-20180717125344-20180717145344-00366.warc.gz | 679,862,482 | 17,509 | # Moment-generating function
In probability theory and statistics, the moment-generating function of a real-valued random variable is an alternative specification of its probability distribution. Thus, it provides the basis of an alternative route to analytical results compared with working directly with probability density functions or cumulative distribution functions. There are particularly simple results for the moment-generating functions of distributions defined by the weighted sums of random variables. However, not all random variables have moment-generating functions.
As its name implies, the moment generating function can be used to compute a distribution’s moments: the nth moment about 0 is the nth derivative of the moment generating function, evaluated at 0.
In addition to real-valued distributions (univariate distributions), moment-generating functions can be defined for vector- or matrix-valued random variables, and can even be extended to more general cases.
The moment-generating function of a real-valued distribution does not always exist, unlike the characteristic function. There are relations between the behavior of the moment-generating function of a distribution and properties of the distribution, such as the existence of moments.
## Definition
The moment-generating function of a random variable X is
${\displaystyle M_{X}(t):=\operatorname {E} \left[e^{tX}\right],\quad t\in \mathbb {R} ,}$
wherever this expectation exists. In other words, the moment-generating function is the expectation of the random variable ${\displaystyle e^{tX}}$. More generally, when ${\displaystyle \mathbf {X} =(X_{1},\ldots ,X_{n})}$T, an n-dimensional random vector, and t is a fixed vector, one uses ${\displaystyle \mathbf {t} \cdot \mathbf {X} =\mathbf {t} ^{\mathrm {T} }\mathbf {X} }$ instead of tX:
${\displaystyle M_{\mathbf {X} }(\mathbf {t} ):=\operatorname {E} \left(e^{\mathbf {t} ^{\mathrm {T} }\mathbf {X} }\right).}$
${\displaystyle M_{X}(0)}$ always exists and is equal to 1. However, a key problem with moment-generating functions is that moments and the moment-generating function may not exist, as the integrals need not converge absolutely. By contrast, the characteristic function or Fourier transform always exists (because it is the integral of a bounded function on a space of finite measure), and for some purposes may be used instead.
The moment generating function is so named as it can be used to find the moments of the distribution.[1] The series expansion of etX is:
${\displaystyle e^{t\,X}=1+t\,X+{\frac {t^{2}\,X^{2}}{2!}}+{\frac {t^{3}\,X^{3}}{3!}}+\cdots +{\frac {t^{n}\,X^{n}}{n!}}+\cdots .}$
Hence:
{\displaystyle {\begin{aligned}M_{X}(t)=\operatorname {E} (e^{t\,X})&=1+t\operatorname {E} (X)+{\frac {t^{2}\operatorname {E} (X^{2})}{2!}}+{\frac {t^{3}\operatorname {E} (X^{3})}{3!}}+\cdots +{\frac {t^{n}\operatorname {E} (X^{n})}{n!}}+\cdots \\&=1+tm_{1}+{\frac {t^{2}m_{2}}{2!}}+{\frac {t^{3}m_{3}}{3!}}+\cdots +{\frac {t^{n}m_{n}}{n!}}+\cdots ,\end{aligned}}}
where mn is the nth moment. Differentiating MX(t) i times with respect to t and setting t = 0, we obtain the ith moment about the origin, mi; see Calculations of moments below.
If X is a continuous random variable, the following relation between its moment generating function MX(t) and the two-sided Laplace transform of its probability density function fX(x) holds:
${\displaystyle M_{X}(t)={\mathcal {B}}\{f\}(-t)}$
as the PDF's two-sided Laplace transform is given as
${\displaystyle {\mathcal {B}}\{f_{X}\}(s)=\int _{-\infty }^{\infty }e^{-sx}f_{X}(x)\,dx,}$
and the moment generating function's definition expands to
${\displaystyle M_{X}(t)=\operatorname {E} \left[e^{tX}\right]=\int _{-\infty }^{\infty }e^{tx}f_{X}(x)\,dx.}$
This is consistent with the characteristic function of X being a Wick rotation of MX(t) when the moment generating function exists, as the characteristic function of a continuous random variable X is the Fourier transform of its probability density function fX(x), and in general when a function f(x) is of exponential order, the Fourier transform of f is a Wick rotation of its two-sided Laplace transform in the region of convergence. See the relation of the Fourier and Laplace transforms for further information.
## Examples
Here are some examples of the moment generating function and the characteristic function for comparison. It can be seen that the characteristic function is a Wick rotation of the moment generating function MX(t) when the latter exists.
Distribution Moment-generating function MX(t) Characteristic function φ(t)
Bernoulli ${\displaystyle \,P(X=1)=p}$ ${\displaystyle \,1-p+pe^{t}}$ ${\displaystyle \,1-p+pe^{it}}$
Geometric ${\displaystyle (1-p)^{k-1}\,p\!}$ ${\displaystyle {\frac {pe^{t}}{1-(1-p)e^{t}}}\!}$
${\displaystyle \forall t<-\ln(1-p)\!}$
${\displaystyle {\frac {pe^{it}}{1-(1-p)\,e^{it}}}\!}$
Binomial B(n, p) ${\displaystyle \,\left(1-p+pe^{t}\right)^{n}}$ ${\displaystyle \,\left(1-p+pe^{it}\right)^{n}}$
Poisson Pois(λ) ${\displaystyle \,e^{\lambda (e^{t}-1)}}$ ${\displaystyle \,e^{\lambda (e^{it}-1)}}$
Uniform (continuous) U(a, b) ${\displaystyle \,{\frac {e^{tb}-e^{ta}}{t(b-a)}}}$ ${\displaystyle \,{\frac {e^{itb}-e^{ita}}{it(b-a)}}}$
Uniform (discrete) U(a, b) ${\displaystyle \,{\frac {e^{at}-e^{(b+1)t}}{(b-a+1)(1-e^{t})}}}$ ${\displaystyle \,{\frac {e^{ait}-e^{(b+1)it}}{(b-a+1)(1-e^{it})}}}$
Normal N(μ, σ2) ${\displaystyle \,e^{t\mu +{\frac {1}{2}}\sigma ^{2}t^{2}}}$ ${\displaystyle \,e^{it\mu -{\frac {1}{2}}\sigma ^{2}t^{2}}}$
Chi-squared χ2k ${\displaystyle \,(1-2t)^{-{\frac {k}{2}}}}$ ${\displaystyle \,(1-2it)^{-{\frac {k}{2}}}}$
Noncentral chi-squared χ2k(λ) ${\displaystyle \,e^{\lambda t/(1-2t)}(1-2t)^{-{\frac {k}{2}}}}$ ${\displaystyle \,e^{i\lambda t/(1-2it)}(1-2it)^{-{\frac {k}{2}}}}$
Gamma Γ(k, θ) ${\displaystyle \,(1-t\theta )^{-k}}$;${\displaystyle \forall t<{\frac {1}{\theta }}}$ ${\displaystyle \,(1-it\theta )^{-k}}$
Exponential Exp(λ) ${\displaystyle \,\left(1-t\lambda ^{-1}\right)^{-1},\,(t<\lambda )}$ ${\displaystyle \,\left(1-it\lambda ^{-1}\right)^{-1}}$
Multivariate normal N(μ, Σ) ${\displaystyle \,e^{t^{\mathrm {T} }\left(\mu +{\frac {1}{2}}\Sigma t\right)}}$ ${\displaystyle \,e^{t^{\mathrm {T} }\left(i\mu -{\frac {1}{2}}\Sigma t\right)}}$
Degenerate δa ${\displaystyle \,e^{ta}}$ ${\displaystyle \,e^{ita}}$
Laplace L(μ, b) ${\displaystyle \,{\frac {e^{t\mu }}{1-b^{2}t^{2}}};{\text{ for }}|t|<1/b\,}$ ${\displaystyle \,{\frac {e^{it\mu }}{1+b^{2}t^{2}}}}$
Negative Binomial NB(r, p) ${\displaystyle \,{\frac {(1-p)^{r}}{\left(1-pe^{t}\right)^{r}}}}$ ${\displaystyle \,{\frac {(1-p)^{r}}{\left(1-pe^{it}\right)^{r}}}}$
Cauchy Cauchy(μ, θ) Does not exist ${\displaystyle \,e^{it\mu -\theta |t|}}$
## Calculation
The moment-generating function is the expectation of a function of the random variable, it can be written as:
Note that for the case where X has a continuous probability density function ƒ(x), MX(−t) is the two-sided Laplace transform of ƒ(x).
{\displaystyle {\begin{aligned}M_{X}(t)&=\int _{-\infty }^{\infty }e^{tx}f(x)\,dx\\&=\int _{-\infty }^{\infty }\left(1+tx+{\frac {t^{2}x^{2}}{2!}}+\cdots +{\frac {t^{n}x^{n}}{n!}}+\cdots \right)f(x)\,dx\\&=1+tm_{1}+{\frac {t^{2}m_{2}}{2!}}+\cdots +{\frac {t^{n}m_{n}}{n!}}+\cdots ,\end{aligned}}}
where mn is the nth moment.
### Linear combination of independent random variables
If ${\displaystyle S_{n}=\sum _{i=1}^{n}a_{i}X_{i}}$, where the Xi are independent random variables and the ai are constants, then the probability density function for Sn is the convolution of the probability density functions of each of the Xi, and the moment-generating function for Sn is given by
${\displaystyle M_{S_{n}}(t)=M_{X_{1}}(a_{1}t)M_{X_{2}}(a_{2}t)\cdots M_{X_{n}}(a_{n}t)\,.}$
### Vector-valued random variables
For vector-valued random variables X with real components, the moment-generating function is given by
${\displaystyle M_{X}(t)=E\left(e^{\langle t,X\rangle }\right)}$
where t is a vector and ${\displaystyle \langle \cdot ,\cdot \rangle }$ is the dot product.
## Important properties
Moment generating functions are positive and log-convex, with M(0) = 1.
An important property of the moment-generating function is that if two distributions have the same moment-generating function, then they are identical at almost all points.[2] That is, if for all values of t,
${\displaystyle M_{X}(t)=M_{Y}(t),\,}$
then
${\displaystyle F_{X}(x)=F_{Y}(x)\,}$
for all values of x (or equivalently X and Y have the same distribution). This statement is not equivalent to the statement "if two distributions have the same moments, then they are identical at all points." This is because in some cases, the moments exist and yet the moment-generating function does not, because the limit
${\displaystyle \lim _{n\rightarrow \infty }\sum _{i=0}^{n}{\frac {t^{i}m_{i}}{i!}}}$
may not exist. The lognormal distribution is an example of when this occurs.
### Calculations of moments
The moment-generating function is so called because if it exists on an open interval around t = 0, then it is the exponential generating function of the moments of the probability distribution:
${\displaystyle m_{n}=E\left(X^{n}\right)=M_{X}^{(n)}(0)=\left.{\frac {d^{n}M_{X}}{dt^{n}}}\right|_{t=0}.}$
That is, with n being a nonnegative integer, the nth moment about 0 is the nth derivative of the moment generating function, evaluated at t = 0.
## Other properties
Jensen's inequality provides a simple lower bound on the moment-generating function:
${\displaystyle M_{X}(t)\geq e^{\mu t},}$
where ${\displaystyle \mu }$ is the mean of X.
Hoeffding's lemma provides a bound on the moment-generating function in the case of a zero-mean, bounded random variable.
When all moments are non-negative, the moment generating function gives a simple, useful bound on the moments:
${\displaystyle M_{X}(t)\geq {\frac {t^{k}E[X^{k}]}{k!}}.}$
The moment-generating function can be used in conjunction with Markov's inequality to bound the upper tail of a real random variable X. Since ${\displaystyle x\mapsto e^{xt}}$ is monotonically increasing for ${\displaystyle t>0}$, we have
${\displaystyle P(X\geq a)=P(e^{tX}\geq e^{ta})\leq e^{-at}E[e^{tX}]=e^{-at}M_{X}(t)}$
for any ${\displaystyle t>0}$ and any a, provided ${\displaystyle M_{X}(t)}$ exists. For example, when X is a standard normal distribution and ${\displaystyle a>0}$, we can choose ${\displaystyle t=a}$ and recall that ${\displaystyle M_{X}(t)=e^{t^{2}/2}}$. This gives ${\displaystyle P(X\geq a)\leq e^{-a^{2}/2}}$, which is within a factor of 1+a of the exact value.
## Relation to other functions
Related to the moment-generating function are a number of other transforms that are common in probability theory:
Characteristic function
The characteristic function ${\displaystyle \varphi _{X}(t)}$ is related to the moment-generating function via ${\displaystyle \varphi _{X}(t)=M_{iX}(t)=M_{X}(it):}$ the characteristic function is the moment-generating function of iX or the moment generating function of X evaluated on the imaginary axis. This function can also be viewed as the Fourier transform of the probability density function, which can therefore be deduced from it by inverse Fourier transform.
Cumulant-generating function
The cumulant-generating function is defined as the logarithm of the moment-generating function; some instead define the cumulant-generating function as the logarithm of the characteristic function, while others call this latter the second cumulant-generating function.
Probability-generating function
The probability-generating function is defined as ${\displaystyle G(z)=E\left[z^{X}\right].\,}$ This immediately implies that ${\displaystyle G(e^{t})=E\left[e^{tX}\right]=M_{X}(t).\,}$ | 3,604 | 11,875 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 74, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-30 | latest | en | 0.800043 |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/edwardcashmere/stack-and-queues-implementations-in-javascript-and-python-3108 | 1,621,354,924,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991288.0/warc/CC-MAIN-20210518160705-20210518190705-00319.warc.gz | 484,002,231 | 30,777 | ## DEV Community is a community of 621,872 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
# Stack and Queues Implementations in JavaScript and Python
## What is a stack?
A stack is a linear data structure that follows a particular order in which the operations are performed
An example is the call stack in JavaScript, it is what a program uses to keep track of method calls. It works based on the LIFO principle i.e., last-in-first-out. This means whatever item or function that is called last is executed first. In this article, we will use SLL(singly linked list) to build a call stack. Look up how the call stack handles recursion here.
There two different approaches we can implement the call stack but one is optimized and will take less time. The first approach is adding a node at the end and removing it from the end also. Here we have adhered to the LIFO principle but when removing we have to loop to the 2nd last node before we remove the last node. if we assume our list is big enough this approach is time expensive O(n). A better approach would be to use a doubly-linked list to avoid using the loop or our 2nd approach.
In this 2nd approach, we will be adding at the beginning and removing from the beginning also our time complexity will always be O(1). Hence this is the approach we will use.
## What is a Queue?
A queue is an abstract data structure, somewhat similar to Stacks. Unlike stacks, a queue is open at both its ends. One end is always used to insert data (enqueue) and the other is used to remove data (dequeue). Queue follows First-In-First-Out methodology, i.e., the data item stored first will be accessed first.
Think about a line at the bank, how you want the system to work is by serving the person who came first always.
Just as the call stack we can implement a queue taking two approaches. Although both approaches will work, only one approach works optimally. That is by adding a node at the end of the list and removing it at the beginning. Try and think about the other approach you can use, and why I have opted not to use it?.
I have implemented both stack and queue using an SLL but you can also use the DLL or even arrays in javascript so long as you adhere to the LIFO and FIFO principles.
Call stack
Stack in JavaScript:
``````class Node{
constructor(val){
this.val = val;
this.next = null;
}
}
class Stack{
constructor(){
this.first = null;
this.last = null;
this.size = 0;
}
push (val){
let newNode = new Node(val);
if(!this.first){
this.first= newNode;
this.last = newNode;
}else{
newNode.next = this.first;
this.first = newNode;
}
this.size++;
return this;
}
pop(){
if(!this.first)return undefined;
let temp = this.first;
this.first = this.first.next;
this.size--;
if(this.size ===0){
this.last = null;
}
return temp;
}
}
let stack = new Stack();
stack.push(19);
stack.push(20);
stack.push(21);
stack.push(22);
stack.push(23);
stack.push(24);
stack.push(25);
``````
In python:
``````class Node:
def __init__(self,val):
self.val = val
self.next = None
class Stack:
def __init__(self):
self.first = None
self.last =None
self.size =0
def push(self,val):
newNode = Node(val)
if self.first == None:
self.first= newNode
self.last = newNode
else:
newNode.next = self.first
self.first = newNode
self.size+=1
return self
def pop(self):
if self.first == None: return
temp = self.first
self.first = self.first.next
self.size-=1
if(self.size == 0):
self.last = None
return temp
``````
Queue in JavaScript:
``````class Node{
constructor(val){
this.val = val;
this.next = null;
}
}
// FIFO First in First out
class Queue{
constructor(){
this.first = null;
this.last = null;
this.size = 0
}
// add at the beginning and remove at the beginning
enqueue(val){
let newNode = new Node(val);
if(!this.first){
this.first= newNode;
this.last = newNode;
}else{
this.last.next = newNode;
this.last = newNode;
}
this.size++
return this
}
dequeue(){
if(!this.first) return undefined
let temp = this.first
this.first = this.first.next
this.size--
if(this.size === 0){
this.last = null
}
return temp
}
}
let queue = new Queue()
queue.enqueue(19)
queue.enqueue(20)
queue.enqueue(21)
queue.enqueue(22)
queue.enqueue(23)
queue.enqueue(24)
queue.enqueue(25)
queue.enqueue(26)
``````
In python:
``````
class Node:
def __init__(self,val):
self.val = val
self.next = None
class Queue:
def __init__(self):
self.first = None
self.last =None
self.size =0
def enqueue(self,val):
newNode = Node(val)
if self.first == None:
self.first = newNode
self.last = newNode
self.size+=1
return self
self.last.next= newNode
self.last = newNode
self.size+=1
return self
def dequeue(self):
if self.first == None: return
temp = self.first
self.first = self.first.next
self.size-=1
if(self.size == 0):
self.last = None
return temp
``````
Conclusion
The implementations of Stack and Queue is simple as long as you adhere to its LIFO(Last-In-First-Out) and FIFO(First-In-First-Out) Principles, you can use arrays in JavaScript or Lists in Python, SLL, DLL the main objective however will be keeping the time complexity at O(1) for optimum performance. | 1,269 | 5,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-21 | longest | en | 0.942934 |
https://www.jiskha.com/display.cgi?id=1304356368 | 1,511,542,489,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808260.61/warc/CC-MAIN-20171124161303-20171124181303-00030.warc.gz | 792,852,462 | 3,806 | # physics
posted by .
What is the period of the hand on a clock that measures the minutes? What is the frequency?
Thank you,
Abbie
• physics -
it makes one cycle per hour.
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https://philoid.com/question/109748-mark-the-correct-alternative-in-the-following-the-value-of-k-which-makes-continuous-at-x-0-is | 1,723,515,123,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641054522.78/warc/CC-MAIN-20240813012759-20240813042759-00039.warc.gz | 362,090,447 | 8,151 | ##### Mark the correct alternative in the following:The value of k which makes continuous at x = 0, is
Formula:-
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Value does not exist for function to be continuous
1 | 80 | 315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-33 | latest | en | 0.937679 |
https://www.jiskha.com/display.cgi?id=1362012951 | 1,503,332,608,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109157.57/warc/CC-MAIN-20170821152953-20170821172953-00153.warc.gz | 923,729,389 | 3,738 | # Math
posted by .
Altogether 178 people attended the feast. There were 44 more men than women who came. How many men and how men women attended?
• Math -
w + w + 44 = 178
2w = 134
2 = 67 women
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More Similar Questions | 520 | 2,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-34 | latest | en | 0.976373 |
http://mathhelpforum.com/advanced-algebra/144495-finding-inverse-matrix-1-a.html | 1,500,866,507,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424683.39/warc/CC-MAIN-20170724022304-20170724042304-00670.warc.gz | 201,851,195 | 10,625 | # Thread: Finding inverse of a matrix (A^-1)
1. ## Finding inverse of a matrix (A^-1)
find the inverse of
1 2 -3
3 2 -1
2 1 3
if there is
2. ## how to find the inverse
The inverse can be found by setting this matrix equal to the identity matrix and row reducing so that the identity matrix is on the other side of equal sign (the inverse matrix will be where the identity matrix is originally).
So
1 2 -3 | 1 0 0
3 2 -1 | 0 1 0
2 1 3 | 0 0 1
Does that help?
3. Originally Posted by swiftshift
find the inverse of
1 2 -3
3 2 -1
2 1 3
if there is
Dear swiftshift,
$\left(\begin{array}{ccc}1&2&-3\\3&2&-1\\2&1&3\end{array}\right)\times\left(\begin{array }{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right)=\left( \begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\r ight)$
By matrix multiplication you could obtain nine equations and solve for the unknowns.
4. Originally Posted by kaelbu
The inverse can be found by setting this matrix equal to the identity matrix and row reducing so that the identity matrix is on the other side of equal sign (the inverse matrix will be where the identity matrix is originally).
So
1 2 -3 | 1 0 0
3 2 -1 | 0 1 0
2 1 3 | 0 0 1
Does that help?
thanks mate
5. Originally Posted by Sudharaka
Dear swiftshift,
$\left(\begin{array}{ccc}1&2&-3\\3&2&-1\\2&1&3\end{array}\right)\times\left(\begin{array }{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right)=\left( \begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\r ight)$
By matrix multiplication you could obtain six equations and solve for the unknowns.
Actually, you get nine equations for the nine unknowns. Probably not the best way to find an inverse! | 567 | 1,629 | {"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": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-30 | longest | en | 0.796332 |
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https://forum.processing.org/two/discussion/7499/can-someone-help-us-with-our-game | 1,591,033,555,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419056.73/warc/CC-MAIN-20200601145025-20200601175025-00112.warc.gz | 342,268,313 | 14,193 | #### Howdy, Stranger!
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# Can someone help us with our game?
edited October 2014
I really need's someone help right now. It's a very easy game for you guys.
Tagged:
This post would be helpful if I knew what help you needed. What game? Do you have code? Images? Can we see what you have already? What still doesn't work? What are you trying to do?
• edited October 2014
``````//Car
Car car1;
b1 b1;
o1 o1;
int xx=1000;
int virtualMouseY;
float y1= random(390, 600);
int c1=0;
int d=0;
int speed=1;
int p=0;
PImage bg;
void setup() {
size(1000, 600);
car1= new Car();
b1= new b1();
o1= new o1();
smooth();
}
void draw() {
println(c1);
loop();
noStroke();
background(bg);
fill(183, 67, 123);
car1.cardisplay();
virtualMouseY = constrain(mouseY, 335, 525);
fill(128);
b1.b1();
o1.o1();
xx-=speed;
fill(108);
textSize(15);
text("CLICK TO PAUSE", 10, 55);
fill(255, 169, 88);
textSize(35);
text("Score:", 10, 35);
text(c1*100, 120, 35);
if (xx<=-20) {
xx=1000;
c1+=1;
y1=(y1*3)+1;
if (y1>1000) {
y1=y1%1000;
}
}
if (((y1-15<=virtualMouseY)&&(y1+15>=virtualMouseY))&&(xx-27<=25)) {
noLoop();
d=1;
}
if (d==1) {
background(128);
textSize(42);
fill(245, 166, 166);
text("GAME OVER", 400, 240);
fill(255);
text("SCORE:", 450, 290);
text(c1*100, 450, 340);
fill(166, 214, 245);
text("CLICK TO RETRY", 350, 390);
}
if (c1>=1) {
speed=(c1/1)+1;
}
}
class Car {
void cardisplay() {
rect(30, virtualMouseY, 100,35);
rect(40, virtualMouseY-20, 60,20);
fill(0);
ellipse(40,virtualMouseY+40,15,15);
ellipse(120,virtualMouseY+40,15,15);
}
}
class b1 {
void b1() {
beginShape();
vertex(xx-1000, 216);
vertex(xx-951, 216);
vertex(xx-951, 247);
vertex(xx-931, 228);
vertex(xx-923, 247);
vertex(xx-923, 150);
vertex(xx-908, 135);
vertex(xx-893, 150);
vertex(xx-893, 179);
vertex(xx-882, 167);
vertex(xx-870, 179);
vertex(xx-870, 195);
vertex(xx-826, 195);
vertex(xx-826, 224);
vertex(xx-807, 198);
vertex(xx-788, 224);
vertex(xx-768, 198);
vertex(xx-749, 224);
vertex(xx-749, 150);
vertex(xx-734, 135);
vertex(xx-719, 150);
vertex(xx-719, 217);
vertex(xx-671, 217);
vertex(xx-671, 248);
vertex(xx-657, 228);
vertex(xx-642, 248);
vertex(xx-642, 195);
vertex(xx-599, 195);
vertex(xx-599, 224);
vertex(xx-579, 200);
vertex(xx-560, 224);
vertex(xx-560, 179);
vertex(xx-548, 168);
vertex(xx-536, 179);
vertex(xx-536, 233);
vertex(xx-493, 233);
vertex(xx-493, 195);
vertex(xx-449, 195);
vertex(xx-449, 151);
vertex(xx-435, 136);
vertex(xx-420, 151);
vertex(xx-420, 179);
vertex(xx-409, 167);
vertex(xx-397, 179);
vertex(xx-384, 167);
vertex(xx-373, 179);
vertex(xx-373, 223);
vertex(xx-353, 198);
vertex(xx-328, 233);
vertex(xx-292, 233);
vertex(xx-292, 214);
vertex(xx-249, 214);
vertex(xx-249, 179);
vertex(xx-236, 168);
vertex(xx-224, 179);
vertex(xx-224, 225);
vertex(xx-204, 200);
vertex(xx-186, 225);
vertex(xx-186, 151);
vertex(xx-171, 137);
vertex(xx-156, 151);
vertex(xx-156, 197);
vertex(xx-114, 197);
vertex(xx-114, 181);
vertex(xx-101, 169);
vertex(xx-89, 181);
vertex(xx-89, 215);
vertex(xx-48, 215);
vertex(xx-23, 181);
vertex(xx, 211);
vertex(xx, 216);
vertex(xx, 216); //1st
vertex(xx, 216);
vertex(xx+49, 216);
vertex(xx+49, 247);
vertex(xx+63, 228);
vertex(xx+77, 247);
vertex(xx+77, 150);
vertex(xx+92, 135);
vertex(xx+107, 150);
vertex(xx+107, 179);
vertex(xx+118, 167);
vertex(xx+130, 179);
vertex(xx+130, 195);
vertex(xx+174, 195);
vertex(xx+174, 224);
vertex(xx+193, 198);
vertex(xx+212, 224);
vertex(xx+232, 198);
vertex(xx+251, 224);
vertex(xx+251, 150);
vertex(xx+266, 135);
vertex(xx+281, 150);
vertex(xx+281, 217);
vertex(xx+329, 217);
vertex(xx+329, 248);
vertex(xx+343, 228);
vertex(xx+358, 248);
vertex(xx+358, 195);
vertex(xx+401, 195);
vertex(xx+401, 224);
vertex(xx+421, 200);
vertex(xx+440, 224);
vertex(xx+440, 179);
vertex(xx+452, 168);
vertex(xx+464, 179);
vertex(xx+464, 233);
vertex(xx+507, 233);
vertex(xx+507, 195);
vertex(xx+551, 195);
vertex(xx+551, 151);
vertex(xx+565, 136);
vertex(xx+580, 151);
vertex(xx+580, 179);
vertex(xx+591, 167);
vertex(xx+603, 179);
vertex(xx+616, 167);
vertex(xx+627, 179);
vertex(xx+627, 223);
vertex(xx+647, 198);
vertex(xx+672, 233);
vertex(xx+708, 233);
vertex(xx+708, 214);
vertex(xx+751, 214);
vertex(xx+751, 179);
vertex(xx+764, 168);
vertex(xx+776, 179);
vertex(xx+776, 225);
vertex(xx+796, 200);
vertex(xx+814, 225);
vertex(xx+814, 151);
vertex(xx+829, 137);
vertex(xx+844, 151);
vertex(xx+844, 197);
vertex(xx+886, 197);
vertex(xx+886, 181);
vertex(xx+899, 169);
vertex(xx+911, 181);
vertex(xx+911, 215);
vertex(xx+952, 215);
vertex(xx+977, 181);
vertex(xx+1000, 211);
vertex(xx+1000, 216); //1st
vertex(xx+1049, 216);
vertex(xx+1049, 247);
vertex(xx+1049, 315);
vertex(xx-1000, 315);
endShape(CLOSE);
}
}
class o1 {
void o1() {
ellipse(xx+50, y1, 50, 40);
ellipse(xx+150, y1+3, 40, 30);
ellipse(xx+200, y1+6, 40, 30);
ellipse(xx+300, y1+9, 40, 30);
ellipse(xx+400, y1+12, 40, 30);
ellipse(xx+500, y1+15, 40, 30);
}
}
void mousePressed() {
if (d==1) {
redraw();
c1=0;
speed=1;
d=0;
xx=1000;
}
if ((d==0)&&(p==0)) {
noLoop();
background(110);
fill(255);
textSize(42);
fill(245, 166, 166);
text("PAUSED", 450, 280);
fill(166, 214, 245);
text("CLICK TO RESUME", 350, 330);
p=1;
}
if ((d==0)&&(p==1)) {
redraw();
p=0;
}
}
``````
• the image is saved as road.jpg
I would like to ask help for the obstacles, which are the tiny stones to be a point. Example, if the car passes one rock then there is a corresponding point.
• The problem in our game is that the points correspond to the set instead of one stone only.
• edited October 2014
It is a convention to write class names with a capital as first letter
I would not recommend to have a one/2 letter class such as b1 - better say Building
Also the same name for class and object looks weird...
The object of that class starts with a small letter by convention so line 3 could be
``````Building building;
``````
or (not as good)
``````Building bldg;
``````
I don't understand why people don't use full names...
Same goes for o1...
I also recommend to have all classes together at the end of the sketch - and not in the middle as you have.
;-)
• edited October 2014
it looks nice btw
the stones shouldn't be allowed to be placed up in the sky...
also you call the method in line 32 /33 which has the same name as the class. Thus it is the constructor. Which shouldn't be called like this at all... This is not good...
name the method in 234 draw() or display() pls
read the tutorial on oop pls....
Car class is much better
• I assume that o1 is the obstacle...
you want to give it random values for x and y - see random in the reference pls
you need to store the values in 5 vars or in an array
then display them from 5 vars or the array
do the collision check Car against 5 vars or the array
• "I don't understand why people don't use full names..."
I fully agree, but it will be easier when the PDE will have auto-completion... | 2,444 | 6,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.488362 |
https://mathoverflow.net/questions/10793/compact-convex-sets-and-extreme-points | 1,701,975,850,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100686.78/warc/CC-MAIN-20231207185656-20231207215656-00715.warc.gz | 433,480,125 | 27,224 | # Compact Convex sets and Extreme Points
There are examples that show the set of extreme points of a compact convex subset of a locally convex topological vector space need not be closed when the real dimension of the space is at least 3. Is it true that the set of extreme points of a compact convex subset must be closed if the locally convex space in question has dimension 2?
• Retagged the question – banach-spaces out, convexity in. Jan 5, 2010 at 14:11
• Thanks for your responses. I deleted my original question, which was rather silly: somehow, in your perfectly clear 2.5 line answer, I missed the part where you used the 2-dimensionality. But I think it's good to have an example of the failure of closedness in higher dimensions. Jan 5, 2010 at 16:55
• @Pete For instance, the convex hull of the four points $(\pm 1, \pm 1, 0)$ and the unit circle in the $x$-$z$ plane. Apr 3, 2011 at 20:43
• There are examples that show the set of extreme points of a compact convex subset of a locally convex topological vector space need not be closed when the real dimension of the space is at least 3 - is that so? Where can one find some example of that set? For me it seems impossible to exists. Nov 3, 2014 at 19:02
[Just a historical remark.] AFAIK, the fact that the set of all extreme points of a compact convex subset of $\mathbb{R}^{2}$ must be closed is due to the legendary American mathematician G. Baley Price (1905-2006), in "On the extreme points of convex sets", Duke Math. J. Volume 3, Number 1 (1937), 56-67 (page 62). | 408 | 1,539 | {"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} | 2.625 | 3 | CC-MAIN-2023-50 | latest | en | 0.920911 |
https://datascience-enthusiast.com/R/Interactive_chord_diagrams_R.html | 1,679,898,347,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948609.41/warc/CC-MAIN-20230327060940-20230327090940-00164.warc.gz | 244,711,797 | 328,615 | A chord diagram is a graphical method of displaying the inter-relationships between data in a matrix. The data are arranged radially around a circle with the relationships between the data points typically drawn as arcs connecting the data (Wikipedia).
The chorddiag package in R allows to create interactive chord diagrams using the JavaScript visualization library D3 from within R using the htmlwidgets interfacing framework.
In this blog post, we will see how to create interactive chord diagrams. You can hover over the arcs for each of the courses/sections below and see how many students from each section are taking each course.
library(chorddiag)
students = data.frame(Math = c(50, 25, 5, 12),
Art = c(10, 55, 5, 20),
Science = c(45,12,29, 20),
PE = c(24,67,27,15))
students = as.matrix(students)
row.names(students) = c("Section A", "Section B", "Section C", "Section D")
chorddiag(students, type = "bipartite", showTicks = F, groupnameFontsize = 14, groupnamePadding = 10, margin = 90)
We can also use chord diagrams to visualize transition matrices. In the dummy data below, let’s assume each row is the number of customers each company has this year. However, next year, let’s say some of the customers will move to new companies (shown in the columns). Assume this matrix as a transition matrix in a marchov-chain. You can hover over the visualization and see which companies are losing the most and which ones are retaining more customers.
suppressMessages(library(tidyverse))
Warning in sample.int(.Machine$integer.max - 1L, 1L): '.Random.seed' is not an integer vector but of type 'NULL', so ignored df = data_frame(Company A = c(800, 200, 100, 50, 140, 200, 140), Company B = c(100, 2000, 300, 400, 50, 0, 290), Company C = c(200, 500, 4000, 80, 120, 320, 600), Company D = c(500, 200, 300, 5000, 250, 140, 450), Company E = c(600, 300, 150, 600, 6000, 30, 0), Company F = c(500, 400, 100, 300, 250, 4500, 140), Company G = c(300, 50, 0, 150, 600, 250, 7000)) df = as.matrix(df) row.names(df) = c(colnames(df)) df Company A Company B Company C Company D Company E Company F Company A 800 100 200 500 600 500 Company B 200 2000 500 200 300 400 Company C 100 300 4000 300 150 100 Company D 50 400 80 5000 600 300 Company E 140 50 120 250 6000 250 Company F 200 0 320 140 30 4500 Company G 140 290 600 450 0 140 Company G Company A 300 Company B 50 Company C 0 Company D 150 Company E 600 Company F 250 Company G 7000 chorddiag(df, type = "directional", showTicks = F, groupnameFontsize = 14, groupnamePadding = 10, margin = 90) ### Chord Diagrams in Shiny we can also embede shiny chord diagrams for more interactivity. The shiny app below is similar to the one shown above, but now we are looking over different markets. The code for the above shiny app is below. ### ui library(shiny) library(chorddiag) shinyUI(fluidPage( br(), br(), radioButtons('select_market',"Select Market",inline = TRUE, choices = c("East","West","South","North"), selected = 'East'), chorddiagOutput("distPlot", height = 600) )) ### server library(shiny) library(chorddiag) set.seed(1) # for reproducibility df_east = matrix(sample(seq(100, 5000, 50), 49), ncol = 7) row.names(df_east) = c("Company A" ,"Company B" ,"Company C" ,"Company D","Company E", "Company F" ,"Company G") colnames(df_east) = row.names(df_east) set.seed(2) df_west = matrix(sample(seq(100, 5000, 50), 49), ncol = 7) colnames(df_west) = row.names(df_east) row.names(df_west) = row.names(df_east) set.seed(3) df_south = matrix(sample(seq(100, 5000, 50), 49), ncol = 7) colnames(df_south) = row.names(df_east) row.names(df_south) = row.names(df_east) set.seed(4) df_north = matrix(sample(seq(100, 5000, 50), 49), ncol = 7) colnames(df_north) = row.names(df_east) row.names(df_north) = row.names(df_east) shinyServer(function(input, output) { output$distPlot <- renderChorddiag({
if(input$select_market =="East"){ chorddiag(df_east, showTicks = F, groupnameFontsize = 14, groupnamePadding = 10, margin = 90) }else if( input$select_market =="West"){
chorddiag(df_west, showTicks = F, groupnameFontsize = 14, groupnamePadding = 10, margin = 90)
} else if(
input\$select_market =="South"){
chorddiag(df_south, showTicks = F, groupnameFontsize = 14, groupnamePadding = 10, margin = 90)
}else{
chorddiag(df_north, showTicks = F, groupnameFontsize = 14, groupnamePadding = 10, margin = 90)
}
})
}) | 1,329 | 4,365 | {"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} | 2.90625 | 3 | CC-MAIN-2023-14 | latest | en | 0.804461 |
https://svn.geocomp.uq.edu.au/escript/trunk/escript/py_src/pdetools.py?sortby=rev&r1=989&r2=990&pathrev=1809 | 1,568,686,765,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573011.59/warc/CC-MAIN-20190917020816-20190917042816-00267.warc.gz | 695,486,064 | 2,810 | # Diff of /trunk/escript/py_src/pdetools.py
revision 989 by gross, Fri Jan 5 00:54:37 2007 UTC revision 990 by ksteube, Wed Feb 21 04:27:52 2007 UTC
404
405 for u and p. The problem is solved with an inexact Uszawa scheme for p: for u and p. The problem is solved with an inexact Uszawa scheme for p:
406
407 M{Q_f (u^{k+1}-u^{k}) = - f(u^{k},p^{k}) M{Q_f (u^{k+1}-u^{k}) = - f(u^{k},p^{k})}
408 M{Q_g (p^{k+1}-p^{k}) = g(u^{k+1})} M{Q_g (p^{k+1}-p^{k}) = g(u^{k+1})}
409
410 where Q_f is an approximation of the Jacobiean A_f of f with respect to u and Q_f is an approximation of where Q_f is an approximation of the Jacobiean A_f of f with respect to u and Q_f is an approximation of
416 """ """
417 initializes the problem initializes the problem
418
419 @parm verbose: switches on the printing out some information @param verbose: switches on the printing out some information
420 @type verbose: C{bool} @type verbose: C{bool}
421 @note: this method may be overwritten by a particular saddle point problem @note: this method may be overwritten by a particular saddle point problem
422 """ """
427 """ """
428 prints text if verbose has been set prints text if verbose has been set
429
430 @parm text: a text message @param text: a text message
431 @type text: C{str} @type text: C{str}
432 """ """
433 if self.__verbose: print "%s: %s"%(str(self),text) if self.__verbose: print "%s: %s"%(str(self),text)
Legend:
Removed from v.989 changed lines Added in v.990 | 492 | 1,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-39 | latest | en | 0.498095 |
https://www.airmilescalculator.com/distance/wnh-to-bsd/ | 1,606,210,380,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176049.8/warc/CC-MAIN-20201124082900-20201124112900-00359.warc.gz | 577,206,878 | 23,669 | # Distance between Wenshan (WNH) and Baoshan (BSD)
Flight distance from Wenshan to Baoshan (Wenshan Puzhehei Airport – Baoshan Yunrui Airport) is 341 miles / 549 kilometers / 296 nautical miles. Estimated flight time is 1 hour 8 minutes.
Driving distance from Wenshan (WNH) to Baoshan (BSD) is 500 miles / 805 kilometers and travel time by car is about 8 hours 35 minutes.
## Map of flight path and driving directions from Wenshan to Baoshan.
Shortest flight path between Wenshan Puzhehei Airport (WNH) and Baoshan Yunrui Airport (BSD).
## How far is Baoshan from Wenshan?
There are several ways to calculate distances between Wenshan and Baoshan. Here are two common methods:
Vincenty's formula (applied above)
• 341.145 miles
• 549.020 kilometers
• 296.447 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 340.745 miles
• 548.376 kilometers
• 296.099 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Wenshan Puzhehei Airport
City: Wenshan
Country: China
IATA Code: WNH
ICAO Code: ZPWS
Coordinates: 23°33′29″N, 104°19′31″E
B Baoshan Yunrui Airport
City: Baoshan
Country: China
IATA Code: BSD
ICAO Code: ZPBS
Coordinates: 25°3′11″N, 99°10′5″E
## Time difference and current local times
There is no time difference between Wenshan and Baoshan.
CST
CST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 75 kg (166 pounds).
## Frequent Flyer Miles Calculator
Wenshan (WNH) → Baoshan (BSD).
Distance:
341
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
341
Round trip? | 515 | 1,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-50 | longest | en | 0.832912 |
http://www.chegg.com/homework-help/questions-and-answers/body-falling-relatively-dense-fluid-oil-example-acted-forces-figure-235-resistive-force-r--q4120771 | 1,387,718,650,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387346051826/warc/CC-MAIN-20131218055411-00042-ip-10-33-133-15.ec2.internal.warc.gz | 332,507,620 | 7,778 | # Ch 2.3 #27
0 pts pending
A body falling in a relatively dense fluid, oil for example, is acted on by three forces (see Figure 2.3.5): a resistive force R, a buoyant force B, and its weight w due to gravity. The buoyant force is equal to the weight of the fluid displaced by the object. For a slowly moving spherical body of radius a, the resistive force is given by Stokes's law, R = 6*pi*u*a*|v|, where v is the velocity of the body, and u is the coefficient of viscosity of the surrounding fluid.
(a) Find the limiting velocity of a solid sphere of radius a and density p falling freely in a medium of density p' and coefficient of viscosity u.
(b) In 1910 R.A. Millikan studied the motion of tiny droplets of oil falling in a electric field. A field of strength E exerts a force E_e on a droplet with charge e. Assume that E has been adjusted so the droplet is held stationary (v=0) and that w and B are as given above. Find an expression for e. Millikan repeated this experiment many times, and from the data he gathered was able to deduce the charge on an electron. | 269 | 1,076 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2013-48 | latest | en | 0.935261 |
https://cs.stackexchange.com/questions/133115/proof-of-the-bubblesort-algorithm/133134 | 1,726,612,510,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00709.warc.gz | 168,100,207 | 41,021 | # Proof of the Bubblesort algorithm
I'm studying The Algorithm Design Manual and I was having some difficulty in the proof exercises, so I asked a question here. Based on the answer I got in that question(which was not the a complete proof),I came up with a proof.
Prove the correctness of the following sorting algorithm.
Bubblesort (A)
for i from n to 1
for j from 1 to i − 1
if (A[j] > A[j + 1])
swap the values of A[j] and A[j + 1]
1.Base case:
An array of length 1 which is by definition sorted.
2.Inductive hypothesis:
We'll assume that for all arrays of length (0 <= m) one iteraion of the outer loop with "n" being the length of the array,
the array gets permutated in such a was that the last element in the array is the biggest.
3.Inductive step:
We want to prove that if our assumption is true for lists of length (0 <= m), then it is also true for lists of length (m)
Let A = a[0], a[1], ..., a[m] (of length m+1)
After (m-1) iterations of the inner loop based on our assumption, in the list A[0:m-1], a[m-1] is the biggest element.
At this point,as (j < m),we have one more iteration to perform.
Before this, j was equal to m-2 and j+1 was equal to m-1. After this loop, j will be equal to m-1 and j+1 will be equal to m.
If A[m-1] and A[m] satisfy the if statement, that is, A[m-1] > A[m], they will swap.If not, then that would mean that A[m] is
the biggest element in the list.
In either case, the last element of the list will be the biggest after one iteration of the outer loop, which was what we were
trying to prove. ▮
I'm not really sure of it's correctness,so I would really appreciate if you could review it and tell me whether its correct or not. Also please note any inadequacies it may have or things I could've done better as I'm quite new to inductive proofs.
Your induction hypothesis is insufficient. In the hypothesis, you are just proving that the last element of the array has the maximum value. However, you have to prove that the resulting array is sorted. Therefore, your induction hypothesis should be the following:
Hypothesis: At the end of 't' iterations of the outer "for" loop, the "n-t" highest elements of the array are in the sorted order and they occupy the indexes from 'n-t+1' to 'n'.
Base case: For 't = 1', the induction hypothesis says that at the end of the first iteration of the outer "for" loop, the algorithm gives the highest element at the index 'n'. And, you can easily prove this statement.
Induction Step: At the end of 't+1' iterations of the outer "for" loop, the "n-t+1" highest elements of the array are in the sorted order and they occupy the indexes from 'n-t' to 'n'. Again, you have to prove this step using the earlier mentioned hypothesis -- for 't' iterations.
This proves the induction hypothesis. And, now you can simply say that at the end of 'n' iterations of the outer "for" loop, all elements of the array are in the sorted order and they occupy the indexes from '1' to 'n'.
• My mistake was thinking that in the main list, which is a recursive object and consists of smaller lists, for all of which we proved that the last element was the biggest, it would be intuitively obvious that the list is sorted(something like: [a], [a,b], [a,b,c], ...). However that's incorrect, there is no such thing as obvious in proof writing. Anyways Thanks for the response. :) Commented Dec 7, 2020 at 20:50 | 875 | 3,387 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-38 | latest | en | 0.965317 |
http://905spmath2011.blogspot.com/2011/11/patricks-math-book-question-24.html | 1,511,541,832,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808260.61/warc/CC-MAIN-20171124161303-20171124181303-00563.warc.gz | 3,735,073 | 14,414 | ## Sunday, November 20, 2011
### Patrick's Math Book Question 2.4
18)A frame measures 30 cm by 20 cm. Can you mount a square picture with an area of 500cm² in the frame ? Explain.
Well first of all, we know that the side of the square needs to be less then 20 cm, or else it wouldn't be able to fit in the frame.
Now we need to find the square root of 500cm², so we can see if the picture will fit or not.
Since the side length of the picture is more then 20cm, the picture will hang off of the frame. | 140 | 506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-47 | longest | en | 0.889813 |
https://brainor.com/math-worksheets/grade-1/addition/addition-to-5/ | 1,722,915,641,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476479.26/warc/CC-MAIN-20240806032955-20240806062955-00502.warc.gz | 106,363,515 | 45,139 | # Addition to 5 Worksheets | 27 Free Pages
These worksheets based on addition to 5 will help your first-graders grow a strong mathematical idea on basic mathematical operations. These worksheets will assist them in developing a mathematical understanding of addition to 5 and create a connection between theoretical knowledge and real-world uses of mathematical operations with our free printable worksheets. Download the worksheets and practice them rigorously.
## Adding on a Frog in Addition to 5 Worksheets
Each worksheet contains nine problems on frogs. Children need to solve the problems and write the results on the frogs.
## Counting Objects and Adding to 5
In this fun activity, students will count the given objects, add them, and write down the results in the selected box.
## Counting the Fingers and Adding
The first graders will like this real-life activity. Make them count the fingers and add the numbers. Then, write down the result in the result box.
## Christmas Addition to 5 Worksheets
As Christmas is the biggest festival, children will love this addition activity with Christmas objects. In these Christmas addition worksheets, you need to count the objects, add them, and write down the result.
## Addition to 5 on School Activities
In this activity, students will count school objects e.g.- school bags, pens, etc., and add the objects. Lastly, place the result in the result box.
## Counting Dice for Addition to 5
Make your students count the dots on the dice, add them, and write the results. Hopefully, they will find the activity helpful.
## Addition to 5 on Ocean Theme
This ocean theme activity will make you have a friendship with the ocean animals. Count the objects, add them, and have fun with the addition.
## Addition to 5 on the Space Theme
By participating in this activity, you will feel to hover in the space. You will also be familiar with the space objects like- spaceships, asteroids, etc. Add the objects and place the results.
## Adding and Coloring Worksheets
This coloring activity is fun. Students will add the numbers given in the required places, find the result, match the instructions, and color the area.
Here, with the help of these worksheets on addition to 5, we have learned how to add numbers within 5 and grow basic number sense by using the worksheets. Here, we have discussed addition on a frog, counting and adding, counting the fingers, Christmas addition, addition on school activities, counting dice, ocean theme addition, space theme addition, and adding & coloring worksheets. These activities should boost the grade 1 student’s basic mathematical skills. | 535 | 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} | 3.375 | 3 | CC-MAIN-2024-33 | latest | en | 0.907327 |
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# Oscilloscope and Function Generator - Lab Report Example
Summary
This was aimed at facilitating the appreciation of the important difference that exists between them. An oscilloscope can be as well be referred to as a…
## Extract of sample"Oscilloscope and Function Generator"
Download file to see previous pages nd in trying to understand the fundamental functioning of the two, the variables that were measured include the period (in seconds), the frequency (in Hz), and the peak-to-peak voltage (in volts).
This experiment verifies the relation between frequency and time as F = 1/T as evidenced by the results. in the functioning of these machines, these variables are used in varied ways. With frequency and time being reciprocals of one another, and this relationship is fundamental when displaying a current waveform or an AC voltage on the screen of an oscilloscope (Witte, 2002). This, in a nutshell, brings out the functions, as well as the difference of the function generator and the oscilloscope.
Having understood the basic functioning of the function generator and the oscilloscope by using the numerical values of the time, period, and voltage variables, this experiment demonstrated the difference between these two machines, but also practically demonstrated each variable as an independent variable in different mathematical contexts. The functional difference, as well as the functional similarity of the function generator and the oscilloscope shed light on the practical functioning of the machines and the different variables they ...Download file to see next pagesRead More
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Let us find you another Lab Report on topic Oscilloscope and Function Generator for FREE!
+16312120006 | 1,342 | 6,443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | longest | en | 0.913824 |
https://brilliant.org/discussions/thread/melodiess-message-board/?sort=new | 1,624,236,406,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488259200.84/warc/CC-MAIN-20210620235118-20210621025118-00442.warc.gz | 148,605,310 | 14,359 | # Melodies's Message Board
hahahah Just saw so many message board notes, so I've decided to create one myself! :)
I am Melodies from Singapore. I'm 16 this year and in Grade 10. I really like Math a lot, especially Olympiad and fun Math stuff :D In particular, I like learning combinatorics and number theory! Geometry is my weakest area though... Do message me if you have any fun facts to share cos I really love fun facts!
Note by Happy Melodies
7 years, 2 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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@Happy Melodies Hi melodies I am Mardokay a beginner in brilliant. May I ask you how you are so good at number theory? Also you write nice solutions.
- 7 years, 1 month ago
Thank you Mardokay. Actually I'm not very good at number theory... But nonetheless, for me, I just like to spot patterns or play with numbers... so I guess that's why number theory is one of my favourite field of math.. :) For writing solutions, I think you have to have the audience in mind, as in you would want to write clear and motivated solutions for others to learn and to understand, just like how I would like to read a solution (when I don't know how to solve). :)
- 7 years, 1 month ago
Thanks for the suggestion @Happy Melodies
- 7 years, 1 month ago
Is your real name Happy Melodies [this is a stupid question, isn't it?]?
- 7 years, 2 months ago
Haha no :)
- 7 years, 2 months ago
Hi, nice to meet you!
Everyone from Singapore is so good at math :P
- 7 years, 2 months ago
Hey Daniel! :) sigh... My math is not really gd... :( And Math Olympiad is coming soon... Really worried
- 7 years, 2 months ago
That's where you're wrong, 'cos I'm also from Singapore but I'm terrible at math. I'm still trying though :D
- 7 years, 2 months ago
Yuxuan, are u in secondary school? If so, are u prepared for SMO???? :((( i m so not!!!
- 7 years, 2 months ago
"I like learning combinatorics and number theory! Geometry is my weakest area though..." Finally! Someone like me! :)
- 7 years, 2 months ago
Agreed!
- 7 years, 2 months ago
hahaha :) ok :D
- 7 years, 2 months ago
same here
- 7 years, 2 months ago | 1,037 | 3,739 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-25 | latest | en | 0.869526 |
https://ch.mathworks.com/matlabcentral/cody/problems/2627-convert-to-binary-coded-decimal/solutions/1826752 | 1,597,503,170,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740848.39/warc/CC-MAIN-20200815124541-20200815154541-00327.warc.gz | 245,225,807 | 15,835 | Cody
# Problem 2627. Convert to Binary Coded Decimal
Solution 1826752
Submitted on 26 May 2019
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
x = 1; y_correct = '0001'; assert(isequal(bin2bcd(x),y_correct))
Undefined function 'bin2bcd' for input arguments of type 'double'. Error in Test1 (line 3) assert(isequal(bin2bcd(x),y_correct))
2 Fail
x = 5; y_correct = '0101'; assert(isequal(bin2bcd(x),y_correct))
Undefined function 'bin2bcd' for input arguments of type 'double'. Error in Test2 (line 3) assert(isequal(bin2bcd(x),y_correct))
3 Fail
x = 12; y_correct = '00010010'; assert(isequal(bin2bcd(x),y_correct))
Undefined function 'bin2bcd' for input arguments of type 'double'. Error in Test3 (line 3) assert(isequal(bin2bcd(x),y_correct))
4 Fail
x = 156; y_correct = '000101010110'; assert(isequal(bin2bcd(x),y_correct))
Undefined function 'bin2bcd' for input arguments of type 'double'. Error in Test4 (line 3) assert(isequal(bin2bcd(x),y_correct)) | 325 | 1,090 | {"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} | 2.6875 | 3 | CC-MAIN-2020-34 | latest | en | 0.346927 |
https://www.physicsforums.com/threads/determining-volume-of-base-for-asa-experiment.828116/ | 1,726,458,002,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00820.warc.gz | 867,657,546 | 19,019 | # Determining Volume of Base for ASA Experiment
• Mary1910
In summary, the conversation discusses an experiment to determine the mass of ASA using a titration method. The relevant information from the experiment includes the Ka value of ASA, the mass of an ASA tablet, and the use of a standardized NaOH solution. The conversation also provides relevant equations and an attempt at calculating the volume of base needed for the titration. The calculations appear to be correct, but it is suggested to be more precise in stating the purpose of the calculation.
Mary1910
1. Homework Statement .
This question is based on an experiment to determine the mass of ASA.
Information from the experiment :
• ASA is a weak acid with a Ka value of 3.2 x 10^-4.
• An ASA tablet has an approximate mass of 500 g.
• The experiment used a titration with a standardized 0.100 mol/L NaOH.
The balanced equation is HC9H7O4(aq) + NaOH(aq) -> NaC9H7O4(aq) + H2O(l)
a) What volume of base will you substitute into the formula when you calculate the number of moles of base used?
2. Relevant equations.
n=m/M
v=n/c
3. The attempt at the solution
First determine molar mass of HC9H7O4
=180.17 g/mol
n=m/M
=(0.500g ) / (180.17g/mol)
=2.78 x 10^-3 mol
v=n/c
=(2.78 x 10^-3) / (0.1mol/L)
=0.0278 mL
=27.8 L
Could someone let me know if I determined the volume of the base correctly? Thanks.
Mary1910 said:
=(2.78 x 10^-3) / (0.1mol/L)
=0.0278 mL
=27.8 L
You mixed up units in the last two lines, but - assuming you meant L first, mL second - your calculations look OK.
But it wouldn't hurt to write more precisely what you are calculating. I am guessing you are checking what should be approximate volume of the base required to titrate ASA from a single tablet, but it doesn't necessarily follow from your post.
Thanks, and sorry for my lack of clarity.
Borek
## 1. How is the volume of base determined in an ASA experiment?
The volume of base is determined by measuring the amount of base needed to neutralize an acidic solution, which can be done using a burette or a graduated cylinder.
## 2. What is the purpose of determining the volume of base in an ASA experiment?
The purpose of determining the volume of base is to accurately determine the concentration of the acidic solution being tested. This information can then be used to calculate other important parameters, such as the molarity of the acid.
## 3. What are the key factors that can affect the accuracy of determining the volume of base in an ASA experiment?
The key factors that can affect the accuracy of determining the volume of base include human error in measuring and recording volumes, air bubbles in the burette or graduated cylinder, and incorrect calibration of equipment.
## 4. How can the accuracy of determining the volume of base be improved in an ASA experiment?
The accuracy of determining the volume of base can be improved by using precise measuring equipment, taking multiple measurements and calculating an average, and ensuring that the equipment is properly calibrated.
## 5. Are there any safety considerations when determining the volume of base in an ASA experiment?
Yes, it is important to handle the acidic and basic solutions with caution and wear appropriate protective equipment, such as gloves and goggles. It is also important to properly dispose of any waste solutions to avoid contamination and potential hazards.
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3K | 889 | 3,609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.874715 |
https://discuss.interviewbit.com/t/a-hint-for-those-pl-getting-baquz-consider-case-when-u-get-a-26-0-instead-o/2195 | 1,611,767,355,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704828358.86/warc/CC-MAIN-20210127152334-20210127182334-00614.warc.gz | 282,611,475 | 3,552 | # A hint for those pl,,getting BAQUZ,,, consider case when u get a%26==0,,instead o
#1
a hint for those pl,getting BAQUZ, consider case when u get a%26==0,instead of use 0,use it as 26,and decrease the num by 26,like a=a-26;,hope understands.
#2
#3
eg: 52 = 26(2)+0 ideally the answer should be B0
since 0 is not valid we are shifting 26 from left hand side which is equivalent to 26(1) +26 => AZ
if you don’t subtract 26 answer will be BZ which corresponds to 78
#4
eg: 52 = 26(2)+0 ideally the answer should be B0
since 0 is not valid we are shifting 26 from left hand side which is equivalent to 26(1) +26 => AZ
if you don’t subtract 26 answer will be BZ which corresponds to (26^1)*2+(26^0)*0 =52 <<<----(U Explain wrong)
#5
Thanks mate I understand and its work😀 | 258 | 783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-04 | latest | en | 0.876174 |
https://www.physicsforums.com/threads/grasp-transformation-matrix.739165/ | 1,529,577,469,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864139.22/warc/CC-MAIN-20180621094633-20180621114633-00265.warc.gz | 900,079,690 | 13,850 | # Grasp transformation matrix
1. Feb 19, 2014
### lwcjoseph
Hello everyone, i am now working on a problem with three fingers robot hand to grab a cube to undergo some motion
however i face some difficulties on deriving the grasp transformation matrix which help to switching the local coordinate frame
at first i was given three point vectors [0 1 0]^T, [1 0.5 0]^T, [0 -1 0]^T (T means transpose)
then to calculate a Grasp transformation matrix
i would like to know the step to achieve the result and any general method for deriving the matrix?
thanks very much for your help !
2. Feb 19, 2014
### maajdl
http://commons.bcit.ca/math/examples/robotics/linear_algebra/
http://elvis.rowan.edu/~kay/papers/kinematics.pdf
http://en.wikipedia.org/wiki/Transformation_matrix#Affine_transformations
http://en.wikipedia.org/wiki/Transformation_matrix
I guess you are familiar with matrices to represent linear transformations like rotations.
The only special thing in these "robotic transformation matrices" is the use of a fourth "homogeneous" coordinate.
This allows to represent the group of rotations and translations at the same time, with matrices.
Last edited: Feb 19, 2014
3. Feb 19, 2014
### lwcjoseph
i have read through the materials ,however, the angle of rotation and any other information concerning the transformation has not been given, that's why i have been stucked for so long
the result is as the pic attached
any help?
#### Attached Files:
• ###### Grasp matrix.jpg
File size:
7.3 KB
Views:
122
4. Feb 20, 2014
### Inner_Peace
Are these vectors the points on the cube that you need to grasp?
Do you have any information about the base frame and the position/ orientation of the hand?
5. Feb 22, 2014
### lwcjoseph
yes they are the point that i need to grasp, however no other information are povided
that's why i have been stucked so long | 468 | 1,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-26 | latest | en | 0.872154 |
hydraev.co.uk | 1,726,382,886,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00446.warc.gz | 266,482,287 | 48,661 | # How much does it cost to run an electric vehicle?
There should be no doubt in anyone’s mind that running an electric vehicle can be considerably cheaper than a similar petrol or diesel powered car and while there are several factors which affect the calculations, running on electric will almost always be cheaper.
Calculating the running cost of a petrol or diesel car is fairly easy and predictable. You probably know what your average fuel consumption is and you know how much your fuel costs as you fill up at the forecourt. You’ll know roughly what your mileage is over any given period and how long that tank fill-up will last.
This is something most drivers have been doing for years and is second nature.
But with electric vehicles the parameters have changed so we need to change the way we look at running costs. Working out how much it costs to fully charge your vehicle’s battery is easy enough, just multiply the battery capacity in kW by the cost of the electric in kWh.
So if your vehicle battery has a total capacity of 50kW and your energy supplier charges 30p per kWh then it is 50 x 30 = 1,500 or £15. If that gives you a range of 180miles, for instance, the cost per mile is £15/180 = 8p per mile.
That is a very simplistic but easy way to calculate costs. The cost per kWh is the largest varying cost. You may pay your domestic energy supplier 30p but plugging into a public charger in a shopping centre car park could cost anywhere between 40p-80p per kWh. This doesn’t necessarily mean your running costs are doubled, though, as you might only plug in for an hour while you’re out shopping then plug in again when you get home so paying 50p or so isn’t going to break the bank, but it is worth bearing in mind.
Obviously there are other factors which can affect running costs. Just like weather, local traffic conditions and driving style will affect petrol consumption, the same applies to electric. Driving in winter at night with the heater on will shorten the range compared to driving in daylight in summer simply because everything is powered from the battery.
Comparing electric vs petrol, even in broadly average terms, show that there is a huge advantage with electric. If we go back to that example above the 180 miles range cost £15 on electric but with Petrol costing £1.70 per litre (at the time of writing) a Ford Focus returning 50mpg would use £27.80 in petrol, and that is being optimistic, frequently driving in traffic or around town is likely to increase the fuel consumption to around 35mpg which pushes costs up to £39.70.
Other factors to bear in mind when considering electric are things like road tax, insurance, parking, congestion charges etc. All of these can be lower for electric cars than similar petrol or diesel models. | 592 | 2,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.956118 |
https://www.jiskha.com/questions/1000987/y-1-2x-3-slope-x-intercept-y-intercept-how-do-i-graph-this | 1,576,467,645,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541315293.87/warc/CC-MAIN-20191216013805-20191216041805-00529.warc.gz | 745,063,557 | 5,732 | # Math
y=1/2x+3
Slope?
x-intercept
y-intercept
How do I graph this?
1. 👍 0
2. 👎 0
3. 👁 95
1. if a line y= ax+b, then a is the slope and b is the y-intercept.
here line is y=1/2 x+3
so slope is 1/2 and y-intercept is 3.
it means that y increases 1 unit for every 3 units of x increases
1. 👍 0
2. 👎 0
posted by gg
2. y = 1/2x + 3
Slope: m = 1/2
y-intercept is when x = 0
y = mx + b
y = 1/2(0)+3
y = 0 + 3
y = 3
y-intercept is 3
x-intercept is when y = 0
0 = 1/2x + 3 > move 1/2x to the left and change sign.
0 -1/2x = 3
-1/2x = 3 > divide on both sides by -1/2
x = -6
x-intercept is -6
Graph:
y=1/2x+3
You can give x a value and plot points, then draw a line.
y=1/2x+3 x = 2 y = 1/2(2)+ 3 y = 4
x = 4 y = 1/2(4)+ 3 y = 5
x = -2 y = 1/2(-2)+3 y = 2
x = -4 y = 1/2(-4)+3 y = 1
(2,4)
(4,5)
(-2,2)
(-4,1)
You can also graph by going to positive 3 on the y-axis. Then go up 1 and to the right 2. From that point, repeat the sequence > up 1, to the right 2.
You can also go to positive 3 on y-axis then go down 1, to the left 2. From that point repeat sequence > down 1, to the left 2.
Then draw a line.
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2. 👎 0
posted by Chelle
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You should try to formalize this intuition, i.e., (i) show that the reduction is correct and (ii) show that there is an efficient algorithm for directed $$s$$-$$t$$ cut. If both check out, you are done. | 412 | 1,547 | {"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": 31, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-26 | latest | en | 0.938606 |
https://oeis.org/A236036 | 1,721,259,017,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00122.warc.gz | 372,767,655 | 4,427 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A236036 a(n) is the smallest (2n+1)-digit prime number whose central digit equals the sum of its other digits. 1
10513, 1005013, 100040201, 10000400021, 1000004000111, 100000040002001, 10000000400000111, 1000000004000100011, 100000000040000000021, 10000000000400010000011, 1000000000004000000000111, 100000000000040000010000101, 10000000000000400000000011001 (list; graph; refs; listen; history; text; internal format)
OFFSET 2,1 COMMENTS Sequence starts at n=2, since no such 3-digit prime exists. LINKS Giovanni Resta, Table of n, a(n) for n = 2..100 EXAMPLE a(3) = 1005013 because the central digit 5 equals the sum of the other digits 1+0+0+0+1+3. MAPLE with(numtheory):for n from 2 to 10 do:m:=2*n-2:ii:=1:ii:=0:for k from 10^m to 10^(m+1)-1 while(ii=0)do:x:=convert(k, base, 10):n1:=nops(x):s:=sum('x[j]', 'j'=1..n1):s1:=s-x[n]:if x[n]=s1 and type(k, prime)=true then ii:=1: printf ( "%d %d \n", n, k):else fi:od:od: MATHEMATICA a[n_] := Catch@Block[{p}, Do[p = Select[ Union[ FromDigits /@ Flatten[ Permutations /@ (IntegerPartitions[d + n - 1, {n}, Range@d] - 1), 1]] + d*10^n + 10^(2*n), PrimeQ, 1]; If[p != {}, Throw@p[[1]]], {d, {4, 5}}]]; a /@ Range[2, 14] (* Giovanni Resta, Jan 20 2014 *) PROG (PARI) isspecial(p, n) = {d = digits(p); s = sumdigits(p); d[n+1] == (s - d[n+1]); } a(n) = {forprime (p = 10^(2*n), 10^(2*n+1), if (isspecial(p, n), return (p)); ); return (0); } \\ Michel Marcus, Jan 19 2014 CROSSREFS Cf. A000040, A235119. Sequence in context: A348758 A214192 A235119 * A216489 A120500 A371623 Adjacent sequences: A236033 A236034 A236035 * A236037 A236038 A236039 KEYWORD nonn,base AUTHOR Michel Lagneau, Jan 18 2014 EXTENSIONS a(9)-a(14) from Giovanni Resta, Jan 20 2014 Name simplified by Jon E. Schoenfield, Sep 09 2017 STATUS approved
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Last modified July 17 19:13 EDT 2024. Contains 374377 sequences. (Running on oeis4.) | 804 | 2,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-30 | latest | en | 0.429575 |
https://tough.forumbee.com/t/81d74?r=62dch | 1,680,292,781,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949678.39/warc/CC-MAIN-20230331175950-20230331205950-00504.warc.gz | 659,838,229 | 10,332 | 0
# TOUGH-HYDRATE - Problems with the maximum mass fraction of methane in aqueous phase
I'm trying to equilibrate a sloping system, containing a AqH phase and a Aqu phase. If methane mass fractions of methane in water are not maximum, hydrate phase dissolves. So, I want TOUGH-HYDRATE to set the maximum mass fraction of methane values in single aqueous phase. To do that I specify a high value for X_mA, than TOUGH calculates the maximum value and replaces the original value to the maximum:
THE SINGLE-PHASE CONDITIONS SPECIFIED AT ELEMENT " 1 1 " ARE ERRONEOUS
At T = 1.92981E+01 C, the CH4 mass fraction in H2O "X_mA" = 9.50000E-01 exceeds the maximum = 5.99297E-03 (T-dependent)
!!! ACTION TAKEN: SET X_mA = Max ==> EXECUTION CONTINUES
Ok, but after that, TOUGH says that this value is wrong, prints a different value for the maximum mass fraction of methane in water and stops execution:
!!!!!! "Phase1_Aqu_Eql": Gas phase evolves at element " 1 1 " ==> P = 8.35121719E+06 Peq_Hydr = 2.69476647E+07 X_mA = 5.99297481E-03 X_mA_max = 1.85578449E-03
!!!!!!!!!! Routine Phase1_Aqu_Eql: ERRONEOUS DATA INITIALIZATION ==> STOP EXECUTION <<<<<<<<<<
Why does this happen? Do I have to change something in the code? I use the licenced version 1.0.
9replies Oldest first
• Oldest first
• Popular
• I suggest you begin the simulation again, but give a value of 1.855e-3 to the mass fraction (i.e., just a bit lower than the one the code gives you). Try it and let me know. If it does not work, please send me your input files.
Like
• With this value, it works. But the problem is that I have a big grid, so I it's hard to give all values manually (these two messages appear for all the cells setted to have aqueous phase).
I have a version of the simulator from a course that sets the mass fraction to 99% of the maximum mass fraction and it works properly. But this version has a limit of 3000 cells, less than what I need to use.
With the licensed version, I think the problem is not only setting the value to 99% of the maximum value (instead of setting to 100% of it) becouse the first message gives a value much bigger than the second one (5.99297E-03 in the first message and 1.85578449E-03 in the second one). When I setted the mass fraction value to 1.855e-3 for the cell 101, the second error message appeared for the cell 102, and so on...
The input file is attached.
Like
• And here, my output file too.
Like
• KHS
• HYDRATE
• KHS
• 4 yrs ago
• Reported - view
Cristiane are you still working with tough +
Like
• KHS Hi! no, I am not. maybe I still can help you. but I do not have much free time and might take a while to answer.
I read your comment about PetraSim. It is useful only for the generation of blocks ELEME and CONNE. Just copy the other blocks from a ready file and you will only have to change some few parameters accordingly to your needs.
Good luck! I agree with you when you say there is not much support here :/
Like
• KHS
• HYDRATE
• KHS
• 4 yrs ago
• Reported - view
aren't these sections already generated when you create mesh file?
petrasim can generate mesh easily, but it can't be exported to be used for running using tough+ hydrate..its in a different format...isn't it?
am I right?
Like
• KHS Hi! yes, they can be generated with MESH. I used PetraSim just because it was easier to visualize and to be sure I was fixing the primary variables on the correct blocks :)
I think the section of Petrasim was called only TOUGH+ and not TOUGH+HYDRATE. Or maybe I used the TOUGH2 section. I can't remember, sorry. But if you can use MESH, just use that, no need for Petrasim. You can not visualize results on it either, you will need TecPlot.
Like
• Cristiane Romio can we visualize the mesh that is made by MM using any software?
Like
• KHS
• HYDRATE
• KHS
• 4 yrs ago
• Reported - view
Cristiane Romio i need your help. | 1,092 | 3,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-14 | latest | en | 0.840401 |
https://www.mathworks.com/matlabcentral/cody/problems/1686-generate-a-melodic-contour-string-matrix/solutions/270371 | 1,506,152,811,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689572.75/warc/CC-MAIN-20170923070853-20170923090853-00093.warc.gz | 848,473,836 | 11,752 | Cody
# Problem 1686. Generate a melodic contour string matrix
Solution 270371
Submitted on 1 Jul 2013 by J.R.! Menzinger
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% str = '*rrr'; melody = '*-*-*-*'; assert(strcmp(parsons(str),melody))
``` ```
2 Pass
%% str = '*du'; melody = ['* *' ' \ / ' ' * ']; assert(strcmp(parsons(str),melody))
``` ```
3 Pass
%% str = '*ruuddduur'; melody = [' * ' ' / \ ' ' * * *-*' ' / \ / ' '*-* * * ' ' \ / ' ' * ']; assert(strcmp(parsons(str),melody))
``` ```
4 Pass
%% str = '*rududdrudud'; melody = [' * * ' ' / \ / \ ' '*-* * * * * ' ' \ / \ / \ ' ' *-* * *']; assert(strcmp(parsons(str),melody))
``` ```
5 Pass
%% str = '*rururddrdrdrd'; melody = [' *-* ' ' / \ ' ' *-* * ' ' / \ ' '*-* *-* ' ' \ ' ' *-* ' ' \ ' ' *-* ' ' \ ' ' *']; assert(strcmp(parsons(str),melody))
``` ``` | 326 | 956 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-39 | latest | en | 0.237388 |
https://au.mathworks.com/matlabcentral/cody/groups/63 | 1,628,022,478,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154471.78/warc/CC-MAIN-20210803191307-20210803221307-00253.warc.gz | 127,330,287 | 20,667 | Cody
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[contact-form-7 id="7458"] | 1,246 | 6,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-26 | latest | en | 0.918331 |
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_INN/TRS/nontermin/CSR/ExIntrod_GM99.trs.Thm12:EMB:SAFE.html.lzma | 1,726,518,206,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00219.warc.gz | 83,689,590 | 2,820 | Term Rewriting System R:
[X, Y, Z]
primes -> sieve(from(s(s(0))))
from(X) -> cons(X, from(s(X)))
tail(cons(X, Y)) -> Y
if(true, X, Y) -> X
if(false, X, Y) -> Y
filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
Innermost Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
PRIMES -> SIEVE(from(s(s(0))))
PRIMES -> FROM(s(s(0)))
FROM(X) -> FROM(s(X))
FILTER(s(s(X)), cons(Y, Z)) -> IF(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
FILTER(s(s(X)), cons(Y, Z)) -> FILTER(s(s(X)), Z)
FILTER(s(s(X)), cons(Y, Z)) -> FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
SIEVE(cons(X, Y)) -> FILTER(X, sieve(Y))
SIEVE(cons(X, Y)) -> SIEVE(Y)
Furthermore, R contains two SCCs.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Narrowing Transformation`
` →DP Problem 2`
` ↳Remaining`
Dependency Pairs:
SIEVE(cons(X, Y)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) -> FILTER(s(s(X)), Z)
SIEVE(cons(X, Y)) -> FILTER(X, sieve(Y))
Rules:
primes -> sieve(from(s(s(0))))
from(X) -> cons(X, from(s(X)))
tail(cons(X, Y)) -> Y
if(true, X, Y) -> X
if(false, X, Y) -> Y
filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
Strategy:
innermost
On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule
FILTER(s(s(X)), cons(Y, Z)) -> FILTER(X, sieve(Y))
one new Dependency Pair is created:
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Nar`
` →DP Problem 3`
` ↳Narrowing Transformation`
` →DP Problem 2`
` ↳Remaining`
Dependency Pairs:
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> FILTER(s(s(X)), Z)
SIEVE(cons(X, Y)) -> FILTER(X, sieve(Y))
SIEVE(cons(X, Y)) -> SIEVE(Y)
Rules:
primes -> sieve(from(s(s(0))))
from(X) -> cons(X, from(s(X)))
tail(cons(X, Y)) -> Y
if(true, X, Y) -> X
if(false, X, Y) -> Y
filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
Strategy:
innermost
On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule
SIEVE(cons(X, Y)) -> FILTER(X, sieve(Y))
one new Dependency Pair is created:
SIEVE(cons(X, cons(X'', Y''))) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Nar`
` →DP Problem 3`
` ↳Nar`
` ...`
` →DP Problem 4`
` ↳Forward Instantiation Transformation`
` →DP Problem 2`
` ↳Remaining`
Dependency Pairs:
SIEVE(cons(X, cons(X'', Y''))) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
SIEVE(cons(X, Y)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> FILTER(s(s(X)), Z)
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
Rules:
primes -> sieve(from(s(s(0))))
from(X) -> cons(X, from(s(X)))
tail(cons(X, Y)) -> Y
if(true, X, Y) -> X
if(false, X, Y) -> Y
filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
Strategy:
innermost
On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule
FILTER(s(s(X)), cons(Y, Z)) -> FILTER(s(s(X)), Z)
two new Dependency Pairs are created:
FILTER(s(s(X'')), cons(Y, cons(Y'', Z''))) -> FILTER(s(s(X'')), cons(Y'', Z''))
FILTER(s(s(X'')), cons(Y, cons(cons(X'''', Y''''), Z''))) -> FILTER(s(s(X'')), cons(cons(X'''', Y''''), Z''))
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Nar`
` →DP Problem 3`
` ↳Nar`
` ...`
` →DP Problem 5`
` ↳Forward Instantiation Transformation`
` →DP Problem 2`
` ↳Remaining`
Dependency Pairs:
FILTER(s(s(X'')), cons(Y, cons(cons(X'''', Y''''), Z''))) -> FILTER(s(s(X'')), cons(cons(X'''', Y''''), Z''))
FILTER(s(s(X'')), cons(Y, cons(Y'', Z''))) -> FILTER(s(s(X'')), cons(Y'', Z''))
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
SIEVE(cons(X, Y)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
SIEVE(cons(X, cons(X'', Y''))) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
Rules:
primes -> sieve(from(s(s(0))))
from(X) -> cons(X, from(s(X)))
tail(cons(X, Y)) -> Y
if(true, X, Y) -> X
if(false, X, Y) -> Y
filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
Strategy:
innermost
On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule
FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
two new Dependency Pairs are created:
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> SIEVE(cons(X'', Y''))
FILTER(s(s(X)), cons(cons(X'', cons(X'''', Y'''')), Z)) -> SIEVE(cons(X'', cons(X'''', Y'''')))
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Nar`
` →DP Problem 3`
` ↳Nar`
` ...`
` →DP Problem 6`
` ↳Forward Instantiation Transformation`
` →DP Problem 2`
` ↳Remaining`
Dependency Pairs:
FILTER(s(s(X)), cons(cons(X'', cons(X'''', Y'''')), Z)) -> SIEVE(cons(X'', cons(X'''', Y'''')))
SIEVE(cons(X, cons(X'', Y''))) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
SIEVE(cons(X, Y)) -> SIEVE(Y)
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> SIEVE(cons(X'', Y''))
FILTER(s(s(X'')), cons(Y, cons(Y'', Z''))) -> FILTER(s(s(X'')), cons(Y'', Z''))
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
FILTER(s(s(X'')), cons(Y, cons(cons(X'''', Y''''), Z''))) -> FILTER(s(s(X'')), cons(cons(X'''', Y''''), Z''))
Rules:
primes -> sieve(from(s(s(0))))
from(X) -> cons(X, from(s(X)))
tail(cons(X, Y)) -> Y
if(true, X, Y) -> X
if(false, X, Y) -> Y
filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
Strategy:
innermost
On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule
SIEVE(cons(X, Y)) -> SIEVE(Y)
two new Dependency Pairs are created:
SIEVE(cons(X, cons(X'', Y''))) -> SIEVE(cons(X'', Y''))
SIEVE(cons(X, cons(X'', cons(X'''', Y'''')))) -> SIEVE(cons(X'', cons(X'''', Y'''')))
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Nar`
` →DP Problem 2`
` ↳Remaining Obligation(s)`
The following remains to be proven:
• Dependency Pairs:
SIEVE(cons(X, cons(X'', cons(X'''', Y'''')))) -> SIEVE(cons(X'', cons(X'''', Y'''')))
SIEVE(cons(X, cons(X'', Y''))) -> SIEVE(cons(X'', Y''))
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> SIEVE(cons(X'', Y''))
FILTER(s(s(X'')), cons(Y, cons(cons(X'''', Y''''), Z''))) -> FILTER(s(s(X'')), cons(cons(X'''', Y''''), Z''))
FILTER(s(s(X'')), cons(Y, cons(Y'', Z''))) -> FILTER(s(s(X'')), cons(Y'', Z''))
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
SIEVE(cons(X, cons(X'', Y''))) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
FILTER(s(s(X)), cons(cons(X'', cons(X'''', Y'''')), Z)) -> SIEVE(cons(X'', cons(X'''', Y'''')))
Rules:
primes -> sieve(from(s(s(0))))
from(X) -> cons(X, from(s(X)))
tail(cons(X, Y)) -> Y
if(true, X, Y) -> X
if(false, X, Y) -> Y
filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
Strategy:
innermost
• Dependency Pair:
FROM(X) -> FROM(s(X))
Rules:
primes -> sieve(from(s(s(0))))
from(X) -> cons(X, from(s(X)))
tail(cons(X, Y)) -> Y
if(true, X, Y) -> X
if(false, X, Y) -> Y
filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
Strategy:
innermost
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Nar`
` →DP Problem 2`
` ↳Remaining Obligation(s)`
The following remains to be proven:
• Dependency Pairs:
SIEVE(cons(X, cons(X'', cons(X'''', Y'''')))) -> SIEVE(cons(X'', cons(X'''', Y'''')))
SIEVE(cons(X, cons(X'', Y''))) -> SIEVE(cons(X'', Y''))
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> SIEVE(cons(X'', Y''))
FILTER(s(s(X'')), cons(Y, cons(cons(X'''', Y''''), Z''))) -> FILTER(s(s(X'')), cons(cons(X'''', Y''''), Z''))
FILTER(s(s(X'')), cons(Y, cons(Y'', Z''))) -> FILTER(s(s(X'')), cons(Y'', Z''))
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
SIEVE(cons(X, cons(X'', Y''))) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
FILTER(s(s(X)), cons(cons(X'', cons(X'''', Y'''')), Z)) -> SIEVE(cons(X'', cons(X'''', Y'''')))
Rules:
primes -> sieve(from(s(s(0))))
from(X) -> cons(X, from(s(X)))
tail(cons(X, Y)) -> Y
if(true, X, Y) -> X
if(false, X, Y) -> Y
filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
Strategy:
innermost
• Dependency Pair:
FROM(X) -> FROM(s(X))
Rules:
primes -> sieve(from(s(s(0))))
from(X) -> cons(X, from(s(X)))
tail(cons(X, Y)) -> Y
if(true, X, Y) -> X
if(false, X, Y) -> Y
filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))
Strategy:
innermost
Innermost Termination of R could not be shown.
Duration:
0:01 minutes | 3,471 | 10,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-38 | latest | en | 0.421938 |
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A set of 3 differentiated worksheets based on finding a quarter of a quantity. Linked to the White Rose Maths Hub planning and they have clear evidence of reasoning and problem solving.
#### Finding half of a quantity or number. Differentiated Worksheets Year 1 (White Rose)
A set of 3 differentiated worksheets based on finding half of a number or quantity. Linked to the White Rose Maths Hub planning and they have clear evidence of reasoning and problem solving.
#### Sharing objects into equal groups Differentiated Worksheets Year1 (White Rose)
A set of 3 differentiated worksheets based on sharing into equal groups. Linked to the White Rose Maths Hub planning and they have clear evidence of reasoning and problem solving.
#### Doubling Numbers to 20 Year 1 Differentiated Worksheets (White Rose)
A set of 3 differentiated worksheets based on doubling numbers up to 20. Linked to the White Rose Maths Hub planning and they have clear evidence of reasoning and problem solving.
#### Simple Images of Arrays Year 1 Differentiated
Arrays for children to identify the related repeated addition and the multiplication number sentences.
#### Adding Equal Groups Differentiated Worksheets Year 1 (White Rose)
A set of 3 differentiated worksheets based on adding equal groups. Linked to the White Rose Maths Hub planning and they have clear evidence of reasoning and problem solving.
#### Multiplying and Dividing by 10 and 100
An effective PowerPoint that takes the pupils through the process of multiplying and dividing by 10 and 100 in a series of simple steps. It shows very clearly how the digits move to make their answers bigger if multiplying and smaller if dividing. Suitable for KS2 pupils as well as a reminder for LA KS3 students.
#### Making Equal Groups of Objects Year 1 Differentiated Worksheets (White Rose)
A set of 3 differentiated worksheets based on identifying and making equal groups of objects. Linked to the White Rose Maths Hub planning and they have clear evidence of reasoning and problem solving.
#### Assessment Pack - Number
End of Access Foundation text book test 2 papers (non calc and calc) including markscheme
#### Maths Applied Problem Solving Worksheet Pack - 4 A5 Sheets (Simplifying Algebra, LCM / HCF, Rounding+Estimation, Powers + Roots)
This is a bundle containing 4 applied A5 worksheets on various topics. Originally designed for high ability year 7, these could easily be used all the way up to KS4 GCSE. Topics Include: Collecting Like Terms / Simplifying Expressions Lowest Common Multiple / Highest Common Factor Powers and Roots / Basic Indices Rounding and Estimation
#### Indices hint sheets
Indices help sheets based on fractional, negative and writing in terms of another number.
#### Number Assessment Pack
Assessment Pack based on Access Foundation chapters 1 - 6 Integers, Order of operations Decimals Approximations Fractions Percentages Assessment and Markschemes
#### Applied Rounding and Estimation - Maths Problem Solving Worksheet + Answers
This is a collection of applied questions for the topic of rounding and estimation. The questions are designed to emphasise student understanding by identifying misconceptions and explaining their answers. Also has questions applying in a worded context. The worksheet is A5 so can fit 2 to a page to save printing and fit better in student books. Was designed for high ability year 7 group, but could be used for any year groups up from this up to GCSE. Full answers provided.
#### Applied Powers and Roots / Indices - Maths Problem Solving Worksheet + Answers
This is a collection of applied questions for the topic of powers and roots. The questions are designed to emphasise student understanding by identifying misconceptions and explaining their answers. Also has questions applying in a geometric context. The worksheet is A5 so can fit 2 to a page to save printing and fit better in student books. Was designed for high ability year 7 group, but could be used for any year groups up from this up to GCSE. Full answers provided.
#### Applied Lowest Common Multiple (LCM) Highest Common Factor (HCF) Problem Solving Worksheet+Answers
This is a collection of applied questions for the topic of lowest common multiple and highest common factors. Has questions applying to worded contexts. The worksheet is A5 so can fit 2 to a page to save printing and fit better in student books. Was designed for high ability year 7 group, but could be used for any year groups up from this up to GCSE. Full answers provided.
#### Applied Collecting Like Terms / Simplifying Algebra - Maths Problem Solving Worksheet + Answers
This is a collection of applied questions for the topic of collecting like terms. Has questions applying the main topic to geometric, worded and fractional context. The worksheet is A5 so can fit 2 to a page to save printing and fit better in student books. Was designed for high ability year 7 group, but could be used for any year groups up from this up to GCSE. Full answers provided.
#### Editable grid method worksheet
Easily editable worksheet for use when teaching grid method multiplication
#### Place Value anchor chart
Activity - role dice and create number. Supporting place value mastery and conversion of numbers to words. These can be laminated to use again and again, hole punched for folder etc…
#### Time Conversion
Micfosoft Excel required A new set of questions every time the file is opened - unlimited worksheets Suitable for KS2 Convert seconds to minutes, minutes to seconds, minutes to hours and hours to days Also included are 5 Time Conversion word problems e.g. Bill spent 190 hours on holiday. How many days is that? Lots of variety and challenges in this resource | 3,593 | 17,749 | {"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} | 3.234375 | 3 | CC-MAIN-2018-22 | latest | en | 0.913237 |
https://playworksheet.com/sheet/cylinders-changing-dimensions | 1,603,614,060,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107888402.81/warc/CC-MAIN-20201025070924-20201025100924-00491.warc.gz | 495,781,608 | 5,200 | Cylinders with Changing Dimensions in Diameter and Height
Matching and sorting are important early childhood skills. Matching is the identification of the same or similar objects according to their common properties. Sorting is the process of arranging objects with similar properties in a specific order or in groups. Matching and sorting are important activities which help children develop their brains, and also build the foundation for later math learning. Children should master the skills of matching and sorting by shapes, sizes, and/or colors.
This worksheet provides an exercise for hands-on matching and sorting skills. It is suitable for kids older than three years old. It consists of four groups of cylinders varying in height and diameter in different ways. The first group contains a row of cylinders decreasing in diameter, but having the same height. The second group contains cylinders decreasing in both diameter and height, and so are either larger or smaller, but always of the same shape. In the third group, the cylinders have the same diameter but vary in height; as the size decreases, the cylinder gradually becomes a disc in form. The last group of cylinders vary in all dimensions, decreasing in diameter but increasing in height, so that the cylinder changes from a disc to a stick in form.
Parents could cut out the cylinders in each group, mix them, and ask kids to fill the cylinders into their original positions. Or you may use the second page of the worksheet, which contains outlines of the cylinders, and ask kids to put the cylinders into the corresponding outlines. This exercise educates kids to distinguish differences in dimensions. Kids could train themselves to make comparisons between objects, form judgements and make decisions. This also is a good exercise for developing fine motor skills.
Exercise for Sorting Shapes by Coloring
Rectangles with Changing Dimensions in Width and Height
Basic Shapes with Increasing/Decreasing Sizes
Squares and Circles with Changing Sizes
Colored Table Chart of Numbers From One to Fifty
High-Contrasting Black and White Flashcards for Newborn Babies
Black and White Images of Fruits for Newborn Babies
Six Different Types of Triangles
Black and White Flashcards of Animals for Newborn Babies
Find and Count Basic Shapes in Compositions
Opposite Words Illustrated with Simple Shapes
Basic Shapes and Colors
High-Contrasting Black and White Images for Newborns
Twelve Irregular Figures of Shapes
Common Symbol Signs - Pictograms
Recognize Basic Shapes on Road Traffic Signs
Shapes: Triangle, Rectangle, Pentagon, Hexagon and Circle
Fruits Pictures: Berries, Melons and More
Fruits Pictures: Pomes, Drupes and Citruses
Flags of International Organizations
Flags of Group of Seven G7 member countries | 543 | 2,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | longest | en | 0.95123 |
http://marinus.pl/kq8nrwq/almost-sure-convergence-vs-convergence-in-probability-90ad89 | 1,656,330,757,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103331729.20/warc/CC-MAIN-20220627103810-20220627133810-00344.warc.gz | 35,566,979 | 13,352 | What information should I include for this source citation? convergence. I've encountered these two examples (used to show how a.s. convergence doesn't imply convergence in Rth mean and visa versa). n → X. iff for every subsequence . In general, almost sure convergence is stronger than convergence in probability, and a.s. convergence implies convergence in probability. For almost sure convergence, we collect all the !’s wherein the convergence happens, and demand that the measure of this set of !’s be 1. If X n are independent random variables assuming value one with probability 1/n and zero otherwise, then X n converges to zero in probability but not almost surely. But, in the case of convergence in probability, there is no direct notion of !since we are looking at a sequence of probabilities converging. Thus, there exists a sequence of random variables Y_n such that Y_n->0 in probability, but Y_n does not converge to 0 almost surely. A brief review of shrinkage in ridge regression and a comparison to OLS. In one case we have a random variable Xn = n with probability $=\frac{1}{n}$ and zero otherwise (so with probability 1-$\frac{1}{n}$).In another case same deal with only difference being Xn=1, not n with probability $=\frac{1}{n}$.Assume Xn's are independent in both. A sequence of random variables $X_1, X_2, \dots X_n$ converges in probability to a random variable $X$ if, for every $\epsilon > 0$, \begin{align}\lim_{n \rightarrow \infty} P(\lvert X_n - X \rvert < \epsilon) = 1.\end{align}. Let us consider a sequence of independent random ariablesv (Z. One thing that helped me to grasp the difference is the following equivalence, $P({\lim_{n\to\infty}|X_n-X|=0})=1 \Leftarrow \Rightarrow \lim_{n\to\infty}({\sup_{m>=n}|X_m-X|>\epsilon })=0$ $\forall \epsilon > 0$, $\lim_{n\to\infty}P(|X_n-X|>\epsilon) = 0$ $\forall \epsilon >0$. In probability theory one uses various modes of convergence of random variables, many of which are crucial for applications. The natural concept of uniqueness here is that of almost sure uniqueness. X. a.s. n. ks → X. Accidentally cut the bottom chord of truss. You obtain $n$ estimates $X_1,X_2,\dots,X_n$ of the speed of light (or some other quantity) that has some true' value, say $\mu$. BFGS is a second-order optimization method – a close relative of Newton’s method – that approximates the Hessian of the objective function. The notation X n a.s.→ X is often used for al-most sure convergence, while the common notation for convergence in probability is X n →p X or Almost sure convergence vs. convergence in probability: some niceties The goal of this problem is to better understand the subtle links between almost sure convergence and convergence in probabilit.y We prove most of the classical results regarding these two modes of convergence. Making statements based on opinion; back them up with references or personal experience. Modes of Convergence in Probability Theory David Mandel November 5, 2015 Below, x a probability space (;F;P) on which all random variables fX ng and X are de ned. Convergence in probability is weaker and merely requires that the probability of the difference Xn(w) X(w) being non-trivial becomes small. Retrieved from This article, published in the Annals of Mathematical Statistics journal, gives a brief but broad overview of high level calculus and statistical concepts Convergence In Probability, free convergence in probability … Relations among modes of convergence. Proposition7.1 Almost-sure convergence implies convergence in probability. ... this proof is omitted, but we include a proof that shows pointwise convergence =)almost sure convergence, and hence uniform convergence =)almost sure convergence. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. However, personally I am very glad that, for example, the strong law of large numbers exists, as opposed to just the weak law. 5. The example I have right now is Exercise 47 (1.116) from Shao: $X_n(w) = \begin{cases}1 &... Stack Exchange Network. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The WLLN (convergence in probability) says that a large proportion of the sample paths will be in the bands on the right-hand side, at time$n$(for the above it looks like around 48 or 9 out of 50). This can be verified using the Borel–Cantelli lemmas. However, recall that although the gaps between the$1 + s$terms will become large, the sequence will always bounce between$s$and$1 + s$with some nonzero frequency. Here, we essentially need to examine whether for every$\epsilon$, we can find a term in the sequence such that all following terms satisfy$\lvert X_n - X \rvert < \epsilon$. You may want to read our, Convergence in probability vs. almost sure convergence, stats.stackexchange.com/questions/72859/…. 4 . In order to understand this lecture, you should first understand the concepts of almost sure property and almost sure event, explained in the lecture entitled Zero-probability events, and the concept of pointwise convergence of a sequence of random variables, explained in the lecture entitled … If you enjoy visual explanations, there was a nice 'Teacher's Corner' article on this subject in the American Statistician (cite below). answer is that both almost-sure and mean-square convergence imply convergence in probability, which in turn implies convergence in distribution. Almost Sure Convergence. So, after using the device a large number of times, you can be very confident of it working correctly, it still might fail, it's just very unlikely. Convergence of Sum of Sums of random variables : trivial? Limits and convergence concepts: almost sure, in probability and in mean Letfa n: n= 1;2;:::gbeasequenceofnon-randomrealnumbers. Convergence in probability means that with probability 1, X = Y. Convergence in probability is a much stronger statement. Example . Said another way, for any$\epsilon$, we’ll be able to find a term in the sequence such that$P(\lvert X_n(s) - X(s) \rvert < \epsilon)$is true. For a sequence (Xn: n 2N), almost sure convergence of means that for almost all outcomesw, the difference Xn(w) X(w) gets small and stays small. I'm looking for a simple example sequence$\{X_n\}$that converges in probability but not almost surely. The hope is that as the sample size increases the estimator should get ‘closer’ to the parameter of interest. The SLLN (convergence almost surely) says that we can be 100% sure that this curve stretching off to the right will eventually, at some finite time, fall entirely within the bands forever afterward (to the right). De nition 5.2 | Almost sure convergence (Karr, 1993, p. 135; Rohatgi, 1976, p. so almost sure convergence and convergence in rth mean for some r both imply convergence in probability, which in turn implies convergence in distribution to random variable X. Relationship between the multivariate normal, SVD, and Cholesky decomposition. We want to know which modes of convergence imply which. You compute the average Thus, the probability that the difference$X_n(s) - X(s)$is large will become arbitrarily small. The following is a convenient characterization, showing that convergence in probability is very closely related to almost sure convergence. Limits are often required to be unique in an appropriate sense. Convergence in probability says that the chance of failure goes to zero as the number of usages goes to infinity. As a bonus, the authors included an R package to facilitate learning. Forums. We can explicitly show that the “waiting times” between$1 + s$terms is increasing: Now, consider the quantity$X(s) = s$, and let’s look at whether the sequence converges to$X(s)$in probability and/or almost surely. Thus, the probability that$\lim_{n \rightarrow \infty} \lvert X_n - X \rvert < \epsilon$does not go to one as$n \rightarrow \infty$, and we can conclude that the sequence does not converge to$X(s)$almost surely. Exercise 1.1: Almost sure convergence: omega by omega - Duration: 4:52. herrgrillparzer 3,119 views. Convergence in probability is a bit like asking whether all meetings were almost full. ... Convergence in Probability and in the Mean Part 1 - Duration: 13:37. Eg, the list will be re-ordered over time as people vote. Example (Almost sure convergence) Let the sample space S be the closed interval [0,1] with the uniform probability … That is, if you count the number of failures as the number of usages goes to infinity, you will get a finite number. = 1 (1) or also written as P lim n!1 X n = X = 1 (2) or X n a:s:! The wiki has some examples of both which should help clarify the above (in particular see the example of the archer in the context of convergence in prob and the example of the charity in the context of almost sure convergence). We only require that the set on which X n(!) Almost sure convergence. I've encountered these two examples (used to show how a.s. convergence doesn't imply convergence in Rth mean and visa versa). almost sure convergence). To assess convergence in probability, we look at the limit of the probability value$P(\lvert X_n - X \rvert < \epsilon)$, whereas in almost sure convergence we look at the limit of the quantity$\lvert X_n - X \rvert$and then compute the probability of this limit being less than$\epsilon$. Assume you have some device, that improves with time. When comparing the right side of the upper equivlance with the stochastic convergence, the difference becomes clearer I think. It's not as cool as an R package. Convergence almost surely implies convergence in probability ... Convergence in probability does not imply almost sure convergence in the discrete case. Consider the sequence Xn of random variables, and the random variable Y. Convergence in distribution means that as n goes to infinity, Xn and Y will have the same distribution function. Here’s the sequence, defined over the interval$[0, 1]: \begin{align}X_1(s) &= s + I_{[0, 1]}(s) \\ X_2(s) &= s + I_{[0, \frac{1}{2}]}(s) \\ X_3(s) &= s + I_{[\frac{1}{2}, 1]}(s) \\ X_4(s) &= s + I_{[0, \frac{1}{3}]}(s) \\ X_5(s) &= s + I_{[\frac{1}{3}, \frac{2}{3}]}(s) \\ X_6(s) &= s + I_{[\frac{2}{3}, 1]}(s) \\ &\dots \\ \end{align}.\endgroup$– user75138 Apr 26 '16 at 14:29 by Marco Taboga, PhD. CHAPTER 1 Notions of convergence in a probabilistic setting In this first chapter, we present the most common notions of convergence used in probability: almost sure convergence, convergence in probability, convergence in Lp- normsandconvergenceinlaw. From a practical standpoint, convergence in probability is enough as we do not particularly care about very unlikely events. Thanks for contributing an answer to Cross Validated! Lp-Convergence. Recall that there is a “strong” law of large numbers and a “weak” law of large numbers, each of which basically says that the sample mean will converge to the true population mean as the sample size becomes large. This type of convergence is equivalently called: convergence with probability one (written X n!X 1 w.p. Can I (should I) change the name of this distribution? (AS convergence vs convergence in pr 1) Almost sure convergence implies convergence in probability. The following result provides insights into the meaning of convergence in dis- tribution. 1 as n!1); convergence almost everywhere (written X n!X 1 a.e. On the other hand, almost-sure and mean-square convergence … Almost surely does. Sure, I can quote the definition of each and give an example where they differ, but I still don't quite get it. We have seen that almost sure convergence is stronger, which is the reason for the naming of these two LLNs. The R code used to generate this graph is below (plot labels omitted for brevity). This type of convergence is similar to pointwise convergence of a sequence of functions, except that the convergence need not occur on a set with probability 0 (hence the “almost” sure). with probability 1) the existence of some finite$n_0$such that$|S_n - \mu| < \delta$for all$n > n_0$(i.e. With the border currently closed, how can I get from the US to Canada with a pet without flying or owning a car? Because now, a scientific experiment to obtain, say, the speed of light, is justified in taking averages. While both sequences converge in probability to zero, only $Y_{n}$ converges almost surely. Convergence in distribution di ers from the other modes of convergence in that it is based not on a direct comparison of the random variables X n with Xbut rather on a comparision of the distributions PfX n 2Ag How does blood reach skin cells and other closely packed cells? In the previous lectures, we have introduced several notions of convergence of a sequence of random variables (also called modes of convergence).There are several relations among the various modes of convergence, which are discussed below and are summarized by the following diagram (an arrow denotes implication in the arrow's … When we say closer we mean to converge. What's a good way to understand the difference? as n!1); convergence almost certainly (written X n!X 1 a.c. as n!1). Is it possible for two gases to have different internal energy but equal pressure and temperature? I know this question has already been answered (and quite well, in my view), but there was a different question here which had a comment @NRH that mentioned the graphical explanation, and rather than put the pictures there it would seem more fitting to put them here. The weak law says (under some assumptions about the$X_n$) that the probability Here is a result that is sometimes useful when we would like to prove almost sure convergence. Is there a particularly memorable example where they differ? I think you meant countable and not necessarily finite, am I wrong? Almost sure convergence is defined based on the convergence of such sequences. X (!) Shouldn't it be MAY never actually attains 0? The Annals of Mathematical Statistics, 43(4), 1374-1379. (something$\equiv$a sequence of random variables converging to a particular value). In contrast, convergence in probability states that "while something is likely to happen" the likelihood of "something not happening" decreases asymptotically but never actually reaches 0. Thus, while convergence in probability focuses only on the marginal distribution of jX n Xjas n!1, almost sure convergence puts … : X n(!) as n!1g and write X n!X 1 a.s. as n!1when this convergence holds. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Using Lebesgue's dominated convergence theorem, show that if (X. n) n2Nconverges almost surely towards X, then it converges in probability towards X. di⁄erent ways to measure convergence: De–nition 1 Almost-Sure Convergence Probabilistic version of pointwise convergence. For example, the plot below shows the first part of the sequence for$s = 0.78$. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. The current definition is incorrect. As you can see, the difference between the two is whether the limit is inside or outside the probability. In one case we have a random variable Xn = n with probability$=\frac{1}{n}$and zero otherwise (so with probability 1-$\frac{1}{n}$).In another case same deal with only difference being Xn=1, not n with probability$=\frac{1}{n}$. It says that the total number of failures is finite. What if we had six note names in notation instead of seven? We live with this 'defect' of convergence in probability as we know that asymptotically the probability of the estimator being far from the truth is vanishingly small. 10. Almost Sure Convergence: We say that (X n: n 1) converges almost surely to X 1 if P(A) = 1, where A= f! The strong law says that the number of times that$|S_n - \mu|$is larger than$\delta$is finite (with probability 1). Convergence almost surely is a bit like asking whether almost all members had perfect attendance. In integer programming what's the difference between using lower upper bound constraints and using a big M constraints? At least in theory, after obtaining enough data, you can get arbitrarily close to the true speed of light. One thing to note is that it's best to identify other answers by the answerer's username, "this last guy" won't be very effective. Chapter Eleven Convergence Types. A sequence of random variables$X_1, X_2, \dots X_n$converges almost surely to a random variable$X$if, for every$\epsilon > 0, \begin{align}P(\lim_{n \rightarrow \infty} \lvert X_n - X \rvert < \epsilon) = 1.\end{align}. X, and let >0. For almost sure convergence, convergence in probability and convergence in distribution, if X n converges to Xand if gis a continuous then g(X n) converges to g(X). Choose some\delta > 0$arbitrarily small. Almost sure convergence vs. convergence in probability: some niceties Uniform integrability: main theorems and a result by La Vallée-Poussin Convergence in distribution: from portmanteau to Slutsky From my point of view the difference is important, but largely for philosophical reasons. When we say closer we mean to converge. J. jjacobs. ... Convergence in probability vs. almost sure convergence. The probability that the sequence of random variables equals the target value is asymptotically decreasing and approaches 0 but never actually attains 0. $$. 2 CONVERGENCE IN DISTRIBUTION . As you can see, each value in the sequence will either take the value s or 1 + s, and it will jump between these two forever, but the jumping will become less frequent as n become large. Now, recall that for almost sure convergence, we’re analyzing the statement. There wont be any failures (however improbable) in the averaging process. converges. This last guy explains it very well. 2 : X n(!) Di erence between a.s. and in probability I Almost sure convergence implies thatalmost all sequences converge I Convergence in probabilitydoes not imply convergence of sequences I Latter example: X n = X 0 Z n, Z n is Bernoulli with parameter 1=n)Showed it converges in probability P(jX n X 0j< ) = 1 1 n!1)But for almost all sequences, lim n!1 x n does not exist I Almost sure convergence … Almost sure convergence does not imply complete convergence. Some people also say that a random variable converges almost everywhere to indicate almost sure convergence. In this section we shall consider some of the most important of them: convergence in L r, convergence in probability and convergence with probability one (a.k.a. By itself the strong law doesn't seem to tell you when you have reached or when you will reach n_0. such that X n˘Bernoulli(1 n);n2IN. Let’s look at an example of sequence that converges in probability, but not almost surely. What is structured fuzzing and is the fuzzing that Bitcoin Core does currently considered structured? Importantly, the strong LLN says that it will converge almost surely, while the weak LLN says that it will converge in probability. The binomial model is a simple method for determining the prices of options. X =)Xn p! Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Some people also say that a random variable converges almost everywhere to indicate almost sure convergence. Casella, G. and R. L. Berger (2002): Statistical Inference, Duxbury. X. i.p. Proof Assume the almost sure convergence of to on (see the section ( Operations on sets and logical ... We can make such choice because the convergence in probability is given. 2.1 Weak laws of large numbers In this paper, we focus on almost sure convergence. Almost sure convergence | or convergence with probability one | is the probabilistic version of pointwise convergence known from elementary real analysis. Example 2.2 (Convergence in probability but not almost surely). In conclusion, we walked through an example of a sequence that converges in probability but does not converge almost surely. Convergence in distribution, convergence in probability, and almost sure convergence of discrete Martingales [PDF]. with convergence in probability). 1, where some famous … Thanks, I like the convergence of infinite series point-of-view! converges has probability 1. In the following we're talking about a simple random walk, X_{i}= \pm 1 with equal probability, and we are calculating running averages, ... = 1: (5.1) In this case we write X n a:s:!X(or X n!Xwith probability 1). ! If almost all members have perfect attendance, then each meeting must be almost full (convergence almost surely implies convergence in probability) As an example, consistency of an estimator is essentially convergence in probability. The R code for the graph follows (again, skipping labels). Advanced Statistics / Probability. The answer is that both almost-sure and mean-square convergence imply convergence in probability, which in turn implies convergence in distribution. Proposition 1. Almost sure convergence is a stronger condition on the behavior of a sequence of random variables because it states that "something will definitely happen" (we just don't know when). X(!)) 2.3K views View 2 Upvoters If you take a sequence of random variables Xn= 1 with probability 1/n and zero otherwise. The reason is that, when $n$ is very high, the probability of observing $X_{n}=1$ remains finite (so that the sum of subsequent probabilities diverges), while the probability of observing $Y_{n}=1$ vanishes to zero (so that the sum of subsequent probabilities converges).$$\sum_{n=1}^{\infty}I(|S_n - \mu| > \delta)$$As he said, probability doesn't care that we might get a one down the road. The example comes from the textbook Statistical Inference by Casella and Berger, but I’ll step through the example in more detail. 2 Convergence in probability Definition 2.1. Almost sure convergence, convergence in probability and asymptotic normality In the previous chapter we considered estimator of several different parameters. "Almost sure convergence" always implies "convergence in probability", but the converse is NOT true. Convergence in Probability 11.1 Introduction/Purpose of the Chapter In probability theory, there exist several different notions of convergence of random … - Selection from Handbook of Probability [Book] The hope is that as the sample size increases the estimator should get ‘closer’ to the parameter of interest. Thus, it is desirable to know some sufficient conditions for almost sure convergence. To learn more, see our tips on writing great answers. The sequence of random variables will equal the target value asymptotically but you cannot predict at what point it will happen. Proof. Note that the weak law gives no such guarantee. In the plot above, you can notice this empirically by the points becoming more clumped at s as n increases. Finite doesn't necessarily mean small or practically achievable. To be more accurate, the set of events it happens (Or not) is with measure of zero -> probability of zero to happen.$$ Prove that X n 6 a:s:!0, by deriving P(fX n = 0;for every m n n 0g) and … This lecture introduces the concept of almost sure convergence. Thus, while convergence in probability focuses only on the marginal distribution of jX n Xjas n!1, almost sure convergence puts restriction on the joint behavior of all random elements in the sequence rev 2020.12.18.38240, The best answers are voted up and rise to the top, Cross Validated works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. As we obtain more data ($n$increases) we can compute$S_n$for each$n = 1,2,\dots$. Let me clarify what I mean by ''failures (however improbable) in the averaging process''. How can massive forest burning be an entirely terrible thing? X. De–nition 2 Convergence in Probability a sequence X n converges in probability to X if 8 > 0 and > 0 9 … Almost surely implies convergence in probability, but not the other way around yah? Almost sure convergence is a stronger condition on the behavior of a sequence of random variables because it states that "something will definitely happen" (we just don't know when). Almost sure convergence. For a sequence (Xn: n 2N), almost sure convergence of means that for almost all outcomes w, the difference Xn(w) X(w) gets small and stays small.Convergence in probability is weaker and merely 3. Welcome to the site, @Tim-Brown, we appreciate your help answering questions here. University Math Help. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In contrast, convergence in probability states that "while something is likely to happen" the likelihood of "something not happening" decreases asymptotically but never actually reaches 0. In convergence in probability or a.s. convergence w.r.t which measure is the probability? The basic idea behind this type of convergence is that the probability of an “unusual” outcome becomes smaller and smaller as the sequence progresses. On an infinite board, which pieces are needed to checkmate? No other relationships hold in general. ˙ = 1: Convergence in probability vs. almost sure convergence: the basics 1. Suppose Xn a:s:! @gung The probability that it equals the target value approaches 1 or the probability that it does not equal the target values approaches 0. Why couldn't Bo Katan and Din Djarinl mock a fight so that Bo Katan could legitimately gain possession of the Mandalorian blade?$\begingroup$@nooreen also, the definition of a "consistent" estimator only requires convergence in probability. Related. So, here goes. Just because$n_0$exists doesn't tell you if you reached it yet. Exercise 5.3 | Almost sure convergence Let fX 1;X 2;:::gbe a sequence of r.v. Or am I mixing with integrals. It only takes a minute to sign up. To assess convergence in probability, we look at the limit of the probability value$P(\lvert X_n - X \rvert < \epsilon)$, whereas in almost sure convergence we look at the limit of the quantity$\lvert X_n - X \rvert$and then compute the probability of this limit being less than$\epsilon$. Simple example wanted:$ X_n $converges to$X$in probability but not almost surely, almost sure convergence and probability of estimator inside a compact set, Countable intersection of almost sure events is also almost sure. Are there cases where you've seen an estimator require convergence almost surely? X. n. k. there exists a subsub-sequence . An important application where the distinction between these two types of convergence is important is the law of large numbers. A type of convergence that is stronger than convergence in probability is almost sure con-vergence. !X 1(!) Remark 1. As Srikant points out, you don't actually know when you have exhausted all failures, so from a purely practical point of view, there is not much difference between the two modes of convergence. X. \frac{S_{n}}{n} = \frac{1}{n}\sum_{i = 1}^{n}X_{i},\quad n=1,2,\ldots. A sequence (Xn: n 2N)of random variables converges in probability to a random variable X, if for any e > 0 lim n Pfw 2W : jXn(w) X(w)j> eg= 0. Why is the difference important? On the other hand, almost-sure and mean-square convergence do not imply each other. (a) Xn a:s:! 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As you can see, the list will be re-ordered over time as people vote:::::. Eg, the difference is important, but largely for philosophical reasons 'm sure. 2020 Stack Exchange Inc ; user contributions licensed under cc by-sa information should I include for this source citation the... Seen that almost sure uniqueness = 0.78$ fX 1 ; X 2 ;::! | 7,308 | 31,683 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2022-27 | latest | en | 0.848077 |
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# How Many Two-digit Numbers Are Divisible by 3? - ICSE Class 10 - Mathematics
ConceptArithmetic Progression - Finding Their General Term
#### Question
How many two-digit numbers are divisible by 3?
#### Solution
The two digit numbers divisible by 3 are as follows:
12, 15, 18, 21, ...., 99
Clearly, this form an A.P with first term a = 12 and common difference d = 3
Last term = nth term = 99
The general term of an A.P is given by
t_n = a + (n -1)d
=> 99 12 + (n - 1)(3)
=> 99 = 12 + 3n - 3
=> 90 = 3n
=> n = 30
Thus 30 two-digit numbers are divisible by 3.
Is there an error in this question or solution?
#### APPEARS IN
Solution How Many Two-digit Numbers Are Divisible by 3? Concept: Arithmetic Progression - Finding Their General Term.
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83_Problem CHAPTER 10
# 83_Problem CHAPTER 10 - PROBLEM 10.73 Using the method of...
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PROBLEM 10.73 Using the method of Section 10.8, solve Problem 10.39. Determine whether the equilibrium is stable, unstable or neutral. ( Hint: The potential energy corresponding to the couple exerted by a torsional spring is 2 1 , 2 K θ where K is the torsional spring constant and θ is the angle of twist.) SOLUTION Potential Energy 2 1 sin 2 V K Pl θ θ = cos dV K Pl d θ θ θ = 2 2 sin d V K Pl d θ θ = + Equilibrium: 0: cos dV
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https://math.stackexchange.com/questions/1752818/problem-with-spectral-theorem-proof | 1,709,124,203,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474715.58/warc/CC-MAIN-20240228112121-20240228142121-00631.warc.gz | 389,979,705 | 38,805 | # Problem with Spectral Theorem Proof
Claim: Let A $\in \mathbb{R}^{n \times n }$ be symmetric. Then there is an orthonormal basis of $\mathbb{R}^n$ consisting of eigenvectors of A.
Sketch of proof:
Induction on n. Claim is clear for n = 1. Let $\lambda$ be an eigenvalue of A, $x$ the associated eigenvector. Let $\{x, v_2, \cdots, v_n \}$ be a basis for $\mathbb{R}^n$.
Gram-Schmidt, then we get an orthonormal basis of $\mathbb{R}^n$, $\{b_1, b_2, \cdots, b_n \}$.
We claim that V = span$\{b_2, b_3, \cdots, b_n \}$ is an invariant subspace under A. (Proof omitted)
Then we define T: $V \to V$ to be $Tv = Av$.
Then we take B $\in \mathbb{R}^{(n-1)\times(n-1)}$ as the the matrix representation of T with respect to $\{b_2, \cdots, b_n \}$ and show that B is symmetrical.
By induction hypothesis, we have an orthonormal basis of $\mathbb{R}^{n-1}$ consisting of eigenvectors of B. Let $\{ u_2, u_3, \cdots, u_n\}$ denote this basis.
We take $w_j = \Sigma_{k=2}^n u_{jk}b_k$ for $j = 2, 3 \cdots, n$ where $u_j = (u_{j2}, u_{j3}, \cdots, u_{jn} ) \in \mathbb{R}^{n-1}$.
Then $\{b_1, w_2, \cdots, w_n\}$ is an orthonormal basis for $\mathbb{R}^n$ consisting of eigenvectors of A.
The parts I am having trouble with are:
• The need for us to show that V is invariant under A (I'm kind of guessing it's to do with us being able to represent T like we did, but would like clarification)
• (My main problem:) Why our selection of $w_j$ satisfies our claim.
Any help would be much appreciated, I've been going a bit crazy going through the details. Thanks!
• Sorry if this is wrong, but isn't $w_j$ just a multiple of $b_j$? In that case, if $w_j$ is an eigenvector of $A$, how is $b_j$ not an eigenvector of $A$? Apr 21, 2016 at 14:26
• Typo, meant to write $b_k$! Sorry Apr 21, 2016 at 14:31
I think this is a very good example where the slightly more abstract proof using linear maps (which you use in any case) instead of matrices simplifies the argument and shows more clearly what is going on.
Let $$(W, \left< \cdot, \cdot \right>)$$ be a finite dimensional real inner product space and let $$T \colon W \rightarrow W$$ be a symmetric operator (so $$T = T^{*}$$). We want to show that $$T$$ has an orthonormal basis of eigenvectors by induction on $$\dim W$$. We have two steps:
1. First, we must show that any symmetric operator has at least one eigenvalue. This is already non-trivial and requires some argument passing through the complex numbers or alternatively a topological argument.
2. Then we perform the induction. The case $$\dim W = 1$$ is clear. Where $$\dim W > 1$$, we take some unit-length eigenvector $$v_1 \in V$$ and consider $$V = \{ v_1 \}^{\perp}$$. We show that $$V$$ is $$T$$-invariant (that is, $$T(V) \subseteq V$$) which implies that we can restrict $$T$$ to $$V$$ and obtain an operator (a linear map from a vector space to itself and not to a different vector space). The operator $$T|_{V} \colon (V, \left< \cdot, \cdot \right>|_{V \times V}) \rightarrow (V, \left< \cdot, \cdot \right>|_{V \times V})$$ is also self-adjoint and $$\dim V = \dim W - 1$$ and so by induction hypothesis there is an orthonormal basis $$v_2, \dots, v_n$$ of eigenvectors of $$T|_V$$. Then the basis $$(v_1, \ldots, v_n)$$ is then an orthonormal basis of eigenvectors of $$T$$ and you don't need to take any linear combination of the $$v_i$$.
1. You need to show that $$V$$ is $$T$$-invariant (or $$A$$-invariant in your case) in order to be able to restrict $$T$$ to $$V$$ and obtain an operator. If you don't show that $$V$$ is $$A$$-invariant, you can still define $$T$$ as $$T \colon V \rightarrow \mathbb{R}^n$$ but then $$T$$ won't be represented by a square $$(n-1)\times(n-1)$$ matrix but by a $$n \times (n-1)$$ matrix and you won't be able to use the induction hypothesis.
2. Denote by $$\mathcal{B} = (b_1, \ldots, b_n)$$ and by $$\mathcal{C} = (e_1, \ldots, e_n)$$ the standard basis. Consider the operator $$T \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$$ defined by $$Tv = Av$$. The matrix $$A$$ represents the operator $$T$$ with respect to $$\mathcal{C}$$ while the matrix $$B$$ represents the action of the operator $$T|_V$$ with respect to the different basis $$\hat{\mathcal{B}} = (b_2, \dots, b_n)$$.
An eigenvector $$u_i$$ of $$B$$ (an element of $$\mathbb{R}^{n-1}$$) can't be an eigenvector of $$A$$ (an element of $$\mathbb{R}^n$$) but it represents an eigenvector of $$A$$ in the sense that $$u_i$$ consists of the coefficients of an eigenvector $$w_i$$ of $$A$$ with respect to the basis $$\hat{B}$$ (the coefficients of $$w_i$$ don't have a $$b_1$$ component since $$w_i \in \{ b_1 \}^{\perp}$$).
Addition: Let $$\mathbf{u} = (u_2, \ldots, u_n) \in \mathbb{R}^{n-1}$$ be an eigenvector of $$B$$ with $$B\mathbf{u} = \lambda \mathbf{u}$$ and set $$\mathbf{w} = \sum_{i=2}^n u_i \mathbf{b}_i$$. We want to show that $$A\mathbf{w} = \lambda \mathbf{w}$$. Since $$(\mathbf{b}_1, \ldots, \mathbf{b}_n)$$ is an orthonormal basis of $$\mathbb{R}^n$$, it is enough to show that
$$\left< A \mathbf{w}, \mathbf{b}_j \right> = \left< \lambda \mathbf{w}, \mathbf{b}_j \right> = \begin{cases} \lambda_j u_j & 2 \leq j \leq n \\ 0 & j = 1 \end{cases}.$$
Since $$V$$ is $$A$$-invariant and $$\mathbf{w} \in V$$ we have $$\left = 0$$. For $$2 \leq j \leq n$$ we have
$$\left< A \mathbf{w}, \mathbf{b}_j \right> = \left< \sum_{i=2}^n u_i A\mathbf{b}_i, \mathbf{b}_j \right> = \sum_{i=2}^n \left< A \mathbf{b}_i, \mathbf{b}_j \right> u_i = \sum_{i=2}^n b_{ij} u_i = \sum_{i=2}^n b_{ji} u_i = (B \mathbf{u})_j = (\lambda \mathbf{u})_j = \lambda_j \mathbf{u}_j$$
where we used the fact that $$b_{ij} = \left< A\mathbf{b}_i, \mathbf{b}_j \right>$$ and that $$B$$ is symmetric.
• I think I'm just doing my head in, need to have a break. Could you show me that $Aw_j = \lambda_j w_j$ for some $\lambda_j$? I just don't seem to be able to follow any arguments today. And I think we did the proof with linear maps later on and I understood that, (thank you again for posting it) I was just wondering why this specific choice of basis works Apr 21, 2016 at 15:40
• I've added the proof. Apr 21, 2016 at 16:19
$V$ needs to be an invariant subspace under $A$ so that $T$ goes from $V$ to $V$. Otherwise, if $V$ was not invariant, $T$ would go from $V$ to $\Bbb{R}^n$ and then it would not have a matrix representation in $\Bbb{R}^{(n-1)x(n-1)}$.
Here's a proof that $w_j$ is an eigenvector of $A$:
$$Bu_j=\lambda_ju_j$$
This is because $u_j$ is an eigenvector of $B$. Now, let $f$ be the function from $V$ to $\Bbb{R}^{n-1}$ that maps a vector in $V$ to its vector representation in $\Bbb{R}^{n-1}$ with respect to the basis $\{b_2, b_3, ..., b_n\}$. (Look at Definition VR on this page.) Take the $f^{-1}$ of both sides.
$$f^{-1}(Bu_j)=f^{-1}(\lambda_ju_j)$$
Let $w_j=f^{-1}(u_j)$ so that $u_j=f(w_j)$. On the right-hand side, we can bring $\lambda_j$ out because $f^{-1}$ is a linear transformation, so we have $\lambda_jf^{-1}(u_j)=\lambda_jw_j$. On the left-hand side, substitute $u_j=f(w_j)$.
$$f^{-1}(Bf(w_j))=\lambda_jw_j$$
Now, look at Theorem FTMR on this page. In our case, $T$ goes from $V$ to $V$, so as per the theorem's variable names, $B=C=\{b_2, b_3, ..., b_n\}$ and $p_B=p_C=f$. Thus, the left-hand side is simply $T(w_j)$ which is the same as $Aw_j$:
$$Aw_j=\lambda_jw_j$$
We have proven $w_j$ is an eigenvector, but we didn't use the same definition as your book. We need to prove these definitions are equivalent. Now, go back to Definition VR from above. According to that and the fact that, as per the definition's variable names, $f=p_B$ and that $B=\{b_2, b_3, ..., b_n\}$, we have:
$$w_j=\sum_{k=1}^{n-1} f(w_j)_kb_{k+1}$$
Now, what's $f(w_j)$? It's $u_j$, as we said above:
$$w_j=\sum_{k=1}^{n-1} u_{jk} b_{k+1}$$
Now, redefine $u_j=(u_{j2}, u_{j3}, ..., u_{jn})$ as you did and adjust the summation:
$$w_j=\sum_{k=2}^{n} u_{jk}b_k$$
Thus, the definitions are indeed equivalent.
• Thank you for this, I have (a different) proof of A being invariant in my notes and understand it – I was just wondering why we need it. My main issue is the second part of my question however. Thank Apr 21, 2016 at 14:52
• You need $V$ to be invariant so that $T$ can go from $V$ to $V$ instead of $V$ to $\Bbb{R}^n$. Once we know $T$ is bijective from $V$ to $V$, we know that there is non-singular (n-1) by (n-1) matrix representation. I am still trying to figure out the second part, too. Apr 21, 2016 at 14:53
• I think I understand the second part now. Can you look at my edit and verify my answer? Apr 21, 2016 at 15:09
• It's getting closer for me, I'm still not thoroughly convinced why it works. What's the original vector behind $u_j$? Thanks again for having a look! Apr 21, 2016 at 15:22
• The original vector behind $u_j$ is the vector in $V$ that has the vector representation of $u_j$ in $\Bbb{R}^{n-1}$ with respect to the basis $\{b_2, b_3, ..., b_n\}$. We can construct a function $f$ from $V$ to $\Bbb{R}^{n-1}$ like this. Such a function is called a vector representation and we can prove that it is bijective, meaning given $u_j$, there has to be a vector in $V$ that maps to $u_j$ under the vector representation. That vector is constructed by summing each component of $u_j$ with the corresponding vector in the basis, which is what they did in $w_j$. Apr 21, 2016 at 15:28 | 3,216 | 9,358 | {"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": 72, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-10 | latest | en | 0.870503 |
https://mathematica.stackexchange.com/posts/56394/revisions | 1,569,121,269,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574765.55/warc/CC-MAIN-20190922012344-20190922034344-00303.warc.gz | 552,701,781 | 17,248 | 2 replaced http://mathematica.stackexchange.com/ with https://mathematica.stackexchange.com/ edited Apr 13 '17 at 12:55 Because of this commentthis comment the following becomes too long to be just a comment so here you go: tt1 = Table[ z /. Solve[z^2 == x^2 y - z, z, Method -> Reduce], {x, 0, 5,1}, {y, 0, 5, 1}]; Max[tt1] Table[{x, y}, {x, 0, 5, 1}, {y, 0, 5, 1}][[Sequence @@ (First@Position[tt1, Max[tt1]])[[;; 2]]]] 1/2 (-1 + Sqrt[501]) {5, 5} Because of this comment the following becomes too long to be just a comment so here you go: tt1 = Table[ z /. Solve[z^2 == x^2 y - z, z, Method -> Reduce], {x, 0, 5,1}, {y, 0, 5, 1}]; Max[tt1] Table[{x, y}, {x, 0, 5, 1}, {y, 0, 5, 1}][[Sequence @@ (First@Position[tt1, Max[tt1]])[[;; 2]]]] 1/2 (-1 + Sqrt[501]) {5, 5} Because of this comment the following becomes too long to be just a comment so here you go: tt1 = Table[ z /. Solve[z^2 == x^2 y - z, z, Method -> Reduce], {x, 0, 5,1}, {y, 0, 5, 1}]; Max[tt1] Table[{x, y}, {x, 0, 5, 1}, {y, 0, 5, 1}][[Sequence @@ (First@Position[tt1, Max[tt1]])[[;; 2]]]] 1/2 (-1 + Sqrt[501]) {5, 5} 1 answered Aug 1 '14 at 11:56 Öskå 7,74533 gold badges2525 silver badges4747 bronze badges Because of this comment the following becomes too long to be just a comment so here you go: tt1 = Table[ z /. Solve[z^2 == x^2 y - z, z, Method -> Reduce], {x, 0, 5,1}, {y, 0, 5, 1}]; Max[tt1] Table[{x, y}, {x, 0, 5, 1}, {y, 0, 5, 1}][[Sequence @@ (First@Position[tt1, Max[tt1]])[[;; 2]]]] 1/2 (-1 + Sqrt[501]) {5, 5} | 649 | 1,501 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2019-39 | latest | en | 0.640052 |
http://www.nag.com/numeric/CL/nagdoc_cl24/examples/source/g03efce.c | 1,438,454,911,000,000,000 | text/plain | crawl-data/CC-MAIN-2015-32/segments/1438042988860.4/warc/CC-MAIN-20150728002308-00286-ip-10-236-191-2.ec2.internal.warc.gz | 594,226,542 | 1,770 | /* nag_mv_kmeans_cluster_analysis (g03efc) Example Program. * * Copyright 1998 Numerical Algorithms Group. * * Mark 5, 1998. * Mark 7, revised, 2001. * Mark 8 revised, 2004. * */ #include #include #include #include #define CMEANS(I, J) cmeans[(I) *tdcmeans + J] #define X(I, J) x[(I) *tdx + J] int main(void) { Integer exit_status = 0, i, *inc = 0, *isx = 0, j, k, m, maxit, n, *nic = 0, nvar; Integer tdcmeans, tdx; NagError fail; char weight[2]; double *cmeans = 0, *css = 0, *csw = 0, *wt = 0, *wtptr, *x = 0; INIT_FAIL(fail); printf( "nag_mv_kmeans_cluster_analysis (g03efc) Example Program Results" "\n\n"); /* Skip heading in the data file */ scanf("%*[^\n]"); scanf("%1s", weight); scanf("%ld", &n); scanf("%ld", &m); scanf("%ld", &nvar); scanf("%ld", &k); scanf("%ld", &maxit); if (n >= 2 && nvar >= 1 && m >= nvar && k >= 2) { if (!(cmeans = NAG_ALLOC((k)*(nvar), double)) || !(css = NAG_ALLOC(k, double)) || !(csw = NAG_ALLOC(k, double)) || !(wt = NAG_ALLOC(n, double)) || !(x = NAG_ALLOC((n)*(m), double)) || !(inc = NAG_ALLOC(n, Integer)) || !(isx = NAG_ALLOC(m, Integer)) || !(nic = NAG_ALLOC(k, Integer))) { printf("Allocation failure\n"); exit_status = -1; goto END; } tdx = m; tdcmeans = nvar; } else { printf("Invalid n or nvar or m or k.\n"); exit_status = 1; return exit_status; } if (*weight == 'W') { for (i = 0; i < n; ++i) { for (j = 0; j < m; ++j) scanf("%lf", &X(i, j)); scanf("%lf", &wt[i]); } wtptr = wt; } else { for (i = 0; i < n; ++i) { for (j = 0; j < m; ++j) scanf("%lf", &X(i, j)); } wtptr = 0; } for (i = 0; i < k; ++i) { for (j = 0; j < nvar; ++j) scanf("%lf", &CMEANS(i, j)); } for (j = 0; j < m; ++j) scanf("%ld", &isx[j]); /* nag_mv_kmeans_cluster_analysis (g03efc). * K-means */ nag_mv_kmeans_cluster_analysis(n, m, x, tdx, isx, nvar, k, cmeans, tdcmeans, wtptr, inc, nic, css, csw, maxit, &fail); if (fail.code != NE_NOERROR) { printf( "Error from nag_mv_kmeans_cluster_analysis (g03efc).\n%s\n", fail.message); exit_status = 1; goto END; } printf("\nThe cluster each point belongs to\n"); for (i = 0; i < n; ++i) printf(" %6ld%s", inc[i], (i+1)%10?"":"\n"); printf("\n\nThe number of points in each cluster\n"); for (i = 0; i < k; ++i) printf(" %6ld", nic[i]); printf("\n\nThe within-cluster sum of weights of each cluster\n"); for (i = 0; i < k; ++i) printf(" %9.2f", csw[i]); printf("\n\nThe within-cluster sum of squares of each cluster\n\n"); for (i = 0; i < k; ++i) printf(" %13.4f", css[i]); printf("\n\nThe final cluster centres\n"); printf(" 1 2 3 4 5\n"); for (i = 0; i < k; ++i) { printf(" %5ld ", i+1); for (j = 0; j < nvar; ++j) printf("%8.4f", CMEANS(i, j)); printf("\n"); } END: NAG_FREE(cmeans); NAG_FREE(css); NAG_FREE(csw); NAG_FREE(wt); NAG_FREE(x); NAG_FREE(inc); NAG_FREE(isx); NAG_FREE(nic); return exit_status; } | 1,065 | 2,775 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2015-32 | latest | en | 0.368036 |
https://www.ednasia.com/recycling-energy-in-dc-motor-drives/ | 1,675,221,544,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499899.9/warc/CC-MAIN-20230201013650-20230201043650-00298.warc.gz | 779,843,740 | 32,468 | # Recycling energy in DC motor drives
#### Article By : Peter Millett
This article describes how energy can be fed back from a motor to a power source from a mechanical system, and how it can be dealt with in motor drive electronics.
When decelerating a moving mass in a permanent magnet motor drive, energy stored in the mechanical system may be returned through the motor driver to the power source. If this energy is not properly accounted for, it can cause damage to the motor driver or the rest of the system by causing an increase in the power supply voltage.
In this article, we will look at ways to dissipate this energy safely. To simplify the examples, a DC brush motor is shown. This applies to brushless motor systems as well.
Conservation of energy
The conservation of energy is a basic principle of physics—energy cannot be created nor destroyed.
When something (such as a mass) is moving or rotating, it accumulates kinetic energy. In a motor system, that kinetic energy comes from a power source that provides electrical energy to a motor, which creates torque to accelerate the mass.
There is energy stored both in the inertia of the motor’s rotor and also in the mechanical system attached to the motor. For simplicity, envision the mechanical system as a flywheel coupled to the motor shaft (Figure 1).
Figure 1 Flywheel example of a mechanical system
The kinetic energy can be calculated with ½ Iω2, where I is the moment of inertia and ω is the angular velocity. The higher the speed or the more inertia there is, the more energy is stored.
This is a very obvious concept—it takes energy to get something moving. What is less obvious is what happens when you want to stop the motion. To stop or slow a moving mass, the stored kinetic energy has to go somewhere. But where?
When you disconnect the power from a spinning motor, the energy stored in the moving mass is dissipated into the mechanical losses in the system. Most of this energy is turned into heat due to friction (Figure 2). Unless there is a great deal of friction, the motor coasts to a stop very slowly. The motor turns into a generator, but since there is no path for current to flow, there is no electromagnetic torque to help stop the motor.
Figure 2 Friction in a stopping motor
If a path is provided for current to flow from this generator by shorting the output of the motor, the current generates a torque in opposition to the direction of rotation (Figure 3). This causes the motor to come to a stop quickly. In this case, the energy is mostly dissipated as heat in the winding resistance of the motor and also in any resistance in the current path, shorting the motor.
Figure 3 Torque in opposition to rotation
This is sometimes called a “short brake.” In reality, a short circuit is usually applied by turning on the low-side MOSFETs of an H-bridge to provide a current path.
When a control system wants to decrease the speed of the motor quickly, the polarity of the current applied to the motor is reversed to provide a torque in opposition to the motion. When this is done, the stored kinetic energy can be driven back through the motor driver circuit into the power supply.
If the power source were a perfect battery, then energy would flow back into the battery and be recycled. However, in the real world, the power source is usually a DC power supply, and unless this supply is specially designed, a DC power supply can only source current. Because it cannot sink current, the only place the energy has to go is into the capacitance that is part of the power supply.
The amount of energy stored in a capacitor can be calculated with ½ CV2, where C is the capacitance and V is the voltage. The voltage across the capacitor must increase as energy flows into it (Figure 4).
Figure 4 Increasing capacitor voltage with increasing energy
If the amount of energy is small (either the speed is low, or the inertia is small), then the voltage increase may be small enough that it will not cause any problems. However, in some cases, if there is too much energy or not enough capacitance, the voltage may rise to destructive levels. This can damage the motor driver circuit or other circuitry connected to the same power supply.
[Continue reading on EDN US: Dissipating the energy] | 897 | 4,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.933961 |
https://documentation.help/Digital-Filter-Design-MathScript/DFMC_firnyquist.html | 1,716,590,731,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058751.45/warc/CC-MAIN-20240524214158-20240525004158-00141.warc.gz | 176,632,611 | 3,497 | # firnyquist (Digital Filter Design Toolkit, MathScript Function)
Owning Class: multirate
## Syntax
b = firnyquist(n, l, rolloff)
b = firnyquist(n, l, rolloff, 'nonnegative')
b = firnyquist('minorder', l, rolloff, ripple)
## Description
Designs a lowpass, finite impulse response (FIR), Nyquist filter.
Examples
## Inputs
Name Description n Specifies the order of the filter. This input also specifies to use the Remez method to design the filter. n is an even, positive number. If you do not specify n, you must specify a valid value for 'minorder'. l Specifies the sampling frequency conversion factor of the multirate filter. l is an integer greater than one. rolloff Specifies the roll off factor. This factor determines the relative transition bandwidth, which equals (transition band)/(2*passband + transition band). rolloff must fall in the range (0, 1). A smaller value of rolloff returns a narrower transition bandwidth if the value of l does not change. 'nonnegative' Specifies that b returns an FIR filter with a nonnegative zero-phase response. If you do not specify 'nonnegative', b might return an FIR filter with negative values in the zero-phase response. 'minorder' Specifies that b returns a filter with the minimum order value that meets the design requirements. This input also specifies to use the Kaiser Window method to design the FIR filter. If you do not specify 'minorder', you must specify a valid value for n. ripple Specifies the maximum ripple in the passband and stopband. ripple is a double-precision, floating-point number that must fall in the range (0, 1).
## Outputs
Name Description b Returns the coefficients of the designed FIR filter. b is a real vector with a length of n+1 or 'minorder'+1.
## Examples
b = firnyquist(20, 4, 0.1);figure;
freqz(b);
b = firnyquist(40, 5, 0.1, 'nonnegative');
fft_mag = abs(fft(b, 16384));
figure;
plot(0:1/8192:1, fft_mag(1:8193));
b = firnyquist('minorder', 5, 0.1, 0.001);
figure;
freqz(b); | 503 | 1,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-22 | latest | en | 0.598616 |
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# V31-07
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Current Student
Joined: 19 Mar 2012
Posts: 4359
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)
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23 Apr 2018, 07:45
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Question Stats:
33% (00:09) correct 67% (02:34) wrong based on 6 sessions
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Critics of the use of trucks to transport nuclear power plant waste are mistaken in claiming that transportation of such cargo poses an undue risk over that of other transportation methods. For example, Yardow Prefecture allows trucks with a typical gross weight greater than 80,000 pounds to transport nuclear waste and despite the significant number of such journeys, no incidents or handling safety violations have been recorded. It is clear that considering Yardow's proven safety record, policy-makers should disregard the critics' concerns and allow trucks to transport nuclear power plant waste as necessary.
Which one of the following, if true, most substantially weakens the argument?
A. Transporting nuclear power plant waste by train does not necessarily require trucks to deliver the nuclear waste to a train's loading depot.
B. The reporting of traffic incidents in Yardow Prefecture was recently transferred to a different regulatory agency.
C. Truck drivers in Yardow Prefecture are required to obtain a specific class of license to drive a truck if the vehicle's gross weight is greater than 80,000 pounds.
D. The roads used for nuclear waste transport in Yardow Prefecture are restricted to trucks driven by drivers with a certain class of license designated for gross loads exceeding 80,000 pounds, which is over 20,000 pounds greater than the typical gross load.
E. Normal gross truck loads are 60,000 pounds, and for any load greater than 80,000 pounds, two trucks are typically used to transport that load.
_________________
Current Student
Joined: 19 Mar 2012
Posts: 4359
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)
### Show Tags
23 Apr 2018, 07:45
Official Solution:
Critics of the use of trucks to transport nuclear power plant waste are mistaken in claiming that transportation of such cargo poses an undue risk over that of other transportation methods. For example, Yardow Prefecture allows trucks with a typical gross weight greater than 80,000 pounds to transport nuclear waste and despite the significant number of such journeys, no incidents or handling safety violations have been recorded. It is clear that considering Yardow's proven safety record, policy-makers should disregard the critics' concerns and allow trucks to transport nuclear power plant waste as necessary.
Which one of the following, if true, most substantially weakens the argument?
A. Transporting nuclear power plant waste by train does not necessarily require trucks to deliver the nuclear waste to a train's loading depot.
B. The reporting of traffic incidents in Yardow Prefecture was recently transferred to a different regulatory agency.
C. Truck drivers in Yardow Prefecture are required to obtain a specific class of license to drive a truck if the vehicle's gross weight is greater than 80,000 pounds.
D. The roads used for nuclear waste transport in Yardow Prefecture are restricted to trucks driven by drivers with a certain class of license designated for gross loads exceeding 80,000 pounds, which is over 20,000 pounds greater than the typical gross load.
E. Normal gross truck loads are 60,000 pounds, and for any load greater than 80,000 pounds, two trucks are typically used to transport that load.
Type: Weaken
Boil It Down: Transporting nuke waste safe in Yardow -> Should be allowed in general
Missing Information: Yardow is a relevant example to the nation as a whole
Goal: We need to find an option that shows there is something than usual about Yardow such that it would be rendered in a relevant example.
A. Nobody's talking about transporting nuclear waste by train so a discussion about the logistics involved in train transport of nuclear waste is entirely irrelevant. This is a classic out of focus option. It's just not clear at all how this information relates in any way to the claim that trucks should be allowed to transport nuclear power plant waste as necessary. For all we know, transporting this waste by train is even more dangerous than it is by truck.
B. We have no way to know whether transferring reporting duties would've affected the results. It would be speculative at best to presume that's the case. For all we know the standards remained exactly the same despite the transfer of reporting duties to a different regulatory agency.
C. This is the statistical runner-up option. At first glance option C appears to show that something is unusual about Yardow; however, the license requirement could be standard across the entire nation. There's no way to know that there's anything than usual about this licensing requirement. Maybe it is, maybe it isn't. Therefore it's not clear whether see would point out a relative difference in Yardow Prefecture's requirements, and this option does not clearly weaken. Compare that to option D, which points to an array of factors that make Yardow special, and therefore likely an unfair example to use in the case to allow trick borne nuclear waste.
D. Yes! This is the correct option because it reveals that something is definitely specific about Yardow's restrictions. The roads in the Yardow example are ONLY travelled on by vehicles in excess of 80,000 pounds, and by specially licensed drivers. Taking every other vehicle off the road and having these trucks operated by specially licensed drivers on these exclusive roads dramatically reduces the risk of an incident happening. If we were to refer to the Yardow case to justify truck driven nuclear power plant waste in general, then we'd be applying an unrealistic, unrepresentative set of circumstances, thus dramatically weakenening the argument that allow trucks to transport nuclear power plant waste as necessary.
E. This option also gets some interest; however, there's no way to assume that for nuclear waste transport this information is even relevant. The option just states what is typically the case. With nuclear waste, how would we know that this information is even relevant to us? Would two trucks or just one truck be used? We have no idea. Furthermore, it's unclear how the use of one truck or two trucks would even impact the safety record. In other words there's just no way to know the relevance of option E.
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Re V31-07 &nbs [#permalink] 23 Apr 2018, 07:45
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# V31-07
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®. | 1,825 | 8,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.889332 |
https://www.mathcelebrity.com/community/threads/the-property-taxes-on-a-boat-were-375-what-was-the-tax-rate-if-the-boat-was-valued-at-75-000.1511/ | 1,718,486,329,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00442.warc.gz | 795,501,678 | 8,832 | # The property taxes on a boat were \$375. What was the tax rate if the boat was valued at \$75,000?
Discussion in 'Calculator Requests' started by math_celebrity, Oct 21, 2018.
Tags:
1. ### math_celebrityAdministratorStaff Member
The property taxes on a boat were \$375. What was the tax rate if the boat was valued at \$75,000?
Tax Rate = Tax Amount / Purchase Price
Tax Rate = 375 / 75,000
Tax Rate = 0.005
Tax Rates are generally expressed in percentages, so the percentage = 0.005 * 100 = 0.5%. | 143 | 504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-26 | latest | en | 0.979361 |
https://www.learn-forextrading.org/2015/03/fibonacci-basic-tutorial.html | 1,601,093,025,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400234232.50/warc/CC-MAIN-20200926040104-20200926070104-00690.warc.gz | 911,550,489 | 52,026 | ## Top Ad unit 728 × 90
My experience makes me a believer in the adage that there is a natural order in the markets that has more geometric symmetry than most traders realize or want to believe. This is a fact, not subjective, because almost all market turning points adhere to a certain numerical sequence that you can prove to yourself by looking at historical charts. This sequence applies to both price and time. The primary tool used for this trade analysis is Fibonacci relationships. They include Fibonacci retracements and extensions, as well as time measurement, pivot dates by ratio and numerical sequence.
There is no need to go into the history of Fibonacci, other than to know it is the force that rules the movement of about anything you can imagine, including the financial markets.
The Fibonacci series is a numerical sequence that expands by adding the previous numbers together as shown here:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, etc.
Fibonacci
The most interesting property of this numerical sequence is that as the series proceeds, any number is 1.618 times the preceding number, and 0.618 of the next number.
Example:
1.618 x 34 = 55 x 1.618 = 89 x 1.618 = 144.
.618 x 55 = 34, .618 x 89 = 55, .618 x 144 = 89.
The .618 and 1.618 numbers are what they call the Golden Numbers. You will also see the sequence root of .618, which is .786, and 1.272, the square root of 1.618, occur frequently in your trading.
Suffice to say, the Golden Numbers are dominant in math, nature and science. Planets revolve around the Golden Number, as does your heartbeat on an EKG, and even the index finger on your hand. Each section of your index finger, from the tip to the base of the wrist, is larger than the preceding one by about the Fibonacci ratio of 1.618. And lastly, they tell us that the Egyptian pyramids are based on the Golden Ratio/Numbers.
Key Point: My source for all of this trivia is goldennumber.net, but you can just go to Google.com and find any number of sites that give you this background in volume. If you believe it, I can teach you how to use it in a pragmatic way trading and investing in the markets.
Key Point: In spite of all the financial, economic and geopolitical events that affect the markets, you will see beyond any doubt that the up and down price fluctuations in all major markets are governed by the Fibonacci Golden Ratio.
Fibonacci is a sequence tool that is excellent by itself and that becomes very powerful when combined with volatility bands and standard deviation, let alone any of the other tools.
Fibonacci Retracement
I use the following Fib retracement ratios:
• .236
• .382
• .50
• .618
• .786
• 1.00
The minor ratio that will come into play sometimes is the .707 Fib retracement (square root of .50).
Retracements depict the potential reversal levels, support and resistance.
To calculate a retracement in an uptrend, which is when a stock rallies and then pulls back to some percentage ratio of the previous swing point low, you would do the following:
• High - Low x Fib Ratio, then subtract it from the High.
• Calculation of .618RT level
60 - 40 = 20 x .618RT = 12.36
60 - 12.36 = 47.64 (.618RT level)
• Complete the other ratios for practice.
• To calculate a retracement in a downtrend, which is when a stock declines and then pulls back to some percentage ratio of the previous swing point high, you would do the following:
• High - Low x Fib Ratio, then add it to the Low.
• Calculation of .382 level
50 - 30 = 20 x .382 = 7.64
30 + 7.64 = 37.64 (.382RT level)
mplete other ratios for practice
• Fibonacci Extensions
I use the following Fib extensions ratios:
• 1.272
• 1.618
• 2.00
• 2.24
• 2.618
• 3.14 (Pi -- Key ratio)
• 4.236
These extensions occur after price exceeds the 1.00 level and makes new lows or new highs beyond the last leg.
A minor extensions ratio that will come into play sometimes is the 1.414.
To calculate a Fib extension after price exceeds the low or high of the last leg, you would:
Extension down
• High - Low x Ratio, then subtract from the High.
Calculation of 1.272 extension down
90 - 80 = 10 x 1.272 = 12.72
90 - 12.72 = 77.28
Extension Up
• High - Low x Ratio, then add to Low.
• Calculation of 1.618 extension up
90 - 80 = 10 x 1.618 = 16.18
80 + 16.18 = 96.18
• We have given you some Fibonacci background, and the basics of calculating the retracements and extensions, in addition to the ratios that I suggest you use. Now we will build on that.
Key Point: When you look at retracements and extensions in terms of price, you should also look for symmetry of time using the same ratios. It is not mandatory that you have both, but it is much better symmetry when you do, which builds your case for higher probability.
Key Point: Once you have identified a high-probability zone, it is the price action at that zone that determines what you will do.
The following SPX charts will demonstrate retracement, extension, time and price action at different zones.
Fibonacci Tutorial
• This SPX weekly chart frames the Fibonacci retracement levels between the 1553 top and 769 bottom. You can see that the .236RT to 1553 of 954 was a major obstacle. The first rally from 776 reached 965, but didn't close above it.
From 965, the SPX declined to the 769 bottom (A) and then rallied to 954, right at the .236RT level, but failed to close above 954. The next leg down (BC) made a 789 low, then reversed to the upside and approached the .236RT level for the third time.
Key Point: The more price trades at resistance or support, the weaker the line is and the probability of penetration increases.
Price breaks above 954 and trades to the 1015 level. The 1.272 Fib extension of the BC leg is 999, and for 12 weeks, the SPX traded sideways with a high close of 998.
Key Point: If price breaks out of a range at a Fib level, the highest probability is that it will seek the next level.
• The .382RT to 1553 is 1068, and that is exactly where the SPX traded and went sideways for four weeks (more detail on next chart) before breaking out and trading to the next Fib zone, which is the .50RT to 1553 of 1161. Notice also that there is a confluence with the 2.24 Fib extension of the BC leg at 1159. At the completion of this course on Jan. 29, 2004, 1155 is the rally high on Jan. 27, the SPX had traded down to 1122.38.
This chart demonstrates the natural order that the SPX has traded just using Fib retracements, extensions and time, which you will see better on the next chart. You have anticipated the key zones in advance.
• On this chart, we will start with the time symmetry that is present in this move.
• The BC leg from 954 - 789 was 14 weeks to the low and 13 weeks (Fib number) to the low close. Any reversal bar setup right there had to be taken. The 789 low week was a reversal bar (Hammer) with a high of 841.39. Sequence traders took entry.
Now it gets more interesting. Once price broke out above 954, you had anticipated the 1.272 zone, but on this chart, you also see that the last low in that zone was week 21 following the 789 low week. 21 is the 1.618 Fib extension of 13. Price then rallied, breaking out of the 1.272 zone range.
Key Point: When looking for time symmetry, you can use the low close or low, and measure it with a high close or high. That will all be considered symmetry.
• After the 1.272 breakout, the SPX traded right up to the .382RT zone at 1088. The first weekly bar in that range was the 34th week, and price went sideways for four to five weeks with a high of 1064. Once again, there was price and time symmetry, as 34 (Fib number) is the 2.618 Fib extension of 13.
The advance above the .782RT level means anticipation of the next Fib RT zone, which would be the .50RT to 1553 at 1160. There is also a .50RT to the 1530 secondary high of 1150, and then the 2.24 Fib extension of the BC leg at 1159. The 1530 high was the 1,2,3 lower top to the 1553 all-time high after the initial decline to below 1350 from 1553. That would make it a significant high to measure a retracement.
From a time perspective, the 1155 high is Week 46 and the high close is Week 45. The 3.14 Fib extension of 14 is 44.
Review these two charts several times so you gain a good idea of the structured way a market trades most of the time and enables you to anticipate high-probability zones where you will take some kind of action, either buy or sell, but it is never nothing.
Fibonacci basic tutorial Reviewed by learn forex trading on March 22, 2015 Rating: 5 | 2,209 | 8,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | longest | en | 0.947388 |
https://www.transtutors.com/questions/to-complete-this-assignment-you-must-submit-a-written-report-and-accompanying-eviden-2942536.htm | 1,604,063,245,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107910815.89/warc/CC-MAIN-20201030122851-20201030152851-00633.warc.gz | 921,578,991 | 14,876 | # To complete this assignment you must submit a written report, and accompanying evidence,... 1 answer below »
Summary
In this assignment you will analyse two algorithms, both theoretically and experimentally, in terms of both
efficiency and correctness. You will be implementing them as executable programs, measuring their runtime
characteristics, and comparing the results with theoretical predictions for these algorithms.
The aim is to directly compare two different algorithms that both perform the same function but have
different efficiencies. In particular, you must ensure that both programs are analysed under exactly the same
conditions, to ensure that the results are meaningfully comparable. You will be required to count the number
of basic operations performed, measure execution times, and produce a clear report describing your findings.
To complete this assignment you must submit a written report, and accompanying evidence, describing the
results of your experiments to compare the efficiency of two given algorithms. The steps you must perform,
and the corresponding summaries required in your written report, are as follows.
1. You must ensure you understand the algorithms to be analysed.
2. You must establish the correctness of the algorithms.
• Your report must include a proof of the partial correctness of each algorithm.
• Your report must also include a proof that both algorithms terminate (total correctness).
3. You must define a common basis for meaningful comparison of the two algorithms, in terms of basic
operations and the size of inputs.
• Your report must describe your choice of ‘basic operation’ applicable to both algorithms.
• Your report must describe your choice of ‘problem size’ applicable to both algorithms.
• Your report must summarise the predicted (theoretical) time efficiency of the two algorithms. As
part of this, you need to argue whether the best-case, average-case and worst-case are different.
If they are, you are only requested to reflect on the average-case efficiency. Of course, you may
still reflect on the best-case and worst-case efficiencies if that can help in your process (as we
have seen in some examples).
You must present the arguments underpinning these predictions, either from first principles or
referenced from a text. It is NOT sufficient merely to state the efficiency class with an
accompanying reference.
Attachments:
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## Recent Questions in Software Engineering
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Sum
0
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The Mode Calculator is a user-friendly online tool designed to calculate the mode of a series of numbers. With its straightforward interface, users can effortlessly obtain the mode by providing a list of comma-separated numbers. This tool does not require any advanced configuration or customization options, making it accessible to users of all levels of expertise. Let's delve into the various scenarios where the Mode Calculator can prove particularly useful for students, teachers, and individuals seeking to calculate the mode in their day-to-day activities.
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# Surf for More Math
## Lesson 5 - Measuring and Comparing Capacity
To encourage your child to have fun on the Web while learning about Measuring and Comparing Capacity, here are some games and interactive activities they can do on their own or in pairs.
### Goal
Estimate, measure, and compare capacities, and determine relationships among units.
### Builds Upon
Student Book pages 312-313
### Instructions for Use
Capacity allows your child to make comparisons and measure the capacity of virtual liquid in several containers.
To use Capacity, click on the button to pour the water from cylinder one to cylinder two. Click the arrow button to pour the water from cylinder one to cylinder three.
Measuring Cylinder prompts your child to estimate and measure the capacity of virtual liquids by turning the tap on, or turning the drain tap.
To use Measuring Cylinder, click on the red marker to show the level marker in the cylinder. Use the control arrows to move the marker or reset it in the center of the scale. To change the value of the scale, click the 'Maximum' buttons up or down. To change the scale interval, click the 'Interval' buttons up or down. Click the control panel on the tap to set the amount of liquid to add to the cylinder, or click the tap to add liquid. To release the liquid in the cylinder, click the control panel at the drain tap and set the amount of liquid to drain, or click the drain tap and release the liquid. Click the 'Reset' button to start again.
Can you fill it? lets your child estimate the amounts containers can hold.
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Absolute events and intersecting events When we were working out the probability of the ball landing in a black or red pocket, we were big business with two separate events, the globe landing in a black pocket after that the ball landing in a burgundy pocket. If two events are commonly exclusive, only one of the two can occur. What about the black and even events? The two events intersect. Brain Power What sort of effect do you think this connection could have had on the probability? Problems at the intersection Calculating the probability of getting a black before even went wrong because we built-in black and even pockets twice. At the outset of all, we found the chance of getting a black pocket after that the probability of getting an constant number. When we added the two probabilities together, we counted the chance of getting a black and constant pocket twice.
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I am conducting a quasi-experimental research in a school, where I have access to 100 students who are chosen as classrooms and who will be the participants of the study. I am interested in the effects of two separate treatments and want to conduct a pre-test/post-test study (with two post-tests). So I have two independent variables (the two treatments) and four dependent variables. I need to form three groups out of the 100 students: one for each experiment and a single control group. Participants will be assigned to these groups randomly.
My questions are:
1. What should I take into consideration when deciding the in what proportions I divide the students to the groups (i.e. how big each group should be). Surely the both treatment groups should have an equal amount of participants, but how about the control group? Would it be better to have 33 in each group, or rather to have for example 40 in each treatment group and only 20 in the control group?
2. What is the best way to calculate the statistical power in such a study that includes three groups?
3. What statistical methods would you consider for this kind of study? Since I have have multiple dependent variables, would it be better to use ANCOVA treating each dependent variable as a independent experiment or to rather use Two-way MANOVA? Or would you suggest something else? | 286 | 1,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.958264 |
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### Exercise :: Series Completion - Section 1
Choose the correct alternative that will continue the same pattern and replace the question mark in the given series.
11.
3, 4, 7, 7, 13, 13, 21, 22, 31, 34, ?
A. 42 B. 43 C. 51 D. 52
Explanation:
The given sequence is a combination of two series :
I. 3, 7, 13, 21, 31, ? and II. 4, 7, 13, 22, 34
The pattern in I is + 4, + 6, + 8, + 10,.....
The pattern in II is + 3, + 6, + 9, + 12,.....
So, missing term = 31 + 12 = 43.
12.
198, 194, 185, 169, ?
A. 92 B. 112 C. 136 D. 144
Explanation:
The pattern is - 4, - 9, - 16,.....i.e. - 22, - 32, - 42,.....
So, missing pattern = 169 - 52 = 169 - 25 = 144.
13.
2, 3, 5,7,11,?, 17
A. 12 B. 13 C. 14 D. 15
Explanation:
Clearly, the given series consists of prime numbers starting from 2. So, the missing
term is the prime number after 11, which is 13.
14.
6, 12, 21, ?, 48
A. 33 B. 38 C. 40 D. 45
Explanation:
The pattern is + 6, + 9, + 12, + 15, .....
So, missing term = 21 + 12 = 33.
15.
Which term of the series 5, 10, 20, 40, ..... is 1280?
A. 10th B. 9th C. 8th D. None of these
Explanation:
Clearly, 5 x 2 = 10, 10 x 2 = 20, 20 x 2 = 40,.....
So, the series is a G.P. in which a = 5 and r = 2.
Let 1280 be the rath term of the series.
Then, 5x2n-1 = 1280 2n-1 = 256 = 28 n - 1 = 8 n = 9. | 566 | 1,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-39 | latest | en | 0.873602 |
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# Calculating Error Rate As A Percentage
## Contents
wikiHow relies on ad money to give you our free how-to guides. Close Yeah, keep it Undo Close This video is unavailable. Normally people use absolute error, relative error, and percent error to represent such discrepancy: absolute error = |Vtrue - Vused| relative error = |(Vtrue - Vused)/Vtrue| Example: You measure the plant to be 80 cm high (to the nearest cm) This means you could be up to 0.5 cm wrong (the plant could be between 79.5 and https://www.mathsisfun.com/numbers/percentage-error.html
## How To Do A Percent Error Calculation
What are the benefits of a 'cranked arrow' delta wing? The absolute value of the error is divided by an accepted value and given as a percent.|accepted value - experimental value| \ accepted value x 100%Note for chemistry and other sciences, Write an Article 125 Imaging the Universe A lab manual developed by the University of Iowa Department of Physics and Astronomy Site Navigation[Skip] Home Courses Exploration of the Solar System General What is the percent error the man made the first time he measured his height?
Percent error = (amount of error)/accepted value amount of error = 6 - 5 = 1 The accepted value is the man's real height or the value he found after he Please enter a valid email address. The graph below is a generic plot of the standard deviation. Calculating Error Rate Running Record Watch Queue Queue __count__/__total__ Find out whyClose Error and Percent Error Tyler DeWitt SubscribeSubscribedUnsubscribe267,547267K Loading...
Sign in 579 29 Don't like this video? How To Find Percent Error With Percents We can do this multiplying both the numerator and the denominator by 20 We get (1 × 20)/(5 × 20) = 20/100 = 20% I hope what I explained above was Try It Out A student analyzing a sample for bromine (Br) makes four trials with the following results: 36.0, 36.3, 35.8, and 36.3. http://astro.physics.uiowa.edu/ITU/glossary/percent-error-formula/ Please try again.
Calculate: the arithmetic mean the percent error for each trial the deviation and percent deviation for each trial the standard deviation Check your work. [Numbers and their Properties] [Numbers in Calculating Percentage Difference This means that you should subtract the real value from the estimated value. We will only use it to inform you about new math lessons. In this case, the real value is 10 and the estimated value is 9.
## How To Find Percent Error With Percents
We can also use a theoretical value (when it is well known) instead of an exact value. Answer this question Flag as... How To Do A Percent Error Calculation Thanks for letting us know. Calculating Percentage Error Chemistry Up next Accuracy and Precision - Duration: 9:29.
For example, you would not expect to have positive percent error comparing actual to theoretical yield in a chemical reaction.[experimental value - theoretical value] / theoretical value x 100%Percent Error Calculation http://iembra.org/calculating-percent/calculating-percentage-error-physics.php This sort of remind me of the difference between a One sample T-test and a paired T-test. If it was possible I would have liked to "convert" the standard error to the new visualization. Generated Thu, 06 Oct 2016 01:51:34 GMT by s_hv996 (squid/3.5.20) Calculating Percentage Error Between Two Values
Did this article help you? Steps 1 Know the formula for calculating percentage error. Jumeirah College Science 66,010 views 4:33 Standard Error - Duration: 7:05. click site Bozeman Science 171,662 views 7:05 How to Calculate Percent Error - Duration: 3:36.
Answer this question Flag as... Calculating Relative Error Tyler DeWitt 209,927 views 10:14 Scientific Notation: Multiplication and Division - Duration: 5:30. BSUBob2008 33,049 views 3:36 Why are Significant Figures Important? - Duration: 7:45.
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Tips Some teachers like the percent error to be rounded to a certain point; most people will be satisfied with the percent error rounded to three significant digits. Yes No Can you please put wikiHow on the whitelist for your ad blocker? The possible solutions, listed in descending order of accuracy and effectiveness, are Compute subject-specific ratios from the experimental data and analyze those separately. Calculating Percent Error Worksheet What are the ages of Apu, and tipu ?
You might also enjoy: Sign up There was an error. Thank you,,for signing up! share|improve this answer edited Jun 12 '12 at 21:48 answered Jun 12 '12 at 19:42 whuber♦ 145k17281540 1 Dear whuber, thanks for your answer. navigate to this website We, however, don't have a stats calculator (well, we do, but we're pretending!), so we have to do it the hard way.
To obtain these results I need to dissect (thus killing) the fly, so I cannot assay the differences in a single individual fly changing overtime, but I can assay the differences Here is how to calculate percent error, with an example calculation.Percent Error FormulaFor many applications, percent error is expressed as a positive value. I thought it could be possible with some kind of mathematical transformation, in a similar way as reported in this document: http://www.census.gov/acs/www/Downloads/data_documentation/Accuracy/PercChg.pdf Many thanks for Your help data-visualization standard-error share|improve this Brian Lamore 46,677 views 18:37 Scientific Notation and Significant Figures (1.7) - Duration: 7:58.
Tyler DeWitt 261,362 views 5:30 How to work out percent error - Duration: 2:12. The student wants to find out the standard deviation for the data set, with particular interest in the range of values from one sigma below the mean to one sigma above By continuing to use our site, you agree to our cookie policy. Richard Thornley 33,145 views 8:30 Understanding Conversion Factors - Duration: 10:14. | 1,376 | 5,982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.860594 |
https://encyclopedia2.thefreedictionary.com/Heigth | 1,603,445,394,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880878.30/warc/CC-MAIN-20201023073305-20201023103305-00180.warc.gz | 316,687,515 | 13,174 | length
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length
1. Prosody Phonetics the metrical quantity or temporal duration of a vowel or syllable
2. the distance from one end of a rectangular swimming bath to the other
3. Prosody the quality of a vowel, whether stressed or unstressed, that distinguishes it from another vowel of similar articulatory characteristics. Thus in English beat is of greater length than in English bit
4. Cricket the distance from the batsman at which the ball pitches
5. Bridge a holding of four or more cards in a suit
Collins Discovery Encyclopedia, 1st edition © HarperCollins Publishers 2005
Length
A one-dimensional extension in space. Length is one of the three fundamental physical quantities, the others being mass and time. It can be measured by comparison with an arbitrary standard; the specific one in most common usage is the international meter. In 1983, at the meeting of the Conférence Général des Poids et Mésures, the meter was redefined in terms of time and the speed of light: “The meter is the length of the path traveled by light in vacuum during a time interval of 1/299 792 458 of a second.” This definition defines the speed of light to be exactly 299 792 458 m/s, and defines the meter in terms of the most accurately known quantity, the second. See Light, Mass, Time
McGraw-Hill Concise Encyclopedia of Physics. © 2002 by The McGraw-Hill Companies, Inc.
The following article is from The Great Soviet Encyclopedia (1979). It might be outdated or ideologically biased.
Length
a numerical characteristic of the extent of curves. The concept of length is defined differently for different cases. (1) The length of a line segment is the distance between its end points, measured by some segment taken as a unit of length. (2) The length of a polygonal line is the sum of the lengths of its components. (3) The length of a simple arc is the limit of the lengths of polygonal lines inscribed in the arc, when the number of components increases indefinitely and the maximum length of the components tends to zero. (4) The length of a continuous curve consisting of a finite number of simple arcs is equal to the sum of the lengths of these arcs. For example, the circumference of a circle can be obtained as the limit of the perimeters of inscribed regular polygons when the number of their sides is doubled indefinitely; it is equal to 2πR where R is the radius of the circle. Any continuous curve has finite or infinite length. If its length is finite, then the curve is said to be rectifiable. The graph (see Figure 1) of the function
for 0 <x ≤ 1 and f(x) = 0 when x = 0 is an example of a nonrectifiable curve; here the lengths of the inscribed polygonal lines increase beyond all bound as the length of the components tend to zero. If the equation of a plane curve has the form y =f(x) (a ^ x ^ b)m rectangular coordinates and the function f(x) has a continuous derivative f’(x), then the length of the curve is given by the integral
The length of a curve given in parametric form and the length of a space curve can be expressed in a similar manner.
Figure 1
To calculate the length of a curve the mathematicians of antiquity essentially used lengths of polygonal lines and passage to the limit. For them, however, such passage to the limit was only a method for calculating the length of a curve and not for defining the concept of the length of a curve, since they apparently perceived the latter as one of the elementary mathematical concepts. The necessity of defining the length of a curve became clear only in the first half of the 19th century. A full elucidation of the problem was achieved by C. Jordan. In differential geometry the length of a curve is also defined on a surface or in an arbitrary Riemannian space.
REFERENCES
Lebesgue, H. Ob izmerenii velichin, 2nd ed. Moscow, 1960. (Translated from French.)
Fikhtengol’ts, G. M. Kurs differentsial’nogo i integral’nogo ischisleniya, 7th ed. vol. 2. Moscow, 1969.
length
[leŋkth]
(mechanics)
Extension in space.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
References in periodicals archive ?
A study evaluating the relative competitiveness of [C.sub.4] physiological grassgrass (Echinochloa crusgalli) with the soybean cultivar BMX Apollo RR showed competitive superiority of the weed in relation to the crop based on the plant heigth of plants (Bastiani et al.
The lowest values were observed for TEC 5718, which is a short- plant heigth cultivar (Table 5).
When corn control was carried out in stage [V.sub.3], TEC 5718 showed a lower grain yeld, most likely because this cultivar is characterized by short plant heigth and determinate growth.
In view of these results, the losses caused by the occurrence of spontaneous corn plants in soybean crops are evident; however, when interference occurs up to the stage [V.sub.3], the losses were higher in cultivars characterized by short plant heigth, such as TEC 5718.
The lift table lowers or raises loads to convenient heigths and the turntable brings the load to the operator.
In Karnataka and Tamil Nadu forests trees grow at elevations of up to 1400 m and there is some evidence that oil formation in the heartwood is optimal where trees are grown at heigths between 600 to 900 m.
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