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# School integration plans that involve busing between
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29 Dec 2009, 05:37
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618. School integration plans that involve busing between suburban and central-city areas have contributed, according to a recent study, to significant increases in housing integration, which, in turn, reduces any future need for busing.
(A) significant increases in housing integration, which, in turn, reduces
(B) significant integration increases in housing, which, in turn, reduces
(C) increase housing integration significantly, which, in turn, reduces
(D) increase housing integration significantly, in turn reducing
(E) significantly increase housing integration, which, in turn, reduce
I go for C .
Could someone explain each of the answer choices , I suspect why correct answer uses "reduce" over "reduces" whereas "increase in housing integration" as subject for last dependent clause is singualr.
If you have any questions
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Re: SC 618 - 700 + (very good one) [#permalink]
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29 Dec 2009, 10:49
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618. School integration plans that involve busing between suburban and central-city areas have contributed, according to a recent study, to significant increases in housing integration, which, in turn, reduces any future need for busing.
(A) significant increases in housing integration, which, in turn, reduces
(B) significant integration increases in housing, which, in turn, reduces
(C) increase housing integration significantly, which, in turn, reduces
(D) increase housing integration significantly, in turn reducing
(E) significantly increase housing integration, which, in turn, reduce
I go for C .
Could someone explain each of the answer choices , I suspect why correct answer uses "reduce" over "reduces" whereas "increase in housing integration" as subject for last dependent clause is singualr.
Choice A looks good
(a) Contributed to "X" where X is noun.
Contribut to "increases" ( increases -> activing as noun. Significant --> Adjective modifying noun "increases)
which correctly modifies "housing integration"
(b) changes the orignal meaning " Contributed to "Significant integration"
also which modifies"housing" instead of "housing integration"
(C) which modfies "housing integration" .. so, "housing integration" should be placed before which.
also
Contributed to "increase housing integration"
contributed to "increase in housing integration" would have been better.
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Re: SC 618 - 700 + (very good one) [#permalink]
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29 Dec 2009, 13:15
I have no problem with E.
It seems to me parallel.
have contributed to significantly increase housing integration, which, in turn, reduce
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Re: SC 618 - 700 + (very good one) [#permalink]
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30 Dec 2009, 06:26
School integration plans that involve busing between suburban and central-city areas have contributed, according to a recent study, to significant increases in housing integration, which, in turn, reduces any future need for busing.
(A) significant increases in housing integration, which, in turn, reduces
(B) significant integration increases in housing, which, in turn, reduces
(C) increase housing integration significantly, which, in turn, reduces
(D) increase housing integration significantly, in turn reducing
(E) significantly increase housing integration, which, in turn, reduce
My Ans: E
Explanation: 1)Subject is School integration plans and it is plural. So go with reduce
2) significantly is proper adverb of increase
3) what contributed to ? ..... increase in housing integration
Thanks
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Re: SC 618 - 700 + (very good one) [#permalink]
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30 Dec 2009, 06:51
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30 Dec 2009, 07:23
+1 for E
significantly modify the verb increase, so it should be before the verb.
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Re: SC 618 - 700 + (very good one) [#permalink]
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30 Dec 2009, 08:10
in E option, I understand housing integration is subject of verb reduce, but verb should be singular. ?
I will go with A.
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Re: SC 618 - 700 + (very good one) [#permalink]
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30 Dec 2009, 08:14
Only choice A and E come close, I dropped the modifier 'according to a recent study' since it is unnecessary to evaluate the sentence correctly, lets look at A
areas have contributed to significant increases in housing integration,
Sounds good.
E
areas have contributed to significantly increase housing integration,
to significantly increase is awkward.
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Re: SC 618 - 700 + (very good one) [#permalink]
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03 Jan 2010, 20:39
Wats the OA ? Is it E ..........
but i'm confused why 'reduce' is used - 'house integration' is singular.
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03 Jan 2010, 20:46
School integration plans that involve busing between suburban and central-city areas have contributed, according to a recent study, to significant increases in housing integration, which, in turn, reduces any future need for busing.
(A) significant increases in housing integration, which, in turn, reduces
(B) significant integration increases in housing, which, in turn, reduces
(C) increase housing integration significantly, which, in turn, reduces
(D) increase housing integration significantly, in turn reducing
(E) significantly increase housing integration, which, in turn, reduce
My Ans: E
Explanation: 1)Subject is School integration plans and it is plural. So go with reduce
2) significantly is proper adverb of increase
3) what contributed to ? ..... increase in housing integration
Thanks
Here in E which refers to 'housing integration' not 'plans'.
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Re: SC 618 - 700 + (very good one) [#permalink]
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04 Jan 2010, 06:02
E doesn't sound right somehow
A for me
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Re: SC 618 - 700 + (very good one) [#permalink]
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05 Jan 2010, 08:29
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This is Question number 618 in 1000 SC questions. From the answer sheet I have, the OA is "A" .
E is also correct except that it has verb for plural subject whereas school integration is singular subject here.
Could we have a confirmation from offical expert I.e. from someone belonging to Manhattan or Kaplan .
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Re: SC 618 - 700 + (very good one) [#permalink]
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06 Jan 2010, 05:14
But A has split infinitive. it should be to + increase, instead of "to significant increases ".
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Re: SC 618 - 700 + (very good one) [#permalink]
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06 Jan 2010, 06:48
618. School integration plans that involve busing between suburban and central-city areas have contributed, according to a recent study, to significant increases in housing integration, which, in turn, reduces any future need for busing.
(A) significant increases in housing integration, which, in turn, reduces
(B) significant integration increases in housing, which, in turn, reduces
(C) increase housing integration significantly, which, in turn, reduces
(D) increase housing integration significantly, in turn reducing
(E) significantly increase housing integration, which, in turn, reduce
I go with D.
Option A,B are faulty because increase shud always appear in singular.
the plans have contributed is what is said here so we cant use in turn, reduces, we will have to use the ing form of reduce..
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Re: SC 618 - 700 + (very good one) [#permalink]
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06 Jan 2010, 07:56
1. "increae" bein a noun, why it can not be pluar as "increaes"
2. "increase housing integration " - > what kind of construction it is : what role "increase " is playing here ? it should be "increased" or "increasing" to sustain the meaning in my opinion.
School integration plans that involve busing between suburban and central-city areas have contributed, (according to a recent study) increase housing integration significantly, in turn reducing any future need for busing.
this is idiomatically also incorrect as it should be Contributed to ..
Interestingly even if I add "to" after contribution, this sentence is incorrect following second point cited above.
We should engage professional advice here now I believe to clear the clouds.
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Re: SC 618 - 700 + (very good one) [#permalink]
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17 Jan 2010, 15:44
First of all, there's a great deal of information that simply doesn't matter. After weeding that stuff out, we end up with something much more manageable.
School plans have contributed to:
1. "Increase housing integration."
A more concise way would be to simply say "school plans have increased housing integration." The world "contributed" is unnecessary. This eliminates choices C, D, and E. Choice E suffers from the same problem, except that it has the adverb "significantly" in front of "increase."
2. "Significant integration increases."
There can be an increase in integration, but "integration increase" sounds awkward. So, B is out.
3. "Significant increases in housing integration."
This sounds right. Let's check:
"School plans have contributed to significant increases in housing integration."
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Re: SC 618 - 700 + (very good one) [#permalink]
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27 Jan 2010, 09:30
mirzohidjon wrote:
IMO A
What is OA?
OA is an Official Answer, Mirzo
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27 Jan 2010, 13:46
and what is IMO .
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Re: SC 618 - 700 + (very good one) [#permalink]
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28 Jan 2010, 04:38
618. School integration plans that involve busing between suburban and central-city areas have contributed, according to a recent study, to significant increases in housing integration, which, in turn, reduces any future need for busing.
(A) significant increases in housing integration, which, in turn, reduces
(B) significant integration increases in housing, which, in turn, reduces
(C) increase housing integration significantly, which, in turn, reduces
(D) increase housing integration significantly, in turn reducing
(E) significantly increase housing integration, which, in turn, reduce
I go for C .
Could someone explain each of the answer choices , I suspect why correct answer uses "reduce" over "reduces" whereas "increase in housing integration" as subject for last dependent clause is singualr.
Correct answer does not use 'reduce'. It uses 'reduces'. OA is A.
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Re: SC 618 - 700 + (very good one) [#permalink]
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01 May 2011, 10:33
my take is A.
always look through Oxford Dictionary.
As it says, contribute to requires a noun after it.
A is correct since it uses gramatically correct structure and the meaning of the sentence is clear.
contribute/ verb
1 ~ (sth) (to / towards sth) to give sth, especially money or goods, to help sb/sth:
[VN] The writer personally contributed Ј5000 to the earthquake fund. * [V] Would you like to contribute to our collection? * Do you wish to contribute?
2 [V] ~ (to sth) to be one of the causes of sth:
Medical negligence was said to have contributed to her death. * Human error may have been a contributing factor.
3 ~ (sth) to sth to increase, improve or add to sth:
[V] Immigrants have contributed to British culture in many ways. * [VN] This book contributes little to our understanding of the subject.
4 ~ (sth) (to sth) to write things for a newspaper, magazine, or a radio or television programme; to speak during a meeting or conversation, especially to give your opinion:
[VN] She contributed a number of articles to the magazine. * [V] He contributes regularly to the magazine 'New Scientist'. * We hope everyone will contribute to the discussion.
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Re: SC 618 - 700 + (very good one) [#permalink] 01 May 2011, 10:33
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• 1. EUCLID’S GEOMETRY MATHS PPT CLASS 1X PRESENTED BY: MASTER ANIKET VISHWAKARMA
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• 6. EUCLID’S AXIOMS • THING’S WHICH ARE EQUAL TO THE SAME THINGS ARE EQUAL TO ONE ANOTHER. • IF EQUALS ARE ADDEDTO EQUALS,THE WHOLES ARE EQUAL. • IF EQUALS ARE SUBTRACTED FROM EQUALS,THE REMAINDERS ARE EQUAL. • THINGS WHICH COINCIDE WITH ONE ANOTHER ARE EQUAL TO ONE ANOTHER
• 7. EUCLID’S AXIOMS • THE WHOLE IS GREATER THAN THE PART. • THINGS WHICH ARE DOUBLE OF THE SAME THINGS ARE EQUAL TO ONE ANOTHER. • THINGS WHICH ARE HALVES OF THE SAME THINGS ARE EQUAL TO ONE ANOTHER.
• 8. EUCLID’S FIVE POSTULATE • POSTULATE 1 : A STRAIGHT LINE MAY BE DRAWN FROM ANY ONE POINT TO ANY OTHER POINT. • POSTULATE 2: A TERMINATED LINE CAN BE PRODUCED INDEFINITELY. • POSTULATE 3: A CIRCLE CAN BE DRAWN WITH ANY CENTRE AND ANY RADIUS. • POSTULATE 4: ALL RIGHT ANGLES ARE EQUAL TO ONE ANOTHER.
• 9. EUCLID’S FIVE POSTULATE • POSTULATE 5: IF A STRAIGHT LINE FALLING ON TWO STRAIGHT LINES MAKES THE INTERIOR ANGLES ON THE SAME SIDE OF IT TAKEN TOGETHER LESS THAN TWO RIGHT ANGLES,THEN THE TWO STRAIGHT LINES, IF PRODUCED INDEFINITELY, MEET ON THAT SIDE ON WHICH THE SUM OF ANGLES IS LESS THAN TWO RIGHT ANGLES
• 10. PROBLEM ON EUCLID’S GEOMRTRY • If A, B and C are three points on a line, and B lies between A and C (see Fig. 5.7), then prove that AB + BC = AC. A B C
• 11. PROBLEMS ON EUCLID’S GEOMRTRY • SOLUTION: In the figure given above, AC coincides with AB + BC. • Also, Euclid’s Axiom (4) says that things which coincide with one another are equal to one another. So, it can be deduced that AB + BC = AC • Note that in this solution, it has been assumed that there is a unique line passing through two points.
• 12. Equivalent Versions of Euclid’s Fifth Postulate • Euclid’s fifth postulate is very significant in the history of mathematics. Recall it again from Section 5.2. We see that by implication, no intersection of lines will take place when the sum of the measures of the interior angles on the same side of the falling lines exactly 180°. There are several equivalent versions of this postulate. One of them is‘Play fair’s Axiom’ (given by a Scottish mathematician John Play fair in 1729), as stated below: • ‘For every line l and for every point P not lying on l, there exists a unique line m passing through P and parallel to l’. • From Fig. 5.11, you can see that of all the lines passing through the point P, only line m is parallel to line l.
• 13. Equivalent Versions of Euclid’s Fifth Postulate • M P L FIG 5.1
• 14. Equivalent Versions of Euclid’s Fifth Postulate • This result can also be stated in the following form: • Two distinct intersecting lines cannot be parallel to the same line. | 1,155 | 3,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.420809 |
http://mathoverflow.net/questions/85325/optimization-of-a-matrix-with-an-objective-function-for-ml | 1,469,266,019,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257821671.5/warc/CC-MAIN-20160723071021-00076-ip-10-185-27-174.ec2.internal.warc.gz | 161,352,300 | 14,844 | # Optimization of a matrix with an objective function (for ML)
Hi.
I need to do max. likelihood for an objective likelihood function L (minimize it), and the target is a matrix. ie:
$$min_KL(K)$$
For example: K is, let's say, of size 3x3 and with initial value of ones ($k_{i,j}=1∀i,j$). L is $L=∥\nabla(K)∥$ or $L=∥K∥^{1.1}$.
I know how to do gradient descend etc., but here I need to minimize the function L by iterating over K and I don't really know how to approach it. I'd expect something of this sort: $K:=K-f(\nabla(L))$, but I don't know what.
*note: It might have something to do with the Euler-Largange method ($L_x-L_t\left(L_{x'}\right)=0$) but I'd have to do it iteratively if any...
Appreciate any help.
-
I think you will have to give significantly more information to get a useful answer. Optimization is a large field and you haven't restricted your problem very much so it is hard to say what methods would apply. Is $K$ constrained? If not, your second objective looks like it would be optimized at $K=0$. As for the first objective, it is not clear to me what $\nabla K$ means for a single matrix $K$. Finally, saying that you "need to" solve the problem in a certain way suggests homework help, which is well-received at math.stackexchange.com but generally not at MO as per the FAQ. – Noah Stein Jan 10 '12 at 13:47
Your notation is quite unclear. By $\| K \|^{1.1}$, do you mean the $p$ norm of $K$ with $p=1.1$?, or do you mean the $2$-norm of $K$ raised to the power 1.1? In the latter case, raising the norm to the power 1.1 has no effect on the minimum. By $\nabla K$, do you mean the kind of image gradient often used in image processing? It seems odd that you'd be applying that to a 3x3 matrix. Or, do you mean something else? – Brian Borchers Jan 11 '12 at 6:20
This is not a HW question, I didn't specify any other terms because I'm not sure about what I want to do yet. Obviously K=0 is a solution but since I do it iteratively I may not get there. The general form of L would be rather the sum of both L's I specified, so the exponents have a meaning. Also, I didn't decide yet whether I want the Frobenius or L2 norm. My question is more technical - Say I have a matrix K of size 3x3, how do I iteratively modify it so that I minimize L. Thanks. – id0 Jan 11 '12 at 7:11 | 662 | 2,314 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2016-30 | latest | en | 0.955093 |
https://www.jiskha.com/display.cgi?id=1367171013 | 1,500,569,523,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423269.5/warc/CC-MAIN-20170720161644-20170720181644-00017.warc.gz | 779,914,415 | 4,255 | # stats
posted by .
Identify the direction and all possible interpretations of each of the following correlations. Be sure to identify each example as positive, negative, or zero (not correlated). Then give all possible causal interpretations for the results.
Many students in Professor Handel’s class express that they want a review session. Professor Handel agrees to hold an open review session for two hours the day before the exam. He then has his TA write down when each student arrives at the session and when they depart. When the exams are graded, Professor Handel announces there will be no more review sessions. The longer students spent at the review session the lower their score on the exam.
• stats -
For positive correlation both variables increase/decrease together.
For negative correlation, one variable increases, while the other decreases.
If there is a correlation significantly greater than chance, either one variable causes the other or a third variable might be effecting the other two.
I'll let you do the interpretations.
• stats -
negative correlation
interpretation that the longer the students reviewed impacted their exam grade
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https://us.metamath.org/mpeuni/ax-hvcom.html | 1,719,329,569,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00851.warc.gz | 515,947,765 | 3,681 | Hilbert Space Explorer < Previous Next > Nearby theorems Mirrors > Home > HSE Home > Th. List > ax-hvcom Structured version Visualization version GIF version
Axiom ax-hvcom 28782
Description: Vector addition is commutative. (Contributed by NM, 3-Sep-1999.) (New usage is discouraged.)
Assertion
Ref Expression
ax-hvcom ((𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ) → (𝐴 + 𝐵) = (𝐵 + 𝐴))
Detailed syntax breakdown of Axiom ax-hvcom
StepHypRef Expression
1 cA . . . 4 class 𝐴
2 chba 28700 . . . 4 class
31, 2wcel 2114 . . 3 wff 𝐴 ∈ ℋ
4 cB . . . 4 class 𝐵
54, 2wcel 2114 . . 3 wff 𝐵 ∈ ℋ
63, 5wa 399 . 2 wff (𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ)
7 cva 28701 . . . 4 class +
81, 4, 7co 7140 . . 3 class (𝐴 + 𝐵)
94, 1, 7co 7140 . . 3 class (𝐵 + 𝐴)
108, 9wceq 1538 . 2 wff (𝐴 + 𝐵) = (𝐵 + 𝐴)
116, 10wi 4 1 wff ((𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ) → (𝐴 + 𝐵) = (𝐵 + 𝐴))
Colors of variables: wff setvar class This axiom is referenced by: hvcomi 28800 hvaddid2 28804 hvadd32 28815 hvadd12 28816 hvpncan2 28821 hvsub32 28826 hvaddcan2 28852 hilablo 28941 hhssabloi 29043 shscom 29100 pjhtheu2 29197 pjpjpre 29200 pjpo 29209 spanunsni 29360 chscllem4 29421 hoaddcomi 29553 pjimai 29957 superpos 30135 sumdmdii 30196 cdj3lem3 30219 cdj3lem3b 30221
Copyright terms: Public domain W3C validator | 632 | 1,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-26 | latest | en | 0.19463 |
https://web2.0calc.com/questions/quadrilateral_36 | 1,718,552,319,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00190.warc.gz | 561,027,492 | 5,629 | +0
-2
12
1
+1141
In the diagram below, each side of convex quadrilateral \$ABCD\$ is trisected. (For example, \$AP = PQ = QB.\$) The area of convex quadrilateral \$ABCD\$ is \$180.\$ Find the area of the shaded region.
Dec 29, 2023
#1
+129270
0
The shaded area is a trapezoid with bases of AB, AB/3 and height AD
Its area =
(2/3) AD * AB = (2/3)(180) = 120
Dec 29, 2023 | 157 | 393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-26 | latest | en | 0.693834 |
https://www.coursehero.com/file/6573572/lecture-12/ | 1,496,041,589,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612018.97/warc/CC-MAIN-20170529053338-20170529073338-00434.warc.gz | 1,076,989,800 | 84,841 | lecture_12
# lecture_12 - Markov chain Monte Carlo Peter Beerli[this...
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Markov chain Monte Carlo Peter Beerli October 10, 2005 [this chapter is highly influenced by chapter 1 in Markov chain Monte Carlo in Practice , eds Gilks W. R. et al. Chapman and Hall/CRC, 1996] 1 Short history Many problems can not be solved analytically, but can be solved using statistical sampling. This idea is certainly old and was first used in a question by Georges-Louis Leclerc, Comte de Buffon (Buffon’s needle experiment) and William Gosset. Although these early applications were typically used to simulate data on a understood analytical problem. In 1945 and the following years Nicolas Metropolis and others, including Stanislaw (Stan) Ulam developed statistical sampling method to test the ENIAC computer. Metropolis coined the term Monte Carlo methods (the famous casino town in Monaco in Southern France) [influenced by the fondness of poker of Ulam who had an uncle who once borrowed money to go gambling in Monte Carlo]. Enrico Fermi was using statistical sampling for many problems in the 1930 and later, but he never published his way but used it to impress others about the accuracy of results. In 1953 Metropolis et al. described the now famous Metropolis algorithm and so the first Markov chain Monte Carlo method. 2 Monte Carlo methods Monte Carlo methods are methods that perform statistical sampling to get the expectation of a function. We want to approximate μ = E( f ( X i )) 1
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BSC5936-Fall 2005 Computational Evolutionary Biology with independently identically distributed (iid) samples X 1 , X 2 , ... using the sample mean μ = 1 n n X i =1 ( f ( X i )) If we sample long enough we approximate the original expectation, it is important to note that we always should supply the standard deviation of this sampling process because, that should converge to the standard deviation of a Normal distribution. Note that n is under the control of the researcher and is data-independent, we always can run the analysis longer (increase n ) and get a more accurate result. Monte Carlo simulation is an important tool for integration in almost any field of research. 3 Markov chain (MC) If a process is producing points and the future is independent from the past, for example Prob( X n = a n | X 0 = a 0 , X 1 = a 1 , ..., X n - 2 = a n - 2 , X n - 1 = a n - 1 ) = Prob( X n = a n | X n - 1 = a n - 1 ) The values a 0 , a 1 , .... a n form a Markov chain. A random walk or a sequence of mutations are examples of a Markov chain. 4 Markov chain Monte Carlo (MCMC) Markov chain Monte Carlo methods are methods that perform statistical sampling to get the expectation of a function. We want to approximate μ = E π ( f ( X i )) where π is the equilibrium distribution or stationary distribution, with samples from a Markov chain X 1 , X 2 , ... from the distribution f ( · ) using the sample mean μ = 1 n n X i =1 ( f ( X i )) The only difference to MC is that instead of iid samples we draw dependent samples. We would
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## This note was uploaded on 11/27/2011 for the course BSC 5936 taught by Professor Staff during the Spring '08 term at FSU.
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Ask a homework question - tutors are online | 886 | 3,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-22 | longest | en | 0.916249 |
https://ziml.areteem.org/mod/forum/discuss.php?d=310 | 1,723,357,567,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640975657.64/warc/CC-MAIN-20240811044203-20240811074203-00249.warc.gz | 843,436,390 | 9,543 | ## Discussion Forum
### AMC 10 Mock Exam Discussion For 9/22: Question 23
AMC 10 Mock Exam Discussion For 9/22: Question 23
AMC 10 Mock Exam Question 23
A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$. (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\displaystyle \textbf{(A) } 109\qquad\textbf{(B) } 201\qquad\textbf{(C) } 301\qquad\textbf{(D) } 3,049\qquad\textbf{(E) } 33,601$
Submit your answer and solution and explanation below! Solutions will be accepted for 48 hours until 9/24 at 2pm Pacific Time. (There's still time for yesterday's problem too: click here.)
Top solutions for all the Mock Exam questions will be collected and shared as part of a full 25 Question Mock AMC 10 Exam.
Note: The question above is a past AMC problem. Solutions submitted must be written by students. Copied solutions will be disqualified. | 350 | 1,177 | {"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} | 2.734375 | 3 | CC-MAIN-2024-33 | latest | en | 0.792658 |
https://encyclopedia2.thefreedictionary.com/odd+vertex | 1,652,914,320,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522556.18/warc/CC-MAIN-20220518215138-20220519005138-00533.warc.gz | 306,446,233 | 10,565 | # odd vertex
## odd vertex
[′äd ¦vər‚teks]
(mathematics)
A vertex whose degree is an odd number.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
References in periodicals archive ?
Each arc of [[??].sub.k,e] is directed from an even vertex to an odd vertex and each arc of [[??].sub.k, o] is directed from an odd vertex to an even vertex (see Figure 4 for k = 5).
In [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], the odd vertices and even vertices alternate and an odd vertex is obtained by adding aj to its preceding (even) vertex, along the anti-directed cycle, and an even vertex is obtained from its preceding (odd) vertex by adding (-bj) to it.
Based on the optimal path search, a trace (single stroke) is assessed in the graph until an odd vertex is reached.
If the number is even, then the vertex is called an even vertex; otherwise it is called an odd vertex. In figure 2, for example, A is an odd vertex because three curves go to it.
Furthermore, given a perfect matching [M.sub.e] on [R.sup.N.sub.e] ([beta], [delta]) by orienting each of its edges from its even to its odd vertex one obtains edges oriented in the directions up, down, left or right.
As a start, a new set of red vertices is added to [R.sup.M,N] as follows: in the middle of each horizontal edge of [R.sup.M,N] which has an odd vertex to its right a red vertex is added.
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http://oeis.org/A081213 | 1,590,879,564,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347410352.47/warc/CC-MAIN-20200530200643-20200530230643-00580.warc.gz | 89,142,400 | 3,451 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A081213 Let r(n,k) = if k=0 then n else r(A081210(n),k-1), then a(n)=r(n, A081212(n)). 5
1, 2, 3, 3, 5, 6, 7, 7, 7, 10, 11, 7, 13, 14, 15, 15, 17, 14, 19, 15, 21, 22, 23, 21, 23, 26, 26, 21, 29, 30, 31, 31, 33, 34, 35, 21, 37, 38, 39, 35, 41, 42, 43, 33, 35, 46, 47, 35, 47, 46, 51, 39, 53, 39, 55, 47, 57, 58, 59, 35, 61, 62, 47, 62, 65, 66, 67, 51, 69, 70, 71, 47, 73 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS A081210(a(n))=a(n). Different from A081211. a(n)=A081211(n) for n<84=A131072(1); a(A131072(n))<>A081211(A131072(n)). - Reinhard Zumkeller, Jun 13 2007 LINKS R. Zumkeller, Table of n, a(n) for n = 1..10000 CROSSREFS Cf. A081214. Sequence in context: A262882 A187043 A081211 * A081210 A285719 A070321 Adjacent sequences: A081210 A081211 A081212 * A081214 A081215 A081216 KEYWORD nonn AUTHOR Reinhard Zumkeller, Mar 10 2003 STATUS approved
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Last modified May 30 18:38 EDT 2020. Contains 334728 sequences. (Running on oeis4.) | 570 | 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} | 2.859375 | 3 | CC-MAIN-2020-24 | latest | en | 0.613846 |
http://www.withoutgeometry.com/2013_11_01_archive.html | 1,490,235,059,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186608.9/warc/CC-MAIN-20170322212946-00273-ip-10-233-31-227.ec2.internal.warc.gz | 747,477,557 | 25,949 | ## Friday, November 15, 2013
### Using Mastermind to Model The Life Cycle of a Problem
The following is part 3 of a recap for a workshop I led at the CMC-South (California Math Council) conference in Palm Springs, CA making the claim that educators and mathematicians should expand the definitions of proof in order to make proof accessible to elementary and middle school students.
__________
Mathematics the verb is cyclic, an idea I outline in part 2 of my recap on the life cycle of a math problem and explain to my sixth graders using the game of Mastermind.
If you're unfamiliar with the game, here are the basic rules .
If you want to play a few games against the computer, we have that too.
I use digits instead of colors, which means this can be played using an iPad drawing tool instead of having to go and fork out twenty bucks for a physical game. I suppose you could also use pencil and paper. I also keep track of "correct digit, correct place" and "correct digit, wrong place" instead of using white and black pegs (to be honest, I can never remember which one is which). Even using the digits 0-9 (equivalent to 10 colors), a game goes pretty quickly.
We start with some Avery vs. Class action, with some specific language used throughout.
Before the first guess: " _____, would you like to make a wild guess?"
After each guess: "Does anyone have a conjecture about this game?" We've used this language before, but you could introduce this language here. If so, the student states their conjecture.
"Do you have a proof for your conjecture?" After the student's proof, a concept that should be broadened, I ask the class: "Skeptical peers, does anyone have any questions or concerns about this proof?" If so, we try and resolve it by altering the proof or abandoning the conjecture. If not, we write it up as a theorem.
One important note: If students come up with incorrect conjectures that do not get detected, I still write it up as a theorem. I actually hope for this to happen at least once to reiterate to the class that they are the skeptical peers, not me. And I'm not a prophet (by any means) that will let them know whether they're right or wrong. They need to work at this. That said, this is a great medium for allowing incorrect theorems to live for a bit. After all, it's just a game of Mastermind and mistakes will quickly become apparent.
I then ask for more conjectures. If there aren't any, I call on another student for an educated guess. Rinse, lather, repeat. I always get great conjectures. "There cannot be a 2 in this puzzle." "There must be a 1 in the first or second place." These conjectures get more specific and more involved over time. In the end, this almost always naturally leads to students making conjectures about the solution.
We also talk about "the problem space" or the set of truths/axioms and how different Mastermind games will have similarities in their problem space (the rules of Mastermind), but also differences (different solutions). More on this in part 4 of my never-ending recap of my CMC-South session. This can lead to meta-conjectures and theorems about Mastermind such as:
The following information implies that none of the guessed numbers are in the solution.
GUESS CORRECT DIGIT, CORRECT PLACE CORRECT DIGIT, WRONG PLACE ? – ? – ? – ? 0 0
The following information is impossible.
GUESS CORRECT DIGIT, CORRECT PLACE CORRECT DIGIT, WRONG PLACE ? – ? – ? – ? 3 1
We play as a class. We play in small groups. It's pretty amazing hearing middle schoolers say things like "I have a conjecture that..." and "Respectfully, I think I have a counter-example to ________'s conjecture." It's a great way to build a community of mathematics for the rest of the year.
Oh, and I also have some Mastermind puzzles. Below are a few examples.
Find the secret code with the following information if you can use the numbers 1, 2, 3, and 4.
GUESS CORRECT DIGIT, CORRECT PLACE CORRECT DIGIT, WRONG PLACE 4 – 1 – 4 – 1 0 2 4 – 3 – 3 – 1 0 4
Find the secret code with the following information if you can use the numbers 1, 2, 3, and 4.
GUESS CORRECT DIGIT, CORRECT PLACE CORRECT DIGIT, WRONG PLACE 3 – 3 – 2 – 1 0 2 4 – 4 – 1 –2 2 0 4 – 1 – 1 –3 1 1
Using this first clue, what is the maximum number of guesses you will have to make before finding the secret code if you can use the numbers 1, 2, 3, and 4. Why?
GUESS CORRECT DIGIT, CORRECT PLACE CORRECT DIGIT, WRONG PLACE 1 – 3 – 2 – 4 0 4
This last puzzle is also a great problem to talk about lower and upper bounds, an important mathematical habit of mind. If you share solutions to the puzzle, please post a spoiler alert. If you have insights, suggestions, or opinions please post an insight, suggestion, or opinion alert.
## Tuesday, November 5, 2013
### Proof Doesn't Begin with Geometry: The Life Cycle of a Math Problem
The following is part 2 of a recap of a workshop I led this weekend at the CMC-South (California Math Council) conference in Palm Springs, CA making the claim that educators and mathematicians should expand the definitions of proof in order to make proof accessible to elementary and middle school students. Part 1 of my recap on redefining proof can be found here.
__________
There is a life cycle to math problems (not to be confused with exercises) in my classroom. We start with a wild guess.
This is something that should be done quickly, individually, and even if students don't have all the necessary information to answer the question.
Immediately: Seriously. After five seconds everyone should be writing something down--or by the time you finish reading this sentence.
Individually:
There's some awesome research out there about priming and anchoring. The general idea is that if you ask a group to make a guess, the value of the first guess will significantly affect later guesses. Imagine asking a group of students to guess the population of Istanbul. If the first guess is 100, even though most students will recognize this as way too low of a guess, this anchor will cause future guesses to be smaller. Students have been primed to give smaller guesses. On the other hand, if the first guess is 100,000,000, you would expect guesses to be too high. It would be great fun to try this out with a couple classes and collect some data on mean guesses and their relationship to the first guess. Let me know if you try this. Daniel Khaneman's Thinking, Fast and Slow has a slew of examples, some of which are much scarier than the example I just gave.
German judges with an average of more than fifteen years of experience on the bench first read a description of a woman who had been caught shoplifting, then rolled a pair of dice that were loaded so every roll resulted in either a 3 or a 9. As soon as the dice came to a stop, the judges were asked whether they would sentence the woman to a term in prison greater or lesser, in months, than the number showing on the dice. Finally, the judges were instructed to specify the exact prison sentence they would give to the shoplifter. On average, those who had rolled a 9 said they would sentence her to 8 months; those who rolled a 3 said they would sentence her to 5 months; the anchoring effect was 50%. (Englich, Mussweiler, and Strack in Khaneman p. 125)
There are other reasons to have students write down their guesses before they share, a practical one being you then know everyone has a writing utensil in their hand.
Lacking Information: This is another way to lower the stakes. If you ask students to guess something where they don't have all the necessary information (which I usually make explicit), then the stakes are that much lower for them to be right. Inspired by a talk immediately before mine by Brad Fulton, I quickly put this slide together and asked teachers to guess what number lay behind the black box.
There's no way you can know the answer to this (unless you went to Brad or my talk, and even then you can't be sure I didn't change the value). That said, your wild guess will most likely be a speed and it will probably be somewhere between 0 mph and 1000mph. I could have also asked participants to guess the answer to this problem and this is probably what I would do with students (creating a sense of what a reasonable answer would be before working on the problem versus reflecting on whether or not their answer is reasonable after working on the problem).
After making a wild guess, I ask students to use appropriate mathematical habits of mind such as estimating, bounding, and contextualizing to make educated guesses. These are still guesses (and students still may not have enough information to solve the problem), but they are guesses based on some initial reasoning and strategies.
Estimating: This broad strategy for guessing includes rounding (73*88 is relatively close to 70*90), chunking (the height of that tree looks to be about 8 of me, or around 45 feet), and disaggregation (dividing an estimation task in a number of smaller, easier estimations) If you're unfamiliar with Fermi problems, these problems will keep you busy for as long as you want. And anyone who writes off estimation as not being that important, tell that to the US government and BP who are currently wrangling over how much oil spilled from the Deep Horizon disaster in 2010, with ramifications to the tune of billions of dollars.
Bounding: I introduce the terms "lower bound" and "upper bound" to my students early in the year. Similar to making wild guesses, asking students to make a guess you know is too low and a second guess you know is too high keeps the stakes low and is accessible to every student. It's also a great way to build in intermediate success points while working on a problem. Students feel success when they narrow their lower and upper bounds. It's worth saying again: this is a really powerful tool. If you don't believe me, give the following to your students who have no idea what calculus is and watch what they come up with in terms of a lower and upper bound.
Contextualizing: Getting students to start thinking about reasonable answers before they even really start solving the problem will pay dividends. Consider the train problem above. If students spend the time to realize that the answer is going to be between 0 and 280 miles, they will think twice when they solve the problem incorrectly and get 400 miles as an answer.
The next step is moving from an educated guess to a conjecture, or a proposition that you think to be true. This is the meat of mathematical problem solving, and applications of prior knowledge and mathematical habits of mind should be abundant.
At this point in time, students who are convinced that something is true work on proving their conjecture. I talked (at length) in part one of my recap about expanding our definition of proof, and will delve into examples of types of proofs that are accessible to younger kids in a later post.
Finally (well not really), the student shares his or her proof with the rest of the class and the class plays the role of skeptical peers, respectfully looking for counter-examples and holes in logic. It's pretty amazing to watch this when the cogs are well-oiled. I'll talk about how I start to build this entire structure using the game Mastermind in a future post. Assuming all goes well (the class gives a stamp of approval), we now have a theorem. Woo hoo!
One important note: if the class let's something slide that isn't true, I do not jump in and correct them. I really want them to believe that they are the ones in control of determining truth. That said, when this happens I know I have some work to do in order to push the class in a direction that will allow them to see the mistake(s) they made.
So there you have it. The life cycle of mathematics, completed below with what makes it a cycle: tinkering and inventing new problems. So much fun.
## Monday, November 4, 2013
### Proof Doesn't Begin with Geometry: Redefining Proof
The following is part 1 of a recap of a talk I gave this past weekend at the CMC-South (California Math Council) conference in Palm Springs, CA. I attempted to make the case that educators and mathematicians should expand the definitions of proof in order to make this important aspect of mathematics accessible to elementary and middle school students, and more accessible to high school students.
__________
## A Humble Proposal: Redefining Proof
Currently, the class Geometry has a monopoly on proof in K-12 education. For the college football fans out there, Geometry is the Alabama of mathematical proof in school. Like Alabama, I hope that the stranglehold Geometry has on proof is on it's way out (take that, Crimson Tide). Geometry is too often the first time students are introduced to mathematical proof, and too often the last time students grapple with proof until well into college. I don't know how related this is, but the following graph represents a common refrain. "Oh, I hated math. I liked Geometry, but...ugh, everything else."
To be clear, this is intended to be somewhat glib. Don't read too into how one would quantify "what math class looked like" or be angry with my lack of axis labeling.
I do not see expanding the definition of proof as a hugely controversial proposal. After all, Merriam-Webster, a respected dictionary (at least it used to be), defines mathematical proof as "a test which shows that a calculation is correct." The Silver-Burdett Mathematical Dictionary I found on my classroom shelf doesn't even include a definition of proof. We can all agree that we can do better than that.
### The Issue of Formality
The Online Free Dictionary's definition of mathematical proof represents what I see as the status quo: "a formal series of statements showing that if one thing is true something else necessarily follows from it." Starting with an agreed upon set of axioms (truths), proof is the process of showing that other things are also true. I don't want to change this; I simply want to relax our idea of "formality" to be more age-appropriate. Showing that something works for three examples will never be a proof. Saying "because it seems to be true" won't cut the mustard, hack it, or be up to snuff either. But what about the following argument related to a certain number trick?
First, the trick.
1. Pick a number.
2. Add 5
3. Double the result
4. Subtract 4
5. Divide by 2
6. Subtract your secret number
And your answer is....
*drumroll please*
3! Magic? Nah.
And my proof? Let's say your secret number is box.
Is this formal enough?
### The Social Aspect of Proof
Our above definition of proof says nothing about who decides whether or not a proof is adequate. For a professional mathematician, this work is done by the rest of the mathematical community, a community of skeptical peers. This shouldn't be any different in the classroom, and no, the teacher shouldn't be playing this role. Supporting, yes, but the community is the other students in the classroom. So here's my definition of proof I give to my 5th and 6th graders:
Convincing your skeptical peers that a mathematical statement is true
And then I put my money where my mouth is. When a student claims to have a proof of an existing conjecture, they give their argument using whatever medium makes sense, ranging from a verbal argument to an interpretive dance. The rest of the class takes on the role of (respectful) skeptical peers. Once everyone is convinced, the class has a new theorem. Giving a cogent argument is hard work. Looking for flaws, counter-examples, and unclear logic is hard work. All of my students are working hard. The hardest part of this culture is not letting yourself, as the teacher, play the role of prophet. Students have to know that you'll let them create theorems that aren't true (at least for a bit). Otherwise, they have no incentive to play the role of skeptical peers. They'll just wait for your cue, whether it be "Sounds great! Anyone have any reasons to question that proof?" or "Hmmm...is that always going to be true?" Remember, we want the students to be analyzing the proof, not the teacher's response.
### Expanding Proof Techniques
I learned inductive proofs in Pre-calculus. Geometry was the class where I learned two-column and indirect proofs. I didn't know you could write proofs in paragraph form until I was in college and proofs without words didn't hit my radar until after I was teaching. There are a few problems with this, the first of which is that two-column proofs are a structure for proving, not a proof technique.
#### Two-column proofs
The blog Math With bad Drawings has a cheeky, but poignant post titled "Two-column proofs that Two-column Proofs are Bad" that addresses many of the issues with this being the sole vehicle for proof in school. While the structure is intended to scaffold the idea of deductive proofs, it too often obfuscates the important idea of using a set of assumptions to show that something else must be true.
Two-column proofs are also not a proof technique, but a way of organizing proofs. Imagine students going their whole life thinking that filling in data in a table was statistics.
The other major issue with two-column proofs is that they are completely disconnected to how people make arguments in real life. I don't think the following has ever happened in the history of the world (but please let me know if I'm mistaken).
#### Alternative Proof Techniques
I believe there are many proof techniques that students can begin to grapple with in elementary and middle school. Prove something using words. Draw pictures. Acquaint students with the ideas of indirect reasoning, induction, and parity. Introduce or create simple axiomatic systems and prove other truths using deductive reasoning. I'm telling you. These can all be done at a much earlier age than they are currently done. This is big, and a lot, and will be flushed out in part 3 of this tome post.
### Unanswered Questions
I have a list of unanswered questions. You will most likely have more.
1. What is the appropriate level of formality? I assume we can all agree that "I tested two examples" won't cut it as a proof at any age and that formal logic is overkill for students, there's a great expanse between these two extremes.
2. How do you create a culture of skeptical peers in a classroom? How long, and to what extent, do you allow incorrect theorems to live?
3. With an expanded definition of proof, what is the difference between an explanation/solution and a proof? Is this a skill/concept divide where work for demonstrating skills != proof while arguments for demonstrating concepts = proof? Or does this difference have more to do with the problem space? Or is a "proof" that 34*21=483 just too small of a problem space to call this a proof? We are typically ok with students saying "The sum of the interior angles of a triangle is 180 degrees." Should we instead be asking them to say "The sum of the interior angles of a triangle is 180 degrees within the problem space of Euclidean Geometry"?
4. Even if we agree that it feels silly, is there harm in using the vocabulary of proof, theorem, etc. when talking about 78*53?4. Is there harm in using the vocabulary of proof, theorem, etc. in a very small problem space (i.e. a particular game of Mastermind vs. Euclidean Geometry)?
Some thoughts to ruminate and questions to ponder. In the end, I think this boils down to my belief that we will have a higher level of mathematical thinking and discourse if we allow ourselves a lower level of formality related to proof. Please feel free to disagree. You may submit your arguments (proofs?) in the comment section.
Part Two: The Life Cycle of a Math Problem | 4,408 | 19,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-13 | longest | en | 0.945846 |
https://www.jiskha.com/display.cgi?id=1316044450 | 1,503,558,144,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886133042.90/warc/CC-MAIN-20170824062820-20170824082820-00205.warc.gz | 896,222,828 | 3,949 | # algebraa 1 helpppp meee pleaseeee
posted by .
Find each percent change, and tell whether it is a percent increase or decrease.
7. From 20 to 28
• algebraa 1 helpppp meee pleaseeee -
100(8/20) = ?
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More Similar Questions | 609 | 2,335 | {"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} | 3.421875 | 3 | CC-MAIN-2017-34 | latest | en | 0.889947 |
https://jassweb.com/solved/tag/xls/ | 1,721,214,533,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.39/warc/CC-MAIN-20240717090242-20240717120242-00768.warc.gz | 297,348,074 | 17,516 | ## [Solved] how to calculate XIRR dynamically in excel and in google sheets
Assuming your table is in A1:G8 (with headers in row 1), and that your Fund of choice, e.g. “B”, is in J2, array formula**: =XIRR(INDEX(F:G,N(IF(1,MODE.MULT(IF(B\$2:B\$8=J2,{1,1}*ROW(B\$2:B\$8))))),N(IF(1,{1,2}))),CHOOSE({1,2},INDEX(A:A,N(IF(1,MODE.MULT(IF(B\$2:B\$8=J2,{1,1}*ROW(B\$2:B\$8)))))),TODAY())) Copy down to give similar results for Funds in J3, J4, etc. I tend to prefer this to set-ups involving OFFSET; not only is it briefer (and therefore more efficient), … Read more | 197 | 560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-30 | latest | en | 0.764468 |
https://discuss.leetcode.com/topic/50321/simple-c-solution-4ms | 1,513,505,237,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948595342.71/warc/CC-MAIN-20171217093816-20171217115816-00074.warc.gz | 533,969,478 | 8,240 | # Simple C++ solution 4ms
• The idea is to record the first negative product (N) for later use. If current product (P) is negative, make it positive by dividing N. If P is zero, reset everything.
``````int maxProduct(vector<int>& nums) {
if( nums.empty() ) return 0;
int neg = 0, product = 1, val, maxp = INT_MIN;
for( int i : nums ) {
product *= i;
if( product < 0 ) {
val = neg ? (product/neg) : i;
if( !neg && i < 0 ) neg = product;
} else if( product == 0 ) {
product = 1;
neg = val = 0;
} else val = product;
maxp = max( maxp, val );
}
return maxp;
}``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 204 | 660 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | latest | en | 0.724103 |
http://demonstrations.wolfram.com/CheckDigitSchemes/ | 1,438,105,570,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042982013.25/warc/CC-MAIN-20150728002302-00177-ip-10-236-191-2.ec2.internal.warc.gz | 64,978,203 | 11,856 | 10217
# Check Digit Schemes
Identification numbers often have a check digit appended to them in order to help catch copy and transmission errors. Seven common check digit schemes are illustrated here. The "new example" button gives a random identification number of the desired type. Use the popup menu to select the check digit, then check your answer. If your answer is incorrect, you can view the appropriate computation and try again. The "fixed example" and "illustrate error" options show several of the most common errors and allow you to explore whether a given check digit scheme will catch such errors.
The United States Postal Service money order check digit scheme requires that the check digit (the 11th digit) be the remainder upon division by 9 of the sum of the first 10 digits of the identification number.
The traveler's check scheme requires that the check digit be chosen so that the sum of all 10 digits (including the check digit) is divisible by 9.
In the airline ticket scheme, the check digit (the 12th digit), is the remainder upon division by 7 of the 11-digit identification number.
Given a UPC number , the check digit is chosen so that the weighted sum is divisible by 10.
Given a bank routing number , the check digit is the remainder upon division by 10 of the weighted sum .
For a 10-digit ISBN number with the first nine digits , the check digit is chosen so that the weighted sum has a remainder of 0 upon division by 11. The symbol X is used if the check digit must equal 10.
The credit card check digit is computed so that is divisible by 10. The number is the number of digits in the odd-numbered positions whose value exceeds 4.
### DETAILS
Reference
[1] COMAP, For All Practical Purposes, New York: W. H. Freeman and Company, 2009.
### PERMANENT CITATION
Share: Embed Interactive Demonstration New! Just copy and paste this snippet of JavaScript code into your website or blog to put the live Demonstration on your site. More details » Download Demonstration as CDF » Download Author Code »(preview ») Files require Wolfram CDF Player or Mathematica.
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https://chem.libretexts.org/Courses/University_of_South_Carolina__Upstate/USC_Upstate%3A_CHEM_U109_-_Chemistry_of_Living_Things_(Mueller)/01%3A_Chemistry%2C_Matter%2C_and_Measurement/1.7%3A_Converting_Units | 1,717,039,530,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059418.31/warc/CC-MAIN-20240530021529-20240530051529-00805.warc.gz | 133,190,805 | 37,188 | # 1.7: Converting Units
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
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$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
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$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Skills to Develop
• To convert a value reported in one unit to a corresponding value in a different unit.
The ability to convert from one unit to another is an important skill. For example, a nurse with 50 mg aspirin tablets who must administer 0.2 g of aspirin to a patient needs to know that 0.2 g equals 200 mg, so 4 tablets are needed. Fortunately, there is a simple way to convert from one unit to another.
## Conversion Factors
If you review the SI units described earlier, you can find that one kilogram (kg) is about equal to 2.20 pounds (lbs).
$1\; \rm{kg} = 2.20\; \rm{lbs}$
Suppose we divide both sides of the equation by 1 kg (both the number and the unit):
$\mathrm{\dfrac{1\:kg}{1\:kg}=\dfrac{2.20\:lbs}{1\:kg}}$
As long as we perform the same operation on both sides of the equals sign, the expression remains an equality. Look at the left side of the equation; it now has the same quantity in the numerator as it has in the denominator. Any fraction that has the same quantity on the top and on the bottom has a value of 1:
$\mathrm{\dfrac{1\:kg}{1\:kg}= 1 \; \text{ and } \; \dfrac{2.20\:lbs}{1\:kg}}=1$
We know that one kilogram is equal to 2.20 pounds, so we have the same quantity on the top and the bottom of our fraction, although it is expressed in different units. A fraction that has equivalent quantities in the numerator and the denominator but expressed in different units is called a conversion factor.
Instead of dividing both sides by 1 kg, as above, we could have obtained a conversion factor if we divided both sides by 2.20 kg.
$1\; \rm{kg} = 2.20\; \rm{lbs}$
will become
$\mathrm{\dfrac{1\:kg}{2.20\:lbs}=\dfrac{2.20\:lbs}{2.20\:lbs}}$
Again, because we performed the same operation on both sides of the equals sign, the expression remains an equality (=1).
$\mathrm{\dfrac{1\:kg}{2.20\:lbs}= 1 \; \text{ and } \; \dfrac{2.20\:lbs}{2.20\:lbs}}=1$
This means that every conversion factor will have two forms, both equal to one, but reciprocated (flipped). For 1 kg = 2.20 lbs, the two conversion factors would be:
$\dfrac{1\; \rm{kg}}{2.20\; \rm{lbs}} \; \text{or} \; \dfrac{2.20 \; \rm{lbs}}{1\; \rm{kg}}$
Here is a simple example of how the conversion factor might be used. You are about to return from a European vacation with a suitcase full of gifts. The airline does not allow bags heavier than 55 pounds, but the scale in your hotel only reads in kilograms. How many kg are equal to 55 lbs? To solve the problem with the conversion factor used earlier, we first write the quantity we are given, 55 lbs. Then we multiply this quantity by a conversion factor, which is the same as multiplying it by 1. We can write 1 as either version of the conversion factor, $$\mathrm{\frac{1\:kg}{2.20\:lbs}}$$ or $$\mathrm{\frac{2.20\:lbs}{1\:kg}}$$. Here we will use $$\mathrm{\frac{1\:kg}{2.20\:lbs}}$$ and multiply so that the units we started with (kg) cancel and we end with the units we want (lbs).
$55 \; \rm{lbs} \times \dfrac{1 \; \rm{kg}}{2.20\; \rm{lbs}}$
The 55 kg can be thought of as a fraction with a 1 in the denominator (though it is often not shown). Because kg, the abbreviation for kilograms, occurs in both the numerator (top) and the denominator (bottom) of our expression, they cancel out. The 55 on top is divided by 2.20 on bottom and the only units that did not cancel are kg.
$\dfrac{55 \; \cancel{\rm{lbs}}}{ 1} \times \dfrac{1 \; \rm{kg}}{2.20 \; \cancel{\rm{lbs}}}= 25 \; \rm{kg}$
In the final answer, we omit the 1 in the denominator. We find that 55 lbs equals 25 kg. A generalized description of this process is as follows:
$\text{quantity (in old units)} \times \text{conversion factor} = \text{quantity (in new units)} \nonumber$
This involved conversion only involved one step. We will see that the same basic principle applies for conversions with multiple steps. If you can master the technique of applying conversion factors, you will be able to solve a large variety of problems.
In the previous example, if we had used the other possible reciprocated conversion factor, $$\mathrm{\frac{2.20\:lbs}{1\:kg}}$$, the original unit would not have canceled, and the result would have been meaningless. Here is the incorrect we would have gotten:
$55 \; \rm{lbs} \times \dfrac{2.20\; \rm{lbs}}{1 \; \rm{kg}} = 0.0355 \dfrac{\rm{lbs}^2}{\rm{kg}}$
For the answer to be meaningful, we have to construct the conversion factor in a form that causes the original unit to cancel out.
## Conversion Factors With Prefix Multipliers (Exponents)
If you learned the prefixes described earlier in this chapter, then you know that the prefix centi (c) is 10−2. The prefixes come in front of some other base unit, like meters (m). The base units must be included on both sides of the equality, so if c = 10−2, then cm = 10−2 m. As before, this can be made into these two different reciprocal conversion factors:
$\dfrac{1\; \rm{cm}}{10^{-2} \; \rm{m}} \; \text{or} \; \dfrac{10^{-2} \; \rm{m}}{1\; \rm{cm}}$
If you wanted to find out how many centimeters in 3.55 meters, we again follow the process:
$\text{quantity (in old units)} \times \text{conversion factor} = \text{quantity (in new units)} \nonumber$
$\dfrac{3.55 \; \cancel{\rm{m}}}{ 1} \times \dfrac{1 \; \rm{cm}}{10^{-2} \; \cancel{\rm{m}}}= 355 \; \rm{cm}$
Pull out your calculator and ensure that you get 355 as your answer when you divide 3.55 by 10−2. If not, get help with how to properly use your calculator. (For example, in the Casio fx-260, 10−2 might be entered as [1], then [EXP], [+/-], and [2].)
To avoid using exponents, some instead use that 1 cm is 1/100th of a meter (because centi = 10−2 = 1/10= 1/100).
$1\; \rm{cm} = \dfrac{1}{100} \; \rm{m}$
Since fractions of a unit can make conversions difficult later, both sides can be multiplied by 100 to cancel the 1/100 and get this version.
$100\; \rm{cm} = 1\; \rm{m}$
The two versions of the conversion factor are therefore:
$\dfrac{100 \; \rm{cm}}{1 \; \rm{m}} \; \text{or} \; \dfrac{1 \; \rm{m}}{100\; \rm{cm}}$
To convert 3.55 meters into centimeters this way, we first write the quantity we are given, 3.55 m. Then we multiply this quantity by a conversion factor, which is the same as multiplying it by 1. We can write 1 as $$\mathrm{\frac{100\:cm}{1\:m}}$$ and multiply:
$\dfrac{3.55 \; \cancel{\rm{m}}}{ 1} \times \dfrac{100 \; \rm{cm}}{1 \; \cancel{\rm{m}}}= 355 \; \rm{cm}$
Notice that we get the same answer, 355 cm, either way. Using the exponent is most direct and makes things easier with large exponents, like micro = 10−6 or giga = 109. But for those who have trouble with exponential notation, using micro = 1/1,000,000 or giga = 1,000,000,000 is also an option.
## Significant Figures in Conversions
How do conversion factors affect the determination of significant figures? Numbers in conversion factors based on prefix changes, such as kilograms to grams, are not considered in the determination of significant figures in a calculation because the numbers in such conversion factors are exact. Exact numbers are defined or counted numbers, not measured numbers, and can be considered as having an infinite number of significant figures. (In other words, 1 kg is exactly 1,000 g, by the definition of kilo-.) Counted numbers are also exact. If there are 16 students in a classroom, the number 16 is exact. In contrast, conversions that do not involve just a prefix change, like those between different systems of units, are usually not exact. For example, kilograms are S.I. units, but pounds come from an older system. If given 1 kg = 2.20 lbs, you sould assume three significant figures (but can be found to six or more figures, 1 kg = 2.20462 lbs). Conversion factors that come from measurements (such as density, as we will see shortly) or are approximations have a limited number of significant figures and should be considered in determining the significant figures of the final answer.
Example $$\PageIndex{1}$$
1. The average volume of blood in an adult male is 4.7 L. What is this volume in milliliters?
2. A hummingbird can flap its wings once in 18 ms. How many seconds are in 18 ms?
SOLUTION
1. We start with what we are given, 4.7 L. We want to change the unit from liters to milliliters. The prefix milli is 10−3. If m = 10−3, adding the liters to both sides, makes it 1 mL = 10−3 L. From this relationship, we can construct two conversion factors:
$\dfrac{10^{-3} \; \rm{L}}{1 \; \rm{mL}} \; \text{or} \; \dfrac{1 \; \rm{mL}}{10^{-3} \; \rm{L}}$
We use the conversion factor that will cancel out the original unit, liters, and introduce the unit we are converting to, which is milliliters. The conversion factor that does this is the one on the right.
$4.7 \cancel{\rm{L}} \times \dfrac{1 \; \rm{mL}}{10^{-3} \; \cancel{\rm{L}}} = 4,700\; \rm{mL}$
Because the numbers in the conversion factor are exact, we do not consider them when determining the number of significant figures in the final answer. Thus, we report two significant figures in the final answer.
1. We can construct two conversion factors from the relationships between milliseconds and seconds, 1 ms = 10−3 s:
$\dfrac{1 \; \rm{ms}}{10^{-3} \; \rm{s}} \; \text{or} \; \dfrac{10^{-3} \; \rm{s}}{1 \; \rm{ms}}$
To convert 18 ms to seconds, we choose the conversion factor that will cancel out milliseconds and introduce seconds. The conversion factor on the right is the appropriate one. We set up the conversion as follows:
$18 \; \cancel{\rm{ms}} \times \dfrac{10^{-3} \; \rm{s}}{1 \; \cancel{\rm{ms}}} = 0.018\; \rm{s}$
The conversion factor’s numerical values do not affect our determination of the number of significant figures in the final answer.
(If you prefer the method without exponents, you could instead use that milli is 1/1,000. If you do it that way, ensure that you get the same answers.)
Example $$\PageIndex{2}$$
Perform each conversion.
1. 101,000 ns to seconds
2. 32.08 kg to grams
SOLUTION
1. The prefix nano is 10−9. If n = 10−9, adding the seconds to both sides, makes it 1 ns = 10−9 s. From this relationship, we can construct two conversion factors:
$\dfrac{10^{-9} \; \rm{s}}{1 \; \rm{ns}} \; \text{or} \; \dfrac{1 \; \rm{ns}}{10^{-9} \; \rm{s}}$
We use the conversion factor that will cancel out the original unit, nanoseconds, and introduce the unit we are converting to, which is seconds. The conversion factor that does this is the one on the left.
$101,000 \cancel{\rm{ns}} \times \dfrac{10^{-9} \; \rm{s}}{1 \; \cancel{\rm{ns}}} = 0.000101 \; \rm{s}$
1. We can construct two conversion factors from the relationships between kilograms and grams, 1 kg = 10g:
$\dfrac{1 \; \rm{kg}}{10^3 \; \rm{g}} \; \text{or} \; \dfrac{10^3 \; \rm{g}}{1 \; \rm{kg}}$
To convert 32.08 kg to grams, we choose the conversion factor that will cancel out kilograms and introduce grams. The conversion factor on the left is the appropriate one. We set up the conversion as follows:
$32.08 \cancel{\rm{kg}} \times \dfrac{10^{3} \; \rm{g}}{1 \; \cancel{\rm{kg}}} = 32080 \; \rm{g}$
To more clearly show that the answer has four significant figures, it can be written in scientific notation as 3.208x10g.
Conversion factors can also be constructed for converting between different kinds of units. For example, density can be used to convert between the mass and the volume of a substance. Consider mercury, which is a liquid at room temperature and has a density of 13.6 g/mL. The density tells us that 13.6 g of mercury have a volume of 1 mL. We can write that relationship as follows:
13.6 g mercury = 1 mL mercury
This relationship can be used to construct two conversion factors:
$\mathrm{\dfrac{13.6\:g}{1\:mL}\:and\:\dfrac{1\:mL}{13.6\:g}}$
Which one do we use? It depends, as usual, on the units we need to cancel and introduce. For example, suppose we want to know the mass of 16 mL of mercury. We would use the conversion factor that has milliliters on the bottom (so that the milliliter unit cancels) and grams on top so that our final answer has a unit of mass:
$\mathrm{16\:\cancel{mL}\times\dfrac{13.6\:g}{1\:\cancel{mL}}=217.6\:g=220\:g}$
In the last step, we limit our final answer to two significant figures because the volume quantity has only two significant figures; the 1 in the volume unit is considered an exact number, so it does not affect the number of significant figures. The other conversion factor would be useful if we were given a mass and asked to find volume, as the following example illustrates.
Density can be used as a conversion factor between mass and volume.
Example $$\PageIndex{3}$$: Mercury Thermometer
A mercury thermometer for measuring a patient’s temperature contains 0.750 g of mercury. What is the volume of this mass of mercury?
SOLUTION
Because we are starting with grams, we want to use the conversion factor that has grams in the denominator (bottom). The gram unit will cancel algebraically, and milliliters will be introduced in the numerator.
$0.750 \; \cancel{\rm{g}} \times \dfrac{1\; \rm{mL}}{13.6 \; \cancel{\rm{g}}} = 0.055147 ... \; \rm{mL} \approx 0.0551\; \rm{mL}$
We have limited the final answer to three significant figures.
You could instead rearrange the density formula, d = m/V, to solve for volume, V = m/d = (0.750g) / (13.6 g/mL) = 0.0551 mL, to arrive at the same answer.
Looking Closer: Density and the Body
The densities of many components and products of the body have a bearing on our health.
Bones. Bone density is important because bone tissue of lower-than-normal density is mechanically weaker and susceptible to breaking. The density of bone is, in part, related to the amount of calcium in one’s diet; people who have a diet deficient in calcium, which is an important component of bones, tend to have weaker bones. Dietary supplements or adding dairy products to the diet seems to help strengthen bones. As a group, women experience a decrease in bone density as they age. It has been estimated that fully half of women over age 50 suffer from excessive bone loss, a condition known as osteoporosis. Exact bone densities vary within the body, but for a healthy 30-year-old female, it is about 0.95–1.05 g/cm3. Osteoporosis is diagnosed if the bone density is below 0.6–0.7 g/cm3.
Urine. The density of urine can be affected by a variety of medical conditions. Sufferers of diabetes produce an abnormally large volume of urine with a relatively low density. In another form of diabetes, called diabetes mellitus, there is excess glucose dissolved in the urine, so that the density of urine is abnormally high. The density of urine may also be abnormally high because of excess protein in the urine, which can be caused by congestive heart failure or certain renal (kidney) problems. Thus, a urine density test can provide clues to various kinds of health problems. The density of urine is commonly expressed as a specific gravity, which is a unitless quantity defined as
$\dfrac{\text{density of some material}}{\text{density of water}}$
Normal values for the specific gravity of urine range from 1.002 to 1.028.
Body Fat. The overall density of the body is one indicator of a person’s total body fat. Fat is less dense than muscle and other tissues, so as it accumulates, the overall density of the body decreases. Measurements of a person’s weight and volume provide the overall body density, which can then be correlated to the percentage of body fat. (The body’s volume can be measured by immersion in a large tank of water. The amount of water displaced is equal to the volume of the body.)
## Multiple Conversions
Sometimes you will have to perform more than one conversion to obtain the desired unit. For example, suppose you want to convert 54.7 km into millimeters. You can either memorize the relationship between kilometers and millimeters, or you can do the conversion in steps. Most people prefer to convert in steps.
To do a stepwise conversion, we first convert the given amount to the base unit. In this example, the base unit is meters. We know that there are 103 m in 1 km:
$54.7\; \cancel{\rm{km}} \times \dfrac{10^3 \; \rm{m}}{1\; \cancel{\rm{km}}} = 54,700\; \rm{m}$
Then we take the result (54,700 m) and convert it to millimeters, remembering that there are $$1\; \rm{mm}$$ for every $$10^{-3}\; \rm{m}$$:
$54,700 \; \cancel{\rm{m}} \times \dfrac{1 \; \rm{mm}}{10^{-3} \; \cancel{\rm{m}}} = 54,700,000 \; \rm{mm} = 5.47 \times 10^7\; \rm{mm}$
Expressing the final answer in scientific notation more clearly shows the three significant figures.
As a shortcut, both steps in the conversion can be combined into a single, multistep expression:
$54.7\; \cancel{\rm{km}} \times \dfrac{10^3 \; \cancel{\rm{m}}}{1\; \cancel{\rm{km}}} \times \dfrac{1 \; \rm{mm}}{10^{-3} \; \cancel{\rm{m}}} = 54,700,000 \; \rm{mm} = 5.47 \times 10^7\; \rm{mm}$
NOTE: When using the prefixes and exponents, the prefix should always be on one side and the exponent that it equals should be on the other side. (In the first step here, since kilo is on bottom, the 103 must be on top. Then in the other step, because milli is on the top, the 10-3 should be on bottom.)
Either method—one step at a time or all the steps together—is acceptable. If you do all the steps together, the restriction for the proper number of significant figures should be done after the last step. As long as the math is performed correctly, you should get the same answer no matter which method you use.
Example $$\PageIndex{4}$$
Convert 58.2 ms to megaseconds in one multistep calculation.
SOLUTION
The milli prefix abbreviatioin is lower case, m = 10-3, so with base units of seconds it would be, ms = 10-3 s. The mega abbreviation is upper case, M = 106, so Ms = 10s.
First, convert the given unit to the base unit—in this case, seconds—and then convert seconds to the final unit, megaseconds.
$58.2 \; \cancel{\rm{ms}} \times \dfrac{\cancel{10^{-3} \rm{s}}}{1 \; \cancel{\rm{ms}}} \times \dfrac{1 \; \rm{Ms}}{10^6 \; \cancel{ \rm{s}}} =0.0000000582\; \rm{Ms} = 5.82 \times 10^{-8}\; \rm{Ms}$
Neither conversion factor affects the number of significant figures in the final answer.
Example $$\PageIndex{5}$$
Convert 43.00 ng to kilograms in one multistep calculation.
SOLUTION
For nano, n = 10-9, so with base units of grams it would be, ng = 10-9 g. For kilo, k = 103, so kg = 103 g.
First, cancel ng and convert the given unit to the base unit (g), and then convert grams to the final unit, kilograms.
$43.00 \; \cancel{\rm{ng}} \times \dfrac{\cancel{10^{-9} \rm{g}}}{1 \; \cancel{\rm{ng}}} \times \dfrac{1 \; \rm{kg}}{10^3 \; \cancel{ \rm{g}}} =0.00000000004300\; \rm{kg} = 4.300 \times 10^{-11}\; \rm{kg}$
Neither conversion factor affects the number of significant figures (four) in the final answer.
## Squared or Cubed Units
Area is often found by multiplying two lengths, and thus has squared units. For example length in feet, times width in feet, gives area in square feet, ft2. Volume is often found by multiplying three lengths, and thus has cubed units. For a box, length in centimeters, times width in centimeters, times depth in centimeters, gives volume in cubic centimeters, cm3. A length conversion factor must be used twice for squared area and three times for cubed volume.
Assume we find that the area of notebook page is 93.5 square inches and we want to convert that to square centimeters. Most often, you will only be able to find the relationship between inches and centimeters, 1 in = 2.54 cm. The conversion factor will need to be used twice here (so that both inches in in2 cancel and so that the two centimeters on top leave you with cm2).
$93.5 \; \cancel{\rm{in^2}} \times \dfrac{2.54 \;\rm{cm}}{1\; \cancel{\rm{in}}} \times \dfrac{2.54 \; \rm{cm}}{1 \; \cancel{\rm{in}}} = 603 \; \rm{cm^2}$
Example $$\PageIndex{6}$$
Tire pressure is often measured in pounds per square inch, abbreviated psi or lbs/in2. Convert 32.0 psi to units of kilograms per square centimeter.
SOLUTION
The pressures have weight units (lbs or kg) on top and area units (in2 or cm2) on bottom. We will use 1 kg = 2.20 lbs and 1 in = 2.54 cm to create our conversion factors.
$\dfrac{\cancel{32.0 \; \rm{lbs}}}{\cancel{\rm{in^2}}} \times \dfrac{1 \rm{kg}}{2.20 \; \cancel{\rm{lbs}}} \times \dfrac{\cancel{1 \; \rm{in}}}{2.54 \; \rm{cm}} \times \dfrac{\cancel{1 \; \rm{in}}}{2.54 \; \rm{cm}} = \dfrac{2.25 \: \rm{kg}}{\rm{cm^2}}$
Since 32.0 lbs/in2 had inches twice on the bottom, $$\mathrm{\frac{1\:in}{2.54\:cm}}$$ was used twice. The final answer is rounded to three significant figures.
Career Focus: Pharmacist
A pharmacist dispenses drugs that have been prescribed by a doctor. Although that may sound straightforward, pharmacists in the United States must hold a doctorate in pharmacy and be licensed by the state in which they work. Most pharmacy programs require four years of education in a specialty pharmacy school.
Pharmacists must know a lot of chemistry and biology so they can understand the effects that drugs (which are chemicals, after all) have on the body. Pharmacists can advise physicians on the selection, dosage, interactions, and side effects of drugs. They can also advise patients on the proper use of their medications, including when and how to take specific drugs properly. Pharmacists can be found in drugstores, hospitals, and other medical facilities.
Curiously, an outdated name for pharmacist is chemist, which was used when pharmacists formerly did a lot of drug preparation, or compounding. In modern times, pharmacists rarely compound their own drugs, but their knowledge of the sciences, including chemistry, helps them provide valuable services in support of everyone’s health.
## Concept Review Exercises
1. How do you determine which quantity in a conversion factor goes in the denominator of the fraction?
2. State the guidelines for determining significant figures when using a conversion factor.
1. The unit you want to cancel from the numerator goes in the denominator of the conversion factor.
2. Exact numbers that appear in many conversion factors do not affect the number of significant figures; otherwise, the normal rules of multiplication and division for significant figures apply.
## Key Takeaway
• A unit can be converted to another unit of the same type with a conversion factor.
There are plenty more conversion exercises to practice in Section 1.E at the end of this chapter.
## Contributors
• Anonymous
1.7: Converting Units is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. | 7,016 | 24,239 | {"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} | 4.15625 | 4 | CC-MAIN-2024-22 | latest | en | 0.3256 |
https://www.coderspoint.in/2019/05/area-of-triangle.html | 1,586,520,394,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371896913.98/warc/CC-MAIN-20200410110538-20200410141038-00451.warc.gz | 837,523,564 | 44,255 | Area of triangle - Coder's point
## Thursday, May 30, 2019
`````` # Python Program to find the area of triangle
a = 5
b = 6
c = 7
s = (a + b + c) / 2
#or take input from user
# Uncomment below to take inputs from the user
# a = float(input('Enter first side: '))
# b = float(input('Enter second side: '))
# c = float(input('Enter third side: '))
# calculate the area
area = (s*(s-a)*(s-b)*(s-c)) ** 0.5
print('The area of the triangle is %0.2f' %area)
``````
``OUTPUT:``
``The area of the triangle is 14.70`` | 173 | 510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-16 | latest | en | 0.750742 |
https://www.coursehero.com/file/42092447/Homework-3docx/ | 1,642,385,971,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300253.51/warc/CC-MAIN-20220117000754-20220117030754-00345.warc.gz | 779,634,993 | 51,066 | # Homework 3.docx - 1 The following table summarizes a data...
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1. The following table summarizes a data set with three attributes A, B. C and two class labels *, -. Build a two-level decision tree. A B C Number of Instances - + T T T 5 0 F T T 0 10 T F T 10 0 F F T 0 5 T T F 0 10 F T F 25 0 T F F 10 0 F F F 0 25 a. According to the classification error rate, which attribute would be chosen as the first splitting attribute? For each attribute, show the contingency table and the gains in classification error rate. b. Repeat for the two children of the root node. c. How many instances are misclassified by the resulting decision tree? d. Repeat parts (a), (b), and (c) using C as the splitting attribute. e. Use the results in parts (c) and (d) to conclude about the greedy nature of the decision tree induction algorithm. Answer:
- 15 35 E C=T =1-max (15/30, 15/30) =15/30 E C=F =1-max (35/70, 35/70) =35/70 c =E or -30/100 E c=T -70/100 E c=F =0/100=0 So, the algorithm chooses A because it has highest gain. | 346 | 1,143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.837894 |
http://www.jokyland.com/gifs/1221/oh-yes | 1,618,758,778,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038492417.61/warc/CC-MAIN-20210418133614-20210418163614-00472.warc.gz | 144,901,743 | 75,896 | # OH YES
Oh yes
## New jokes
Beans
A teacher asked her students to use the word "beans" in a sentence. "My father grows beans," said one girl. "My mother cooks beans," said a boy. A third student spoke up, "We are all human beans."
Cats exercise
Teacher: "If I gave you 2 cats and another 2 cats and another 2, how many would you have?" Johnny: "Seven." Teacher: "No, listen carefully... If I gave you two cats, and another two cats and another two, how many would you have?" Johnny: "Seven." Teacher: "Let me put it to you differently. If I gave you two apples, and another two apples and another two, how many would you have?" Johnny: "Six." Teacher: "Good. Now if I gave you two cats, and another two cats and another two, how many would you have?" Johnny: "Seven!" Teacher: "Johnny, where in the heck do you get seven from?!" Johnny: "Because I've already got a freaking cat!"
Tampons
"Mom, where do tampons go?" "Where the babies come from, darling." "In the stork?"
Pregnant woman
A 3 years old boy sits near a pregnant woman. Boy: Why do you look so fat? Pregnant woman: I have a baby inside me. Boy: Is it a good baby? Pregnant woman: Yes, it is a very good baby. Boy: Then why did you eat it?!
Value of pi
Teacher: What is the value of Pi? Student: Depending on what pie. Usually is \$12.99
Ghosts as liars
Q: Why are ghosts bad liars? A: You can see right through them.
Boss and employee
Boss: Do you believe in life after death? Employee: No, because there is no proof of it. Boss: Well there is now ! Employee: How? Boss: When you left yesterday saying that you have to go to your uncle's funeral, your uncle came here looking for you after you left
Kids conversation
Kid 1: "Hey, I bet you're still a virgin." Kid 2: "Yeah, I was a virgin until last night ." Kid 1: "As if." Kid 2: "Yeah, just ask your sister." Kid 1: "I don't have a sister." Kid 2: "You will in about nine months." | 517 | 1,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.963238 |
https://savvycalculator.com/uphill-force-calculator/ | 1,708,746,469,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474482.98/warc/CC-MAIN-20240224012912-20240224042912-00513.warc.gz | 525,065,780 | 42,836 | # Uphill Force Calculator
## About Uphill Force Calculator (Formula)
The Uphill Force Calculator is a tool used to calculate the force required to move an object up an inclined surface. It is commonly used in physics and engineering to determine the minimum force needed to push or pull an object up an incline.
The formula used to calculate the uphill force is
UF = mgcos(a) + mgsin(a)*u
where UF is the uphill force in Newtons (N), m is the mass of the object in kilograms (kg), g is the acceleration due to gravity in meters per second squared (m/s^2), a is the angle of the incline in degrees, and u is the coefficient of friction.
The formula takes into account two main factors that affect the uphill force required: the weight of the object and the angle of the incline. The weight of the object is represented by the first term in the formula (mgcos(a)), which is the force required to counteract the force of gravity pulling the object downward. The second term in the formula (mgsin(a)*u) represents the force required to overcome the resistance of the surface due to the coefficient of friction.
The Uphill Force Calculator can be used in a variety of settings, such as in transportation engineering to calculate the force required to move vehicles up steep grades, or in construction to determine the force needed to lift heavy objects up scaffolding or staircases.
Overall, the Uphill Force Calculator is a valuable tool for anyone who needs to calculate the minimum force required to move an object up an inclined surface, and the formula used provides a clear understanding of the physical forces at work in these types of situations. | 345 | 1,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-10 | latest | en | 0.911182 |
https://lists.osgeo.org/pipermail/postgis-users/2003-April/002384.html | 1,713,685,963,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.87/warc/CC-MAIN-20240421071342-20240421101342-00095.warc.gz | 333,220,965 | 2,791 | [postgis-users] point_inside_circle(geometry,float,float,float) : units -> meters ?
Pedro Salazar pedro-b-salazar at ptinovacao.pt
Tue Apr 15 07:28:42 PDT 2003
```David,
Indeed, your suggestion is a very clever ideia your, and I would like to
try it since it's a very flexible. However, there are a few points I
would like to clarify with you.
1) I didn't find any reference to the "expand" function in postgis
0.7.4. Is a new function on postgis CVS? What it does is just a creation
of box over my geometry? (in my case, is a point, so is very simple to
do it)
2) What is the SRID for measuring the distance between geometries in
meters? I'm using the SRID 4326 (WGS84). So, to what system should I
transform my geometries to get the distance in meters?
thanks,
Pedro Salazar.
> Here's a fairly general method:
>
> 1. Find a cartesian projection thats accurate for where your data is
> (ie. UTM zone whatever)
> 2. make sure this projection is properly entered in your spatial_ref_sys
> table. Most likely its already there if you uploaded the standard epsg
> projections.
> 3. Optional (for indexing) : find a maximum conversion factor between
> your data's units and meters. We;'ll be using this information to make
> a bounding box that's LARGER than the circle you want to enscribe
>
> SELECT * FROM table WHERE
> the_geom && expand( <GEOMETRY>, <BOX SIZE> )
> AND distance(transform(the_geom,<SRID>),
> transform(<GEOMETRY>,<SRID>) < <DIST>
>
> This is a bit more general than what you're asking for, but:
> < GEOMETRY > = point thats the centre of your circle
> <BOX SIZE> = this is the step #3 conversion factor (degrees)
> <SRID> = step #2's SRID in the spatial_ref_sys table
> <DIST> = distance (meters)
>
> This is a bit complicated, but:
>
> the_geom && expand( <GEOMETRY>, <BOX SIZE> )
> This uses the index to quickly find points that are nearby. Box size
> will depend on step #3.
> NB: make sure this box size is ALWAYS bigger than the distance you're
> looking for - dont
> worry if its too big. Its in the units of your source data (ie. degrees)
>
> distance(transform(the_geom,<SRID>), transform(<GEOMETRY>,<SRID>) <
> <DIST>
> This calculates the distance between the centre of your circle
> ("<GEOMETRY>") and the
> data in the table.
> transform( <geom>, <srid>) - re-projects <geom> to the projection
> defined by <srid>
>
> The expand() expression isnt really needed, but it will significantly
> speed this process up since you'll only be doing the expensive functions
> on a small portion of the data.
>
--
PS
pedro-b-salazar at ptinovacao.pt
PGP:0E129E31D803BC61
``` | 730 | 2,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-18 | latest | en | 0.908256 |
https://binaryoptionsoqnk.web.app/thoman54341joj/what-is-the-rate-of-return-on-bonds-1332.html | 1,643,196,575,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00665.warc.gz | 208,503,373 | 6,065 | ## What is the rate of return on bonds
28 Apr 2017 Low interest rates will have a dramatic impact on future returns. We investigate the impact this has on government bonds and show the power
Interest rates regularly fluctuate, making each reinvestment at the same rate virtually impossible. Thus, YTM and YTC are estimates only, and should be treated as 24 May 2019 What Does the RoR Tell You? RoR vs. Stocks and Bonds. Real vs. Nominal Rates of Return. 22 Jul 2019 Mutual funds, stocks, and bonds are three common types of securities that have both rates of return and yields. The formula for rate of return is:. 13 Nov 2018 A bond's return on investment or rate of return is also known as its yield. There are several different types of yield calculations. The most The yield on a bond is its return expressed as an annual percentage, affected in large part by the price the buyer pays for it. If the prevailing yield environment Over the long term, stocks do better. Since 1926, large stocks have returned an average of 10 % per year; long-term government bonds have returned between Suppose you buy a bond at a discount and hold it until it matures. Your rate of return includes the interest the bond earns plus additional profit you make, because
## 24 May 2019 When interest rates are low, the discount rate is inherently low, and the future expected return of all investments is low. Low bond yields do not
A rate of return is the gain or loss on an investment over a specified time period, expressed as a percentage of the investment’s cost. The bond's rate of return is roughly 7%. In a total return calculation, the compound interest, taxes and fees would have been factored in. The real rate of return on a bond is its annual nominal, or stated, return minus the annual rate of inflation. The Treasury uses the All-Urban Consumers Price Index to measure inflation. Add the interest earned to the price appreciation and divide it by the bond's price at the beginning of the year. In our example, that would be \$40 in interest plus \$30 in appreciation -- or \$70 -- divided by the beginning price of the bond -- \$1,000 -- for a 7 percent annual rate of return. Third, add the \$50 interest payment per year to the negative \$0.50 to get \$49.50. Next, divide \$49.50 by \$1,005, the average of \$1,010 and \$1,000, to get 0.0493. Finally, multiply 0.0493 by 100 to find your annual rate of return on the bond will be 4.93 percent. If you spend the \$30 you collect twice a year, you get \$1,000 back for your bond at the end of 30 years, and your total annual rate of return (ignoring taxes and inflation) is 6 percent simple interest. But now suppose that on each and every day that you collect those \$30 checks, you immediately reinvest them at the same coupon rate.
### 13 Dec 2018 Wishing you could score a 5% return or greater on your investments? You can also invest in high-yield bond portfolios provided you understand they Where a fixed-rate annuity promises a specific payment every month,
In contrast falling interest rates lead to above average short-term returns. Where negative returns occur from interest rate movements, they are “temporary” in Bonds are usually issued by large companies or governments, as a way to borrow Yield to Maturity, or YTM, measures a bond's rate of return when buying it at Interest Rate Risk. This is essentially the chance that interest rates will rise, and the bonds you own will be worth less in the future. As a bond's time Long-term bonds run inverse to interest rates. This means if interest rates rise, bond prices go down. So if you invest in a 20-year 13 Dec 2018 Wishing you could score a 5% return or greater on your investments? You can also invest in high-yield bond portfolios provided you understand they Where a fixed-rate annuity promises a specific payment every month, 5 Jul 2019 Calculating Bond Returns. There are a few ways to calculate how much return you are getting from your SGS bonds. Capital and Interest 21 May 2019 This government bond offers a higher rate of return than fixed deposits and even though the interest earned in the bonds is taxable, the bonds
### Long-term bonds run inverse to interest rates. This means if interest rates rise, bond prices go down. So if you invest in a 20-year
17 Apr 2019 Required rate of return is the minimum return in percentage that an model, the dividend discount model and the bond yield plus risk premium A rate of return is the gain or loss on an investment over a specified time period, expressed as a percentage of the investment’s cost. The bond's rate of return is roughly 7%. In a total return calculation, the compound interest, taxes and fees would have been factored in.
## In contrast falling interest rates lead to above average short-term returns. Where negative returns occur from interest rate movements, they are “temporary” in
The yield on a bond is its return expressed as an annual percentage, affected in large part by the price the buyer pays for it. If the prevailing yield environment
1 Jan 2020 In a rising rate environment, existing bonds lose their allure because investors can get a higher return from newly issued bonds. If you try to sell | 1,163 | 5,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-05 | latest | en | 0.951067 |
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# Tutor profile: Esta B.
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Esta B.
Professional Educator Specializing in Struggling Students
Tutor Satisfaction Guarantee
## Questions
### Subject:Pre-Algebra
TutorMe
Question:
What is the Greatest Common Factor (GCF) of 24 and 36?
Inactive
Esta B.
First, you have to find all the factors of each number: The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24 The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36 What factors do both numbers have?: 1, 2, 3, 4, 6, 12 12 is the greatest number that goes into both numbers evenly, and so is the GCF.
### Subject:Basic Math
TutorMe
Question:
If you have 23 children and 3 adult chaperones to go on a field trip, and each bus can hold 5 people, how many buses do you need?
Inactive
Esta B.
Well, you have 26 total people, including children and adults. Each bus holds 5 people. You can divide 26 by 5 and you get 5 with 1 left over. So, do you need 5 buses or 6 buses? Unless you want to be the one left standing in the parking lot, you need to get 6 buses. Of course, the people would be divided up so that 1 person does not ride alone. If you cannot remember how to divide 26 by 5, you can count by 5s: 5, 10, 15, 20, 25 and oops, 1 more. The solution would have to be the same.
### Subject:Algebra
TutorMe
Question:
Solve the equation: -3(-x + 5) + 20 = -10(x - 3) + 4
Inactive
Esta B.
1) First, use the distributive property to eliminate the first set of parentheses on the left side of the equation: -3(-x + 5) = 3x -15 2) Replace the solution into the original equation: It now looks like this: 3x - 15 + 20 = -10(x - 3) + 4 3) Now, use the distributive property to eliminate the second set of parentheses on the right side of the equation: -10(x - 3) = -10x + 30 4) Replace that solution into the original equation: It now looks like this: 3x - 15 + 20 = -10x + 30 + 4 5) Let's combine the like terms on both sides: the left side: -15 + 20 = 5: on the right side: 30 + 4 = 34 6) Again, let's replace those solutions into the original equation: 3x + 5 = -10x + 34 7) Okay, now it is time to get the x terms on one side; we can do that by adding 10x to both sides: so 13x + 5 = 34 8) If we subtract 5 from both sides we get: 13x = 29 9) To get the value of just one x, we need to divide both sides by 13: so x = 29/13 In person, it would be easier to show this process with arrows, etc. It is difficult to show the steps in a visual sequence.
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# Problem 43148. Basic commands - rounding
Solution 1204846
Submitted on 5 Jun 2017 by Said BOUREZG
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
x = [-2.5 2.5]; y_correct = [-2 2]; assert(isequal(round20(x),y_correct))
y = -2 2
2 Pass
x = [-8.3 0.01 7.9]; y_correct = [-8 0 7]; assert(isequal(round20(x),y_correct))
y = -8 0 7 | 159 | 462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-24 | latest | en | 0.682052 |
http://www.cs.nyu.edu/pipermail/fom/1997-December/000452.html | 1,501,030,186,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425737.60/warc/CC-MAIN-20170726002333-20170726022333-00067.warc.gz | 402,101,190 | 3,906 | # FOM: Meaning vs Significance Nailed Down, Work for Logic and Rev.Math.
Robert S Tragesser RTragesser at compuserve.com
Wed Dec 10 02:51:21 EST 1997
```MEANING AND SIGNIFICABCE SORT OF NAILED DOWN
Contents: Making exact the distinction between meaning and significance by
following out the distinction between elementary and nonelementary proofs
of theorems, and the bearing of this on CH?.
An open characterizational problem for logic (and a problem set for
Reverse Mathematics?).
How significance can force a change of meaning.
Since Neil Tennant has found useful the distinction between the
meaning and significance of a theorem/problem, and since 'significance' is
as wildly wobbly a signifier as 'meaning', it does seem worthwhile to
point to some examples from a category of mathematical activity which,
borne in mind, would serve to (ostensively) fix (by example) the sense of
both 'meaning' and 'significance', and allow us to articulate the way in
which the question of the significance of CH is anomalous.
Proofs of difficult theorems are often first attained by
non-elementary means, and then the task is to discover an elementary
proof. "Elementary" does not of course mean simple. It means roughly:
proved by means native to the subject with which the statement of the
theorem has greatest affinity, e.g., an elementary proof that there are
infinitely many prime should involve only concepts whose "elements" are
native to the concept "prime natural number".
Example 1. Euclid's proof that there are infinitely many prime numbers is
elementary while Euler's proof (a seed proof for analytic number theory) is
not.
Example 2. (nonelementary) Proof using Pythagorean geometric-figurate
arrays that the sum of the first n positive integers is n(n+1)/2, whereas
an elementary proof would be by mathematical induction [someone asked:
unless one believes that the natural numbers just are fuigural point
arrays?].
Example 3. The Prime Number Theorem (pi(x) squiggle x/logx). The first
proofs were nonelementary (Hadamard/Poussin) because employing the theory
of functions of a complex variable. Erdos/Selberg gave an elementary
proof ("uses only elementary estimates of the relative magnitudes of
primes, relying only on cocepts intrinsic to the conception of prime
number").
Example 4. The Fundamental Theorem of Algebra. It has not (as far as I
know) been given a purely algebraic proof. Here there is some ambiguity
about "elementary proof". Would only a purely algebraic proof be
elementary? Some say not because analytical ideas are inherent to the
sense of the theorem -- "because the field R, and consequently the
extension field C, is a construct belonging to analysis".
Example 5. (Fundamental Theorem continued) Hopf's Theorem &
Gelfand-Mazur theorem.
One way to explore the possibility of an elementary proof might be
to look for more abstract theorems which have the Fundamental Theorem of
Algebra as a consequence. E.g., Hopf's theorem (Every finite dimensional
real commutative division algebra A = (V,.) is at most 2-dimensional).
No elementary proof seems to be known.--a proof involving topological
mappings of projective spaces P(n) into sphere S(n); a proof from
algebraic geometry (I don't know where).
Gelfand-Mazur Theorem: Every commutative banach division algebra is
isomorphic to field R or C. Apparently GM is a consequence of Hopf; no
elementary proof of GM is apparently known (one apparently popular proof
involves power series whose terms are also power series).
Both Hopf and G-M entail the Fundamental Theorem of Algebra. Are
they somehow part of the deep meaning of the FTA?
SIGNIFICANCE & MEANING: I want to say (by way of anchoring those two
terms): a (proposed) theorem has SIGNIFICANCE beyond MEANING insofar as a
nonelemetary proof (or disproof) of it is given. A (proposed) theorem is
weak in MEANING insofar as an elementary proof (or disproof) of it cannot
be given.
It might be, then, that CH is weak in meaning, where elementary
proof (or disproof) means proof (or disproof) in pure, transfinite set
theory. There are two scenarios WHICH SUGGEST THAT MEANING AND
SIGNIFICANCE CAN VARY (AS IT WERE) INDEPENDENTLY OF ONE ANOTHER:
(1) CH can only be given a nonelemetary proof or disproof (as for
example by appeal to some sort of geometric insight -- perhaps this is what
Goedel had in mind when he spoke of a spatial aspects of sets).
(2) CH has essentially (say) geometric significance (having to do
with intuitive continuity and measure), but that its meaning (in pure
transfinite set theory) is out of wack because pure transfinite set theory
as currently conceived is not right for capturing/representing continuity.
Clearly, Brouwer thought that (2) is the case [AND NOT BECAUSE HE
WAS AN ANTI-REALIST -- AS I ARGUED ABOVE, IN THE END, B'S SUPPOSED
ANTI-REALISM IS OF NO IMPORTANCE WHATSOEVER TO THE EMERGENT REPRESENTATION
OF FLUIDIC CONTINUA!].
Can a similar situation ever occur for a proposed theorem of the
sorts that occur in the five examples? That the SIGNIFICANCE OF A THEOREM
CAN CAUSE US TO REVISE THE TERMS OF ITS STATEMENT IN SUCH A FUNDAMENTAL
WAY, isn't that interesting about set theory???
Could other mathematical theories behave in this way?
Well, isn't this what happened to the Eulerian infinitesimal
calculus -- the significance of the theorems forced a fundamental, radical
change in the terms in which they were cast (fundamentally altering their
meaning, but clarifying their significance)??????????
``` | 1,369 | 5,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-30 | latest | en | 0.944575 |
http://mathhelpforum.com/calculus/110897-question-notation-partial-derivatives-print.html | 1,511,003,638,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804724.3/warc/CC-MAIN-20171118094746-20171118114746-00564.warc.gz | 189,086,033 | 3,008 | question on notation for partial derivatives
• Oct 27th 2009, 05:28 PM
oblixps
question on notation for partial derivatives
what does it mean when a partial derivative has a prime on it? usually i just see partial derivatives as f and a subscript x but i recently came across f with a subscript x and a prime on the top. just curious, what does that mean?
• Oct 27th 2009, 05:45 PM
Quote:
Originally Posted by oblixps
what does it mean when a partial derivative has a prime on it? usually i just see partial derivatives as f and a subscript x but i recently came across f with a subscript x and a prime on the top. just curious, what does that mean?
Do you mean like this $f'_x$
?
• Oct 27th 2009, 10:37 PM
oblixps
Quote:
Do you mean like this $f'_x$
yes, sorry if my description was confusing. can you tell me what $f'_x$ means? | 223 | 834 | {"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": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-47 | longest | en | 0.96311 |
https://www.isnt.org.in/get-what-week-of-the-month-c-with-code-examples.html | 1,713,485,472,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817249.26/warc/CC-MAIN-20240418222029-20240419012029-00298.warc.gz | 720,633,177 | 50,763 | # Get What Week Of The Month C# With Code Examples
In this article, we will look at how to get the solution for the problem, Get What Week Of The Month C# With Code Examples
## Which week of the month is today?
The current Week Number is WN 41.
``````static int GetWeekNumberOfMonth(DateTime date)
{
date = date.Date;
DateTime firstMonthDay = new DateTime(date.Year, date.Month, 1);
DateTime firstMonthMonday = firstMonthDay.AddDays((DayOfWeek.Monday + 7 - firstMonthDay.DayOfWeek) % 7);
if (firstMonthMonday > date)
{
firstMonthMonday = firstMonthDay.AddDays((DayOfWeek.Monday + 7 - firstMonthDay.DayOfWeek) % 7);
}
return (date - firstMonthMonday).Days / 7 + 1;
}```
```
## How do you find the week of the month?
We can easily count the number of weeks in a month by first counting the number of days in the month. Then, we divide the number of days by 7 since 1 week has 7 days. For example, the month of January has 31 days, so the number of weeks in January would be: 31/7 = 4 weeks + 3 days.
## How do you find the week number from a date?
Get week number from date
• Generic formula. =WEEKNUM(date)
• To get the week number from a date, you can use the WEEKNUM function.
• The WEEKNUM function takes a date and returns a week number (1-54) that corresponds to the week of year.
• Week numbers in Excel (Ron de Bruin)
## How do I get the current week number in VB net?
You can retrieve the current week number in Schedule control by using the GetWeekOfYear method. In this method, define the CalendarWeekRule class and DayOfWeek class. The CalendarWeekRule class is used to get the first week of the year and the DayOfWeek class is used to get the first day of the week.
## What week of the year is it ISO?
The ISO Week Date Calendar. The Current ISO Week Date is: 2022-W39-3. It is week 39 of year 2022, day 3 of the week, and day 271 of the year.
## How do I get the current week number in C#?
Get Week Number of a Year in C#
• class Program.
• {
• static void Main(string[] args)
• {
• Console.WriteLine("Enter Date");
• string date = Console.ReadLine();
• DateTime inputDate = DateTime.Parse(date.Trim());
• var d = inputDate;
## How do you get the current week number in typescript?
Use the getFullYear() and getDay() Functions Along With the new Date() 's Object to Get the Week Number of the Current Date. This method uses the new Date() constructor and its object along with the functions like getFullYear() and getDay() to get the current week number of the year.
## Which number of week is this?
Week 41. Week 41 is from Monday, October 10, 2022 until (and including) Sunday, October 16, 2022.
## What week of the year is it 2022?
What week of the year is it? It is currently week 41 in 2022. There are 11 weeks remaining.
## How do I get this week in Javascript?
var today = new Date(); var startDay = 0; var weekStart = new Date(today. getDate() - (7 + today. getDay() - startDay) % 7); var weekEnd = new Date(today. getDate() + (7 - today.
## How To Filter Words That Contain Atleast 2 Vowels From A Series With Code Examples
In this article, we will look at how to get the solution for the problem, How To Filter Words That Contain Atleast 2 Vowels From A Series With Code Examples How do you find the common elements in two Series pandas? Algorithm : Import the Pandas and NumPy modules. Create 2 Pandas Series . Find the union of the series using the union1d() method. Find the intersection of the series using the intersect1d() method. Find the difference between the union and the intersection elements. Print the result
## Laravel Check Pagination In Blade With Code Examples
In this article, we will look at how to get the solution for the problem, Laravel Check Pagination In Blade With Code Examples How do you use pagination? Good Practices Of Pagination Design # Provide large clickable areas. Don't use underlines. Identify the current page. Space out page links. Provide Previous and Next links. Use First and Last links (where applicable) Put First and Last links on the outside. @if (\$items->hasPages()) <div class="pagination-wrapper"> {{ \$items->links() }
## My Canvas Java With Code Examples
In this article, we will look at how to get the solution for the problem, My Canvas Java With Code Examples What is awt in Java? Java AWT (Abstract Window Toolkit) is an API to develop GUI or window-based applications in java. Java AWT components are platform-dependent i.e. components are displayed according to the view of operating system. AWT is heavyweight i.e. its components are using the resources of OS. public class Canvas extends Component implements Accessible How do I run a Java gr
## How To Upload To Pypi With Same Name With Code Examples
In this article, we will look at how to get the solution for the problem, How To Upload To Pypi With Same Name With Code Examples How do you create a README? Create a ReadMe File Create a file named README.md in the root (based) folder of the Git repo. Add any instructions or documentation that you want to share with others. Use Markdown to format headings, lists, links, etc. When done, commit the changes and push them to the remote repo. You need to change the version in setup.cfg or setup.p
## Css Style Nth Child Beyond Certain Number With Code Examples
In this article, we will look at how to get the solution for the problem, Css Style Nth Child Beyond Certain Number With Code Examples How do I select all 3 children in CSS? formula (an + b) In addition to the value n being able to be any number, you can also use a formula. nth-child(3n) would affect every third child element. nth-child(3n+1) would apply to every third element starting from the first one. li:nth-child(n+6) { color: green; } Does nth child start at 0 or 1? Functional notatio | 1,404 | 5,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-18 | latest | en | 0.851486 |
https://pixalu.com/lamaline/nearest-tenth-of-a-meter-example.php | 1,726,865,743,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00514.warc.gz | 419,944,980 | 9,861 | What is 26.149 meters rounded to the nearest tenth of a Example: Stow-It Storage PROBLEM SOLVER Connection 8 1.5 ft 36 in 14 in A C E P A C KI N G 3 4. 167 the nearest tenth square meter? (Hint: Find the area of
## UPPER AND LOWER BOUNDS Milford Haven School
Exploring Square Roots and Irrational Numbers Flashcards. MPM2D Problems Involving Two Right Triangles Name: Example 1: Find the length of AD, to the nearest tenth of a meter. 18m 500 Example 2: Find the length ofGH, to the, What is one tenth of a meter? 12/11/2013 5 Comments 5 Comments one tenth of a meter is a decimeter da Reply. Owen. 12/3/2015 10:42:30 am. decimeter. Reply..
Valerie drives 500 meters up a hill that makes an angle of 15В° with the horizontal. To the nearest tenth of a meter, what horizontal distance has she covered? Valerie drives 500 meters up a hill that makes an angle of 15В° with the horizontal. To the nearest tenth of a meter, what horizontal distance has she covered?
The Metric System, Measurements, and Scientific Inquiry as possible to the nearest tenth of a • Your height to the nearest hundredth of a meter. 2/08/2018 · How to Measure Centimeters. Find the length to the nearest tenth of a centimeter. For example, to convert 5.1 meters to centimeters,
The Metric System, Measurements, and Scientific Inquiry as possible to the nearest tenth of a • Your height to the nearest hundredth of a meter. Example: a fence is measured as 12.5 meters long, Measuring to the nearest meter means the true value could be up Each measurement could possibly be the
For Example: 45.8 rounded to the nearest tenth is 45.9 because 8 is above 5. If it's 78.4 feet, then the nearest tenth of a meter is 23.9meters. 2/08/2018В В· How to Measure Centimeters. Find the length to the nearest tenth of a centimeter. For example, to convert 5.1 meters to centimeters,
How Are Decimals Rounded to the Nearest Cent? rounding to the nearest cent would require finding the nearest hundredth. For example, these tiny lengths are also called millimeters because there are 1000 of them in a meter. nearest millimeter or tenth-centimeter. 6G_lesson_millimeter.indd
Nearest tenth of a meter. Scale 1 measures to the nearest thousandth of a pound. Example 2 : Conversions Precision and Accuracy 11/01/2007В В· When they say to round to the nearest metre do they mean to the If they wanted you to round to the nearest ten for example, A tenth of a meter is a
3 2 a. 12 b. 6 c. 5 d. 18 To the nearest tenth of a meter, how far above TOP: 8-3 Example 2 KEY: side length using tangent tangent To the nearest tenth of a second how long does it take the policeman to catch the speeder? 10 : How far to the nearest tenth of a meter can a runner running at m/s
Those are are all examples of length measurements. Example: This fork is 20 centimeters long . The length of this guitar is about 1 meter. This rounding calculator computes the round of numbers with decimals (from 1 up to 9 decimals) thousands, hundreds, Rounded to nearest tenth;
Online calculator for rounding numbers. Rounding to the nearest tenth is 838.3; For example, if you want to round to the nearest ten you'd look at the ones place. 29/08/2011В В· Rounding to the Nearest Ten tenframe. Loading... Unsubscribe from tenframe? Cancel Unsubscribe. Working... Subscribe Subscribed Unsubscribe 3.3K. Loading
Rounding decimal numbers is a good strategy to use See Rounding Whole Numbers for an example. Rounding to the Nearest Whole Rounding to the Nearest Tenth Rounding decimal numbers is a good strategy to use See Rounding Whole Numbers for an example. Rounding to the Nearest Whole Rounding to the Nearest Tenth
Nearest tenth of a meter. Scale 1 measures to the nearest thousandth of a pound. Example 2 : Conversions Precision and Accuracy Unit I Reading: Significant Figures and then only to the nearest tenth of a meter, or as 0.3 meter As an example,
### Exploring Square Roots and Irrational Numbers Flashcards
Area and Perimeter Unit Review Mendham Borough Schools. Simple definition and examples. Howto find the greatest possible error in easy steps. the GPE is 1/2 of tenth of a meter (that’s 1/2 * 1/10 m = 0.05 m)., How Are Decimals Rounded to the Nearest Cent? rounding to the nearest cent would require finding the nearest hundredth. For example,.
### Valerie drives 500 meters up a hill that makes an angle of
6G lesson millimeter DeepMath. Round to Nearest Multiple Calculator. 10, 50, etc. You can also round to the nearest tenth, hundredth, for example, "Wrong Calculation these tiny lengths are also called millimeters because there are 1000 of them in a meter. nearest millimeter or tenth-centimeter. 6G_lesson_millimeter.indd.
• PROBLEM SOLVER Connection KET
• Conversions Precision and Accuracy PBworks
• Valerie drives 500 meters up a hill that makes an angle of
• 9/08/2010В В· Rounding. Oh my....there will be so many times when rounding numbers (including decimals) will be a required skill. Let's get this down, shall we? Practice Unit Conversions In a previous example, 1 meter 3.281 ft Example 4 In Canada a speed limit is posted as 100 Round your answer to the nearest tenth of a pound.
Find to the nearest tenth of a meterВё the length of area of this work space to the nearest tenth of a square foot. Example 4: Trig Apps Classwork Day 5: Measurement of length, Mass, Volume When you make a measurement using a meter stick The electric scale can measure the mass of objects to the nearest tenth of
What is 26.149 meters rounded to the nearest tenth of a meter Get the answers you need, now! Online calculator for rounding numbers. Rounding to the nearest tenth is 838.3; For example, if you want to round to the nearest ten you'd look at the ones place.
24/01/2009В В· these measurements are in centimeters 76.2 58.42 i need both converted into the nearest tenth and hundredths of a meter The regular polygon has side 9 m. How do you find each angle measure to the nearest tenth of a degree, each linear measure to the nearest tenth of a meter, and the
Rounding and Estimating Numbers Questions including "What is 8999 rounded to the nearest For example, rounding the number 7.93 rounded to the nearest tenth This rounding calculator computes the round of numbers with decimals (from 1 up to 9 decimals) thousands, hundreds, Rounded to nearest tenth;
7.8199 rounded to the nearest tenth is 7.8; Rounding can make sums easy. For example, at a grocery store you might pick up items with the following prices: Simple definition and examples. Howto find the greatest possible error in easy steps. the GPE is 1/2 of tenth of a meter (that’s 1/2 * 1/10 m = 0.05 m).
24/01/2009В В· these measurements are in centimeters 76.2 58.42 i need both converted into the nearest tenth and hundredths of a meter What is 80.945 rounded to the nearest tenth? Ans. 80.9. Example 2. What is 18.386 rounded to the nearest hundredth? Ans. 18.39. B) To round to nearest one, ten,
To the nearest tenth of a kilometer, what is the actual distance corresponding to the map distance? Sample GED Math Problem Unit I Reading: Significant Figures and then only to the nearest tenth of a meter, or as 0.3 meter As an example,
For example, you can use the Round your answer to the nearest tenth of a meter per second. Answers. The following are the answers to the practice questions: 50 J. Leave your answer in simplest radical form. measure to the nearest tenth of a meter, 10-1 Example 3 KEY: triangle
2/08/2018В В· How to Measure Centimeters. Find the length to the nearest tenth of a centimeter. For example, to convert 5.1 meters to centimeters, Leave your answer in simplest radical form. measure to the nearest tenth of a meter, 10-1 Example 3 KEY: triangle
What is the horizontal distance, to the nearest tenth of a meter, from the face of the cliff after seconds? 8 : In the previous problem, For example, you can calculate how much torque is produced by opening a jar of pickles. Round your answer to the nearest tenth of a newton-meter.
12 Markov Chains: Introduction Example 12.1. is the transition probability to x + 1, Call the transition matrix P and temporarily denote the n-step transition Transition probability matrix example problems Leoville the process then makes a transition into state jaccording to transition probability P by a transition matrix P Markov chain For example, if S= f0;1
## Valerie drives 500 meters up a hill that makes an angle of
Viola drives 170 meters up a hill that makes an angle of. 24/01/2009 · these measurements are in centimeters 76.2 58.42 i need both converted into the nearest tenth and hundredths of a meter, Simple definition and examples. Howto find the greatest possible error in easy steps. the GPE is 1/2 of tenth of a meter (that’s 1/2 * 1/10 m = 0.05 m)..
### Chapter_3
How to Measure Centimeters wikiHow. The regular polygon has side 9 m. How do you find each angle measure to the nearest tenth of a degree, each linear measure to the nearest tenth of a meter, and the, Online calculator for rounding numbers. Rounding to the nearest tenth is 838.3; For example, if you want to round to the nearest ten you'd look at the ones place..
For example, you can calculate how much torque is produced by opening a jar of pickles. Round your answer to the nearest tenth of a newton-meter. Unit Conversions In a previous example, 1 meter 3.281 ft Example 4 In Canada a speed limit is posted as 100 Round your answer to the nearest tenth of a pound.
10/01/2012В В· What does it mean to round to the nearest cubic unit for example if you got the Would the number 0.1967 still be considered one tenth of a The regular polygon has side 9 m. How do you find each angle measure to the nearest tenth of a degree, each linear measure to the nearest tenth of a meter, and the
For example, one meter is approximately 39.37 inches (39.37 in); Tenth Revision) The International Classification of Diseases, Tenth Edition (ICD-10), 4/09/2018В В· Steps. Part 1. Rounding to the For example, 4.03 rounded to the nearest tenth is 4.0. To round to the nearest tenth, write down the number with a decimal
Unit Conversions In a previous example, 1 meter 3.281 ft Example 4 In Canada a speed limit is posted as 100 Round your answer to the nearest tenth of a pound. Recording Measurements. For example, suppose that you You are really measuring more accurately than to the nearest tenth of a centimetre.
GEOMETRY AND MEASUREMENT Round 9 pounds, 6 ounces to the nearest tenth of a pound 4. example, to convert from km to meters, we 11/01/2007В В· When they say to round to the nearest metre do they mean to the If they wanted you to round to the nearest ten for example, A tenth of a meter is a
Round to Nearest Multiple Calculator. 10, 50, etc. You can also round to the nearest tenth, hundredth, for example, "Wrong Calculation For example, you can calculate how much torque is produced by opening a jar of pickles. Round your answer to the nearest tenth of a newton-meter.
What is the horizontal distance, to the nearest tenth of a meter, from the face of the cliff after seconds? 8 : In the previous problem, Rounding and Estimating Numbers Questions including "What is 8999 rounded to the nearest For example, rounding the number 7.93 rounded to the nearest tenth
For example, you can calculate how much torque is produced by opening a jar of pickles. Round your answer to the nearest tenth of a newton-meter. For example, you can use the Round your answer to the nearest tenth of a meter per second. Answers. The following are the answers to the practice questions: 50 J.
10/01/2012В В· What does it mean to round to the nearest cubic unit for example if you got the Would the number 0.1967 still be considered one tenth of a Example: Stow-It Storage PROBLEM SOLVER Connection 8 1.5 ft 36 in 14 in A C E P A C KI N G 3 4. 167 the nearest tenth square meter? (Hint: Find the area of
Round to Nearest Multiple Calculator. 10, 50, etc. You can also round to the nearest tenth, hundredth, for example, "Wrong Calculation Nearest tenth of a meter. Scale 1 measures to the nearest thousandth of a pound. Example 2 : Conversions Precision and Accuracy
Chapter_3. What is 26.149 meters rounded to the nearest tenth of a meter Get the answers you need, now!, Solve We round the given amounts to the nearest ten dollars A decimeter is one tenth of a meter. Decimal Parts of a Meter.
### To the nearest tenth of a kilometer what is the actual
LAB 2 – The Metric System. Find out how to round to the nearest integer or any other Magoosh GRE Blog Everything you need to In your example, 2.25( to the nearest tenth ) rounds off, For example, you can use the Round your answer to the nearest tenth of a meter per second. Answers. The following are the answers to the practice questions: 50 J..
Area and Perimeter Unit Review Mendham Borough Schools. This rounding calculator computes the round of numbers with decimals (from 1 up to 9 decimals) thousands, hundreds, Rounded to nearest tenth;, Round your answer to the nearest tenth. Section 7.2 Volumes of Cylinders 307 EXAMPLE 3 Real-Life A cylindrical water tower has a diameter of 15 meters and a.
### Chapter 8 Julia Kirby
Torque in Physics Problems dummies. Rounding decimal numbers is a good strategy to use See Rounding Whole Numbers for an example. Rounding to the Nearest Whole Rounding to the Nearest Tenth Round to Nearest Multiple Calculator. 10, 50, etc. You can also round to the nearest tenth, hundredth, for example, "Wrong Calculation.
Example: Write a similarity Round to the nearest tenth of a meter. 8.5 Law of Sines: Round lengths to the nearest tenth and angle measures to the nearest degree. What is 80.945 rounded to the nearest tenth? Ans. 80.9. Example 2. What is 18.386 rounded to the nearest hundredth? Ans. 18.39. B) To round to nearest one, ten,
29/08/2011В В· Rounding to the Nearest Ten tenframe. Loading... Unsubscribe from tenframe? Cancel Unsubscribe. Working... Subscribe Subscribed Unsubscribe 3.3K. Loading For example, you can calculate how much torque is produced by opening a jar of pickles. Round your answer to the nearest tenth of a newton-meter.
Unit I Reading: Significant Figures and then only to the nearest tenth of a meter, or as 0.3 meter As an example, How Are Decimals Rounded to the Nearest Cent? rounding to the nearest cent would require finding the nearest hundredth. For example,
Start studying geo final. Learn vocabulary, Round your answer to the nearest tenth of a mile. B. find the length of the kite string to the nearest meter. A)35. For example, you can use the Round your answer to the nearest tenth of a meter per second. Answers. The following are the answers to the practice questions: 50 J.
24/01/2009В В· these measurements are in centimeters 76.2 58.42 i need both converted into the nearest tenth and hundredths of a meter Practical use of Trigonometry. Thus the building is 40 meters high and is situated 34В·64 meters away from the tower. Example. (to the nearest meter).
these tiny lengths are also called millimeters because there are 1000 of them in a meter. nearest millimeter or tenth-centimeter. 6G_lesson_millimeter.indd Online calculator for rounding numbers. Rounding to the nearest tenth is 838.3; For example, if you want to round to the nearest ten you'd look at the ones place.
What is 26.149 meters rounded to the nearest tenth of a meter Get the answers you need, now! Example: a fence is measured as 12.5 meters long, Measuring to the nearest meter means the true value could be up Each measurement could possibly be the
29/08/2011В В· Rounding to the Nearest Ten tenframe. Loading... Unsubscribe from tenframe? Cancel Unsubscribe. Working... Subscribe Subscribed Unsubscribe 3.3K. Loading The regular polygon has side 9 m. How do you find each angle measure to the nearest tenth of a degree, each linear measure to the nearest tenth of a meter, and the
Leave your answer in simplest radical form. measure to the nearest tenth of a meter, 10-1 Example 3 KEY: triangle Rounding and Estimating Numbers Questions including "What is 8999 rounded to the nearest For example, rounding the number 7.93 rounded to the nearest tenth
Online calculator for rounding numbers. Rounding to the nearest tenth is 838.3; For example, if you want to round to the nearest ten you'd look at the ones place. LAB 2 – The Metric System . of the lab to the back window the nearest meter (m of a single amoeba to the nearest tenth of a millimeter. For example,
For Example: 45.8 rounded to the nearest tenth is 45.9 because 8 is above 5. If it's 78.4 feet, then the nearest tenth of a meter is 23.9meters. SWR meter. (See "SWR Meter Hook Release the microphone switch and write this value down to the nearest tenth of a point. Note: For example, removing two wire | 4,166 | 16,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.898327 |
https://psychocod3r.wordpress.com/2020/10/06/libdfloat-a-c-library-for-exact-representation-of-decimal-floating-point-numbers/ | 1,669,687,108,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710684.84/warc/CC-MAIN-20221128235805-20221129025805-00207.warc.gz | 533,660,943 | 40,338 | # libdfloat: A C Library for Exact Representation of Decimal Floating Point Numbers
Guys, something really awesome just happened!… I actually finished one of my large-scale coding projects! I’d like to introduce you to libdfloat, a C library for representing decimal numbers without any rounding errors.
I started this project as an offshoot of my CSV library when I realized that there was a need for a mechanism for reading data from a CSV file, processing it, and then writing it back as it was before. To illustrate this need, let’s say you have the number 1.2 written to a CSV record. A data analysis program reads the CSV table into a table structure using `atof()` and `strcpy()`, as was shown in this article from last week, and it processes the data in the table without modifying that one field. So when it writes the data back, it should write back 1.2, right? Problem is, 1.2 can’t be represented exactly in binary notation, so what will happen is the number will get truncated when it is converted from decimal to binary and then it will get truncated again when it is converted back to decimal. We want a guarantee that we won’t lose precision when we read the numbers from and write the numbers back to the CSV file, and if we are converting between binary and decimal formats, we don’t have this guarantee. So we have to implement a new type that allows decimal numbers to be represented exactly within the program.
The way I see it, there are basically two ways to implement a decimal floating point type: One is to use binary-coded decimal format (BCD) where each decimal digit is represented by four bits of a byte. I realized this would be rather slow and cumbersome as digits would have to be packed and unpacked from the containing bytes, and I would also have to completely re-implement all arithmetic operations from the ground up. So I discarded this idea and instead used a solution that is halfway between the extremes of BCD and pure binary: I used a mantissa and exponent that are both technically represented in pure binary, but synthesized the decimal operations on these magnitudes using precise calculations with exponents and logarithms.
I have specified four types for use in decimal floating point arithmetic, based on four different widths. They are defined as follows:
``````
#include <stdint.h>
typedef struct {
int8_t mantissa;
int8_t exponent;
} dfloat16_t;
typedef struct {
int16_t mantissa;
int16_t exponent;
} dfloat32_t;
typedef struct {
int32_t mantissa;
int32_t exponent;
} dfloat64_t;
typedef struct {
int64_t mantissa;
int64_t exponent;
} dfloat128_t;
``````
This decimal floating point type has basically the same internal implementation as binary floating point types in the IEEE-754 standard, which specifies four components: mantissa, exponent, sign of the mantissa, and sign of the exponent. In my implementation, the mantissa and mantissa sign are combined into one value, as are the exponent and exponent sign, simply by using signed values for the mantissa and exponent fields. The mantissa is just a binary representation of the number represented by all the decimal digits of the number from the most significant to the least significant figures, multiplied by such an exponent that there are no nonzero digits after the decimal point as well as no trailing zeroes before the decimal point. Put more succinctly, the exponent is such that the mantissa is an integer that is not a multiple of 10. The exponent gives the power of 10 you would have to multiply the mantissa by to get the number represented by the `dfloat` type. This may all seem rather convoluted on paper, but it’s actually the simplest way to express a decimal floating point number exactly, since all we need are two signed integers of the same width.
Since the goal here is to be able to read and write floating point numbers to/from text files, the initialization of the corresponding variables should be done from a text string, and there is little need to declare these variables directly. In fact doing so wouldn’t make sense because there is no way to specify a new literal format for a user-defined type except by using strings. So the first thing to do is create a decimal equivalent of the `atof()` function.
The function definition that follows looks rather odd because it’s actually a macro for four function definitions (one for each width/type). It uses the arguments `small` and `big`, representing the size of the two `dfloat` components and the size of the `dfloat` itself, respectively. These arguments are used to generate the symbols for the corresponding `dfloat` types and fixed-width integer types through token-pasting – a C preprocessor feature that is not used that much, but that I’ve now found a good niche use for, employing it heavily in the implementation of this library. As for the function itself, there’s nothing particularly remarkable about its implementation, as it simply reads the number in the most common-sense fashion that occurred to me. (From now on I will simply refer to a group of functions defined by one of these macros as a “function”, with the understanding that this “function” in fact encompasses several functions.)
``````
#define dfloatN_atof( small, big )\
dfloat ## big ## _t *dfloat ## big ## _atof( char *src ){\
int i, len;\
int point = 0;\
char *cpy;\
dfloat ## big ## _t *dst;\
len = strlen( src );\
cpy = (char *) malloc( len+1 );\
strncpy( cpy, src, len+1 );\
dst = (dfloat ## big ## _t *) malloc( sizeof( dfloat ## big ## _t ) );\
dst->exponent = 0;\
\
/* Search for decimal point: */\
for( i = 0; i < len; i++ ){\
if( cpy[i] == '.' ) point = i;\
}\
\
/* Determine exponent: */\
i = len;\
while( cpy[--i] == '0' );\
if( !point ){\
dst->exponent = len - i - 1;\
}\
else if( i == point ){\
while( cpy[--i] == '0' )\
dst->exponent++;\
}\
else{\
dst->exponent = point - i;\
}\
\
/* Determine mantissa: */\
if( point ){\
for( i = point; i < len; i++ )\
cpy[i] = cpy[i+1];\
len--;\
}\
i = len;\
while( cpy[--i] == '0' )\
cpy[i] = '\0';\
dst->mantissa = (int ## small ## _t) atoi( cpy );\
\
free( cpy );\
return dst;\
}
dfloatN_atof( 8, 16 )
dfloatN_atof( 16, 32 )
dfloatN_atof( 32, 64 )
dfloatN_atof( 64, 128 )
``````
The function for writing a `dfloat` back to a string is a lot more convoluted and difficult to understand and implement. This is due to the litany of different possibilities for combinations of mantissa and exponent out there. There are a number of different classes of such combinations, each demanding a different method for printing the value.
``````
#define dfloatN_ftoa( small, big )\
char *dfloat ## big ## _ftoa( dfloat ## big ## _t *src ){\
int ## small ## _t size1, size2;\
int ## small ## _t whole_part, frac_part;\
double shift_factor;\
int ## small ## _t whole_magnitude, frac_magnitude;\
int i;\
int zeros, zeros_end;\
char *buf;\
size1 = ceil( log10( abs( src->mantissa ) ) );\
if( abs( src->mantissa ) == 1 )\
/* Accounts for exact powers of 10 */\
size1++;\
if( src->mantissa == 0 ){\
buf = (char *) malloc( 2 );\
sprintf( buf, "0\0" );\
}\
else if( src->exponent == 0 ){\
/* This code simply prints the mantissa. */\
size2 = size1 + 2;\
buf = (char *) malloc( size2 );\
sprintf( buf, "%d\0", src->mantissa );\
}\
else if( src->exponent > 0 ){\
/* This code prints the mantissa and then a number of zeros */\
/* equal to the exponent. */\
size2 = size1 + src->exponent;\
buf = (char *) malloc( size2 + 2 );\
sprintf( buf, "%d", src->mantissa );\
if( src->mantissa < 0 ){\
/* Accounts for a minus sign at the beginning */\
size1++;\
size2++;\
}\
for( i = size1; i < size2; i++ ){\
buf[i] = '0';\
}\
buf[size2] = '\0';\
}\
else if( src->exponent < 0 ){\
/* This code shifts the fractional part out of the mantissa */\
/* to get the whole part, then shifts in zeros and subtracts*/\
/* from the original mantissa to get the fractional part. */\
shift_factor = pow( 10, src->exponent );\
whole_part = abs( src->mantissa ) * shift_factor;\
frac_part = abs( src->mantissa ) - whole_part / shift_factor;\
\
whole_magnitude = whole_part?(ceil( log10( whole_part ) )):1;\
frac_magnitude = ceil( log10( frac_part ) );\
\
/* This code accounts for exact powers of 10. */\
if( whole_magnitude == log10( whole_part ) )\
whole_magnitude++;\
if( frac_magnitude == log10( frac_part ) )\
frac_magnitude++;\
\
/* This code accounts for fractional parts less than 0.1 */\
zeros = -(src->exponent) - frac_magnitude;\
\
if( src->mantissa < 0 )\
/* Add one character for the minus sign */\
whole_magnitude++;\
size2 = whole_magnitude + frac_magnitude + 2;\
buf = (char *) malloc( size2 );\
sprintf( buf, "%s%d.", (src->mantissa < 0)?"-":"", whole_part );\
i = whole_magnitude + 1;\
if( zeros > 0 ){\
zeros_end = zeros + whole_magnitude + 1;\
for( i = whole_magnitude+ 1; i < zeros_end; i++ )\
buf[i] = '0';\
}\
sprintf( buf + i, "%d\0", frac_part );\
}\
return buf;\
}
dfloatN_ftoa( 8, 16 )
dfloatN_ftoa( 16, 32 )
dfloatN_ftoa( 32, 64 )
dfloatN_ftoa( 64, 128 )
``````
Basically this function computes the number of decimal digits it needs for the number’s textual representation by taking its logarithm, then it goes into one of four branches depending on the value of the exponent. If the mantissa is zero, that means the number is zero, so it can just print “0” and skip over everything else. If not, it has a different branch for a zero exponent, positive exponent, or negative exponent. A zero exponent means you can simply print the mantissa as it is, because the value of the number is literally equal to its mantissa. A positive exponent means there are zeroes at the end, so the mantissa is printed first, followed by a number of zeroes equal to the exponent. The case of a negative exponent (which simply means there are digits past the decimal point) is the trickiest of all – it involves breaking the number down into its whole and fractional components and then printing each in turn, separated by a decimal point and however many leading zeros there are past the decimal point. I had a hell of a time getting this function to work properly, again, because there were so many different possibilities to account for.
The only other function that took any great effort to implement was the addition function. This was due to the need to shift the mantissas so that the exponents are equal, thus allowing you to add the mantissas properly. Like the former function, this function took quite a bit of trial and error to get right.
``````
void dfloat ## big ## _add( dfloat ## big ## _t *dst, dfloat ## big ## _t *src ){\
int i;\
long log_max, log_diff;\
/* log_max gives the maximum number of times */\
/* you can multiply by 10 before overflowing. */\
int ## small ## _t src_mantissa, dst_mantissa;\
int ## small ## _t src_magnitude, dst_magnitude;\
int ## small ## _t smaller_exponent;\
int ## small ## _t larger_magnitude;\
int ## small ## _t target_mantissa, target_exponent;\
int ## small ## _t larger_mag_exponent;\
src_mantissa = src->mantissa;\
dst_mantissa = dst->mantissa;\
src_magnitude = ceil( log10( abs( src->mantissa ) * pow( 10, src->exponent ) ) );\
dst_magnitude = ceil( log10( abs( dst->mantissa ) * pow( 10, dst->exponent ) ) );\
/* magnitude is the number of digits before the decimal point */\
/* or the number of zeros after the decimal point if negative */\
if( abs( src_mantissa ) == 1 )\
src_magnitude++;\
if( abs( dst_mantissa ) == 1 )\
dst_magnitude++;\
/* Increment accounts for exact powers of 10 */\
smaller_exponent = (src->exponent < dst->exponent)?src->exponent:dst->exponent;\
/* A lower exponent indicates a higher degree of precision for a number. */\
larger_magnitude = (src_magnitude > dst_magnitude)?src_magnitude:dst_magnitude;\
larger_mag_exponent = (src_magnitude > dst_magnitude)?src->exponent:dst->exponent;\
\
/* Next part figures out the target exponent for each number to */\
/* be shifted to so they can be added. */\
/* If log_diff <= log_max then the exponent should be log_diff. */\
/* Otherwise (if there's overflow and some digits need to be cut */\
/* off) the exponent should be equal to the exponent of the number */\
/* with the larger magnitude, minus abs(log_diff-log_max). */\
log_max = ceil( log10( ((int ## small ## _t) 1) << (small-3) ) );\
log_diff = larger_magnitude - smaller_exponent;\
target_exponent = (log_diff <= log_max)?smaller_exponent:larger_mag_exponent-abs(log_diff-log_max)+1;\
\
/* Shift both src and dst until they have the same exponent: */\
/* Changes to mantissa and exponent should cancel each other out. */\
src_mantissa *= pow( 10, src->exponent - target_exponent );\
dst_mantissa *= pow( 10, dst->exponent - target_exponent );\
\
target_mantissa = src_mantissa + dst_mantissa;\
\
/* Count number of trailing zeros and adjust */\
/* mantissa and exponent accordingly: */\
for( i = 0; i < log_max; i++ ){\
if( target_mantissa % 10 ) break;\
target_mantissa /= 10;\
target_exponent++;\
}\
dst->mantissa = target_mantissa;\
dst->exponent = target_exponent;\
}
``````
Like the number printing function, the addition function uses logarithms to calculate the sizes of the decimal representations. Basically we want to know the distance between the most significant figure and the least significant figure among the two numbers, because this tells us how much space we need for the result. This distance is represented by the difference between the larger magnitude and the smaller exponent. If the difference between these two numbers is less than the maximum number of decimal digits representable by that width, then we know we can add them without any data loss, simply by shifting the larger number until its exponent equals that of the smaller number. We do this by multiplying by a positive power of 10 while subtracting the equivalent amount from the exponent (note that the changes in mantissa and exponent must cancel each other out if the value of the number is to be preserved). If the difference is greater than the maximum size, then this means the result is too big to fit into that width, so some of the least significant decimal digits have to be shifted out. This is done by dividing the smaller number’s mantissa by a certain amount after multiplying the larger number’s mantissa until it reaches its size limit. The two shift amounts (meaning their absolute values) must add up to the difference between the two exponents for this to work.
Once we have implemented addition, the other three arithmetic operations, as well as the one comparison operation, are fairly easy to implement. The subtraction function simply calls the addition function with the second mantissa negated. The multiplication function multiplies the mantissas and adds the exponents. The division function divides the mantissas and subtracts the exponents. The division function also artificially inflates the numerator beforehand so it doesn’t get truncated to zero, and it has an additional precision argument that determines how many decimal places the result will come to. The comparison function subtracts the second operand from the first and looks at the sign of the result.
``````
#define dfloatN_sub( small, big )\
void dfloat ## big ## _sub( dfloat ## big ## _t *dst, dfloat ## big ## _t *src ){\
dfloat ## big ## _t *tmp;\
tmp = (dfloat ## big ## _t *) malloc( sizeof( dfloat ## big ## _t ) );\
dfloat ## big ## _cpy( tmp, src );\
tmp->mantissa *= -1;\
dfloat ## big ## _add( dst, tmp );\
free( tmp );\
}
dfloatN_sub( 8, 16 )
dfloatN_sub( 16, 32 )
dfloatN_sub( 32, 64 )
dfloatN_sub( 64, 128 )
#define dfloatN_mul( small, big )\
void dfloat ## big ## _mul( dfloat ## big ## _t *dst, dfloat ## big ## _t *src ){\
int i;\
int ## small ## _t log_max;\
log_max = ceil( log10( ((int ## small ## _t) 1) << (small-3) ) );\
dst->mantissa *= src->mantissa;\
dst->exponent += src->exponent;\
\
/* Count number of trailing zeros and adjust */\
/* mantissa and exponent accordingly: */\
for( i = 0; i < log_max; i++ ){\
if( dst->mantissa % 10 ) break;\
dst->mantissa /= 10;\
dst->exponent++;\
}\
}
dfloatN_mul( 8, 16 )
dfloatN_mul( 16, 32 )
dfloatN_mul( 32, 64 )
dfloatN_mul( 64, 128 )
#define dfloatN_div( small, big )\
void dfloat ## big ## _div( dfloat ## big ## _t *dst, dfloat ## big ## _t *src, int precision ){\
int i;\
/* First section shifts the destination mantissa */\
/* so it doesn't become zero when divided. */\
long log_max;\
int ## small ## _t dst_magnitude;\
int ## small ## _t shift_factor;\
log_max = ceil( log10( ((int ## small ## _t) 1) << (small-3) ) );\
dst_magnitude = ceil( log10( dst->mantissa * pow( 10, dst->exponent ) ) );\
shift_factor = pow( 10, log_max - dst_magnitude );\
dst->mantissa *= shift_factor;\
\
dst->mantissa /= src->mantissa;\
dst->exponent -= (src->exponent+precision);\
dst->mantissa /= (shift_factor/pow( 10, precision ));\
\
/* Count number of trailing zeros and adjust */\
/* mantissa and exponent accordingly: */\
for( i = 0; i < log_max; i++ ){\
if( dst->mantissa % 10 ) break;\
dst->mantissa /= 10;\
dst->exponent++;\
}\
}
dfloatN_div( 8, 16 )
dfloatN_div( 16, 32 )
dfloatN_div( 32, 64 )
dfloatN_div( 64, 128 )
#define dfloatN_cmp( small, big )\
int dfloat ## big ## _cmp( dfloat ## big ## _t *df1, dfloat ## big ## _t *df2 ){\
dfloat ## big ## _t *cpy;\
int result;\
cpy = (dfloat ## big ## _t *) malloc( sizeof( dfloat ## big ## _t ) );\
dfloat ## big ## _cpy( cpy, df1 );\
dfloat ## big ## _sub( cpy, df2 );\
if( cpy->mantissa > 0 )\
return 1;\
if( cpy->mantissa < 0 )\
return -1;\
return 0;\
}
dfloatN_cmp( 8, 16 )
dfloatN_cmp( 16, 32 )
dfloatN_cmp( 32, 64 )
dfloatN_cmp( 64, 128 )
``````
There are two other functions in the library, and they include one for copying a source operand to a destination operand, and one for casting a `dfloat` to another size. They are as follows:
``````
#define dfloatN_cpy( small, big )\
void dfloat ## big ## _cpy( dfloat ## big ## _t *dst, dfloat ## big ## _t *src ){\
dst->mantissa = src->mantissa;\
dst->exponent = src->exponent;\
}
dfloatN_cpy( 8, 16 )
dfloatN_cpy( 16, 32 )
dfloatN_cpy( 32, 64 )
dfloatN_cpy( 64, 128 )
#define dfloatM_castN( smallM, bigM, smallN, bigN )\
dfloat ## bigN ## _t *dfloat ## bigM ## _cast ## bigN ( dfloat ## bigM ##_t *src ){\
dfloat ## bigN ## _t *dst;\
dst = (dfloat ## bigN ## _t *) malloc( sizeof( dfloat ## bigN ## _t ) );\
dst->mantissa = (int ## smallN ## _t) src->mantissa;\
dst->exponent = (int ## smallN ## _t) src->exponent;\
return dst;\
}
dfloatM_castN( 8, 16, 16, 32 )
dfloatM_castN( 8, 16, 32, 64 )
dfloatM_castN( 8, 16, 64, 128 )
dfloatM_castN( 16, 32, 8, 16 )
dfloatM_castN( 16, 32, 32, 64 )
dfloatM_castN( 16, 32, 64, 128 )
dfloatM_castN( 32, 64, 8, 16 )
dfloatM_castN( 32, 64, 16, 32 )
dfloatM_castN( 32, 64, 64, 128 )
dfloatM_castN( 64, 128, 8, 16 )
dfloatM_castN( 64, 128, 16, 32 )
dfloatM_castN( 64, 128, 32, 64 )
``````
Aaaand that’s pretty much the whole library. If you want to see the full code, along with the license and installation instructions, I have posted this project to my GitHub, and you can access its repository here. This is the first complete project I’ve posted there, so needless to say I’m very excited about this. See you next time and happy coding!
## 5 thoughts on “libdfloat: A C Library for Exact Representation of Decimal Floating Point Numbers”
1. Very clever implementation, indeed!
This nicely solves issues for importing from, say, CSV files.
I still wonder: Maybe it tries to solve a non-existent problem?
There is still the problem with number representations in general: 1/3 or exp(2) can’t be expressed exactly, neither as a binary number nor as a decimal. Therefore, as long as we’re doing numerical work, we will always have to deal with rounding errors. This library deals with a small subset of the issues that derive from this problem: Import data from sources where numbers are represented as decimals. In most cases this will be financial and/or statistical data.
With this library, you can import your figures as exact numbers. Still, almost any mathematical operation (besides addition and multiplication) will convert it into a number with no exact representation as a p-adic number.
However, when we are dealing with financial or statistical data with 2,3 or 6 decimals and represent our data as, say, 64 or 128 bit floats, rounding errors are simply not an issue when we convert them back into the original format.
Just my 2¢. Let me know what you think.
Liked by 1 person
2. Yeah, I’m not sure how much of an issue it will be practically. I’m just reasoning that since floating point numbers are truncated when they are converted to a base that doesn’t have an exact representation, reading and writing decimal numbers in CSV files is likely to result in weird repeating decimals where you would have originally had something like 1.7. I guess this is more of an annoyance than anything else, and this library was largely created out of OCD/autism. Still it’s annoying enough to me that I felt the need to create a library to combat the problem.
Like
3. My personal solution to this issue is: Whenever I write numbers which are supposed to be looked at by humans, I always round them to the desired precision, so 1.7000000000012 becomes 1.70.
As long as only computers deal with them, I let them in their native format.
Liked by 1 person
1. I guess that works too. But I wanted to make sure that if you read a number from a file and don’t modify it, you will write it back to the file exactly as it was. Unless you keep track of how many digits it had past the decimal point, you don’t always know what precision to use when rounding. Again, I’m being anal-retentive and autistic about this for the most part.
Like
4. Dealing with numbers is always tricky. I remember, when I was teaching maths to engineering students, and they turned in their homework, they would write at the end of a calculation:
The result is log(3) = 1.09861228867
It drove me near crazy…
Liked by 1 person | 6,144 | 22,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-49 | latest | en | 0.944427 |
http://intermath.coe.uga.edu/topics/datanlss/stats/r01.htm | 1,505,970,797,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687642.30/warc/CC-MAIN-20170921044627-20170921064627-00306.warc.gz | 175,316,413 | 3,033 | Home | Data Analysis | Statistics | Recommended Investigations | Finding Averages
John's Way of Finding Averages
John claims he has found an easier way of finding the average (mean) of a set of numbers when you only need an estimate. It's easier because you work with smaller numbers. To illustrate his method, let's use these numbers:
31,25,35,18,14
John first takes the smallest number in the set and subtracts it from the other numbers in the set.
(31-14)=17, (25-14)=11, (35-14)=21, (18-14)=4.
Then he uses the "standard" procedure to average those numbers.
(17+11+21+4 + 0)/5 = ~10.6 (approximately 10.6)
He then adds the smallest number from the original set to this average (mean).
14 + 10.6 = 24.6
This average (mean) is the average (mean) of the original set of numbers (24.6).
Does John's way of finding the average (mean) of a set of numbers always work? Why or why not? Would his method work if you did the first step AGAIN with the smallest number in the new set (4), found the average (mean) of this new set of numbers, and then added this average (mean) with 14 and 4? Why or why not?
Extensions
Would John's method work if he chose any other number in the data set to subtract from the other numbers? Explain.
Submit your idea for an investigation to InterMath. | 333 | 1,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-39 | latest | en | 0.901518 |
https://dev.to/dattran1999/how-neural-networks-work-dma | 1,685,778,070,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649177.24/warc/CC-MAIN-20230603064842-20230603094842-00739.warc.gz | 247,109,486 | 21,674 | dattran1999
Posted on
# Teaching Philosophy
I know that there are A LOT of tutorials/blog posts on neural networks already (some of my favourites include 3B1B series on YouTube),
but I am a big advocate of learning by doing. So this series will not just present a bunch of information to you,
but actually asking you to implement the things we covered in each post.
# Introduction to Neural Network
The inspiration of Artificial Neural Networks (or neural network for short) comes from Biological Neural Networks. But I haven't had a biology class
since high school so I have no idea how a biological neural network works :) but I bet it looks something like this:
For this tutorial, we will go through the primitive building block of an Artificial Neural Networks, which is a perceptron.
## Assumed maths knowledge
• Functions
• Coordinates geometry
# Perceptron
Perceptron and its learning rule is not popular anymore, but it is a great start for building an understanding of how everything works.
The goal of perception is to classify sets of points.
## How perceptron works
Definition: A perceptron is a function that takes several inputs, and produces one output.
Formula:
where w's are the weights, f is the activation function (explained below), x's are the inputs, and y is the output.
This is basically putting a polynomial into some function called activation function.
And the goal of perceptron is to classify sets of points.
## Weights of perceptron and Classification
To understand the importance of weights, it's useful to think about the case where we only have 2 inputs.
Considering only the part where we multiply inputs by weights and sum them up, we have:
Notice that this equation is very similar to the standard form of linear equation, which is of the form:
### Example
Consider the following diagram, where we want to classify point A and point B (i.e. finding a way to separate them).
In the diagram, the line has equation
Looking visually, it's clear that the line separates the two points. Below is the mathematical explanation.
From coordinate geometry, we know that any points to the "above" or "to the left" of the line (e.g. point A) will satisfy
and any points to the "below" or "to the right" of the line (e.g. point B) will satisfy
With that straight line, we have successfully classified point A and point B into 2 classes.
But that only works visually, not mathematically yet. To make it work mathematically, we need the activation function.
### Importance of Weights
It is important to note that if the weights are different, we might not be able to classify point A and point B. One such example is the line
, which is a horizontal line passing through (0,0)
So a question to ask it how do we find the weights that will correctly classify the points we have. The answer to that is through perceptron learning, and we will cover that in the next post.
## Activation function
More often that not, we want the output in the range 0 to 1 only, to notate if that certain perceptron is activated or not.
So we need some function, called activation function, to do that for us.
One simple way to achieve that is to use the heaviside function, which converts all negative numbers to 0, and all positive numbers (including 0) to 1.
Coming back to our example, for point A, it satisfies
Hence putting that in the heaviside function will output 1. With the similar approach, putting B in the heaviside function outputs 0.
Therefore, we have correctly classify points A and B mathematically.
# Sum Up
Weights will determine if a straight line (or plane in higher dimension) can separate the points into classes. Only a set of weights will be able to separate the points.
Activation function is just a function that generalize all points that fit certain criteria.
# Exercise
Write a function that takes a list of pairs of coordinates, and a list of classes, determine if the given weights will be able to classify the classes.
``````def is_correct_weights(coords, classes, w_0, w_1, w_2) -> bool:
pass
# example from above
coords = [(-0.7, 2.7), (1.5, 1.1)]
# classes[i] is class of coords[i]
classes = [0, 1]
is_correct_weights(coords, classes, -1, -2, 1) # True
is_correct_weights(coords, classes, -1, 0, 1) # False
``````
NOTE: do it in any language you want, but it is recommended to use Python, since we will use Python much more later on. | 1,006 | 4,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-23 | latest | en | 0.940713 |
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September 20, 2004, 13:12 FOAM #1 CFDtoy Guest Posts: n/a Hello All, Has anyone used FOAM Software for simulating unsteady flows? Thanks CFDtoy
September 20, 2004, 21:57 Re: FOAM #2 luiz Guest Posts: n/a Yes. I have. LESimulation.
September 21, 2004, 07:41 Re: FOAM #3 baal Guest Posts: n/a is it a free software? how can i get it? thanks.
September 21, 2004, 09:40 Re: FOAM #4 matej Guest Posts: n/a you can have a free license for academic use; surf to www.nabla.co.uk, or you can search this forum, I gues the link is here about 10 times already. You may also find google or other search engine to be usefull. matej
September 21, 2004, 10:44 Re: FOAM #5 CFDtoy Guest Posts: n/a Hi Luiz, I would like to know if you were able to access the source code as claimed by FOAM. Were you able to implement any of your own algorithms or additional subroutines ? CFDtoy
September 21, 2004, 14:09 Re: FOAM #6 luiz Guest Posts: n/a Yes. The source code is really available, and yes, I have done some interesting things myself. I think the main advantage of FOAM is its OO philosophy, which makes it very easy to implement new algoritms. Since there are many differential operators readily available (both spatial and temporal), you can mount new equations/variables/models adding a couple of lines to the original code, without having to worry about the lower layers... For instance, if you want to solve a completely new equation for a tensor T, you could just add: fvMatrix T { fvm::ddt(T) - fvm::div(T) == fvc::curl(U) } T.solve() This little piece of code is almost self explanatory: 1) build a matrix T to be solved; 2) obeying the equation ddt(T) -div(T) = curl(U) (That is just an example meaning nothing, just to show you are not tied to navier stokes eq); here fvm is a class of implicit operators, while fvc is for explicit ones...; 3) then solve the eq. This way, you can QUICKLY implement new models, or even new algorithms, focusing only on what you need. And I think that is why they are able to offer so many turbulence models, multiphase, etc ready to use with their package. I also expect them to be quicker to release new algorithms, models, etc than other commercial codes. As you can see, I am enjoying a lot the code, and I am becoming a big fan of C++ and FOAM... good luck
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All times are GMT -4. The time now is 02:29. | 874 | 3,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-39 | longest | en | 0.913482 |
http://blog.targettestprep.com/gmat-official-guide-ds-what-is-the-tens-digit/ | 1,501,187,665,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549429485.0/warc/CC-MAIN-20170727202516-20170727222516-00373.warc.gz | 41,494,308 | 8,806 | # GMAT OFFICIAL GUIDE DS – What is the tens digit…
## Solution:
We need to determine the tens digit of the positive integer x.
Statement One Alone:
x divided by 100 has a remainder of 30.
Using the information in statement one, we can test some numerical values for x.
x = 30
30/100 = 0 remainder 30
We see that x has a tens digit of 3.
x = 130
130/100 = 1 remainder 30
We see that x has a tens digit of 3.
x = 230
230/100 = 2 remainder 30
We see that x has a tens digit of 3.
We see that regardless of which value we select for x, when x is divided by 100 and yields a remainder of 30, x will always have a tens digit of 3. Statement one is sufficient to answer the question. We can eliminate answer choices B, C and E.
Statement Two Alone:
x divided by 110 has a remainder of 30.
Using the information in statement two we can test some numerical values for x.
x = 30
30/110 = 0 remainder 30
We see that x has a tens digit of 3.
x = 140
140/110 = 1 remainder 30
We see that x has a tens digit of 4.
Statement two is not sufficient to answer the question. | 297 | 1,080 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2017-30 | latest | en | 0.91371 |
https://studylib.net/doc/10503624/math-311-topics-in-applied-mathematics-i-lecture-38- | 1,726,047,987,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00505.warc.gz | 513,938,821 | 13,306 | # MATH 311 Topics in Applied Mathematics I Lecture 38:
```MATH 311
Topics in Applied Mathematics I
Lecture 38:
Review for the final exam.
Topics for the final exam: Part I
Elementary linear algebra (L/C 1.1–1.5, 2.1–2.2)
• Systems of linear equations: elementary
operations, Gaussian elimination, back substitution.
• Matrix of coefficients and augmented matrix.
Elementary row operations, row echelon form and
reduced row echelon form.
• Matrix algebra. Inverse matrix.
• Determinants: explicit formulas for 2×2 and
3×3 matrices, row and column expansions,
elementary row and column operations.
Topics for the final exam: Part II
Abstract linear algebra (L/C 3.1–3.6, 4.1–4.3)
• Vector spaces (vectors, matrices, polynomials, functional
spaces).
• Subspaces. Nullspace, column space, and row space of a
matrix.
• Span, spanning set. Linear independence.
• Bases and dimension.
• Rank and nullity of a matrix.
• Coordinates relative to a basis.
• Change of basis, transition matrix.
• Linear transformations.
• Matrix transformations.
• Matrix of a linear mapping.
• Change of basis for a linear operator.
• Similarity of matrices.
Topics for the final exam: Part III
Advanced linear algebra (L/C 5.1–5.6, 6.1–6.3)
• Eigenvalues, eigenvectors, eigenspaces
• Characteristic polynomial
• Bases of eigenvectors, diagonalization
•
•
•
•
•
Euclidean structure in Rn (length, angle, dot product)
Inner products and norms
Orthogonal complement, orthogonal projection
Least squares problems
The Gram-Schmidt orthogonalization process
Topics for the final exam: Part IV
Vector analysis (L/C 8.1–8.4, 9.1–9.5, 10.1–10.3,
11.1–11.3)
•
•
•
•
•
•
•
•
•
•
•
•
Fubini’s Theorem
Change of coordinates in a multiple integral
Geometric meaning of the determinant
Length of a curve
Line integrals
Green’s Theorem
Conservative vector fields
Area of a surface
Surface integrals
Gauss’ Theorem
Stokes’ Theorem
Problem. Let V be the vector space spanned by
functions f1 (x) = x sin x, f2(x) = x cos x,
f3 (x) = sin x, and f4 (x) = cos x.
Consider the linear operator D : V → V ,
D = d /dx.
(a) Find the matrix A of the operator D relative to
the basis f1 , f2, f3, f4.
(b) Find the eigenvalues of A.
(c) Is the matrix A diagonalizable?
A is a 4×4 matrix whose columns are coordinates of
functions Dfi = fi ′ relative to the basis f1 , f2, f3, f4.
f1′ (x) = (x sin x)′ = x cos x + sin x = f2 (x) + f3(x),
f2′ (x) = (x cos x)′ = −x sin x + cos x
= −f1(x) + f4 (x),
f3′ (x) = (sin x)′ = cos x = f4 (x),
f4′ (x) = (cos x)′ = − sin x = −f3 (x).
0 −1 0 0
1 0 0 0
Thus A =
1 0 0 −1.
0 1 1 0
Eigenvalues of A are roots of its characteristic
polynomial
−λ −1
0
0
1 −λ
0
0
det(A − λI ) =
1
0 −λ −1
0
1
1 −λ
Expand the determinant by the 1st row:
1
0
0
−λ
0
0
det(A − λI ) = −λ
0 −λ −1 − (−1) 1 −λ −1
0
1 −λ
1
1 −λ
= λ2 (λ2 +1)+(λ2 +1) = (λ2 +1)2 = (λ−i )2(λ+i )2.
The roots are i and −i , both of multiplicity 2.
One can show that both eigenspaces of A are one-dimensional.
The eigenspace for i is spanned by (0, 0, i , 1) and the
eigenspace for −i is spanned by (0, 0, −i , 1). It follows that
the matrix A is not diagonalizable in the complex vector space
C4 (let alone real vector space R4 ).
There is also an indirect way to show that A is not
diagonalizable. Assume the contrary. Then A = UPU −1 ,
where U is an invertible matrix with complex entries and
i 0 0 0
0 i
0 0
P=
0 0 −i
0
0 0 0 −i
(note that P should have the same characteristic polynomial
as A). This would imply that A2 = UP 2 U −1 . But P 2 = −I
so that A2 = U(−I )U −1 = −I .
Let us check if A2 = −I .
2
0 −1 0 0
−1 0 0 0
1 0 0 0
= 0 −1 0 0.
A2 =
1 0 0 −1
0 −2 −1 0
0 1 1 0
2 0 0 −1
Since A2 6= −I , we have a contradiction. Thus the
matrix A is not diagonalizable in C4 .
Problem. Consider a linear operator L : R3 → R3
defined by L(v) = v0 × v, where
v0 = (3/5, 0, −4/5).
(a) Find the matrix B of the operator L.
(b) Find the range and kernel of L.
(c) Find the eigenvalues of L.
(d) Find the matrix of the operator L2015 (L applied
2015 times).
L(v) = v0 × v, v0 = (3/5, 0, −4/5).
Let v = (x, y , z) = xe1 + y e2 + ze3 . Then
e1 e2 e3
L(v) = v0 × v = 3/5 0 −4/5
x y
z
0 −4/5
3/5 −4/5
3/5 0
e1 −
e2 +
e
y
z
x
z
x y 3
= 54 y e1 − 54 x + 35 z e2 + 53 y e3 = 45 y , − 54 x − 35 z, 35 y .
=
In particular, L(e1 ) = 0, − 54 , 0 , L(e2) =
L(e3 ) = 0, − 53 , 0 .
3
4
5 , 0, 5
,
0
4/5
0
Therefore B = −4/5 0 −3/5.
0
3/5
0
The range of the operator L is spanned by columns
of the matrix B. It follows that Range(L) is the
plane spanned by v1 = (0, 1, 0) and v2 = (4, 0, 3).
The kernel of L is the nullspace of the matrix B,
i.e., the solution set for the equation Bx = 0.
0
4/5
0
1 0 3/4
−4/5 0 −3/5 → 0 1 0
0
3/5
0
0 0 0
=⇒ x + 34 z = y = 0 =⇒ x = t(−3/4, 0, 1).
Alternatively, the kernel of L is the set of vectors
v ∈ R3 such that L(v) = v0 × v = 0.
It follows that this is the line spanned by
v0 = (3/5, 0, −4/5).
Characteristic polynomial of the matrix B:
−λ 4/5
0
det(B − λI ) = −4/5 −λ −3/5
0
3/5 −λ
= −λ3 −(3/5)2λ−(4/5)2λ = −λ3 −λ = −λ(λ2 +1).
The eigenvalues are 0, i , and −i .
The matrix of the operator L2015 is B 2015.
Since the matrix B has eigenvalues 0, i , and −i , it is
diagonalizable in C3 . Namely, B = UDU −1, where
U is an invertible matrix with complex entries and
0 0 0
D = 0 i 0 .
0 0 −i
Then B 2015 = UD 2015U −1
. We have that D 2015 =
= diag 0, i 2015, (−i )2015 = diag(0, −i , i ) = −D.
Hence
0 −0.8 0
B 2015 = U(−D)U −1 = −B = 0.8 0 0.6.
0 −0.6 0
Problem. Find a quadratic polynomial that is the
best least squares fit to the function f (x) = |x| on
the interval [−1, 1].
The best least squares fit is a polynomial q(x) that
minimizes the distance relative to the integral norm
ˆ 1
1/2
kf − qk =
|f (x) − q(x)|2 dx
−1
over all polynomials of degree 2.
The norm k · k is induced by the inner product
ˆ 1
hg , hi =
g (x)h(x) dx.
−1
Therefore kf − pk is minimal if p is the orthogonal
projection of the function f on the subspace P3 of
Suppose that p0 , p1, p2 is an orthogonal basis for
P3 . Then
hf , p1i
hf , p2i
hf , p0i
p0 (x) +
p1(x) +
p2(x).
q(x) =
hp0 , p0i
hp1 , p1i
hp2 , p2i
An orthogonal basis can be obtained by applying the
Gram-Schmidt orthogonalization process
to the basis 1, x, x 2:
p0 (x) = 1,
p1 (x) = x −
hx, 1i
hx, p0i
p0 (x) = x −
= x,
hp0 , p0i
h1, 1i
hx 2 , p1i
hx 2 , p0i
p0(x) −
p1 (x)
p2 (x) = x −
hp0 , p0i
hp1 , p1i
2
= x2 −
hx 2 , 1i hx 2 , xi
1
−
x = x2 − .
h1, 1i
hx, xi
3
Problem. Find a quadratic polynomial that is the
best least squares fit to the function f (x) = |x| on
the interval [−1, 1].
Solution:
hf , p0i
hf , p1i
hf , p2i
q(x) =
p0(x) +
p1(x) +
p2 (x)
hp0 , p0i
hp1 , p1i
hp2 , p2 i
15
1
= p0 (x) + p2 (x)
2
16
1 15 2 1 3
= +
x −
= (5x 2 + 1).
2 16
3
16
``` | 2,902 | 6,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-38 | latest | en | 0.679279 |
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# Class 6 NSO Sample Paper 8
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1.
Mohit is twice as old as his little brother. Which expression can be used to find Mohits age it x represents his brother’sage?
A.
2+x
B.
2x
C.
x+2
D.
x-2
2.
The given table shows the colours of the houses on Seth’s street and the fraction of the total number of houses paintedeach colour.Houses on Seth’s StreetColour PaintedFraction PaintedThat ColourWhite31Tan61Gray181Yellow94Which colour was used for the smallest fraction of houses on Seth’s street?
A.
White
B.
Tan
C.
Gray
D.
Yellow
3.
Side BC of a AABC has been produced to point D.If LACD = 1150 and
A = 45° then
B is
A.
70
o
B.
80°
C.
105°
D.
51°
4.
walltheof muchday.Howsecondthewalltheof 41anddayfirstthewallaof 83 paintedSunil
is remained to paint?
A.
85
B.
81
C.
91
D.
83
5.
A number is divisible by 9 if,
A.
The sum of the digits is divisible by 3
B.
The sum of the digits is divisible by 9
C.
The last digit is divisible by 9
D.
Both (B) and (C)
6.
Complete the following figures from the given set of alternatives
A.B.C.D.7.
A number is always divisible by 24 if
A.
It is divisible by both 6 and 4
B.
It is divisible by both 12 and 2
C.
It is divisible by both 8 and 3
D.
The number formed by the tens and units digits of the given number is divisible by 24
8.
The table shows the temperature on four winter mornings in the NorthEast Mountains. Which day had the warmestmorning?Winter temperaturein the Berkshire MountainsDayTemperature at 6: 00 A MThursday-9°CFriday-10°CSaturday-18°CSunday-12°C
A.
Thursday
B.
Friday
C.
Saturday
D.
Sunday
9.
of__?valuetheiswhat,16 41__241 If
A.
4
B.
6
C.
8
D.
14
10.
Which number equals (2)
-4
?
A.
-8
B.
161
C.
161
D.
81
11.
Find the area of the adjoining triangle.
A.
50 cm
2
B.
15 cm
2
C.
25 cm
2
D.
45 cm
2
12.
Find the perimeter of the adjoining figure
A.
28 cm
B.
56 cm
C.
36 cm
D.
112 cm
13.
Which figure has more than 5 sides?
A.
Triangle
B.
Pentagon
C.
Parallelogram
D.
Hexagon
14.
Nadia is putting beads on a piece of string, as shown below. Each bead is 6 mm longWhat is the greatest number of beads ? | 774 | 2,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2015-35 | latest | en | 0.867326 |
https://management.ind.in/forum/jamia-hamdard-university-bca-319154.html | 1,679,733,064,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00239.warc.gz | 470,041,289 | 18,402 | 2022 2023 Student Forum Jamia Hamdard University BCA
#1
20th January 2017, 04:51 PM
Unregistered Guest
Jamia Hamdard University BCA
Recently I have taken admission in BCA 3rd Semester Course of Jamia Hamdard University. I need syllabus of 3rd Semester Course, so will you provide complete syllabus of BCA 3rd Semester Course of Jamia Hamdard University?
#2
23rd January 2017, 08:29 AM
Super Moderator Join Date: Aug 2012
Re: Jamia Hamdard University BCA
As you want syllabus of BCA 3rd Semester Course of Jamia Hamdard University, so here I am providing detailed syllabus:
Jamia Hamdard University BCA 3rd Semester Syllabus
BCA 301 (Numerical Methods and Statistical Techniques Numerical Methods)
Numerical methods versus numerical analysis, Errors and Measure of Errors.
Non-linear Equations, Iterative solutions, multiple roots and other difficulties,
interpolation methods, BI-section, false position methods, Newton Raphson-Methods.
Simultaneous Solutions of Equations, Gauss elimination Methods, Gauss Jordan
methods, Gauss Siedel methods.
Interpolations and curve fitting, lagrangian polynomials, Newton’s methods: Forward
Difference methods, Backward Difference methods, Divided difference methods.
Numerical Integration: Trapezoidal Rule, Simpson 1/3 Rule Simpson’s 3/8 Rule.
Numerical Differentiation by polynomial Fit.
Statistical techniques:
Measure of central tendency, Preparing frequency distribution table. Mean, arithmetic
mean, Harmonic Mean. Median , mode.
Measure of dispersion, skewness and kurtosis Ranges, Mean deviation. Standard
deviation, co-efficiency of variation, Moments, skewness, kurtosis.
BCA 302 (Fundamental Concepts of Operating Systems)
Operating systems overview: Computer System Structure, operating systems structure,
OS functions, facilities; Processes: introduction, concurrency, inter process
communication, classical problems, process scheduling, Memory management:
swapping, virtual memory segmentation. File systems: files, directories, file system
implementation, security, and protection mechanism. Input/output: principles of
input/output hardware and software, disks, clocks, terminals. Deadlocks: introduction,
detection, recovery, and prevention; Coordinated Case Study of Unix and Windows.
BCA 303 (Introduction to Object Programming using C++)
OOP Programming methodologies: concepts of structured and object oriented programming;
advantage of OOP methodologies, characteristics of OOP languages: objects, classes,
Data Abstraction, Encapsulation, inheritance, reusability, polymorphism and operator
Programming in C++
functions, friend functions, in line functions. Constructors and destructors, derived
base class, pointers and arrays, pointers and functions, support for OOP.
BCA 304 (System Programming Concepts & Design)
Mathematical preliminaries, sets, relations and functions, graphs and trees, strings, theory
of automata, DFA, NFA, acceptability of a string by finite automata, minimization of
finite automata, applications of finite automata-lexical analysis, text editors etc.
Introduction to formal languages-regular grammars, context free grammar, context
sensitive grammar. Evolution of the Components of a Programming System, compilers,
Macros, Variety of software tools, Text editors, Interpreters and program generators
Debug Monitor.
Compilers: Basic concepts, compilers and interpreters, pass of a compilers, phases-lexical
phase, syntax phase. Semantic analysis phase, parser, top down, bottom up parsing,
translation schemes, type analysis and type checking, code generation phase and
optimization, Symbol table management, error handling.
Jamia Hamdard University BCA 3rd Semester Syllabus
#3
26th February 2023, 12:10 PM
rayshant Guest
Re: Jamia Hamdard University BCA
can i get 2018 to 2020 previous year papers for bca 102 and bca 103 1sem question papers
Message:
Options | 870 | 3,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-14 | latest | en | 0.827 |
https://www.jiskha.com/questions/549594/three-resistors-of-10-15-and-20-ohms-are-connected-in-parallel-to-a-source-of-current | 1,558,649,311,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257396.96/warc/CC-MAIN-20190523204120-20190523230120-00128.warc.gz | 836,135,501 | 5,005 | # physics
three resistors of 10, 15, and 20 ohms are connected in parallel to a source of current.
In which resistor is the current greatest
How do I set this up?
1. 👍 0
2. 👎 0
3. 👁 59
1. current greatest? in the lowest resistance.
1. 👍 0
2. 👎 0
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two resistors are connected in parallel. Their resistances are 6 ohms and 10 ohms. What is the resistance, (R), of the combination? R=-----ohms?
asked by Nancy on April 5, 2012
More Similar Questions | 977 | 3,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.951438 |
https://isabelle.in.tum.de/repos/isabelle/rev/207a7811faa9 | 1,669,455,633,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446706285.92/warc/CC-MAIN-20221126080725-20221126110725-00568.warc.gz | 372,234,766 | 3,044 | author nipkow Sun, 02 Nov 1997 13:47:58 +0100 changeset 4067 207a7811faa9 parent 4066 7b508ac609f7 child 4068 99224854a0ac
Documented split_t_case' thm genearted by datatype.
doc-src/Logics/HOL.tex file | annotate | diff | comparison | revisions
--- a/doc-src/Logics/HOL.tex Sat Nov 01 13:03:00 1997 +0100
+++ b/doc-src/Logics/HOL.tex Sun Nov 02 13:47:58 1997 +0100
@@ -1398,7 +1398,7 @@
case $e$ of [] => $a$ | $$x$$\#$$xs$$ => b
\end{center}
is defined by translation. For details see~\S\ref{sec:HOL:datatype}. There
-is also a case splitting rule \tdx{expand_list_case}
+is also a case splitting rule \tdx{split_list_case}
$\begin{array}{l} P(\mathtt{case}~e~\mathtt{of}~\texttt{[] =>}~a ~\texttt{|}~ @@ -1631,6 +1631,22 @@ Violating this restriction results in strange error messages. \end{warn} +To perform case distinction on a goal containing a \texttt{case}-construct, +the theorem \texttt{split_}t\texttt{_case} is provided: +\[ +\begin{array}{@{}rcl@{}} +P(t_\mathtt{case}~f@1~\dots~f@m~e) &=& +((\forall x@1 \dots x@{k@1}. e = C@1~x@1\dots x@{k@1} \to + P(f@1~~x@1\dots x@{k@1})) \\ +&& ~\land~ \dots ~\land \\ +&&~ (\forall x@1 \dots x@{k@m}. e = C@m~x@1\dots x@{k@m} \to + P(f@m~~x@1\dots x@{k@m}))) +\end{array} +$
+where $t$\texttt{_case} is the internal name of the \texttt{case}-construct.
+(see~\S\ref{subsec:HOL:case:splitting}).
+
\subsubsection{The function \cdx{size}}\label{sec:HOL:size}
Theory \texttt{Arith} declares an overloaded function \texttt{size} of type
@@ -1698,7 +1714,7 @@
\end{ttdescription}
\begin{warn}
Induction is only allowed on a free variable that should not occur among
- the premises of the subgoal. Exhaustion is works for arbitrary terms.
+ the premises of the subgoal. Exhaustion works for arbitrary terms.
\end{warn}
\bigskip
` | 679 | 1,786 | {"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": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-49 | latest | en | 0.600388 |
https://blog.csdn.net/weixin_42870739/article/details/104906081 | 1,638,560,911,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362918.89/warc/CC-MAIN-20211203182358-20211203212358-00391.warc.gz | 217,512,435 | 21,993 | 2 篇文章 0 订阅
# 关于SQL中,连续天数问题的求解
## 问题的求解
1 将数据按照天数排序，并利用数据库row_number()函数为销售人员的销售记录编号排序。
2 用当前时间减去随意给定的一个时间，得到一个日期间差值，在这里为了表述方便，我们可以记这个差值为subtraction1。这里减去的这个日期是可以随意指定的，但是个人认为最好取今天now会好一些。
3 利用步骤1中，得到的row_number列，减去2中的差值subtraction1，即可得到另一列的差值subtraction2。
4 根据cust_id，和列subtraction2分组，即可得到每个人员的连续天数。
drop table if exists consecutive_task ;
create table consecutive_task
(
cust_id varchar(10)
,date_col date
,sell_amount int
,primary key(cust_id,date_col)
)
;
insert into consecutive_task values('c001','2019-01-01','100');
insert into consecutive_task values('c001','2019-01-03','20');
insert into consecutive_task values('c001','2019-01-04','76');
insert into consecutive_task values('c001','2019-01-05','80');
insert into consecutive_task values('c001','2019-01-06','75');
insert into consecutive_task values('c002','2019-01-01','58');
insert into consecutive_task values('c002','2019-01-02','60');
insert into consecutive_task values('c003','2019-01-01','49');
insert into consecutive_task values('c003','2019-01-02','51');
insert into consecutive_task values('c003','2019-01-03','60');
insert into consecutive_task values('c004','2019-01-01','100');
insert into consecutive_task values('c005','2019-01-01','30');
insert into consecutive_task values('c006','2019-01-01','51');
insert into consecutive_task values('c006','2019-01-02','52');
insert into consecutive_task values('c006','2019-01-03','50');
insert into consecutive_task values('c007','2019-01-01','78');
insert into consecutive_task values('c007','2019-01-02','80');
insert into consecutive_task values('c007','2019-01-03','69');
insert into consecutive_task values('c007','2019-01-04','66');
insert into consecutive_task values('c007','2019-01-05','30');
insert into consecutive_task values('c008','2019-01-01','20');
insert into consecutive_task values('c008','2019-01-04','50');
insert into consecutive_task values('c008','2019-01-05','52');
insert into consecutive_task values('c008','2019-01-07','53');
insert into consecutive_task values('c008','2019-01-08','55');
drop table if exists consecutive_task_tmp1;
create table consecutive_task_tmp1
(
cust_id varchar(10)
,date_col date
,sell_amount int
,row_num int
)
;
insert into consecutive_task_tmp1
(
cust_id
,date_col
,sell_amount
,row_num
)
select t1.cust_id
,t1.date_col
,t1.sell_amount
,row_number() over(partition by t1.cust_id order by t1.date_col asc)
from consecutive_task as t1
where t1.sell_amount >= 50
;
drop table if exists consecutive_task_tmp2;
create table consecutive_task_tmp2
(
cust_id varchar(10)
,date_col date
,sell_amount int
,row_num int
,subtraction1 int
,subtraction2 int
)
;
insert into consecutive_task_tmp2
(
cust_id
,date_col
,sell_amount
,row_num
,subtraction1
,subtraction2
)
select t1.cust_id
,t1.date_col
,t1.sell_amount
,t1.row_num
,datediff(t1.date_col,now())
,t1.row_num - datediff(t1.date_col,now())
from consecutive_task_tmp1 as t1
-- order by t1.cust_id,t1.date_col
;
select a.cust_id,a.subtraction2,count(1)
from consecutive_task_tmp2 as a
group by a.cust_id,a.subtraction2
having count(1) > 1
;
## 数学原理
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http://hosting131287.a2e1f.netcup.net/gdalnmy/e9e9a3-repair-rate-formula | 1,621,079,139,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991801.49/warc/CC-MAIN-20210515100825-20210515130825-00026.warc.gz | 23,876,038 | 5,809 | Failure prediction is one of the key challenges that have to be mastered for a new arena of fault tolerance techniques: the proactive handling of faults. Failure Rate = 1 / 11.25; Failure Rate = 0.08889 Failure rate per hour would be 0.08889.. ADVERTISEMENTS: Machine Hours Rate: Formula, Calculation, Problems and Solutions! Let’s say you have a very expensive piece of medical equipment that is responsible for taking important pictures of healthcare patients.. Over the last year, it has broken down a total of five times. While standards & codes are focused on defetcts acceptance criteria and sampling extension criteria, no requirements are stated for acceptable welding repair rate (n° of repaired joints/ n° of total joints). Repair Rate models are based on counting the cumulative number of failures over time A different approach is used for modeling the rate of occurrence of failure incidences for a repairable system. Tech Hourly Rate Gross Profit Factor Labor Rate \$20.00 30% \$66.67 If your technicians were available 8 hours and produced 5 billable hours, you would calculate your effective labor rate as follows: Hourly Labor Rate Hours Billed Revenue \$66.67 5 \$333.35 Revenue Available Hours Effective Labor Rate \$333.35 8 \$41.67 Taking the derivative of this gives the repair rate model \(m(t)\). So if we add up his repair percentages it comes up to 3+2 = 5, but if we add up total no. Please advice which is the regular practice in doing repair calculation, whether its like adding up individual dia repair percentage or total no. Granted, this rate may increase as your business grows over time, as it should, but to get started you need a foundation to build on. Repair rate models are defined by first picking a functional form for \(M(t)\), the expected number of cumulative failures by time \(t\). The handbook also identifies which metrics are leading indicators (predictive) and which are lagging indicators (historical). Hence the system is more stable!! To get back to my statement about repair / reject rates being B/S. Below is the step by step approach for attaining MTBF Formula. a detailed description of each metric, a formula to calculate the metric, and an explanation of the metric’s importance and relationship to other metrics. Here’s an example. Formula: The formula used in computing the rate is: ADVERTISEMENTS: Factory overhead/Machine hours If factory overhead is Rs 3, 00,000 and total machine […] The machine hour rate is similar to the labour hour rate method and is used where the work is performed primarily on machines. In this chapter, these rates are called repair rates (not to be confused with the length of time for a repair, which is not discussed in this chapter). of joints ie 10" and 6 " and then take total percentage repair rate come only up to 1.3 Percentage. It is also a guide for data investigation. As a definition, prediction is a statement about what will happen or might happen in the future. In the construction of petrochemical plant an important issue is welding quality control. The time that each repair took was (in hours), 3 hours, 6 hours, 4 hours, 5 hours and 7 hours respectively, making a total maintenance time of 25 hours. Knowing what to charge for your labor rate is a critical step to solidifying the long term success of your auto repair shop. In the next three sections we will describe three models, of increasing complexity, for \(M(t)\). Explanation. Repair rate come only up to 3+2 = 5, but if add... Definition, prediction is a critical step to solidifying the long term success of auto... To get back to my statement about repair / reject rates being B/S comes to! Step to solidifying the long term success of your auto repair shop rate is similar to the labour hour is! 1 / 11.25 ; Failure rate per hour would be 0.08889 sections we will describe three models, increasing... Doing repair calculation, whether its like adding up individual dia repair percentage or no. What will happen or might happen in the future labour hour rate is a critical to... Rate come only up to 1.3 percentage and is used where the work is performed primarily on.! Percentage repair rate model \ ( m ( t ) \ ) to charge for your labor is... Total percentage repair rate come only up to 1.3 percentage success of your auto repair shop and 6 and. Per hour would be 0.08889 might happen in the future ie 10 '' and 6 `` and take... 6 `` and then take total percentage repair rate model \ ( m t., calculation, whether its like adding up individual dia repair percentage or no., but if we add up total no a critical step to the. His repair percentages it comes up to 1.3 percentage 6 `` and take! Then take total percentage repair rate model \ ( m ( t ) \ ) adding individual! Comes up to 1.3 percentage 0.08889 Failure rate = 1 / 11.25 ; Failure rate = Failure... For attaining MTBF Formula three models, of increasing complexity, for (... Which are lagging indicators ( predictive ) repair rate formula which are lagging indicators ( predictive ) and which are lagging (... Predictive ) and which are lagging indicators ( predictive ) and which are lagging indicators ( predictive ) and are. ; Failure rate = 1 / 11.25 ; Failure rate per hour would be 0.08889 total no practice... To 1.3 percentage rate = 1 / 11.25 ; Failure rate = 1 / 11.25 Failure! Below is the step by step approach for attaining MTBF Formula will happen might. Like adding up individual dia repair percentage or total no to solidifying the long term success of auto. Is performed primarily on machines a statement about repair / reject rates being B/S lagging (. 11.25 ; Failure rate = 1 / 11.25 ; Failure rate per hour would be 0.08889 would be 0.08889 is. Machine Hours rate: Formula, calculation, whether its like adding up individual repair. Happen or might happen in the next three sections we will describe three models of... Is similar to the labour hour rate is a critical step to solidifying the term! 3+2 = 5, but if we add up his repair percentages it comes up to 3+2 5... Will describe three models, of increasing complexity, for \ ( m ( t ) \ ) /. The labour hour rate method and is used where the work is performed primarily on.... Where the work is performed primarily on machines hour would be 0.08889 and are. Method and is used where the work is performed primarily on machines ie 10 '' and 6 `` and take. Or might happen in the future knowing what to charge for your rate! ( t ) \ ) is used where the work is performed primarily on machines increasing... Up his repair percentages it comes up to 3+2 repair rate formula 5, but we... Models, of increasing complexity, for \ ( m ( t ) \ ) auto repair shop 3+2. 11.25 ; Failure rate = 1 / 11.25 ; Failure rate = 0.08889 Failure rate = Failure. 11.25 ; Failure rate = repair rate formula / 11.25 ; Failure rate = 0.08889 Failure rate = 1 11.25... Doing repair calculation, whether its like adding up individual dia repair percentage or total.... Model \ ( m ( t ) \ ) of your auto repair shop and 6 and... ) and which are lagging indicators ( historical ) advertisements: Machine Hours rate:,. To 3+2 = 5, but if we add up his repair percentages it comes up to 3+2 =,... For your labor rate is similar to the labour hour rate method and is used where the is. It comes up to 1.3 percentage is the step by step approach attaining! Comes up to 1.3 percentage advice which is the step by step approach for attaining MTBF Formula step step! Repair percentages it comes up to 1.3 percentage 1 / 11.25 ; repair rate formula! Complexity, for \ ( m ( t ) \ ) the rate... Rate method and is used where the work is performed primarily on machines '' and 6 and. Auto repair shop a definition, prediction is a critical step to solidifying the long term success of your repair. Gives the repair rate model \ ( m ( t ) \ ) ( historical ) three we! Hour would be 0.08889 is a statement about what will happen or might in! A critical step to solidifying the long term success of your auto shop... And which are lagging indicators ( historical ), Problems and Solutions come only up to percentage! To charge for your labor rate is similar to the labour hour rate is similar to the labour rate. Predictive ) and which are lagging indicators ( predictive ) and which are lagging indicators ( predictive ) and are. Add up his repair percentages it comes up to 1.3 percentage this gives the repair come. Per hour would be 0.08889 step by step approach for attaining MTBF Formula and! Calculation, whether its like adding up individual dia repair percentage or total.. ) \ ) we add up his repair percentages it comes up to 3+2 =,. Critical step to solidifying the long term success of your auto repair shop we describe... About what will happen or might happen in the future prediction is a statement about will... T ) \ ) ( predictive ) and which are lagging indicators historical... Like adding up individual dia repair percentage or total no of this gives the repair model. The regular practice in doing repair calculation, whether its like adding up individual dia repair percentage or total.! 0.08889 Failure rate per hour would be 0.08889 for \ ( m ( t \... The next three sections we will describe three models, of increasing,. Machine hour rate method and is used where the work is performed primarily on machines whether its like up! Model \ ( m ( t ) \ ) auto repair shop repair percentages it comes up to 1.3.! Rate per hour would be 0.08889 up total no but if we add up his percentages. Step by step approach for attaining MTBF Formula joints ie 10 '' and 6 `` and then take percentage! Calculation, whether its like adding up individual dia repair percentage or total no percentage. Term success of your auto repair shop ( m ( t ) \ ) is used the... In the next three sections we will describe three models, of complexity! Long term success of your auto repair shop long term success of your repair! Which are lagging indicators ( predictive ) and which are lagging indicators ( predictive ) and which lagging... Which is the step by step approach for attaining MTBF Formula and which lagging! A statement about repair / reject rates being B/S the Machine hour rate method and used., calculation, whether its like adding up individual dia repair percentage or total no about! The derivative of this gives the repair rate come only up to 3+2 = 5, but if we up... Total no prediction is a statement about what will happen or might happen in the future adding up individual repair! Hour rate is similar to the labour hour rate method and is used the... By step approach for attaining MTBF Formula and is used where the work is performed primarily machines. Is the regular practice in doing repair calculation, Problems and Solutions prediction is a about. Is the regular practice in doing repair calculation, Problems and Solutions ) and which are lagging indicators predictive. ) and which are lagging indicators ( predictive ) and which are lagging indicators ( historical ) indicators... Then take total percentage repair rate model \ ( m ( t \! Total percentage repair rate come only up to 3+2 = 5, if... Repair calculation, whether its like adding up individual dia repair percentage or total no, for \ m. The long term success of your auto repair shop \ ) identifies which are! To solidifying the long term success of your auto repair shop model \ ( m ( t ) \.! For your labor rate is similar to the labour hour rate is a critical step solidifying... Step approach for attaining MTBF Formula / 11.25 ; Failure rate per hour would be 0.08889 per hour would 0.08889! Up total no and Solutions come only up to 1.3 percentage to for... M ( t ) \ ) definition, prediction is a critical step to solidifying the long term of. About repair / reject rates being B/S whether its like adding up individual dia percentage! \ ( m ( t ) \ ) of joints ie 10 '' and 6 `` and take! The derivative of this gives the repair rate come only up to 1.3 percentage handbook also identifies metrics... Definition, prediction is a critical step to solidifying the long term success of your auto repair shop repair rate formula... | 2,731 | 12,108 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.9179 |
https://essaywriting.help/2021/07/16/finance-questions-accounting-homework-help/ | 1,628,047,574,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154500.32/warc/CC-MAIN-20210804013942-20210804043942-00657.warc.gz | 244,646,152 | 10,092 | # Finance questions | Accounting homework help
Homework chapter 6
1- Zane Perelli currently has \$100 that he can spend today on polo shirts costing \$25 each. Alternatively, he could invest the \$100 in a risk-free U.S. Treasury security that is expected to earn a 9% nominal rate of interest. The consensus forecast of leading economists is a 5% rate of inflation over the coming year.
a. How many polo shirts can Zane purchase today?
b. How much money will Zane have at the end of 1 year if he forgoes purchasing the polo shirts today?
c. How much would you expect the polo shirts to cost at the end of 1 year in light of the expected inflation?
d. Use your findings in parts b and c to determine how many polo shirts (fractions are OK) Zane can purchase at the end of 1 year. In percentage terms, how many more or fewer polo shirts can Zane buy at the end of 1 year?
e. What is Zane’s real rate of return over the year? How is it related to the percentage change in Zane’s buying power found in part d? Explain.
2-Bond interest payments before and after taxes Charter Corp. has issued 2,500 debentures with a total principal value of \$2,500,000. The bonds have a coupon interest rate of 7%.
a. What dollar amount of interest per bond can an investor expect to receive each year from Charter?
b. What is Charter’s total interest expense per year associated with this bond issue?
c. Assuming that Charter is in a 35% corporate tax bracket, what is the company’s net after-tax interest cost associated with this bond issue?
3 – Valuation Fundamentals: Imagine that you are trying to evaluate the economics of purchasing an automobile. You expect the car to provide annual after-tax cash benefits of \$1,200 at the end of each year and assume that you can sell the car for after tax proceeds of \$5,000 at the end of the planned 5-year ownership period. All funds for purchasing the car will be drawn from your savings, which are currently earning 6% after taxes.
A. Identify the cash flows, their timing, and the required return applicable to valuing the car.
B. What is the maximum price you would be willing to pay to acquire the car? Explain.
4- Midland Utilities has outstanding a bond issue that will mature to its \$1,000 par value in 12 years. The bond has a coupon interest rate of 11% and pays interest annually.
a. Find the value of the bond if the required return is
(1) 11%,
(2) 15%, and
(3) 8%.
b. Plot your findings in part a on a set of “required return (x axis)–market value of bond (y axis)” axes.
c. Use your findings in parts a and b to discuss the relationship between the coupon interest rate on a bond and the required return and the market value of the bond relative to its par value.
d. What two possible reasons could cause the required return to differ from the coupon interest rate?
5- The Salem Company bond currently sells for \$955, has a 12% coupon interest rate and a \$1,000 par value, pays interest annually, and has 15 years to maturity.
a. Calculate the yield to maturity(YTM) on this bond.
b. Explain the relationship that exists between the coupon interest rate and yield to maturity and the par value and market value of a bond.
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How it works? | 1,066 | 4,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.948948 |
https://sites.duke.edu/probabilityworkbook/category/basic-probability/testing/ | 1,718,722,699,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00067.warc.gz | 468,602,335 | 10,872 | # Category Archives: Testing
## Clinical trial
Let $$X$$ be the number of patients in a clinical trial with a successful outcome. Let $$P$$ be the probability of success for an individual patient. We assume before the trial begins that $$P$$ is unifom on $$[0,1]$$. Compute
1. $$f(P \mid X)$$
2. $${\mathbf E}( P \mid X)$$
3. $${\mathbf Var}( P \mid X)$$
## Random Digit
Let $$D_i$$ be a random digit chosen uniformly from $$\{0,1,2,3,4,5,6,7,8,9\}$$. Assume that each of the $$D_i$$ are independent.
Let $$X_i$$ be the last digit of $$D_i^2$$. So if $$D_i=9$$ then $$D_i^2=81$$ and $$X_i=1$$. Define $$\bar X_n$$ by
$\bar X_n = \frac{X_1 + \cdots+X_n}{n}$
1. Predict the value of $$\bar X_n$$ when $$n$$ is large.
2. Find the number $$\epsilon$$ such that for $$n=10,000$$ the chance that you prediction is off by more than $$\epsilon$$ is about 1/200.
3. Find approximately the least value of $$n$$ such that your prediction of $$\bar X_n$$ is correct to within 0.01 with probability at least 0.99 .
4. If you just had to predict the first digit of $$\bar X_{100}$$, what digit should you choose to maximize your chance of being correct, and what is that chance ?
[Pitman p206, #30]
## Mark-recapture
A common problem in ecology, social networks, and marketing is estimating the population of a particular species or type. The mark-recapture method is a classic approach to estimating the population.
Assume we want to estimate the population of sturgeon in a section of the Hudson river. We use the following procedure:
1. Capture and mark $$h$$ sturgeons
2. Recapture $$n$$ sturgeon and you find that $$y$$ of them are marked
3. The estimated sturgeon population is $$N = \frac{h n}{y}$$.
Motivate statement $$3$$ using the hypergeometric distribution.
## Leukemia Test
A new drug for leukemia works 25% of the time in patients 55 and older, and 50% of the time
in patients younger than 55. A test group has 17 patients 55 and older and 12 patients younger than 55.
1. A uniformly random patient is chosen from the test group, and the drug is administered and it is a success. What is the probability the patient was 55 and older?
2. A subgroup of 4 patients are chosen and the drug is administered to each. What is the probability that the drug works in all of them? | 644 | 2,289 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-26 | latest | en | 0.84428 |
http://www.investopedia.com/university/economics/economic-basics-measuring-economic-activity.asp | 1,508,536,775,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824357.3/warc/CC-MAIN-20171020211313-20171020231313-00042.warc.gz | 481,414,799 | 47,495 | One important part of macroeconomics is measuring the economy. Knowing if the economy is growing and by how much is an important metric for policy makers, financial professionals, corporate strategy and everyday citizens. Here we will bring up just a few of the most important measures of economic activity at the national level.
## Gross Domestic Product (GDP)
Gross domestic product, or GDP, is one of the main indicators used to measure a country's economic activity. It represents the total aggregate dollar value of all goods and services produced in a country each year, and is often equated with the size of the economy. GDP is often measured quarterly, but expressed as an annualized figure. For example, if 3rd quarter GDP is reported to be up 3%, this tells us that economy has grown by 3% over the last year starting from the 3rd quarter.
GDP is tabulated either by adding up what everyone working within a county (citizens or non-citizens) earned over the course of a year (the income approach), or else by adding up what everyone spent (expenditure method). In theory, both measures should arrive at roughly the same total since your spending is somebody else’s income.
The income approach, which is sometimes referred to as GDP(I), is calculated by adding up total compensation to employees, gross profits for incorporated and non-incorporated firms, and taxes less any subsidies and government transfers (such as welfare checks). The expenditure method is the more common approach and it is calculated by adding up total consumption (C), investment (I), government spending (G), and the net difference between imports and exports (X-M). Sometimes economists express this as the GDP equation, where Y is the national income, or GDP.
## Y = C + I + G + (X - M)
Consumption is typically the largest GDP component in the economy, consisting of private expenditures on the wants and needs of a nation’s citizens. Investment is what businesses spend on things like equipment purchases or new construction of factories. Government spending includes items such as salaries of civil servants and government contractors, purchase of weapons for the military, and any investment expenditure by a government. Exports are the goods produced in a country but sold abroad, and imports are goods produced abroad but purchased here.
When the economy is healthy and growing, you will typically see steady increases in a county’s GDP. If GDP falls, the economy is contracting. Investors worry about negative GDP growth, which is one of the factors economists use to determine whether an economy is in a recession. The rule of thumb is that two consecutive quarters of shrinking GDP is the signal for a recession.
## Unemployment
The unemployment rate measures how many people in a country are out of work. It is the share of the labor force that is jobless, expressed as a percentage. Unemployment generally rises or falls in response to changing economic conditions, making it a lagging indicator. When the economy is in poor shape and jobs are scarce, the unemployment rate will rise. When the economy is growing at a healthy rate and jobs are relatively plentiful, it can be expected to fall.
To calculate the unemployment rate, the number of unemployed people is divided by the number of people in the labor force, where the labor force consists of all employed and unemployed people. The ratio is expressed as a percentage. This represents the so-called headline unemployment figure, or U3 unemployment. Some have criticized this measure for not accurately reflecting the employment picture of a country. This is because it includes people who are working part time but would rather work full time, and more importantly because it excludes people who are no longer looking for work – and therefore are no longer considered in the labor force. Some discouraged workers that have given up looking for work would probably like to work but have lost hope. A more inclusive unemployment measure that includes discouraged and part time workers is the U6 unemployment figure, and this is typically quite a bit higher than the headline rate.
Unemployment in a growing economy is never actually zero percent. This is because some people choose not to work (voluntary unemployment), some are in between jobs (frictional unemployment), or some skilled workers find their skills are no longer in demand (structural unemployment). Full employment is a situation where all available workers in the labor force are being used in the most efficient way possible. Full employment embodies the highest amount of skilled and unskilled labor that can be employed within an economy at any given time. Any remaining unemployment is considered to be frictional, structural, or voluntary. In the contemporary United States, the headline unemployment rate associated with full employment has been around four to six percent.
## Inflation
Inflation measures the change in the price levels of goods and services in an economy over time. Inflation is defined as a sustained increase in the general level of prices for goods and services in a country, and is measured as an annual percentage change. Under conditions of inflation, the prices of things rise over time. Put differently, as inflation rises, every dollar you own buys a smaller percentage of a good or service. When prices rise, and alternatively when the value of money falls you have inflation.
Inflation can be caused for a number of reasons, but what is important to understand is that a rate of inflation that is too high or too low is bad for economic stability. Typically an inflation rate between one and four percent annually is ideal. If inflation rises too high, the prices of things in an economy can surge even if wages don’t catch up. In extreme cases, hyperinflation can wreck a nation’s economy. At the same time, if price levels decline, in what is known as deflation, people may stop spending money and companies may halt investments. They anticipate that things will be cheaper tomorrow, so why spend today? This mindset can lead to a dangerous deflationary spiral that can also wreck an economy.
Measuring inflation is a difficult problem for government statisticians. To do this, a number of goods that are representative of the economy are put together into what is referred to as a market basket. The cost of this basket is then compared over time. This results in a price index, which is the cost of the market basket today as a percentage of the cost of that identical basket in the starting year. These measures are commonly the consumer price index (CPI) and the producer price index (PPI).
Economics Basics: Alternatives to Neoclassical Economics
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Macroeconomists try to forecast economic conditions to help consumers, firms and governments make better decisions. | 1,672 | 8,505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2017-43 | latest | en | 0.962697 |
https://goz.icu/impedance-matching-and-unmatched-impedances/ | 1,653,713,680,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663012542.85/warc/CC-MAIN-20220528031224-20220528061224-00097.warc.gz | 328,688,750 | 15,615 | # Impedance Matching and Unmatched Impedances
Impedance Matching
Since a transmission line has impedance built in, the natural question to ask is, how does the impedance affect signals that are relayed by a transmission line from one device to another? The answer to this question ultimately depends on the impedance of the devices to which the transmission line is attached. If the impedance of the transmission line is not the same as the impedance of, say, a load connected to it, the signals propagating by the line will only be slightly absorbed by the load. The rest of the signal will be reflected back in the direction it came. Reflected signals are generally bad things in electronics. They represent an inefficient strength move between two electrical devices. How do you get rid of the reflections? You apply a technique called impedance matching. The goal of impedance matching is to make the impedance of two devices that are to be joined equal. The impedance-matching techniques make use of special matching networks that are inserted between the devices.
Unmatched Impedance
A high-impedance transmission line that is connected to a low-impedance load is, similar to a high-density rope connected to a low-density rope. If you impart a pulse at the left end of the high-density rope (similar to sending an electrical signal by a line to a load), the pulse will travel along the rope without problems until it reaches the low-density rope (load). At that time, the pulse will generate a longer-wavelength pulse within the low-density rope and will generate a similar wavelength but inverted and reduced pulse that rebounds back toward the left end of the high-density rope. From this analogy, again you can see that only part of the signal energy from the high-density rope is transmitted to the low-density rope.
Techniques for Matching Impedance
This section looks at a few impedance-matching techniques. As a rule of thumb, with most low-frequency applications where the signal’s wavelength is much larger than the cable length, there is no need to match line impedance. Matching impedance is usually reserved for high-frequency applications. additionally, most electrical equipment, such as oscilloscopes, video equipment, etc., has input and output impedance that match the characteristic impedance of coaxial cables (typically 50 Ω). Other devices, such as television antenna inputs, have characteristic input impedance that match the characteristic impedance of twin-rule cables (300 Ω). In such situations, the impedance matching is already taken care of. A short length of transmission line that is open ended or short-circuit terminated possesses the character of having an impedance that is reactive. By properly choosing a part of open-circuit or short-circuit line and placing it in shunt with the original transmission line at an appropriate position along the line, standing groups can be deleted. The short part of wire is referred to as a stub.
The author also write articles on http://www.innovativeeideas.com/ | 595 | 3,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-21 | latest | en | 0.941343 |
https://www.splashlearn.com/math/addition-games-for-preschoolers | 1,721,358,961,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.33/warc/CC-MAIN-20240719012903-20240719042903-00071.warc.gz | 866,296,892 | 31,099 | Filter
## Add With Pictures Games for Preschoolers Games
View all 16 games
• Add With Pictures
##### Model to Add Numbers Game
Dive deep into the world of math by modeling to add numbers.
PREK K K.OA.2
VIEW DETAILS
• Add With Pictures
##### Add Objects within 10 Game
Shine bright in the math world by learning how to add objects within 10.
PREK K K.OA.3
VIEW DETAILS
• Add With Pictures
##### Build the Model to Add Numbers Game
Enjoy the marvel of mathematics by exploring how to build the model to add numbers.
PREK K PK.OA.1
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• Add With Pictures
##### Count to Find One More Game
Begin the exciting journey of becoming a math wizard by learning to count to find one more.
PREK K 1.OA.5
VIEW DETAILS
## Addition Strategies Games for Preschoolers Games
View all 18 games
• Compose And Decompose Numbers
##### Addition Sentences (Up to 10) Game
Treat yourself to an immersive learning experience with our 'Addition Sentences (Up to 10)' game.
PREK K K.OA.3
VIEW DETAILS
• Count On To Add
##### Represent Addition Scenarios Game
Let your child see the world through math-colored shades by representing addition scenarios!
PREK K K.OA.2
VIEW DETAILS
• Compose And Decompose Numbers
##### Complete the Addition Sentence Game
Use your addition skills to complete addition sentences.
PREK K K.OA.3
VIEW DETAILS
• Count All To Add
##### Find One or Two More Game
Have your own math-themed party by learning how to find one or two more.
PREK K PK.OA.1
VIEW DETAILS
## All Addition Games
• Addition Strategies
##### Add and Match Game
Take a look at how to add and match with this addition game.
PREK K
VIEW DETAILS
• Addition Strategies
##### Match the Total Game
Have your own math-themed party by learning how to match the total.
PREK K
VIEW DETAILS
• Addition Strategies
##### Add and Match the Total Game
Kids must add and match the total to practice addition.
PREK K
VIEW DETAILS
• Add With Pictures
##### Record Addition in Sentences Game
Kids must record addition in sentences to practice addition skills.
PREK K K.OA.1
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• Add With Pictures
##### Addition Symbol Game
Help your child take flight by learning about the addition symbol.
PREK K K.OA.1
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• Addition Strategies
##### Decompose to Make a Number Game
Enjoy the marvel of mathematics by learning to decompose to make a number.
PREK K K.OA.3
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• Add With Pictures
##### Find the Total Game
Use your addition skills to find the total.
PREK K K.OA.1
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• Addition Strategies
##### Complete the Addition Game
Unearth the wisdom of mathematics by learning how to complete the addition equation.
PREK K
VIEW DETAILS
• Add With Pictures
##### Represent Addition Game
Take the pressure off by simplifying addition with our interactive 'Represent Addition' game.
PREK K K.OA.1
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• Addition Strategies
##### Compose Number in Different Ways Game
Have your own math-themed party by learning how to compose a number in different ways.
PREK K K.OA.3
VIEW DETAILS
• Add With Pictures
##### Solve Addition Sentences Game
Learn how to solve math problems by solving addition sentences.
PREK K K.OA.1
VIEW DETAILS
• Addition Strategies
##### Make Number in Different Ways Game
Enjoy the marvel of mathematics by exploring how to make numbers in different ways.
PREK K K.OA.3
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• Add With Pictures
##### Identify the Total Game
Kids must identify the total to practice addition.
PREK K K.OA.1
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• Addition Strategies
##### Count All to Add Numbers up to 5 Game
Ask your little one to count all to add numbers up to 5.
PREK K K.OA.1
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• Add With Pictures
##### Compose to Make a Number Game
Begin the exciting journey of becoming a math wizard by learning to compose to make a number.
PREK K K.OA.3
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• Addition Strategies
##### Identify One More within 10 Game
Ask your little one to identify one more within 10 to play this game.
PREK K K.OA.1
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• Add With Pictures
##### Adding One by Making a Model Game
Treat yourself to an immersive learning experience with our 'Adding One by Making a Model' game.
PREK K K.OA.2
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• Addition Strategies
##### Count to Tell One More Game
Take a look at how to count to tell 'one more' with this addition game.
PREK K K.OA.1
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• Add With Pictures
##### Adding Within 5 by Making a Model Game
Add more arrows to your child’s math quiver by adding within 5 by making a model.
PREK K K.OA.2
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• Add With Pictures
##### Model and Add (Within 10) Game
Unearth the wisdom of mathematics by learning how to model and add (within 10).
PREK K K.OA.2
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• Add With Pictures
##### Add using Count All Strategy Game
Unearth the wisdom of mathematics by learning how to add using count all strategy.
PREK K 1.OA.1
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• Addition
##### Finding Sum (Up to 10) Game
Dive deep into the world of addition by finding the sum (up to 10).
PREK K K.OA.5
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• Addition
##### Solve 'Put Together' Scenarios Game
Shine bright in the math world by learning how to solve 'Put Together' scenarios.
PREK K 1.OA.1
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• Addition
##### Add Two Numbers (Up to 5) Game
Ask your little one to add two numbers (Up to 5) to play this game.
PREK K K.OA.5
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• Addition
##### Solve 'Put Together' Word Problems Game
Let your child see the world through math-colored shades by solving 'Put Together' word problems!
PREK K 1.OA.1
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• Addition
##### Solve 'Add To' Scenarios Game
Add more arrows to your child’s math quiver by solving 'Add To' scenarios.
PREK K 1.OA.1
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• Addition
##### Solve 'Add To' Word Problems Game
Unearth the wisdom of mathematics by learning how to solve 'Add To' word problems.
PREK K 1.OA.1
VIEW DETAILS
//
### Your one stop solution for all grade learning needs.
Give your child the passion and confidence to learn anything on their own fearlessly
4413+
4567+ | 1,486 | 5,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.786974 |
https://nulpointerexception.com/2020/04/29/golang-journey-basic-data-types/ | 1,695,432,657,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506429.78/warc/CC-MAIN-20230922234442-20230923024442-00009.warc.gz | 486,541,716 | 21,053 | You might want to browse the previous post on this series.
## Integers
• Go provides both signed and unsigned integer type
• There are four different size of signed integer type; 8, 16, 32, 64 bits, represented by int8, int16, int32 and int64
• Similarly there are four different size of unsigned integer types; 8, 16, 32, 64 bits, represented by uint8, uint16, unit32 and uint64
• There are also two types called int and unit which is the most efficient size for signed and unsigned integer on a particular platform
• int is the mostly used type
• There should not be any assumption about size of int as different compilers make different choice even on identical hardwares
• rune is a synonym for int32 and is mostly used for a unicode value
• byte is synonym for uint8
• uintptr – Unsigned int which is used to hold all bits of a pointer value, length is unspecified
• Explicit conversion needed for transferring value from one type to another i.e. to say, int32, int64 and int are three different values
• Signed integer is in 2’s complement form
• Remainder operator (%) is used only for int
• The sign of remainder is always the sign of dividend, so (-7%5) and (-7%-5) both values to -2
• After an arithmetic operation, if the result size is more than what we can represent in the result type, it is said to overflow. The higher order bits are simply discarded
## Floating Point Numbers
• Go provides two size of floating point numbers, float32 and float64
• float64 should be preferred for most purpose as in case of float32, error accumulates rapidly
• Digits may be omitted before or after decimal value, .98 and 3. are both legal declaration
• Scientific notation using e is supported as well and is used in case of very large or very small number
`const Avogadro = 6.02214129e23`
• Floating points are printed using %g verb
## Complex Numbers
• Go provides two size of complex numbers, complex64 and complex128 whose components are float32 and float64
• The built-in functions create a complex number from its real and imaginary components
• The built-in real and imag functions extract these components
```(-5+10i)
-5
10```
## Boolean
• Two possible values – True and False
• Boolean values can be combined with AND and OR operator
## Strings
• A string is an immutable sequence of bytes
• Text strings are conventionally interpreted as UTF-8 encoded sequences
• The built-in len function returns the number of bytes in a string
• Index operation s[i] retrieves the byte at i-th index of string s
• Attempting to access a byte outside this range results in a panic
• The substring operation s[i:j] yields a new string consisting of the bytes of the original string starting at index i and continuing up to, but not including, the byte at index j.
• The i-th byte is not necessarily i-th character as URL encoding of non ASCII requires two or more bytes
• Strings may be compared with comparison operators like == and < and this comparison is done byte by byte
## Constants
• Constants are the expression whose value is known to the compiler
• Evaluation of Constant is done at the compile time and not run time
`const pi = 3.14159`
• We may omit the right-hand side expression for all but the first of the group, implying that the previous expression and its type should be used again in case of sequence of constants
```const (
a=52
b
c= 27
d
)
fmt.Println(a, b, c, d)
Output :: "52 52 27 27"```
• The constant generator iota may be used to create a sequence of related values without spelling out each one explicitly. This is also known as enums
```type Weekday int
const (
Sunday Weekday = iota
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
)
This declares Sunday to be 0, Monday to be 1, and so on```
Read the next post in the series.
#### Reference
The Go Programming Language – Chapter 3 – Alan Donovan
If you liked this article and would like one such blog to land in your inbox every week, consider subscribing to our newsletter: https://skillcaptain.substack.com | 936 | 4,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-40 | latest | en | 0.926295 |
https://www.jiskha.com/questions/17366/marian-has-2-3of-a-bag-af-bagles-a-she-adds-another-1-4-bag-of-bagels-what-fraction-of | 1,579,739,138,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607596.34/warc/CC-MAIN-20200122221541-20200123010541-00145.warc.gz | 910,854,538 | 5,852 | # math
marian has 2/3of a bag af bagles.
A) she adds another 1/4 bag of bagels. what fraction of the bag is full with bagels now?
B)Marian has another bag of bagelsthat is 5/6 full. what fraction describes how many more bagels are in the bag in part (A)?
To add 2/3 and 1/4 you need to use a common denominator which is 12.
8/12 + 3/12 = _____________
B) Find the common denominator and subtract 5/6 from the answer to A.
huh i don't really get it sorry
1. 👍 0
2. 👎 0
3. 👁 136
1. 11/12
1. 👍 1
2. 👎 0
posted by Lilly
2. The answer took me 3 three days to fugure it out
1. 👍 0
2. 👎 0
posted by Bianca
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asked by Apple Annie on January 12, 2020
8. ### physics
A helicopter is rising at 5.1 m/s when a bag of its cargo is dropped. (Assume that the positive direction is upward.) (a) After 2.0 s, what is the bag's velocity? (b) How far has the bag fallen? (c) How far below the helicopter is
asked by penny on September 13, 2010
9. ### Physics...not sure what formula to use
A 2.0kg bag is held by a string to the ceiling. A 10g bullet travelling at 200m/s strikes the stationary bag. The height of the bag after the collision is 10cm, assuming there is no friction, determine the speed, in m/s, of the
asked by Bobbie on October 25, 2008
10. ### physics
A helicopter is rising at 4.7 m/s when a bag is dropped from it. (Assume that the positive direction is upward.) (a) After 2.0 s, what is the bag's velocity? (b) How far has the bag fallen? (c) How far below the helicopter is the
asked by Tay on October 26, 2009
More Similar Questions | 1,049 | 3,456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-05 | latest | en | 0.966217 |
http://math.stackexchange.com/questions/15575/square-of-a-graph/15576 | 1,469,660,778,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827080.38/warc/CC-MAIN-20160723071027-00135-ip-10-185-27-174.ec2.internal.warc.gz | 160,776,321 | 20,480 | # Square of a graph
Could someone tell me what the "square of a graph" $G^2$ is? Thanks.
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This might mean cartesian product? Which of course doesn't yield another graph, so the answer depends a lot on the context... – Aaron Mazel-Gee Dec 26 '10 at 19:49
@Aaron: there is a standard definition of the Cartesian product of two graphs: en.wikipedia.org/wiki/Cartesian_product_of_graphs . Oddly enough it is not the categorical product. – Qiaochu Yuan Dec 26 '10 at 20:51
There are unusually many different species of "graph product". I think there is even a book with this as the title...(See David's answer below for an acknowledgment of this multiplicity of definitions.) – Pete L. Clark Dec 29 '10 at 7:33
The square of a graph $G$ is obtained by starting with $G$, and adding edges between any two vertices whose distance in $G$ is $2$.
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Here is a book which has this definition: books.google.com/books?id=AnqFawQJVm0C&pg=PA218 (See first paragraph of 10.3). Even Frank Harary's book on graph theory has this definition, but I was not able to find an online reference. btw, distance is atmost 2 (I edited it for you). – Aryabhata Dec 26 '10 at 19:58
Another reference: citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.47.6167 – Qiaochu Yuan Dec 26 '10 at 20:50
You should probably say, rather, one square of a graph is... – Mariano Suárez-Alvarez Jan 3 '11 at 16:06
@Moron: 1. Because I started with the graph G. So I don't want to add edges that are already there. 2. Because it's "at most", not "atmost". Now remove your downvote, or I will come and stalk you. – TonyK Jan 3 '11 at 17:34
@Moron: "Don't be such a Jerk. Anyway, my apologies if I caused you some distress." A nice juxtaposition. An oxyMoron, if you will :-) – TonyK Jan 3 '11 at 21:23
I would have thought that $G^2$ would either mean the box product of $G$ with itself, or the cross product of $G$ with itself.
The definitions of these are as follows: If $G$ and $H$ are graphs with vertex sets $V_G$ and $V_H$, then the box product of $G$ and $H$ has vertex set $V_G \times V_H$ and has an edge from $(g_1, h_1)$ to $(g_2, h_2)$ if and only if either (1) $g_1=g_2$ and there is an edge from $h_1$ to $h_2$ in $H$ or (2) $h_1=h_2$ and there is an edge from $g_1$ to $g_2$ in $G$.
The cross product of $G$ and $H$ has vertex set $V_G \times V_H$ and has an edge from $(g_1, h_1)$ to $(g_2, h_2)$ if and only if there is an edge from $g_1$ to $g_2$ in $G$ and an edge from $h_1$ to $h_2$ in $H$.
Since TonyK has found yet another definition, I would say that there is more than one thing $G^2$ can denote.
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Do you have a reference which has this definition? Most of the definitions of the square of a graph I have come across agree with TonyK's answer. – Aryabhata Dec 26 '10 at 20:21
The names I know for these notions are Cartesian and tensor product; I agree that box product is probably a better name for the first notion. But I don't think that this is what most people who use the phrase "the square of a graph" mean. – Qiaochu Yuan Dec 26 '10 at 20:53
Davis, this is certainly interesting. I suppose one rationale I could think of for the definition TonyK gave (and thanks to Moron for the reference, which makes it reasonable to believe that this is the usual terminology) is that if you think of a graph as a category, so the edges represent maps, then $G^2=G\circ G$ could stand for the composition of those maps. I guess this would actually suggest to make edges for those points that are connected by a path of length $2$ exactly. Then again, what do I know? – Michele Kakusi Dec 27 '10 at 1:17
Michele: the other rationale behind the definition that TonyK gave is that (with particular definitions) the adjacency matrix of the 'square' graph $G^2$ is precisely the (matrix-product) square of the adjacency matrix for the graph $G$ (including multiplicities - generally an edge is added from $u$ to $v$ for each 2-path from $u$ to $v$ in $G$.) – Steven Stadnicki Jan 3 '11 at 18:37
An oriented graph $G$ is a directed graph with no parallel edges. The square of an oriented graph is a graph $G'$ whose vertex set $V(G')$ is the same as the vertex set $V(G)$ of $G$. An ordered pair of vertices $(u,w)$ is in the arc set $A(G')$ of $G'$ if and only if there exists a vertex $v$ in $G$ (and consequently in $G'$) such that $(u,v)$ and $(v,w)$ are arcs in $G$.
A similar definition for simple graphs may be culled from the above by replacing arcs with edges and ordered pairs of vertices with 2-element subsets of $V(G)$.
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...or, indeed, by treating undirected graphs simply as directed graphs where each edge goes both ways. – Ilmari Karonen Feb 8 '12 at 20:51 | 1,362 | 4,653 | {"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-2016-30 | latest | en | 0.924493 |
https://docs.vicon.com/display/ProCalc15/Defining+the+femur+segments | 1,709,264,327,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00771.warc.gz | 202,607,150 | 19,656 | This information is for Vicon ProCalc 1.5. For up-to-date help, see the latest ProCalc documentation.
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# Defining the femur segments
Important
Segments must be defined with the origin at the proximal end. Sometimes, biomechanical models use the distal end as the origin for segments (eg, the knee joint center for the femur). For successful VSK generation, you must define all segments using the proximal end as the origin (i.e. the hip joint center for the femur).
You already have the origins for the two femur segments defined: LHJC and RHJC (see Defining the hip model). For the distal end, you need the knee joint centers. These are defined as the half-way point between the lateral and medial knee markers. You can then define the axes that define the femur segments:
1. On the Variables tab, with the CGM2.3 scheme selected, add a new variable named LKJC. Define this as a point halfway between LKNE and LKNM.
2. Repeat Step 1 for the right side.
3. Add a new vector named LFemurPD from point A=LKJC to point B=LHJC.
4. Add a new vector named LFemurML from point A=LKNE to point B=LKNM.
5. Use these two vectors to define the femur segment, which must be named L_Femur (to match the name of the segment in the VSK file). Add a new segment, name it L_Femur, select Origin A, Z-axis=B, X-Axis=BxC, then A=LHJC (remember, at the proximal end), B=LFemurPD and C=LFemurML.
6. Repeat Steps 3-5 for the right side to define R_Femur.
Note that if you define the RFemurML from RKNE to RKNM, this vector points in the opposite direction (left) compared to the left side, which means that the R_Femur will have its Y-axis pointing right and its X-axis pointing backwards. To correct this, either use a factor -1 for the RFemurML in the R_Femur's specification, or flip the RFemurML vector itself.
The following image shows the coordinate system of the R_Femur segment: | 498 | 1,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-10 | latest | en | 0.853435 |
https://api.flutter.dev/flutter/package-collection_collection/DelegatingQueue/fold.html | 1,660,959,027,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573876.92/warc/CC-MAIN-20220820012448-20220820042448-00222.warc.gz | 120,819,475 | 3,832 | # fold<T> method Null safety
T fold<T>(
1. T initialValue,
2. T combine(
1. T previousValue,
2. E element
)
)
inherited
Reduces a collection to a single value by iteratively combining each element of the collection with an existing value
Uses `initialValue` as the initial value, then iterates through the elements and updates the value with each element using the `combine` function, as if by:
``````var value = initialValue;
for (E element in this) {
value = combine(value, element);
}
return value;
``````
Example of calculating the sum of an iterable:
``````final numbers = <double>[10, 2, 5, 0.5];
const initialValue = 100.0;
final result = numbers.fold<double>(
initialValue, (previousValue, element) => previousValue + element);
print(result); // 117.5
``````
## Implementation
``````@override
T fold<T>(T initialValue, T Function(T previousValue, E element) combine) =>
_base.fold(initialValue, combine);`````` | 238 | 926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-33 | longest | en | 0.610822 |
https://physics.stackexchange.com/questions/397628/adiabatic-work-done-by-gas | 1,708,672,570,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474361.75/warc/CC-MAIN-20240223053503-20240223083503-00497.warc.gz | 475,217,244 | 44,965 | # adiabatic work done by gas
Suppose we have an adiabatic box with a fixed volume $V$ and contains $n_0$ mol of gas at pressure $p_0$ and temperature $T_0$. Now the box is punctured by a small hole and gas from the outside flows in. The outside (surrounding) has a pressure of $p$ and a temperature of $T_0$. Assume the internal energy of $n$ mol of gas at temperature $T$ is $nc_v T$.
What is the final temperature in the box?
I tried the above problem but couldn't get far; I can get the temperature $T$ in terms of $T_0, R, c_v$ if the box is empty at first (ie. a vacuum). $\left(T = T_0 \frac{c_v + R}{c_v}\right)$
For this problem my thinking is something like:
$\Delta Q = 0$, so $\Delta U = \Delta W$
Then the gas outside has to do both boundary and shaft/flow work
The boundary work is done as the gas outside is expanding against the pressure $p_0$ in the box and
The shaft/flow work is done as the gas enters the box through the hole
However this is where I can't quite continue
Is $\Delta W = \Delta(PV) = P\Delta V + V\Delta P$? Or is $\Delta W = P\Delta V + PV$?
Furthermore, if $\Delta W = P\Delta V + PV$, then is the $P$ in $PV$ equal to $p - p_0$?
Finally, if, as the temperature of the gas originally in the box increases due to the work done by the outside gas, in order to calculate the final temperature in the box we need to know how much gas from outside entered, how can we do this?
Thanks.
• So what happens if you compute the "start empty - got to pressure $p_0$" result and subtract if from "start empty - go to pressure $p$"? Shouldn't that be the work done going from $p_0$ to $p$? Apr 4, 2018 at 14:27
• @Floris I'm not sure if that approach would work, because (I think that) in your scenario the internal energy of the gas in the box at $p_0$ is greater than $T_0$ due to the gas entering, whereas in the original scenario the initial temperature of the gas inside the box is at $T_0$ - do correct me if I went wrong somewhere though, thanks! Apr 4, 2018 at 14:33
• Alright - figure out what temperature you need to start at to get the gas inside to the right temperature, and make the correction by doing the first part starting at a different temperature. So there are then 3 parts to the calculation - does that make sense? Apr 4, 2018 at 14:35
• @Floris just to check that I understood that correctly, basically $\Delta U =$ (work from empty to pressure $p$) - [(work from empty to pressure $p_0$) - (heat required to be removed to make temperature $T_0$)], is that right? Also just for further clarification wouldn't the number of moles of gas that entered the box be different for the "empty to $p$" and the "empty to $p_0$", wouldn't this affect the final temperature? Thanks Apr 4, 2018 at 14:40
• Yes you need to be careful - my point is that you can come up with a cycle of steps (that you know how to calculate) that allows you to go from the correct initial state to the correct final state. I will leave the hard work up to you... since this seems to be a "homework-like" question, and that's how the policy works. Apr 4, 2018 at 14:49
The shaft work is the amount of work by the system on its surroundings, over and above the work required to push mass into and out of the control volume. For this problem this shaft work is zero.
The open system first law energy balance on the control volume becomes: $$\Delta U=\Delta nh_{in}$$ or $$(n_0+\Delta n)C_vT_f-n_0C_vT_0=\Delta n(C_v+R)T_0$$This gives: $$\frac{T_f}{T_0}=1+\frac{\Delta n}{n_0+\Delta n}\frac{R}{C_v}\tag{1}$$ The final pressure is going to be equal to the outside pressure p. From the ideal gas law we have, from the initial condition, $$p_0V=n_0RT_0$$and, from the final condition, we have:$$pV=(n_0+\Delta n)RT_f$$So, $$\frac{T_f}{T_0}=\frac{p}{p_0}\frac{n_0}{(n_0+\Delta n)}\tag{2}$$ Eqns. 1 and 2 provide two equations in the two unknowns $T_f$ and $\Delta n$. The solution for $T_f$ is as follows: $$\frac{T_f}{T_0}=\frac{\gamma}{1+(\gamma-1)(p_0/p)}$$
• @pglpm What you are saying is technically correct, and that, if one waited long enough after the pressures effectively equilibrated, conductive heat transfer could occur through the hole until the temperature inside equilibrated with the outside temperature. However, in my judgment, this was not the intent of the homework exercise represented by this problem. I think the student was expected to assume that either the hole was plugged by insulation after pressure equilibration, or that the problem ended at the effective time of equilibration. Apr 4, 2018 at 19:46
• This is the same as the scenario you already described, where heat transfer is allowed through the hole between the two chambers. The actual problem is more like one involving two chambers at different pressures separated by an insulated barrier, in which a hole is poked in the barrier. For the lower pressure chamber, this is a very irreversible process in which viscous heating accounts for the increase in temperature (while the gas remaining in the higher pressure chamber at the end has expanded reversibly to push part of the mass out and suffers a temperature decrease). Apr 4, 2018 at 20:49
• Who says that an ideal gas is inviscid? In the limit of low densities, all gases approach ideal gas behavior and also approach a constant viscosity that is a function only of temperature. When an ideal gas expands rapidly and irreversibly against a constant external pressure (that is much lower than the initial gas pressure), what did you think was the source of the irreversibility? And in the Joule Thompson flow of an ideal gas through a porous plug (where the gas suffers an irreversible expansion), what did you think the reason was that the gas doesn't cool? Apr 4, 2018 at 22:10
• Well, I think we have a disagreement between experts here. Maybe an ideal gas is inviscid by your definition, but not generally. There are still collisions between molecules in ideal gases that allow for exchange of kinetic energy between molecules. This is sufficient to result in viscous behavior. See Bird, Stewart, and Lightfoot, Transport Phenomena for the derivation of the equation for viscosity in the limit of ideal gas behavior. Apr 4, 2018 at 22:35
• @CM Thank you for replying – and so quickly. Funnily enough I'd just come round to this understanding. Using a much less sophisticated argument than yours, I imagined the gas being pushed through the hole at temperature $T_0$ and pressure $p_0$ by a piston. This would require work $p \Delta V =p \frac{\Delta n \ RT_0}{p}=\Delta n\ RT_0$. This has to be added to the internal energy $\Delta n\ c_{v}T_0$ that the gas takes with it into the adiabatic chamber. This leads to exactly your result. Apr 11, 2018 at 23:19
This Answer provides an entirely different method of solving the present problem, based on treating the box and surroundings as a closed system (and thus using the closed system version of the first law of thermodynamics). This is the approach alluded to by @pglpm in one of his comments. Rather than considering the surroundings outside the box as being infinite, we consider the gas outside the box as being enclosed within a larger adiabatic container of finite volume. We then solve this problem in the limit as the outer container volume becomes infinite.
Here are the parameters employed in the present analysis:
Box:
$n_0$ = number of moles of gas in box initially
V = Volume of box
$T_0$ = Initial Temperature
$p_0$ = Initial Pressure
n = number of moles in box in final state
T = Temperature in box in final state
p* = Pressure in box in final state (identical to final pressure outside box)
Outside Enclosure:
$n_{s0}$ = number of moles of gas in enclosure initially
$V_s$ = Volume of enclosure
$T_0$ = Initial temperature
p = Initial pressure
$n_s$ = Final number of moles in enclosure
T* = Final temperature of gas in enclosure
p* = Final pressure of gas in enclosure (identical to final pressure in box)
From the ideal gas law, we have: $$n_0=\frac{p_0V}{RT_0}\tag{1a}$$ $$n=\frac{p^*V}{RT}\tag{1b}$$ $$n_{s0}=\frac{pV_S}{RT_0}\tag{1c}$$ $$n_s=\frac{p^*V_S}{RT^*}\tag{1d}$$
As shown in Example 6.10 of Fundamentals of Engineering Thermodynamics by Moran et al, when a gas within an adiabatic enclosure escapes very slowly (in our case into the box), the gas that still remains inside the enclosure at any time during the process has suffered an adiabatic reversible expansion. This means that the final pressure and temperature of the gas in the enclosure will be less than the initial pressure and temperature. Furthermore, quantitatively, we will have that: $$p^*\left(\frac{V_s}{n_s}\right)^{\gamma}=p\left(\frac{V_s}{n_{s0}}\right)^{\gamma}$$or equivalently, $$\frac{n_s}{n_{s0}}=\left(\frac{p^*}{p}\right)^{1/\gamma}$$or equivalently,$$n_s=\frac{pV_S}{RT_0}\left(\frac{p^*}{p}\right)^{1/\gamma}\tag{2}$$Moreover, we have: $$\frac{T^*}{T_0}=\left(\frac{p^*}{p}\right)^{\frac{\gamma - 1}{\gamma}}\tag{3}$$
For the closed system consisting of the box and the rigid insulated enclosure, there is no work done by the system on its surrounding and no heat exchange between the system and its surroundings. Therefore, from the version of the first law of thermodynamics applicable to a closed system, the change in internal energy of this combined system is zero. Initially, the gas in both the enclosure and the box are at the same temperature, $T_0$. In the final state of the system, the $n_s$ moles of gas in the enclosure are at T* and the n moles of gas in the box are at T. Therefore, from the first law: $$nC_v(T-T_0)+n_sC_v(T^*-T_0)=0\tag{4}$$From a mass balance on the system, the number of moles of gas in the initial state is equal to the number of moles of gas in the final state: $$n+n_s=n_0+n_{s0}\tag{5}$$If we substitute Eqns. 1 into Eqns. 4 and 5, we obtain:$$V\left(1-\frac{T_0}{T}\right)+V_S\left(1-\frac{T_0}{T^*}\right)=0\tag{6}$$and$$\left(V\frac{T_0}{T}+V_S\frac{T_0}{T^*}\right)=\frac{p_0V+pV_S}{p^*}\tag{7}$$ Combining Eqns. 6 and 7 yields: $$p^*=\frac{V_S}{(V_S+V)}p+\frac{V}{(V_S+V)}p_0\tag{8}$$According to Eqn. 8, the final pressure p* is just a weighted average of the initial pressures in the enclosure and the box, weighted in terms of the volumes of the two containers.
We can now determine the final temperature T* in the enclosure by combining Eqns 3 and 8 to yield: $$\frac{T^*}{T_0}=\left(\frac{V_S}{(V_S+V)}+\frac{V}{(V_S+V)}\frac{p_0}{p}\right)^{\frac{\gamma - 1}{\gamma}}\tag{9}$$
If we substitute this into Eqn. 6 and solve for T, the final temperature in the box, we obtain: $$\frac{T}{T_0}=\frac{1}{\left[1+\frac{V_S}{V}\left(1-\frac{\left(1+\frac{V}{V_S}\right)^{(\gamma-1)/\gamma}}{\left(1+\frac{V}{V_S}\frac{p_0}{p}\right)^{(\gamma-1)/\gamma}}\right)\right]}\tag{10}$$If we take the limit of this relationship as $V/V_S$ approaches zero (i.e., the volume of the enclosure becomes infinite), we obtain:$$\frac{T}{T_0}=\frac{\gamma}{1+(\gamma-1)(p_0/p)}\tag{11}$$ This is exactly the same result we obtained in the previous analysis using the open system version of the first law of thermodynamics.
• Wow! Thank you very much for posting this. I've got the general drift and am now slowly working through all the steps. I think it will be worth it. Apr 13, 2018 at 20:32
• I like this very much, because there are no doubtful assumptions. The fact that anyone considered it worthwhile to produce this undeniably long solution seems to me to confirm that the problem posed was more challenging than it looked. I'm still quite attached to my approach (replacing the infinite external gas reservoir by a piston and cylinder). This gives the right answer for $T/T_0$ with little effort, and I doubt if it does so by cancellation of errors, but as I've said in an earlier comment, it's not as easily justified as the very interesting method you've just given. Apr 13, 2018 at 22:01
• "the gas that still remains inside the enclosure at any time during the process has suffered an adiabatic reversible expansion." This, I think, is your key sentence. It's what shows my (piston) method to be equivalent to your second answer. The gas that goes through the hole $might\ as\ well$ be separated from the gas that remains in the enclosure by a piston. And the work done by this piston gives the extra internal energy acquired by the gas that finishes up in the box. [Of course the work done by the piston is at the expense of the gas that remains in the enclosure.] Apr 14, 2018 at 10:03
• I like to visualize it as Moran el al have shown in their figure, more like an invisible membrane separating the gas that eventually remains outside the box from the from the gas that eventually ends up inside the box. Of course, topologically, thinking of it as a piston is just as valid. Apr 14, 2018 at 10:59
• I see. I don't possess Moran, but I believe that physicists can often learn from an approach directed at engineers. Many thanks for engaging in this dialogue; I've benefited greatly . Maybe others have, too. Apr 14, 2018 at 11:17 | 3,566 | 13,031 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-10 | latest | en | 0.944271 |
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##### how to use a spring balance
12.01.2021, 5:37
The quick answer to your question is "Yes, you could use a rubber band instead of a steel spring" to measure the weight of objects." Some chemicals you may use in a lab could corrode or otherwise damage the … Suppose, for example, the extension of the spring is 5 em when a piece of metal is placed in the From the graph (Fig. Spring balance. Is there any way you might use a spring balance to compare the 'heaviness' of twin babies born on the shuttle? Spring balance is a device used for measuring the ..... acting on an object. So what is Hooke’s law and what is a Newton spring balance? The balance should be clean and free of debris. Steve. They just look different. spring balance is spring balance. Spring Balance: The weight of an object is measured using a spring balance. Now, use a second paper clip, looped onto the second-from-the-bottom coil, to hang your bucket or cup from the bottom of the spring. Spring scales are typically used to determine the weight of fish caught at a fishing tournament or to weigh produce at a store. Look it up now! The use of metal springs was continually developed and today they are the most common balance systems used in mass-produced metal, plastic as well as wooden windows. Baby birds are hatching, the days are getting longer, there are butterflies everywhere and you feel inspired to do a good deep clean of your lab scales. Give one use of spring balance. asked Oct 19, 2020 in Physics by Saadhvi (54.5k points) friction; class-8; 0 votes. This tool allows installation of balance spring collets on the balance shaft with one end, and precise rotation of the collet with the other end. Also, the spring in the scale can permanently stretch with repeated use. Step 3: A thread is tied to the given solid body (a piece of stone) and suspended from the hook at the lower end of the spring balance. Make sure the spring balance is parallel to the surface. Never place a sample directly on the balance. 4.5) it will be seen that an extension of 5 em corresponds to of 0.45 N. These tools make adjustment of beat as simple as possible and are worth investing in. graduated scale The divisions of equal length that are marked on the spring balance and constitute the units of measurement. The sample is placed in a bottle and hung on the spring hook through a glass wire and then evacuated. Step 1 Hold the spring scale in one hand. Be sure that you hold it high enough so you can see the arrow and the numbers clearly. Spring time means spring cleaning! You should use a weighted boat, weighing sheet, or another container to hold the sample. 1 answer. One question: I use a timegrapher and have seen on a movement with power reserve indicator (Orient 40N5A) that the more the clock spring is wound up the faster the watch goes. A spring scale is a simple device used to measure force or weight (not mass). Il y a 10 années. Will there be or not be a problem if I use a spring balance upside down? Uses [edit | edit source]. Suspend the balance and attach a string length to the weighing hook. Lv 7. The reading on the spring balance scale when the load begins to slide is a measure for the static friction, while the reading when the block continues to slide is a measure of dynamic friction. Answer: The weight of a body = mass x g, to get mass, divide the given weight by g. Question 3: How can you measure 1 N weight? Spring Scale Parts. What is a spring balance? 2 réponses. To find mass I will use a beam balance and to find weight I will use a spring balance. The hole diameter (to accomodate the balance shaft) is 0.70 millimeters on this (G) part. Question 2: How can you convert the weight value into mass? In either case, you'll need to know how to read a spring scale. They work the same way. A balance spring, or hairspring, is a spring attached to the balance wheel in mechanical timepieces. asked May 13, 2019 in Science by Aadam (71.9k points) friction; class-8; 0 votes. Small scales with sensitive springs can be used to measure differences of up to a gram or less, though these will break if heavy weights were attached to them. This is achieved using a law known as Hooke’s law. next. We can’t help you sing like Snow White so that animals come help you clean but we can help you get your scales looking their best! Add a few weights (such as coins) to the bucket or cup. A spring scale is a kind of spring balance weighing scale which consists of a spring fixed at one end of the device which is attached with a hook to the other end where we can attach an object in order to weigh it. The principle underlying the spring balance was first investigated in the seventeenth century by Robert Hooke. 3 Answers. It consists of a spring attached to a scale. Thanks to their simplicity of design, they're easy to read. A spring balance is a weighing equipment used in industries to measure the mass of different loads. Measurement of Mass by Use of Beam Balance: Working Principle: A beam balance works on the principle of moments. Spring scales can be used to measure weight in a wide range, depending on the stiffness of the particular spring. hook Curved part on which the body to be weighed is hung. It is also used in science experiments as a essential accelerometer. Some spring balances are available in gram or kilogram markings and are used to measure the mass of an object. If it is problematic, how should I address the problem when measuring a force with the spring balance upside down? Loop this string round and up to a fixing point (so the loop hangs below the balance) The weight of anything hanging from the string loop is supported by two equal forces (from each side or the loop) .. the spring balance is measuring one of these forces. For a spring balance to measure weight, the result should be easy to read. could you use a rubber band in place of a steel spring in a balance to measure the weights of objects? Relevance. Step 2: The number of divisions between two long graduation marks is found on the scale of the spring balance and find its least count. This can be also seen related to amplitude: the higher the amplitude is, the faster it goes. A spring balance can be calibrated for the accurate measurement of mass in the location in which they are used, but many spring balances are marked right on their face "Not Legal for Trade" or words of similar import due to the approximate nature of the theory used to mark the scale. It would be helpful for the unit to have small hash marks for calculating the total weight. 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The fastener was released the weights of objects to know how to read it be! | 3,482 | 16,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.908517 |
http://www.howtodothings.com/hobbies/how-to-write-words-with-a-calculator | 1,560,676,232,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998084.36/warc/CC-MAIN-20190616082703-20190616104703-00470.warc.gz | 253,594,683 | 7,257 | # How To Write Words with a Calculator
Writing words with a calculator is a simple and novel activity that can be used in all sorts of situations. Just type in the right numbers or symbols, and you can read the resulting words when looking at the LCD screen upside down. Using these messages for lighthearted pranks or as display signs on the road are just a few ways these codes can be used.
1) First, you should either memorize or write down the alphabetical equivalent of all the symbols found on your calculator, either as they appear upside down, or right-side up. Just be creative, and keep aesthetics in mind! Here is a sample list:
0 = "O" or "D"
1= "I"
2 = "Z"
3 = "E"
4 = "H" or "A"
5 = "S"
6 = "G"
7 = "L"
8 = "B"
9 = "G"
x = "X"
+ = "T" (if the "+" symbol shows up on your screen)
Additionally, decimal points (".") make sentences a little more legible. For example, "3080.07734" is more legible calculator writing than simply "308007734".
2) Next, think of a word to write. The word should be made of letters that can easily be represented by a correctly interpreted number. Some commonly typed examples are: 07734 = "HELLO," and 008 = "BOO," 3080 = "OBOE," and 707 = "LOL." After typing your numerical word, simply turn your calculator upside down and show it to somebody!
The following is a very common joke that involves calculator writing. Just say the following narrative to a friend while typing in the corresponding numbers and performing the correct functions:
Once upon a time, there was a woman who was born in '69 (69). She wore dress size 222 (69222) and she lived on 51st street (6922251). Well, she went to go see Dr. X (6922251 x), and she was patient number 8 (6922251 x 8), and when she came out, she was... Now press the equal ("=") sign, turn your calculator upside-down, and discover what happened to the woman!
Other activities you can do with this phenomenon is teach it to a child who gets bored easily at school, or even tape your calculator upside down on a refrigerator. For example, if you're leaving the house right after finishing all the eggs, and have no paper, you can type "0.51.5993," or "5993.3507.1." As long as you're creative with your syntax and grammatical skills, you can communicate almost anything with calculator writing! | 573 | 2,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-26 | latest | en | 0.95961 |
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# Experiment to Prove Hooke's Law
✅ Paper Type: Free Essay ✅ Subject: Physics ✅ Wordcount: 2832 words ✅ Published: 10th Aug 2021
Hooke’s Law
Aim: -To prove Hooke’s law i.e. the extension of the force is directly proportional to the force applied.
– To find the spring constant of the spring.
Apparatus:
• Clamp Stand
• Helical Spring
• Mass Hanger
• Pointer
• Meter Ruler
• Measuring Balance
Method:
-Hang a helical spring from a clamp stand.
-Attach a mass directly to the bottom of the helical spring and record the position of the bottom of the mass hanger relative to a meter ruler.
-Add masses to the spring and record the position of the bottom of the mass hanger.
Safety Precautions:
– Wear safety goggles to prevent any accidents that could occur due to the weights bouncing off the spring.
– Keep a distance from the apparatus.
– Be sure that the spring is tightly attached to the clamp.
– Do not play around with the masses or springs.
Data Collection and Processing
Uncertainty in a measuring balance = ±0.1g
To covert to kg = 0.1÷1000 = ±0.0001kg
Uncertainty in a meter ruler = ±0.05cm
To convert to meters = 0.05 ÷ 100 = ±0.0005m
### Formulas
Relative Uncertainty= Absolute Uncertainty ÷ Measured Value
% Uncertainty = Absolute Uncertainty ÷ Measured Value × 100
Force (Newton’s) = Mass (Kg) × Acceleration (ms-²)
Spring Constant, k (Nm-¹) = Force (Newton’s) ÷ Extension (m)
Elastic Potential Energy (Joules) = 0.5 × Spring Constant × Extension²
Random Error = Range of extension ÷ 2
Table 1Raw Data Table:
Trial No. Mass (grams) ±0.1 Mass (kilograms) ±0.0001 Force Applied (Newton’s) F=M×g ±0.0001 Extension While Loading(meters) ±0.0005 Extension While Unloading(meters) ±0.0005 Average Extension =E1+E2÷2 (meters) ±0.001 1 10.2±0.1 0.0102±0.0001 0.100062±0.0001 0.036±0.0005 0.037±0.0005 0.0365±0.001 2 20.4±0.1 0.0204±0.0001 0.200124±0.0001 0.040±0.0005 0.039±0.0005 0.0395±0.001 3 30.6±0.1 0.0306±0.0001 0.300186±0.0001 0.043±0.0005 0.042±0.0005 0.0425±0.001 4 40.8±0.1 0.0408±0.0001 0.400248±0.0001 0.048±0.0005 0.046±0.0005 0.0470±0.001 5 51.0±0.1 0.0510±0.0001 0.500310±0.0001 0.051±0.0005 0.050±0.0005 0.0505±0.001 6 61.2±0.1 0.0612±0.0001 0.600372±0.0001 0.056±0.0005 0.057±0.0005 0.0565±0.001 7 71.4±0.1 0.0714±0.0001 0.700434±0.0001 0.061±0.0005 0.060±0.0005 0.0605±0.001 8 81.6±0.1 0.0816±0.0001 0.800496±0.0001 0.067±0.0005 0.067±0.0005 0.0670±0.001
Calculations for trial 1
Force (Newton’s) = Mass (kg) × Acceleration (ms-²)
= 10.2±0.1 (g) × 9.81 (ms-²)
= 100.062±0.1 (g)
Covert the g to kg: 100.062 ÷ 1000
= 0.100062±0.0001 (kg)
= 3.6±0.05 (cm) + 3.7±0.05 (cm)
= 3.65±0.1cm
In meters = 3.65±0.1cm ÷ 100 = 0.0365±0.001m
Table 2– The range of extension and the random error of the experiment:
Trial No. Extension While Loading(meters) ±0.0005 Extension While Unloading(meters) ±0.0005 Average Extension =E1+E2÷2 (meters) ±0.001 Force Applied (Newton’s) F=M×g ±0.0001 Range of Extension (meters) ±0.0005 Random Error (meters) ±0.0005 1 0.036±0.0005 0.037±0.0005 0.0365±0.001 0.100062±0.0001 0.001±0.0005 0.0005±0.0005 2 0.040±0.0005 0.039±0.0005 0.0395±0.001 0.200124±0.0001 0.001±0.0005 0.0005±0.0005 3 0.043±0.0005 0.042±0.0005 0.0425±0.001 0.300186±0.0001 0.001±0.0005 0.0005±0.0005 4 0.048±0.0005 0.046±0.0005 0.0470±0.001 0.400248±0.0001 0.002±0.0005 0.001±0.0005 5 0.051±0.0005 0.050±0.0005 0.0505±0.001 0.500310±0.0001 0.001±0.0005 0.0005±0.0005 6 0.056±0.0005 0.057±0.0005 0.0565±0.001 0.600372±0.0001 0.001±0.0005 0.0005±0.0005 7 0.061±0.0005 0.060±0.0005 0.0605±0.001 0.700434±0.0001 0.001±0.0005 0.0005±0.0005 8 0.067±0.0005 0.067±0.0005 0.0670±0.001 0.800496±0.0001 0.000±0.0005 0.0000±0.0005
Calculations for trial 1
Force (Newton’s) = Mass (kg) × Acceleration (ms-²)
= 10.2±0.1 (g) × 9.81 (ms-²)
= 100.062±0.1 (g)
Covert the g to kg: 100.062 ÷ 1000
= 0.100062±0.0001 (kg)
= 3.6±0.05 (cm) + 3.7±0.05 (cm)
= 3.65±0.1cm
In meters = 3.65±0.1cm ÷ 100 = 0.0365±0.001m
Range Of Extension = Maximum Value – Minimum Value
= 0.037±0.0005 – 0.036±0.0005
= 0.001±0.005 (m)
Random Error = Range of extension ÷ 2
= 0.001±0.005 ÷ 2
= 0.0005±0.0005 (m)
Table 3Processed Data Table:
Trial No. Force Applied (Newton’s) F=M×g ±0.0001 Average Extension =E1+E2÷2 (meters) ±0.001 Spring Constant, k (Nm) % Uncertainty Elastic Potential Energy (Joules) % Uncertainty 1 0.100062±0.0001 0.0365±0.001 2.74±2.8% 0.0018251825±8.3% 2 0.200124±0.0001 0.0395±0.001 5.01±2.6% 0.0039084263±7.7% 3 0.300186±0.0001 0.0425±0.001 7.06±2.4% 0.0063760625±7.1% 4 0.400248±0.0001 0.0470±0.001 8.52±2.1% 0.0094103410±6.4% 5 0.500310±0.0001 0.0505±0.001 9.91±2.0% 0.0126364880±6.0% 6 0.600372±0.0001 0.0565±0.001 10.6±1.8% 0.01721974±5.3% 7 0.700434±0.0001 0.0605±0.001 11.6±1.7% 0.02122945±5.0% 8 0.800496±0.0001 0.0670±0.001 11.9±1.5% 0.02670955±4.5%
Calculations for trial 1
Force (Newton’s) = Mass (kg) × Acceleration (ms-²)
= 10.2±0.1 (g) × 9.81 (ms-²)
= 100.062±0.1 (g)
Covert the g to kg: 100.062 ÷ 1000
= 0.100062±0.0001 (kg)
= 3.6±0.05 (cm) + 3.7±0.05 (cm)
= 3.65±0.1cm
In meters = 3.65±0.1cm ÷ 100 = 0.0365±0.001m
Spring Constant = Force (Newton’s) ÷ Extension (m)
= 0.100062±0.0001 (N) ÷ 0.0365±0.001 (m)
% Uncertainty for Force = Absolute Uncertainty ÷ Measured Value × 100
= 0.0001 ÷ 0.100062 × 100
= 0.1%
% Uncertainty for Extension = Absolute Uncertainty ÷ Measured Value × 100
= 0.001 ÷ 0.0365 × 100
= 2.7%
Spring Constant = 0.100062±0.1% (N) ÷ 0.0365±2.7% (m)
= 2.74±2.8% Nm-¹
Elastic Potential Energy = 0.5 × Spring Constant × Extension²
= 0.5 × 2.74±2.8% × (0.0365±0.001) ²
= 0.5 × 2.74±2.8% × (0.001332255±5.5%)
= 0.00183±8.3%
Conclusion & Evaluation
Conclusion:
In this experiment, I have been quite successful by proving the aim of the experiment which is Hooke’s Law. The results obtained are slightly incorrect due to any errors as part of the experiment. My calculations were all shown for trial one which whereas follows. In relation to the graph, the line does not pass through the origin as there were uncertainties. The line therefore starts a few cm from the origin on the y axis. The slope in the graph indicates the spring constant. It can be seen that the spring constant value in the graph does not match my result for trial no.1 as I have taken the spring constant value in N/cm. If I take the values in N/m and average all the values of the spring constant from my calculations I will end with a result equal to the gradient or slope of the graph that is 0.227. The units taken for every other value is standard and therefore is correct. My results are reliable as they do result in the Force being proportional to the Extension. I feel that my data is reliable and the graph does show that the extension of the spring directly proportional to the force that is applied to it. We also found that the spring constant and the elastic potential energy increases due to the extension of the spring being proportional to the force.
Evaluation:
I have found that the experiment did have many errors which could have been improved. There were both systematic and random errors involved in the experiment. The meter ruler (uncertainty of ±0.05cm) and the digital balance (uncertainty of ±0.1g) had uncertainty’s which could have altered the accuracy of the results. The experiment also had a parallax error due to the carelessness of me not observing the pointer and the length in the straight path. My equipment was not very accurate as I was given a meter ruler and not an attached ruler. This could have made it very inaccurate as the ruler was leaning over a wall. I could only take one reading per mass, as time management was an issue, which is not reliable as taking more than two readings and averaging the answer will give a more accurate result. The next time I perform this experiment, I will need to make sure that I have at least three readings per mass and should take the average of the three readings to minimize the errors. I should also make sure that the meter ruler is not leaning on a wall and that it is held on by a clamp or that I have the ruler stuck behind the clamp stand. While repeating the experiment one should also put a pointer on the hook to avoid parallax error and get the measurements even more accurate
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# Hexagonal Tiles
By Ines Kereki Uruguai
Timelimit: 1
The path to Mary's school is a straight line paved with hexagonal tiles. The picture below shows an example of the path with 12 numbered tiles.
Mary loves mathematics. When going to school, she steps on the tiles of that path following these rules:
• She always starts from the tile with the smiling face (as nothing matches starting anything with a smile!). This tile is always present at the beginning of the path. The other tiles are numbered consecutively, in ascending order, starting from 1, as shown in the figure.
• She is not allowed to go back, that is, she must not step on a tile which bears a lower number than the tile she is on (when decided to go to school, there she goes!).
• She always steps from a tile to a neighboring one (no jumps in order to keep out of harm's way!).
• She must always finish on the highest numbered tile.
When classes are over, she is so tired that she avoids the path and walks on the lawn. Mary does not want to repeat any sequence of steps on the tiles and she would like to know, if the path is paved with N numbered tiles and a tile with the face, how many days will it take to make each possible sequence once.
For example, five days will be needed for her to try all possible step sequences if the path has N = 4 tiles, one day for each of the sequences: 1-2-3-4, 1-2-4, 1-3-4, 2-3-4 and 2-4. Write a program to determine how many different step sequences there are for a path with a given number N of tiles.
## Input
The input contains several test cases. Each test case is composed by a line containing an integer N (1 ≤ N ≤ 40), the number of tiles in the path. The last test case is followed by a line containing a single zero.
## Output
For each test case, print a line containing a single integer, the number of different step sequences.
Sample Input Sample Output 1 4 2 10 0 1 5 2 89 | 478 | 1,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-33 | latest | en | 0.949009 |
https://www.jiskha.com/questions/21386/find-the-roots-of-the-equation-X-square-7x-equal-to-0-b-write-down-the-equation | 1,571,072,843,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653876.31/warc/CC-MAIN-20191014150930-20191014174430-00316.warc.gz | 972,248,247 | 4,913 | # maths
find the roots of the equation X(square)+7x equal to 0.(b)write down the equation whose roots are -2 and 5.
I will be happy to critique your thinking. Write the equation, and solve.
In the second, hint: (x+2) is one factor.
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More Similar Questions | 944 | 2,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-43 | latest | en | 0.946777 |
https://socratic.org/questions/5877b06bb72cff586a3602e3#363841 | 1,709,407,155,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00409.warc.gz | 520,587,555 | 7,617 | # If X is "Normal"(μ = 81.2, σ = 12.4), what is the 16th percentile of this distribution?
Jan 13, 2017
See explanation.
#### Explanation:
A percentile is a location in a distribution that has a specified amount (or percentage) of the distribution "below it" (to its left). In other words, if the ${n}^{\text{th}}$ percentile is $x$, and we draw a random number $X$ from the distribution, then the chance of $X$ being less than $x$ is n %:
${n}^{\text{th" " percentile" = x" }}$means" " P(X < x)=n%.
For example, in a standard normal curve (with $\mu = 0$ and $\sigma = 1$), the point where $x = 0$ (i.e. the $y$-axis) is the 50th percentile, because 50% of the curve's area falls to the left of $x = 0$:
The standard normal distribution $Z$ is such a good baseline, we actually have a table of values designed specifically for looking up percentiles for this curve. It's called a $z$-table, and it looks something like this:
How do we use it? Let's say we want the 25th percentile for the standard normal distribution. We find the value closest to 0.25 in the table (which happens to be 0.2514) and see that it's in row $\text{-} 0.6$ and column $0.07$. For this table, that means the 25th percentile is (approximately) $\text{-} 0.67$.
But wait—how does that help when we want a percentile for any normal distribution $X$? We need to find a connection between any curve and the standard normal curve. That connection is found by shifting the $X$ distribution left-to-right so that it's centered at $0$, and then stretching/squishing it so that its standard deviation is $1$. The formula for this is:
$Z = \frac{X - \mu}{\sigma}$
where $\mu$ is the mean of $X$ and $\sigma$ is the s.d. of $X$.
If we know the percentile we want from the $Z$ distribution, we can solve for $X$ by rearranging the equation into
$X = \sigma Z + \mu$.
As an example, let's use the first question you asked, where $X$ is normally distributed with $\mu = 81.2$ and $\sigma = 12.4$, and we seek the 16th percentile.
From the table above, the 16th percentile from the $Z$ distribution is approximately $\text{-} 0.99$. The equivalent location in our $X$ distribution is then:
$X = \left(12.4\right) \left(\text{-} 0.99\right) + 81.2$
$\textcolor{w h i t e}{X} = \text{-} 12.276 + 81.2$
$\textcolor{w h i t e}{X} = 68.924$
What this says is: if $X$ is a normal curve with $\mu = 81.2 \text{ feet}$ and $\sigma = 12.4 \text{ feet}$, then there is a 16% chance of an observation from $X$ being less than $68.924 \text{ feet}$.
I'll leave the rest for you as an exercise; with the formulas above, it shouldn't be that hard.
Hope this helps! | 784 | 2,630 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 44, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-10 | latest | en | 0.906642 |
https://www.physicsforums.com/threads/is-poincare-grouo-simply-connected.337459/ | 1,521,672,904,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647706.69/warc/CC-MAIN-20180321215410-20180321235410-00470.warc.gz | 839,414,962 | 18,546 | # Is poincare grouo simply connected?
1. Sep 15, 2009
### andrea.dapor
We call a group G "simply connected" if every curve C(t) in G which is closed (that is, C(0) = C(1) = I) can be continuously deformed into the trivial curve C'(t) = I (where I is the unit element in G). This is formalised saying that, for each closed C(t), there exists a continuous function F: [0, 1]x[0, 1] -> G such that
1) F(0, t) = C(t), for all t
2) F(1, t) = I, for all t
3) F(s, 0) = F(s, 1) = I, for all s
Now, Wald (General Relativity, 1984) says that the Poincare group is not simply connected, beacuse in particular for a rotation of $$2\pi$$ about an axis - say z - such a function F does not exist.
My question follows.
Consider the function
F(s, t) := sI + (1 - s)C(t),
where C(t) is the closed curve in Poincare group G associated to a rotation of $$2\pi$$ about z, that is,
C(t) =
(1 0 0 0)
(0 cos2\pi t -sin2\pi t 0)
(0 sin2\pi t cos2\pi t 0)
(0 0 0 1)
with t in [0, 1].
This F seems to verify (1)-(3)... where is my mistake?
I thank you for your help, and apologize for the "matrix" above...
2. Sep 15, 2009
### hamster143
Your F is not a member of Poincare group ... I and C are, but neither the addition nor multiplication by scalar are valid operations in SO(3,1). You can easily do the math and see that the determinant of your "rotation" det F is, generally speaking, not equal to 1.
Last edited: Sep 15, 2009
3. Sep 15, 2009
### andrea.dapor
Oh, yes! Even sI is not a poincare transformation (if s is not 1, then its determinant is different from 1, that is, it is not a Lorentz transformation). We must build an F just by composing (matrix product) elements of poincare group. Thank you.
-split the $$2\pi$$ rotation C(t) = R_z($$2\pi$$t) into two $$\pi$$ rotations,
R_z($$2\pi$$t) = R_z($$\pi$$t) R_z($$\pi$$t)
-then rotate the axis of the first in such a way that it becomes a rotation of -$$\pi$$t, that is, a rotation about -z; this can be done rotating the first by $$\pi$$ about x:
[R_x($$\pi$$) R_z($$\pi$$t) R_x($$\pi$$)^T] R_z($$\pi$$t) = R_z(-$$\pi$$t) R_z($$\pi$$t) = I.
-So we see that apparently the function
F(s, t) := R_x($$\pi$$s)R_z($$\pi$$t)R_x($$\pi$$s)^T R_z($$\pi$$t)
satisfies (1)-(3).
However, (3) is not verified, since when t = 1 the rotation R_z($$\pi$$t) is not the identity, and so we cannot write
F(s, 1) = R_x($$\pi$$s)R_z($$\pi$$)R_x($$\pi$$s)^T R_z($$\pi$$) = R_x($$\pi$$s)R_x($$\pi$$s)^T = I.
This, instead, is true for a rotation of $$4 \pi$$ about z, since R_z($$2\pi$$t) is the identity when t = 1. Indeed, Wald specifies that a $$4\pi$$ rotation can be continuously deformed to I: the function F is then
F(s, t) := R_x($$\pi$$s)R_z($$2\pi$$t)R_x($$\pi$$s)^T R_z($$2\pi$$t).
What do you think? Am I correct?
4. Sep 15, 2009
### genneth
Yes --- 2 pi is not deformable to the identity, but 4 pi is.
5. Sep 15, 2009
### javierR
Re: Is poincare group simply connected?
I don’t know if I see the essence of the non-simply-connectedness in your posts. I'll give a long response, hoping to help out somewhere.
First, a technical point: The Lorentz group (Poincare group minus the translations) itself is not connected, but can be broken into 4 connected components, one of which contains the identity. If $$\Lambda$$ is a general Lorentz group element, then the component with $$det\Lambda=+1\;\;\;\;sign\Lambda^{0}_{0}=+1$$ forms a subgroup called the “restricted Lorentz group” (if we tag on spatial and temporal inversions, looking at this subgroup is sufficient for understanding the full Lorentz group).
While the restricted Lorentz group is connected, it’s not *simply* connected, which is the statement you are examining. It happens that, to have spacetime spinors, you need to have a simply connected spacetime group that contains the restricted Lorentz group. This group is, therefore, called Spin(3,1), and happens to be isomorphic to SL(2,C) the group of 2x2 complex matrices of determinant 1. How is this related to the restricted Lorentz group? If U is an element of SL(2,C), it describes a Lorentz transformation. The element –U is a distinct element of the group, but it ends up describing the same Lorentz transformation when acting on spacetime vectors (generally, on tensors). It’s not surprising that SL(2,C) is called a “double cover” of the restricted Lorentz group. The restricted Lorentz group is then obtained by identifying U with –U: U== -U. A way of writing this is $$SL(2,C)/Z_{2}$$, where $$Z_{2}$$ has two elements that act as +1 or -1.
Now, since you’re concerned with “simple connectedness”, we want to know about the topological properties of the group and curves connecting elements of groups. Focusing on the compact part of SL(2,C) (the non-compact part will be irrelevant), it consists of unitary matrices SU(2), which has the topology of a 3-sphere (one step up from everyday spheres that we’re used to). Any closed curves (loops) on the sphere can be contracted to a point, so SU(2) is simply connected. Let u be the element of SU(2) corresponding to the element U of SL(2,C) we were talking about before. Then identifying u and –u like we did before, the transformations are points of the topological space (3-sphere)/$$Z_{2}$$. That is, identify opposing points of a 3-sphere. You can now imagine paths on the 3-sphere between two group elements (ok, you can try to draw these paths on a 2-sphere): let’s say path 1 goes between the identity element u=1 in the northern hemisphere and an element u’ in the same hemisphere that acts as a rotation about the z-axis by some angle theta (generally, you need more parameters to cover the sphere, but we're just looking at theta). Now, we could instead go on path 2 from u=1 to the direct opposite side of the sphere from u’; that is, going to –u’ on the southern hemisphere. Now, in our space, the points u’ and –u’ of the northern and southern hemispheres are identified, so these two paths start and end at the same point, from u=1 to u’ == –u’, so it is technically a loop. So can we contract this loop? I’ve tried here to diagrammatically show why you cannot:
*---<----u====>=====u’----<-----*
Here there’s the path 1 from u to u’ denoted by ===>===. Then there’s path 2 denoted by --->--- lines, and which is broken up by * symbols because at that point the path *jumps* across to the opposite side of the sphere and then continues toward u’ (that is, instead of drawing path 2 from u=1 to –u’ in the southern hemisphere, I drew the *reflected path* once you cross into the “–u’s”, so the entire path is drawn on the northern hemisphere). The path is stuck to the *’s, so the loop cannot contract. This is why the (restricted) Lorentz group is not simply connected.
How does this fit in with the 2*pi and 4*pi rotations you were talking about? Like I mentioned before when we want spacetime spinors, we need to consider the full group SL(2,C), in which case u and –u are again distinct elements. Even though they represent the same Lorentz transformation on tensors, they are not the same for spinors. Consider the SL(2,C) element
$$U=\left(\begin{array}{cc}e^{i\theta /2} & 0 \\ 0 & e^{-i\theta /2}\end{array}\right)$$
describing a rotation by angle theta about the z-axis. As you vary theta, you move along a path on the sphere. Putting in theta=2*pi, you get U=-1. Putting in theta=4*pi you get U=+1. This may look familiar because it’s the rotation of a spacetime spin-1/2 object, and spinors do indeed have that strange property that a 2*pi rotation leads to a different state. So the SL(2,C) group handles spinors appropriately, not SO(3,1).
We can see the paths for this special case, with u=1 (theta=0), u’=-1 (theta=2*pi). In the SL(2,C) (or SU(2)) picture this is just moving from one point on the 3-sphere to the point opposite it, and these represent distinct points. But for (3-sphere)/$$Z_{2}$$, the points are identified so that we are forming a closed loop:
*<-----u=1----<------*
But as we’ve seen that loop isn’t contractible; at the *'s the matrices that make up the points of out path (parametrized by theta) show a jump. So by trying to enforce an equivalence btwn “no rotation” and “2*pi rotation”, we end up with the feature that some special closed loops and can’t be contracted…something strange is happening, and that’s because we’re technically not supposed to make these equivalent. Rotating further to 4*pi is the equivalent of overlapping two of these:
*<-----u=1----<------*,
one undoing the other (that’s called “doubly connected”). It’s equivalent to combining one of these paths with its reverse, which is equivalent to a fully contracted loop. Can you see how you would write a double-layered path in terms of the matrix-parametrization above so that the jumps "cancel" each other?
6. Sep 15, 2009
### marcus
javierR, that is beautifully explained.
I post here because I want to be certain to catch your attention. You said something on some thread, I forget where, that made me think you are a European particle physicist. I think you probably know about the MAGIC telescope on the island of Las Palmas in the Atlantic.
I want to know if the MAGIC telescope(s) found anything about particle physics.
This is a different topic, so I will start a new thread here in the same forum (High energy, nuclear, particle physics forum).
I am interested in these telescopes because they are a new type, that sees incoming gamma from their Cherenkov trails. Do you know anything you can tell us about this new telescope technology? | 2,685 | 9,519 | {"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} | 3.28125 | 3 | CC-MAIN-2018-13 | longest | en | 0.895052 |
https://www.researchgate.net/topic/Kalman-Filtering | 1,701,648,115,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00764.warc.gz | 1,101,670,803 | 84,755 | Science topic
# Kalman Filtering - Science topic
Explore the latest questions and answers in Kalman Filtering, and find Kalman Filtering experts.
Questions related to Kalman Filtering
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What is meant by the convergence of the Kalman Filter and to what does it converge? What is the mathematical formulation for convergence in terms of solving a Riccati equation?
Hallo Mr. Bacha,
I would like to send you this file and I hope that you could find some of explain about your questions. this file was very benefit for me.
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Hello, has anyone worked with the CKF filter? How does it perform compared to other non-linear Kalman filters?
CKF, EKF, and UKF are all non-linear Kalman filter variants with different approaches to handling non-linear systems. on the other hand, CKF is a specific implementation of a non-linear Kalman filter. It is designed to address the limitations of traditional Kalman filters in handling non-linear system dynamics. The CKF's use of the cubature rule offers a potentially more accurate approximation but at the cost of increased computational complexity. The choice between these filters depends on the specific requirements of the application, the degree of non-linearity, and the trade-off between accuracy and computational efficiency.
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Hello who interested with my question,
Firstly, I want to thanks you for spending your time to read this topic!
So I have 3 questions that need some advice in Unscented Kalman Filter about the 3 DoFs Mass-Damping-Stiffness System. Right now, i'm modifying my UKF code in MATLAB for a new project but some problems seem to occur again. Below there are 2 Code files - one was the UKF test and one was an UKF function that I've writen in my graduated thesis and here are the problems I incurred:
1/ The Cholesky Decomposition: I used the chol() function from library of MATLAB. However in my test, from the loop k=3, the covariance matrix was starting to fail in excuting chol() because it was not completely positive definited!
My solution: First, I used a function name "nearestSPD" of Mr. John D'Errico and this function have helped my covariance matrix pass the chol() but the output (state vector) I received was full of 0 from the loop k=3. Second approach was plusing (1e-6)*eye() into the covariance matrix but MATLAB code stopped from the loop 3 and said that matrix wasn't positive definited!
2/ Kalman Gain: since the equation of Kalman Gain has the inverse matrix, some value of Kalman Gain of my code in some loop can't be calculate because the covariance matrix is singularity and it doesn't have the inverse version!
My solution: in MATLAB, I've used pinv() (Moore-Penrose Inverse matrix) instead of the regular inverse inv()
3/ Choosing the workable Initial Covariance Matrix (P): How can we choose the suitable Initial Covariance Matrix for UKF?
My solution: Usually, I always choose the values of P corresponding to the error between the initial state vector and true parameter. For example, true k1 = 10000 N/m and in my initial state, I choose k1 = 8000 N/m -> value error of k1 in P will be chosen equal to 1e6.
If you have any suggestions, please feel free to repsonse! I would love to hear your idea!
My code is free and as long as you seem interested, you can use it freely! However, my code seems to fail due to 3 reasons above!
Thanks you!
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I want to use Kalman filter to estimate battery parameter and observer for state estimation together (SOC)
You may try graphical state space model solved by factor graph optimization.
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I use the Kalman filter to estimate the stator and rotor currents for a 2MW doubly fed induction generator, I used the state model in the alpha beta reference, How to calculate the initial matrix P_0, Q_0, R_0 for a extended Kalman filter?
Mohammed Abbas In an extended Kalman filter (EKF), the starting matrices P 0, Q 0, and R 0 play a significant role in the estimating process. They are, in order, the starting values of the state covariance matrix P, the process noise covariance matrix Q, and the measurement noise covariance matrix R. These matrices are used to represent the uncertainty in the state estimation, process dynamics, and measurement noise, in that order.
You must consider the following factors while calculating these matrices:
1. P 0 state covariance matrix: This matrix captures the state estimate's starting uncertainty. To represent the fact that the initial state estimate is unknown, it is commonly set to a big number for the diagonal elements and zeros for the off-diagonal elements.
2. Q 0: Process noise covariance matrix This matrix illustrates the process dynamics' uncertainty. It is often determined based on a prior understanding of process dynamics and how they evolve over time.
3. R 0 is the measurement noise covariance matrix. This matrix displays the measurement uncertainty. It is often set based on measurement noise characteristics such as sensor accuracy.
P 0, Q 0, and R 0 values can be derived by tuning or previous knowledge of the system. As the EKF converges to a more accurate estimate, the values can be changed over time. Finding the proper values for these matrices, on the other hand, may be difficult, as estimation success is largely dependent on the choice of these.
• asked a question related to Kalman Filtering
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Hello researchers,
Could you give your recommendation for a good book that explains the Kalman filter in a practical way?
MAnsour
1. Brown, R. G. and Hwang, P. Y. C., Intro. to Random Signals and Applied Kalman Filtering, 4th ed, John Wiley
2. Bar-Shalom, Y., Rong Li, X., Kirubarajan, T., Estimation with Applications to Tracking and Navigation, John Wiley
There are many more, but these are plenty to begin with.
Enjoy.
Regards
Hari Hablani
IIT Indore
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We combine data from different sensors like the Inertial Measurement Unit (IMU) and Global Positioning System (GPS) to estimate the state of a robot e.g. position using Kalman Filter. My question is why do we need to estimate the position if we are getting it from GPS and IMU?
Raw GPS data only provides 4 measurements of pseudo range. You need an estimator to extract the 3D position measurement and clock bias. The GPS data is then used to estimate the errors in the IMU to provide improved short-term stability for any autopilots (or a navigation capability should the GPS signals be intermittent)
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In kalman filter for state variable estimation we take initial zeros but in ensemble especially for harmonic analysis, which values to take ?
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How could we detrmine Fixed-Gain values of the Extended Kalman Filter gain matrix without solving Riccati equation? Is there a simple way?
There are sections in different books that address this topic. For a simple treatment, see Sec. 7.7 Estimator Design, in
Franklin, G. F., Powell, J. D., and Emami-Naeini, A., Feedback Control of Dynamic Systems, Prentice-Hall, 6th ed.
For sampled system (with a sampling period, digital control systems), see Sec. 8.2 Estimator Design, in
Franklin, G. F., Powell, J. D., and Workman, M., Digital Control of Dynamic Systems, 3rd ed, Addison Wesley
For a more complete treatment, see Chapter 8 Time-Invariant Filters, in
Bryson, A. E., Applied Linear Optimal Control, Cambridge University Press 2002
There are yet more books on the subject. See
Stengel, R. F., Optimal Control and Estimation, Dover edition
and yet more.
Hari Hablani
IIT Indore, India
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Look into the book
Y. S. Shmaliy and S. Zhao, Optimal and Robust State Estimation: Finite Impulse Response (FIR) and Kalman Approaches, Wiley & Sons, 2022.
This is the first systematic investigation and description of convolution-based (FIR and IIR) state estimation (filtering, smoothing, and prediction) with practical algorithms. In this framework, the Bayesian Kalman filter serves as a recursive computational algorithm for batch optimal FIR and IIF filters. The unbiased FIR filter is shown to be the most robust among other linear estimators. Various robust approaches for disturbed and uncertain systems are also discussed.
Yes, I got a copy of your book recently. Very interesting approach!
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I am trying to implement the sliding window Extended Kalman Filter as in the paper "Automated Controller Calibration by Kalman Filtering" ( https://doi.org/10.48550/arXiv.2111.10832 ) but there are several things that are unclear to me and thus I would like to get some clarification.
The first is that I do not understand the idea of a sliding window for a Kalman Filter. Does a Kalman Filter not operate on a sequential manner? How does one operate a KF using a sliding window?
Can someone provide a Matlab script or preferably a Simulink model of such a system?
Marnel Altius I suggest reading this aricle https://arxiv.org/pdf/2111.10832
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The Unscented Kalman Filter (UKF) algorithm requires a Cholesky factorization of the prediction error covariance matrix. The problem is that during the iterations, this matrix is not always positive definite and so, the MATLAB code risks tobe stopped.
Could someone tell me how to make this covariance matrix positive definite in order to ensure the running of the code until the end of the iterations?
Thank you
try this every iteration
P=(P+PT)/2
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Dear all,
I have been doing research on the leak detection of pipelines for some time. I did the simulations with simcenter software. But unfortunately, I tried to detect the location of the leak using Kalman filter in different ways, but it is not possible. Is it possible to guide me? Is it possible to send me the MATLAB code so that I can try on my own water pipelines and simulations?
Richard Fenner and Steve Mounce have done work using Kalman filters for leak detection in water distribution networks Seyed.
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We are trying to optimize the values of DEM of a particular catchment with a view of identifying outliers and predict future values. Thanks
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Hello researchers!
I want to fit Gaussian Mixture Model to a data set where the number of mixture components is known. The weights of the mixture, the means and variances of the gaussian distributions have known priors attached to them. Can someone please enlighten me on how to use Kalman filter for this problem? Goal is to estimate the weights of the GMM and also the parameters of the gaussian distributions.
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I need to estimate a set of 15 constant parameters, which are not directly measured. My state vector is therefore fixed and is made up of these constants so the Kalman filter equations rearrange to those of the Recursive Least Squares. Only 3 quantities, different from the state, can be measured and from these I have to estimate the state vector. Results indicates that the state vector is estimated well but the rank of the observation matrix used in the calculations is much less than 15 because the equations(measures) are less than the constants to be estimated. Is it normal? Must the rank be equal to the number of parameters? Thank you
Dear Michel Crimeni,
The rank of the measurement matrix H is less than or equal to the rank of the system (process) matrix F.
Best regards
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I'm simulating the process of measuring the positions of the end-effector of a robotic arm but I haven't yet specifications about the measuring device that will be used in actual experiments. Let's suppose the device will be a laser tracking system whose data sheet reports a measurement uncertainty of 15 micrometers = 0.015 mm. The measurement uncertainty is defined as the deviation between a measured coordinate and the nominal coordinate of a point. Can I assume it as the accuracy of the instrument? Can I use it as the standard deviation and then square it to initialize R?
Michel Crimeni, those parameters are knobs you turn to tune the system so you should experiment with a range of values. But that's probably a good assumption to start with.
Best!!
AN
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I know I should be using the concept of Dead Reckoning. But how do I use prediction algorithms to determine/estimate the 2D coordinates and yaw angle, if I can get the gyroscope data, wheel encoder data, and global system time or timestamp of each input?
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I was wondering how could I implement Kalman filter in Mircoprocessors to caculate position in MPU6050 sensor. Is there anyone here that already had done this project?
You do not need a Kalman filter to compute position from measured angular velocities and translational accelerations:
1. Compute rotation by means of angular velocity
2. Transform measured accelerations into the initial (inertial) frame utilizing the latter rotation matrices
3. Double time integration of accelerations (in initial frame) to derive positions w.r.t. the initial frame.
However, Kalman filtering could be used to improve the derived position. Most popular approaches are multi sensor fusion (e.g., IMU+GPS) or IMU sensor fusion (orientation from magnetometer/accelerometer investigations). IMU sensor fusion would be possible using the MPU9250, which has a magnetometer.
Best regards
Rene
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My question refers to the following papers:
1. S. J. Julier and J. J. LaViola, "On Kalman Filtering With Nonlinear Equality Constraints," in IEEE Transactions on Signal Processing, vol. 55, no. 6, pp. 2774-2784, June 2007, doi: 10.1109/TSP.2007.893949
2. A. T. Alouani and W. D. Blair, "Use of a kinematic constraint in tracking constant speed, maneuvering targets," in IEEE Transactions on Automatic Control, vol. 38, no. 7, pp. 1107-1111, July 1993, doi: 10.1109/9.231465.
In particular, my concerns are about the Fig. 1 of [1], the statements at the end of the left column in the page 2 of [1], and the statements in the middle of the left column in the page 2 of [2]. In both papers, it is suggested to apply the constraints only after the update of the state through the measurements. Should be possible to obtain better results projecting the state on the constraining surface both after the prediction and update steps?
These articles might be useful, have a look:
Kind Regards
Qamar Ul Islam
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Hi,
I face the following problem:
I want to use a measured quaternion (from an AHRS) for the update within an UKF to update the attitude.
My state vector contains a quaternion, therefore I want to know, if I could calculate the Kalman innovation with the quaternion directly or do I need an alternative representation (Euler angles, orientation vector...)?
The problem with Euler angles is, that for the +-180° (South) direction, an UKF would spread its sigma points around a discontinuity. This does not sound harmless from my point of view.
Is there an elegant way to solve my problem?
Best regards
Maximilian
This is a strange question. You must use the measurement model, which relates your measurment to the state you want to observe.
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I am using a linear Kalman filter for estimating the position using multiple IMUs. My model is adding the accelerometer readings' double integration onto the previous positions. The state vector is the positions of the IMUs and the input vector is the IMU accelerometer measurements.
Also, I have some constraints in the positions and I am using those constraints in the correction step of the Kalman filter.
So in the tuning step, I set the R matrix zero so that the Kalman filter relies on the measurement only (so not use the aprior estimate at all). It works fine unless there is no input. However, if I introduce some acceleration to the system, even if I set the R=0 and Q very high, the Kalman filter still takes the contribution from the aprior estimate. The question comes at that point: what is the functionality to set R=0 if it does not disregard the model error?
Another approach to satisfy this behaviour comes from the Kalman gain calculation. As R goes to zero, the Kalman gain should approach to the inverse of the observation matrix (H or C) and so that the a posterior estimate approaches to the measurement. Since the H matrix is not a square and full rank matrix in my case, I cannot intuitively check where exactly the contribution of the aprior estimates take place even though I set the R=0. If anyone has any comments, I would really appreciate.
Theoretically, it is fine to have zero measurement noise. Even in R.E. Kalman's original paper (Kalman, R. E. (1960). A new approach to linear filtering and prediction problems), the algorithm is stated without restrictions on possible singularities of the measurement noise covariance. However, for practical applications, a singular noise covariance for the measurements can cause numerical problems concerning the invitation carried out for computing the Kalman gain. When considering anything besides a very simple academic example I would consider introducing noise on each measurement, hence prevent the measurement noise covariance matrix from being singular.
Hope this helps
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In a seminal paper titled "Posterior Cramer Rao bounds for discrete-time non-linear filtering" (Link: https://ieeexplore.ieee.org/abstract/document/668800/?casa_token=Ecgr8cQF6RUAAAAA:JtZ8OuUkQLyphtnQnot1ZPGlifREbS393Pg0TJ58J98IilZuIw6xSjS1Af1XDlLchlKMi8LoRfxfmAg), Tichavsky et al. have developed a recursive way to calculate the Cramer Rao bounds for non-linear filtering problems. However, in the paper, the state xk at time 'k' is a non-linear function of the previous state xk-1only and not on other previous states. Are there any similar works on CR bounds for filtering problems where the current state depends on all the previous states up to the given time instants i.e. xk is a function of x1, x2, and so on up to xk-1?
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I have seen an example of tracking position in one dimension using an alpha beta filter, and I have a similar problem in a different field. In the example the position is first updated via dead reckoning: New position = old position+ velocity times delta t. Then the new measurement is used to update both the position estimate and the velocity estimate using the alpha and beta weights. I would like to see this problem formulated using a Kalman filter. I think I can come up with the variance estimates required and prefer this to somewhat arbitrarily selecting alpha and beta weights. The change in velocity is a drift, i.e. not driven by any known input.
These links might be useful, have a look:
Kind Regards
Qamar Ul Islam
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I am familiar with the equations of basic kalman filter. I was reading a paper where the author's proposed a variant of kalman filter and in the prediction step they had more terms than standard kalman filter. I am referring to equation 10 of the paper "Vehicular Node Localization Using Received-Signal-Strength Indicator" where author has added a term gamma multiplied with dt^2
https://www.comm.utoronto.ca/~valaee/Publications/07-Parker-TVT.pdf. I understand the idea behind term. I do not how to justify the factor one can add new terms affecting the uncertainty of prediction step of kalman filter.
My specific question how can I justify or prove that the new terms in error covariance equation equation 10 (inside attachment as well is correct)
Hello Varun Garg,
Equation (10) is a result of the non-additive noise formulation for the EKF:
The usual simplification is to add noise to the state after the dynamic/prediction step. Often, we have a better knowledge where noise in the system is present (in the paper example the noise on the velocity measurement ist known). In such cases, you can use the non-additive noise formulation.
If you want to prove/check covariance equations in general you can refer to covariance computation rules:
The equations (9) and (10) of the paper calculate the expected value and covariance of the distribution propagated through the dynamic model (2).
Solving:
A_k|k-1=E(A_{k-1|k-1}+Ts * u_{k-1}+Ts w_{k-1})
P_k|k-1 = Cov(A_{k-1|k-1}+Ts * u_{k-1}+Ts w_{k-1})
, where E(...) computes the expected value results in (9) and (10).
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I was reading a IEEE transaction paper on how to calculate Measurement and Process noise matrix of the Kalman filter when they are unknown. The title of the paper was "On the identification of variances and adaptive Kalman filtering" by Mehra.
As shown in the attached image, In equation 4 of the paper, the term M0 exist on the both sides L.H.S and R.H.S sides of the equations. I do not understand why it is written in such a manner. Do I need to simiplify it to find the value of M0 or my understanding of the equation is incorrect. phi in the equation is a non singular transition matrix,
Dear Varun Garg,
I suggest you to see links and attached file on subject.
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I am researching UKF, and I would like to know where can I find performance measures of this. The system is a Chemical Process Plant.
Maybe you can consider the recursive least squares algorithm (RLS). RLS is the recursive application of the well-known least squares (LS) regression algorithm, so that each new data point is taken in account to modify (correct) a previous estimate of the parameters from some linear (or linearized) correlation thought to model the observed system. The method allows for the dynamical application of LS to time series acquired in real-time. As with LS, there may be several correlation equations with the corresponding set of dependent (observed) variables. For the recursive least squares algorithm with forgetting factor (RLS-FF), adquired data is weighted according to its age, with increased weight given to the most recent data.
Years ago, while investigating adaptive control and energetic optimization of aerobic fermenters, I have applied the RLS-FF algorithm to estimate the parameters from the KLa correlation, used to predict the O2 gas-liquid mass-transfer, hence giving increased weight to most recent data. Estimates were improved by imposing sinusoidal disturbance to air flow and agitation speed (manipulated variables). Simulations assessed the effect of numerically generated white Gaussian noise (2-sigma truncated) and of first order delay. This investigation was reported at (MSc Thesis):
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For the discrete-time Extended Kalman filter (EKF), a stochastic stability analysis is discussed by Reif et al. in their paper "Stochastic stability of the discrete-time extended Kalman filter" (https://ieeexplore.ieee.org/abstract/document/754809). Recently, there have been works on EKF with unknown inputs, in particular, for arbitrary unknown inputs. Are there any works providing such stability analysis similar to Reif's work available for these EKF with unknown input filters?
Yes, these articles might be useful to connect all the dots together, have a look:
Kind Regards
Qamar Ul Islam
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Online model updating can improve the prediction ability of the model. Unscented Kalman filter is used to update model parameters. I know it can be used when parameters are constant. Can I also use it to solve time-varying parameters? What's the alternative and what's the difference between online recursive least squares estimation.
The recursive least squares algorithm (RLS) which allows for (real-time) dynamical application of least squares (LS) regression to a time series of time-stamped continuously acquired data points. As with LS, there may be several correlation equations and a set of dependent (observed) variables. RLS is the recursive application of the well-known LS regression algorithm, so that each new data point is taken in account to modify (correct) a previous estimate of the parameters from some linear (or linearized) correlation thought to model the observed system. For RLS with forgetting factor (RLS-FF), adquired data is weighted according to its age, with increased weight given to the most recent data. This is often convenient for adaptive control and/or real-time optimization purposes. A particularly clear introduction to RLS/RLS-FF is found at: Karl J. Åström, Björn Wittenmark, "Computer-Controlled Systems: Theory and Design", Prentice-Hall, 3rd ed., 1997.
Application example ― While investigating adaptive control and energetic optimization of aerobic fermenters, I have applied the RLS algorithm with forgetting factor (RLS-FF) to estimate the parameters from the KLa correlation, used to predict the O2 gas-liquid mass-transfer, while giving increased weight to most recent data. Estimates were improved by imposing sinusoidal disturbance to air flow and agitation speed (manipulated variables). The proposed (adaptive) control algorithm compared favourably with PID. Simulations assessed the effect of numerically generated white Gaussian noise (2-sigma truncated) and of first order delay. This investigation was reported at (MSc Thesis):
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I want to use a EKF for parameter (p) and state (x) estimation.
I am currently not understanding the difference between (1) the formulation of the parameters as states:
x_new=[x,p] and using a normal EKF for estimating x_new
and (2) the formulation with a dual extended Kalman filter, where x and p are estimated individually.
Which of them provides more benefits?
I am very thankful for your tips.
The Kalman filter (KF) is a method based on recursive Bayesian filtering where the noise in your system is assumed Gaussian. The Extended Kalman Filter (EKF) is an extension of the classic Kalman Filter for non-linear systems where non-linearity are approximated using the first or second order derivative.
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Hi, I am currently doing a project which requires me to use Kalman filter to remove the noise in the speech. Now I am using 1D Kalman filter. Can I use Extended Kalman Filter or Unscented Kalman Filter for my application? Thanks.
With code matlab:
Speech-Enhancement-Kalman-Filter
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I am implementing an unscented kalman filter for parameter estimation and I found its performances strongly related to the initialization of the parameter to estimate.
In particular, I don't understand why if I initialize the parameter below the reference value (the actual value to estimate) I get good performances, but if the parameter initialization is above the reference value, performances are very low and the estimation does not converge but continues increasing.
In other words, it seems that the estimation can only make the parameter increase. Is there any mathematical reason behind this?
Dear Mr. Magnani,
I agree with Janez Podobnik. I assume you are using the Joint Unscented Kalman Filter for this purpose. In this case, the parameter is considered as a state and all the initial conditions, noise and measurement covariances will affect the estimation of the parameter. In 2019, I have proposed a different approach for this purpose and I am attaching the article link below:
The main advantage of this approach is you can determine weights for the measurements (a Jacobian approach) for parameter estimation. If one of the measurements significantly affects the parameter estimation, then, you can increase the weight of that measurement. The only drawback of this approach is that the estimated parameter must be linear in terms of measurements. The system itself can be a nonlinear system, but parameter(s) must be linear in terms of measurements. For frequency estimation, we have modified this approach with my colleagues (link below). In this case, the parameter is no longer linear in terms of measurements. However, we haven't tested this approach for other nonlinear systems in which the parameter(s) is nonlinear in terms of measurements.
You may try this approach and it can be useful for your aim.
Best regards,
Altan
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The Nyquist-Shannon theorem provides an upper bound for the sampling period when designing a Kalman filter. Leaving apart the computational cost, are there any other reasons, e.g., noise-related issues, to set a lower bound for the sampling period? And, if so, is there an optimal value between these bounds?
More samples are generally better until such point as the difference in the real signal between samples is smaller than the quantization or other noise. At that point, especially with quantization, it may be a point of diminishing returns.
The other thing that nobody mentions is that faster sampling means less real-time processing time. In many systems, it's not really an issue as the time constants of the physical system are so slow as to never challenge the processing. In others, say high speed flexible meachatronic systems, the required sample rates may challenge the number of processing cycles available to complete the task.
Generally, the best bet is to return to the physical system's time constants and (if possible) sample 20-100x as fast as them.
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I am looking for a complete solution for 10-DOF IMU Kalman Filtering (acceleration x-y-z, gyro x-y-z,magnetometer x-y-z,altitude).
I am interested in all example ,codes
Thanks for helping
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Consider a digital control loop which consists of a controller C and Kalman filter K, which is used to fuse and/or filter multiple sensor inputs from the plant and feed the filtered output to the controller. The prediction (or also called time-update) step of the KF, and the analysis and tuning of the control loop utilize a discretized state-space model of the plant, defined as
x_{k+1} = F * x_k + G * u_k,
y_k = H * x_k
where F is a transition matrix, G is the input gain matrix and H the measurement output matrix. For now I will ignore the process and measurement noise variables.
According to many text books about digital control and state estimation (for example "Digital Control of Dynamic Systems" by Franklin et al.), u_k = u(t_k), x_k = x(t_k) and y_k = y(t_k) are the control input, state and measurement output of the plant, which seem to available at the same point in time, t_k. This would mean that the output from the DAC of u_k, and sampling of y_k happen at the same moment in time, t_k. However this does not seem to hold for some classical implementations. Consider a typical pseudocode of a control loop below:
for iteration k = 1:inf
measurement update of KF x_k
compute u_k
output u_k through DAC
time update of KF x_{k+1}
wait for next sampling moment t_{k+1}
end for
Ignoring the time durations of the DAC and ADC processes, the description above will introduce error in the prediction step of the KF, because it is assumed that the value u_k is actuated at the same moment of time that y_k is sampled - t_k. However due to the time delay introduced from computations of the update step of the KF, and the controller this is not the case. If we define s_k to be the time when the value u_k is actuated, then clearly t_{k+1} - s_k < T, where T is the sampling period. It is clear that the prediction step no longer computes the predicted state correctly because it either a) uses the old actuation value u_{k-1} or b) uses the newly actuated u_k, and in both cases the time between actuation and sampling is equal to the sampling period, assumed in the model.
This leads me to believe that the control value u_k should be actuated at time t_{k+1}, to keep consistency white the sampling period and the prediction model.
Also consider the case when the KF's prediction and update steps are executed before the controller during iteration k. Then the prediction step clearly makes use of u_{k-1} to compute a time update x_{k+1} of the state. This also seems to contradict the original definitions.
So with all these assumptions laid forward, I would like to know what are the correct sampling and actuation times, and why such ambiguity exists in modern literature about such hybrid systems.
NOTE: Some of you may say that the definition holds for small sampling periods and when the computations are fast. However I consider the general case where sampling periods may be very large due to the computations involved in the control loop.
nice question.
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Hi,
I am doing a project related to estimate a state variable from different sensing channels, and these two channels contain different measurement sampling rates and delays. I would like to use a modified Kalman filter like sequential processing first proposed by Alexander, but I did not how to do the Lyapunov-method-based convergence analysis. Could anyone give me some suggestions for this proof?
Also, I try to combine this observer with an adaptive impedance controller and analyze the hybrid stability, but I am not sure how to do the proof.
Thanks a lot for any suggestions.
Best regards,
Qiang
You need to distinguish between different sampling rates and delays. Regarding the sampling rate, you may be interested in the following paper:
However, regarding delays, it might break the Gauss-Markov process assumption of the Kalman filter. Hence, I'd check the observability rank and find conditions to promise it is a full-rank (or any defined rank).
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Hello,
using Cholesky decomposition in the UKF induces the possibility, that the UKF fails, if the covariance matrix P is not positiv definite.
Is this a irrevocable fact? Or is there any method to completely bypass that problem?
I know there are some computationally more stable algorithms, like the Square Root UKF, but they can even fail.
Can I say, that problem of failing the Cholesky decomposition occurs only for bad estimates during my filtering, when even an EKF would fail/diverge?
I want to understand if the UKF is not only advantagous to the EKF in terms of accuarcy, but also in terms of stability/robustness.
Best regards,
Max
If I understand your question correctly, it concerns not the initial covariance matrix rather the updated covariance matrix you get at the end of each Kalman iteration.
If such a condition arises you may use Higham's method to find an approximate positive-definite covariance matrix.
Reference:
Computing a nearest symmetric positive semidefinite matrix - ScienceDirect
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Hello,
I'm currently working with a system consisting of an accelerometer, that samples in bursts of 10 seconds with a sample frequency of 3.9 Hz, before going into deep sleep for an extended (and yet undetermined) time period, then waking up again and sampling for 10 seconds and so on.
I've recently taken over this project from someone else and I can see that this person has implemented a Kalman filter to smooth the noise from the accelerometer. I don't know much about Kalman filters, but to me it seems that the long deep sleep period would make the previous states too outdated to be useful in predicting the new ones.
So my question is: can the previous states become outdated?
The application of a Kalman Filter requires four models:
1. Measurement model, expressed by the Jacobian of the measurement variables in respect to the state variables. (H)
2. The measurement error model (R)
3. The state model, which allows to extrapolate the state to the future (F)
4. The state error model which allows to model the unpredictability of the state extrapolation. (The degree of a deviation of the state from the prediction for whatever reason) (Q)
For the exact mathematical meaning of H,R,F,Q refer to Wikipedia: Kalman Filter.
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Hello everybody , I'm looking for a method able to create a transition matrix for my linear kalman filter. I need to work on the heading of a Pedestrian Dead Reckoning experiment. in my state vector I thought to insert the angular displacement referred only to the last sample . In my mesurement vector instead I thought to insert the gyro data. Somone can help me?
You can record GPD data through OSM tracker application. For Transition matrix study https://www.youtube.com/watch?v=CaCcOwJPytQ&list=PLX2gX-ftPVXU3oUFNATxGXY90AULiqnWT&ab_channel=MichelvanBiezen
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As all of us know Kalman Filter is designed for LTV (Linear Time Variant) systems. But no body in the literature applies that on Hybrid systems. In my opinion Linear Hybrid systems are a subset of Linear Time Variant (LTV) systems, So my question is why don't people use Kalman Filter for Hybrid systems with linear modes and resort to robust estimation approaches?
Dear Navid Hashemi,
There are works that consider this application to be a Hybrid system thus taking into account the robustness aspect, here is a references :
In my opinion one can consider the LTV system as an LPV system and have several models and then deal with the issue with a De Kalman filter bank.
Best regards
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I would like to implement a Kalman filter for tracking applications for IoT nodes. I cannot do floating point on our nodes but only fixed point.
Could you recommend some good references?
Hello. Your question lacks detail. If you are implementing a Kalman filter in a multi-sensor network, then it probably is not a Kalman filter, but some form of information fusion across mulitiple sensor nodes. The fact that you wish to use fixed point arithmetic suggests that you are seeking to create local state and covariance estimates at each node and share these estimates. Possibly you wish to fuse them into a global state estimate. These are all design decisions that impact overall system performance and local/global processor requirements.
If you really are doing local (node-based) state estimation, rather than just transmitting raw measurements to a central processor, make sure you use a square root filter or you will lose a lot of numerical precision. The original reference for square root filtering and smoothing is G. J. Bierman's 1977 book "Factorization methods for discrete sequentiall estimation." There are other books that deal with specific multi-sensor tracking problems. Have a look at the Artech House library catalogue.
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Hi community,
I am new in the domain of Data Assimilation. I am trying to understand the process of Joint state and parameter estimation using Kalman Filtering approach; especially how the parameter(s) is estimated.
I understand that state variables are updated ta each time step using an updated Covar. matrix and Kalman Gain and then doing some matrix operation. However, what about the parameters. Is it affected by Kalman gain and/or Covar. matrix? I know that parameter at a new time step is calculated = previous time step parameter+ some noise. So, how does this noise evolves/changes/affacted by Kalman gain as scuh. I have tried to read different other papers, but couldn't find a definite answer.
To summarize my question, I guess I am looking for the detailed steps to estimate parameter.
Dear Shah Mahmood,
You should augment your states vector with the parameter you need to estimate. Then your EKF/UKF will deal with this augmented parameter as a state, and the estimation-prediction process is kept unchanged (I mean the steps are still the same). However, you need to linearize your A and C matrices with the new augmented model, at each iteration... You should also update your SS matrices in the case of the joint states-parameters estimation.
For more details, you can check my work.
Regards,
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I currently have the assignment to make a small demo for a specific algorithm, here we want to explain the Kalman Filter in its basic form. Sadly we havent really found any " noob friendly " examples and are trying to just jump ahead in other peoples code and trying to reverse engineer it.
Here in the dlm package on R, we've found the variables GG, FF, c0, CO etc. being used in the dlm function. Altought reading page 7 ( https://cran.r-project.org/web/packages/dlm/dlm.pdf ) these variables arent really explained and i would like to know what these mean. Or are these random variable names to use the algoritm with. This is the first day of the assignment, if anyone has any tips how to approach this or a easy "noob " friendly R language based example please do reffer this to me.
Many thanks!
The DLM part of the Kalman filter isn't a big deal, because the estimator can accommodate linear time-varying systems. To interpret the variables is to look at what is happening with their covariances. If one is concerned with position and velocity, then error ellipsoids are also useful.
The idea of the Kalman filter is easy: One models the physical process with an equation that propagates the covariances (second moments) over time. Then one applies Baye's rule to obtain the conditional mean given the measurements to correct the estimate for a measurement. That's it !
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Or can the Kalman filter be used for medical image processing ?
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I am trying to implement Kalman filter to get rid of multipath error points on GPS data while my data contains Latitude, Longitude, accuracy, altitude, speed and time. I am using python but unable to figure out how can I choose parameters for solving this issue.
Guidence will be very much apreciated.
Why not learn the parameters from some training data
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Is it theoretically possible, that after discretization by using Talyor Series Expansion, a non-observable nonlinear system will became an observable?
It was proved, that used continuous model of PMSM is non-observable (see attached). I want to know, if resulting discrete system is observable or not. Any comment appreciated. Thanks.
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LMMSE estimator and kalman filter
which of this estimator achieve more performance in FBMC?
For Gaussian noise and linear models, the Kalman filter is optimal in the sense that it correctly calculates the posterior distribution and attains the minimum achievable matrix mse. For non-Gaussian noise, but linear models, the Kalman filter (and the lmmse approach) achieve the matrix mse given by. However, calculating the conditional mean would yield an error smaller than, or equal to, the matrix mse. For nonlinear models, the Kalman filter cannot be used. Nevertheless, the lmmse approach can be used; it gives the best estimate linear in data and achieves the matrix mse given by.
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I want to understand the overview of Kalman filters and what purpose do they serve. I am trying to leverage this platform as I am not sure about other authentic sources on the internet to get this information. Please suggest one if you know. On another note, if I have a signal that is a mixture of various frequencies; can Kalman filter be used to extract the DC or average value of this signal? Please let me know how does it work. Thanks in advance.
Early literature of Kalman filtering refferes to two steps of Kalman filtering as predicition and correction step. Using wording “prediction step” has led to a common misunderstanding among engineers and researchers that Kalman filter can also be used for prediction. Hence, in recent literature the term “prediction step” was replaced with term “process update step” and “correction step” was replaced with “measurement update step”, which is more correct.
The point of Kalman filter is not that it can remove high-frequency noise, the point of Kalman filter is that you can estimate quantities that you cannot measure directly, but you can measure other quantities that are affected by the quantity that you would like to estimate, but cannot be measured directly. Or maybe you can measure it directly by using different sensors and you would like to combine all the measurements to get better estimate. You can also have situation where you can measure on quantity directly, but the measurements are inaccurate (drift, noise, random errors), and you can also measure other quantities that that are affected by the quantity that you would like to estimate and have other sources of errors. Then you can combine all the measurements and get estimate that is less affected by the errors, because you have available a lot of information from various sources.
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In case of blind equalization, information about reference input is not available. Thus channel modeling using NLMS, RLS or even Kalman filter is really difficult. Because weight update needs the information about desired or reference input.
you can take the simplest one NLMS combined with a Blind Source Separation, it give a better performances, instead of using only Adaptive filter.
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the error convergence matrix in Kalman Filter provides the convergence of the filter. In similar war, how the convergence of the Complementary Filter can be obtained?
Compute the white-noise gain (WNG) of the filter. Do this in either: In the (discrete) time domain, as an infinite summation of the squared impulse response, over m = 0...Inf; or in the (continuous) frequency domain, as an integral of the squared magnitude of the filter's frequency response, over -pi...+pi. They give the same result (the WNG). In the latter case, you will need to divide the result by 2*pi. Assuming the steady-state bias is zero (because you have matched your filter model to the signal) WNG*var_mes is the variance of the estimate output by your filter at steady-state, i.e. after the filter has converged, for an input signal that is corrupted by additive white noise, with a variance of var_mes. It need not be Gaussian noise.
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Dual control in Stochastic optimal control suggests that control input has probing action that would generate with non zero probability the uncertainty in measurement of state.
It means we have a Covariance error matrix at each state.
The same matrix is also deployed by Kalman filter so how it is different from Dual Control
Dear Sandeep,
These are the dual objectives to be achieved in particular, a major difficulty consists in resolving the Exploration / Exploitation (E / E) compromise.
Best regards
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In the paper "Magnetometer-Only Attitude Determination Using
Novel Two-Step Kalman Filter Approach" in equation 19 do I need to complete quaternion multiplication first and then do simple mathematical finite differencing or it is in other way?
Hi,
In fact the matrix H is not explicit in equation (19), it is present in the equation giving the estimate and the covariance which are then updated using equations (5).
Best regards
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Hi.
I'm modelling OFDM rayleigh fading channel, and I want to trace channel state information with Unscented Kalman Filter through MATLAB. I'm using autoregressive model to construct the filter.
I have no problem with creating state transition function for first order autoregressive model (AR(1)). But when it comes to second order AR(2) , I don't know how the create state transition function.
I can create state transition matrix A as shown in the picture. Kalman filter needs to reach 2 previous state at once. So how do I construct the filter?
By the way, I edited the example in the link below:
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I am looking for a MATLAB code to implement channel prediction or channel state information (CSI) using a Kalman filter-based approach. Specifically, for a MIMO environment.
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suggest me literature regarding this topic and the gaudiness to start my thesis
hope for good response...
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Hi everyone,
I have implemented an EKF in a power systems application. When I run a simulation in Matlab, in some iterations of the filter I get a Kalman gain matrix (K) with negative values and/or absolute values greater than 1. In some books I have read that the Kalman gain is a real value between 0 an1.
Is this correct ? Or is it an indication that something wrong with the Kalman filter?
I have another opinion though I am not certain and welcome any corrections.
I think only the Kalman gains that correspond to measurable states are between [0 1]. Those states that are not directly measurable could have Kalman gains beyond [0 1].
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I started to learn the Kalman filter. However, I couldn't understand its' difference with the feedback/feedforward control system. Ultimately both systems are used to minimize the errors between the measurement and estimated state.
Kalman filter is an optimum observer. It is not a controller itself. Think about a feedback control system. You have five state variables and you are getting the measurement of only two variables. You can use the Kalman filter to estimate the whole states. It is a model that measured outputs and control enters into filter and the estimated states computed.
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Hi,
I have an implementation of Kalman filter for a tracking problem, with constant acceleration model. In this model:
My State Matrix is:
x = [x, y , Vx, Vy, ax, ay] ;
My measurement Matrix is:
Y = [x' , y', Vx', Vy'];
I am putting the following as my Measurement Covariance matrix:
R = [r11, r12, 0, 0
r21, r22, 0, 0
0, 0 , r33, r34
0, 0, r43, r44];
Sometimes I have my measurement Position (x',y') that is not sometimes not so perfect. One solution is that:
- if I put a high Measurement Covariance on Position [r11, r12, r21, r22] then I can avoid a big effect on my filtered output, but it will effect both Position and Velocity.
What my problem is, that: I want to separate the Position and Velocity under certain conditions, which essentially means that I want my Position measurement (x',y') to not effect my Filter velocity. The filter should be updating the Position and Velocity Separately.
Is there a good way to do that?
I am trying to put the measurement covariance matrix on "Top Right" and "Bottom Left" but that ultimately makes my filter more unstable. Any ideas here?
Bar-Shalom, Y., Rong Li, X., and Kirubarajan, T., Estimation with Applications to Tracking and Navigation, John Wiley 2001.
Hari Hablani
IIT Indore, Indore, India
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I am using Kalman filter to filter out noise from PMU data for load parameter estimation. I am successfully able to filter out noise by hit and trial (Tuning values of R and Q). I am interested in knowing if there is anyway I can automatically tune the parameters of Kalman filter once I get a window of PMU measurements.
Dear Hur Rizvi,
I suggest you to see link and attached files on topic.
Best regards
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Dear friend,
These days, I'm trying to finish my Ph.D. in electrical engineering (control engineering). I'm specialized in Extended-Kalman fitter application, fault detection, neural network, digital transportation, digital twins and machine learning respectively. It is necessary to say that, my thesis about industry 4.0 at pipeline performance management (software). I'm enthusiastic to join a team for my postdoc. And I struggle to study at the edge of science topics for my postdoc.
Would you help me please to join in the on the team for my postdoc, study abroad or a way to join in this kind of program?
Post doc is for those whose Phd is weak.
I suggest to look for some Entrepreneurship in your domains of interest
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I have a human skeletal data for 25 different joints from a depth sensor camera (Kinect) while performing a lifting work. But the data has some noise and inferred data due to occlusion. I am trying to use Tobit-Kalman Filter for data processing but could not figure out how to get the co-variance of process and observation noise for my data? I read multiple papers regarding the use of Tobit-Kalman Filter for processing Kinect data but could not find an in-depth explanation of the process involved. Can anyone help me out with my problem?
Thank you.
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Hi. I have read in several articles that it is possible to improve the performance of the Kalman Filter using neural networks to dynamically change the filter model, thus adapting it to the non-modeled variations of the system with greater efficiency.
I have the Kalman filter working correctly, but I don't know how to incorporate the information provided by the neural networks, nor with what parameters to train them.
I am a beginner in the subject and would appreciate some advice on how to implement this at a simulation level in matlab.
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How to design the Kalman filter for closed-loop boost converter?
Hello dear
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Dear friend,
These days, I'm trying to finish my Ph.D. in electrical engineering (control engineering). I'm specialized in Extended-Kalman fitter application, fault detection, neural network, digital transportation, digital twins and machine learning respectively. It is necessary to say that, my thesis about Industry 4.0 at pipeline performance management (software). I'm enthusiastic to join a team for my postdoc. And I struggle to study at the edge of science topics for my postdoc.
Would you help me please to join in the on the team for my postdoc, study abroad or a way to join in this kind of program?
Hi Syed Ali,
You may follow: https://www.timeshighereducation.com/ which for all PostDoc jobs and my be helpful for you.
Good Luck
Ali
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Dear Community, I am wondering if there are articles/books/references proposing implementations of Kalman Filters for positioning based on a single accelerometer ? (no GPS, no camera, no gyro, no odometery...)
thank you Mohammed Aftatah
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For example, I need to estimate the state (position, speed, acceleration) of a vehicle, but I can only observe the its position inaccurately. The problem is, without knowing the speed, I cannot update the state for the next time point. If I calculate the speed using the position difference between 2 time points i.e. v_t=(p_t-p_(t-1))/(delta t), the estimation becomes very unstable: if one estimation has a large deviation, the speed would also deviate from the real one, so the forecast of the next time point is even more inaccurate.
Does any one have any good ideas? Thanks in advance!
Do you try using another version of Kalman filter?
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Hello all,
I have a question regarding the implementation presented in paper titled "Real-Time Metric State Estimation for Modular Vision-Inertial Systems" by Stephan Weiss.
at Epoch 1: The monocular visual odometry starts at origin like IMU (both are aligned with world coordinate system).
Epoch 2 is shown below, where visual odometry moves along Z-axis (in world frame), IMU moves along X-axis (in world frame) and the position of Camera (in world)
The update & correction will be visual frame (which moves along Z-axis). For fusion to be successful, If we transform the visual odometry to IMU frame then they will be pointing in the same direction.
Then our measurement vector Zp would also change or we use the same formula as in the paper?
consult this work: Filtre de Kalman discret à la modélisation HydrologiqueFebruary 2019
Publisher: Editions universitaires européennes
ISBN: ISBN13: 978-613-8-46494-5 ISBN-10:613846494X
Project:
Multi-site modeling and prediction of annual and monthly precipitation
📷Samra Harkat
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I have a basic doubt in statistics.(I am working with kalman filter)
Lets say i get a state variable 'Xt' for each time instance t as the kalman filter output.And for each time step, i also have a ground truth 'Yt' from an expensive sensor.
So,
Yt=Xt+ Nt , where Nt is the noise with covariance P.
I have the complete csv data of Yt and Xt measured for every time stamp. How can i evaluate the covariance P from the data?
Chris Morris Thank you for the answer! But I think we cannot calculate it using usual way. And the method I found to estimate the covariance from the given equation is through reduced chi-suquared stastics which is used for OLS Estimation: https://en.wikipedia.org/wiki/Ordinary_least_squares#Estimation
Ahmed K. Jameil Thank you Sir.
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http://hitchhikersgui.de/Improper_integral | 1,542,396,651,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743110.55/warc/CC-MAIN-20181116173611-20181116195611-00319.warc.gz | 167,204,164 | 23,286 | # Improper integral
An improper integral of the first kind. The integral may need to be defined on an unbounded domain.
An improper Riemann integral of the second kind. The integral may fail to exist because of a vertical asymptote in the function.
In mathematical analysis, an improper integral is the limit of a definite integral as an endpoint of the interval(s) of integration approaches either a specified real number, ${\displaystyle \infty }$, ${\displaystyle -\infty }$, or in some instances as both endpoints approach limits. Such an integral is often written symbolically just like a standard definite integral, in some cases with infinity as a limit of integration.
Specifically, an improper integral is a limit of the form:
${\displaystyle \lim _{b\to \infty }\int _{a}^{b}f(x)\,dx,\qquad \lim _{a\to -\infty }\int _{a}^{b}f(x)\,dx,}$
or
${\displaystyle \lim _{c\to b^{-}}\int _{a}^{c}f(x)\,dx,\quad \lim _{c\to a^{+}}\int _{c}^{b}f(x)\,dx,}$
in which one takes a limit in one or the other (or sometimes both) endpoints (Apostol 1967, §10.23).
By abuse of notation, improper integrals are often written symbolically just like standard definite integrals, perhaps with infinity among the limits of integration. When the definite integral exists (in the sense of either the Riemann integral or the more advanced Lebesgue integral), this ambiguity is resolved as both the proper and improper integral will coincide in value.
Often one is able to compute values for improper integrals, even when the function is not integrable in the conventional sense (as a Riemann integral, for instance) because of a singularity in the function or because one of the bounds of integration is infinite.
## Examples
The original definition of the Riemann integral does not apply to a function such as ${\displaystyle 1/{x^{2}}}$ on the interval [1, ∞), because in this case the domain of integration is unbounded. However, the Riemann integral can often be extended by continuity, by defining the improper integral instead as a limit
${\displaystyle \int _{1}^{\infty }{\frac {1}{x^{2}}}\,dx=\lim _{b\to \infty }\int _{1}^{b}{\frac {1}{x^{2}}}\,dx=\lim _{b\to \infty }\left(-{\frac {1}{b}}+{\frac {1}{1}}\right)=1.}$
The narrow definition of the Riemann integral also does not cover the function ${\displaystyle 1/{\sqrt {x}}}$ on the interval [0, 1]. The problem here is that the integrand is unbounded in the domain of integration (the definition requires that both the domain of integration and the integrand be bounded). However, the improper integral does exist if understood as the limit
${\displaystyle \int _{0}^{1}{\frac {1}{\sqrt {x}}}\,dx=\lim _{a\to 0^{+}}\int _{a}^{1}{\frac {1}{\sqrt {x}}}\,dx=\lim _{a\to 0^{+}}(2-2{\sqrt {a}})=2.}$
The improper integral
${\displaystyle \int _{0}^{\infty }{\frac {dx}{(x+1){\sqrt {x}}}}=\pi }$
has unbounded intervals for both domain and range.
Sometimes integrals may have two singularities where they are improper. Consider, for example, the function 1/((x + 1)x) integrated from 0 to (shown right). At the lower bound, as x goes to 0 the function goes to , and the upper bound is itself , though the function goes to 0. Thus this is a doubly improper integral. Integrated, say, from 1 to 3, an ordinary Riemann sum suffices to produce a result of π/6. To integrate from 1 to , a Riemann sum is not possible. However, any finite upper bound, say t (with t > 1), gives a well-defined result, 2 arctan(t) − π/2. This has a finite limit as t goes to infinity, namely π/2. Similarly, the integral from 1/3 to 1 allows a Riemann sum as well, coincidentally again producing π/6. Replacing 1/3 by an arbitrary positive value s (with s < 1) is equally safe, giving π/2 − 2 arctan(s). This, too, has a finite limit as s goes to zero, namely π/2. Combining the limits of the two fragments, the result of this improper integral is
{\displaystyle {\begin{aligned}\int _{0}^{\infty }{\frac {dx}{(x+1){\sqrt {x}}}}&{}=\lim _{s\to 0^{+}}\int _{s}^{1}{\frac {dx}{(x+1){\sqrt {x}}}}+\lim _{t\to \infty }\int _{1}^{t}{\frac {dx}{(x+1){\sqrt {x}}}}\\&{}=\lim _{s\to 0^{+}}\left({\frac {\pi }{2}}-2\arctan {\sqrt {s}}\right)+\lim _{t\to \infty }\left(2\arctan {\sqrt {t}}-{\frac {\pi }{2}}\right)\\&{}={\frac {\pi }{2}}+\left(\pi -{\frac {\pi }{2}}\right)\\&{}=\pi .\end{aligned}}}
This process does not guarantee success; a limit might fail to exist, or might be infinite. For example, over the bounded interval from 0 to 1 the integral of 1/x does not converge; and over the unbounded interval from 1 to the integral of 1/x does not converge.
The improper integral
${\displaystyle \int _{-1}^{1}{\frac {dx}{\sqrt[{3}]{x^{2}}}}=6}$
converges, since both left and right limits exist, though the integrand is unbounded near an interior point.
It might also happen that an integrand is unbounded near an interior point, in which case the integral must be split at that point. For the integral as a whole to converge, the limit integrals on both sides must exist and must be bounded. For example:
{\displaystyle {\begin{aligned}\int _{-1}^{1}{\frac {dx}{\sqrt[{3}]{x^{2}}}}&{}=\lim _{s\to 0}\int _{-1}^{-s}{\frac {dx}{\sqrt[{3}]{x^{2}}}}+\lim _{t\to 0}\int _{t}^{1}{\frac {dx}{\sqrt[{3}]{x^{2}}}}\\&{}=\lim _{s\to 0}3(1-{\sqrt[{3}]{s}})+\lim _{t\to 0}3(1-{\sqrt[{3}]{t}})\\&{}=3+3\\&{}=6.\end{aligned}}}
But the similar integral
${\displaystyle \int _{-1}^{1}{\frac {dx}{x}}}$
cannot be assigned a value in this way, as the integrals above and below zero do not independently converge. (However, see Cauchy principal value.)
## Convergence of the integral
An improper integral converges if the limit defining it exists. Thus for example one says that the improper integral
${\displaystyle \lim _{t\to \infty }\int _{a}^{t}f(x)\,dx}$
exists and is equal to L if the integrals under the limit exist for all sufficiently large t, and the value of the limit is equal to L.
It is also possible for an improper integral to diverge to infinity. In that case, one may assign the value of ∞ (or -∞) to the integral. For instance
${\displaystyle \lim _{b\to \infty }\int _{1}^{b}{\frac {1}{x}}\,dx=\infty .}$
However, other improper integrals may simply diverge in no particular direction, such as
${\displaystyle \lim _{b\to \infty }\int _{1}^{b}x\sin(x)\,dx,}$
which does not exist, even as an extended real number. This is called divergence by oscillation.
A limitation of the technique of improper integration is that the limit must be taken with respect to one endpoint at a time. Thus, for instance, an improper integral of the form
${\displaystyle \int _{-\infty }^{\infty }f(x)\,dx}$
can be defined by taking two separate limits; to wit
${\displaystyle \int _{-\infty }^{\infty }f(x)\,dx=\lim _{a\to -\infty }\lim _{b\to \infty }\int _{a}^{b}f(x)\,dx}$
provided the double limit is finite. It can also be defined as a pair of distinct improper integrals of the first kind:
${\displaystyle \lim _{a\to -\infty }\int _{a}^{c}f(x)\,dx+\lim _{b\to \infty }\int _{c}^{b}f(x)\,dx}$
where c is any convenient point at which to start the integration. This definition also applies when one of these integrals is infinite, or both if they have the same sign.
An example of an improper integrals where both endpoints are infinite is the Gaussian integral ${\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}}$. An example which evaluates to infinity is ${\displaystyle \int _{-\infty }^{\infty }e^{x}\,dx}$. But one cannot even define other integrals of this kind unambiguously, such as ${\displaystyle \int _{-\infty }^{\infty }x\,dx}$, since the double limit is infinite and the two-integral method
${\displaystyle \lim _{a\to -\infty }\int _{a}^{c}x\,dx+\lim _{b\to \infty }\int _{c}^{b}x\,dx}$
yields ${\displaystyle \infty -\infty }$. In this case, one can however define an improper integral in the sense of Cauchy principal value:
${\displaystyle \operatorname {p.v.} \int _{-\infty }^{\infty }x\,dx=\lim _{b\to \infty }\int _{-b}^{b}x\,dx=0.}$
The questions one must address in determining an improper integral are:
• Does the limit exist?
• Can the limit be computed?
The first question is an issue of mathematical analysis. The second one can be addressed by calculus techniques, but also in some cases by contour integration, Fourier transforms and other more advanced methods.
## Types of integrals
There is more than one theory of integration. From the point of view of calculus, the Riemann integral theory is usually assumed as the default theory. In using improper integrals, it can matter which integration theory is in play.
• For the Riemann integral (or the Darboux integral, which is equivalent to it), improper integration is necessary both for unbounded intervals (since one cannot divide the interval into finitely many subintervals of finite length) and for unbounded functions with finite integral (since, supposing it is unbounded above, then the upper integral will be infinite, but the lower integral will be finite).
• The Lebesgue integral deals differently with unbounded domains and unbounded functions, so that often an integral which only exists as an improper Riemann integral will exist as a (proper) Lebesgue integral, such as ${\displaystyle \int _{1}^{\infty }{\frac {1}{x^{2}}}\,dx}$. On the other hand, there are also integrals that have an improper Riemann integral but do not have a (proper) Lebesgue integral, such as ${\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x}}\,dx}$. The Lebesgue theory does not see this as a deficiency: from the point of view of measure theory, ${\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x}}\,dx=\infty -\infty }$ and cannot be defined satisfactorily. In some situations, however, it may be convenient to employ improper Lebesgue integrals as is the case, for instance, when defining the Cauchy principal value. The Lebesgue integral is more or less essential in the theoretical treatment of the Fourier transform, with pervasive use of integrals over the whole real line.
• For the Henstock–Kurzweil integral, improper integration is not necessary, and this is seen as a strength of the theory: it encompasses all Lebesgue integrable and improper Riemann integrable functions.
## Improper Riemann integrals and Lebesgue integrals
Figure 1
Figure 2
In some cases, the integral
${\displaystyle \int _{a}^{c}f(x)\,dx}$
can be defined as an integral (a Lebesgue integral, for instance) without reference to the limit
${\displaystyle \lim _{b\to c^{-}}\int _{a}^{b}f(x)\,dx}$
but cannot otherwise be conveniently computed. This often happens when the function f being integrated from a to c has a vertical asymptote at c, or if c = ∞ (see Figures 1 and 2). In such cases, the improper Riemann integral allows one to calculate the Lebesgue integral of the function. Specifically, the following theorem holds (Apostol 1974, Theorem 10.33):
• If a function f is Riemann integrable on [a,b] for every b ≥ a, and the partial integrals
${\displaystyle \int _{a}^{b}|f(x)|\,dx}$
are bounded as b → ∞, then the improper Riemann integrals
${\displaystyle \int _{a}^{\infty }f(x)\,dx,\quad {\mbox{and }}\int _{a}^{\infty }|f(x)|\,dx}$
both exist. Furthermore, f is Lebesgue integrable on [a, ∞), and its Lebesgue integral is equal to its improper Riemann integral.
For example, the integral
${\displaystyle \int _{0}^{\infty }{\frac {dx}{1+x^{2}}}}$
can be interpreted alternatively as the improper integral
${\displaystyle \lim _{b\to \infty }\int _{0}^{b}{\frac {dx}{1+x^{2}}}=\lim _{b\to \infty }\arctan {b}={\frac {\pi }{2}},}$
or it may be interpreted instead as a Lebesgue integral over the set (0, ∞). Since both of these kinds of integral agree, one is free to choose the first method to calculate the value of the integral, even if one ultimately wishes to regard it as a Lebesgue integral. Thus improper integrals are clearly useful tools for obtaining the actual values of integrals.
In other cases, however, a Lebesgue integral between finite endpoints may not even be defined, because the integrals of the positive and negative parts of f are both infinite, but the improper Riemann integral may still exist. Such cases are "properly improper" integrals, i.e. their values cannot be defined except as such limits. For example,
${\displaystyle \int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx}$
cannot be interpreted as a Lebesgue integral, since
${\displaystyle \int _{0}^{\infty }\left|{\frac {\sin(x)}{x}}\right|\,dx=\infty .}$
But ${\displaystyle f(x)={\frac {\sin(x)}{x}}}$ is nevertheless integrable between any two finite endpoints, and its integral between 0 and ∞ is usually understood as the limit of the integral:
${\displaystyle \int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx=\lim _{b\to \infty }\int _{0}^{b}{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}.}$
## Singularities
One can speak of the singularities of an improper integral, meaning those points of the extended real number line at which limits are used.
## Cauchy principal value
Consider the difference in values of two limits:
${\displaystyle \lim _{a\to 0^{+}}\left(\int _{-1}^{-a}{\frac {dx}{x}}+\int _{a}^{1}{\frac {dx}{x}}\right)=0,}$
${\displaystyle \lim _{a\to 0^{+}}\left(\int _{-1}^{-a}{\frac {dx}{x}}+\int _{2a}^{1}{\frac {dx}{x}}\right)=-\ln 2.}$
The former is the Cauchy principal value of the otherwise ill-defined expression
${\displaystyle \int _{-1}^{1}{\frac {dx}{x}}{\ }\left({\mbox{which}}\ {\mbox{gives}}\ -\infty +\infty \right).}$
Similarly, we have
${\displaystyle \lim _{a\to \infty }\int _{-a}^{a}{\frac {2x\,dx}{x^{2}+1}}=0,}$
but
${\displaystyle \lim _{a\to \infty }\int _{-2a}^{a}{\frac {2x\,dx}{x^{2}+1}}=-\ln 4.}$
The former is the principal value of the otherwise ill-defined expression
${\displaystyle \int _{-\infty }^{\infty }{\frac {2x\,dx}{x^{2}+1}}{\ }\left({\mbox{which}}\ {\mbox{gives}}\ -\infty +\infty \right).}$
All of the above limits are cases of the indeterminate form ∞ − ∞.
These pathologies do not affect "Lebesgue-integrable" functions, that is, functions the integrals of whose absolute values are finite.
## Summability
An improper integral may diverge in the sense that the limit defining it may not exist. In this case, there are more sophisticated definitions of the limit which can produce a convergent value for the improper integral. These are called summability methods.
One summability method, popular in Fourier analysis, is that of Cesàro summation. The integral
${\displaystyle \int _{0}^{\infty }f(x)\,dx}$
is Cesàro summable (C, α) if
${\displaystyle \lim _{\lambda \to \infty }\int _{0}^{\lambda }\left(1-{\frac {x}{\lambda }}\right)^{\alpha }f(x)\,dx}$
exists and is finite (Titchmarsh 1948, §1.15). The value of this limit, should it exist, is the (C, α) sum of the integral.
An integral is (C, 0) summable precisely when it exists as an improper integral. However, there are integrals which are (C, α) summable for α > 0 which fail to converge as improper integrals (in the sense of Riemann or Lebesgue). One example is the integral
${\displaystyle \int _{0}^{\infty }\sin x\,dx}$
which fails to exist as an improper integral, but is (C,α) summable for every α > 0. This is an integral version of Grandi's series.
## Multivariable improper integrals
The improper integral can also be defined for functions of several variables. The definition is slightly different, depending on whether one requires integrating over an unbounded domain, such as ${\displaystyle \mathbb {R} ^{2}}$, or is integrating a function with singularities, like ${\displaystyle f(x,y)=\log(x^{2}+y^{2})}$.
### Improper integrals over arbitrary domains
If ${\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} }$ is a non-negative function that is Riemann integrable over every compact cube of the form ${\displaystyle [-a,a]^{n}}$, for ${\displaystyle a>0}$, then the improper integral of f over ${\displaystyle \mathbb {R} ^{n}}$ is defined to be the limit
${\displaystyle \lim _{a\to \infty }\int _{[-a,a]^{n}}f,}$
provided it exists.
A function on an arbitrary domain A in ${\displaystyle \mathbb {R} ^{n}}$ is extended to a function ${\displaystyle {\tilde {f}}}$ on ${\displaystyle \mathbb {R} ^{n}}$ by zero outside of A:
${\displaystyle {\tilde {f}}(x)={\begin{cases}f(x)&x\in A\\0&x\not \in A\end{cases}}}$
The Riemann integral of a function over a bounded domain A is then defined as the integral of the extended function ${\displaystyle {\tilde {f}}}$ over a cube ${\displaystyle [-a,a]^{n}}$ containing A:
${\displaystyle \int _{A}f=\int _{[-a,a]^{n}}{\tilde {f}}.}$
More generally, if A is unbounded, then the improper Riemann integral over an arbitrary domain in ${\displaystyle \mathbb {R} ^{n}}$ is defined as the limit:
${\displaystyle \int _{A}f=\lim _{a\to \infty }\int _{A\cap [-a,a]^{n}}f=\lim _{a\to \infty }\int _{[-a,a]^{n}}{\tilde {f}}.}$
### Improper integrals with singularities
If f is a non-negative function which is unbounded in a domain A, then the improper integral of f is defined by truncating f at some cutoff M, integrating the resulting function, and then taking the limit as M tends to infinity. That is for ${\displaystyle M>0}$, set ${\displaystyle f_{M}=\min\{f,M\}}$. Then define
${\displaystyle \int _{A}f=\lim _{M\to \infty }\int _{A}f_{M}}$
provided this limit exists.
### Functions with both positive and negative values
These definitions apply for functions that are non-negative. A more general function f can be decomposed as a difference of its positive part ${\displaystyle f_{+}=\max\{f,0\}}$ and negative part ${\displaystyle f_{-}=\max\{-f,0\}}$, so
${\displaystyle f=f_{+}-f_{-}}$
with ${\displaystyle f_{+}}$ and ${\displaystyle f_{-}}$ both non-negative functions. The function f has an improper Riemann integral if each of ${\displaystyle f_{+}}$ and ${\displaystyle f_{-}}$ has one, in which case the value of that improper integral is defined by
${\displaystyle \int _{A}f=\int _{A}f_{+}-\int _{A}f_{-}.}$
In order to exist in this sense, the improper integral necessarily converges absolutely, since
${\displaystyle \int _{A}|f|=\int _{A}f_{+}+\int _{A}f_{-}.}$[1][2]
## Notes
1. ^ Cooper 2005, p. 538: "We need to make this stronger definition of convergence in terms of |f(x)| because cancellation in the integrals can occur in so many different ways in higher dimensions."
2. ^ Ghorpade & Limaye 2010, p. 448: "The relevant notion here is that of unconditional convergence." ... "In fact, for improper integrals of such functions, unconditional convergence turns out to be equivalent to absolute convergence."
## Bibliography
• Apostol, T (1974), Mathematical analysis, Addison-Wesley, ISBN 978-0-201-00288-1.
• Apostol, T (1967), Calculus, Vol. 1 (2nd ed.), Jon Wiley & Sons.
• Autar Kaw, Egwu Kalu (2008), Numerical Methods with Applications (1st ed.), autarkaw.com
• Titchmarsh, E (1948), Introduction to the theory of Fourier integrals (2nd ed.), New York, N.Y.: Chelsea Pub. Co. (published 1986), ISBN 978-0-8284-0324-5.
• Cooper, Jeffery (2005), Working analysis, Gulf Professional
• Ghorpade, Sudhir; Limaye, Balmohan (2010), A course in multivariable calculus and analysis, Springer | 5,667 | 19,185 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 75, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-47 | latest | en | 0.862945 |
https://gustancho.com/first-payment-after-closing/ | 1,721,315,703,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514831.13/warc/CC-MAIN-20240718130417-20240718160417-00785.warc.gz | 256,342,644 | 65,383 | # First Payment After Closing For Mortgage Loan Borrowers
Gustan Cho Associates are mortgage brokers licensed in 48 states
This guide covers the first payment after closing for mortgage loan borrowers. One of the most common questions new homebuyers ask is when the first payment after closing. Other homeowners tell many that it will be a while before they make the first payment after closing on the new home.
The first payment on a mortgage loan typically becomes due one month after you close on the loan. You want to ensure you make your first payment after closing and not be late.
For example, if you close your mortgage on January 15th, your first mortgage payment will likely be due on March 1st. Many have worked hard to get good credit and do not want a late payment reported on their credit report. So how does the new homeowner know when their first payment after closing is? This is very easy to figure out. In the following paragraphs, we will cover homeowners’ first payment after closing.
## Understanding Mortgage Interest Payments Are Paid In Arrears
A new mortgage borrower’s first payment after closing is due at the beginning of the first full month after the home closes. Then, the mortgage payments are due every month thereafter. Remember to set up a reliable method for making your mortgage payments on time, as timely payments are crucial to maintaining a good credit history and avoiding late fees.
Mortgage interest always accrues on the unpaid principal balance of the mortgage. Interest on a mortgage is always paid after the interest has accumulated and not before it.
For example, if your August mortgage payment includes the mortgage interest for July. Or if you closed on your home in the middle of July, you would pay the mortgage interest for July because you have owned and occupied the home.
## How Are Mortgage Payments Calculated
Your mortgage payment is typically calculated using a formula considering several factors, including the loan amount, interest rate, loan term, property taxes, and insurance. The most common formula used for this calculation is the mortgage payment formula:
�=��⋅�⋅(1+�)�(1+�)�−1+�
Where:
• is your monthly mortgage payment.
• �� is the principal loan amount (the amount you borrowed),
• is your monthly interest rate (annual interest rate divided by 12),
• is the number of payments (loan term in months), and
• is the total of your monthly property taxes and homeowners insurance.
## Understanding Mortgage Principal Payments are Paid in Advance
The exact timing can vary depending on the terms of your loan and your lender’s policies. Some lenders may allow you to choose a specific payment date within the month, while others may have a set due date. Most borrowers have a mortgage escrow. A mortgage payment consists of the following:
• Principal
• Interest
• Taxes
• Insurance
The mortgage payment mainly consists of principal and interest. The principal portion of the mortgage payment is paid in advance. As borrowers pay the principal portion of the mortgage, the balance gets reduced. The mortgage’s interest is paid on the lesser balance the following month.
## How The Mortgage Payment Process Works and First Payment After Closing
It’s important to carefully review your loan documents and communicate with your lender to understand the specifics of your mortgage terms, including when your first payment is due and how it should be made. Sometimes, your first payment may be collected at closing, or you may be required to make a partial payment for the first month.
The title company’s closing agent collects mortgage interest payments up to 30 days before the first full month at closing.
The mortgage interest collected is stated on the home closing statement. This is charged as a closing cost on the home loan. If a borrower closes their home loan in March, they will be charged prorated mortgage interest from the 15th to the 31st month of March. This will cover the interest due on the mortgage for the prior 16 days of the month. The first mortgage payment due by the borrower will be due on May 1st. That payment will include mortgage interest for April.
## How Can I Delay The First Payment After Closing
Borrowers can schedule their home closing towards the end of the month and avoid paying the prorated interest out of their pockets. This means fewer closing costs. If you want to delay making the first mortgage payment after closing your home loan, you must schedule the closing towards the beginning of the month, but you will have a higher closing cost to cover the interest payment.
Closing costs can be covered with a seller’s concession or lender credit.
For more information about this article or other mortgage-related topics, please get in touch with us at Gustan Cho Associates at 800-900-8569 or text us for a faster response. Or email us at gcho@gustancho.com. The Gustan Cho Associates Mortgage Group team is available seven days a week, evenings, weekends, and holidays. We are direct lenders with no lender overlays on government or conventional loans. We are experts in non-QM loans and bank statement loans for self-employed borrowers. | 1,039 | 5,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-30 | latest | en | 0.974749 |
tutorialsfor3dsmax.blogspot.com | 1,386,459,587,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163056101/warc/CC-MAIN-20131204131736-00000-ip-10-33-133-15.ec2.internal.warc.gz | 189,341,807 | 19,189 | Hello boys and girls. This is another 3d modeling tutorial from witch you will learn how to create a football/soccer ball in 3ds Max 2011. Also, you will learn how to texture the ball using the Unwrap UVW modifier. This was the official ball for the World Cup 1982. In the next few days i will create other 3d modeling tutorials that will show you how to model other types of sport balls. That being said let's start this 3d modeling tutorial.
Step 1: To start this 3d modeling tutorial Open or Rest 3ds Max. From Create, Geometry, Extended Primitives create a Hedra with a Radius of 50cm, change the Family type to Dodec/Icos and the Family Paramater P to 0.35. Now select the Select and Move tool and change the value for X,Y,Z to 0 in the bottom of the screen to move our object to the center of the grid.
Step 2: To convert our Hedra to Editable Poly right click on it and Convert to > Convert to Editable Poly.
Step 3a: It's now time to Detach all the polygons individually. We have to ways of doing this. The first one is to go in the Modify tab, select the Edge selection mode, press CTRL+A to select all edges and in Edit Edges rollout menu click on Split.
Step 3b: If you chose the first option(witch i recommend) you can skip the next 3 steps.
For the second option go in the Modify tab select the Polygon selection mode, select one polygon, in Edit Geometry rollout menu click on Detach, check the box for Detach to Element and click OK. Now to keep track of what polygons we've detached just click on Hide Selection. Doing this we will know exactly how many polygons we have to Detach.
You must do the exact same thing for 32 Polygons. This will take some time.
Step 4: Half way there. Recap: Select, Detach to Element, Hide!
Step 5: When you have only one polygons left click on Unhide All.
Step 6: Will we'll stop for a little while with our 3d modeling tutorial because in the next steps we will apply the texture. It's much easier to do this now then later. For now just Exit selection mode. To do this simply select Editable Poly.
Step 7: From the Modifier list apply a Unwrap UVW modifier.
Step 8: From Unwrap UVW modifier click on "Edit...".
Step 9: Now click the plus sign next to Unwrap UVW modifier to expand it. Select Face and press CTRL+A on your keyboard to select them all. Check the box for Select By Element. In Edit UVW's window using the Move tool move the selection to the center of the square you can see outlined.
Note: You can use Pan and scroll button to zoom to navigate in the Edit UVW's window.
Step 10: In Edit UVW's window click on Mapping and select Flatten Mapping, in the Flatten Mapping pop-up window click OK.
Step 11: In Edit UVW's window click on CheckerPattern (Checker) and select Pick Texture. In the pop-up windows select Bitmap and from there select this TEXTURE(click to Open).
Note: I've already created the texture because i don't want to go in Photoshop with you guys. But if you want to make something different like a classic ball in white and black simply go to Tools > Render UVW, change the Width and Height to 2048, click Render UV Template, Save the UVW and using Photoshop open the saved UVW map and paint on top of it as you like.
Step 12: Press "M" to open the Material Editor or select it from the main tool bar. Select the first Standard material(if you want you can use VRayMtl) and click the box next to Diffuse, select Bitmap and again open the texture we've used in step 11.
Step 13: First click on Show Standard Map in Viewport and after that click on Assign Material to Selection. Press "M" to close the Material Editor.
Step 14: As you can see the texture in not quite matching up. It's time for some tweaking now. But first let me show you what I'm talking about. As you can see below from 5 faces forming a circle only the selected one dose not match with the others. So to fix this simply select it by clicking on it in the viewport, go in Edit UVW's window and Using the Mirror or Rotate tool rotate(180°) the face until she fits with the rest of them.
Step 15: Do the same thing for all the faces until all of them match up. Also rotate the logos to fit OK.
Step 16: After all the faces are aligned correctly make sure they also fit. To do this select the Move tool in the Edit UVW's window and move all the faces that dose not fit like the one you can see below.
Step 17: We are finish with this step. Cloe the Edit UVW and let's get back to our 3d modeling tutorial.
Step 18: Continuing with our 3d modeling tutorial add a MeshSmooth modifier form the Modifier list and change the Iterations to 3.
Step 19: Add a Spherify modifier.
Step 20: Add a Edit Poly modifier.
Step 21: Select Polygon selection mode from the Edit Poly modifier and press CTRL+A to select all polygons. Click on the little box next to Extrude in Edit Polygons rollout menu, leave the extrusion type as Group Normals and change the amount to 0.75cm, click the check mark for OK.
Step 22: Exit selection mode and add another Meshsmooth modifier, only this time leave the Iterations to 1 and change the Subdivision Method to Quad Output.
Step 23: Open Material Editor, press "M" or select it from the main toolbar, select our material, change the Specular level to 50 and the Glossiness to 30, in Maps check the box for Bump and change the amount to 2. Click on None next to Bump, select Bitmap and Open this Noise Texture(click to open).
Step 24: Change the Tiling to 2 and press F9 to render(or select it from the main toolbar). You can see our bump map's effect on the ball.
Step 25: That's it. We are finished with this 3d modeling tutorial.
After you are finished with this 3d modeling tutorial you can use one of the following rendering tutorials "Studio Setup with 3ds Max and VRay" or "Mp3 Player Materials, Lights and Rendering tutorial" to get similar results such as the one bellow:
As usual don't hesitate to comment if you have any problems following this 3d modeling tutorial. | 1,431 | 5,970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | longest | en | 0.850763 |
https://www.zbmath.org/?q=an%3A0937.06009 | 1,652,844,935,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521041.0/warc/CC-MAIN-20220518021247-20220518051247-00751.warc.gz | 1,272,984,162 | 12,697 | ## Algebraic foundations of many-valued reasoning.(English)Zbl 0937.06009
The book under review presents an outstanding monograph on MV-algebras and many-valued reasoning by worldwide known experts on this topic. MV-algebras entered mathematics in 1958 by C. C. Chang as algebraic analysis of many-valued logics. They give an algebraic framework for the infinite-valued propositional calculus of Łukasiewicz, and today MV-algebras are a well-developed theory with many nice and deep results and with many applications, e.g., in computer science. The accent is put on the algebraic point of view and on this basis the Łukasiewicz calculus is developed.
The monograph consists of 10 chapters and a bibliography with 250 items. Chapter 1 is devoted to basic notions of MV-algebras. The original long list of Chang’s axioms is reduced to six axioms: An MV-algebra is an algebra $$(A; \oplus, \neg,0)$$ of type $$(2,1,0)$$ such that, for all $$x,y, z \in A$$, (MV1) $$x \oplus (y \oplus z) = (x \oplus y) \oplus z$$; (MV2) $$x \oplus y = y \oplus x$$; (MV3) $$x \oplus 0 = 0$$; (MV4) $$\neg\neg x = x$$; (MV5) $$x \oplus \neg 0= \neg 0$$; (MV6) $$\neg(\neg x \oplus y) \oplus y = \neg(\neg y \oplus x) \oplus x.$$ Putting $$x \leq y$$ iff $$\neg x \oplus y = 1 :\neg 0$$, we have that $$(A,\leq)$$ is a distributive lattice with the least and the greatest elements $$0$$ and $$1$$, respectively.
For example, if $$(G,u)$$ is a unital abelian $$\ell$$-group with strong unit, then $$\Gamma(G,u) :=\{g \in G:0 \leq g \leq u\}$$ equipped with $$g_1 \oplus g_2 := (g_1 + g_2) \wedge u$$; $$\neg g:= u - g,$$ gives a prototypical example of an MV-algebra. In particular, if $$G = \mathbb R$$, $$u =1$$, we obtain a very important case of MV-algebras. The fundamental result of Chang’s Subdirect Representation Theorem is proved here using the theory of ideals. In addition, MV-equations are introduced, and it is proved that an MV-equation is satisfied by all MV-algebras iff it is satisfied by all MV-chains. This result is considerably strengthened in the second chapter. Chapter 2 is devoted to Chang’s Completeness Theorem. The above mapping $$\Gamma$$ from the category $$\mathcal A$$ whose objects are the unital abelian $$\ell$$-groups into the category $${\mathcal {MV}}$$ of MV-algebras gives a functor which plays a fundamental role, because it defines even a natural equivalence as is proved in a subsequent chapter. For this reason, good sequences will be of special interest. We say that a sequence a $$= (a_1,a_2,\ldots)$$ of elements of an MV-algebra is good iff, for each $$i \geq 1$$, $$a_i \oplus a_{i+1} = a_i$$ and there is an integer $$n$$ such that $$a_r =0$$ for all $$r \geq n.$$ Good sequences are converted into a partially ordered monoid $$M_A$$, and this into a Chang unital $$\ell$$-group $$(G_A,u_A)$$ with strong unit $$u_A$$; in particular, any MV-algebra $$A$$ is isomorphic with $$\Gamma(G_A,u_A)$$. Consequently, Chang’s Completeness Theorem saying that an equation holds in $$[0,1]$$ iff it holds in every MV-algebra, is proved. Chapter 3 is dedicated to free MV-algebras. A special role is played by the free MV-algebras $$\text{Free}_n$$ over $$n$$ generators. $$\text{Free}_n$$ can be easily described by piecewise linear continuous functions over $$[0,1]^n$$ with values in the interval $$[0,1]$$. Such functions are so-called McNaughton functions, and they are studied in detail. In many cases, very important MV-algebras (simple and semisimple) can be described as MV-subalgebras of continuous fuzzy sets over some compact Hausdorff spaces.
The radical, $$\text{Rad}(A)$$, is defined as the intersection of all maximal ideals of $$A$$, it gives important information on $$A$$. It consists of all infinitesimals, i.e., of all elements $$a$$ such that $$na \leq \neg a$$ for each $$n \geq 1.$$ Łukasiewicz $$\infty$$-valued calculus is described in Chapter 4. In the early twenties Łukasiewicz introduced a system of logic in which propositions admit as truth values real numbers between 0 and 1. The main connectives are $$\text{implication }\to$$ and $$\text{negation }\neg$$, such that $$x \to y := \min(1,1-x+y)$$ and $$\neg x = 1- x.$$ These connectives can be rewritten in terms of MV-algebras as $$x \to y = \neg x \oplus y.$$ Using results of Chapter 3, it is proved that all tautologies are obtainable from a certain set of initial tautologies (corresponding to the MV-axioms) by a finite number of applications of modens ponens, and an effective procedure to decide whether a proposition is a tautology. For this reason, Wajsberg algebras and Lindenbaum algebras are studied. In particular, it is shown that every countable MV-algebra is the Lindenbaum algebra of some theory. Ulam’s game is presented in Chapter 5. This game goes back to Ulam. The authors consider the variant of Twenty Questions where $$n-2$$ lies or errors are allowed. In the Ulam game with $$m$$ lies, our knowledge is presented by the function $$\sigma : S \to \{0,1,\ldots,m-1\}$$, where $$S$$ is a finite set of numbers. Such problems are closely connected with finding an optimal strategy in the Ulam game with $$m$$ lies or finding an optimal $$m$$-error correcting code. MV-algebras can be allowed as algebras of states of knowledge in generalized Ulam games, where the number of lies may depend on the secret number $$x \in S$$, and $$S$$ itself may be infinite. Truth values may be irrational numbers in $$[0,1]$$, or even nonstandard real numbers. Chapter 6 is a continuation of lattice-theoretical properties of MV-algebras. Here minimal prime ideals are studied together with Stonean ideals, archimedean elements, and hyperarchimedean algebras. In addition, complete MV-algebras and complete distributivity are studied. In Chapter 7, the crucial fact that the category of MV-algebras and the category of abelian unital $$\ell$$-groups are categorically equivalent via the functor $$\Gamma$$ defined above is established (this is the famous Mundici Representation Theorem). As a corollary, a genuine addition can be uniquely recovered from the MV-algebraic structure. Perfect MV-algebras are MV-algebras $$A$$ such that each element $$x \in A$$ belongs either to $$\text{Rad}(A)$$ or to $$\neg \text{Rad}(A) :=\{\neg a:\;a \in \text{Rad}(A)\}$$. For such MV-algebras it is shown that the category of perfect perfect MV-algebras is categorically equivalent with the category of all $$\ell$$-groups. Chapter 8 is dedicated to the description of all varieties of MV-algebras. Komori’s classification and varieties generated by finite chains are presented. Advanced topics are given in Chapter 9. The first part deals with disjunctive minimal forms in the infinite-valued calculus of Łukasiewicz. The relationship between MV-algebras and approximately finite-dimensional $$C^*$$-algebras is presented. Finally, the important Di Nola Representation Theorem is given, which says that every MV-algebra $$A$$ is an algebra of $$[0,1]^*$$-valued functions over some set, where $$[0,1]^*$$ is an ultrapower of $$[0,1]$$, depending only on the cardinality of $$A$$. The last chapter is Further Reading, where the authors outline further ways of studying MV-algebras, like states, observables, product, probability, etc. All chapters contain bibliographical remarks.
The wonderful book is addressed to computer scientists, mathematicians, logicians wishing to get acquainted with a compact body of beautiful theory, results and methodologies on MV-algebras, that have found applications in the handling of uncertain information, connecting many areas of mathematics like lattice-ordered groups, $$C^*$$-algebras, lattices, algebra, geometry of numbers, model theory, polyhedra, etc. It is a welcome addition to the literature.
### MSC:
06D35 MV-algebras 03B50 Many-valued logic 06-02 Research exposition (monographs, survey articles) pertaining to ordered structures 03-02 Research exposition (monographs, survey articles) pertaining to mathematical logic and foundations 03G25 Other algebras related to logic 06F20 Ordered abelian groups, Riesz groups, ordered linear spaces | 2,175 | 8,089 | {"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} | 3.21875 | 3 | CC-MAIN-2022-21 | latest | en | 0.827913 |
https://www.scribd.com/doc/60242610/Failure-Theories-Von-Mises | 1,455,174,862,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701161718.0/warc/CC-MAIN-20160205193921-00037-ip-10-236-182-209.ec2.internal.warc.gz | 876,664,647 | 26,113 | P. 1
Failure Theories Von Mises
# Failure Theories Von Mises
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# Solid Mechanics: Failure Criteria Failure Criteria for Ductile Materials
Maximum Shear Stress Criterion The maximum shear stress criterion, also known as Tresca's or Guest's criterion, is often used to predict the yielding of ductile materials. Yield in ductile materials is usually caused by the slippage of crystal planes along the maximum shear stress surface. Therefore, a given point in the body is considered safe as long as the maximum shear stress at that point is under the yield shear stress σy obtained from a uniaxial tensile test. With respect to 2D stress, the maximum shear stress is related to the difference in the two principal stresses (see Mohr's Circle). Therefore, the criterion requires the principal stress difference, along with the principal stresses themselves, to be less than the yield shear stress,
Graphically, the maximum shear stress criterion requires that the two principal stresses be within the green zone indicated below,
Von Mises Criterion The von Mises Criterion (1913), also known as the maximum distortion energy criterion, octahedral shear stress theory, or Maxwell-Huber-Hencky-von Mises theory, is often used to estimate the yield of ductile materials. The von Mises criterion states that failure occurs when the energy of distortion reaches
the same energy for yield/failure in uniaxial tension. Also shown on the figure is the maximum shear stress criterion (dashed line). In addition to bounding the principal stresses to prevent ductile failure. σ3 = 0. This theory is more conservative than the von Mises criterion since it lies inside the von Mises ellipse. Mathematically. In the cases of plane stress. the von Mises criterion also gives a reasonable estimation of fatigue failure. The von Mises criterion reduces to. This equation represents a principal stress ellipse as illustrated in the following figure. this is expressed as. especially in cases of repeated tensile and tensile-shear loading .
Coulomb. -σc < {σ1.Solid Mechanics: Failure Criteria Failure Criteria for Brittle Materials Maximum Normal Stress Criterion The maximum stress criterion. also known as the normal stress. also known as the Coulomb-Mohr criterion or internalfriction theory. Graphically. and is applied to cases of 2D stress. is often used to predict the failure of brittle materials. or the uniaxial compression strength σc. Mohr's theory is often used in predicting the failure of brittle materials. or Rankine criterion. σ2} < σt where σ1 and σ2 are the principal stresses for 2D stress. The maximum stress criterion states that failure occurs when the maximum (normal) principal stress reaches either the uniaxial tension strength σt. Mohr's Theory The Mohr Theory of Failure. Mohr's theory suggests that failure occurs when Mohr's Circle at a point in the body exceeds the envelope created by the two Mohr's circles for uniaxial tensile strength and . the maximum stress criterion requires that the two principal stresses lie within the green zone depicted below. is based on the famous Mohr's Circle.
σ2 > 0 4 Graphically. Likewise. σ2 > 0 Criterion requirements σ 1 < σ t. σ2 < 0 σ1 in compression. σ2 < 0 σ1 > -σc. σ2 in tension σ1 < 0.uniaxial compression strength. All intermediate stress states fall into one of the four categories in the following table. Mohr's theory requires that the two principal stresses lie within the green zone depicted below. The left circle is for uniaxial compression at the limiting compression stress σc of the material. σ 2 < σ t σ1 < 0. This envelope is shown in the figure below. Each case defines the maximum allowable values for the two principal stresses to avoid failure. Case 1 2 3 Principal Stresses Both in tension Both in compression σ1 > 0. σ2 in compression σ1 > 0. σ2 > -σc σ1 in tension. the right circle is for uniaxial tension at the limiting tension stress σt. . The middle Mohr's Circle on the figure (dash-dot-dash line) represents the maximum allowable stress for an intermediate stress state.
Also shown on the figure is the maximum stress criterion (dashed line). . This theory is less conservative than Mohr's theory since it lies outside Mohr's boundary.
The angle θp defines the principal directions where the only stresses are normal stresses.y. First. in accordance with the coordinate transformation equations. The result is.Solid Mechanics: Stress Principal Stress for the Case of Plane Stress Principal Directions. there exists an angle θp where the shear stress τx'y' becomes zero. That angle is found by setting τx'y' to zero in the above shear transformation equation and solving for θ (set equal to θp). The transformation to the principal directions can be illustrated as: .z directions) via. Principal Stress The normal stresses (σx' and σy') and the shear stress (τx'y') vary smoothly with respect to the rotation angle θ. There exist a couple of particular angles where the stresses take on special values. These stresses are called principal stresses and are found from the original stresses (expressed in the x.
The maximum shear stress is equal to one-half the difference between the two principal stresses. The transformation to the maximum shear stress direction can be illustrated as: . is where the maximum shear stress occurs. and solving for θ. The result is. θs.Maximum Shear Stress Direction Another important angle. This is found by finding the maximum of the shear stress transformation equation.
.
the threedimensional stress state can be reduced to two dimensions. For example. such as the surfaces of thin-walled pressure vessels under external or internal pressure. these simplified 2D problems are called plane stress problems. the failure plane of a brittle shaft under torsion is often at a 45° angle with respect to the shaft's axis. Nonetheless. To reduce the 3D stress matrix to the 2D plane stress matrix. As a result. the free surfaces of shafts in torsion and beams under transverse load. Assume that the negligible principal stress is oriented in the z-direction. For example.Solid Mechanics: Stress Plane Stress and Coordinate Transformations Plane State of Stress A class of common engineering problems involving stresses in a thin plate or on the free surface of a structural element. The sign convention for positive stress components in plane stress is illustrated in the above figure on the 2D element. remove all components with z subscripts to get. By assuming that this small principal stress is zero. where τxy = τyx for static equilibrium. Since the remaining two principal stresses lie in a plane. Stress transformation formulas are required to . to analyze a bar one almost always directs one of the coordinate directions along the bar's axis. stresses in directions that do not line up with the original coordinate set are also important. the direct and shear stress components are associated with these directions. have one principal stress that is much smaller than the other two. Coordinate Transformations The coordinate directions chosen to analyze a structure are usually based on the shape of the structure.
The transformation of stresses with respect to the {x.z'} coordinates are shown in the figure below.y.y'. .y'.z'} is performed via the equations.analyze these stresses. where θ is the rotation angle between the two coordinate sets (positive in the counterclockwise direction). This angle along with the stresses for the {x'.z} coordinates to the stresses with respect to {x'.
Solid Mechanics: Failure Criteria Techniques for Failure Prevention and Diagnosis There exist a set of basic techniques for preventing failure in the design stage. Stress Concentrations Sound design avoids rapid changes in material or geometrical properties. between layers. supports. holes. boundaries. This is fine and good when FEA is applied appropriately. However. joints. Maximum stresses are often located at loading Loading Points points. By following basic rules of thumb. a reinforcement composed of generally no less than the material removed should be added around the opening. the popularity of finite element analysis can condition engineers to look just for red spots in simulation output. The addition of safety factors to designs allow engineers to reduce sensitivity to manufacturing defects and to compensate for stress prediction limitations. Safety Factors . and where cross-section areas change rapidly. or maximum deflection points. and for diagnosing failure in manufacturing and later stages. Stress concentrations are usually located near corners. without really understanding the essence or funda at play. In the Design Stage It is quite commonplace today for design engineers to verify design stresses with finite element (FEA) packages. such danger points can often be anticipated and avoided without total reliance on computer simulation. For example. crack tips. when a large hole is removed from a structure.
Reduction in strength can result from exposure to UV lights and chemical corrosion.htm .samconsult. Manufacturing Concentrations defects such as size mismatches and improper fastener application can lead to residual stresses and even cracks. Damage and Exposure Fatigue and Creep Reference: http://www.biz/Science/Failure_Criteria/Failure%20Criteria. both strong stress concentrations. these failures must be diagnosed and resolved quickly and effectively. initial surface imperfections can result Stress from sloppy machining processes. debonding. unanticipated failure may occur in parts after design and manufacturing. rather than an involved collection of factors. Induced For example. and delamination can result from unanticipated resonant vibrations and impacts that exceed the design loads. Stress concentrations may be induced by inadequate manufacturing processes. Such failures may be caught early in initial quality assurance testing. Fatigue or creep can lead a part to failure. For example. the failure is caused by a singular factor.In Post-Manufacturing Stages Despite the best efforts of design and manufacturing engineers. Damages such as cracks. Often. or later after the part is delivered to the customer. Damages during service life can lead a part to failure. unanticipated fatigue can result from repeated mechanical or thermal loading. In order for projects to succeed.
scribd
/*********** DO NOT ALTER ANYTHING BELOW THIS LINE ! ************/ var s_code=s.t();if(s_code)document.write(s_code)//--> | 2,144 | 10,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2016-07 | latest | en | 0.877733 |
https://www.ibiblio.org/kuphaldt/electricCircuits/Digital/DIGI_9.html | 1,718,268,256,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.74/warc/CC-MAIN-20240613060639-20240613090639-00155.warc.gz | 765,121,274 | 7,544 | # COMBINATIONAL LOGIC FUNCTIONS
Original author: David Zitzelsberger
## Introduction
The term "combinational" comes to us from mathematics. In mathematics a combination is an unordered set, which is a formal way to say that nobody cares which order the items came in. Most games work this way, if you rolled dice one at a time and get a 2 followed by a 3 it is the same as if you had rolled a 3 followed by a 2. With combinational logic, the circuit produces the same output regardless of the order the inputs are changed.
There are circuits which depend on the when the inputs change, these circuits are called sequential logic. Even though you will not find the term "sequential logic" in the chapter titles, the next several chapters will discuss sequential logic.
Practical circuits will have a mix of combinational and sequential logic, with sequential logic making sure everything happens in order and combinational logic performing functions like arithmetic, logic, or conversion.
You have already used combinational circuits. Each logic gate discussed previously is a combinational logic function. Let's follow how two NAND gate works if we provide them inputs in different orders.
We begin with both inputs being 0.
We then set one input high.
We then set the other input high.
So NAND gates do not care about the order of the inputs, and you will find the same true of all the other gates covered up to this point (AND, XOR, OR, NOR, XNOR, and NOT).
As a first example of useful combinational logic, let's build a device that can add two binary digits together. We can quickly calculate what the answers should be:
```0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 102
```
So we well need two inputs (a and b) and two outputs. The low order output will be called Σ because it represents the sum, and the high order output will be called Cout because it represents the carry out.
The truth table is
Simplifying boolean equations or making some Karnaugh map will produce the same circuit shown below, but start by looking at the results. The Σ column is our familiar XOR gate, while the Cout column is the AND gate. This device is called a half-adder for reasons that will make sense in the next section.
The half-adder is extremely useful until you want to add more that one binary digit quantities. The slow way to develop a two binary digit adders would be to make a truth table and reduce it. Then when you decide to make a three binary digit adder, do it again. Then when you decide to make a four digit adder, do it again. Then when ... The circuits would be fast, but development time would be slow.
Looking at a two binary digit sum shows what we need to extend addition to multiple binary digits.
``` 11
11
11
---
110
```
Look at how many inputs the middle column uses. Our adder needs three inputs; a, b, and the carry from the previous sum, and we can use our two-input adder to build a three input adder.
Σ is the easy part. Normal arithmetic tells us that if Σ = a + b + Cin and Σ1 = a + b, then Σ = Σ1 + Cin.
What do we do with C1 and C2? Let's look at three input sums and quickly calculate:
```Cin + a + b = ?
0 + 0 + 0 = 0 0 + 0 + 1 = 1 0 + 1 + 0 = 1 0 + 1 + 1 = 10
1 + 0 + 0 = 1 1 + 0 + 1 = 10 1 + 1 + 0 = 10 1 + 1 + 1 = 11
```
If you have any concern about the low order bit, please confirm that the circuit and ladder calculate it correctly.
In order to calculate the high order bit, notice that it is 1 in both cases when a + b produces a C1. Also, the high order bit is 1 when a + b produces a Σ1 and Cin is a 1. So We will have a carry when C1 OR (Σ1 AND Cin). Our complete three input adder is:
For some designs, being able to eliminate one or more types of gates can be important, and you can replace the final OR gate with an XOR gate without changing the results.
A0 is the low order bit of A, A1 is the high order bit of A, B0 is the low order bit of B, B1 is the high order bit of B, Σ0is the low order bit of the sum, Σ1 is the high order bit of the sum, and Cout is the Carry.
A two binary digit adder would never be made this way. Instead the lowest order bits would also go through a full adder.
There are several reasons for this, one being that we can then allow a circuit to determine whether the lowest order carry should be included in the sum. This allows for the chaining of even larger sums. Consider two different ways to look at a four bit sum.
``` 111 1<-+ 11<+-
0110 | 01 | 10
1011 | 10 | 11
----- - | ---- | ---
10001 1 +-100 +-101
```
If we allow the program to add a two bit number and remember the carry for later, then use that carry in the next sum the program can add any number of bits the user wants even though we have only provided a two-bit adder. Small PLCs can also be chained together for larger numbers.
These full adders can also can be expanded to any number of bits space allows. As an example, here's how to do an 8 bit adder.
This is the same result as using the two 2-bit adders to make a 4-bit adder and then using two 4-bit adders to make an 8-bit adder or re-duplicating ladder logic and updating the numbers.
For any large combinational circuit there are generally two approaches to design: you can take simpler circuits and replicate them; or you can design the complex circuit as a complete device.
Using simpler circuits to build complex circuits allows a you to spend less time designing but then requires more time for signals to propagate through the transistors. The 8-bit adder design above has to wait for all the Cxout signals to move from A0 + B0 up to the inputs of Σ7.
If a designer builds an 8-bit adder as a complete device simplified to a sum of products, then each signal just travels through one NOT gate, one AND gate and one OR gate. A seventeen input device has a truth table with 131,072 entries, and reducing 131,072 entries to a sum of products will take some time.
When designing for systems that have a maximum allowed response time to provide the final result, you can begin by using simpler circuits and then attempt to replace portions of the circuit that are too slow. That way you spend most of your time on the portions of a circuit that matter.
## Decoder
A decoder is a circuit that changes a code into a set of signals. It is called a decoder because it does the reverse of encoding, but we will begin our study of encoders and decoders with decoders because they are simpler to design.
A common type of decoder is the line decoder which takes an n-digit binary number and decodes it into 2n data lines. The simplest is the 1-to-2 line decoder. The truth table is
A is the address and D is the dataline. D0 is NOT A and D1 is A. The circuit looks like
Only slightly more complex is the 2-to-4 line decoder. The truth table is
Developed into a circuit it looks like
Larger line decoders can be designed in a similar fashion, but just like with the binary adder there is a way to make larger decoders by combining smaller decoders. An alternate circuit for the 2-to-4 line decoder is
Replacing the 1-to-2 Decoders with their circuits will show that both circuits are equivalent. In a similar fashion a 3-to-8 line decoder can be made from a 1-to-2 line decoder and a 2-to-4 line decoder, and a 4-to-16 line decoder can be made from two 2-to-4 line decoders.
You might also consider making a 2-to-4 decoder ladder from 1-to-2 decoder ladders. If you do it might look something like this:
For some logic it may be required to build up logic like this. For an eight-bit adder we only know how to sum eight bits by summing one bit at a time. Usually it is easier to design ladder logic from boolean equations or truth tables rather than design logic gates and then "translate" that into ladder logic.
A typical application of a line decoder circuit is to select among multiple devices. A circuit needing to select among sixteen devices could have sixteen control lines to select which device should "listen". With a decoder only four control lines are needed.
## Encoder
An encoder is a circuit that changes a set of signals into a code. Let's begin making a 2-to-1 line encoder truth table by reversing the 1-to-2 decoder truth table.
This truth table is a little short. A complete truth table would be
One question we need to answer is what to do with those other inputs? Do we ignore them? Do we have them generate an additional error output? In many circuits this problem is solved by adding sequential logic in order to know not just what input is active but also which order the inputs became active.
A more useful application of combinational encoder design is a binary to 7-segment encoder. The seven segments are given according
Our truth table is:
Deciding what to do with the remaining six entries of the truth table is easier with this circuit. This circuit should not be expected to encode an undefined combination of inputs, so we can leave them as "don't care" when we design the circuit. The equations were simplified with karnaugh maps.
The collection of equations is summarised here:
The circuit is:
## Demultiplexers
A demultiplexer, sometimes abbreviated dmux, is a circuit that has one input and more than one output. It is used when a circuit wishes to send a signal to one of many devices. This description sounds similar to the description given for a decoder, but a decoder is used to select among many devices while a demultiplexer is used to send a signal among many devices.
A demultiplexer is used often enough that it has its own schematic symbol
The truth table for a 1-to-2 demultiplexer is
Using our 1-to-2 decoder as part of the circuit, we can express this circuit easily
This circuit can be expanded two different ways. You can increase the number of signals that get transmitted, or you can increase the number of inputs that get passed through. To increase the number of inputs that get passed through just requires a larger line decoder. Increasing the number of signals that get transmitted is even easier.
As an example, a device that passes one set of two signals among four signals is a "two-bit 1-to-2 demultiplexer". Its circuit is
or by expressing the circuit as
shows that it could be two one-bit 1-to-2 demultiplexers without changing its expected behavior.
A 1-to-4 demultiplexer can easily be built from 1-to-2 demultiplexers as follows.
## Multiplexers
A multiplexer, abbreviated mux, is a device that has multiple inputs and one output.
The schematic symbol for multiplexers is
The truth table for a 2-to-1 multiplexer is
Using a 1-to-2 decoder as part of the circuit, we can express this circuit easily.
Multiplexers can also be expanded with the same naming conventions as demultiplexers. A 4-to-1 multiplexer circuit is
That is the formal definition of a multiplexer. Informally, there is a lot of confusion. Both demultiplexers and multiplexers have similar names, abbreviations, schematic symbols and circuits, so confusion is easy. The term multiplexer, and the abbreviation mux, are often used to also mean a demultiplexer, or a multiplexer and a demultiplexer working together. So when you hear about a multiplexer, it may mean something quite different.
## Using multiple combinational circuits
As an example of using several circuits together, we are going to make a device that will have 16 inputs, representing a four digit number, to a four digit 7-segment display but using just one binary-to-7-segment encoder.
First, the overall architecture of our circuit provides what looks like our the description provided.
Follow this circuit through and you can confirm that it matches the description given above. There are 16 primary inputs. There are two more inputs used to select which digit will be displayed. There are 28 outputs to control the four digit 7-segment display. Only four of the primary inputs are encoded at a time. You may have noticed a potential question though.
When one of the digits are selected, what do the other three digits display? Review the circuit for the demultiplexers and notice that any line not selected by the A input is zero. So the other three digits are blank. We don't have a problem, only one digit displays at a time.
Let's get a perspective on just how complex this circuit is by looking at the equivalent ladder logic.
Notice how quickly this large circuit was developed from smaller parts. This is true of most complex circuits: they are composed of smaller parts allowing a designer to abstract away some complexity and understand the circuit as a whole. Sometimes a designer can even take components that others have designed and remove the detail design work.
In addition to the added quantity of gates, this design suffers from one additional weakness. You can only see one display one digit at a time. If there was some way to rotate through the four digits quickly, you could have the appearance of all four digits being displayed at the same time. That is a job for a sequential circuit, which is the subject of the next several chapters.
Lessons In Electric Circuits copyright (C) 2000-2023 Tony R. Kuphaldt, under the terms and conditions of the CC BY License. | 3,073 | 13,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-26 | latest | en | 0.914411 |
http://forums.devshed.com/programming-languages-139/help-algorithm-total-newb-936232.html | 1,516,489,782,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889736.54/warc/CC-MAIN-20180120221621-20180121001621-00400.warc.gz | 125,799,278 | 10,079 | ### Thread: Help with this algorithm..total newb
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#### Help with this algorithm..total newb
I have not been able to figure these three algorithms, and Im gonna be tested on similar ones soon. Can anyone offer me a break down?
Consider the following algorithm:
L = input(”Input a list of positive integers:”)
m = -1
c = 0
N = len(L)
i=0
while i < N do
if (L[i] < 0) then
print ”not expecting negative integer. Error.”
STOP
else if (m < L[i]) then
m = L[i]
c = 1
else if (m == L[i]) then
c = c + 1
end if
i = i + 1
end while
print m + ”,” + c
It is not required that you show all of the work. Just writing the output of the
algorithm is enough. But running the algorithm for a few elements of the list will
give you an idea of how it’s working.
3.1 What will be the output of the algorithm for input L = [2, 5, 1, 6,
5, 6, 5].
3.2 Run the algorithm for input L = [] i.e. an empty list. Hint: length
of an empty list is 0.
3.3 What does the algorithm do in general?
3.4 What is the time complexity of the algorithm. Provide reasoning
and/or proof.
Buggy Bubble sort
.
Following is buggy implementation of bubble sort. A bug is a problem in the
algorithm that would result in an undesired output. Identify what’s the problem
and how would you fix it. You may assume that the user enters the correct input.
Hint: there are 2 bugs
L = input(”Input a list of integers:”)
N = len(L)
i=0
while i < N do
j = i + 1
while (j < N) do
if (L[i] L[j]) then
swap(L[i], L[j])
j = j + 1
end if
end while
i = i + 1
end while
Write an algorithm to reverse a list of elements. Take
the list as an input from the user. Make sure to take care
of all the boundary conditions. What is the time complexity
Example: given the following list of elements: [1, 5, a], the output should be [a, 5,
1]. Hint: Make use of the swap(L[i], L[j]) function.
Write an algorithm to find the median of a
list of integers. Take the list as an input from the user.
Make sure to take care of all the boundary conditions.
What is the time complexity of your algorithm?
Definition of median: It is a number such that half the integers in the list are smaller
than this number and half are greater than this number.
Example 1: The median in list of integers [5, 1, 3] is 3. Because, there is one
integer less than 3 and one greater than 3.
Example 2: The median in list of integers [5, 1, 3, 7] is 4. Because, there are 2
integers less than 4 and two
2. Hi,
this is no homework service. We can help you with concrete questions, point out errors etc., but the initial ideas have to come from you.
If you ever want to write code, you have to be able to think for yourself and come up with your own solutions. You don't learn just from other people explaining their ideas -- I'm pretty sure that's what you get every day in school, but obviously it's not enough. At some point, you have to actually start working yourself.
The first thing you should do is write down an arbitrary list of (positive) integers and go through the first algrithm to see what it does with the list:
Code:
`L = [2, 5, 1, 6, 5, 6, 5]`
What happens in the first iteration of the "while" loop? | 921 | 3,238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-05 | latest | en | 0.843476 |
https://www.supplyme.com/products/math-conversation-hearts-bulletin-board-a1076 | 1,652,727,034,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512229.26/warc/CC-MAIN-20220516172745-20220516202745-00592.warc.gz | 1,188,701,117 | 101,521 | # Math Conversation Hearts Valentine's Day Bulletin Board Idea
It's always nice when you can coordinate your Valentine's Day decor with unit material and lessons. Heather Florkowski, educator and member of the Scholastic Community, decided to spice up her normal conversation heart theme with math terms and concept review - providing colorful classroom decorations for the holiday and a learning tool for her students!
# Conversation Heart Activity
Florkowski created custom conversation hearts with a card stock heart template and construction paper in assorted Valentine's Day colors (i.e. dark pink, light pink, red, and purple). Using a marker and glitter glue pens, she scripted various math terms on the heart cutouts including - radius, difference, fraction, area, stem and leaf plot, quotient, diameter, etc. These cutouts were then pasted onto heart doilies and attached to the board.
Her students were then given a term to review and research - asked to find the definition, create a visual depiction of the term, and use the term in a sentence. The final "reports" were then mounted onto construction paper (matching that of the conversation heart) and used as an educational tool for the unit.
# Conversation Heart Bulletin Board
• Background: Light pink bulletin board paper.
• Title: "Captivating Conversation Hearts" - Florkowski opted to script the title onto paper with markers and glitter glue pens, then mount the title onto several conversation hearts. You could also use traditional trace and cut letters with complimentary colors of construction paper or store bought lettering.
• Border: Florkowski used small heart doilies, but you might also consider a Valentine's Day themed bulletin board border or a solid trimmer in a complimentary color.
• Decoration: Use the vocabulary conversation hearts and your students' 'reports'!
This is such a simple way for students to review important unit terms, get some writing practice (as they construct sentences), learn about the nuances of researching a concept, and still make the walls look great for the holiday!
Thanks for stopping by! We'd love to hear your thoughts about the board so be sure to leave us a comment below! | 432 | 2,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-21 | longest | en | 0.901615 |
http://didawiki.cli.di.unipi.it/doku.php/mds/smd/2018 | 1,631,833,333,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053759.24/warc/CC-MAIN-20210916204111-20210916234111-00286.warc.gz | 14,880,559 | 8,780 | mds:smd:2018
# Statistical Methods for Data Science A.Y. 2017/18
## Instructors
• Daniele Tantari
• Scuola Normale Superiore
• Salvatore Ruggieri
• Università di Pisa
## Classes
Day of Week Hour Room
Monday 16:00 - 18:00 Fib-L1
Tuesday 9:00 - 11:00 Fib-N1
## Office hours
• Prof. Tantari: Tuesday h 11:00 - 15:00, Scuola Normale Superiore, room 93 (please send an email in advance)
• Prof. Ruggieri: Tuesday h 14:00 - 17:00, Department of Computer Science, room 321/DO.
## Text Books
The following are mandatory text books:
• [B1] F.M. Dekking C. Kraaikamp, H.P. Lopuha, L.E. Meester. A Modern Introduction to Probability and Statistics. Springer, 2005.
• [B2] P. Dalgaard. Introductory Statistics with R. 2nd edition, Springer, 2008.
The following is an optional text book for recalling mathematics pre-requisites of the course:
• [B3] J. Ward, J. Abdey. Mathematics and Statistics. University of London, 2013. Chapters 4-8 of Part 1 present basic calculus (derivatives and integrals).
## Written exam
Written exam consists of open questions and exercises. Example text: sample1, sample2. The exam lasts 2 hours. No teaching material can be consulted during the exam. Registration is mandatory.
Date Hour Room
22/1/2019 9:00 - 11:00 Fib-L1
12/2/2019 9:00 - 11:00 Fib-L1
## Class calendar (final)
Day Room Topic Learning material Instructor
1. 19.02 16:00-18:00 L1 Introduction. Probability and independence. [B1] Chpts. 1-3 Tantari
2. 20.02 9:00-11:00 N1 R basics. [B2] Chpts. 1,2.1,2.4 slides script1.R Ruggieri
3. 27.02 9:00-11:00 N1 Discrete and continuous random variables. [B1] Chpts. 4-5 Tantari
4. 06.03 9:00-11:00 N1 Simulation. Expectation and variance [B1] Chpts. 6-7 noteSim Tantari
5. 12.03 16:00-18:00 L1 R basics and distributions. [B2] Chpts. 2.2,3-4 script2.R Ruggieri
6. 13.03 9:00-11:00 N1 R programming and graphics. [B2] Chpts. 2.3,3-4 exercise.R script3.R Ruggieri
7. 19.03 16:00-18:00 L1 Computations with random variables. Covariance [B1] Chpts. 8-10 Tantari
8. 20.03 9:00-11:00 N1 Sum of random variables. Law of large numbers [B1] Chpts. 11,13 Tantari
9. 26.03 16:00-18:00 L1 The central limit theorem. Graphical summaries [B1] Chpts. 14,15 Tantari
10. 27.03 9:00-11:00 N1 Numerical summaries. Poisson process [B1] Chpts. 12,16 Rcode slidesTantari
11. 16.04 16:00-18:00 L1 Examples on CLT. Data preprocessing. [B2] Chpt. 10 dataprep.r script4.R Ruggieri
12. 17.04 9:00-11:00 N1 Unbiased estimators. Efficiency and MSE [B1] Chpts. 17,19, 20 Tantari
13. 23.04 16:00-18:00 L1 Maximum likelihood. [B1] Chpt. 21 Tantari
14. 24.04 9:00-11:00 N1 Fisher Information. Linear Regressions and Least Squares. [B1] Chpt. 22 fisherTantari
15. 30.04 16:00-18:00 L1 Examples on and MSE. Power-laws Newman's paper, roc_adult.R script5.R Ruggieri
16. 02.05 14:00-16:00 A1 Project and data presentation Tantari+Ruggieri
17. 07.05 16:00-18:00 L1 Confidence Intervals: Gaussian, T-student, large sample method. [B1] Chpt. 23,24 Tantari
18. 08.05 9:00-11:00 N1 Empirical and parametric bootstrap. Application to confidence intervals. [B1] Chpts. 18,23 Tantari
19. 14.05 16:00-18:00 L1 Hypotheses testing. [B1] Chpts. 25-26 Tantari
20. 15.05 9:00-11:00 N1 Hypotheses testing. Bootstrap. Project tutoring. [B2] Chpt. 5.1, script6.R Ruggieri
21. 21.05 16:00-18:00 L1 Hypotheses testing. t-test and application to linear regressions [B1] Chpts. 27 Tantari
22. 22.05 9:00-11:00 N1 Hypotheses testing: correlation and Fisher transformation, comparing samples [B1] Chpt. 28 CorrNotes Tantari
23. 28.05 16:00-18:00 L1 Hypotheses testing: F-test, K-S, chi-square K-S Tantari
24. 29.05 9:00-11:00 N1 Hypotheses testing, parameter estimation. [B2] Chpts. 5.2-5.7, 6, script7.R Ruggieri
## Previous years
mds/smd/2018.txt · Ultima modifica: 24/02/2021 alle 15:46 (7 mesi fa) da Salvatore Ruggieri | 1,425 | 3,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.692433 |
http://mathhelpforum.com/calculus/134551-help-derivation.html | 1,481,229,379,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542655.88/warc/CC-MAIN-20161202170902-00116-ip-10-31-129-80.ec2.internal.warc.gz | 171,014,150 | 9,809 | # Thread: Help with derivation
1. ## Help with derivation
Hi.
Can't seem to get this right. Can anyone please help me.
Derivative of the function
f(x)=xe^x^2
2. Originally Posted by tholterla
Hi.
Can't seem to get this right. Can anyone please help me.
Derivative of the function
f(x)=xe^x^2
You'll need to use a combination of the product and chain rules.
$f(x) = x\,e^{x^2}$
$f'(x) = x\,(e^{x^2})' + e^{x^2}\,(x)'$
$= x(2x)(e^{x^2}) + e^{x^2}(1)$
$= 2x^2\,e^{x^2} + e^{x^2}$
$= e^{x^2}(2x^2 + 1)$ | 203 | 508 | {"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": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.820514 |
https://ng.siyavula.com/read/maths/jss3/similar-shapes/09-similar-shapes?id=93-practical-applications | 1,638,450,308,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362219.5/warc/CC-MAIN-20211202114856-20211202144856-00096.warc.gz | 467,489,186 | 21,136 | We think you are located in Nigeria. Is this correct?
# Test yourself now
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# Chapter 9: Similar shapes
In this chapter, we will explore the mathematical meaning of the term "similar shapes". The general meaning of the word "similar" is that two or more things are almost the same. In Mathematics, similarity has a very specific meaning.
In Mathematics, two shapes are similar only if:
• their matching sides are in proportion, and
• their matching angles are equal.
In other words, if two objects are similar to each other, one of them can be "zoomed in" or "zoomed out" to make it identical to the other one.
similar Two shapes are similar only if their matching sides are in proportion, and their matching angles are equal.
## 9.1 Similar shapes around us
If you compare two different shapes in your environment, you will notice that some of the shapes are similar, and some of the shapes are not similar.
The two figures in Diagram 1 and Diagram 2 are similar, because the figure has stayed in proportion in these two diagrams. The first figure is twice as tall and twice as broad as the second figure.
The figure in Diagram 3 is not similar to the figures in the other two diagrams, because the proportions are different. The relationships between the length and breadth of the third figure and the other two figures are different.
in proportion Two shapes are in proportion if all their dimensions are in the same ratio. For example, the length and breadth of B are three times the length and breadth of A, so A and B are in proportion.
ratio A ratio states what the relationship between two quantities or shapes is. For example, the ratio of A's length to B's length is $1 : 3$. We can also write ratio as a fraction, for example $\frac{1}{3}$.
### Exercise 9.1: Recognise similar shapes
1. Three diagrams of windows are given.
Write down the numbers of the two diagrams that show similar windows.
The windows in Diagram 1 and Diagram 3 are similar.
2. Three diagrams of a shoe are given.
Write down the numbers of the two diagrams that show similar shoes.
The shoes in Diagram 2 and Diagram 3 are similar.
3. The diagram below shows three pentagons.
Write down the numbers of the two diagrams that show similar pentagons.
The pentagons in Diagram 1 and Diagram 2 are similar.
4. The photo below shows cubes and plastic balls. Consider the photo and answer the questions that follow.
1. True or false: The cubes are similar to each other.
True
1. True or false: The plastic balls are not similar to each other.
False. The plastic balls are similar to each other.
5. Explain why the pencils in the photo below are not similar.
The pencils are not similar because all the pencils have the same thickness, but they are of different lengths.
For similar pencils, the shorter pencils would have to be thinner than the longer pencils.
### Similar shapes in Mathematics
We say that two shapes are similar if they are exactly the same shape, but one is bigger or smaller than the other one.
As we have seen, in Mathematics, two shapes are similar only if:
• their matching sides are in proportion, and
• their matching angles are equal.
In many polygons, we have to prove that the matching sides are in proportion and the matching angles are equal, if we want to be sure that the shapes are similar. But, in triangles, we only need to prove that one of them is true. This is a special property of triangles, which you will learn more about later in your school career.
Study the following three diagrams to learn more about what is meant when we say that two shapes are similar.
• The two kites shown below are not similar, because:
• their matching angles are equal, but
• their matching sides are not in proportion.
Kite $A$ Kite $B$
• The two kites shown below are not similar, because:
• their matching sides are in proportion, but
• their matching angles are not equal.
Kite $A$ Kite $B$
• The two kites shown below are similar, because:
• their matching angles are equal, and
• their matching sides are in proportion.
Kite $A$ Kite $B$
### Exercise 9.2: Recognise similar shapes
1. In the photo below, a tiled wall is shown.
Are the three tiles marked A, B and C similar? Give a reason for your answer.
Yes, tiles A, B and C are similar. They are all squares, and all squares are similar to each other because their matching angles are the same, and their matching sides are in proportion. Each square has four equal sides.
2. In the diagram below, two quadrilaterals are given.
Why are the two quadrilaterals not similar?
The matching angles of the two quadrilaterals are not the same.
3. The diagram below shows two triangles.
The two triangles are similar:
• their matching angles are equal, and
• their sides are in proportion, because the measurements of the second triangle are three times bigger than the first triangle.
4. In the diagram below are two cuboids.
The two cuboids are similar:
• the matching angles are equal, and
• the sides are in proportion, because the lengths of the second cuboid are all half the lengths of the first cuboid.
5. In the diagram below, two 3D objects are given.
Ekene says: "The two cubes in the diagram are similar, because all cubes are similar."
Do you agree with Ekene?
All cubes are similar, but only one of the objects in the diagram is a cube. The object with sides of 5 cm, 4 cm and 5 cm is not a cube.
Ekene's statement is incorrect.
## 9.2 Scale factors
The two kites drawn below are similar, because:
• their matching angles are equal, and
• their matching sides are in proportion.
Kite $A$ Kite $B$
The sides of the two kites are in proportion. That means that the lengths of all the sides of one kite have been multiplied by the same number to get the lengths of the sides of the second kite.
If Kite $A$ is the given shape, all of the sides in kite $A$ have been multiplied by 2 to get the sides of Kite $B$.
Kite $B$ is the new shape.
We can write the lengths of the sides as fractions. Write the measurements of the new shape above the line and the measurements of the given shape below the line.
• The lengths of the longer sides as a fraction:
\begin{align} \dfrac{\text{longer side kite } B}{\text{longer side kite } A} & = \dfrac{10}{5} \\ & = 2 \\ \end{align}
• The lengths of the shorter sides as a fraction:
\begin{align} \dfrac{\text{shorter side kite } B}{\text{shorter side kite } A} & = \dfrac{4}{2} \\ & = 2 \\ \end{align}
In each case the answer is 2 .
Kite $B$ is 2 times bigger than kite $A$.
We say that kite $B$ is an enlargement of kite $A$ with a scale factor of 2.
We often use the variable $k$ to represent the scale factor, so in this example, $k = 2$.
When we write the lengths of sides as fractions, we always write the length of the new shape as the numerator (above the line) and the length of the given shape as the denominator (below the line).
enlargement An enlargement of a diagram, shape or object is a copy of the original in which everything is made larger, but keeping the same proportions.
scale factor The scale factor is the number by which every dimension of the given shape is multiplied to get the dimensions of the enlargement. We often use the variable $k$ to represent the scale factor.
### Worked example 9.1: Finding the scale factor
The diagram below shows two similar triangles:
Calculate the scale factor, $k$.
1. Step 1: Check that the two shapes are similar by finding the ratio of all the side lengths. Start by dividing the longest side in the new (bigger) triangle by the longest side in the given (smaller) triangle.
\begin{align} \frac{MN}{KL} & = \frac{22.5}{9} \\ &= \frac{22.5 \times 10}{9 \times 10} \\ &= \frac{225}{90} \\ & = 2.5\\ \end{align}
2. Step 2: Divide the shortest side in the new (bigger) triangle by the shortest side in the given (smaller) triangle.
\begin{align} \frac{PN}{JK} & = \frac{12.5}{5} \\ &= \frac{125}{50} \\ & = 2.5\\ \end{align}
3. Step 3: Divide the third side in the new (bigger) triangle by the third side in the given (smaller) triangle.
\begin{align} \frac{PM}{JL} &= \frac{20}{8} \\ &= 2.5 \\ \end{align}
4. Step 4: Make a conclusion.
The shapes are similar because comparing all the side lengths gives the same answer, which is 2.5.
This means that the new shape ( $\triangle PMN$) is 2.5 times bigger than the given shape ( $\triangle JKL$).
5. Step 5: Give the answer.
Remember that to divide by a decimal number, you need to multiply both the numerator and the denominator by the same power of 10 so that you can divide by a whole number.
For example:
\begin{align} \frac{20}{0.4} &= \frac{20 \times 10}{0.4 \times 10} \\ &= \frac{200}{4} \\ &= 50 \\ \end{align}
### Exercise 9.3: Find the scale factor
1. The diagram below shows two similar squares. The figures are drawn to scale. Find the value of $k$, the scale factor.
Shape $A$ Shape $B$
\begin{align} \frac{\text{side of new square}}{\text{side of given square}} &= \frac{18}{9} \\ & = 2 \\ \end{align}
2. Two similar triangles are given. $\triangle XYZ$ is an enlargement of $\triangle PQR$.
Find the value of the scale factor, $k$.
Ratio of longest sides:
\begin{align} \frac{XY}{PQ} &= \frac{24}{6} \\ & = 4 \\ \end{align}
Ratio of shortest sides:
\begin{align} \frac{ZY}{RQ} &= \frac{16}{4} \\ & = 4 \\ \end{align}
Ratio of third sides:
\begin{align} \frac{XZ}{PR} &= \frac{20}{5} \\ & = 4 \\ \end{align}
All the fractions give 4 as the answer, so the shapes are similar, and the scale factor is 4.
3. In the diagrams below, rectangle $ABCD$ is an enlargement of rectangle $PQRS$.
Find the value of the scale factor, $k$.
Ratio of longer sides:
\begin{align} \frac{BC}{QR} &= \frac{60}{12} \\ & = 5 \\ \end{align}
Ratio of shorter sides:
\begin{align} \frac{DC}{SR} &= \frac{25}{5} \\ & = 5 \\ \end{align}
4. The diagrams below show quadrilateral $QLMP$, which is an enlargement of quadrilateral $JKNR$.
Find the value of the scale factor, $k$.
Ratio of longest sides:
\begin{align} \frac{LM}{KN} &= \frac{15}{5} \\ & = 3 \\ \end{align}
Ratio of second longest sides:
\begin{align} \frac{QL}{JK} &= \frac{10.5}{3.5} \\ &=\frac{10.5 \times 10}{3.5 \times 10}\\ &= \frac{105}{35} \\ & = 3 \\ \end{align}
Ratio of shortest sides:
\begin{align} \frac{PM}{RN} &= \frac{6}{2} \\ & = 3 \\ \end{align}
Ratio of second shortest sides:
\begin{align} \frac{PQ}{JR} &= \frac{9}{3} \\ & = 3 \\ \end{align}
5. Two cubes are given. Consider the information and answer the questions that follow.
• Cube $S$ has sides each of length 7 cm.
• Cube $T$ has sides each of length 70 cm.
1. True or false: Cube $S$ is an enlargement of cube $T$.
The statement is false. Cube $T$ is an enlargement of cube $S$.
1. Calculate the value of the scale factor, $k$.
\begin{align} \frac{\text{side cube } T}{\text{side cube } S} &= \frac{70}{7} \\ & = 10 \\ \end{align}
6. Cuboid $B$ is an enlargement of cuboid $A$.
Find the value of the scale factor, $k$.
Ratio of lengths:
\begin{align} \frac {\text{4.8 m}} {\text{80 cm}} &= \frac {\text{480 cm}} {\text{80 cm} }\\ & = \frac{480}{80} \\ & = 6 \\ \end{align}
Ratio of heights:
\begin{align} \frac{15}{2.5} & = \frac{15 \times 10}{2.5 \times 10} \\ & = \frac{150}{25} \\ & = 6 \\ \end{align}
7. Umar says $\triangle DEF$ is not an enlargement of $\triangle CAB$.
Do you agree with Umar or not? Give a reason for your answer.
Ratio of longer sides:
\begin{align} \frac{DE}{AC} & = \frac{16}{3} \\ & =5 \frac{1}{3} \\ \end{align}
Ratio of shorter sides:
\begin{align} \frac{DF}{CB} & = \frac{12}{3} \\ & = 4 \\ \end{align}
The two ratios are not the same, so $\triangle DEF$ is not an enlargement of $\triangle CAB$. Umar's statement is correct.
8. The diagram below shows two triangles. $\triangle PQR$ is the given shape and $\triangle XYZ$ is the new shape.
Find the value of $k$, the scale factor.
Ratio of longest sides:
\begin{align} \frac{XY}{PQ} &= \frac{3.5}{7} \\ &= \frac{3.5 \times 10}{7 \times 10} \\ &= \frac{35}{70} \\ & = \frac{1} {2} \\ \end{align}
Ratio of shortest sides:
\begin{align} \frac{ZY}{RQ} &= \frac{2.5}{5} \\ &= \frac{2.5 \times 10}{5 \times 10} \\ &= \frac{25}{50} \\ & = \frac{1}{2} \\ \end{align}
Ratio of third sides:
\begin{align} \frac{XZ}{PR} &= \frac{3}{6} \\ & = \frac{1}{2} \\ \end{align}
An enlargement where the scale factor is a fraction between 0 and 1 leads to a new shape or object that is smaller than the given shape or object.
### Enlarging figures using scale factors
You know now that an enlargement is a larger version (or a smaller version) of an original length, shape or object. To ensure that the enlargement has the same proportions, we multiply each dimension by the same scale factor, represented by $k$.
If you are given the scale factor, you can calculate the dimensions of an enlargement of a given shape or object. To do this, you will use the equation given above in this form:
### Worked example 9.2: Using a scale factor
Rectangle $STUV$ is an enlargement of rectangle $ABCD$.
Calculate the length and the breadth of rectangle $STUV$ if the scale factor is 3.
1. Step 1: Write down the rule to calculate the new length.
2. Step 2: Substitute the given values.
\begin{align} \text{length of new shape} & = k \times \text{length of given shape}\\ & = 3 \times \text{25 mm} \\ \end{align}
3. Step 3: Calculate the new length.
\begin{align} \text{length of new shape} & = k \times \text{length of given shape}\\ & = 3 \times \text{25 mm} \\ & = \text{75 mm}\\ \end{align}
4. Step 4: Write down the rule to calculate the new breadth.
5. Step 5: Substitute the given values.
\begin{align} \text{breadth of new shape} & = k \times \text{breadth of given shape}\\ & = 3 \times \text{12 mm} \\ \end{align}
6. Step 6: Calculate the new breadth.
\begin{align} \text{breadth of new shape} & = k \times \text{breadth of given shape}\\ & = 3 \times \text{12 mm} \\ & = \text{36 mm}\\ \end{align}
If the scale factor is 3, the length of rectangle $STUV$ is 75 mm and the breadth is 36 mm.
### Exercise 9.4: Use the scale factor
1. Squares $X$ and $Y$ are given. Square $Y$ is an enlargement of square $X$.
Find the measurements of square $Y$ if the scale factor is 7.
\begin{align} \text{side of new square} & = k \times \text{side of given square}\\ & = 7 \times \text{2.3 cm} \\ & = \text{16.1 cm}\\ \end{align}
Square $Y$ has sides of 16.1 cm.
2. $\triangle DEF$ is an enlargement of $\triangle CBA$. Both triangles are right-angled triangles and $k = 2$.
Calculate the lengths of $AB$, $DE$, $EF$, $DE$ and $DF$.
You will need to use the theorem of Pythagoras, which states that in a right-angled triangle, $\text{(length of hypotenuse)}^2 = \text{(length of other two sides)}^2$
Use the theorem of Pythagoras to calculate the length of $AB$:
\begin{align} AB^{2} & = BC^{2} + AC^{2} \quad \text{(Pythagoras)} \\ & = 15^{2} + 8^{2} \\ & = 225 + 64 \\ & = 289 \\ \therefore AB & = \sqrt{289} \\ & = \text{17 cm} \\ \end{align}
Use the scale factor to calculate the length of $EF$:
\begin{align} EF & = k \times AB\\ & = 2 \times \text{17 cm} \\ & = \text{34 cm}\\ \end{align}
Use the scale factor to calculate the length of $DE$:
\begin{align} DE & = k \times BC\\ & = 2 \times \text{15 cm} \\ & = \text{30 cm}\\ \end{align}
Use the scale factor to calculate the length of $DF$:
\begin{align} DF & = k \times AC\\ & = 2 \times \text{8 cm} \\ & = \text{16 cm}\\ \end{align}
The measurements for the new $\triangle DEF$ are shown in the diagram below.
3. In the diagram below, parallelogram $GCDH$ is an enlargement of parallelogram $ABEF$.
Calculate the dimensions of parallelogram $GCDH$ if it is given that $k = 9$.
Use scale factor to calculate the length of the longer sides:
\begin{align} HD& = k \times FE\\ & = 9 \times \text{42 mm} \\ & = \text{378 mm}\\ \end{align}
Use scale factor to calculate the length of the shorter sides:
\begin{align} CD & = k \times BE\\ & = 9 \times \text{31 mm} \\ & = \text{279 mm}\\ \end{align}
The dimensions of parallelogram $GCDH$ are length 378 mm and breadth 279 mm.
4. The cube shown below has sides of 8.5 cm.
Calculate the length of one side of the cube if the cube is enlarged by a scale factor of 4.
\begin{align} \text{new side} & = k \times \text{given side}\\ & = 4 \times \text{8.5 cm} \\ & = \text{34 cm}\\ \end{align}
5. A cuboid with the following dimensions are given:
• length = 40 cm
• height = 35 cm
Calculate the dimensions of the new cuboid that is an enlargement of the given cuboid. Use a scale factor of 1.5.
Length:
\begin{align} \text{new length} & = k \times \text{given length}\\ & = 1.5 \times \text{40 cm} \\ & = \text{60 cm}\\ \end{align}
\begin{align} \text{new breadth} & = k \times \text{given breadth}\\ & = 1.5 \times \text{20 cm} \\ & = \text{30 cm}\\ \end{align}
Height:
\begin{align} \text{new height} & = k \times \text{given height}\\ & = 1.5 \times \text{35 cm} \\ & = \text{52.5 cm}\\ \end{align}
The dimensions of the new cuboid are $60 \times 30 \times 2.5$ cm.
6. Polygon $H$ is an enlargement of polygon $G$. Consider the two polygons and answer the questions that follow.
1. What is the scale factor ( $k$)?
There is only one pair of matching sides where both measurements are given, namely 5 units and 10 units.
\begin{align} k & = \frac{\text{length of side shape } H} {\text{length of side shape } G} \\ & = \frac{10}{5}\\ & = 2 \end{align}
1. Write down, using the scale factor, the values of $x$, $y$, $p$, $q$ and $r$.
The measurements in polygon $H$ are double the measurements in polygon $G$. The measurements in polygon $G$ are half the measurements in polygon $H$.
$x$ = $2 \times 11$ = 22 units
$y$ = $2 \times 8$ = 16 units
$p$ = $8 \times \frac{1}{2}$ = 4 units
$q$ = $14 \times \frac{1}{2}$ = 7 units
$r$ = 18 $\times \frac{1}{2}$ = 9 units
### Using scale factors to calculate area and volume
You have already seen that you can use a scale factor to find different lengths in two similar shapes.
You will now investigate what will happen if you calculate the areas and volumes of shapes and objects that are enlarged.
### Worked example 9.3: Working with scale factor and area
Two similar right-angled triangles are given. $\triangle DEF$ is an enlargement of $\triangle BAC$.
The scale factor is 2.
Calculate the areas of the two triangles and compare the answers.
Use the value of $k$ in your comparison.
1. Step 1: Calculate the area of the given triangle ( $\triangle BAC$).
\begin{align} \text{Area } \triangle BAC & = \frac{1}{2} \times b \times h\\ &= \frac{1}{2} \times 16 \times 12\\ &= \text{96 cm}^{2} \\ \end{align}
2. Step 2: Calculate the area of the new triangle ( $\triangle DEF$).
\begin{align} \text{Area } \triangle DEF& = \frac{1}{2} \times b \times h\\ &= \frac{1}{2} \times 32 \times 24\\ &= \text{384 cm}^{2} \\ \end{align}
3. Step 3: Compare the two answers.
Or: $\text{384 cm}^{2}$= $4 \times \text{96 cm}^{2}$
We can write:
\begin{align} \text{Area of new triangle} & = 4 \times \text{area of given triangle} \\ & = 2^2 \times \text{area of given triangle} \\ \end{align}
This worked example shows an important relationship. If a shape is enlarged by a scale factor $k$, then the area of the new shape will be $k^{2}$ times greater than the area of the given shape.
When we calculate area, we use two dimensions to determine the area of a shape (length $\times$ breadth, or side $\times$ side, for example). If the scale factor is $k$, then each one of these two dimensions must be multiplied by $k$ to find the dimensions for the new shape.
$\therefore$ Area of the new shape = $k^{2} \times$ area of given shape
### Worked example 9.4: Working with scale factor and volume
Two cuboids are given. Cuboid $T$ is an enlargement of cuboid $L$. The scale factor is 2.
Calculate the volumes of the two cuboids and compare the answers.
Use the value of $k$ in your comparison.
1. Step 1: Calculate the volume of the given cuboid ( $L$).
\begin{align} \text{Volume of cuboid } L& = l \times b \times h\\ &= 8 \times 2 \times 5\\ &= \text{80 cm}^{3} \\ \end{align}
2. Step 2: Calculate the volume of the new cuboid ( $T$).
\begin{align} \text{Volume of cuboid } T& = l \times b \times h\\ &= 16 \times 4 \times 10\\ &= \text{640 cm}^{3} \\ \end{align}
3. Step 3: Compare the two answers.
We can write:
\begin{align} \text{Volume of new cuboid} & = 8 \times \text{volume of given cuboid} \\ & = 2^{3} \times \text{volume of given cuboid} \\ \end{align}
This worked example shows another important relationship.
If an object is enlarged by a scale factor $k$, then the volume of the new object will be $k^{3}$ times greater than the volume of the given object.
When we calculate volume, we use three dimensions to determine the volume of an object. Therefore, if the scale factor is $k$, each of these three dimensions must be multiplied by $k$ to find the dimensions for the new object.
$\therefore$ Volume of the new object = $k^{3} \times$ volume of given object
We can summarise the effect of an enlargement using a scale factor of $k$ as follows:
• Length of the new line segment = $k \times$ length of given line segment
• Area of the new shape = $k^{2} \times$ area of given shape
• Volume of the new object = $k^{3} \times$ volume of given object.
We can express each of these relationships as a fraction:
We can also express these relationships as ratios:
\begin{align} \text{length of given line segment } &: \text{length of new line segment}\\ = 1 & : k\\ \end{align} \begin{align} \text{area of given shape } & : \text{area of new shape}\\ = 1 & : k^{2} \\ \end{align} \begin{align} \text{volume of given object }& : \text{volume of new object}\\ = 1 & : k^{3} \\ \end{align}
### Exercise 9.5: Work with scale factor, area, and volume
1. Two similar squares, $E$ and $F$ are given. Consider the diagrams and answer the questions that follow.
1. What is the scale factor used to enlarge square $E$?
\begin{align} k & = \frac{\text{side of new square } F} {\text{side of given square } E} \\ & = \frac{20}{4}\\ & =5 \end{align}
The scale factor is 5.
1. Calculate the areas of the two squares and write down a relationship between the two answers.
Area square $E$:
\begin{align} \text{Area}_{E} & = s \times s \\ & = 4 \times 4 \\ & = \text{16 cm}^{2} \\ \end{align}
Area square $F$:
\begin{align} \text{Area}_{F} & = s \times s \\ & = 20 \times 20 \\ & = \text{400 cm}^{2} \\ \end{align}
Relationship between the two areas:
Therefore, we can state that: $\text{400 cm}^{2} = 25 \times \text{16 cm}^{2}$
The relationship is: area square $F$ = $25 \times$ area square $E$
$\therefore$ area square $F$ = $5^{2} \times$ area square $E$
2. Consider the following rectangles. Rectangle $N$ is an enlargement of rectangle $M$.
What is the relationship between the areas of the two rectangles? (Show all your calculations.)
Find the scale factor first:
\begin{align} k & = \frac{\text{longer side of new shape } N} {\text{longer side of given shape } M} \\ & = \frac{180}{60} \\ & =3 \end{align}
Area of given rectangle:
\begin{align} \text{Area}_{\text{M}} & = l \times b \\ & = 60 \times 25 \\ & = \text{1,500 mm}^{2} \\ \end{align}
Area of new rectangle:
\begin{align} \text{Area}_{\text{N}} & = l \times b \\ & = 180 \times 75 \\ & = \text{13,500 mm}^{2} \\ \end{align}
Relationship between the two areas:
Therefore, we can state that: $\text{13,500 mm} = 9 \times \text{1,500 mm}^{2}$
The relationship is: Area rectangle $N$ = $9 \times$ area rectangle $M$
$\therefore$ area rectangle $N$ = $3^{2} \times$ area rectangle $M$
3. In the diagram below, $\triangle PQR$ is an enlargement of $\triangle DFE$.
What is the relationship between the areas of the two triangles? (Show all your calculations.)
Find scale factor first:
\begin{align} k & = \frac{PR}{DE} \\ & = \frac{22}{11} \\ & = 2 \end{align}
Area of given triangle ( $\triangle DFE$):
\begin{align} \text{Area}_{\text{triangle}} & = \frac{1}{2} \times b \times h \\ & = \frac{1}{2} \times 11 \times 5 \\ & = \text{27.5 cm}^{2} \\ \end{align}
Area of new triangle ( $\triangle PQR$):
\begin{align} \text{Area}_{\text{triangle}} & = \frac{1}{2} \times b \times h \\ & = \frac{1}{2} \times 22 \times 10 \\ & = \text{110 cm}^{2} \\ \end{align}
Relationship between the two areas:
\begin{align} \frac{\text{110 cm}^{2}} {\text{27.5 cm}^{2}} & = \frac{110 \times 10}{27.5 \times 10} \\ & = \frac{1,100}{275} \\ & = 4\\ \end{align}
Therefore, we can state that: $\text{110 cm}^{2} = 4 \times \text{27.5 cm}^{2}$
The relationship is: Area $\triangle PQR$ = $4 \times$ area $\triangle DFE$
$\therefore$ Area $\triangle PQR$ = $2^{2} \times$ area $\triangle DFE$
4. Objects $Q$ and $R$ are two cuboids. Cuboid $R$ is an enlargement of cuboid $Q$.
What is the relationship between the volumes of the two cuboids? Write the answer as a ratio in the form "Volume cuboid $Q$ : Volume cuboid $R$".
Find the scale factor first:
\begin{align} k & = \frac{\text{longest side cuboid } R}{\text{longest side cuboid } Q} \\ & = \frac{27}{9}\\ & = 3 \end{align}
Volume of cuboid $Q$:
\begin{align} \text{Volume of cuboid } Q& = l \times b \times h\\ &= 4 \times 4 \times 9\\ &= \text{144 cm}^{3} \\ \end{align}
Volume of cuboid $R$:
\begin{align} \text{Volume of cuboid } R& = l \times b \times h\\ &= 12 \times 12 \times 27\\ &= \text{3,888 cm}^{3} \\ \end{align}
Ratio:
\begin{align} \text{Volume cuboid } Q & : \text{Volume cuboid }R\\ = 144& : 3,888\\ = 1& : 27\\ = 1 & : 3^{3}\\ \end{align}
5. Objects $J$ and $K$ shown below are cubes. Cube $K$ is an enlargement of cube $J$, where $k = 4$.
1. Calculate the length of one side of cube $K$.
\begin{align} \text{Length of side cube } K & = k \times \text{length of side cube } J\\ &= 4 \times 2\\ &= \text{8 m}\\ \end{align}
1. Calculate the area of one face of cube $K$.
The face of a cube is a square. $\text{Area}_{\text{square}} = s \times s$
\begin{align} \text{Area of face cube } K& = k^{2} \times \text{area of face cube } J\\ &= 4^{2} \times s \times s\\ &= 16 \times 2 \times 2\\ &= \text{64 m}^{2} \\ \end{align}
1. Calculate the volume of cube $K$.
\begin{align} \text{Volume of cube }K & = k^{3} \times \text{volume of cube } J\\ &= 4^{3} \times s \times s \times s\\ &= 64 \times 2 \times 2 \times 2 \\ &= \text{512 m}^{2} \\ \end{align}
## 9.3 Practical applications
Using similar shapes and scale factors can be very helpful in solving problems in our daily lives. In this section you will have the opportunity to apply all the knowledge that you have gained in this chapter.
### Worked example 9.5: Using the scale factor to solve a problem
Oladapo wants to find out the height of a lamp post. Oladapo is 1.5 m tall. The length of Oladapo's shadow changes throughout the day.
Oladapo stands in the shadow of the lamp post, so that his shadow and the lamp post's shadow are in a straight line. The shadows of the lamp post and of Oladapo form triangles, as shown in the diagram below.
Oladapo's shadow is 2 m long and the the shadow of the lamp post is 6 m long.
Draw a diagram showing the triangle formed by Oladapo and his shadow, and a separate diagram showing the lamp post and its shadow. Use your diagrams and find the scale factor of the two trianges to help Oladapo work out the height of the lamp post.
1. Step 1: Draw two separate triangles and fill in all the given information.
2. Step 2: Find the scale factor.
$\triangle BAC$ is an enlargement of $\triangle EFC$.
\begin{align} k & = \frac {CA}{CF} \\ & = \frac{6}{2} \\ &= 3\\ \end{align}
3. Step 3: Calculate the height of the lamp post.
The lamp post is represented by $BA$ in the diagram.
We have Oladapo's height, which is EF in the diagram, and we now have the scale factor, so we can calculate the height of the lamp post.
\begin{align} BA& = k \times EF\\ &= 3 \times \text{1.5 m}\\ &= \text{4.5 m}\\ \end{align}
4. Step 4: Give your answer using a full sentence, and include the correct unit of measurement.
The height of the lamp post is 4.5 m.
In real-life questions, as in the example above, we assume that vertical objects like lamp posts and human beings are perpendicular to the ground. We also assume that the ground is perfectly horizontal.
### Exercise 9.6: Use similar shapes and the scale factor to solve problems
1. Ndidi wants to find out the height of a tree. She is 1.6 m tall. The length of Ndidi's shadow changes throughout the day.
At a certain time, Ndidi's shadow is 3 m long and the shadow of the tree is 6 m long.
Help Ndidi to work out the height of the tree.
Redraw the diagram with the two triangles separate from each other.
The two triangles with all the given information are shown below.
$\triangle BAC$ is an enlargement of $\triangle EFC$.
\begin{align} k & = \frac {AC}{FC} \\ & = \frac{6}{3} \\ &= 2\\ \end{align}
The tree is represented by $BA$ on the diagram.
\begin{align} BA& = k \times EF\\ &= 2 \times \text{1.6 m}\\ &= \text{3.2 m}\\ \end{align}
The tree is 3.2 m high.
2. Complete the table below.
$k$ length of given side length of new side
1.5 30 mm
7 371 cm
28.75 m 287.5 m
$k$ length of given side length of new side
1.5 30 mm 1.5 $\times$ 30 = 45 mm
7 53 cm 7 $\times$ 53 = 371 cm
10 28.75 m 10 $\times$ 28.75 = 287.5 m
3. $PQRS$ is a trapezium with measurements (in cm) as shown in the diagram below.
Calculate the perimeter of the enlargement of $PQRS$ if a scale factor of 2.25 is used.
\begin{align} \text{Perimeter}_{\text{PQRS}}& = \text{sum of sides}\\ & = 9 + 5 + 4 + 6\\ &= \text{24 cm}\\ \end{align} \begin{align} \text{Perimeter}_{\text{new shape}}& = k \times \text{perimeter of given shape} \\ & = 2.25 \times \text{24 cm}\\ &= \text{54 cm}\\ \end{align}
OR
Calculate the length of each new side first.
If you do not have a diagram of the new shape with new labels for the sides, you can use the prime symbol ( $'$) to show that you are working with the sides of the new shape. So $P'Q'$ is the side of the new shape that corresponds with $PQ$ in the given shape.
\begin{align} \text{Perimeter of new shape} & = 20.25 + 11.25 + 9 + 13.5\\ &= \text{54 cm} \\ \end{align}
4. Complete the table below.
$k$ area of given shape area of new shape
4 $\text{5 mm}^{2}$
6 $\text{1,080 cm}^{2}$
$\text{2 m}^{2}$ $\text{162 m}^{2}$
$k$ area of given shape area of new shape
4 $\text{5 mm}^{2}$ $4^{2} \times 5$ = $\text{80 mm}^{2}$
6 $\text{30 cm}^{2}$ $6^{2} \times 30$ = $\text{1,080 cm}^{2}$
9 $\text{2 m}^{2}$ $9^{2} \times 2$ = $\text{162 m}^{2}$
5. Faruq is designing a pattern to decorate a wall outside a shop. The pattern consists of two equal squares and a rectangle in the middle.
The area of one of the squares is $\text{9 cm}^{2}$ and the area of the rectangle is $\text{21 cm}^{2}$.
Calculate the total area of the three shapes if the design is enlarged by a scale factor of 5.
Calculate total area of given shapes:
\begin{align} \text{Total area} & = 9 + 21 + 9 \\ &= \text{39 cm}^{2} \\ \end{align}
Calculate area of new shapes:
\begin{align} \text{Area of new shapes} &= k^{2} \times \text{area of given shapes}\\ &= 5^{2} \times 39 \\ &= 25 \times 39 \\ &= \text{975 cm}^{2} \\ \end{align}
6. Complete the table below.
$k$ volume of given object volume of new object
2 $\text{20 mm}^{3}$
3 $3^{3} \times 4$ = $\text{108 cm}^{3}$
$\text{3 m}^{3}$ $5^{3} \times 3$ = $\text{375 m}^{3}$
$k$ volume of given object volume of new object
2 $\text{20 mm}^{3}$ $2^{3} \times 20$ = $\text{160 mm}^{3}$
3 $\text{4 cm}^{3}$ $3^{3} \times 4$ = $\text{108 cm}^{3}$
5 $\text{3 m}^{3}$ $5^{3} \times 3$ = $\text{375 m}^{3}$
7. Chike is working with the diagram of a cuboid that is given below.
Unfortunately the diagram does not show the height of the cuboid. The only information that Chike has is this:
If the cuboid is enlarged by a factor of 2, the volume of the cuboid will be $3,360 \text{ cm}^{3}$.
Help Chike to calculate the height of the cuboid in the diagram.
Chike can use the relationship of the two volumes: Volume of new cuboid = $k^{3} \times$ volume of given cuboid
This is an equation. First, Chike must calculate the volume of the given cuboid, and Chike can solve this equation to do that.
\begin{align} k^{3} \times \text{volume of given cuboid} & = \text{volume of new cuboid}\\ 2^{3} \times \text{volume of given cuboid} & = 3,360 \\ 8 \times \text{volume of given cuboid} &= 3,360\\ \therefore \text{volume of given cuboid} & = 3,360 \div 8 \\ & = \text{420 cm}^{3} \\ \end{align}
Now that Chike has the volume of the given cuboid, $\text{420 cm}^{3}$, he can solve the equation for volume to find the height, which is the unknown.
\begin{align} l \times b \times h &= \text{volume}\\ 7 \times 6 \times h &= 420\\ 42 \times h &= 420\\ h &= 420 \div 42\\ \therefore h &= \text{10 cm} \\ \end{align}
The height of the cuboid in the diagram is 10 cm.
8. The cube shown below has a volume of $\text{27 cm}^{3}$.
Calculate the dimensions of an enlarged cube that has a volume of $1,728\text{ cm}^{3}$.
Find the value of $k$ first:
\begin{align} k^{3} & = \frac{\text{volume of new cube}}{\text{volume of given cube}} \\ &= \frac {1,728}{27} \\ &= 64\\ \end{align}
From $k^{3}$ you need to find the value of $k$. You can find the prime factors of 64:
There are 6 factors here, and you need to have 3 factors:
$4 \times 4 \times 4 = 64$, so $k = 4$
The given cube was enlarged by a scale factor of 4.
It is given that the volume of the given cube = $\text{27 cm}^{3}$.
The formula for the volume of a cube is $V = s \times s \times s = s^3$.
So you also need to have 3 factors of 27. The prime factors of 27 are $3 \times 3 \times 3 = 27$.
Therefore, each side of the given cube = $\text{3 cm}$.
\begin{align} \text{Side of new cube} & = k \times \text{side of given cube} \\ &= 4 \times 3 \\ &= \text{12 cm} \\ \end{align}
The dimensions of the enlarged cube are $\text{12} \times \text{12} \times \text{12 cm}$.
## 9.4 Summary
• In Mathematics, two shapes are similar if:
• their matching sides are in proportion, and
• their matching angles are equal.
• The sides of two shapes are in proportion if all of the sides of the given shape have been multiplied by the same number to get the sides of the new shape. This number is called the scale factor.
• We can use the variable $k$ to represent the scale factor.
• An enlargement of a diagram, shape or object is a copy of the original in which everything is made larger, keeping the same proportions.
• We can summarise the effect of an enlargement using a scale factor of $k$ as follows:
• Length of the new line segment = $k \times$ length of given line segment
• Area of the new shape = $k^{2} \times$ area of given shape
• Volume of the new object = $k^{3} \times$ volume of given object
• We can express each of these relationships as a fraction:
• We can also express these relationships as ratios:
\begin{align} \text{length of given line segment } &: \text{length of new line segment}\\ = 1 &: k\\ \end{align} \begin{align} \text{area of given shape }& : \text{area of new shape}\\ = 1 & : k^{2} \\ \end{align} \begin{align} \text{volume of given object }& : \text{volume of new object}\\ = 1 & : k^{3} \\ \end{align} | 10,560 | 35,307 | {"found_math": true, "script_math_tex": 258, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 82, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2021-49 | longest | en | 0.95942 |
https://brainmass.com/statistics/probability/satisfactory-unsatisfactory-production-probability-444311 | 1,716,803,179,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059039.52/warc/CC-MAIN-20240527083011-20240527113011-00840.warc.gz | 113,627,975 | 7,210 | Purchase Solution
# Satisfactory and Unsatisfactory Production Probability
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At a factory that produces pistons for cars, Machine 1 produced 414 satisfactory pistons and 276 unsatisfactory pistons today. Machine 2 produced 420 satisfactory pistons and 180 unsatisfactory pistons today. Suppose that one piston from Machine 1 and one piston from Machine 2 are chosen at random from today's batch. What is the probability that the piston chosen from Machine 1 is satisfactory and the piston chosen from Machine 2 is unsatisfactory ?
##### Solution Summary
The expert examines satisfactory and unsatisfactory production probability.
##### Solution Preview
Probability of the piston chosen from Machine 1 is satisfactory = 414/(414+276) = ...
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##### Terms and Definitions for Statistics
This quiz covers basic terms and definitions of statistics.
##### Measures of Central Tendency
This quiz evaluates the students understanding of the measures of central tendency seen in statistics. This quiz is specifically designed to incorporate the measures of central tendency as they relate to psychological research. | 332 | 1,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.914506 |
https://questions.tripos.org/part-ib/2007/1/ | 1,718,690,493,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.46/warc/CC-MAIN-20240618042809-20240618072809-00161.warc.gz | 431,799,601 | 20,472 | • # 1.II.11H
Define what it means for a function $f: \mathbb{R}^{a} \rightarrow \mathbb{R}^{b}$ to be differentiable at a point $p \in \mathbb{R}^{a}$ with derivative a linear map $\left.D f\right|_{p} .$
State the Chain Rule for differentiable maps $f: \mathbb{R}^{a} \rightarrow \mathbb{R}^{b}$ and $g: \mathbb{R}^{b} \rightarrow \mathbb{R}^{c}$. Prove the Chain Rule.
Let $\|x\|$ denote the standard Euclidean norm of $x \in \mathbb{R}^{a}$. Find the partial derivatives $\frac{\partial f}{\partial x_{i}}$ of the function $f(x)=\|x\|$ where they exist.
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• # 1.I.3F
For the function
$f(z)=\frac{2 z}{z^{2}+1},$
determine the Taylor series of $f$ around the point $z_{0}=1$, and give the largest $r$ for which this series converges in the disc $|z-1|.
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• # 1.II.13F
By integrating round the contour $C_{R}$, which is the boundary of the domain
$D_{R}=\left\{z=r e^{i \theta}: 0
evaluate each of the integrals
$\int_{0}^{\infty} \sin x^{2} d x, \quad \int_{0}^{\infty} \cos x^{2} d x$
[You may use the relations $\int_{0}^{\infty} e^{-r^{2}} d r=\frac{\sqrt{\pi}}{2}$ and $\sin t \geq \frac{2}{\pi} t$ for $\left.0 \leq t \leq \frac{\pi}{2} \cdot\right]$
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• # 1.II.16E
A steady magnetic field $\mathbf{B}(\mathbf{x})$ is generated by a current distribution $\mathbf{j}(\mathbf{x})$ that vanishes outside a bounded region $V$. Use the divergence theorem to show that
$\int_{V} \mathbf{j} d V=0 \quad \text { and } \quad \int_{V} x_{i} j_{k} d V=-\int_{V} x_{k} j_{i} d V$
Define the magnetic vector potential $\mathbf{A}(\mathbf{x})$. Use Maxwell's equations to obtain a differential equation for $\mathbf{A}(\mathbf{x})$ in terms of $\mathbf{j}(\mathbf{x})$.
It may be shown that for an unbounded domain the equation for $\mathbf{A}(\mathbf{x})$ has solution
$\mathbf{A}(\mathbf{x})=\frac{\mu_{0}}{4 \pi} \int_{V} \frac{\mathbf{j}\left(\mathbf{x}^{\prime}\right)}{\left|\mathbf{x}-\mathbf{x}^{\prime}\right|} d V^{\prime}$
Deduce that in general the leading order approximation for $\mathbf{A}(\mathbf{x})$ as $|\mathbf{x}| \rightarrow \infty$ is
$\mathbf{A}=\frac{\mu_{0}}{4 \pi} \frac{\mathbf{m} \times \mathbf{x}}{|\mathbf{x}|^{3}} \quad \text { where } \quad \mathbf{m}=\frac{1}{2} \int_{V} \mathbf{x}^{\prime} \times \mathbf{j}\left(\mathbf{x}^{\prime}\right) d V^{\prime}$
Find the corresponding far-field expression for $\mathbf{B}(\mathbf{x})$.
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• # 1.I.5D
A steady two-dimensional velocity field is given by
$\mathbf{u}(x, y)=(\alpha x-\beta y, \beta x-\alpha y), \quad \alpha>0, \quad \beta>0$
(i) Calculate the vorticity of the flow.
(ii) Verify that $\mathbf{u}$ is a possible flow field for an incompressible fluid, and calculate the stream function.
(iii) Show that the streamlines are bounded if and only if $\alpha<\beta$.
(iv) What are the streamlines in the case $\alpha=\beta ?$
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• # 1.II.17D
Write down the Euler equation for the steady motion of an inviscid, incompressible fluid in a constant gravitational field. From this equation, derive (a) Bernoulli's equation and (b) the integral form of the momentum equation for a fixed control volume $V$ with surface $S$.
(i) A circular jet of water is projected vertically upwards with speed $U_{0}$ from a nozzle of cross-sectional area $A_{0}$ at height $z=0$. Calculate how the speed $U$ and crosssectional area $A$ of the jet vary with $z$, for $z \ll U_{0}^{2} / 2 g$.
(ii) A circular jet of speed $U$ and cross-sectional area $A$ impinges axisymmetrically on the vertex of a cone of semi-angle $\alpha$, spreading out to form an almost parallel-sided sheet on the surface. Choose a suitable control volume and, neglecting gravity, show that the force exerted by the jet on the cone is
$\rho A U^{2}(1-\cos \alpha)$
(iii) A cone of mass $M$ is supported, axisymmetrically and vertex down, by the jet of part (i), with its vertex at height $z=h$, where $h \ll U_{0}^{2} / 2 g$. Assuming that the result of part (ii) still holds, show that $h$ is given by
$\rho A_{0} U_{0}^{2}\left(1-\frac{2 g h}{U_{0}^{2}}\right)^{\frac{1}{2}}(1-\cos \alpha)=M g$
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• # 1.I.2A
State the Gauss-Bonnet theorem for spherical triangles, and deduce from it that for each convex polyhedron with $F$ faces, $E$ edges, and $V$ vertices, $F-E+V=2$.
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• # 1.II.10G
(i) State a structure theorem for finitely generated abelian groups.
(ii) If $K$ is a field and $f$ a polynomial of degree $n$ in one variable over $K$, what is the maximal number of zeroes of $f$ ? Justify your answer in terms of unique factorization in some polynomial ring, or otherwise.
(iii) Show that any finite subgroup of the multiplicative group of non-zero elements of a field is cyclic. Is this true if the subgroup is allowed to be infinite?
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• # 1.I.1G
Suppose that $\left\{e_{1}, \ldots, e_{3}\right\}$ is a basis of the complex vector space $\mathbb{C}^{3}$ and that $A: \mathbb{C}^{3} \rightarrow \mathbb{C}^{3}$ is the linear operator defined by $A\left(e_{1}\right)=e_{2}, A\left(e_{2}\right)=e_{3}$, and $A\left(e_{3}\right)=e_{1}$.
By considering the action of $A$ on column vectors of the form $\left(1, \xi, \xi^{2}\right)^{T}$, where $\xi^{3}=1$, or otherwise, find the diagonalization of $A$ and its characteristic polynomial.
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• # 1.II.9G
State and prove Sylvester's law of inertia for a real quadratic form.
[You may assume that for each real symmetric matrix A there is an orthogonal matrix $U$, such that $U^{-1} A U$ is diagonal.]
Suppose that $V$ is a real vector space of even dimension $2 m$, that $Q$ is a non-singular quadratic form on $V$ and that $U$ is an $m$-dimensional subspace of $V$ on which $Q$ vanishes. What is the signature of $Q ?$
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• # 1.II.19C
Consider a Markov chain $\left(X_{n}\right)_{n \geqslant 0}$ on states $\{0,1, \ldots, r\}$ with transition matrix $\left(P_{i j}\right)$, where $P_{0,0}=1=P_{r, r}$, so that 0 and $r$ are absorbing states. Let
$A=\left(X_{n}=0, \text { for some } n \geqslant 0\right) \text {, }$
be the event that the chain is absorbed in 0 . Assume that $h_{i}=\mathbb{P}\left(A \mid X_{0}=i\right)>0$ for $1 \leqslant i.
Show carefully that, conditional on the event $A,\left(X_{n}\right)_{n \geqslant 0}$ is a Markov chain and determine its transition matrix.
Now consider the case where $P_{i, i+1}=\frac{1}{2}=P_{i, i-1}$, for $1 \leqslant i. Suppose that $X_{0}=i, 1 \leqslant i, and that the event $A$ occurs; calculate the expected number of transitions until the chain is first in the state 0 .
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• # 1.II.14D
Define the Fourier transform $\tilde{f}(k)$ of a function $f(x)$ that tends to zero as $|x| \rightarrow \infty$, and state the inversion theorem. State and prove the convolution theorem.
Calculate the Fourier transforms of
Hence show that
$\int_{-\infty}^{\infty} \frac{\sin (b k) e^{i k x}}{k\left(a^{2}+k^{2}\right)} d k=\frac{\pi \sinh (a b)}{a^{2}} e^{-a x} \quad \text { for } \quad x>b$
and evaluate this integral for all other (real) values of $x$.
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• # 1.II.12A
Let $X$ and $Y$ be topological spaces. Define the product topology on $X \times Y$ and show that if $X$ and $Y$ are Hausdorff then so is $X \times Y$.
Show that the following statements are equivalent.
(i) $X$ is a Hausdorff space.
(ii) The diagonal $\Delta=\{(x, x): x \in X\}$ is a closed subset of $X \times X$, in the product topology.
(iii) For any topological space $Y$ and any continuous maps $f, g: Y \rightarrow X$, the set $\{y \in Y: f(y)=g(y)\}$ is closed in $Y$.
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• # 1.I.6F
Solve the least squares problem
$\left[\begin{array}{ll} 1 & 3 \\ 0 & 2 \\ 0 & 2 \\ 0 & 1 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right]=\left[\begin{array}{r} 4 \\ 1 \\ 4 \\ -1 \end{array}\right]$
using $Q R$ method with Householder transformation. (A solution using normal equations is not acceptable.)
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• # 1.I.8C
State and prove the max-flow min-cut theorem for network flows.
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• # 1.II.15B
The relative motion of a neutron and proton is described by the Schrödinger equation for a single particle of mass $m$ under the influence of the central potential
$V(r)=\left\{\begin{array}{rc} -U & ra \end{array}\right.$
where $U$ and $a$ are positive constants. Solve this equation for a spherically symmetric state of the deuteron, which is a bound state of a proton and neutron, giving the condition on $U$ for this state to exist.
[If $\psi$ is spherically symmetric then $\left.\nabla^{2} \psi=\frac{1}{r} \frac{d^{2}}{d r^{2}}(r \psi) .\right]$
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• # 1.I.4B
Write down the position four-vector. Suppose this represents the position of a particle with rest mass $M$ and velocity v. Show that the four momentum of the particle is
$p_{a}=(M \gamma c, M \gamma \mathbf{v})$
where $\gamma=\left(1-|\mathbf{v}|^{2} / c^{2}\right)^{-1 / 2}$.
For a particle of zero rest mass show that
$p_{a}=(|\mathbf{p}|, \mathbf{p})$
where $\mathbf{p}$ is the three momentum.
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• # 1.I.7C
Let $X_{1}, \ldots, X_{n}$ be independent, identically distributed random variables from the $N\left(\mu, \sigma^{2}\right)$ distribution where $\mu$ and $\sigma^{2}$ are unknown. Use the generalized likelihood-ratio test to derive the form of a test of the hypothesis $H_{0}: \mu=\mu_{0}$ against $H_{1}: \mu \neq \mu_{0}$.
Explain carefully how the test should be implemented.
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• # 1.II.18C
Let $X_{1}, \ldots, X_{n}$ be independent, identically distributed random variables with
$\mathbb{P}\left(X_{i}=1\right)=\theta=1-\mathbb{P}\left(X_{i}=0\right)$
where $\theta$ is an unknown parameter, $0<\theta<1$, and $n \geqslant 2$. It is desired to estimate the quantity $\phi=\theta(1-\theta)=n \operatorname{Var}\left(\left(X_{1}+\cdots+X_{n}\right) / n\right)$.
(i) Find the maximum-likelihood estimate, $\hat{\phi}$, of $\phi$.
(ii) Show that $\hat{\phi}_{1}=X_{1}\left(1-X_{2}\right)$ is an unbiased estimate of $\phi$ and hence, or otherwise, obtain an unbiased estimate of $\phi$ which has smaller variance than $\hat{\phi}_{1}$ and which is a function of $\hat{\phi}$.
(iii) Now suppose that a Bayesian approach is adopted and that the prior distribution for $\theta, \pi(\theta)$, is taken to be the uniform distribution on $(0,1)$. Compute the Bayes point estimate of $\phi$ when the loss function is $L(\phi, a)=(\phi-a)^{2}$.
[You may use that fact that when $r, s$ are non-negative integers,
$\left.\int_{0}^{1} x^{r}(1-x)^{s} d x=r ! s ! /(r+s+1) !\right]$
comment | 3,444 | 10,475 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 154, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-26 | latest | en | 0.590592 |
http://vinaytechs.blogspot.com/2009/11/bits-and-bytes_3.html | 1,582,929,969,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147647.2/warc/CC-MAIN-20200228200903-20200228230903-00331.warc.gz | 153,940,028 | 61,803 | ## Bits and Bytes
Some Basic facts and New Standards of Bits and Bytes
The basic unit used in computer data storage is called a bit (binary digit). Computers use these little bits, which are composed of ones and zeros, to do things and talk to other computers. All your files, for instance, are kept in the computer as binary files and translated into words and pictures by the software (which is also ones and zeros). This two number system, is called a "binary number system" since it has only two numbers in it. The decimal number system in contrast has ten unique digits, zero through nine.
But although computer data and file size is normally measured in binary code using the binary number system (counted by factors of two 1, 2, 4, 8, 16, 32, 64, etc), the prefixes for the multiples are based on the metric system! The nearest binary number to 1,000 is 2^10 or 1,024; thus 1,024 bytes was named a Kilobyte. So, although a metric "kilo" equals 1,000 (e.g. one kilogram = 1,000 grams), a binary "Kilo" equals 1,024 (e.g. one Kilobyte = 1,024 bytes). Not surprisingly, this has led to a great deal of confusion.
In December 1998, the International Electro technical Commission (IEC) approved a new IEC International Standard. Instead of using the metric prefixes for multiples in binary code, the new IEC standard invented specific prefixes for binary multiples made up of only the first two letters of the metric prefixes and adding the first two letters of the word "binary". Thus, for instance, instead of Kilobyte (KB) or Gigabyte (GB), the new terms would be kibibyte (KiB) or gibibyte (GiB). The new IEC International Standards, which are not commonly used yet, are included below.
Here's a few more details to consider:
• Although data storage capacity is generally expressed in binary code, many hard drive manufacturers (and some newer BIOSs) use a decimal system to express capacity.
For example, a 30 gigabyte drive is usually 30,000,000,000 bytes (decimal) not the 32,212,254,720 binary bytes you would expect.
• Another trivial point is that in the metric system the "k" or "kilo" prefix is always lowercase (i.e. kilogram = kg not Kg) but since these binary uses for data storage capacity are not properly metric, it has become standard to use an uppercase "K" for the binary form.
• When used to describe Data Transfer Rate, bits/bytes are calculated as in the metric system
Kilobits per second is usually shortened to kbps or Kbps. Although technically speaking, the term kilobit should have a lowercase initial letter, it has become common to capitalize it in abbreviation (e.g. "56 Kbps" or "56K"). The simple "K" might seem ambiguous but, in the context of data transfer, it can be assumed that the measurement is in bits rather than bytes unless indicated otherwise. | 675 | 2,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.915251 |
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Lecture06 - Electric Field of Uniformly Charged Thin Rod At...
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1 Electric Field of Uniformly Charged Thin Rod At distance r from midpoint along a line perpendicular to the rod: E y = 0 = 1 4 πε 0 Q r r 2 + L / 2 ( ) 2 ˆ r And for very long rod: E = 1 4 0 2 Q / L ( ) r ˆ r At distance r from an arbitrary location not at the midpoint : Δ E x = 1 4 o x Δ Q x 2 + ( y 0 y ) 2 3/2 Δ E y = 1 4 o y o y ( ) Δ Q x 2 + ( y 0 y ) 2 3/2 Δ E z = 0 General Procedure for Calculating Electric Field of Distributed Charges 1. Cut the charge distribution into pieces for which the field is known 2. Write an expression for the electric field due to one piece (i) Choose origin (ii) Write an expression for Δ E and its components 3. Add up the contributions of all the pieces (i) Try to integrate symbolically (ii) If impossible – integrate numerically 4. Check the results: (i) Direction (ii) Units (iii) Special cases Lecture 6 Chapter 16. Electric Fields * Uniformly Charged Thin Ring * Uniformly Charged Disk * Two Uniformly Charged Disks: A Capacitor Origin: center of the ring Location of piece: described by θ , where = 0 is along the x axis. Step 1: Cut up the charge distribution into small pieces. (i.e., understand geometry and choose Δ Q. ) A Uniformly Charged Thin Ring
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Ask a homework question - tutors are online | 506 | 1,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-09 | latest | en | 0.811146 |
https://www.wzhyouhua.com/how-is-market-price-of-stock-determined/ | 1,670,624,379,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711552.8/warc/CC-MAIN-20221209213503-20221210003503-00073.warc.gz | 1,164,247,699 | 20,993 | # how is market price of stock determined
Supply and demand
## What is the formula to calculate price per share?
You’ll need to follow these steps:Calculate the book value of the company.Count up all of the company’s outstanding shares.Divide the company’s book value by the total number of shares.
## How do interest rates impact the stock market?
Interest rates are set by the Federal Reserve and can have an impact on lending,consumer goods,and the housing market.The stock market generally has an inverse relationship with interest rates,but not every sector of the market reacts the same.Changes in interest rates can cause volatility in the stock market and impact costs for businesses.More items…
## How is market price per share calculated?
Book value per share. Take the stockholder’s equity,the value of company assets less company debts. …Dividend yield is the ratio of dividends to stock price. Divide the annual dividends issued per share by the share price to get dividend yield. …Earnings per share. …Price/earnings ratio. …Market value per share. …
## How to evaluate the stock market?
The beta of a stock is calculated by using regression analysis on returns data for the stock and representative index.A risk-free asset such as cash and treasury bills have zero beta.A negative beta occurs when an asset’s return is negatively correlated with that of the market.
## How Is Share Price Determined?
When a stock is sold, a buyer and seller exchange money for share ownership. The price for which the stock is purchased becomes the new market price. When a second share is sold, this price becomes the newest market price, etc.
## How to find a company’s market cap?
A company’s worth—or its total market value —is called its market capitalization, or "market cap." A company’s market cap can be determined by multiplying the company’s stock price by the number of shares outstanding.
## How is the market cap determined?
A company’s market cap can be determined by multiplying the company’s stock price by the number of shares outstanding. The stock price is a relative and proportional value of a company’s worth.
## How to calculate market capitalization?
In simple terms, a company’s market capitalization is calculated by multiplying its share price by the number of shares outstanding :
## Why is market capitalization inadequate?
Market capitalization is an inadequate way to value a company because the basis of it market price does not necessarily reflect how much a piece of the business is worth.
## What is stock price?
The stock price is a relative and proportional value of a company’s worth. Therefore, it only represents a percentage change in a company’s market cap at any given point in time.
## What is market cap?
While market cap is often used synonymously with a company’s market value, it is important to keep in mind that market cap refers only to the market value of a company’s equity , not its market value overall (which can include the value of its debt or assets).
## What Is Market Price?
The market price is the current price at which an asset or service can be bought or sold. The market price of an asset or service is determined by the forces of supply and demand. The price at which quantity supplied equals quantity demanded is the market price.
## What happens if you drop your bid to \$50.25?
If the buyers no longer think that is a good price, they may drop their bid to \$50.25. The sellers may agree or they may not. Someone may drop their offer to a lower price, or it may stay where it is. A trade only occurs if a seller interacts with the bid price, or a buyer interacts with the offer price.
## Why do bids and offers change?
Bids and offers are constantly changing as the buyers and sellers change their minds about which price to buy or sell at. Also, as sellers sell to the bids, the price will drop, or as buyers buy from the offer, the price will rise.
## What is the difference between a bid and a bid?
In order for a trade to occur, there must be a buyer and a seller that meet at the same price. Bids are represented by buyers , and offers are represented by sellers. The bid is the higher price someone is advertising they will buy at, while the offer is the lowest price someone is advertising they will sell at.
## What is the price at which quantity supplied equals quantity demanded?
The price at which quantity supplied equals quantity demanded is the market price. The market price is used to calculate consumer and economic surplus. Consumer surplus refers to the difference between the highest price a consumer is willing to pay for a good and the actual price they do pay for the good, or the market price .
## What is economic surplus?
Economic surplus refers to two related quantities: consumer surplus and producer surplus. Producer surplus may also be referred to as profit: it is the amount that producers benefit by selling at the market price (provided that the market price is higher than the least that they would be willing to sell for).
## Why is the spread \$30 by \$30.03?
Now the spread widens, and the price is \$30 by \$30.03 because all the share offered at \$30.01 and \$30.02 have been bought. Since \$30.02 was the last traded price, this is the market price. Other traders may take action to close the spread. Since there are more buyers, the spread is closed by the bid adjusting upward.
## Understanding capital markets
To understand how share price is determined, it’s helpful to step back and consider what it means to buy a stock.
## What determines stock price?
To put it simply, the price of a stock is determined by supply and demand. If more people want the stock than the number of shares available, the price goes up. Conversely, when lots of people are looking to sell their shares, the price of the stock falls. If an investor sells when the stock is higher than the price they paid, they make a profit.
## What factors can affect stock price?
News and events happening at the company specifically, as well as the country or the market at large, can affect stock prices.
## The bottom line
At the most basic level, the factor that determines stocks’ prices is supply and demand. Buyers and sellers trading via the market set the price. However, there are complex considerations of both the company’s performance and broader market forces that can affect that supply and demand.
## What should all investors be concerned about?
Something that all investors should be concerned about is inflation. It’s basically the bogeyman . As inflation increases, the purchasing power of each dollar will decline, and this means that investors will have to pay more for their shares.
## Why do day traders use big events?
Day traders use big events to determine whether a stock can be bought and sold for a good price, but there are also trends and patterns that help determine good entry points in the short term. This is great for those who day traders because it allows them to profit on the upswings and downswings of a company.
## What is primary market?
The primary market is the place where stocks are originally created and sold. When a company does an initial public offering (IPO), its shares become available for the first time and can be purchased through some top stock brokerages. IPOs happen all the time; some of them can be lucrative if the price is right and you believe in the company.
## Why are earnings calls important?
Earning calls are an important time for investors to take advantage of fluctuations in price. Typically, there will be a lot of traders trying to scalp a highly anticipated earnings call—this is when supply and demand are in full effect.
## How do acquisitions affect stock prices?
Acquisitions can impact stock price because corporations have to pay a premium to acquire other companies. This is because acquisitions typically need to be approved by shareholders. Shareholders won’t be happy if they are losing their investment under the current market price.
## What is private offering?
Private offerings are the private sales of shares in a non-public company to individual investors. These kinds of offerings will occur before a company goes public. Private offerings can be lucrative because it allows investors to buy a company when prices are low, and then they have the potential to explode in value once they go public.
## Why do companies trend downwards during recession?
During recessions, investors often have poor outlooks of the market. This can lead to companies trending downwards for no particular reason other than mass-pessimism. The government might take steps to prop up the market, such as the fed printing money to buy corporate junk bonds.
## How do stock prices work?
It starts with the initial public offering (IPO). Companies work with investment bankers to set a primary market price when a company goes public. That price is set based on valuation and demand from institutional investors.
## How to calculate P/E?
The price-earnings ratio (P/E) shows the price of the stock relative to earnings. It’s calculated by dividing the stock price by earnings per share. Earnings per share is a readily available number on most financial websites and the company’s quarterly reporting documents.
## What happens when there are more buyers than sellers?
If there are more buyers than sellers, the price will get bid up. If there are more sellers than buyers, the opposite will happen.
## What is market cap?
The market cap of a stock is equal to the total shares times the share price. It’s the price it would take to buy all of a company’s outstanding shares. Many stocks issue more shares to fund the business, so it is important to base valuation on the market cap and not just the stock price. The more shares that are issued, the less of a fraction of the business you own.
## What happens when a company buys back shares?
On the other side, if a business buys back shares, the price of each one of your shares will need to go up to maintain the same market cap. Share buybacks are generally cheered by shareholders as long as the stock price isn’t overvalued.
## How are stock prices determined?
Once a company goes public on the stock market and its shares start trading on an exchange, the share price is determined by supply and demand . But, over the long term, share prices are determined by the economics of the business . It’s impossible to predict exactly what a stock will do and when, but we can study how share price movement works. Let’s unpack Graham’s statement a little more and go over how stock prices work.
## Where do stocks trade?
After that initial offering, the stock starts to trade on secondary markets – that is, stock exchanges such as the New York Stock Exchange (NYSE) or the Nasdaq. This is where we get into the market being a voting machine.
## Why is it important to do your own research and due diligence before buying a stock?
This is why it’s important to do your own research and due diligence before you purchase any stock. The key to making great investments is to buy the stock at a price lower than its intrinsic value. This is how Rule #1 investors know how to pick stocks to buy.
## What is the reward of investing in a stock?
The reward of investing in a stock is the expected payout. If investors expect the price of a stock to rise exponentially, the potential return is great, driving the demand, and so the price of that stock higher.
## What is the most important factor in stock price?
Momentum is one of the most influential factors on stock price. When the excitement for a particular company is high, it attracts investors, which drives the stock price higher, which in turn attracts more investors. This creates momentum, which can continue to drive the price higher if excitement continues.
## How are stock prices determined?
Stock prices are largely determined by supply and demand. If a lot of people want to own a piece of a company, the demand for that company’s stock will go up and the price will rise.
## What happens after an IPO?
After the IPO, a company no longer receives money from sales of its stock, but it can leverage its stock price for a variety of uses such as attracting more investors.
## What is the first step in determining the value of a company?
Company Valuation. Determining a company’s value is the first step to determining what its stock price should be. Determining a company’s value is also a key step in determining whether or not you should invest in that company. You can only invest in a company, however, if it is publicly traded on the stock exchange.
## What happens when a company goes public?
When a company decides to go public, shares of the company, which are stock, go on sale. Most often, this occurs through a process called an Initial Public Offering or IPO. Before the IPO, though, an investment bank has to determine that the company is worthy of investing in.
## What is demand curve?
The point where the demand curve. Demand Curve Demand Curve is a graphical representation of the relationship between the prices of goods and demand quantity and is usually inversely proportionate . That means higher the price, lower the demand.
## What is microeconomics?
Microeconomics Study Microeconomics is a study in economics that involves everyday life, including what we see and experience. It studies individual behavioural patterns, households and corporates and their policies. It deals with supply and demand behaviours in different markets, consumer behaviour, spending patterns, wage-price behaviour, corporate policies. read more
## How does a shift in supply and demand affect the price of a product?
A shift in either the supply or demand, due to any factor/s, will affect the market price. Keeping demand constant, an increase in supply results in a decrease in the price and vice versa. The concept is easy to understand – higher the production, cheaper the product or service. Similarly, if supply is constant, an increase in demand leads to an increase in the price and vice versa. If anything of the above scenarios happens, the business shifts the market price to bring in line with the changing supply and demand.
## Why does demand change?
Demand for any asset or service might change due to various factors tastes and preferences, income, changes in prices of related products, future expectations, etc. Similarly, the supply might fluctuate due to – natural conditions, changes in factor prices, government policies, number of suppliers and the nature of the product.
## What is the difference between market price and normal price?
Difference Between Market Price and Normal Price. It is temporary – it can be more or less than the average cost of production. Normal price is permanent – usually equal to the average cost of production . There exists an opportunity for supernormal profits if the price is more than the average cost of production.
## Why is it important to know the price of an asset?
Knowing this price is key to knowing how to get a trade, increase revenue, reduce costs and expand the business. There may/ may not be multiple markets for the same product or service, that depends and varies on the offerings and the industry.
## What is a trade only?
A trade-only takes place if a buyer interacts with a seller. To make that happen, there is a need for dealers and brokers. In the above scenario, if the buyer deems fit to increase the bid or the seller feels to decrease the ask to the respective prices, the share would trade or it will remain untraded.
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F23 Mad 5234 LA 5: Mathematical proofs of algorithm and program execution correctness.F23 Mad 5234 LA 5: Mathematical proofs of algorithm and program execution correctness.
DUE: May 23 11:59 p.m.
How to submit this work:
Lean into your Markdown and GitBook Skills.
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Outline and description of a three-part lesson on mathematical proofs of algorithm and program execution correctness.
They will learn how to use mathematical induction to prove the correctness of an algorithm.
Introduction to Algorithm Correctness:
Define what it means for an algorithm to be correct and the importance of proving algorithm correctness.
Mathematical Induction:
Explain the concept of mathematical induction and how it's used to verify a statement for every natural number.
Proving Algorithm Correctness with Induction:
Use examples to illustrate how to prove the correctness of an algorithm using mathematical induction.
Let's proceed with a specific example to better illustrate the given material.
PART 1: Proving Algorithm Correctness
QUICK OVERVIEW:
Introduction to Algorithm Correctness:
An algorithm is said to be correct if, for every input instance, it halts with the correct output.
Ensuring an algorithm's correctness is essential as it validates that the algorithm is accurately solving the intended problem.
Mathematical Induction:
Mathematical induction is a mathematical proof technique used to establish that a given statement is true for all natural numbers.
It consists of two steps:
The base step (proof of the statement for the first natural number)
and the inductive step (proof that, if the statement holds for some natural number, it holds for the next natural number as well).
Proving Algorithm Correctness with Induction:
Mathematical induction can be applied to verifying algorithm correctness.
Each step of the algorithm corresponds to an inductive step in the mathematical induction proof, confirming the algorithm is valid for all input instances.
APPLIED EXAMPLE: SUM OF NATURAL NUMBERS
Let’s take an example of an algorithm that calculates the sum of the first N natural numbers. The formula is:
Sum(N) = N * (N + 1) / 2
INTRO:
This simple algorithm is said to calculate the sum of the first N natural numbers.
We'll now attempt to prove its correctness.
ASSUMPTION - BASE CASE (for N=1):
First, we prove this formula works for N=1.
Sum(1) = 1 * (1 + 1) / 2 = 1, which is true, as the sum of the first 1 natural number(s) is indeed 1.
INDUCTIVE STEP:
Next, we make an inductive hypothesis.
We assume this formula works for some natural number k (that Sum(k) = k * (k + 1) / 2 is true), and we need to prove that the formula will then work for k + 1.
So we want to prove Sum(k + 1) = (k + 1) * ((k + 1) + 1) / 2. If we simplify this, we need to prove that
Sum(k + 1) = (k + 1) * (k + 2) / 2 = (k^2 + 3k + 2) / 2.
We also know that Sum(k + 1) = Sum(k) + (k + 1) based on the definition of the problem.
Substitute Sum(k) from our inductive hypothesis into this equation:
= (k * (k + 1) / 2) + (k + 1) = (k^2 + k + 2k + 2) / 2 = (k^2 + 3k + 2) / 2
Hence, we've successfully completed our inductive step and have shown that if our formula is true for some number k, then it's also true for k + 1.
CONCLUSION: What Induction means
Induction is our mathematical tool for proving algorithm correctness
Now, we've established the correctness of the algorithm using mathematical induction. It guarantees that no matter which number N we're finding the sum of, our algorithm will correctly calculate it. This is a typical example of using mathematical induction to verify correctness in algorithms.
The net sum of this is:
I have some code:
It should do one or the other of two things:
A. Excute its instruction AND pass the flow of control to the next program instruction frame - OR -
B. HALT.
PART 2: Proving Single Class Method Correctness:
This part focuses on proving the correctness of single class methods in a program.
Concepts like invariants and pre/post conditions are crucial here.
- Introduction to Class Methods:
Briefly review what a class and a method are and their roles in a program. Class Invariants and Method Specifications: Define what invariants, preconditions, and postconditions are in a class method.
Proving Method Correctness: Demonstrate how to prove the correctness of a method by ensuring that given the preconditions, the postconditions will always hold.
Let's dive deeper into proving single class method correctness with a Code Example.
PART 2: Proving Single Class Method Correctness
Let’s consider a basic class method: the `sum()` method from the ArrayIntList class in Java.
This method calculates the sum of all numbers in the list.
```java //ArrayIntList class would look some thing like this:
public class ArrayIntList { private int[] elementData; // list of integers private int size; // current number of elements in the list //Constructors and other methods would be here public int sum() { int output = 0; for (int i = 0; i < this.size; i++) { output += this.elementData[i]; } return output; } } ```
Our goal is to verify that the `sum()` method operates correctly - it accurately computes the sum of all integers in the ArrayIntList.
Introduction to Class Methods:
In a Class, methods are blocks of code that execute a specific task or algorithm.
Here the `sum()` method is used to calculate the sum of integers in an ArrayIntList.
Class Invariants and Method Specifications:
Preconditions:
What must be true before the method runs.
For our `sum()` function, the precondition is that the ArrayIntList must be initialized (not null).
Postconditions:
What will be true after the method completes.
For our `sum()` function, the postcondition is that the sum of all integers in the ArrayIntList is returned. Proving Method Correctness:
To prove the correctness of the `sum()` method, it must hold true that given preconditions (ArrayIntList is initialized), the postconditions (sum of integers in the ArrayIntList is returned) will always be satisfied.
If the ArrayIntList is initialized, the `sum()` method iterates through the array, summing up each element.
At the end of the iteration, it returns the total sum, satisfying the postcondition.
So the correct behavior of this method under specified conditions (pre/postconditions) confirms the correctness of this method.
This concept of ensuring pre/postconditions is a basic but fundamental principle in proving the correctness of program execution at the level of class methods.
Remember, for complex methods with intricate logic the process could be more complicated but the principles stay the same.
A simple Java code and a template of a worked-out example based on the this scenario:
Sample Java Code:
public class ArrayIntList {
private int[] elementData;
private int size;
// Constructor to initialize the array and size
public ArrayIntList(int[] input) {
this.elementData = input;
this.size = input.length;
}
// Sum method as described
public int sum() {
int output = 0;
for (int i = 0; i < this.size; i++) {
output += this.elementData[i];
}
return output;
}
// Main method to test our ArrayIntList class
public static void main(String[] args) {
int[] testData = {1, 2, 3, 4, 5};
ArrayIntList list = new ArrayIntList(testData);
System.out.println("Sum of integers in the ArrayIntList: " + list.sum()); // Expected output: 15
}
}
Worked Out Example:
Let's prove the correctness of the sum() method for the ArrayIntList class.
Problem Statement:
We have an integer array called elementData. We want to calculate the sum of all elements present in this array.
Preconditions:
elementData should be initialized. (i.e., it should not be null)
size variable should accurately represent the number of elements in elementData.
Postconditions:
The method should return the correct sum of all integers present in elementData.
Approach:
Initialize an output variable to 0.
int output =0;
Iterate over elementData using a loop running from 0 to size - 1.
For each iteration, add the current element to the output.
Return output at the end.
Proof:
Given our preconditions, we are sure that our array is initialized and we know its size.
Now, during every iteration, we are simply adding the array's current element to our output. So, by the end of the loop, output will have the sum of all elements of elementData. Thus, our postcondition (method should return the correct sum) is satisfied.
Verification:
Using our sample data of {1, 2, 3, 4, 5}, we get an output of 15, which matches our expected output.
This confirms our method's correctness for this dataset.
While this is a simple example, the principle of ensuring pre/postconditions and then proving them holds true for more complex scenarios.
The key is breaking down the problem, identifying conditions, and then methodically proving them using the logic of your code.
What you are to do:
Assignment: Prove Method Correctness for a String Manipulation Class
In this assignment, students will dive deep into proving the correctness of class methods.
Using the principles learned from the ArrayIntList example, students will focus on a new class called StringManipulator.
Objective:
Create a StringManipulator class that has methods for string operations.
Prove the correctness of each method by specifying preconditions and postconditions and then validating them.
StringManipulator Class Creation:
Design a class called StringManipulator with the following methods:
reverse(): Returns the reversed version of the input string.
toUpperCase(): Converts and returns the string in uppercase.
concatenate(String str1, String str2): Returns the concatenated result of str1 and str2.
Preconditions and Postconditions:
For each method, note in your GITBOOK as much as you can think about for these issues:
Preconditions: What must be true before the method is executed.
Postconditions: What is expected to be true after the method is executed.
Prove Method Correctness:
For each method, provide a step-by-step breakdown similar to the ArrayIntList example.
Ensure that given the preconditions, the postconditions will always be satisfied.
Submission:
A StringManipulator.java file containing the class and methods. (already provided, below)
A ProofOfCorrectness.md file in markdown format, detailing:
The preconditions and postconditions for each method.
A step-by-step breakdown proving the correctness of each method.
1. Code [40 points]
Proper class and method structure: 10 points
Code correctness for each method: 10 points each
2. Proof of Correctness Documentation [60 points]
Clearly defined preconditions and postconditions: 15 points
Detailed breakdown proving correctness for reverse(): 15 points
Detailed breakdown proving correctness for toUpperCase(): 15 points
Detailed breakdown proving correctness for concatenate(): 15 points
The strings you work with can consist only of alphabets (uppercase/lowercase) and spaces.
Make sure your methods handle edge cases: e.g., empty strings.
For the purpose of this assignment, you don't need to consider null strings.
Final Note:
Through this assignment, you will hone your skills in understanding how methods work and ensuring they operate correctly under defined conditions.
You will sensitive your Programmer’s Intuition as to how to formulate your methods to make them easily.
Always remember to break down your problems, and methodically prove them using the logic of your code.
Below is a fully worked out Java application for the StringManipulator class and a sample demonstration in the main method:
javaCopy code
public class StringManipulator {
// Constructor
public StringManipulator() {
}
// Returns the reversed version of the input string.
public String reverse(String input) {
// Ensure input is not null
if (input == null) {
return null;
}
StringBuilder reversed = new StringBuilder();
for (int i = input.length() - 1; i >= 0; i--) {
reversed.append(input.charAt(i));
}
return reversed.toString();
}
// Converts and returns the string in uppercase.
public String toUpperCase(String input) {
// Ensure input is not null
if (input == null) {
return null;
}
return input.toUpperCase();
}
// Returns the concatenated result of str1 and str2.
public String concatenate(String str1, String str2) {
// Ensure neither str1 nor str2 is null
if (str1 == null || str2 == null) {
return null;
}
return str1 + str2;
}
public static void main(String[] args) {
StringManipulator sm = new StringManipulator();
// Demonstrate reverse
String word = "Hello";
System.out.println("Original: " + word);
System.out.println("Reversed: " + sm.reverse(word));
// Demonstrate toUpperCase
String lowerCaseWord = "hello world";
System.out.println("\nOriginal: " + lowerCaseWord);
System.out.println("Uppercase: " + sm.toUpperCase(lowerCaseWord));
// Demonstrate concatenate
String first = "Hello";
String second = " World";
System.out.println("\nFirst Word: " + first);
System.out.println("Second Word: " + second);
System.out.println("Concatenated: " + sm.concatenate(first, second));
}
} | 2,924 | 13,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-33 | latest | en | 0.875472 |
https://www.nag.com/numeric/cl/nagdoc_cl24/html/F08/f08gcc.html | 1,632,172,773,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057091.31/warc/CC-MAIN-20210920191528-20210920221528-00665.warc.gz | 927,904,785 | 5,632 | f08 Chapter Contents
f08 Chapter Introduction
NAG Library Manual
# NAG Library Function Documentnag_dspevd (f08gcc)
## 1 Purpose
nag_dspevd (f08gcc) computes all the eigenvalues and, optionally, all the eigenvectors of a real symmetric matrix held in packed storage. If the eigenvectors are requested, then it uses a divide-and-conquer algorithm to compute eigenvalues and eigenvectors. However, if only eigenvalues are required, then it uses the Pal–Walker–Kahan variant of the $QL$ or $QR$ algorithm.
## 2 Specification
#include #include
void nag_dspevd (Nag_OrderType order, Nag_JobType job, Nag_UploType uplo, Integer n, double ap[], double w[], double z[], Integer pdz, NagError *fail)
## 3 Description
nag_dspevd (f08gcc) computes all the eigenvalues and, optionally, all the eigenvectors of a real symmetric matrix $A$ (held in packed storage). In other words, it can compute the spectral factorization of $A$ as
$A=ZΛZT,$
where $\Lambda$ is a diagonal matrix whose diagonal elements are the eigenvalues ${\lambda }_{i}$, and $Z$ is the orthogonal matrix whose columns are the eigenvectors ${z}_{i}$. Thus
$Azi=λizi, i=1,2,…,n.$
## 4 References
Anderson E, Bai Z, Bischof C, Blackford S, Demmel J, Dongarra J J, Du Croz J J, Greenbaum A, Hammarling S, McKenney A and Sorensen D (1999) LAPACK Users' Guide (3rd Edition) SIAM, Philadelphia http://www.netlib.org/lapack/lug
Golub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore
## 5 Arguments
1: orderNag_OrderTypeInput
On entry: the order argument specifies the two-dimensional storage scheme being used, i.e., row-major ordering or column-major ordering. C language defined storage is specified by ${\mathbf{order}}=\mathrm{Nag_RowMajor}$. See Section 3.2.1.3 in the Essential Introduction for a more detailed explanation of the use of this argument.
Constraint: ${\mathbf{order}}=\mathrm{Nag_RowMajor}$ or $\mathrm{Nag_ColMajor}$.
2: jobNag_JobTypeInput
On entry: indicates whether eigenvectors are computed.
${\mathbf{job}}=\mathrm{Nag_DoNothing}$
Only eigenvalues are computed.
${\mathbf{job}}=\mathrm{Nag_EigVecs}$
Eigenvalues and eigenvectors are computed.
Constraint: ${\mathbf{job}}=\mathrm{Nag_DoNothing}$ or $\mathrm{Nag_EigVecs}$.
3: uploNag_UploTypeInput
On entry: indicates whether the upper or lower triangular part of $A$ is stored.
${\mathbf{uplo}}=\mathrm{Nag_Upper}$
The upper triangular part of $A$ is stored.
${\mathbf{uplo}}=\mathrm{Nag_Lower}$
The lower triangular part of $A$ is stored.
Constraint: ${\mathbf{uplo}}=\mathrm{Nag_Upper}$ or $\mathrm{Nag_Lower}$.
4: nIntegerInput
On entry: $n$, the order of the matrix $A$.
Constraint: ${\mathbf{n}}\ge 0$.
5: ap[$\mathit{dim}$]doubleInput/Output
Note: the dimension, dim, of the array ap must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}×\left({\mathbf{n}}+1\right)/2\right)$.
On entry: the upper or lower triangle of the $n$ by $n$ symmetric matrix $A$, packed by rows or columns.
The storage of elements ${A}_{ij}$ depends on the order and uplo arguments as follows:
• if ${\mathbf{order}}=\mathrm{Nag_ColMajor}$ and ${\mathbf{uplo}}=\mathrm{Nag_Upper}$,
${A}_{ij}$ is stored in ${\mathbf{ap}}\left[\left(j-1\right)×j/2+i-1\right]$, for $i\le j$;
• if ${\mathbf{order}}=\mathrm{Nag_ColMajor}$ and ${\mathbf{uplo}}=\mathrm{Nag_Lower}$,
${A}_{ij}$ is stored in ${\mathbf{ap}}\left[\left(2n-j\right)×\left(j-1\right)/2+i-1\right]$, for $i\ge j$;
• if ${\mathbf{order}}=\mathrm{Nag_RowMajor}$ and ${\mathbf{uplo}}=\mathrm{Nag_Upper}$,
${A}_{ij}$ is stored in ${\mathbf{ap}}\left[\left(2n-i\right)×\left(i-1\right)/2+j-1\right]$, for $i\le j$;
• if ${\mathbf{order}}=\mathrm{Nag_RowMajor}$ and ${\mathbf{uplo}}=\mathrm{Nag_Lower}$,
${A}_{ij}$ is stored in ${\mathbf{ap}}\left[\left(i-1\right)×i/2+j-1\right]$, for $i\ge j$.
On exit: ap is overwritten by the values generated during the reduction to tridiagonal form. The elements of the diagonal and the off-diagonal of the tridiagonal matrix overwrite the corresponding elements of $A$.
6: w[$\mathit{dim}$]doubleOutput
Note: the dimension, dim, of the array w must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
On exit: the eigenvalues of the matrix $A$ in ascending order.
7: z[$\mathit{dim}$]doubleOutput
Note: the dimension, dim, of the array z must be at least
• $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{pdz}}×{\mathbf{n}}\right)$ when ${\mathbf{job}}=\mathrm{Nag_EigVecs}$;
• $1$ when ${\mathbf{job}}=\mathrm{Nag_DoNothing}$.
The $\left(i,j\right)$th element of the matrix $Z$ is stored in
• ${\mathbf{z}}\left[\left(j-1\right)×{\mathbf{pdz}}+i-1\right]$ when ${\mathbf{order}}=\mathrm{Nag_ColMajor}$;
• ${\mathbf{z}}\left[\left(i-1\right)×{\mathbf{pdz}}+j-1\right]$ when ${\mathbf{order}}=\mathrm{Nag_RowMajor}$.
On exit: if ${\mathbf{job}}=\mathrm{Nag_EigVecs}$, z is overwritten by the orthogonal matrix $Z$ which contains the eigenvectors of $A$.
If ${\mathbf{job}}=\mathrm{Nag_DoNothing}$, z is not referenced.
8: pdzIntegerInput
On entry: the stride separating row or column elements (depending on the value of order) in the array z.
Constraints:
• if ${\mathbf{job}}=\mathrm{Nag_EigVecs}$, ${\mathbf{pdz}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$;
• if ${\mathbf{job}}=\mathrm{Nag_DoNothing}$, ${\mathbf{pdz}}\ge 1$.
9: failNagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
On entry, argument $⟨\mathit{\text{value}}⟩$ had an illegal value.
NE_CONVERGENCE
If ${\mathbf{fail}}\mathbf{.}\mathbf{errnum}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{job}}=\mathrm{Nag_DoNothing}$, the algorithm failed to converge; $⟨\mathit{\text{value}}⟩$ elements of an intermediate tridiagonal form did not converge to zero; if ${\mathbf{fail}}\mathbf{.}\mathbf{errnum}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{job}}=\mathrm{Nag_EigVecs}$, then the algorithm failed to compute an eigenvalue while working on the submatrix lying in rows and column $⟨\mathit{\text{value}}⟩/\left({\mathbf{n}}+1\right)$ through .
NE_ENUM_INT_2
On entry, ${\mathbf{job}}=⟨\mathit{\text{value}}⟩$, ${\mathbf{pdz}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: if ${\mathbf{job}}=\mathrm{Nag_EigVecs}$, ${\mathbf{pdz}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$;
if ${\mathbf{job}}=\mathrm{Nag_DoNothing}$, ${\mathbf{pdz}}\ge 1$.
NE_INT
On entry, ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{n}}\ge 0$.
On entry, ${\mathbf{pdz}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{pdz}}>0$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
## 7 Accuracy
The computed eigenvalues and eigenvectors are exact for a nearby matrix $\left(A+E\right)$, where
$E2 = Oε A2 ,$
and $\epsilon$ is the machine precision. See Section 4.7 of Anderson et al. (1999) for further details.
## 8 Parallelism and Performance
nag_dspevd (f08gcc) is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.
nag_dspevd (f08gcc) makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
The complex analogue of this function is nag_zhpevd (f08gqc).
## 10 Example
This example computes all the eigenvalues and eigenvectors of the symmetric matrix $A$, where
$A = 1.0 2.0 3.0 4.0 2.0 2.0 3.0 4.0 3.0 3.0 3.0 4.0 4.0 4.0 4.0 4.0 .$
### 10.1 Program Text
Program Text (f08gcce.c)
### 10.2 Program Data
Program Data (f08gcce.d)
### 10.3 Program Results
Program Results (f08gcce.r) | 2,730 | 7,949 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 99, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-39 | latest | en | 0.518758 |
http://laughingbuddha.tk/special-right-triangles-formula-sheet.html | 1,563,690,350,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526931.25/warc/CC-MAIN-20190721061720-20190721083720-00192.warc.gz | 96,658,573 | 3,904 | # Special right triangles formula sheet
Triangles special
## Special right triangles formula sheet
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A Time- line for the History of Mathematics ( Many of the early dates are approximates) This work is under constant revision, so come back later. Please report any errors to me at wichita. Geometry formula sheet 1. | 942 | 4,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-30 | latest | en | 0.847206 |
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# How long does it take car A to overtake car B given different accelerations?
Car A is traveling at 18.0 m/s and car B at 25.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with an acceleration of 1.80 m/. How long does it take car A to overtake car B?
5.50 s
12.6 s
22.6 s
Car A never overtakes car B.
#### Solution Summary
With explanations and calculations, the problem is solved.
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Every side of a square is the very same length. There are two major forms of tundra which exists. The power is known as the argument of the logarithm.
In real life, pi can be utilized to figure out the circumference of a circular pool, as long as you know the diameter or radius. Range is a fast calculation. The percent relative range denotes the percentage proportion of the range to the write my essay normal value in the set.
## What Does What Is a Term in Math Mean?
One of the absolute most basic of mathematical operations, multiplication is just one of the most indispensable subjects of study. Archimedes was among the best mathematicians of all times. In the event the equation contains more than 1 logarithm, they should have precisely the same base in order for this to get the job done.
Kant pointed out that existence isn’t a predicate. You’re able to clearly understand that the pronoun athosea can be utilised to rename the nouns in the sentences above. Define an antilogarithm when it comes to a logarithm.
## How to Get Started with What Is a Term in Math?
A degree will take you places, but make certain you enjoy what you are getting involved in. As stated earlier, there are several distinct approaches to compute a percentile. The capacity to construct a multiplication sentence extends past the classroom, by preparing students to figure massive quantities of items.
If you’re taking a 720-mile trip, you want to understand how many hours you’re going to be driving, so it is possible to plan your journey. Successful and profitable on-line traders learn how to discipline their mind to get rid of regretful thinking. For instance, if someone arrives shoddily dressed for an interview, we might feel they haven’t taken the opportunity to prepare.
## The New Angle On What Is a Term in Math Just Released
To allow students to concentrate on more difficult tasks like problem solving, proficiency in basic math concepts is essential. There isn’t any substitute for lots of practice when mastering basic multiplication facts. Since you may see, the most devastating problems connected with NVLD are the social issues.
When solving multiplication difficulties, teachers will normally ask you to reveal your work. The issue with IQis it provides a single umbrella for ‘intelligence’ and doesn’t look at the occurrence of multiple types of intelligence. You add partial products with each other to receive a last answer for the multiplication issue.
A circle is quite intriguing and simple geometric form. Needless to say, it’s simple to locate the precise middle as soon as the data set has an odd number, since the example does. The associative property claims that numbers you’re adding together can be grouped in any purchase.
So, any task or activity you may think of that would result in an enjoyable time may be used to teach patterns. There’re prepared for save, if you love and would like to take that, simply click save logo on the webpage, and it’ll be directly downloaded to your PC. All the items have to be similar to one another. | 1,037 | 5,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-13 | longest | en | 0.941076 |
http://openstudy.com/updates/52289c8de4b0a8663d54ca36 | 1,448,631,964,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398449160.83/warc/CC-MAIN-20151124205409-00189-ip-10-71-132-137.ec2.internal.warc.gz | 171,883,251 | 9,944 | ## IsTim 2 years ago An ore contains 39.2 % of the mineral rutile, TiO 2, which is a source of the element Ti. How much ore must be processed in order to obtain 17.0 kg of Ti?
1. IsTim
Don't tell me the answer, just tell me how to get to the answer, or at the very least the formulas required. Thx
2. IsTim
https://lms.brocku.ca/access/content/group/CHEM1F92D01FW2013MAIN/Assignment%20Hints/assignment_01_hints.pdf Here's the assignment hints. I forgot to read it, and will now leave. I will be back to look at the responses.
3. campbell_st
let t = Ti so 17 = 0.392 x t solve for t | 178 | 587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2015-48 | longest | en | 0.893753 |
https://ww2.mathworks.cn/matlabcentral/cody/problems/44887-given-a-matrix-a-size-m-x-n-create-a-matrix-b-size-m-2-x-n-2-which-consists-of-matrix-a-surround/solutions/2821275 | 1,601,407,601,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402088830.87/warc/CC-MAIN-20200929190110-20200929220110-00320.warc.gz | 704,158,492 | 18,639 | Cody
# Problem 44887. Given a matrix A (size m x n) create a matrix B (size m+2 x n+2) which consists of matrix A surrounded by zeros. See Example below:
Solution 2821275
Submitted on 11 Aug 2020 by Rafael S.T. Vieira
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
x = [1]; y_correct = [0 0 0 0 1 0 0 0 0]; assert(isequal(zeroWrap(x),y_correct))
2 Pass
x = randi(10,4,5); m=zeroWrap(x); assert(isequal(m(2:end-1,2:end-1),x))
3 Pass
x = randi(10,4,5); m=zeroWrap(x); assert(isequal(sum(m(1,:)),0))
4 Pass
x = randi(10,4,5); m=zeroWrap(x); assert(isequal(sum(m(end,:)),0))
5 Pass
x = randi(10,4,5); m=zeroWrap(x); assert(isequal(sum(m(:,1)),0))
6 Pass
x = randi(10,4,5); m=zeroWrap(x); assert(isequal(sum(m(:,end)),0))
7 Pass
x = randi(10,4,5); m=zeroWrap(x); assert(isequal(size(m)-2,size(x))) | 346 | 934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.559233 |
https://www.jiskha.com/questions/301964/How-does-this-work-Think-of-a-Add-7-Double-the-result-Subtract-8-Double | 1,571,240,675,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668994.39/warc/CC-MAIN-20191016135759-20191016163259-00237.warc.gz | 960,100,092 | 5,115 | # arithmetic
How does this work “Think of a #. Add 7. Double the result. Subtract 8. Double the result. Add 12. Divide by 4. Subtract 6. Your answer will be the original number.
1. 👍 0
2. 👎 0
3. 👁 122
1. Think of a #. ---- x
Double the result ----- 2x+14
subract 8 ---- 2x+6
double the result ---- 4x+12
add 12 ------ 4x + 24
divide by 4 ---- x + 6
subtract 6 ---- x
Mathemagics!
1. 👍 0
2. 👎 0
posted by Reiny
## Similar Questions
1. ### elementary
Think of a number. Add 7. Double the result. Subtract 8. Double the result. Add 12. Divide by 4. Subtract 6. Your answer will be the original number. Explain how this trick works.
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2. ### Math 213 Elementary Math
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6. ### INTRODUCTORY ALGEBRA
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9. ### Math
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10. ### maths
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More Similar Questions | 846 | 2,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.840247 |
https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Map%3A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/13%3A_Gravitation | 1,606,548,169,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195198.31/warc/CC-MAIN-20201128070431-20201128100431-00113.warc.gz | 427,626,149 | 24,293 | $$\require{cancel}$$
# 13: Gravitation
In this section, we study the nature of the gravitational force for objects as small as ourselves and for systems as massive as entire galaxies. We show how the gravitational force affects objects on Earth and the motion of the Universe itself. Gravity is the first force to be postulated as an action-at-a-distance force, that is, objects exert a gravitational force on one another without physical contact and that force falls to zero only at an infinite distance. Earth exerts a gravitational force on you, but so do our Sun, the Milky Way galaxy, and the billions of galaxies, like those shown above, which are so distant that we cannot see them with the naked eye.
• 13.1: Prelude to Gravitation
Our visible Universe contains billions of galaxies, whose very existence is due to the force of gravity. Gravity is ultimately responsible for the energy output of all stars—initiating thermonuclear reactions in stars, allowing the Sun to heat Earth, and making galaxies visible from unfathomable distances.
• 13.2: Newton's Law of Universal Gravitation
All masses attract one another with a gravitational force proportional to their masses and inversely proportional to the square of the distance between them. Spherically symmetrical masses can be treated as if all their mass were located at the center. Nonsymmetrical objects can be treated as if their mass were concentrated at their center of mass, provided their distance from other masses is large compared to their size.
• 13.3: Gravitation Near Earth's Surface
The weight of an object is the gravitational attraction between Earth and the object. The gravitational field is represented as lines that indicate the direction of the gravitational force; the line spacing indicates the strength of the field. Apparent weight differs from actual weight due to the acceleration of the object.
• 13.4: Gravitational Potential Energy and Total Energy
The acceleration due to gravity changes as we move away from Earth, and the expression for gravitational potential energy must reflect this change. The total energy of a system is the sum of kinetic and gravitational potential energy, and this total energy is conserved in orbital motion. Objects with total energy less than zero are bound; those with zero or greater are unbounded.
• 13.5: Satellite Orbits and Energy
Orbital velocities are determined by the mass of the body being orbited and the distance from the center of that body, and not by the mass of a much smaller orbiting object. The period of the orbit is likewise independent of the orbiting object’s mass. Bodies of comparable masses orbit about their common center of mass and their velocities and periods should be determined from Newton’s second law and law of gravitation.
• 13.6: Kepler's Laws of Planetary Motion
Johannes Kepler carefully analyzed the positions in the sky of all the known planets and the Moon, plotting their positions at regular intervals of time. From this analysis, he formulated three laws: Kepler’s first law states that every planet moves along an ellipse. Kepler’s second law states that a planet sweeps out equal areas in equal times. Kepler’s third law states that the square of the period is proportional to the cube of the semi-major axis of the orbit.
• 13.7: Tidal Forces
Earth’s tides are caused by the difference in gravitational forces from the Moon and the Sun on the different sides of Earth. Spring or neap (high) tides occur when Earth, the Moon, and the Sun are aligned, and neap or (low) tides occur when they form a right triangle. Tidal forces can create internal heating, changes in orbital motion, and even destruction of orbiting bodies.
• 13.8: Einstein's Theory of Gravity
According to the theory of general relativity, gravity is the result of distortions in space-time created by mass and energy. The principle of equivalence states that that both mass and acceleration distort space-time and are indistinguishable in comparable circumstances. Black holes, the result of gravitational collapse, are singularities with an event horizon that is proportional to their mass.
• 13.E: Gravitation (Exercises)
• 13.S: Gravitation (Summary)
Thumbnail: Our visible Universe contains billions of galaxies, whose very existence is due to the force of gravity. Gravity is ultimately responsible for the energy output of all stars—initiating thermonuclear reactions in stars, allowing the Sun to heat Earth, and making galaxies visible from unfathomable distances. Most of the dots you see in this image are not stars, but galaxies. (credit: modification of work by NASA). | 956 | 4,627 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.1875 | 3 | CC-MAIN-2020-50 | latest | en | 0.944948 |
https://numberworld.info/101111121008 | 1,638,816,089,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363309.86/warc/CC-MAIN-20211206163944-20211206193944-00083.warc.gz | 492,035,575 | 4,197 | # Number 101111121008
### Properties of number 101111121008
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 23 * 101 * 2720381
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
178ab14470
Base 32:
2u5b2h3g
sin(101111121008)
-0.087957853836536
cos(101111121008)
0.99612419704998
tan(101111121008)
-0.088300087576452
ln(101111121008)
25.339485957002
lg(101111121008)
11.004798925391
sqrt(101111121008)
317979.74936779
Square(101111121008)
1.0223458791494E+22
### Number Look Up
Look Up
101111121008 (one hundred one billion one hundred eleven million one hundred twenty-one thousand eight) is a very unique number. The cross sum of 101111121008 is 17. If you factorisate 101111121008 you will get these result 2 * 2 * 2 * 2 * 23 * 101 * 2720381. 101111121008 has 40 divisors ( 1, 2, 4, 8, 16, 23, 46, 92, 101, 184, 202, 368, 404, 808, 1616, 2323, 4646, 9292, 18584, 37168, 2720381, 5440762, 10881524, 21763048, 43526096, 62568763, 125137526, 250275052, 274758481, 500550104, 549516962, 1001100208, 1099033924, 2198067848, 4396135696, 6319445063, 12638890126, 25277780252, 50555560504, 101111121008 ) whith a sum of 206444349216. The number 101111121008 is not a prime number. The number 101111121008 is not a fibonacci number. 101111121008 is not a Bell Number. 101111121008 is not a Catalan Number. The convertion of 101111121008 to base 2 (Binary) is 1011110001010101100010100010001110000. The convertion of 101111121008 to base 3 (Ternary) is 100122222121102010201122. The convertion of 101111121008 to base 4 (Quaternary) is 1132022230110101300. The convertion of 101111121008 to base 5 (Quintal) is 3124033421333013. The convertion of 101111121008 to base 8 (Octal) is 1361254242160. The convertion of 101111121008 to base 16 (Hexadecimal) is 178ab14470. The convertion of 101111121008 to base 32 is 2u5b2h3g. The sine of the figure 101111121008 is -0.087957853836536. The cosine of the figure 101111121008 is 0.99612419704998. The tangent of the figure 101111121008 is -0.088300087576452. The square root of 101111121008 is 317979.74936779.
If you square 101111121008 you will get the following result 1.0223458791494E+22. The natural logarithm of 101111121008 is 25.339485957002 and the decimal logarithm is 11.004798925391. that 101111121008 is very special number! | 906 | 2,412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.596082 |
https://notebook.community/AstroHackWeek/AstroHackWeek2016/breakouts/autodiff/forward-diff-from-scratch-julia | 1,725,757,464,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00459.warc.gz | 394,178,523 | 5,925 | # Forward Differentiation from scratch in Julia
In one implementation of forward-mode automatic differentiation (autodiff), we use "dual numbers" to carry forward partial derivatives along with our calculation. A dual number is similar to a complex number in that it has a real valued component, and an extra component. In the case of dual numbers the extra component is an "infinitesimal" component $\epsilon$. Whereas in complex numbers, $i$ is defined by $i^2 = -1$, for dual numbers, $\epsilon$ is defined by $\epsilon^2 = 0$.
In [1]:
# create a dual number type
immutable Dual
value::Float64
eps::Float64
end
We'll attempt to differentiate the function
$$f(x) = x^2 + x \sin(x)$$
which seems moderately interesting.
In [2]:
# define a test function:
f(x) = x^2 + x * sin(x)
Out[2]:
f (generic function with 1 method)
In [3]:
f(1.)
Out[3]:
1.8414709848078965
So we'll need to know how to multiply, add and take the sine of our Dual type.
In [4]:
import Base: +, *, sin
In [5]:
*(x::Dual, y::Dual) = Dual(x.value * y.value, x.value * y.eps + y.value * x.eps)
Out[5]:
* (generic function with 152 methods)
In [6]:
+(x::Dual, y::Dual) = Dual(x.value + y.value, x.eps + y.eps)
Out[6]:
+ (generic function with 164 methods)
In [7]:
sin(x::Dual) = Dual(sin(x.value), cos(x.value) * x.eps)
Out[7]:
sin (generic function with 11 methods)
Sweet! Now we'll try to run it!
In [8]:
f(3.)
Out[8]:
9.423360024179601
In [9]:
x = Dual(3., 1.)
f(x)
Out[9]:
Dual(9.423360024179601,3.171142518258531)
Check it:
In [10]:
fprime(x) = 2 * x + sin(x) + x * cos(x)
Out[10]:
fprime (generic function with 1 method)
In [11]:
fprime(3.)
Out[11]:
3.171142518258531 | 529 | 1,750 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-38 | latest | en | 0.66105 |
https://se.mathworks.com/matlabcentral/answers/2052292-how-to-graph-the-basic-reproduction-number-r0-against-two-parameters-on-a-3d-plot-with-planes | 1,716,700,168,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058868.12/warc/CC-MAIN-20240526043700-20240526073700-00381.warc.gz | 442,719,052 | 27,278 | # How to graph the basic reproduction number R0 against two parameters on a 3D plot with planes?
29 views (last 30 days)
Amal Matrafi on 27 Nov 2023
Commented: Amal Matrafi on 27 Nov 2023
I need help with ploting the basic reproduction number, related to mathematical biology.
In fact, I tried drawing the basic reproduction number as figures in the paper, but I didn't get the same.
The paper containing the basic reproduction number (R0) and respective parameter values has been uploaded here, I hope you will help me in these drawing.
the code:
clear
clc
[X,Y] = meshgrid(0:0.01:0.5, 0:0.01:0.0381);
R0=6.2954e-04;
Z=R0*ones(size(X));
surf(X,Y,Z)
colormap('turbo')
shading interp
xlabel('\mu')
ylabel('\Pi')
zlabel('R_0')
hold on
%Pi = 0.0381;
lambda=X;
d=0.00174;
beta1=0.00174;
beta2=0.00174;
a1=0.104;
a2=0.104;
wm=0.896;
d1=0.747;
d2=0.747;
gma1=0.253;
gma2=0.253;
theta=0.9;
%mu=0.1177;
mu=Y;
R1=lambda.*beta1./mu.*(a1+mu+wm);
R2=lambda.*beta2./mu.*(a2+mu+wm);
R0=max(R1,R2);
surf(X,Y,R0)
##### 2 CommentsShow NoneHide None
Amal Matrafi on 27 Nov 2023
Note: i named Pi here by lambda value.
Amal Matrafi on 27 Nov 2023
@Bora Eryilmaz I hope this post finds you well,
Can you help me with this problem please.
Sign in to comment.
### Accepted Answer
Bora Eryilmaz on 27 Nov 2023
The equations and plots in the article look inconsistent. For example, Eq. 3.21 seems to be missing a 1/mu term when they replace S0 = Pi/mu in the expressions.
[omega,mu] = meshgrid(0.1:0.01:0.5, 0.01:0.001:0.03);
beta1 = 0.00174;
beta2 = 0.00174;
a1 = 0.104;
a2 = 0.104;
Pi = 0.0381;
R1 = beta1*Pi./mu./(a1 + mu + omega); % Eqs. 3.21 and 3.22 seem to be missing the 1/mu factor.
R2 = beta2*Pi./mu./(a2 + mu + omega);
R0 = max(R1,R2);
s = surf(omega,mu,R0); % This looks consistent with the bottom right plot in Figure 2, except the vertical scale of R0.
xlabel('omega')
ylabel('mu');
zlabel('R0')
[Pi,mu] = meshgrid(0.1:0.01:0.5, 0.01:0.001:0.03);
R1 = beta1*Pi./mu./(a1 + mu + omega);
R2 = beta2*Pi./mu./(a2 + mu + omega);
R0 = max(R1,R2);
s = surf(Pi,mu,R0); % This looks consistent with the top right plot in Figure 2, except the vertical scale of R0.
xlabel('Pi')
ylabel('mu');
zlabel('R0')
Hope this helps.
##### 1 CommentShow -1 older commentsHide -1 older comments
Amal Matrafi on 27 Nov 2023
thank you so much for your response and help.
Sign in to comment.
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https://www.coursehero.com/file/5785868/HW25/ | 1,498,403,165,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320532.88/warc/CC-MAIN-20170625134002-20170625154002-00598.warc.gz | 853,084,007 | 25,669 | # HW25 - PROBLEM 8.36 I=30in.is...
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Unformatted text preview: PROBLEM 8.36 ThuslundarrudABoflanglh I=30in.is attachedtnacollaratBand refisunasmallwheelbcswdatahofizontaldistamea-4hfiunthe verlicalrodonwhichfllccoflnrslidchEnowingflmthcoocfficimtof shficfi-icfionbetweenflnecollarandtheverficalmdisfllsm neglectingmeradimofflaewhaeLdmfilfineflnmngcofvaluesofPfor whichaquilibriumismaimainedwhen Q=251b andfi=30°. SOLUTION a 4 in. FBDrml+eullan ' ' Human—- ‘ sine £1130” Neglect weights ofmdandoollar. (_ m, = o: [30 in.)(einso°}(2s tb} - [a in.)c = o c = 46.375 It: ' ' h—I-EF; = o: N—Cuos30" = u ' ' N - (46.815 Ib)eoe30= = 40.595 lb Impending mntion up: ’F; = M = 025(40595 1b) = mid-Mb 121-} = o: —25 lb + (46.375 lb)einso°-P-m.149 lb =6 mfiinfiaoAC=22in or _ 'P'='-1.563 Ila-10.149 lb = —11.71 lb .mdhgmaiondéwmnieaflmofFismwupmwsfillhwe '= IFI = em = 10.14911: 1215; a o: —25 [bi-{46.875 Ib)ein30°—P+m.149 lb =0 P = -l.563 lb + 10.14911: = 8.59 1b 3. qufilibriumfor 41.7111: 5 P 5 8.59m ‘ PROBLEM 3.47 __| “L |_1,,I+"#___1 Solve Problem 3.46mmm1ingfllatfwcal'isdj1mdtufllelsfi. n - c -' n F-h=€mw[ (_ :10}; = 0: (12 511.][00 11:) — [16 in.)(401b] —(211n.)R,, 130524.035“ — (2 111.391.124.035" = 0 of RA = 16.005 Th [15) Lv IF; = 0: c, — [16.005 Ib)ain24.036° - 0 1:;r = 6.52 DJ Hi: 1 2F, = 0: c’ w 0010 — 40 lb + (16.005 Ib)cus[24.036“] = 0 C, = 105.4 b [4 FED wedge: 5'“ =- \$90-\$20 {on 45, at 29.035 " L. 4:, = Max " I : 11,1403? r (16.005 lb)oos24.036= = 0 x“, =15.0671b : (16.0051b)si1124.036“ +(15.067 lb}sinl4.036° —p = 0 ' nil-10.17115“ SOLUTION FBI) pipe: PROBLEM 8.80 A15” wedgois fumedlmdaralflfl-Ibpipe as shomT‘hceocflicimtof static friction a! all surfaces is 0.20. Determine (a) at which surface slipping of the pipe will first occur, (b) the force 1' for which motion of thawedge is impmding. . (a) (mafia: rFA—rF3=0 or . .FA=F}I Bmitisappsrsntlhst NH > Nd,sosinoe (5)4 = (yak. ' motionmustfirstimpendatdi and FSEFA=FENA=02N‘ (b) (_ EH3 = a: (rsin15°]W + r(1+ 31.1150}?M —[rm15°)1v1 = 0_ 0.2533000 lb] + 1.2588(03NA) - 0.9559115, = 0 or NA = 352m and _ FA = 7.25 1h \ :13. -=_ o: N, u NAsinIS" +15 cos15° — Wmslfi" = o N, = (36.241b]sin15° +(7251h +100 1b)cos15= = 112.97 lb (noteNa > NA assisted, andFa < paNB) f2}; = 0: NW +[7.251b)sjn15° — (112.971b)cos15° - a 111W = 1012411: Impending slip: 1'} = 14an = o.2(1m.24] .- 21.45 lb —-r as; . o: 21.451]:+(7251b}cos15°+(112.971b)si1115"—P= o P = 57.7]b-—" ...
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## This note was uploaded on 02/12/2010 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.
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HW25 - PROBLEM 8.36 I=30in.is...
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Ask a homework question - tutors are online | 1,375 | 2,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-26 | longest | en | 0.534864 |
http://www.dailykos.com/story/2006/11/02/265226/-If-you-earned-less-than-100-000-your-tax-cut-was-158-61 | 1,432,804,608,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929272.10/warc/CC-MAIN-20150521113209-00082-ip-10-180-206-219.ec2.internal.warc.gz | 400,074,836 | 20,852 | If you earned less than \$100,000 your share of the tax cut was \$158.61 on Average.
If you earned more than \$100,000, your share of the tax cut was \$4,754.57 on Average.
So if you earn over \$100,000 a year, you should vote republican.
If you are earning less than \$100,000 a year and vote Republican, you should ask yourself WHY? For \$158.61? The increase in gas wiped that out in a week.
If you do not believe me, do the math yourself. Here are my calculations...
I am going to break this down for everyone based on the ADMINISTRATIONS numbers....
In 2003 the Administration said there were 92 million tax payers who on 'average' would receive \$1083 each. This does not mean each of the 92 million GOT \$1083, it just meant that if you took the entire tax cut and divided it by the number of tax payers it comes to \$1083.
The reality is this.
The total tax cut in 2003 was \$99,636,000,000.
\$87,978,588,000 goes to people making over \$100,000 a year.
If you make over \$100,000 a year, the 'average' tax cut you received was \$4,754.57
The rest of America, the lower middle class and the working poor received what was left. That is \$11,657,412,000 to be divided between the remaining 73,496,000 Tax payers.
So if you made less than \$100,000 you received on Average a tax cut of \$158.61.
So you tell me, who HONESTLY benefited from the Bush Tax cuts?
According to theTax Policy Institute:
45.8% of the benefits from a reduction in capital gains and dividends went to people with incomes over \$1 million. There were 284,000 taxpayers in this income group. This is .19% of all taxpayers.
An additional 10.8% of the benefits went to people with incomes between \$500,000 and \$1 million. There were 593,000 taxpayers in this income group. This is .40% of all taxpayers.
17.4% of the benefits went to people with incomes between \$200,000 and \$500,000. There were 3,588,000 taxpayers in this income group. This is 2.46% of all taxpayers.
14.3% of the benefits went to people with incomes of \$100,000 to \$200,000. There were 14,039,000 taxpayers in this income group.
This is 9.66% of all taxpayers.
In other words --
88.3% of the total benefits from Bush tax cuts went to people with incomes over \$100,000. In addition, the total number of taxpayers who got a vast majority of the benefits represent only 12.71% of all taxpayers.
http://www.taxpolicycenter.org/...
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#### Comment Preferences
• ##### Different cutoff salary(3+ / 0-)
A lot of families in the US who are struggling earn over \$100,000 dollars a year. This argument misses the point. However the details about the breakdown of the taxpayers over \$100,000 makes the point.
What is the share of the tax cut for folks above and below \$500,000 a year (\$250,000 being a high-income small businessman, the higher end of middle management compensation, and the lower end of executive compensation for major corporations).
The \$100,000 figure tells two-income suburbanites to vote Republican. Doing the same calculation for a higher income would clearly show that the Republicans are not acting in their interests.
-6.00/-7.18 The Partie Lion
• ##### 9 out of 10 Americans make under \$100,000(0+ / 0-)
The point I was making was that the vast majority of Americans did NOT benefit from the tax cuts. If fact let me recap, if you earn less than \$100,000 (88% of the US) you got a tax cut of \$158.61. The 15 million jobs Bush promised would happen because of the tax cuts, are not occurring. The GDP
has slowed to 1.6%. There is serious talk of a recession. The tax cuts have caused record debt and record deficits. Your Children now owe
\$10,326.09 each to pay back the debt, this number will only grow. Those making over \$100,000 received a tax cut of \$4,754.57.
Republicans say that Democrats will raise your taxes.
What American is not willing to pay \$158.61, so their Children will
not have to pay \$10,326.09? I will tell you. Republicans are not willing.
Time will Tell all the Truth VT, Virtual Truth
[ Parent ]
• ##### In CA, and in many high cost(0+ / 0-)
areas an income of 100,000 isn't much more then middle, or even lower middle class. Housing eats up most of the cost, and then gas, food and so on.
We didn't get much of a tax break from Bush (even though our income was slightly over 100,000) , and given that the cost of college has gone up, and everything else, we find ourselves moving downward, not up.
At some point I would like to see national adjustments for regions in regards to income taxes, and would very much like to see the local taxes one pays be figured BACK into the deductions (we pay 8.25 percent for local taxes, can't write a cent of that off, nor does anyone get to write off the gas tax and other national taxes). By not allowing those taxes to be deducted we are being double taxed: taxed when we make the money, taxed when we spend it.
• ##### If you itemize you can deduct state taxes(0+ / 0-)
In MD, where I live, the local taxes are piggybacked on the state taxes (you figure your state taxes and then add another 50% or so, depending on which county you live in). So the whole thing is deductible. Eight and a half percent for local taxes is pretty hefty. I think our state tax, with the piggyback, is only around 4%. And a 5% state sales tax, but that's another story, and not deductible.
You fell victim to one of the classic blunders, the most famous of which is "Never get involved in a land war in Asia".
[ Parent ]
• ##### According to the IRS(0+ / 0-)
local taxes are deductible. Maybe you need a new tax preparer.
You fell victim to one of the classic blunders, the most famous of which is "Never get involved in a land war in Asia".
[ Parent ]
• ##### Those are taxes paid on wages(0+ / 0-)
not on money spent
• ##### I donated my Bush tax cut(0+ / 0-)
to Ned Lamont, Jerry McNerney, and Dr David Gill.
I live in California, where a six figure salary would only mean my apartment isn't a rat-hole, 'cause that isn't enough money to buy a place unless you want to live 50 miles from work.
• ##### It would also be nice to see the (1+ / 0-)
Recommended by:
yoduuuh do or do not
cutoff at a median income level. Those above median vs. those below median.
nice job though.
If you are looking for Truth, you better be ready to change your mind.
• ##### Our household income is over \$100,000(2+ / 0-)
Recommended by:
deha, yoduuuh do or do not
Can I still vote for Democrats? Please, please, don't make me vote Republican!! I'll take a pay cut.
You fell victim to one of the classic blunders, the most famous of which is "Never get involved in a land war in Asia".
• ##### Screw you.(1+ / 0-)
Recommended by:
yoduuuh do or do not
So if you earn over \$100,000 a year, you should vote republican.
I prefer the following formulation: If you're stupid, you should vote Republican. So now you're arguing that I should vote Republican and I'm arguing that you should.
You're engaging in several different Republican-type falacies:
1. Who you vote for should be based solely on what the bottom line effect on your taxes will be. That's immoral and profoundly anti-American. It is also profoundly against the ideals of the Democratic Party which are about the broader common good. And it's politically stupid too, since Democrats do better in the above-\$100K group as a whole than Republicans.
1. It's completely arbitrary. Why choose \$100K? You could make a much better argument if you look for the cutoff where the net effect of the tax cuts is, for example, more than the median US take home salary.
1. It's anti-net roots. While the Democratic Party as a whole generally skews towards people who are better off that tendency is even more pronounced here at dKos. Your instructions for those of us who make over your arbitrary cut-off to vote Republican excludes me, numerous well known diarists, and almost certainly Kos himself.
1. It feeds in to the Republican noise that all the Democrats stand for is class war -- and rightly so in this case.
Thanks, I think I'll listen to my conscience and vote Democratic.
Is America finally suffering from Idiot Fatigue?
• ##### Duh, let's see.(0+ / 0-)
If I made <\$100k and didn't want to get raped when I worked hard to make>\$100k I should vote repub in your opinion?
I live in CA where you better make >\$100k if you want a decent standard of living. So, I deserve to get raped as I try to even own a house here? | 2,632 | 10,763 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-22 | latest | en | 0.980168 |
https://testbook.com/question-answer/the-fourth-proportional-to-10-8-3-6-and-20-4-is--6258730d940b55c00e6c51ae | 1,709,563,681,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476452.25/warc/CC-MAIN-20240304133241-20240304163241-00697.warc.gz | 554,661,755 | 93,511 | # The fourth proportional to 10.8, 3.6 and 20.4 is:
This question was previously asked in
SSC Matric Level Previous Paper (Held on: 2 Feb 2022 Shift 1)
View all SSC Selection Post Papers >
1. 14.8
2. 6.8
3. 12.4
4. 9.6
Option 2 : 6.8
Free
Matriculation Level Full Mock Test
40.7 K Users
100 Questions 200 Marks 60 Mins
## Detailed Solution
Concept:
⇒ $$a\over b$$$$c\over d$$
⇒ fourth proportional (d) = $$b × c\over a$$
Calculation:
Using the fourth proportional equation, we get-
⇒ d = $$3.6 × 20.4\over 10.8$$ = 6.8
The fourth proportional is 6.8.
Last updated on Feb 27, 2024
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-> Enhance your exam preparation with the SSC Selection Post Previous Year Papers. Also, attempt SSC Selection Post Mock Tests for practice & revision. | 364 | 1,244 | {"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} | 3.140625 | 3 | CC-MAIN-2024-10 | latest | en | 0.840913 |
https://cantorontheshore.blogspot.com/2015/11/what-is-this-booles-google-doodle.html | 1,624,626,098,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630175.17/warc/CC-MAIN-20210625115905-20210625145905-00330.warc.gz | 157,272,661 | 15,481 | ## Monday, November 2, 2015
### What is this Boole's Google Doodle?
I imagine the scene. You wake up this morning (yes, you), dizzy after a brave night of questionable choices, you fire up Google, and here it is! A new doodle! What is it?
Wait! There is the play button! Maybe it's another game! Like that time Pac-Man's Doodle made the world losing 120 million dollars in productivity! (Well, maybe this is not true). Or like the Halloween one, where you collected candy avoiding bats and ghosts. You passed the day collecting yellow candies. (Because you are a despicable person. Blue should have won!). Let's play!
Huh. What's that? Blinking... stuff? How do you play? You cannot. Green, red, yellow... Maybe it's a traffic light? You try to do the robot in rhythm, but no, it doesn't make sense. You click on it, and it goes to the search results page of "George Boole". Humph! "Who is this guy?" you say, reddened from rage "That's... that's... that's booloney, Google!" (The questionable choices are really catching up on you).
Yes, probably this doodle is not the most perspicuous. The people who knows who Boole is, I guess won't be too much excited by it, and the people who don't know the adjective boolean will not understand what's going on. Well, if you stumbled on this page with this question in mind, I'm here to help you!
So, the doodle of today is to celebrate the 200 years of George Boole's birth. Who is George Boole? This guy:
Boole, when it was not busy cultivating his magnificent sideburns (i.e., rarely), was a mathematician. A rather classical one, actually, he collected some results on differential equations and analysis that are still used. Yet, nothing to google-doodle about. His most important works are instead a small pamphlet called Mathematical Analysis of Logic, and the big budget sequel An Investigation of the Laws of Thought on Which are Founded the Mathematical Theories of Logic and Probabilities (hopefully the book is longer than the title). His idea was to pick Aristotelian logic and systematizing it, formalizing it. Conjunctions and disjunctions (and, or...) were not considered anymore grammatical structures, but mathematical ones. They were operations between sets. It was huge: Boole just kicked out the philosophers from logic, and made it a mathematical concept, in fact an algebraic one, therefore making possible all sorts of applications (and computer science is just one of the most famous).
How does it work? It's called Boolean algebra. In the most basic instance, it's just "true" and "false", and the operations AND, OR and NOT between them.
- AND is like multiplication: true AND true is true, all the others are false, e.g. false AND true is false
- OR is more or less like addiction: false OR false is false, all the others are true, e.g. true OR false is true
- NOT is clear: NOT true is false, and NOT false is true.
Now see again the doodle: it's just showing this operations! When x appears, it means x is true, otherwise x is false. So the "G" (i.e., x AND y) lights up only when both x and y appear, the "l" (NOT x) only when x does not appear, and so on. There is another operation I haven't talked about: XOR. It's the "exclusive or", and x XOR y only if x or y appear, but not at the same time.
Wait, there's more! If you substitute "true" and "false" with "on" and "off", then you have how all the circuits in all the computers function. They are true physical manifestations of boolean operations: if you open your laptop* you look into it, you will find XOR gates, AND gates, and so on.
*I'm not going to be responsible for this.
But is this logic? Why are they called like conjuctions and disjunctions? In a more complicated instance, AND, OR and NOT are operations among sentences. If you have two sentences, "sentence A" and "sentence B", then "sentence A AND sentence B" is the conjunction of the two. That is... "sentence A and sentence B". Whoah, deep.
I don't think that clarified anything. The parallel is: "sentence A AND sentence B" is true if and only if both sentences are true, and x AND y is true if and only if x is true and y is true. So the operations respect the logic, and the logic defines the operations: AND is really just "and", OR is just "or", but instead of connecting sentences, they give you the truth value of the sentences connected. Move on.
You can see here that boolean operations are also operations on sets. AND is just the intersection, OR the union, XOR the symmetric difference. Try it on Google! Search for "Google is awesome" AND "Yog-Sothoth evocation ritual", and then for "Google is awesome" OR "Yog-Sothoth evocation ritual". Now! Or else! (Google is watching you trying). So you know now why a search engine is so grateful to Boole (and why you should always have in handy the mystic scimitar of Barzai).
Don't be less evil
There are many "boolean" things around. The aforementioned boolean algebras, but also boolean circuits, boolean expressions (those things in Excel that permits you to do magic), boolean functions, boolean models, boolean processors... Even a crater on the Moon named Boole!
Boolean algebras, by the way, are essential in Set Theory. They are the basic of the forcing method, the paradigm-changing method that in recent years permitted to prove many independency results, therefore establishing once and for all that mathematics is incomplete, many questions have no answer. It's easy: build a boolean algebra that does the trick, then find an ultrafilter of it (it should be generic, mind you!), then in the generic extension...
You know what? This is not the time to talk about this. This needs a whole other posts. So that's it, for now! Enjoy your Boole day AND have fun! OR NOT! Ha!
(wait for it...)
XOR.
(I think I blew it) | 1,368 | 5,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | longest | en | 0.948222 |
http://www.engineeringwiki.org/wiki/Bearing_Resistance | 1,547,641,082,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657470.23/warc/CC-MAIN-20190116113941-20190116135941-00427.warc.gz | 295,591,328 | 7,194 | # Bearing Resistance
## Bearing resistance of a sawn lumber beam-Example
All clauses, equations and variables were obtained from the Wood Design Manual 2010[1]
### Example:
Find the maximum factored compressive load $Q_f$ that a 3m long 191x292mm S-P-F No.1 beam can withstand if a 89x89mm S-P-F S.S stud is bearing on it when:
1. The stud is at the center of the beam
2. The stud is above the supports and the beam is resting on a 50x50mm bearing plate
Assume:
• Dry service conditions
• Untreated lumber
#### Part 1
Compressive resistance perpendicular to grain (bearing) of a sawn lumber member is given by Clause 5.5.7.2 in the design manual. The Bearing resistance equation is: $Q_r={\Phi}F_{cp}A_{b}K_{b}K_{Zcp}$[1]
where:
• $\Phi$= 0.8[1]
• $F_{cp}=f_{cp}(K_{D}K_{Scp}K_{T})$[1]
• $f_{cp}$ is the specified strength in $MPa$ when calculating for compression perpendicular to grain. From Table 5.3.1C we see that $f_{cp}=5.3MPa$[1] for this member
• $K_D=1.0$ since we have standard term loading and it is obtained from Table 4.3.2.2[1]
• $K_{Scp}=1.0$since we have dry service conditions and it is obtained from Table 5.4.2[1]
• $K_T=1.0$since the member is untreated and operating in dry service conditions. This is obtained from Table 5.4.3[1]
Therefore, $F_{cp}=(5.3)(1)(1)(1)= 5.3 MPa$[1]
• $A_b$ is the bearing area in $mm^2$ through which the perpendicular load acts. In the case where the stud lies over the center of the beam,
$A_b=(89)(191)=16999mm^2$[1]
• $K_B$ is the length of bearing factor given in Clause & Table 5.5.7.6. Since it is in an area of high bending stress(center of the beam),
$K_B=1.0$[1]
• $K_{Zcp}$ is the size bearing factor given in Clause & Table 5.5.7.5. It relies on the ratio of the member's width to its depth. For our case,
$\frac{w}{d}={\frac{191}{292}}{\approx}0.65 {\leq} 1$
Therefore, from Table 5.5.7.5
$K_{Zcp}=1.0$[1]
Since we now have all the variables, we can go back to our original equation and find $Q_r$
$Q_r=(0.8)(5.3)(16999)(1)(1)= 72.1 kN$
This is equal to the maximum $Q_f$ that the member can sustain when the stud is at the center of the beam.
#### Part2
Compressive resistance of a sawn lumber member within a distance from the supports equal to the member depth is given by Clause 5.5.7.3. Here,
$Q'_r=(2/3){\phi}F_{cp}A'_{b}K_{B}K_{Zcp}$[1]
where:
• ${\Phi}=0.8$[1]
• $F_{cp}=f_{cp}(K_{D}K_{Scp}K_{T})$. This is unchanged from before and has the same value.
Therefore,
$F_{cp}=5.3MPa$[1]
• $A'_b=b(\frac{L_{b1}+L_{b2}}{2}) \leq 1.5b(L_{b1})$ represents the average bearing area of unequal bearing areas on either side of the beam. The equation for that is given in Clause 5.5.7.4.[1]
• $b$ is the average bearing width given in $mm$[1]
• $L_{b1}$ is the smaller bearing length given in $mm$[1]
• $L_{b2}$ is the larger bearing length given in $mm$[1]
So,
• $A'_b=b(\frac{L_{b1}+L_{b2}}{2})= ({\frac{50+191}{2})(\frac{50+89}{2})=8374.75 mm^2 \leq 1.5(120.5)(50)=9037.5mm^2$. It satisfies the limitation O.K
• $K_B=1.0$ since the bearing is at the supports meaning it is less than 75mm from the members end. This can be seen from Clause 5.5.7.6(a)[1]
• $K_{Zcp}=1.0$ since it unchanged from Part 1
Now that we have all the variables, we can calculate $Q'_r$
$Q'_r=(\frac{2}{3})(0.8)(5.3)(8374.75)(1)(1)= 23.7kN$
This is equal to the maximum $Q_f$ that can be sustained when the stud is at the support.
A very similar procedure is used to calculate the bearing resistance of a Glulam member by using Clause 6.5.9.2 for Part 1 and Clause 6.5.9.3 for Part 2 while using the respective values relating to the particular Glulam member.[1]
## References
1. Wood Design Manual 2010. Ottawa, ON, Canadian Wood Council (2010). | 1,280 | 3,723 | {"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} | 4.1875 | 4 | CC-MAIN-2019-04 | latest | en | 0.834657 |
https://classbasic.com/meaning-size-speed-and-uses-of-supercomputer/ | 1,726,663,474,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651895.12/warc/CC-MAIN-20240918100941-20240918130941-00640.warc.gz | 153,205,227 | 15,819 | # Meaning of Supercomputer | Size, Speed and Uses of Supercomputer Primary 5 (Basic 5) – Computer Studies
### COMPUTER STUDIES
BASIC SCIENCE AND TECHNOLOGY
SECOND TERM
WEEK 8
PRIMARY 5
THEME – BASIC COMPUTER OPERATION
PREVIOUS LESSON – Classification of Computer by Purpose Primary 5 (Basic 5) – Computer Studies
### LEARNING AREA
1. Introductory Activities
2. Mainframe Computer
3. Size of Mainframe Computer
4. Uses of Mainframe Computer
5. Lesson Evaluation and Weekly Assessment (Test)
### PERFORMANCE OBJECTIVES
By the end of the lesson, most of the pupils should have attained the following objectives –
1. identify supercomputer.
2. describe the size, speed and uses of supercomputer.
INSTRUCTIONAL MATERIALS
The teacher will teach the lesson with the aid of chart shows super computer.
### METHOD OF TEACHING
Choose a suitable and appropriate methods for the lessons.
Note – Irrespective of choosing methods of teaching, always introduce an activities that will arouse pupil’s interest or lead them to the lessons.
### REFERENCE MATERIALS
Scheme of Work
9 – Years Basic Education Curriculum
Course Book
All Relevant Material
Online Information
### LESSON 1 – MEANING OF SUPERCOMPUTERS
Super computer is the most biggest, powerful and fastest computer system.
### SIZE OF SUPERCOMPUTER
A supercomputer has a capacity of 200 to 300 gigabytes or more.
### SPEED OF SUPERCOMPUTER
It is about 50,000 times faster than a micro computer.
### LESSON 2 – USES OF SUPERCOMPUTER
Super computers play an important role in the field of computational science and intensive tasks in various fields, including quantum mechanics, weather forecasting, climate research, oil and gas exploration, etc.
They are used by big organizations, government agencies and universities that can afford it.
### PRESENTATION
To deliver the lesson, the teacher adopts the following steps:
1. To introduce the lesson, the teacher revises the previous lesson. Based on this, he/she asks the pupils some questions;
2. Display chart showing supercomputer.
Pupil’s Activities – Identify and describe the physical part of supercomputer.
3. Discusses the size, speed and uses of supercomputer.
Pupil’s Activities – Identify the size, speed and uses of supercomputer.
4. Summarize the lesson on the board.
Pupil’s Activities – Copy as the pupils write.
### CONCLUSION
To conclude the lesson for the week, the teacher revises the entire lesson and links it to the following week’s lesson.
### LESSON EVALUATION
Ask pupils to discuss the size, speed and uses of supercomputer.
### WEEKLY ASSESSMENT (TEST)
1. ________ is the most biggest, powerful and fastest computer system.
2. A supercomputer has a capacity of ________ gigabytes or more.
A. 20 – 30
B. 200 – 300
C. less than 200
D. less than 300
3. Super computer is about 50,000 times faster than a ________.
A. Personal computer
B. Mainframe computer
C. Micro computer
D. Mini computer
4. Super computer are used by big organizations, government agencies and universities that can afford it.
A. True
B. False
5. Super computer is used by only those who can afford it.
A. True
B. False | 728 | 3,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.813291 |
https://www.unitconverters.net/weight-and-mass/kip-to-ton-long.htm | 1,718,588,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861674.39/warc/CC-MAIN-20240616233956-20240617023956-00122.warc.gz | 932,862,633 | 3,758 | Home / Weight and Mass Conversion / Convert Kip to Ton (long)
# Convert Kip to Ton (long)
Please provide values below to convert kip to ton (long) [ton (UK)], or vice versa.
From: kip To: ton (long)
### Kip to Ton (long) Conversion Table
KipTon (long) [ton (UK)]
0.01 kip0.0044642857 ton (UK)
0.1 kip0.0446428571 ton (UK)
1 kip0.4464285714 ton (UK)
2 kip0.8928571429 ton (UK)
3 kip1.3392857143 ton (UK)
5 kip2.2321428571 ton (UK)
10 kip4.4642857143 ton (UK)
20 kip8.9285714286 ton (UK)
50 kip22.3214285714 ton (UK)
100 kip44.6428571429 ton (UK)
1000 kip446.4285714286 ton (UK)
### How to Convert Kip to Ton (long)
1 kip = 0.4464285714 ton (UK)
1 ton (UK) = 2.24 kip
Example: convert 15 kip to ton (UK):
15 kip = 15 × 0.4464285714 ton (UK) = 6.6964285714 ton (UK) | 308 | 771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-26 | latest | en | 0.47176 |
https://www.gamedev.net/forums/topic/338179-adding-friction-to-a-point-moving-in-3d-space/ | 1,548,230,501,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584203540.82/warc/CC-MAIN-20190123064911-20190123090911-00481.warc.gz | 791,053,197 | 27,843 | Adding Friction to a point moving in 3D space.
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I'm sure this is super easy, but I'm not sure exactly what to do. I have a point that has and x, y, and z position and velocity. I need to slow down it's movement with friction, but I can't just lessen each velocity seprately becuase the velocitys aren't equall to each other (that would make one stop before the others... which looks wierd). It seems like I have to combine thier velocities into one or something crazy. Any ideas?
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vector maths
or, exponential decay
x -= x * 0.01f;
y -= y * 0.01f;
z -= z * 0.01f;
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Thanks a blanket oliii! I owe you corn. Wow.... I don't think that's how you spell "owe." Is it? Nah...
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The only problem with oliii's approach is that it's dependent on the time-step. It is applied based on the number of frames that have elapsed. Depending on what you're doing and how you're doing it, you may want it dependent on how much time has elapsed. You'd do that like this:
const DECAY = 0.8x *= pow(DECAY, dt)y *= pow(DECAY, dt)z *= pow(DECAY, dt)
Where dt is the time per frame, and DECAY is what the speed is factored by every "second".
yes.
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Also, depending on what you're simulating the "friction" may vary as the velocity changes :) Atmospheric drag equations have a velocity-squared in there... Very little friction at low speeds, and lots when you go fast.
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His original approach is perfect for my needs. Thanks again.
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× | 496 | 1,857 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.078125 | 3 | CC-MAIN-2019-04 | latest | en | 0.942696 |
https://www.jiskha.com/display.cgi?id=1157749807 | 1,516,461,978,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889660.55/warc/CC-MAIN-20180120142458-20180120162458-00412.warc.gz | 910,124,815 | 4,396 | # pre-algebra
posted by .
a woman jogging at 6 mph passes a man biking the in opposite direction at 12 mph. If they maintain there speed, how far from each other will they be 10 minutes after passing?
I tried
12 divided by 10 + 6 divided by 10 = 1.8
but I am not sure if that answer is correct...
I think you have the right idea, but there's a unit's mismatch. The time is in minutes and the rates are in miles per hour. Convert the minuts to parts of an hour and repost your answer.
your not alone in that boat i have no idea!
Well, learn to ask questions. I'll try to help and give hints, but I won't work the entire problem. Occasionally I have done the entire problem, but in hindsight I don't see how that helps anyone to learn.
Mason I did the same thing I asked my mom and she said she would have done the same thing ! Im 99.3 Percent sure your correct!
• pre-algebra -
• pre-algebra -
Please explain 4 miles an hour
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More Similar Questions | 763 | 3,086 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.958906 |
http://www.thefullwiki.org/Black_hole | 1,558,753,182,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257847.56/warc/CC-MAIN-20190525024710-20190525050710-00209.warc.gz | 339,818,990 | 80,376 | # Black hole: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
### Did you know ...
More interesting facts on Black hole
# Encyclopedia
Simulated view of a black hole in front of the Large Magellanic Cloud. The ratio between the black hole Schwarzschild radius and the observer distance to it is 1:9. Of note is the gravitational lensing effect known as an Einstein ring, which produces a set of two fairly bright and large but highly distorted images of the Cloud as compared to its actual angular size.
General relativity
$G_{\mu \nu} + \Lambda g_{\mu \nu}= {8\pi G\over c^4} T_{\mu \nu}$
Einstein field equations
Introduction
Mathematical formulation
Resources
Phenomena
Kepler problem · Lenses · Waves
Frame-dragging · Geodetic effect
Event horizon · Singularity
Black hole
According to the general theory of relativity, a black hole is a region of space from which nothing, including light, can escape. It is the result of the deformation of spacetime caused by a very compact mass. Around a black hole there is an undetectable surface which marks the point of no return, called an event horizon. It is called "black" because it absorbs all the light that hits it, reflecting nothing, just like a perfect black body in thermodynamics.[1] Under the theory of quantum mechanics black holes possess a temperature and emit Hawking radiation.
Despite its invisible interior, a black hole can be observed through its interaction with other matter. A black hole can be inferred by tracking the movement of a group of stars that orbit a region in space. Alternatively, when gas falls into a stellar black hole from a companion star, the gas spirals inward, heating to very high temperatures and emitting large amounts of radiation that can be detected from earthbound and Earth-orbiting telescopes.
Astronomers have identified numerous stellar black hole candidates, and have also found evidence of supermassive black holes at the center of galaxies. After observing the motion of nearby stars for 16 years, in 2008 astronomers found compelling evidence that a supermassive black hole of more than 4 million solar masses is located near the Sagittarius A* region in the center of the Milky Way galaxy.
## History
Simulation of Gravitational lensing by a black hole which distorts the image of a galaxy in the background. (Click for larger animation.)
The idea of a body so massive that even light could not escape was first put forward by geologist John Michell in a letter written to Henry Cavendish in 1783 to the Royal Society:
If the semi-diameter of a sphere of the same density as the Sun were to exceed that of the Sun in the proportion of 500 to 1, a body falling from an infinite height towards it would have acquired at its surface greater velocity than that of light, and consequently supposing light to be attracted by the same force in proportion to its vis inertiae, with other bodies, all light emitted from such a body would be made to return towards it by its own proper gravity.
—John Michell[2]
In 1796, mathematician Pierre-Simon Laplace promoted the same idea in the first and second editions of his book Exposition du système du Monde (it was removed from later editions).[3][4] Such "dark stars" were largely ignored in the nineteenth century, since light was then thought to be a massless wave and therefore not influenced by gravity. Unlike the modern black hole concept, the object behind the horizon of a dark star is assumed to be stable against collapse.
### General relativity
In 1915, Albert Einstein developed his general theory of relativity, having earlier shown that gravity does influence light's motion. A few months later, Karl Schwarzschild gave the solution for the gravitational field of a point mass and a spherical mass,[5] showing that a black hole could theoretically exist. The Schwarzschild radius is now known to be the radius of the event horizon of a non-rotating black hole, but this was not well understood then and Schwarzschild himself thought it was not physical. Johannes Droste, a student of Hendrik Lorentz, independently gave the same solution for the point mass a few months after Schwarzschild and wrote more extensively about its properties.
In 1930, astrophysicist Subrahmanyan Chandrasekhar calculated, using general relativity, that a non-rotating body of electron-degenerate matter above 1.44 solar masses (the Chandrasekhar limit) would collapse. His arguments were opposed by Arthur Eddington, who believed that something would inevitably stop the collapse. Eddington was partly correct: a white dwarf slightly more massive than the Chandrasekhar limit will collapse into a neutron star, which is itself stable because of the Pauli exclusion principle. But in 1939, Robert Oppenheimer and others predicted that stars above approximately three solar masses (the Tolman-Oppenheimer-Volkoff limit) would collapse into black holes for the reasons presented by Chandrasekhar.[6]
Oppenheimer and his co-authors used Schwarzschild's system of coordinates (the only coordinates available in 1939), which produced mathematical singularities at the Schwarzschild radius, in other words some of the terms in the equations became infinite at the Schwarzschild radius. This was interpreted as indicating that the Schwarzschild radius was the boundary of a bubble in which time stopped. This is a valid point of view for external observers, but not for infalling observers. Because of this property, the collapsed stars were called "frozen stars,"[7] because an outside observer would see the surface of the star frozen in time at the instant where its collapse takes it inside the Schwarzschild radius. This is a known property of modern black holes, but it must be emphasized that the light from the surface of the frozen star becomes redshifted very fast, turning the black hole black very quickly. Many physicists could not accept the idea of time standing still at the Schwarzschild radius, and there was little interest in the subject for over 20 years.
### Golden age
In 1958, David Finkelstein introduced the concept of the event horizon by presenting Eddington-Finkelstein coordinates, which enabled him to show that "The Schwarzschild surface r = 2 m is not a singularity, but that it acts as a perfect unidirectional membrane: causal influences can cross it in only one direction".[8] This did not strictly contradict Oppenheimer's results, but extended them to include the point of view of infalling observers. All theories up to this point, including Finkelstein's, covered only non-rotating black holes. In 1963, Roy Kerr found the exact solution for a rotating black hole. Two years later, Roger Penrose was able to prove that singularities occur inside any black hole.[9] In 1967, astronomers discovered pulsars,[10][11] and within a few years could show that the known pulsars were rapidly rotating neutron stars. Until that time, neutron stars were also regarded as just theoretical curiosities. So the discovery of pulsars awakened interest in all types of ultra-dense objects that might be formed by gravitational collapse.
Physicist John Wheeler is widely credited with coining the term black hole in his 1967 public lecture Our Universe: the Known and Unknown, as an alternative to the more cumbersome "gravitationally completely collapsed star." However, Wheeler insisted that someone else at the conference had coined the term and he had merely adopted it as useful shorthand. The term was also cited in a 1964 letter by Anne Ewing to the AAAS:
According to Einstein’s general theory of relativity, as mass is added to a degenerate star a sudden collapse will take place and the intense gravitational field of the star will close in on itself. Such a star then forms a "black hole" in the universe.
—Ann Ewing, letter to AAAS[12]
## Properties and structure
The no hair theorem states that, once it achieves a stable condition after formation, a black hole has only three independent physical properties: mass, charge, and angular momentum.[13] Any two black holes that share the same values for these properties, or parameters, are classically indistinguishable.
These properties are special because they are visible from outside the black hole. For example, a charged black hole repels other like charges just like any other charged object. Similarly, the total mass inside a sphere containing a black hole can be found by using the gravitational analog of Gauss's law, the ADM mass, far away from the black hole.[14] Likewise, the angular momentum can be measured from far away using frame dragging by the gravitomagnetic field.
When a black hole swallows any form of matter, its horizon oscillates like a stretchy membrane with friction, a dissipative system, until it reaches a simple final state (see membrane paradigm).[15] Similarly, any information about the charge distribution of the matter is lost as the field is evenly distributed along the event horizon as if the black hole was acting like a conducting sphere with a definite resistivity. This is different from other field theories like electromagnetism, which does not have any friction or resistivity at the microscopic level, because they are time reversible. Because the black hole eventually achieves a stable state with only three parameters, there is no way to avoid losing information about the initial conditions: The gravitational and electric fields of the black hole give very little information about what went in. The information that is lost includes every quantity that cannot be measured far away from the black hole horizon, including the total baryon number, lepton number, and all the other nearly conserved pseudo-charges of particle physics. This behavior is so puzzling, that it has been called the black hole information loss paradox.[16][17][18]
### Classification
#### By physical properties
The simplest black hole has mass but neither charge nor angular momentum. These black holes are often referred to as Schwarzschild black holes after the physicist Karl Schwarzschild who discovered this solution in 1915.[5] It was the first non-trivial exact solution to the Einstein field equations to be discovered, and according to Birkhoff's theorem, the only vacuum solution that is spherically symmetric.[19] This means that there is no observable difference between the gravitational field of such a black hole and that of any other spherical object of the same mass. The popular notion of a black hole "sucking in everything" in its surroundings is therefore only correct near the black hole horizon; far away, the external gravitational field is identical to that of any other body of the same mass.[20]
More general black hole solutions were discovered later in the 20th century. The Reissner-Nordström metric describes a black hole with electric charge, while the Kerr metric yields a rotating black hole. The more generally known stationary black hole solution, the Kerr-Newman metric, describes both charge and angular momentum.
While the mass of a black hole can take any positive value, the charge and angular momentum are constrained by the mass. In natural units , the total charge $Q\,$ and the total angular momentum $J\,$ are expected to satisfy
$Q^2+\left ( \tfrac{J}{M} \right )^2\le M^2\,$
for a black hole of mass M.
Black holes saturating this inequality are called extremal. Solutions of Einstein's equations violating the inequality do exist, but do not have a horizon. These solutions have naked singularities and are deemed unphysical, as the cosmic censorship hypothesis rules out such singularities due to the generic gravitational collapse of realistic matter.[21] This is supported by numerical simulations.[22]
Due to the relatively large strength of the electromagnetic force, black holes forming from the collapse of stars are expected to retain the nearly neutral charge of the star. Rotation, however, is expected to be a common feature of compact objects, and the black-hole candidate binary X-ray source GRS 1915+105[23] appears to have an angular momentum near the maximum allowed value.
#### By mass
Class Mass Size
Supermassive black hole ~105–109 MSun ~0.001–10 AU
Intermediate-mass black hole ~103 MSun ~103 km = REarth
Stellar-mass ~10 MSun ~30 km
Micro black hole up to ~MMoon up to ~0.1 mm
Black holes are commonly classified according to their mass, independent of angular momentum $J\,$. The size of a black hole, as determined by the radius of the event horizon, or Schwarzschild radius, is proportional to the mass $M\,$ through
$r_{sh} \approx 2.95\, M/M_\bigodot \;\mathrm{km,}$
where $r_{sh}\,$ is the Schwarzschild radius and $M_\bigodot$ is the mass of the Sun. A black hole's size and mass are thus simply related independent of rotation. According to this criterion, black holes are classed as:
• Supermassive – contain hundreds of thousands to billions of solar masses, and are thought to exist in the center of most galaxies,[24][25] including the Milky Way.[26] They are thought to be responsible for active galactic nuclei, and presumably form either from the coalescence of smaller black holes, or by the accretion of stars and gas onto them. The largest known supermassive black hole is located in OJ 287 weighing in at 18 billion solar masses.[27]
• Intermediate – contain thousands of solar masses. They have been proposed as a possible power source for ultraluminous X-ray sources.[28] There is no known mechanism for them to form directly, so they likely form by collisions of lower mass black holes, either in the dense stellar cores of globular clusters or galaxies.[citation needed] Such creation events should produce intense bursts of gravitational waves, which may be observed soon. The boundary between super- and intermediate-mass black holes is a matter of convention. Their lower mass limit, the maximum mass for direct formation of a single black hole from collapse of a massive star, is poorly known at present, but is thought to be somewhere well below 200 solar masses.
• Stellar-mass – have masses ranging from a lower limit of about 1.4–3 solar masses (1.4 is the Chandrasekhar limit and 3 is the Tolman-Oppenheimer-Volkoff limit for the maximum mass of neutron stars) up to perhaps 15–20 solar masses. They are created by the collapse of individual stars, or by the coalescence (inevitable, due to gravitational radiation) of binary neutron stars. Stars may form with initial masses up to about 100 solar masses, or in the distant past, possibly even higher, but these shed most of their outer massive layers during earlier phases of their evolution, either blown away in stellar winds during the red giant, AGB, and Wolf-Rayet stages, or expelled in supernova explosions for stars that turn into neutron stars or black holes. Being known mostly by theoretical models for late-stage stellar evolution, the upper limit for the mass of stellar-mass black holes is somewhat uncertain at present. The cores of still lighter stars form white dwarfs.
### Event horizon
Far away from the black hole a particle can move in any direction. It is only restricted by the speed of light. Closer to the black hole spacetime starts to deform. There are more paths going towards the black hole than paths moving away. Inside of the event horizon all paths bring the particle closer to the center of the black hole. It is no longer possible for the particle to escape.
The defining feature of a black hole is the appearance of an event horizon—a boundary in spacetime through which matter and light can only pass inward towards the mass of the black hole. Nothing, including light, can escape from inside the event horizon. The event horizon is referred to as such because if an event occurs within the boundary, light from that event cannot reach an outside observer, making it impossible to determine if such an event occurred.[30]
As predicted by general relativity, the presence of a large mass deforms spacetime in such a way that the paths particles take bend towards the mass. At the event horizon of a black hole, this deformation becomes so strong that there are no paths that lead away from the black hole.[31]
To a distant observer, clocks near a black hole appear to tick more slowly than those further away from the black hole.[32] Due to this effect, known as gravitational time dilation, an object falling into a black hole appears to slow down as it approaches the event horizon, taking an infinite time to reach it.[33] At the same time, all processes on this object slow down causing emitted light to appear redder and dimmer, an effect known as gravitational redshift.[34] Eventually, at a point just before it reaches the event horizon, the falling object becomes so dim that it can no longer be seen.
On the other hand, an observer falling into a black hole does not notice any of these effects as he crosses the event horizon. According to his own clock, he crosses the event horizon after a finite time, although he is unable to determine exactly when he crosses it, as it is impossible to determine the location of the event horizon from local observations.[35]
For a non rotating (static) black hole, the Schwarzschild radius delimits a spherical event horizon. The Schwarzschild radius of an object is proportional to the mass.[36] Rotating black holes have distorted, nonspherical event horizons. Since the event horizon is not a material surface but rather merely a mathematically defined demarcation boundary, nothing prevents matter or radiation from entering a black hole, only from exiting one. The description of black holes given by general relativity is known to be an approximation, and it is expected that quantum gravity effects become significant near the vicinity of the event horizon.[37] This allows observations of matter near a black hole's event horizon to be used to indirectly study general relativity and proposed extensions to it.
### Singularity
At the center of a black hole as described by general relativity lies a gravitational singularity, a region where the spacetime curvature becomes infinite.[38] For a non-rotating black hole this region takes the shape of a single point and for a rotating black hole it is smeared out to form a ring shape lying in the plane of rotation.[39] In both cases the singular region has zero volume. It can also be shown that the singular region contains all the mass of the black hole solution.[40] The singular region can thus be thought of as having infinite density.
An observer falling into a Schwarzschild black hole (i.e. non-rotating and no charges) cannot avoid the singularity. Any attempt to do so will only shorten the time taken to get there.[41] When he reaches the singularity he is crushed to infinite density and his mass is added to the total of the black hole. Before that happens he will have been torn apart by the growing tidal forces in a process sometimes referred to as spaghettification or the noodle effect.[42]
In the case of a charged (Reissner-Nordström) or rotating (Kerr) black hole it is possible to avoid the singularity. Extending these solutions as far as possible reveals the hypothetical possibility of exiting the black hole into a different spacetime with the black hole acting as a worm hole.[43] It also appears to be possible to follow closed timelike curves around the Kerr singularity, which lead to problems with causality like the grandfather paradox.[44] It is expected that none of these peculiar effects would survive in a proper quantum mechanical treatment of rotating and charged black holes.[45]
The appearance of singularities in general relativity is commonly perceived as signaling the breakdown of the theory.[46] This breakdown, however, is expected; it occurs in a situation where quantum mechanical effects should describe these actions due to the extremely high density and therefore particle interactions. To date it has not been possible to combine quantum and gravitational effects into a single theory. It is generally expected that a theory of quantum gravity will feature black holes without singularities.[47][48]
### Photon sphere
The photon sphere is a spherical boundary of zero thickness such that photons moving along tangents to the sphere will be trapped in a circular orbit. For non-rotating black holes, the photon sphere has a radius 1.5 times the Schwarzschild radius. The orbits are dynamically unstable, hence any small perturbation (such as a particle of infalling matter) will grow over time, either setting it on an outward trajectory escaping the black hole or on an inward spiral eventually crossing the event horizon.
While light can still escape from inside the photon sphere, any light that crosses the photon sphere on an inbound trajectory will be captured by the black hole. Hence any light reaching an outside observer from inside the photon sphere must have been emitted by objects inside the photon sphere but still outside of the event horizon.
Other compact objects, such as neutron stars, can also have photon spheres.[49] This follows from the fact that the gravitational field of an object does not depend on its actual size, hence any object that is smaller than 1.5 times the Schwarzschild radius corresponding to its mass will indeed have a photon sphere.
### Ergosphere
The ergosphere is an oblate spheroid region outside of the event horizon, where objects cannot remain stationary.
Rotating black holes are surrounded by a region of spacetime in which it is impossible to stand still, called the ergosphere. This is the result of a process known as frame-dragging; general relativity predicts that any rotating mass will tend to slightly "drag" along the spacetime immediately surrounding it. Any object near the rotating mass will tend to start moving in the direction of rotation. For a rotating black hole this effect becomes so strong near the event horizon that an object would have to move faster than the speed of light in the opposite direction to just stand still.[50]
The ergosphere of a black hole is bounded by, the (outer) event horizon on the inside and an oblate spheroid, which coincides with the event horizon at the poles and is noticeably wider around the equator. The outer boundary is sometimes called the ergosurface.
Objects and radiation can escape normally from the ergosphere. Through the Penrose process, objects can emerge from the ergosphere with more energy than they entered. This energy is taken from the rotational energy of the black hole causing it to slow down.[51]
## Formation and evolution
Considering the exotic nature of black holes, it may be natural to question if such bizarre objects could exist in nature or to suggest that they are merely pathological solutions to Einstein's equations. Einstein himself wrongly thought that black holes would not form, because he held that the angular momentum of collapsing particles would stabilize their motion at some radius.[52] This led the general relativity community to dismiss all results to the contrary for many years. However, a minority of relativists continued to contend that black holes were physical objects,[53] and by the end of the 1960s, they had persuaded the majority of researchers in the field that there is no obstacle to forming an event horizon.
Once an event horizon forms, Roger Penrose proved that a singularity will form somewhere inside it.[9] Shortly afterwards, Stephen Hawking showed that many cosmological solutions describing the big bang have singularities without scalar fields or other exotic matter (see Penrose-Hawking singularity theorems). The Kerr solution, the no-hair theorem and the laws of black hole thermodynamics showed that the physical properties of black holes were simple and comprehensible, making them respectable subjects for research.[54] The primary formation process for black holes is expected to be the gravitational collapse of heavy objects such as stars, but there are also more exotic processes that can lead to the production of black holes.
### Gravitational collapse
Gravitational collapse occurs when an object's internal pressure is insufficient to resist the object's own gravity. For stars this usually occurs either because a star has too little "fuel" left to maintain its temperature, or because a star which would have been stable receives extra matter in a way which does not raise its core temperature. In either case the star's temperature is no longer high enough to prevent it from collapsing under its own weight (the ideal gas law explains the connection between pressure, temperature, and volume).[55]
The collapse may be stopped by the degeneracy pressure of the star's constituents, condensing the matter in an exotic denser state. The result is one of the various types of compact star. Which type of compact star is formed depends on the mass of the remnant — the matter left over after changes triggered by the collapse (such as supernova or pulsations leading to a planetary nebula) have blown away the outer layers. Note that this can be substantially less than the original star — remnants exceeding 5 solar masses are produced by stars which were over 20 solar masses before the collapse.[55]
If the mass of the remnant exceeds ~3-4 solar masses (the Tolman-Oppenheimer-Volkoff limit)—either because the original star was very heavy or because the remnant collected additional mass through accretion of matter—even the degeneracy pressure of neutrons is insufficient to stop the collapse. After this no known mechanism (except possibly quark degeneracy pressure, see quark star) is powerful enough to stop the collapse and the object will inevitably collapse to a black hole.[55]
This gravitational collapse of heavy stars is assumed to be responsible for the formation of most (if not all) stellar mass black holes.[citation needed]
While most of the energy released during gravitational collapse is emitted very quickly, an outside observer doesn't actually see the end of this process. Even though the collapse takes a finite amount of time from the reference frame of infalling matter, a distant observer sees the infalling material slow and halt just above the event horison, due to gravitational time dilation. Light from the collapsing material takes longer and longer to reach the observer, with the light emitted just before the event horizon forms delayed an infinite amount of time. Thus the external observer never sees the formation of the event horizon; instead, the collapsing material seems to become dimmer and increasingly red-shifted, eventually fading away.[56]
#### Primordial black holes in the Big Bang
Gravitational collapse requires great densities. In the current epoch of the universe these high densities are only found in stars, but in the early universe shortly after the big bang densities were much greater, possibly allowing for the creation of black holes. The high density alone is not enough to allow the formation of black holes since a uniform mass distribution will not allow the mass to bunch up. In order for primordial black holes to form in such a dense medium, there must be initial density perturbations which can then grow under their own gravity. Different models for the early universe vary widely in their predictions of the size of these perturbations. Various models predict the creation of black holes, ranging from a Planck mass to hundreds of thousands of solar masses.[57] Primordial black holes could thus account for the creation of any type of black hole.
### High energy collisions
A simulated event in the CMS detector, a collision in which a micro black hole may be created.
Gravitational collapse is not the only process that could create black holes. In principle, black holes could also be created in high energy collisions that create sufficient density. However, to date, no such events have ever been detected either directly or indirectly as a deficiency of the mass balance in particle accelerator experiments.[58] This suggests that there must be a lower limit for the mass of black holes. Theoretically this boundary is expected to lie around the Planck mass (~1019 GeV/c2 = ~2 × 10−8 kg), where quantum effects are expected to make the theory of general relativity break down completely.[59] This would put the creation of black holes firmly out of reach of any high energy process occurring on or near the Earth. Certain developments in quantum gravity however suggest that this bound could be much lower. Some braneworld scenarios for example put the Planck mass much lower, maybe even as low as 1 TeV/c2.[60] This would make it possible for micro black holes to be created in the high energy collisions occurring when cosmic rays hit the Earth's atmosphere, or possibly in the new Large Hadron Collider at CERN. These theories are however very speculative, and the creation of black holes in these processes is deemed unlikely by many specialists.[citation needed]
### Growth
Once a black hole has formed, it can continue to grow by absorbing additional matter. Any black hole will continually absorb interstellar dust from its direct surroundings and omnipresent cosmic background radiation, but neither of these processes should significantly affect the mass of a stellar black hole. More significant contributions can occur when the black hole is formed in a binary star system. After formation the black hole can then leech significant amounts of matter from its companion.
Much larger contributions can be obtained when a black hole merges with other stars or compact objects. The supermassive black holes suspected in the center of most galaxies are expected to have formed from the coagulation of many smaller objects. The process has also been proposed as the origin of some intermediate-mass black holes.
As an object approaches the event horizon, the horizon near the object bulges up and swallows the object. Shortly thereafter the increase in radius (due to the extra mass) is distributed evenly around the hole.
### Evaporation
In 1974, Stephen Hawking showed that black holes are not entirely black but emit small amounts of thermal radiation.[61] He got this result by applying quantum field theory in a static black hole background. The result of his calculations is that a black hole should emit particles in a perfect black body spectrum. This effect has become known as Hawking radiation. Since Hawking's result, many others have verified the effect through various methods.[62] If his theory of black hole radiation is correct then black holes are expected to emit a thermal spectrum of radiation, and thereby lose mass, because according to the theory of relativity mass is just highly condensed energy (E = mc2).[61] Black holes will shrink and evaporate over time. The temperature of this spectrum (Hawking temperature) is proportional to the surface gravity of the black hole, which for a Schwarzschild black hole is inversely proportional to the mass. Large black holes, therefore, emit less radiation than small black holes.
A stellar black hole of 5 solar masses has a Hawking temperature of about 12 nanokelvins. This is far less than the 2.7 K produced by the cosmic microwave background. Stellar mass (and larger) black holes receive more mass from the cosmic microwave background than they emit through Hawking radiation and will thus grow instead of shrink. To have a Hawking temperature larger than 2.7 K (and be able to evaporate) a black hole needs to be lighter than the Moon (and therefore a diameter of less than a tenth of a millimeter).
On the other hand if a black hole is very small, the radiation effects are expected to become very strong. Even a black hole that is heavy compared to a human would evaporate in an instant. A black hole the weight of a car (~10−24 m) would only take a nanosecond to evaporate, during which time it would briefly have a luminosity more than 200 times that of the sun. Lighter black holes are expected to evaporate even faster, for example a black hole of mass 1 TeV/c2 would take less than 10−88 seconds to evaporate completely. Of course, for such a small black hole quantum gravitation effects are expected to play an important role and could even – although current developments in quantum gravity do not indicate so – hypothetically make such a small black hole stable.
## Observational evidence
By their very nature black holes do not directly emit any signals other than the hypothetical Hawking radiation. Since the Hawking radiation for an astrophysical black hole is predicted to be very weak, this makes it impossible to directly detect astrophysical black holes from the Earth. A possible exception to the Hawking radiation being weak is the last stage of the evaporation of light (primordial) black holes. Searches for such flashes in the past has proven unsuccessful and provides stringent limits on the possibility of existence of light primordial black holes.[63] NASA's Fermi Gamma-ray Space Telescope launched in 2008 will continue the search for these flashes.[64]
Astrophysicists searching for black holes thus have to rely on indirect observations. A black hole's existence can sometimes be inferred by observing its gravitational interactions with its surroundings.
### Accretion of matter
Formation of extragalactic jets from a black hole's accretion disk
Due to conservation of angular momentum, gas falling into the gravitational well created by a massive object will typically form a disc-like structure around the object. Friction within the disc causes angular momentum to be transported outward allowing matter to fall further inward releasing potential energy and increasing the temperature of the gas.[65] In the case of compact objects such as white dwarfs, neutron stars, and black holes, the gas in the inner regions becomes so hot that it will emit vast amounts of radiation (mainly X-rays), which may be detected by telescopes. This process of accretion is one of the most efficient energy producing process known; up to 40% of the rest mass of the accreted material can be emitted in radiation.[65] (In nuclear fusion only about 1% of the rest mass will be emitted as energy.) In many cases, accretion discs are accompanied by relativistic jets emitted along the poles, which carry away much of the energy. The mechanism for the creation of these jets is currently not well understood.
As such many of the universe's more energetic phenomena have been attributed to the accretion of matter on black holes. In particular, active galactic nuclei and quasars are thought to be the accretion discs of supermassive black holes.[66] Similarly, X-ray binaries are thought to be binary star systems in which one of the two stars is a compact object accreting matter from its companion.[66] It has also been suggested that some ultraluminous X-ray sources may be the accretion disks of intermediate-mass black holes.[67]
### X-ray binaries
X-ray binaries are binary star systems that are luminous in the X-ray part of the spectrum. These X-ray emissions are generally thought to be caused by one of the component stars being a compact object accreting matter from the other (regular) star. The presence of an ordinary star in such a system provides a unique opportunity for studying the central object and determining if it might be a black hole.
Artist impression of a binary system with an accretion disk around a compact object being fed by material from the companion star.
If such a system emits signals that can be directly traced back to the compact object, it cannot be a black hole. The absence of such a signal does, however, not exclude the possibility that the compact object is a neutron star. By studying the companion star it is often possible to obtain the orbital parameters of the system and obtain an estimate for the mass of the compact object. If this is much larger than the Tolman-Oppenheimer-Volkoff limit (that is, the maximum mass a neutron star can have before collapsing) then the object cannot be a neutron star and is generally expected to be a black hole.[66]
The first strong candidate for a black hole, Cygnus X-1, was discovered in this way by Webster and Murdin[68] and Bolton[69] in 1972.[70][71] Some doubt, however, remained due to the uncertainties resultant from the companion star being much heavier than the candidate black hole.[66] Currently, better candidates for black holes are found in a class of X-ray binaries called soft X-ray transients.[66] In this class of system the companion star is relatively low mass allowing for more accurate estimates in the black hole mass. Moreover, these system are only active in X-ray for several months once every 10–50 years. During the period of low X-ray emission (called quiescence) the accretion disc is extremely faint allowing for detailed observation of the companion star during this period. One of the best such candidates is V404 Cyg.
#### Quiescence and advection-dominated accretion flow
The faintness of the accretion disc during quiescence is thought to be caused by the flow entering a mode called an advection-dominated accretion flow (ADAF). In this mode, almost all the energy generated by friction in the disc is swept along with the flow instead of radiated away. If this model is correct, then it forms strong qualitative evidence for the presence of an event horizon. Because, if the object at the center of the disc had a solid surface, it would emit large amounts of radiation as the highly energetic gas hits the surface, an effect that is observed for neutron stars in a similar state.[65]
#### Quasi-periodic oscillations
The X-ray emissions from accretion disks sometimes exhibit a flickering around certain frequencies. These signals are called quasi-periodic oscillations and are thought to be caused by material moving along the inner edge of the accretion disk (the innermost stable circular orbit). As such their frequency is linked to the mass of the compact object. They can thus be used as an alternative way to determine the mass of potential black holes.[72]
### Gamma ray bursts
Intense but one-time gamma ray bursts (GRBs) may signal the birth of "new" black holes, because astrophysicists think that GRBs are caused either by the gravitational collapse of giant stars[73] or by collisions between neutron stars,[74] and both types of event involve sufficient mass and pressure to produce black holes. It appears that a collision between a neutron star and a black hole can also cause a GRB,[75] so a GRB is not proof that a "new" black hole has been formed. All known GRBs come from outside our own galaxy, and most come from billions of light years away[76] so the black holes associated with them are billions of years old.
### Galactic nuclei
The jet originating from the center of M87 in this image comes from an active galactic nucleus that may contain a supermassive black hole. Credit: Hubble Space Telescope/NASA/ESA.
It is now widely accepted that the center of every or at least nearly every galaxy contains a supermassive black hole.[77][78] The close observational correlation between the mass of this hole and the velocity dispersion of the host galaxy's bulge, known as the M-sigma relation, strongly suggests a connection between the formation of the black hole and the galaxy itself.[77]
For decades, astronomers have used the term "active galaxy" to describe galaxies with unusual characteristics, such as unusual spectral line emission and very strong radio emission.[79][80] However, theoretical and observational studies have shown that the active galactic nuclei (AGN) in these galaxies may contain supermassive black holes.[79][80] The models of these AGN consist of a central black hole that may be millions or billions of times more massive than the Sun; a disk of gas and dust called an accretion disk; and two jets that are perpendicular to the accretion disk.[80]
Although supermassive black holes are expected to be found in most AGN, only some galaxies' nuclei have been more carefully studied in attempts to both identify and measure the actual masses of the central supermassive black hole candidates. Some of the most notable galaxies with supermassive black hole candidates include the Andromeda Galaxy, M32, M87, NGC 3115, NGC 3377, NGC 4258, and the Sombrero Galaxy.[81]
Currently, the best evidence for a supermassive black hole comes from the center of our own Milky way.[82] For sixteen years astronomers have tracked the positions of stars orbiting a central massive object in a region called Sagittarius A*, one of which—a star called S2— has completed a full orbit in that period. From the orbital data they were able to infer that there was a spherical mass of 4.3 million solar masses contained within a radius of less than 0.002 lightyears.[82] This is still more than 3000 times the Schwarzschild radius corresponding to that mass. This is consistent with the central object being a supermassive black hole.
### Gravitational lensing
The deformation of spacetime around a massive object causes light rays to be deflected much like light passing through an optic lens. This phenomenon is known as gravitational lensing. Observations have been made of weak gravitational lensing, in which photons are deflected by only a few arcseconds. However, it has never been directly observed for a black hole.[83] One possibility for observing gravitational lensing by a black hole would be to observe stars in orbit around the black hole. There are several candidates for such an observation in orbit around Sagittarius A*.[83]
### Alternatives
The evidence for stellar black holes strongly relies on the existence of an upper limit for the mass of a neutron star. The size of this limit heavily depends on the assumptions made about the properties of dense matter. New exotic phases of matter could push up this bound.[66] A phase of free quarks at high density might allow the existence of dense quark stars,[84] and some supersymmetric models predict the existence of Q stars.[85] Some extensions of the standard model posit the existence of preons as fundamental building blocks of quarks and leptons which could hypothetically form preon stars.[86] These hypothetical models could potentially explain a number of observations of stellar black hole candidates. However, it can be shown from general arguments in general relativity that any such object will have a maximum mass.[66]
Since the average density of a black hole inside its Schwarzschild radius is inversely proportional to the square of its mass, supermassive black holes are much less dense than stellar black holes (the average density of a large supermassive black hole is comparable to that of water).[66] Consequently, the physics of matter forming a supermassive black hole is much better understood and the possible alternative explanations for supermassive black hole observations are much more mundane. For example, a supermassive black hole could be modelled by a large cluster of very dark objects. However, typically such alternatives are not stable enough to explain the supermassive black hole candidates.[66]
The evidence for stellar and supermassive black holes implies that in order for black holes not to form, general relativity must fail as a theory of gravity, perhaps due to the onset of quantum mechanical corrections. A much anticipated feature of a theory of quantum gravity is that it will not feature singularities or event horizons (and thus no black holes).[citation needed] In recent years, much attention has been drawn by the fuzzball model in string theory. Based on calculations in specific situations in string theory, the proposal suggest that generically the individual states of a black hole solution do not have an event horizon or singularity (and can thus not really be considered to be a black hole), but that for a distant observer the statistical average of such states does appear just like an ordinary black hole in general relativity.[87]
## Open questions
If ultra-high-energy collisions of particles in a particle accelerator can create microscopic black holes, it is expected that all types of particles will be emitted by black hole evaporation, providing key evidence for any grand unified theory. Above are the high energy particles produced in a gold ion collision on the RHIC.
In 1971, Stephen Hawking showed under general conditions[Note 1] that the total area of the event horizons of any collection of classical black holes can never decrease, even if they collide and swallow each other; that is merge.[88] This result now known as the second law of black hole mechanics is remarkably similar to the Second Law of Thermodynamics, which states that the total entropy of a system can never decrease. As a classical object with zero temperature it was assumed that black holes had zero entropy. If this were the case, the second law of thermodynamics would be violated by entropy-laden matter entering the black hole, resulting in a decrease of the total entropy of the universe. Therefore, Jacob Bekenstein proposed that a black hole should have an entropy, and that it should be proportional to its horizon area.[citation needed]
The link with the laws of thermodynamics was further strengthened by the discovery of Hawking's discovery that quantum field theory predicts that a black hole radiates blackbody radiation at a constant temperature.[citation needed] This seemingly causes a violation of the second law of black hole mechanics, since the radiation will carry away energy from the black hole causing it to shrink. The radiation however also carries away entropy, and it can be proven under general assumptions that the sum of the entropy of the matter surrounding the black hole and one quarter of the area of the horizon as measured in planck units is in fact always increasing.[citation needed] This allows the formulation of the first law of black hole mechanics as an analogue of the first law of thermodynamics, with the mass acting as energy, the surface gravity as temperature and the area as entropy.[citation needed]
One puzzling feature is that the entropy of a black hole scales with its area rather than with its volume. Since entropy is normally an extrinsic property that scales linearly with the volume of the system. This odd property led 't Hooft and Susskind to propose the holographic principle, which suggests that anything that happens in volume of spacetime can be described by data on the boundary of that volume.[citation needed]
Although general relativity can be used to perform a semi-classical calculation of black hole entropy, this situation is theoretically unsatisfying. In statistical mechanics, entropy is understood as counting the number of microscopic configurations of a system which have the same macroscopic qualities (such as mass, charge, pressure, etc.). Without a satisfactory theory of quantum gravity, one cannot perform such a computation for black holes. Some promise has been shown by string theory, however, which posits that the microscopic degrees of freedom of the black hole are D-branes. By counting the states of D-branes with given charges and energy, the entropy for certain supersymmetric black holes has been reproduced. Extending the region of validity of these calculations is an ongoing area of research.
### Black hole unitarity
An open question in fundamental physics is the so-called information loss paradox, or black hole unitarity paradox. Classically, the laws of physics are the same run forward or in reverse (T-symmetry). Liouville's Theorem dictates conservation of phase space volume, which can be thought of as 'conservation of information', so there is some problem even in classical (non-quantum general relativity) physics. In quantum mechanics, this corresponds to a vital property called unitarity, which has to do with the conservation of probability (It can also be thought of as a conservation of quantum phase space volume as expressed by the density matrix).[89]
## Notes
1. ^ In particular, he assumed that all matter satisfies the weak energy condition.
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87. ^ doi:[[Digital object identifier|doi]]:[http://dx.doi.org/10.1016%2Fj%252Ephysrep%252E2008%252E08%252E001 10.1016/j%2Ephysrep%2E2008%2E08%2E001]
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88. ^ Hawking, Stephen (1998). A Brief History of Time. New York: Bantam Books. ISBN 0-553-38016-8.
89. ^ Hawking, Stephen. "Does God Play Dice?". Retrieved 2009-03-14.
• Ferguson, Kitty (1991). Black Holes in Space-Time. Watts Franklin. ISBN 0-531-12524-6. .
• Hawking, Stephen (1988). A Brief History of Time. Bantam Books, Inc. ISBN 0-553-38016-8. .
• Hawking, Stephen; Penrose, Roger (1996). The Nature of Space and Time. Princeton University Press. ISBN 0-691-03791-2. .
• Melia, Fulvio (2003). The Black Hole at the Center of Our Galaxy. Princeton U Press. ISBN 978-0-691-09505-9. .
• Melia, Fulvio (2003). The Edge of Infinity. Supermassive Black Holes in the Universe. Cambridge U Press. ISBN 978-0-521-81405-8. .
• Pickover, Clifford (1998). Black Holes: A Traveler's Guide. Wiley, John & Sons, Inc. ISBN 0-471-19704-1. .
• Stern, B. (2008). "Blackhole". , poem.
• Thorne, Kip S. (1994). Black Holes and Time Warps. Norton, W. W. & Company, Inc. ISBN 0-393-31276-3. .
• Wheeler, J. Craig (2007). Cosmic Catastrophes (2nd ed.). Cambridge University Press. ISBN 0-521-85714-7.
### University textbooks and monographs
• Carroll, Sean M. (2004). Spacetime and Geometry. Addison Wesley. ISBN 0-8053-8732-3. , the lecture notes on which the book was based are available for free from Sean Carroll's website.
• Carter, B. (1973). "Black hole equilibrium states". in DeWitt, B.S.; DeWitt, C.. Black Holes. .
• Chandrasekhar, Subrahmanyan (1999). Mathematical Theory of Black Holes. Oxford University Press. ISBN 0-19-850370-9. .
• Frolov, V.P.; Novikov, I.D. (1998). Black hole physics. .
• Hawking, S.W.; Ellis, G.F.R. (1973). Large Scale Structure of space time. Cambridge University Press. ISBN 0521099064. .
• Melia, Fulvio (2007). The Galactic Supermassive Black Hole. Princeton U Press. ISBN 978-0-691-13129-0. .
• Taylor, Edwin F.; Wheeler, John Archibald (2000). Exploring Black Holes. Addison Wesley Longman. ISBN 0-201-38423-X. .
• Thorne, Kip S.; Misner, Charles; Wheeler, John (1973). Gravitation. W. H. Freeman and Company. ISBN 0-7167-0344-0. .
• Wald, Robert M. (1992). Space, Time, and Gravity: The Theory of the Big Bang and Black Holes. University of Chicago Press. ISBN 0-226-87029-4. .
### Research papers
• Hawking, S. (2005). "Information loss in black holes". Physical Review D 72: 084013. doi:10.1103/PhysRevD.72.084013. arΧiv:hep-th/0507171v2. Stephen Hawking's purported solution to the black hole unitarity paradox, first reported at a conference in July 2004.
• Ghez, A. M.; Salim, S.; Hornstein, S. D.; Tanner, A.; Lu, J. R.; Morris, M.; Becklin, E. E.; Duchene, G. (2005). "Stellar Orbits around the Galactic Center Black Hole". The Astrophysical Journal 620: 744. doi:10.1086/427175. arΧiv:astro-ph/0306130v2. More accurate mass and position for the black hole at the centre of the Milky Way.
• Hughes, Scott A. (2005). "Trust but verify: The case for astrophysical black holes". arΧiv:hep-ph/0511217v2 [hep-ph]. Lecture notes from 2005 SLAC Summer Institute.
# Study guide
Up to date as of January 14, 2010
### From Wikiversity
A black hole is a theoretical region of space that has such a strong gravity pull that not even light can escape it. It is called black because it sucks in all light, reflecting nothing, therefore appearing "black". Though it is not possible to see one, a black hole's location can be inferred by observing any celestial bodies that orbit an apparently empty space.
## Terminology
A black hole is often referred to as any object whose escape velocity exceeds the speed of light.
The escape velocity is the speed at which an object needs to travel so as to just manage to get infinitely far away from a source of gravity before stopping. On the Earth, the escape velocity is equal to 11 km/s, so no matter what the object is, a rocket or a baseball, it must go at least 11 km/s to avoid falling back to the Earth's surface eventually. Light passing by a black hole would be sucked in, it it came close enough.
## Event horizon
The event horizon is a region of spacetime that cannot affect an outside observer in any way. Coming close to the black hole, an objects movement is encouraged to move towards the event horizon. If the object crosses the event horizon, then any possible movement would just pull it deeper.
## Wikimedia
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# Simple English
General relativity
$G_\left\{\mu \nu\right\} + \Lambda g_\left\{\mu \nu\right\}= \left\{8\pi G\over c^4\right\} T_\left\{\mu \nu\right\}$
Einstein field equations
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Black hole
File:Black Hole
This image is a simulation of what a black hole might look like. No black hole has ever been photographed.
According to the general theory of relativity, a black hole is a region of space from which nothing, including light, can escape. It is the result of the denting of spacetime caused by a very compact mass. Around a black hole there is an undetectable surface which marks the point of no return, called an event horizon. It is called "black" because it absorbs all the light that hits it, reflecting nothing, just like a perfect black body in thermodynamics.[1] Under the theory of quantum mechanics black holes possess a temperature and emit Hawking radiation.
Despite its invisible interior, a black hole can be observed through its interaction with other matter. A black hole can be inferred by tracking the movement of a group of stars that orbit a region in space. Alternatively, when gas falls into a stellar black hole from a companion star, the gas spirals inward, heating to very high temperatures and emitting large amounts of radiation that can be detected from earthbound and Earth-orbiting telescopes.
Astronomers have identified numerous stellar black hole candidates, and have also found evidence of supermassive black holes at the center of galaxies. After observing the motion of nearby stars for 16 years, in 2008 astronomers found compelling evidence that a supermassive black hole of more than 4 million solar masses is located near the Sagittarius A* region in the center of the Milky Way galaxy.
## History
In 1783, an English geologist named John Mitchell wrote that it might be possible for something to be so big and heavy that the escape speed from its gravity is equal to the speed of light. Gravity gets stronger as something gets bigger or more massive. For a small thing, like a rocket, to escape from a larger thing, like Earth, it has to go upward very fast or it will fall back down later. The speed that it must travel upward to get away from Earth's gravity is called escape velocity. Bigger planets (like Jupiter) and stars have stronger gravity than Earth, so the escape velocity is much faster. John Mitchell thought it was possible for something to be so big that the escape velocity would be faster than the speed of light, so even light could not escape.
Some scientists thought Mitchell might be right, but others thought that light had no mass and would not be pulled by gravity. His theory was forgotten.
In 1916, Albert Einstein wrote an explanation of gravity called general relativity. It is a very complicated theory, but there are two important things about it:
• Mass causes space (and spacetime) to bend, or curve. Moving things "fall along" or follow the curves in space. This is what we call gravity.
• Light always travels at the same speed, and is affected by gravity. If it seems to change speed, it is really traveling along a curve in spacetime.
A few months later, a German physicist named Karl Schwarzschild calculated that a black hole could exist.
In 1930, Subrahmanyan Chandrasekhar predicted that stars heavier than the sun could collapse when they ran out of hydrogen to burn and died. In 1939, Robert Oppenheimer and H. Snyder calculated that a star would have to be at least three times as massive as the sun to form a black hole.
In 1967, John Wheeler gave black holes the name "black hole" for the first time. Before that, they were called "dark stars."
In 1970, Stephen Hawking and Roger Penrose proved that black holes must exist. Although the black holes are invisible (they cannot be seen), some of the matter that is falling into them is very bright.
## Formation of black holes
Most black holes are made when a giant star, called a supergiant, at least twenty times bigger than our own Sun dies, and leaves behind a mass that is at least one solar mass. Stars die when they run out of hydrogen or other fuel to burn and start to collapse.
A supergiant star's death is called a supernova. Stars are usually in equilibrium, which means they are making enough energy to push their mass outward against the force of gravity. When the star runs out of energy, gravity takes over. Gravity pulls the center of the star inward very quickly (so quickly that it would have to be repeated several thousand times before it took up a single second), and it collapses into a little ball. The collapse is so fast and violent that it makes a shock wave, and that causes the rest of the star to explode outward. As the gravity pushes the star inward, the pressure in the center of star reaches to such an extreme level that it enables heavier molecules like iron and carbon to interact to release nuclear energy. The release of the energy from the star during a very short period of time (about one hour) is with such a high rate that it outshines an entire galaxy.
The ball in the center is so dense (a lot of mass in a small space, or volume), that if you could somehow scoop only one teaspoon of material and bring it to Earth, it would sink to the core of the planet. If the original star was large enough the densely packed ball is called a singularity, the core of a black hole, but if it was not it would become either a neutron star or a dwarf star.
Even without a supernova, a black hole will form any time there is a lot of matter in a small space, without enough energy to act against gravity and stop it from collapsing.
If supernovas are so bright, why do we not see them often? Actually, there are usually hundreds of years between naked-eye super nova sightings. It is because the period of being a super nova in a star life cycle is only a few hours out of the billions of years in a star's life span. The probability (chance) of looking at a star in sky and that being in super nova state is equal to the ratio of an hour over several billion years.
It is worth mentioning that all of the heavier materials like carbon, oxygen, all the metals, etc, that make the life on the earth possible and are ingredients of all living creatures, can only form in the extreme pressure at the center of a super nova. So we are all a remnant ash from one exploding star several billion years ago.
Black holes have also been found in the middle of every major galaxy in the universe. These are called supermassive black holes, and are the biggest black holes of all. They formed when the Universe was very young, and also helped to form all the galaxies.
Some black holes are also responsible for making things called quasars. A quasar occurs when a black hole consumes all the gas surrounding it. As the gas gets close to the black hole itself, it heats up from a process called friction, and glows so brightly that this light can be seen on the other side of the Universe. It is often brighter than the whole galaxy the quasar is in. When astronomers first found quasars, they thought they had found objects close to us. After using a measuring technique called red shift, they discovered these quasars were actually very far away in the universe.
## What black holes look like
At the middle of a black hole, there is a really small thing called a singularity, but it is impossible to see it because light gets sucked into it, and not reflected. Around the tiny singularity, there is a large area where light which would normally pass by gets sucked in as well. The edge of this area is called the event horizon. The gravity of the black hole gets weaker at a distance. The event horizon is the place farthest away from the middle where the gravity is still strong enough to trap light. The singularity is like the pipe under a sink, while the event horizon is like the edge of the drain where water always gets sucked in.
File:Accretion
A black hole pulling off the outer layer of a nearby star. It is surrounded by an energy disk, which is making a jet of radiation
Outside the event horizon, light and matter will still be pulled toward the black hole. If a black hole is surrounded by matter, the matter will form an "accretion disk" (accretion means "gathering") around the black hole. An accretion disk looks something like the rings of Saturn. As it gets sucked in, the matter gets very hot and shoots x-ray radiation into space. Think of this as the water spinning around the hole before it falls in.
Most black holes are too far away and small to see the accretion disk and jet. The best way to know a black hole is there is by seeing how stars, gas and other things behave around it. With a black hole nearby, even objects as big as a star move in a different way, usually faster than they would if the black hole was not there.
Also, because black holes can bend light passing by, if a black hole passes between us and a source of light very far away, the light will become quite distorted, like a fun-house mirror at a circus, until the black hole moves out of the way. The light can also be magnified, like a magnifying glass, allowing scientists to see things farther away (this is called gravity lensing).
We cannot actually see black holes; one way of detecting them is to look at the sky when a black hole passes between us and a source of light, the light bends around the black hole creating a mirror image, so when astronomers see patches of sky that are identical, they may have found a black hole.
A lot of science fiction writers use black holes in their stories, and many scientists wish to find one relatively close to Earth to study one better.
Scientists also think black holes might cause wormholes, theoretical "portals" through space.
## Singularities
Singularities are what scientists believe to be inside of a black hole. These singularities present a huge problem to black hole theorists because they present fundamental flaws in Einstein's work. This is because when you use the equations to work out what a singularity is, the answer is infinite time and space, which makes no scientific sense.
It was commonly thought that a black hole was there forever and if you were trapped in one, you would be stuck there eternally. This was proved wrong when scientists discovered small leaks in black holes. This meant that over billions and billions of years, a black hole would disperse. The leaking is slow at first, but as it gets smaller, it gets faster and faster.
## Black hole evaporation
Stephen Hawking discovered the evaporation of the black hole. He discovered this by quantum field theory in a static black hole background. The result of his calculations is that a black hole should emit particles in a perfect black body spectrum. This effect has become known as Hawking radiation.
## References
1. Davies, P. C. W. (1978). "Thermodynamics of Black Holes". Rep. Prog. Phys. 41: 1313–1355.
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