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# TEXTBOOK SOLUTIONS FOR Problem Solving and Programming Concepts 9th Edition
• 371 step-by-step solutions
• Solved by publishers, professors & experts
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Over 90% of students who use Chegg Study report better grades.
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PROBLEM
Chapter: Problem:
Susan has an audiotape collection that she wants to catalog. She would like to create the following reports:
a. An alphabetical listing of the names of the tapes and associated information (artist and favorite listening place)
b. An alphabetical listing of the artists and their tapes
c. A listing of the tapes organized by where she most likes to listen to them (the car, bedroom, family room, while walking or outside)
d. A listing of all music selections and the tape on which each is found Plan an RDBMS for this problem.
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 4
To plan an RDBMS for Susan. She wants to make a catalog in which she wants to make alphabetical listing of tapes according to name and artist name. She also wants to make a list of artist and his/her tapes. She also wants a listing of favorite places according to tapes she prefer to listen there. She also wants to make a list of all music selections such as jazz, sufi according to the tapes on which they are found.
RDBMS stands for relational database management system, most important concept of RDBMS is normalization, list of reports , query and schema development.
Database normalization involves processing the tables in such a way that the redundancy and dependency, these two factors are minimized. It generally involves dividing tables into smaller tables and thus defining relationships between these newly constructed tables. It helps in easy updating, deleting and adding data.
There are three forms of normalization.
1. First normalization
2. Second normalization
3. Third normalization
Codd defined the first normalization. It handles the update anomaly also called as avoiding unflat tables.
For example in a skill table there are three fields name, skill, address has to be updated a number of times as a person can have more skills thus 1 address for his/her each skill. By first normalization we endeavor to remove such anomalies.
• Chapter , Problem is solved.
Corresponding Textbook
Problem Solving and Programming Concepts | 9th Edition
9780133001785ISBN-13: 0133001784ISBN: Authors: | 516 | 2,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2016-07 | latest | en | 0.942526 |
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Post Patron
Hi,
I have this measure -> GP = - 1800 + SUM(vw_PivotVisitsInvoiced[Income Generated])
Calc Column -> Income Generated = vw_PivotVisitsInvoiced[Visits_NoHrs] * 6.25
Why is my total below not right?
Thanks,
Gerry
1 ACCEPTED SOLUTION
Microsoft Employee
@android1
Your GP is a measure. At the grand total, it will just sum all data together first, then minus 1800. If you want your expression be evaluated on each row, you should use SUMX().
`GP =SUMX(vw_PivotVisitsInvoiced,vw_PivotVisitsInvoiced[Income Generated]-1800)`
See sample below:
Regards,
5 REPLIES 5
Microsoft Employee
Community Champion
what are you expecting the number to be and why is it wrong? Your snip has no context for us to determine the source of the error.
Proud to be a Super User!
Post Patron
Hi kcantor,
Was hoping the total of GP, €3157.81 would be -€16642.2 (Ths sum of the GP totals)
Microsoft Employee
@android1
Your GP is a measure. At the grand total, it will just sum all data together first, then minus 1800. If you want your expression be evaluated on each row, you should use SUMX().
`GP =SUMX(vw_PivotVisitsInvoiced,vw_PivotVisitsInvoiced[Income Generated]-1800)`
See sample below:
Regards,
Post Patron
Hi Simon,
Thanks very much.
See the GP column. 1800 is subtracted from Income Generated each week. (The total of -1000 is correct)
How would I show GP to start at -1800 & add Income Generated to it each week.
So in attched,
I would like to see:
Wk 10 GP = -1712.5
Wk 11 GP = - 1603.12 ( - 1712.5 +109.38)
Wk 12 GP = - 1499.99 ( - 1603.12 + 103.13)
Gerry | 511 | 1,765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-30 | latest | en | 0.856215 |
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# Method for Setting the Resistive Reach of
## Quadrilateral Characteristics of Distance Relays
Elmer Sorrentino Eliana Rojas Jesus Hernandez
Universidad Simon Bolivar, Venezuela ABB, Venezuela Seneca, Venezuela
elmersor@usb.ve eliana.rojas@gmail.com jesusr.hernandez@laedc.com.ve
Abstract-A new method for setting the resistive reach on -The time delays for zones 2, 3 and 4 are assumed to be
quadrilateral characteristics of distance relays is proposed in predefmed and fixed. Additionally, it is considered that there
this article. The method is based on: a) analysis of the are not teleprotection schemes (communication-assisted
impedances seen by the relay (apparent impedances); and b) trip), breaker-failure protection (50BF), line differential
explicit definition of the protection desirable attributes for each protection (87L), or other line protection functions. These
analyzed zone (selectivity, sensitivity). In the proposed method,
considerations affect the criteria for setting the relay reaches.
the resistive reach setting is calculated assuming that the
reactive reach setting has been previously defined. The -The proposed method for setting the resistive reach (R0
proposed method was applied in an example with 18 distance can be adapted to the different ways of setting XR. The
relays and its solution was compared with a conventional criteria used in this article for XR setting are relatively
simplified solution. The conventional simplified solution complex. This helps to better explain the proposed method.
consists in setting the resistive reach by multiplying the reactive
reach by a constant factor. The result obtained with the B. Analyzed quadrilateral characteristics
proposed method is different since there is not a constant factor The quadrilateral characteristics may have different
among the settings of the reactive and resistive reaches. shapes; Fig. I shows the first quadrant for 3 cases. It will be
assumed that the settings are defmed by the first quadrant in
Index Terms-Distance relay setting, Quadrilateral the R-X plane. For the sake of simplicity, shape of Fig. la
developed here can be applied to other characteristic shapes
I. INTRODUCTION
(such as those shown in Figures Ib and lc) by using an
Traditionally, the distance relay zones have been set adaptation to the geometry of each particular characteristic.
according to simple rules [1-4]. The non-traditional options
can be grouped according to their conceptual basics: based
on expert systems, mathematical optimisation, adaptive
protection or probabilistic methods [5]. The well-known
traditional setting rules have been developed to have a
specific reactive reach for solid faults [1-5]. In the case of
relays with quadrilateral characteristics, the reactive and
resistive reaches can be set independently. For these relays,
it is desirable to defme the resistive reach by an analysis of
faults through impedance. Some traditional setting methods
consider faults through impedance, but usually just a typical Fig. I. Examples of different shapes for quadrilateral characteristics.
resistance value is considered [1,2]. Such methods usually C. Setting the zone 1 resistive reach
do not consider that the apparent impedances are affected by C.l. Criterion used/or the reactive reach
diverse factors [6]. A still more simplistic option is to set the The first criterion states that zone I only has to operate for
resistive reach by multiplying the reactive reach by a faults on the line since this zone is instantaneous. Zone I
constant factor [7]. should not operate for faults at the remote bus, by
method for setting the resistive reach on quadrilateral the reactance of the protected line (X L+): XR1= 0.8 XL+.
characteristics of distance relays. The proposed method is
based on: a) analysis of the impedances seen by the relay C.2. Criterion used/or the resistive reach
(apparent impedances); and b) explicit definition of the According to the previous paragraph, zone 1 resistive
protection desirable attributes for each analyzed zone reach (RRI) must be set in a way that assures that the relay
(selectivity and sensitivity). first zone will not trip for faults at the remote bus.
Considering the effect of the fault resistances (~) on the
II. PROPOSED SETIING CRITERIA apparent impedance (ZAP), there are 3 cases:
a) Faults at the remote bus whose ZAP tends to fall within
A. Preliminary basic considerations zone I (Fig. 2a). To have a safety margin, the resistive
-It is considered that the reactive setting (XR) has been setting will be limited to the value of the real part of ZAP
evaluated previously by traditional rules. Such rules are not (RRI-A) where the imaginary part of ZAP is 90% of XL+.
universal, especially for the delayed zones; due to this fact, b) Faults at the remote bus whose ZAP tends to be parallel
the rule used for each zone in this article will be specifically to XR1 (Fig. 2b). It will be assumed that the possible error of
described here. measurement of the relay is proportional to the ZAP module.
For this reason, when the ZAP imaginary part minus 5% of
ZAP module is 85% of X L+, the corresponding real part of setting for the zone 2 reactive reach (X Z-MIN-Z) will be
ZAP limits the resistive setting (~I-B)' computed as 110% erx., (Xz_MIN_z=I ,IXL+)'
c) Faults at the remote bus whose ZAP tends to separate of -If X Z-AVG is greater than X Z-MIN-Z, there is not conflict
X RI (Fig. 2c) : RRI is not limited by ZAP. between those values and the setting is: XR2=XZ-AVG.
RRI setting will be the smaller value of ~I-A and RRI-B, if -IfXz-AVG is less than XZ-MIN-Z, it will be assumed that it
both situations can happen. If the RRI setting is not limited is not possible to guaranty selectivity with that short
by ZAP, RRI could be set to a very high value. adjacent line and the setting will be: XR2=XZ-MIN-Z.
Actually, the solution for these cases is to implement a
unit protection scheme for the short adjacent line (line
a
differential and/or a scheme with teleprotection) and/or
a change in the zone 2 delay for the line in study . The
If .Im{ZAP}=0,9XL+, analysis of such solutions is beyond the scope of the
=> RR I .A = ~e{ ZA P } present work .
## ~:: : : ".':') .------------~
protected line
- - - - - - - - - - - - -- , If [.Im{ZAP }-0,05IZAPI1=0,85XL+, R
XR1=0,8XL+ :
I
=> RRI .B =~e{ ZA P } RZ.MIN.Z RZ.MIN.l
I
I
I
j) Zona I of the adja cent
line, as it is viewed by
c the analyzed relay
RRI is not limited by ZAP RZ· MAX=0' 9RZ·RI·ADY IFaults out of. the I
protected lme
R
R Fig. 3. Limits for the settingof the zone2 resistive reach.
Fig. 2. Criteriaused for settingthe zone 1 resistive reach.
D.2. Criterion usedfor the resistive reach
D. Setting the zone 2 resistive reach The setting criterion of zone 2 resistive reach (RR2) is
D.l. Criterion usedfor the reactive reach similar to the criterion described for X R2 . A desirable
It will be considered that the main objective of zone 2 is to sensitivity will be defined to cover faults in 100% of the
cover the sector of the line that is not covered by zone I . protected line, with the typical fault resistance value (R F-TyP)
This implies that the reactive reach should be set to cover multiplied by a safety factor (F sl) . An allowable minimum
more than 100% of the protected line impedance, in order to sensitivity will be defmed using a smaller safety factor (Fsz):
guaranty sensitivity for internal faults . This criterion is Fsz<F sl; RFI=(Fsl)RF_TYP; RFZ=(Fsz)RF-TYP (thus, RFZ< R FI).
frequently used; however, it is usually necessary to take The desirable and allowable minimum settings (R Z-MIN-I
precautions that guaranty selectivity when there are adjacent and R Z-MIN-Z, respectively) correspond to the real part of the
short lines at the remote bus . This is because the beginning apparent impedance seen by the relay for faults at 100% of
of zone 2 of the relay of the adjacent short line could overlap the protected line, with the fault resistances that consider the
with the zone 2 of the relay in study . The setting of the zone aforementioned safety factors (R FI and R FZ, Fig. 3a).
2 reactive reach (Xd will be done thus: The desirable maximum setting for the zone 2 resistive
-The desirable minimum setting for the zone 2 reactive reach (R Z-MAX) is evaluated with resistive faults at the
reach (X Z-MIN-I) will be computed as 120% of the reactance beginning of adjacent lines that are out of the zone I of the
of the protected line: XZ-MIN-I = 1.2 XL+. adjacent line. The maximum fault resistance (Rp-L1M-ADJ) that
-The desirable maximum setting for the zone 2 reactive is able to see the relay of the adjacent line at the beginning
reach (X Z-MAX) will be computed as 80% of the total of its zone 1 will be found. With this fault resistance, the real
reactance seen by the relay for a fault at the beginning of part of the apparent impedance seen by the relay of the line
zone 2 of the adjacent line protection at the remote bus in study, for faults at 100% of the protected line (RZ-RI-ADY),
(XL+,ADJ,SHORT). The case with the smaller additional will be computed: it will be considered that the desirable
reactance will be used: X Z-MAX = 0.8 (X L++0 ,8XL+, ADJ,SHORT). maximum adjustment corresponds to 90% of this value (R z_
-If X Z-MAX is greater than X Z-MIN-" then there is not conflict MAX=0.9 RZ-RI-ADY, Fig. 3b). The algorithm to set RR2 is
between those values and the setting will be : X R2 = XZ-MIN-I . similar to the described one for X RZ:
-If X Z-MAX is less than XZ-MIN-" the desired sensitivity is -If RZ-MAX is greater than RZ-MIN- h then there is not conflict
not possible without a lack of selectivity, and : between those values and the setting will be: RR2=Rz-MIN-I.
-The average of the previous values will be computed -If RZ-MAX is less than RZ-MIN-" the desired sensitivity is not
(XZ-AVG=[XZ-MIN-I+XZ-MAX]/2). The allowable minimum possible without a lack of selectivity, and:
-If RZ- MAX is greater than RZ-MIN-Z, then there is not of the apparent impedance (ZAP) seen by the relay in study is
conflict between those values and the setting will be: 110% of XR3, or if the imaginary part of the apparent
RR2=Rz-MAX • impedance minus 5% of the ZAP module is 105% of XRJ , the
-If RZ-MAX is less than RZ-MIN-Z, then it is not possible to corresponding real part of ZAP will be the value of resistive
guaranty selectivity for some values of fault resistance reach (Rd. As zone 1, if the RR3 setting is not limited by
and the setting will be: RR2=Rz-MIN-Z. ZAP, then RRJ could be set to a very high value.
In both setting criteria, if the first condition is satisfied jX (FINFEED)ZL+,AD.J
~
I ZAP(varyIng
. R)
F
I
(X Z-MAX>XZ-MIN-" or RZ-MAX>Rz-MIN-I), then a different action
could be taken, in order to increase still more the zone 2 XR3=O,8(X L++(f' INFEED)XR2-ADJ,SHORT)
sensitivity. For example, in such cases the setting could be
If:
the maximum value instead of the minimum, or an average (Im( ZAP}=I,IXR3)
of both values. An analysis of those options is outside the OR
scope of the present work; however, Fig. 4 helps to illustrate ([Im( ZAP} -O,051ZAPI]=I,05XR3),
R
this concept. In the example of Fig. 4, it is assumed that the ~ RR3=>\'f(ZAP}
resistive setting (Rd has been limited by the allowable
minimum sensitivity (RZ-MIN-Z) ' In such case, an increase of Fig. 5. Criteria used for setting the zone 3 resistive reach.
the reactive reach sensitivity (to use XR2-CASEZ, instead of
XR2-CASEI) would imply a greater lack of selectivity for III. SYSTEM USED AS EXAMPLE
resistive faults in the adjacent line if the fault is out of the
A. Power system description
zone 1 of the adjacent line protection.
Figure 6 shows the power system used as example and its
data are in Tables I, II and III.
C I:26km
jx XR2-CASE2
~~ ~~~~~~~~~~ -~ - ~ ~ ~ ~ ~ ~ ~ B Increase er x, implies
~
, XR2-CASEI : a lack of selectivity if
ZL+ : ZAP .ISIn. th'ISregIon
.
eL+ i R
RR2-MIN-2 LCA C I:3.37km
PLM
Fig. 6. Power system used as example (II5kV).
Fig. 4. Example of a lack of selectivity by increasing XR2 sensitivity .
E. Setting the zone 3 resistive reach TABLE I: LINE PARAMETERS (r,x in Q/km; b in umho/km),
£.1 . Criterion usedfor the reactive reach r, x+ b+ ro Xo bo
It will be assumed that the main objective of zone 3 is to Cl 0.1211 0.4959 3.347 0.3160 1.102 1,938
operate as backup protection for faults in adjacent lines [8]. C2 0.1714 0.4928 3.421 0.3630 1.151 1.860
However, selectivity between zones 3 of different lines will TABLE II ' EQUIVALENT GENERATOR DATA
have priority because zone 3 is the faster backup function. X+=K(Q) Xo(Q) P(MW) Q(MVAR)
This criterion presupposes that the faults non-covered by a LCA 7.3 3.3 Slack Slack
zone 3 as backup will be covered by its zone 4, that is more GUA 15.9 15.9 120 74.37
sensitive (zone 4 has a greater reach or it is simply a LM 120,0 53.0 20 12.39
directional function). TABLE Ill' Loxn DATA
Zone 3 reactive reach (XR3) should be set at 80% of the LCA LM LA PMT LR PLM GUA
lowest total apparent reactance seen by the relay in study for P(MW) 73 48 31 38 56 38 30
faults at the end of zone 2 of the relays that protect adjacent cos(<!» 0.900 0.900 0.900 0.900 0.936 0.900 0.850
lines. Worst case combines the smaller zone 2 reactive reach
of the relays of the adjacent lines (XR2-ADJ,SHORT) with the B. Ground distance function description
smaller infeed (FINFEED, due to the current contributions at The apparent impedance seen by ground distance function
the remote bus): XR3=0.8 (XL++ (FINFEED) XR2-ADJ,SHORT). (ZPh-G) depends on its polarisation method [9]. It is assumed
To fmd the previous value may not be simple. For the sake that the relay uses the following form of polarisation:
of simplicity, XRJ was set to 75% of the smallest total ZPh-G = VPh-G / (IPh+KoI~ (I)
apparent reactance for faults at the end of the adjacent lines VPh-G: Phase-to-earth voltage of the faulted phase.
to the remote bus: XR3 = 0.75 (XL++XAP-ADJ-LOWEST). Base Iph: Current of the faulted phase.
Case of load flow was used; therefore, the FINFEED values IR: Residual current (IA + IB + Ie).
correspond to the Base Case. Ko: Residual compensation factor.
It is assumed that Ko is set exactly to see the positive
E.2. Criterion usedfor the resistive reach sequence line impedance for solid faults:
Zone 3 resistive reach (RRJ, Fig. 5) is set similarly to RRI. Ko = (ZLO - ZL+) / (3 ZL+) (2)
The resistive faults were computed at the end of the same ZL+: Positive sequence line impedance.
adjacent line used for the XR3 setting. If the imaginary part ZLO: Zero sequence line impedance.
c. Pre-fault loadflow The settings obtained for zone 2 are in Table VI. RR2 is, in
With fault resistance (R F), apparent impedance depends on some cases, less than RRI (it happens at bus 2 of the lines
the pre-fault load flow, measured in the relay locality [6],[9]. LM-LA, PLM-LR, LCA-PLM, LCA-LR). This result is
The determination of the worst possible condition for each illustrated in Fig. 7, it is not conventional and it happens
zone of each relay is outside the scope of the present article. since RR2 must be limited to reduce the risk of lack of
By simplicity, a simple preliminary analysis of the system in selectivity (since X R2 is greater than X R1).
study suggested the use of the following conditions of pre- The settings obtained for zone 3 are in Table VII. RR3 is,
fault load flow: in some cases, less than RR2 (it happens at bus 1 of the lines
-Base Case: It is the system described in section III-A. GUA-LM, LA-PMT, and at bus 2 of LA-PMT, LR-PMT).
-Case 1: It is the Base Case without one transmission line. This result is similar to the result described for zone 2. If RR3
For the system in study, the approximated load flow values is less than RR2' then zone 2 is more sensitive for faults with
at the relay localities are in Table IV (QMAX). fault resistance. In such cases, as zone 3 is not so sensitive,
-Case 2: This case is as Case 1 and, additionally, this case the backup function for faults with a high value of fault
presupposes that the system operators can control the resistance is the zone 4. By this reason, zone 4 would have
reactive power flow. For the system in study, a half of the to be sufficiently sensitive in these cases.
previous reactive power value was assumed; the values are In the cases where RR2 is less than RRI (or RR3 is less than
in Table IV (QMIN). RR2), this nonconventional result could be avoided if the
These cases were used thus: a) For setting zone 1, Case 1 desired condition were imposed. Such condition is: the
was used when the pre-fault load flow is positive and Case 2 greater the value of the reactive setting, the greater must be
when it is negative; for the system in study, by coincidence, the value of the zone resistive reach. There are different
P and Q have the same sign in the simulated cases. b) For ways for imposing such condition. Special care should be
setting zones 2 and 3, Base Case was used. taken to update the resistance setting values since there are
dependences among the reach settings; for example, Fig. 3b
D. Typical groundfault resistance value shows how the RR2 setting of a relay depends on the value of
The ground fault resistance value depends on multiple RRI of the relay of an adjacent line.
factors. Each R F value has a probability of occurrence [10];
however, a typical value is required in the present work in TABLE IV: PRE-FAULTLOAD FLOW, AT THE LOCALITY OF THE RELAYS IN
order to compute the desirable zone 2 reach. Such value was STUDY, FOR THE CASES 1 AND 2 (QMAX AND QMIN, RESPECTIVELY)
supposed arbitrarily (R F_TYP=50). Using safety factors, the Line Line Prefault load flow (line in study)
results are: RFl=(Fsl)RF-TYP=200; RF2=(Fs2)RF-TYP= 100. in out of PandQ P OMAX OMIN
## study service direction MW MVAR MVAR
IV. RESULTS FOR THE RELAY SETTINGS GUA-LM PLM-LCA GUA->LM 90 56 28.0
The settings obtained for zone 1 are in Table V. The X R1 LM-LA PMT-LR LM->LA 70 33 16.5
values are identical to those obtained in another study [11] LA-PMT PMT-LR LA->PMT 38 18 9.0
since the used criterion is exactly the same. However, in that LA-PMT LM-LA LA->PMT -31 -15 -7.5
study a unique factor of RIJX R was used (RIJXR=2). Table 5 PMT-LR PMT-LR LR->PMT 70 33 16.5
shows clearly that RIJX R is not constant with the developed LR-PLM PLM-LCA PLM->LR 87 36 18.0
method: RIJX R varies between 0.77 and 33.41.
LR-PLM LCA-LR PLM->LR -38 -18 -9.0
On the other hand, the maximum value of fault resistance,
PLM-LCA LCA-LR LCA->PLM 125 54 27.0
at the remote bus, for which the adjustment of the RRI was
defmed, varies between 0.970 and 17.390. However, there LCA-GUA GUA-LM LCA->GUA 90 56 28.0
are 5 cases where the resistive reach was limited by the line LCA-LM GUA-LM LCA->LM 68 33 16.5
thermal capability and not by the apparent impedance locus. LCA-LR PLM-LCA LCA->LR 125 54 27.0
TABLE V : ZONE SETTINGS. THE FAULT RESISTANCE THAT DEFINED THE RESISTIVE REACH SETTING (FIG. 2) IS SHOWN. VALUES IN PRIMARY OHMS.
Line Bus 1 Bus 2
(Bus 1-
Bus2J XR1 RRl R F MAX RR1/ XRl XR1 RRl R F MAX RR1/ XRl
GUA-LM 10.31 12.11 3.86 1.17 10.31 18.06 5.56 1.75
LM-LA 4.81 7.75 8.71 1.61 4.81 88* N/A 18.30
LA-PMT 2.41 4.39 4.98 1.82 2.41 5.06 5.92 2.10
LR-PMT 2.71 4.95 5.67 1.83 2.71 88* N/A 32.49
PLM-LR 1.34 2.98 2.70 2.23 1.34 104.79 17.39 78.38
LCA-PLM 3.95 8.30 8.62 2.10 3.95 132* N/A 33.41
LCA-GUA 10.31 16.60 6.16 1.61 10.31 7.89 0.97 0.77
LCA-LM 8.73 14.17 8.77 1.62 8.73 132* N/A 15.12
LCA-LR 3.98 8.39 8.54
. .
2.11 3.98 132* N/A 33.17
(*): These values were only limited by the line thermal capability (Fig. 2c).
TABLE VI: ZONE 2 SETIINGS (SEE FIG. 3). VALUES INPRlMARY OHMS. the variations in the results. On the other hand, the effect of
Line Bus 1. Bus 2 the inclusion of more cases for the pre-fault load flows
(Bus 1.- should be studied .
Bus2J X R2 RR2 X R2 RR2
GUA-LM 14.82 25.28 15.47 46.42 REFERENCES
LM-LA 6.98 15.75 7.21 40.39 [I] Areva T&D, "Network protection & automation guide," 2002.
LA-PMT 3.61 42.03 3.61 27.97 [2] G. Ziegler , "Numerical distance protection . Principles and
4.06 11.14 3.92 88* application s," Siemens AG, 1999.
LR-PMT
[3] ABB, "Protective relaying. Theory and applicat ions," Marcel Dekker
PLM-LR 2.01 19.34 2.01 31.33 Inc, 1994.
LCA-PLM 5.47 13.77 5.93 42.20 [41 R. Mason, 'T he art and science of protective relaying," John Wiley &
Sons Inc, 1956.
LCA-GUA 14.47 31.42 14.47 60.14 [5] V. De Andrade, E. Sorrentino, "Revision bibliografica sobre los
LCA-LM 12.83 22.19 12.49 132* metodos para ajustar el alcance de los reles de distancia" (in
LCA-LR 5.51 14.62 5.97 76.59 Spanish), Proceedings of the I CVREE , Lecheria, Venezuela, 2007.
[6] T. Rodolakis , D. Crevier , "Effect of loads, shunts and system
(") : These values were only limited by the line thermal capability.
uncerta inties on short circuit relay settings," IEEE Trans. on PAS,
Dec. 1981, pags, 4701-4709 .
.I·x X R2 [7] ABB Relay, "Distance Relay Type Razoa," 1985.
i~: - - - - - - - - - - - - - - - - - - - - - - - - - - : [8] S. Horowitz , A. Phadke, "Third zone revisited," IEEE Trans. on
PWRD,Jan. 2006, pags. 23-29 .
~
- - - -x;~ - - - - - - - - - - - - - - - - - - - - t - - - - - - - -: [9] E. Sorrentino , "Polarizacion de la funcion de distancia ante fallas a
tierra y su efecto sobre el alcance resistivo en zonas cuadrilateras" (in
I
I
I
I
Spanish), Proceedings of the XII ERIAC, Foz do Iguazu, Brasil,
8 L+ : : R 2007.
[10] J. Barnard, A. Pahwa, "Determination of the impacts of high
R R2 R RI impedance faults on protection of power distribution systems using a
probabilistic model ," EPSR, 1993, pags. 11-18.
Fig. 7. Example of a nonconventional result: XR2 can be smaller than XRl. [11] E. Rojas, "Coordinacion de las protecciones de distancia del sistema
a 115 kV de Seneca incluyendo El Guamache " (in Spanish), Final
TABLE VII: ZONE 3 SETIINGS(SEE FIG. 5), VALVES INPRlMARY OHMS. Project for BEE, Universidad Simon Bolivar, Venezuela, 2007.
Linea Ext. 1. Ext. 2
(Ext. 1.-
Ext.2J XR3 RR3 X R3 RR3
## GUA-LM 18.01 19.58 21.92(1) 132*
LM-LA 7.55 19.88 44.89 88*
LA-PMT 5.18 11.32 7.36 15.32
LR-PMT 5.33 14.28 4.39 39.25
PLM-LR 8.26 43.46 4.19 132*
LCA-PLM 5.57 132* 8.43(1) 132*
LCA-GUA 21.92(1) 50.24 33.16 132*
LCA-LM 22.34 67.41 13.47(1) 132*
LCA-LR 6.05 132* 8.43(1) 132*
(*): These values were only limited by the line thermal capability.
section Il-E, since it was not being possible (the reactances would
be negative). To find them, the smaller reactance ofadjacent lines
was added to XL+ and the total value was multiplied by O.75.
V. CONCLUSION
-A novel method for setting the resistive reach of
quadrilateral characteristics in distance relays was presented.
The method is based on the analysis of the apparent
impedance seen by the relay, and the explicit definition of
the protection desirable attributes for each analyzed zone.
-The proposed method was applied to an example with 18
distance relays and the resistive reaches for 3 relay zones
were calculated. The results obtained with the proposed
method were compared with a conventional simplified
solution. The conventional simplified solution is to set the
resistive reach by multiplying the reactive reach by a
constant factor . The results obtained with the proposed
method are substantially different since there are particular
solutions for each relay location .
-In the future, this work could be complemented of diverse
ways. For example, an analysis of other criteria to set the
reactive and resistive reaches could be done in order to study | 6,954 | 23,560 | {"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} | 2.703125 | 3 | CC-MAIN-2020-10 | latest | en | 0.913902 |
https://jennalaib.wordpress.com/2019/03/18/mathematical-connections-in-a-kindergarten-science-unit/ | 1,571,091,096,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655310.17/warc/CC-MAIN-20191014200522-20191014224022-00506.warc.gz | 551,622,832 | 27,813 | “I’m sorry, we’re only loosely doing math today,” Jessica apologized as I walked into her kindergarten classroom. “We’re deep into our structures unit in science, and I’d like to keep going.”
Yes!
I love the structures unit in kindergarten. There are so many opportunities for students to play with shapes, devise and revise plans, and make mathematical connections. Mathematical thinking is not confined to the “math block” on a teacher’s schedule. Of course, it helps for a teacher to look with a mathematical lens, to ask questions that elicit mathematical thinking, and help students attend to their own mathematical observations about the world.
Yesterday, groups of students were given a photograph of a structure — from the Petronas Towers in Kuala Lumpur, to the Eiffel Tower, to an elaborate tree house and a water tower. They were also given building materials like Magnatiles, foam blocks, Cuisenaire rods, and straws. Students were then tasked with building a replica of their structure using their given building material.
Jessica had already posted photographs of yesterday’s structures on the bulletin board in the hallway. Students had clearly done a lot of mathematical thinking! Look at this group’s replica of the water tower. They attended to those long “legs” of the power, and build them with a square base. These legs supported a round basin, which they recreated with beautifully circular frames. There was a lovely interplay between the angular and round shapes, coming together to form a deceptively simple design.
Another group built a large castle, with 3 turrets that stretched higher than the surrounding wall. They were attending to quantities, to size, to shape. Even from a 2 dimensional image, the students knew that they wanted the surrounding wall to be rectangular.
Today, students would be given the opportunity to create a replica of the structure they worked with yesterday — but with new building materials. Some materials were easy to use for construction, like these foam blocks. These students recreated a castle with three visible turrets, and a large surrounding wall.
Some materials more challenging, like the Cuisenaire rods, which students struggled to keep upright.
I sat down with a group working on the Qutb Minar complex, a UNESCO World Heritage site in Delhi, India that contains both the minaret and the Quwwat-Ul-Islam Mosque.
Yesterday, they had used Magnatiles to build their structure. “It was easy to build. The blocks click,” Makayla explained.
“I made the tower,” Vihaan stated proudly, as he traced the edges.
Today, they were given the straws and links tool that the water tower group had used the day before. The three students sat down together to make sense of the materials, each fiddling with the connectors and the straws until they finally pushed together. I assumed that Makayla, shown here on the left, was making the round base of the tower, and that Eleanor, who was starting with the square base in the upper right corner of the image, was working on rectangular mosque, but it turns out they were building two different interpretations of the same part of the structure.
“The building is round!” Makayla insisted, pointing to the dome of the mosque. “So I’m making a circle!”
“It’s flat!” Eleanor shot back. “It doesn’t need a circle!”
Vihaan shrugged, as he continued to build a long, straight line of straws. “It’s the tower.”
As an adult observer, this felt like such a clear peanut-butter-and-chocolate moment. The circle and the rectangle go together! The base is rectangular, while the dome is a round hemisphere! But the girls couldn’t see it.
“Hmm. How can we make these parts into one structure?”
“I’m making the tower!” Vihaan reminded them.
The girls frowned at one another’s shapes.
“Where do you see something circular?” I asked Eleanor, who was mostly ignoring me to build up her rectangular base.
“It’s mostly a rectangle, so I think it should be a rectangle.”
“Do you see anything that reminds you of a rectangle?” I asked Makayla, who was sheltering her circular base with a protective hand.
“I guess. But the circle is important!”
Creating a replica is all about extracting the important thing. It reminded me of “The Important Book.” The Important Thing about building the Qutb Minar is that it has a tall tower, that it has a small building with flat sides, and that the small building has a round dome top.
I pulled out my phone to examine the kindergarten standards for geometry.
K.G.B.5
Model shapes in the world by building shapes from components (e.g., sticks and clay balls) and drawing shapes.
Well! We were certainly doing that!
Then I saw this:
K.G.B.4
Analyze and compare two- and three-dimensional shapes, in different sizes and orientations, using informal language to describe their similarities, differences, parts (e.g., number of sides and vertices/”corners”) and other attributes (e.g., having sides of equal length).
I wanted to hear students talking about the shapes — what the sides should look like, what the corners should look like, how many sides they need to build, what the shape of the base is, etc. There is also a standard (K.G.1) about describing relative position, which seemed like it might be helpful as students join different constructions.
Makayla continued to build, adding a few straws to her base to make it a larger circle. “It’s too hard to put together,” she lamented of the smaller circumference, which kept breaking under the stress.
Meanwhile, Vihaan and Eleanor teamed up forces. Vihaan released that his long, thin rope of straws — the important thing about the tower is that it has long edges! — might work well in conjunction with Eleanor’s square base.
“Let’s make it longer.”
“No, taller.”
“It needs to be straight.”
“It needs a triangle on top.”
Soon, they had build a tall tower. Meanwhile, Makayla was struggling with her circular base. The straws popped out of the connectors every time she tried to add another layer.
At left is their final product. The had a big tower to the left, like in the picture.
“It’s really tall.”
“It’s big.”
“It’s straight.”
“It has triangles.” Any other shapes? “It has squares.”
The smaller building, the masjid, was harder for them to describe.
“It’s shorter.”
“It’s attached to the tower.” So it’s next to it? “Yes, it is next to it.”
What shapes do you see? “I don’t know. It’s supposed to be like a circle.”
I asked the students whether the magnatiles or the straws were better building materials, and they all agreed that while the Magnatiles were easier to build with, the plastic straw structure looks like more like the structure in their picture.
Meanwhile, across the room, three students were working to build a replica of this structure. (It is the Broadway Tower inWorcestershire, England, the second highest point in the Cotswolds.)
.
The Important Thing about this tower is that it has 3 turrets, and a balcony (see on the right side of the structure, close to the photograph). Originally, the central turret had an extra 15 magnatile squares stacked up in its base. As an adult, I assume that the turrets are actually of equal height, but the students do not know that. I did ask them about the height.
“These two are the same. This one is taller.”
How much taller? “A little.”
How about in your structure? Oh. Hmm…. How can you change it to match the photograph better? We can take some squares out of the middle! Then it will be shorter!
I was impressed with how student were analyzing their shapes, and attending to the geometric features in their structure.
I saw Jessica across the room. She was asking a group about their treehouse made out of cuisenaire rods. She probed their thinking about whether to make their replica flat against the table, or build it up, nicely hitting common core standard K.G.3 without even realizing it. She asked them about some of the important features, and what shape she made. She listened intently to students, with her head tilted and her brow furrowed.
After the lesson, Jessica and I debriefed briefly about all of the mathematical ideas present in the lesson — some inherent, and some because she drew them out. She made beautiful connections that helps students improve their mathematical observations and thinking about their structures.
I looked up the geometry standards once again, and sent Jessica a follow up e-mail with some additional lines of questions — although her intuition was already spot on.
As teachers, we can model how to mathematize the world. (Dan Meyer (@ddmeyer) has been writing about this lately both on twitter and his blog.) Some of the thinking here is explicitly about content (e.g. what shape is the base of the structure) and some of it was about important avenues of mathematical thinking, like the process of revising our work.
What is the value in mathematizing outside of “math class”? When do we draw attention to these mathematical ideas, and how do we improve the clarity of our connections? | 1,924 | 9,032 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | longest | en | 0.967768 |
https://www.hsslive.co.in/2022/05/ap-board-class-10th-maths-chapter-6-progressions-ex-6-1-textbook-solutions-pdf.html | 1,701,828,497,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100575.30/warc/CC-MAIN-20231206000253-20231206030253-00458.warc.gz | 899,492,440 | 44,773 | # HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board
## AP Board Class 10 Maths Chapter 6 Progressions Ex 6.1 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 6 Progressions Ex 6.1 Book Answers
AP Board Class 10 Maths Chapter 6 Progressions Ex 6.1 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 6 Progressions Ex 6.1 Book Answers
## Andhra Pradesh State Board Class 10th Maths Chapter 6 Progressions Ex 6.1 Books Solutions
Board AP Board Materials Textbook Solutions/Guide Format DOC/PDF Class 10th Subject Maths Chapters Maths Chapter 6 Progressions Ex 6.1 Provider Hsslive
2. Click on the Andhra Pradesh Board Class 10th Maths Chapter 6 Progressions Ex 6.1 Answers.
3. Look for your Andhra Pradesh Board STD 10th Maths Chapter 6 Progressions Ex 6.1 Textbooks PDF.
## AP Board Class 10th Maths Chapter 6 Progressions Ex 6.1 Textbooks Solutions with Answer PDF Download
Find below the list of all AP Board Class 10th Maths Chapter 6 Progressions Ex 6.1 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:
### 10th Class Maths 6th Lesson Progressions Ex 6.1 Textbook Questions and Answers
Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
i) The taxi fare after each km when the fare is Rs. 20 for the first km and rises by Rs. 8 for each additional km.
Fare for the first km = Rs. 20 = a
Fare for each km after the first = Rs. 8 = d
∴ The fares would be 20, 28, 36, 44, …….
The above list forms an A.P.
Since each term in the list, starting from the second can be obtained by adding ‘8’ to its preceding term.
ii) The amount of air present in a cylinder when a vacuum pump removes 14th of the air remaining in the cylinder at a time.
Let the amount of air initially present in the cylinder be 1024 lit.
First it removes 14th of the volume
i.e., 14 × 1024 = 256
∴ Remaining air present in the cylinder = 768
At second time it removes 14th of 768
i.e., 14 × 768 = 192
∴ Remaining air in the cylinder = 768 – 192 = 576
Again at third time it removes 14th of 576
i.e., 14 × 576 = 144
Remaining air in the cylinder = 576 – 144 = 432
i.e., the volume of the air present in the cylinder after 1st, 2nd, 3rd,… times is 1024, 768, 576, 432, …..
Here, a2 – a1 = 768 – 1024 = – 256
a3 – a2 = 576 – 768 = – 192
a4 – a3 = 432 – 576 = – 144 .
Thus the difference between any two successive terms is not equal to a fixed number.
∴ The given situation doesn’t show an A.P.
iii) The cost of digging a well, after, every metre of digging, when it costs ? 150 for the first metre and rises by ? 50 for each subsequent metre.
Cost for digging the first metre = Rs. 150
Cost for digging subsequent metres = Rs. 50 each.
i.e.,
The list is 150, 200, 250, 300, 350, ……..
Here d = a2 – a1 = a3 – a2 = a4 – a3 = ……. = 50
∴ The given situation represents an A.P.
iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8 % per annum.
Amount deposited initially = P = Rs. 10,000
Rate of interest = R = 8% p.a [at C.I.]
∴ 𝐴=𝑃(1+𝑅100)𝑛
The terms 10800, 11664, 12597.12, ……. a2 – a1 = 800
Here, a = 10,000 a3 – a2 = 864
But, a2 – a1 ≠ a3 – a2 ≠ a4 – a3 a4 – a3 = 953.12
∴ The given situation doesn’t represent an A.P.
Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
i) a = 10, d = 10
ii) a = -2, d = 0
iii) a = 4, d = – 3
iv) a = – 1, d = 1/2
v) a = – 1.25, d = – 0.25
Question 3.
For the following A.Ps, write the first term and the common difference:
i) 3, 1, – 1, – 3,….
ii) – 5, – 1, 3, 7,….
iii) 13, 53, 93, 133, ……..
iv) 0.6, 1.7, 2.8, 3.9,…
Question 4.
Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
i) 2, 4, 8, 16, …….
ii) 2, 52, 3, 72, …….
iii) – 1.2, – 3.2, – 5.2, – 7.2,……
iv) -10,-6, -2, 2, …….
v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …….
vi) 0.2, 0.22, 0.222, 0.2222, ……
vii) 0, -4, -8, -12, …….
viii) –12, –12, –12, –12
ix) 1, 3, 9, 27,…..
x) a, 2a, 3a, 4a,….
xi) a, a2, a3, a4, …..
xii) √2, √8, √18, √32, …….
xiii) √3, √6, √9, √12, …….
## Andhra Pradesh Board Class 10th Maths Chapter 6 Progressions Ex 6.1 Textbooks for Exam Preparations
Andhra Pradesh Board Class 10th Maths Chapter 6 Progressions Ex 6.1 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 10th Maths Chapter 6 Progressions Ex 6.1 exam preparation. The AP Board STD 10th Maths Chapter 6 Progressions Ex 6.1 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 6 Progressions Ex 6.1 Books State Board syllabus with maximum efficiency.
## FAQs Regarding Andhra Pradesh Board Class 10th Maths Chapter 6 Progressions Ex 6.1 Textbook Solutions
#### Can we get a Andhra Pradesh State Board Book PDF for all Classes?
Yes you can get Andhra Pradesh Board Text Book PDF for all classes using the links provided in the above article.
## Important Terms
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Share: | 1,939 | 5,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-50 | latest | en | 0.875001 |
https://www.foxtrot-fab.ru/solve-transportation-problem-3376.html | 1,623,672,166,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612154.24/warc/CC-MAIN-20210614105241-20210614135241-00476.warc.gz | 705,254,657 | 9,178 | # Solve Transportation Problem
Allocate this minimum of supply/ demand in the place of odd valued ACVs at first in the AT formed in Step-4. In case of same ACVs, select the ACV where minimum allocation can be made. Again in case of same allocation in the ACVs, choose the minimum cost cell which is corresponding to the cost cells of TT formed in Step-1 (i.e. (2012) The Impact of Transportation Cost on Potato Price: A Case Study of Potato Distribution in Bangladesh.
Allocate this minimum of supply/ demand in the place of odd valued ACVs at first in the AT formed in Step-4. In case of same ACVs, select the ACV where minimum allocation can be made. Again in case of same allocation in the ACVs, choose the minimum cost cell which is corresponding to the cost cells of TT formed in Step-1 (i.e. (2012) The Impact of Transportation Cost on Potato Price: A Case Study of Potato Distribution in Bangladesh.
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This is a special kind of the network optimization problems in which goods are transported from a set of sources to a set of destinations subject to the supply and demand of the source and destination, respectively, such that the total cost of transportation is minimized. Arc (i, j) joining source i to destination j carries two pieces of information: the transportation cost per unit, c Considering the above notations, the transportation problem can be stated mathematically as a linear programming problem as: Minimize:. The constraint j in the second set of constraints ensures that the total units transported to the destination j is greater than or equal to its demand. Proposed Approach to Find an Initial Basic Feasible Solution In the proposed approach, an allocation table is formed to find the solution for the transportation problem.
The basic transportation problem was originally developed by Hitchcock in 1941 [1] . That’s why this method is named as Allocation Table Method (ATM) and the method is illustrated below: ・ Step-1: Construct a Transportation Table (TT) from the given transportation problem.
The model we are going to solve looks as follows in Excel.
What is the overall measure of performance for these decisions? Explanation: The SUM functions calculate the total shipped from each factory (Total Out) to each customer (Total In).
In this paper, we consider a class of transportation problems which arises in sample surveys and other areas of statistics.
The associated cost matrices of these transportation problems are of special structure.
In this article, a new approach is proposed to find an initial basic feasible solution for the transportation problems. In this paper, a new algorithm is proposed to find an initial basic feasible solution for the transportation problems.
The method is also illustrated with numerical examples. [21] -[23] , Pandian & Natarajan [24] , Reinfeld & Vogel [25] , Sayedul Anam et al. A comparative study is also carried out by solving a good number of transportation problems which shows that the proposed method gives better result in comparison to the other existing heuristics available in the literature. Network Representation and Mathematical Model of Transportation Problem Generally the transportation model is represented by the network in Figure 1. | 724 | 3,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-25 | latest | en | 0.885331 |
https://mathoverflow.net/questions/379877/question-on-the-algebraic-structure-of-the-set-mathcalp-sum-i-1n-a | 1,643,151,865,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304876.16/warc/CC-MAIN-20220125220353-20220126010353-00665.warc.gz | 444,025,411 | 25,184 | # Question on the algebraic structure of the set $\mathcal{P} = \{ \sum_{i=1}^n a_i x^i = 0, a_i \in \pm \mathbb{P}~or~0\}$
It is well-known that the set $$\mathcal{A} = \bigg\{ x\in \Bbb C: \sum_{i=1}^n a_i x^i = 0, a_i \in \mathbb{Z}\text{ and } n \text{ is a positive integer}\bigg\},$$ is the set of all algebraic numbers, so by a heuristic argument, now let's consider the set $$\mathcal{P} = \bigg\{ x\in \Bbb C : \sum_{i=1}^n a_i x^i = 0, a_i \in \pm \mathbb{P}\cup\{ 0\}\text{ and }n\text{ is a positive integer}\bigg \},$$ where $$\mathbb{P}$$ is the set of all prime numbers. Now the question is: does the set $$\mathcal{P}$$ has a good algebraic structure?
• By "algebraic structure" I guess you mean to say that the first set $\mathcal{A}$ is also a field. The operations on $\mathcal{A}$ transfer computatively to the $a_i$'s. Hence you can check whether $\mathcal{P}$ has an algebraic structure (group, ring, ...) by hand. Or, more likely, you can compute explicit counter-examples to any such structure induced by the usual operations. Dec 28 '20 at 10:36
• Out of curiosity, is there a particular motivation behind considering $\mathcal{P}$? Dec 28 '20 at 10:56
• Can you prove $\mathcal P$ is not the set of all algebraic numbers? For example is $\sqrt{6} \in \mathcal P$? Dec 28 '20 at 12:59 | 451 | 1,311 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-05 | latest | en | 0.812292 |
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# Using accumulate algorithm with a set of pairs
P: n/a I have a set of pairs defined as follow: set< pair > numberOfFreqSlotsToAdd; and I need to iterate through all the elements of the set in order accumulate the second field of the pair (that is double). Is there any way I can use the algorithm accumulate to accomplish this? The following line of code double unit = accumulate( numberOfFreqSlotsToAdd.begin(), numberOfFreqSlotsToAdd.end(), 0.0); of course, won't work. Can you suggest anything similar using the algorithm accumulate? Thanks Francesco Feb 12 '06 #1
5 Replies
P: n/a cesco sade: I have a set of pairs defined as follow: set< pair > numberOfFreqSlotsToAdd; and I need to iterate through all the elements of the set in order accumulate the second field of the pair (that is double). Is there any way I can use the algorithm accumulate to accomplish this? The following line of code double unit = accumulate( numberOfFreqSlotsToAdd.begin(), numberOfFreqSlotsToAdd.end(), 0.0); of course, won't work. Can you suggest anything similar using the algorithm accumulate? template T accumulate (InputIterator first, InputIterator last, T init, BinaryOperation binary_op); Where foreach element: init = binary_op(init,*first); -- TB @ SWEDEN Feb 12 '06 #2
P: n/a In article <11**********************@z14g2000cwz.googlegroups .com>, "cesco" wrote: I have a set of pairs defined as follow: set< pair > numberOfFreqSlotsToAdd; and I need to iterate through all the elements of the set in order accumulate the second field of the pair (that is double). Is there any way I can use the algorithm accumulate to accomplish this? The following line of code double unit = accumulate( numberOfFreqSlotsToAdd.begin(), numberOfFreqSlotsToAdd.end(), 0.0); of course, won't work. Can you suggest anything similar using the algorithm accumulate? I gave an answer to this in comp.learn.c-c++. Here I will give several answers: class Object { }; int main() { typedef std::set > MySet; MySet mySet; double unit = accumulate(mySet.begin(), mySet.end(), 0.0, plus_second() ); } The 'plus_second' functor is defined as follows: template struct plus_second : std::binary_function { typename Pair::second_type operator()( const typename Pair::second_type& x, const Pair& y ) const { return x + y.second; } }; Or something slightly more generic: class Object { }; int main() { typedef std::set > MySet; MySet mySet; double unit = accumulate(mySet.begin(), mySet.end(), 0.0, apply_to_second, MySet::value_type>()); } Which uses something like (though I can't say I like the name much:) template < typename Op, typename Pair > class apply_to_second : std::binary_function< typename Pair::second_type, Pair, typename Pair::second_type > { Op fn; public: typename Pair::second_type operator()( const typename Pair::second_type& x, const Pair& y ) const { return fn( x, y.second ); } }; Whether you use one of the above, the one I gave in the other group, or the simplest one: class Object { }; typedef std::set > MySet; double fn(double x, const MySet::value_type& y) { return x + y.second; } int main() { MySet mySet; double unit = accumulate(mySet.begin(), mySet.end(), 0.0, ptr_fun(&fn)); } Is up to you. I'll leave you with the XP mantra, "use the simplest thing that works, refactor mercilessly." -- Magic depends on tradition and belief. It does not welcome observation, nor does it profit by experiment. On the other hand, science is based on experience; it is open to correction by observation and experiment. Feb 13 '06 #3
P: n/a cesco wrote: I have a set of pairs defined as follow: set< pair > numberOfFreqSlotsToAdd; and I need to iterate through all the elements of the set in order accumulate the second field of the pair (that is double). Is there any way I can use the algorithm accumulate to accomplish this? The following line of code double unit = accumulate( numberOfFreqSlotsToAdd.begin(), numberOfFreqSlotsToAdd.end(), 0.0); of course, won't work. Can you suggest anything similar using the algorithm accumulate? You can use an iterator that gets a first or second pair's member for you. http://groups.google.com/group/comp....b344960588d3ac Feb 13 '06 #4
P: n/a I'm facing a similar problem: I need to transform the set of pairs by scaling the second element (that is a double) by a certain factor. The following line of code: // typedef set SetOfFsDouble; // SetOfFsDouble numberOfFreqSlotsToAdd transform(numberOfFreqSlotsToAdd.begin(), numberOfFreqSlotsToAdd.end(), numberOfFreqSlotsToAdd.begin(), f_gx(bind2nd(std::multiplies(), unit), select2nd())); reports the following error: 1>c:\program files\microsoft visual studio 8\vc\include\algorithm(650) : error C2679: binary '=' : no operator found which takes a right-hand operand of type 'double' (or there is no acceptable conversion) 1> c:\program files\microsoft visual studio 8\vc\include\utility(55): could be 'std::pair<_Ty1,_Ty2> &std::pair<_Ty1,_Ty2>::operator =(const std::pair<_Ty1,_Ty2> &)' 1> with 1> [ 1> _Ty1=UserEquipment *, 1> _Ty2=double 1> ] 1> while trying to match the argument list '(std::pair<_Ty1,_Ty2>, double)' 1> with 1> [ 1> _Ty1=UserEquipment *, 1> _Ty2=double 1> ] while the following approach: transform(numberOfFreqSlotsToAdd.begin(), numberOfFreqSlotsToAdd.end(), numberOfFreqSlotsToAdd.begin(), ::make_pair(select1st() , (bind2nd(std::multiplies(), unit), select2nd())) ); where I try to overcome the previous problem by creating a Pair to give back to the set, reports the following error: 1>c:\program files\microsoft visual studio 8\vc\include\algorithm(650) : error C2064: term does not evaluate to a function taking 1 arguments 1> c:\program files\microsoft visual studio 8\vc\include\algorithm(685) : see reference to function template instantiation '_OutIt std::_Transform::iterator,_Out It,_Fn1,std::_Iter_random_helper<_Cat1,_Cat2>::_It er_random_cat>(_InIt,_InIt,_OutIt,_Fn1,_InOutItCat ,std::_Range_checked_iterator_tag)' being compiled 1> with 1> [ 1> _OutIt=std::_Tree,SinrBasedWithoutPowerScaling::NumberOfFr equencySlotsComparison,std::allocator>,false>>::iterator, 1> _Traits=std::_Tset_traits,SinrBasedWithoutPowerScaling::NumberOfFr equencySlotsComparison,std::allocator>,false>, 1> _Fn1=std::pair>,select2nd>>, 1> _Cat1=std::_Tree,SinrBasedWithoutPowerScaling::NumberOfFr equencySlotsComparison,std::allocator>,false>>::iterator::iterator_category, 1> _Cat2=std::_Tree,SinrBasedWithoutPowerScaling::NumberOfFr equencySlotsComparison,std::allocator>,false>>::iterator::iterator_category, 1> _InIt=std::_Tree,SinrBasedWithoutPowerScaling::NumberOfFr equencySlotsComparison,std::allocator>,false>>::iterator, 1> _InOutItCat=std::_Iter_random_helper,SinrBasedWithoutPowerScaling::NumberOfFr equencySlotsComparison,std::allocator>,false>>::iterator::iterator_category,st d::_Tree,SinrBasedWithoutPowerScaling::NumberOfFr equencySlotsComparison,std::allocator>,false>>::iterator::iterator_category>:: _Iter_random_cat 1> ] Do you have any suggestion on how to solve this problem? Feb 13 '06 #5
P: n/a cesco wrote: I'm facing a similar problem: I need to transform the set of pairs by scaling the second element (that is a double) by a certain factor. Use boost::transform_iterator for that. Feb 13 '06 #6
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Replies have been disabled for this discussion. | 1,885 | 7,462 | {"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} | 3.1875 | 3 | CC-MAIN-2019-43 | latest | en | 0.843451 |
https://fireboatduwamish.org/graphing-sine-and-cosine-worksheet/ | 1,653,237,393,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545875.39/warc/CC-MAIN-20220522160113-20220522190113-00497.warc.gz | 312,407,057 | 11,440 | # Graphing Sine And Cosine Worksheet
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http://mathematica.stackexchange.com/questions/43885/fast-way-of-finding-position-of-interval-containing-an-integer/43890 | 1,466,897,046,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783393997.50/warc/CC-MAIN-20160624154953-00198-ip-10-164-35-72.ec2.internal.warc.gz | 188,083,651 | 24,803 | # Fast way of finding position of Interval containing an Integer
I have a large list (about 2.2 million) of pairs of integers. Each pair defines an Interval[{}]. Given some integer x, I need to find the position of the interval in my list that bounds x (or alternatively, return the element of the list).
None of the intervals overlap, thus x can belong to only one of the intervals. You can assume the large list is sorted. The challenge is that I need to run this "look-up" millions of times, so even a small gain in speed would be very beneficial
I tried a few obvious brute force approaches. First I created a list of intervals from the list of integer pairs, and then tried
Pick[myBigList,
Map[IntervalMemberQ[#, x] &, myBigList]]; //AbsoluteTiming
4.196432
and then
Select[myBigList, IntervalMemberQ[#,x]& ]; // AbsoluteTiming
8.814063
Assuming the list is ordered, then
LengthWhile[mySortedBigList, #[[1]] <= x &]
3.556826
All of these methods seem too slow. Any suggestions would be much appreciated.
(For those of you interested, here's the background on this problem. I need a fast way of translating an IP address to an approximate set of latitude-longitude coordinates. Each pair of integers in the list above corresponds to a range of IP addresses that have been converted from the zz.zz.zz.zz format to an integer. The interval defined by each pair of integers is associated with a set of lat-long coordinates. Again, this needs to be run millions or tens of millions of time on a regular basis. When this look-up was done on a 64-bit Windows desktop system running standard SQL, it took about 8 hours to geocode ten million IPs)
Thanks,
Mark
-
Since you can sufficiently narrow down the lat-long with just the first three octets (sometimes, even with two), can you not create a hash table (downvalues/dispatch) for ~16.7 million IPs (first three octets) to the lat-long coordinates? – R. M. Mar 12 '14 at 18:44
Will a bounding interval exist for every input? – Dr. belisarius Mar 12 '14 at 18:53
Are the intervals "touching", as in $[0,1)$, $[1,4)$, $[4,6)$, etc., or are there gaps inbetween? It sounds like good application for a binary search. – Szabolcs Mar 12 '14 at 19:37
Here is a solution based on binary search (compiled).
### Implementation
First, this is a version of a binary search, which would return the position of a maximal number in a list, smaller or equal to yours, and -1 if no such is found:
ClearAll[bsminComp];
bsminComp=
Compile[
{{lst,_Integer,1},{elem,_Integer}},
Module[{n0=1,n1=Length[lst],m=0,pos=-1},
While[
n0<=n1,
m=Floor[(n0+n1)/2];
If[lst[[m]]==elem,pos=m;Break[];];
If[lst[[m]]<elem,n0=m+1,n1=m-1]
];
If[pos==-1,pos=If[lst[[m]]<elem,m,m-1]];
pos
]
]
Here is then the main function:
ClearAll[makePositionFunction];
makePositionFunction[ints_List]:=
Module[{starts=ints[[All,1]],ends=ints[[All,2]]},
Function[
num,
Module[{},
With[{pos=bsminComp[starts,num]},
If[pos==-1||pos==0,Return[-1,Module]];
If[ends[[pos]]>=num,pos,-1]
]
]
]
];
### How it works
Basically, the strategy is simple: find the lower end of the interval possibly containing your number, using binary search on the lower ends of your intervals. Then, find the upper end of the interval and check if it is larger or equal to the element in question. If yes, return the position found by a binary search. If no, return -1, meaning that the element is not within any of the intervals.
### Tests and benchmarks
ints = Partition[Sort@RandomSample[Range[20], 10], 2]
(* {{1, 3}, {4, 5}, {9, 13}, {14, 15}, {17, 18}}
pfun = makePositionFunction[ints];
pfun /@ Range[15]
(* {1, 1, 1, 2, 2, -1, -1, -1, 3, 3, 3, 3, 3, 4, 4} *)
Larger test:
largeInts = Partition[Sort@RandomSample[Range[10000000],2000000],2];
pfLarge = makePositionFunction[largeInts];
pfLarge/@Range[1000]//Short//AbsoluteTiming
(* {0.018322,{-1,1,1,1,1,1,2,2,2,3,<<980>>,104,104,-1,105,105,105,105,105,105,-1}} *)
### Module vs With - a cautionary tale
You could've noticed that I used Module to localize some of the variables, as:
Module[{starts = ints[[All, 1]], ends = ints[[All, 2]]}, ...]
Question is, why I did not use With (since nothing changes here), and what would happen if I did. You can try yourself, replacing Module -> With, and you will find that the function will still work, but be thousands times slower. You will also find that it would have a huge ByteCount, if you apply ByteCount to the resulting pure function (for large lists).
Now, here is a general caution: when you embed a large enough list (or other expression) into your pure function, using With, you are asking for trouble. The way Mathematica symbolic engine works is that when you apply such a function, it has to apply this huge expression (it is huge because Function makes its internals fully transparent to the evaluator, and if the body happens to contain huge expressions, Function will be a huge expression). By using Module, I delay the actual computation of the list until run-time, and make the body of my Function tiny - which makes all the difference here. This trick saved me more than once, and is IMO essential to know for anyone who wants to routinely use closures in Mathematica.
-
Heh, I was too lazy to implement a binary search function, but I should've remembered that I probably could've found it in quite a few of your answers... would've been simpler to steal it ;) – R. M. Mar 12 '14 at 19:28
@rm-rf Yeah, right - I was luckier since I also happen to use these for work, so all I had to do was to go to the right source file, which was quick :). – Leonid Shifrin Mar 12 '14 at 19:30
@rm-rf Actually, I personally view the last remark on Module vs With as the most valuable part of this answer, since I doubt that many people were aware of this subtlety. And there are real-world cases when this is critical. At the same time, this isn't very easy to discover on your own - I've spent a lot of time more than once figuring this out. This can be particularly bad when you have a large and complex code, because it is very hard to track the slowdowns to pure functions (they don't have a name after all, and don't exist as a part of global state). – Leonid Shifrin Mar 12 '14 at 19:31
I agree. I didn't know that either. I don't think I've run into such a use case, but I know that if I came across a need, I would've definitely used With there. Good to know! – R. M. Mar 12 '14 at 19:34
@rm-rf The second part of this story, which I did not tell above, is that for such pure functions with captured Module - variables (mutable), these variables will be garbage-collected all right, but only if \$HistoryLength was set to 0 before those were produced (in the interactive session), otherwise those variables won't be garbage-collected. IIRC, they are garbage-collected just fine when they are not part of any Out result, in any case. But this is important to know, of course, and we are lucky with this - otherwise, the practical uses for such closures would've been very limited. – Leonid Shifrin Mar 12 '14 at 19:38
An alternative approach, which uses binary search under the hood, is to have a zero-order interpolating function that maps from the jth value in the list to j. I'll illustrate using the same example as from my other response.
SeedRandom[1111];
ll = DeleteDuplicates[RandomInteger[10^9, 3*10^6]];
If[Mod[Length[ll], 2] == 1, ll = Most[ll]];
myBigList = Partition[Sort[ll], 2];
vals = Map[Round[Mean[#]] &, myBigList];
vals2 = RandomSample[vals];
That sets up the data for the example. Next is the off-line preprocessing.
Timing[interp =
Interpolation[Transpose[{Flatten[myBigList], Range[Length[ll]]}],
InterpolationOrder -> 0];]
(* Out[12]= {3.756000, Null} *)
Here we time how long to locate all query points.
Timing[res =
Table[nbr = interp[vals2[[j]]];
If[OddQ[nbr], myBigList[[(nbr + 1)/2]], myBigList[[nbr/2]]], {j,
Length[vals2]}];]
(* Out[13]= {14.288000, Null} *)
Now check that the sorted result is the same as myBigList.
Timing[resb = Sort[res];]
(* Out[14]= {3.644000, Null} *)
resb===myBigList
(* Out[16]= True *)
Somewhat faster than the other method I showed (factor of 2-3 or so). Not sure how it compares to Leonid's compiled explicit binary search.
-
+1. I did not fully optimize my function, though: first, binary search can be compiled to C, and then, if we have many elements in a list (rather than, say, separate function calls for each element), it is possible to write a compiled vectorized binary search (which will also compile the loop over the searched elements) - I've done that before - and also vectorize the top-level closure. If someone else with (too much) free time on their hands would come up with a detailed timing comparisons between all posted solutions, I will make and post the vectorized version :). – Leonid Shifrin Mar 12 '14 at 20:56
@Leonid Shifrin What about setting the compiled search to be listable via RuntimeAttributes? That should handle the vectorizing part more or less for free. – Daniel Lichtblau Mar 12 '14 at 21:00
Unfortunately, upon a superficial look at least, I think this won't work, in such a direct form at least, because we need it Listable in only the second argument, but not the first (which also is a List), while setting RuntimeAttributes -> Listable would affect both arguments. I could compile the binary search with embedded list, first, and then inline / use that in another compiled function which I can make runtime - Listable - this might work. I'll test this when I have more time. – Leonid Shifrin Mar 12 '14 at 21:09
@Leonid Shifrin Right, you'd feed in the pairs list and return a listable compiled function that does binary search of that list on input query points. – Daniel Lichtblau Mar 12 '14 at 21:24
You can use Nearest to find possible pair endpoints, then do post-processing to get the actual intervals.
Here is an example using around 1.5 million intervals.
SeedRandom[1111];
Timing[ll = DeleteDuplicates[RandomInteger[10^9, 3*10^6]];
If[Mod[Length[ll], 2] == 1, ll = Most[ll]];]
(* Out[25]= {0.356000, Null} *)
Length[ll]
(* Out[26]= 2995472 *)
Timing[myBigList = Partition[Sort[ll], 2];]
(* Out[27]= {0.612000, Null} *)
Timing[nf = Nearest[ll];]
(* Out[28]= {10.096000, Null} *)
We'll use the interval midpoints for our sample queries, and shuffle them so that we know order is not playing a role in the speed.
vals = Map[Round[Mean[#]] &, myBigList];
vals2 = RandomSample[vals];
Now for each query we find the three closest neighbors. We could find the two closest but in rare cases where intervals are separated by one unit we might bet the wron result.
Timing[res = Map[nf[#, 3] &, vals2];]
(* Out[45]= {26.920000, Null} *)
Timing[res2 = Sort[Map[Sort, res]];]
(* Out[46]= {1.804000, Null} *)
Now figure out whether the correct interval is from the first two or last two points. This next line is really part of the preprocessing.
Timing[Map[(interval[#] = True)&, myBigList];]
(* Out[90]= {4.664000, Null} *)
Timing[res3 = Map[If[TrueQ[interval[Most[#]]], Most[#], Rest[#]] &, res
2];]
(* Out[91]= {6.476000, Null} *)
Check that we get back that original pair list; the example is rigged to do exactly that.
res3===myBigList
(* Out[93]= True *)
-
Since the integers are all ordered, one approach would be to put them all in a big list and then use a NearestFunction to find the closest one. As a simple example, here's a sorted list of integers and a nearest function nf' based on that list.
sorted = Sort[RandomInteger[{0, 100000}, 500]];
nf = Nearest[sorted];
To find the closest integer to 20000 in the list, use nf[20000], which returns the nearest value. You would then need to do some post-processing to figure out whether the actual match was below, in, or above the relevant interval.
-
Because the value to be looked up can only belong to one interval, a trivial optimization is to use the Select variant which picks only the first element for which IntervalMemberQ is True:
Select[myBigList, IntervalMemberQ[#,x]&, 1]
`
-
This won't change the linear complexity though, and for lists of this size (millions of elements), this is a big deal. – Leonid Shifrin Mar 12 '14 at 20:18 | 3,274 | 12,191 | {"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} | 3.015625 | 3 | CC-MAIN-2016-26 | latest | en | 0.893431 |
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Solve this equation using Mathematica: $2^x + 113^y = z^2$, $\operatorname{gcd}(x, y) = 1$. | 539 | 2,027 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-51 | latest | en | 0.911355 |
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Microsoft Excel is one of the most popular spreadsheet software used in various industries. It is widely used to manage data, perform calculations, and create charts and graphs. One of the essential functions in Excel is subtraction. The subtraction function in Excel is straightforward, but there are many ways to perform it. In this article, we will discuss pengurangan Excel rumus or the subtraction function in Excel in detail.
## What is Pengurangan Excel Rumus?
Source: bing.com
Pengurangan Excel rumus or the subtraction function in Excel is a mathematical operation that subtracts one number from another. It is denoted by the minus sign (-). The subtraction function is used to find the difference between two numbers in Excel.
## How to Perform Pengurangan Excel Rumus?
Source: bing.com
Performing pengurangan Excel rumus or the subtraction function in Excel is easy. You can subtract two or more numbers in Excel using various methods. The most common method is to use the subtraction formula, which is:
= number1 – number2
You can also use the SUM function to subtract numbers in Excel. For example, you can use the formula:
=SUM(number1, -number2)
Another way to subtract numbers in Excel is to use the Paste Special function. To do this, select the cell where you want to paste the result, right-click on it, and select Paste Special. In the Paste Special dialog box, select Values, and then select Subtract from the Operation section. Finally, click OK to subtract the values.
## Using Pengurangan Excel Rumus with Cell References
Source: bing.com
One of the advantages of using pengurangan Excel rumus or the subtraction function in Excel is that you can use cell references instead of actual numbers. This way, you can easily update the values in the cells, and the subtraction formula will automatically update the result.
For example, if you want to subtract the value in cell A2 from the value in cell A1, you can use the formula:
=A1-A2
Similarly, if you want to subtract a range of cells, you can use the SUM function with cell references. For example, to subtract the values in cells A1 to A5 from the value in cell A6, you can use the formula:
=SUM(A1:A5)-A6
## Using Pengurangan Excel Rumus with Absolute Cell References
Source: bing.com
When using cell references in the subtraction formula, it is essential to use absolute cell references to ensure that the formula works correctly. Absolute cell references are denoted by the dollar sign (\$).
For example, if you want to subtract the value in cell A2 from the value in cell A1 and want to copy the formula to other cells, you should use absolute cell references. The formula should be:
=A\$1-A\$2
Similarly, if you want to subtract a range of cells and want to copy the formula to other cells, you can use absolute cell references. For example, to subtract the values in cells A1 to A5 from the value in cell A6 and copy the formula to other cells, you can use the formula:
=SUM(A\$1:A\$5)-A\$6
## Using Pengurangan Excel Rumus with Named Ranges
Source: bing.com
Named ranges are a useful feature in Excel that allows you to give a name to a range of cells. This makes it easy to refer to the range in formulas and functions. You can use named ranges in the subtraction formula to make it easier to read and understand.
To create a named range in Excel, select the range of cells, right-click on it, and select Define Name. In the Define Name dialog box, enter a name for the range, and then click OK.
Once you have created a named range, you can use it in the subtraction formula. For example, if you have named the range A1:A5 as “Sales,” and you want to subtract the sales from the value in cell A6, you can use the formula:
=SUM(Sales)-A6
## Using Pengurangan Excel Rumus with Functions
Source: bing.com
Excel has many built-in functions that you can use to perform complex calculations. You can use these functions in the subtraction formula to make it more powerful and flexible.
For example, you can use the IF function to subtract one value from another based on a condition. The formula for this would be:
=IF(condition, number1 – number2, 0)
You can also use the ROUND function to round the result of the subtraction formula to a specific number of decimal places. For example, to round the result to two decimal places, you can use the formula:
=ROUND(number1 – number2, 2)
## Conclusion
Pengurangan Excel rumus or the subtraction function in Excel is a simple yet powerful function that allows you to subtract one number from another. You can use various methods to perform subtraction in Excel, including the subtraction formula, the SUM function, and the Paste Special function. You can also use cell references, absolute cell references, named ranges, and functions in the subtraction formula to make it more powerful and flexible. By mastering the pengurangan Excel rumus or the subtraction function in Excel, you can perform complex calculations and manage data more efficiently.
## Related video of Pengurangan Excel Rumus: A Comprehensive Guide
##### Mustakim ID
Seorang blogger enthusiasm & starter copy writer. Sangat tertarik dengan perkembangan dunia teknologi.
Articles: 766 | 1,150 | 5,247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-40 | latest | en | 0.824399 |
https://www.coursehero.com/file/p3gpm2/%C2%B2-m-%C2%B2-1-y-t-j-%C2%B2-m-%C2%B2-2-y-tM-2-But-with-breaks-always-in-the-same-place-y-t-k/ | 1,555,900,074,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578533774.1/warc/CC-MAIN-20190422015736-20190422041736-00321.warc.gz | 655,263,731 | 235,957 | # ² m ² 1 y t j ² m ² 2 y tm 2 but with breaks
• Notes
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This preview shows page 6 - 9 out of 11 pages.
² m ² 1 ' , y t , j ² m ² 2 ' ,..., y tM ' , 2 But: with breaks always in the same place, y t ' k would be too correlated with y t ' k ² 1 Solution(!): (1) For k ± 1, set 2 k " 1 and y ' k " 1 to initial guesses (2) Set j ± 0 (3) Draw m ' L Poisson( 5 ) If m ' ² 1 ³ M , set m ± M otherwise, set m ± m ' ² 1 (4) Draw y tj ' k , y t , j ² 1 ' k ,..., y t , j ² m ' k | y , 2 , y ti ' ± y t , i ' k " 1 for i ³ £ j , j ² 1,..., j ² m ¤ and t ± 1,2,..., T (5) Set j ± j ² m ² 1. If j ³ M , go to (6) else, go to (3)
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7 (6) Draw 2 k | y , y ' k (7) Set k ± k ² 1 and go to (2) How to do step (6)? Conditional on y ' , this is a more standard problem Example: Cox-Ingersol-Ross model dy t ± ¡ ) " * y t ¢ dt ² @ y t dW t Let x t ± log y t
8 Ito’s Lemma: if dy t ± 6 y t , t dt ² @ y t , t dW t x t ± f y t with f . twice continuously differentiable, then dx t ± a y t , t dt ² b y t , t dW t a y t , t ± 6 y t , t ² f y ² y y ± y t ² 1/2 ¡ @ y t , t ¢ 2 ² 2 f y ² y 2 y ± y t b y t , t ± @ y t , t ² f y ² y y ± y t Here f y ± log y dy t ± ¡ ) " * y t ¢ dt ² @ y t dW t ² f / ² y ± 1/ y ² 2 f / ² y 2
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• Winter '09
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174
30
In my earlier project I used this device as a IR Transmitter and promised to upload this project description in next instructables.
So here I present you IR Transmitter using 555 Timer.
Last project in which this remote
We want to design a astable multivibrator of 38KHz. This can be done by using 555 Timer.
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Schematic Diagram
In the above circuit, 555 Timer is wired as an Astable Multivibrator. The 100μF capacitor (C1) is used to reduce ripples in the power supply. 1st and 8th pins of 555 are used to give power Vcc and GND respectively. 4th pin is the reset pin which is active low input, hence it is connected to Vcc. 5th pin is the Control Voltage pin which is not used in this application, hence it is grounded via a capacitor to avoid high frequency noises through that pin. Capacitor C2, Resistors R1, R2 determines the time period of oscillation. Capacitor C2 charges to Vcc via resistors R1 and R2. It discharges through Resistor R2 and 7th pin of 555. The voltage across capacitor C2 is connected to the internal comparators via 2nd and 6th pins of 555. Output is taken from the 3ed pin of the IC. Charging time constant of the capacitor (output HIGH period) is determined by the expression 0.693(R1+R2)C2 and discharging time constant (output LOW period) is determined by 0.693R2C2. They are approximately equal.
You can use the RESET pin of 555 for transmitting binary data.
## Step 2: RequirementS
1. 9V battery( I used an old 9V battery)
2. 100uF capacitor ( optional )
3. 0.001uf capacitor
4. 0.1uf capacitor
5. 1 K resistor
6. 100 Ohms resistor
7. 20 K resistor
8. 1 or 2 IR LED's
9. Switch
10. NE555 Timer IC
## Step 3: Finished Product
these are some picture of finished product how it look alike.
I an also adding the PCB layout for those who don't want use wires.
Last project in which this remote
Hope you like it.
Note : Output frequency of above circuit is about 35.2KHz. As per our experiment TSOP1738 is detecting it but you will get more range if you use exact 38KHz. You may also use 18K resistor instead of 20K which will produce 39KHz. Better you can try a preset for accurate 38KHz.
## Step 4: Some Calculation
Since we are using this circuit in Astable mode and we need 38 khz then we have to use R1=1.025k , R2=18.47k and c1= 1nf or we can say 0.001uF. Since we can't get 18.47 k and 1.025 k resistor then we used here is 20k and 1 k resistor after using these resistor we get 35.188 khz. If we will use exact 18 K and 1 k resistors in the circuit it will give 38.992 khz. Since 5th pin is the Control Voltage pin which is not used in this application, hence it is grounded via a capacitor to avoid high frequency noises through that pin. C3=0.01uF will not going to effect on the calculation part. so you can remove it. It totally depends on you.
## Step 5: What Next
Participated in the
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Participated in the
Full Spectrum Laser Contest 2016
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## 30 Discussions
Hello, i am asking question not related to this project but question is regarding ir receiver circuit, which i have made using ic hcf1407, which control electrical appliances but the relay in it us automatically switching on and off how to correct it
Hey, Can I change the sending values of this circuit. And I want to add more buttons.
yes you can change the sending values and also add more buttons but for this you have to use a programmable chip like arduino
hi
i will make it today ... but what will be this range
can u reply me url of ir reciever
For receiving I used TSOP 1738 IR receiver which has around 10mtr. Range.
http://www.micropik.com/PDF/tsop17xx.pdf
You've made some cool projects! Keep it up!
I'm not sure if I under stand this or not but with this would I b about to make a dimmer or even a switch if I get a receiver and enter the right code? BTW I'm a rookie at a lot of this
6 replies
sandy 65 i did all the things tight so why not my circuit is working
Can you do one on a dimmer that shows a digital speed selection number for 110v ac motor
Do one what? 8-/
And 1 More thing why don't you just check my another project on 555 timer IC. I think you will understand the difference of this circuit and that circuit.
Ya, But I can make a dimmer circuit for DC motor of voltage 0-12V. But I will let you know that you can make it for 110V AC motor. First I have to study about the dimmer.
But you can switch ON/OFF your FAN with it
Did you really did that.. but nice joke bro.. After all photo editing I think you learnt something from my project.
Did you really did that
Did you really ask that? ;)
After all photo editing I think you learnt something from my project.
Don't worry, it was just a few minutes, but yes, I learned that obsessive placement of a clothes patch is really distracting and doesn't help at all.
Then I actually looked at your schematic... And learned that it's possible to make an error for every 2 components used.
The 100μF capacitor (C1) is used to reduce ripples in the power supply.
Why do you wanna do that and how did you select that value?
5th pin is the Control Voltage pin which is not used in this
application, hence it is grounded via a capacitor to avoid high
frequency noises through that pin.
Why do you wanna do that?
How did you select the value of R3 (and why not put it in the parts list)?
Do you honestly believe that the poor 555 will put out 250..300mA @9V?
You didn't mention using high power IR-LED(s) (guess you meant 1 to 2 IR-LEDs, rather than half a LED as you wrote), how much current do you think an average IR-LED can handle?
Given the frequency of a 555 is (ln(2) x (R1+2 x R2) x C)^-1, your 38kHz is actually 35.2kHz, which means that the range is roughly halved - What changes will it take to get the frequency right?
If you go read the datasheet of the 555 and that of the IR-LED you used and then try to answer the questions I put forward, there's a good chance that you might learn a thing or two!
Oh, BTW. the TSOP1738 that you want to use at the receiving end is not made for CW
Quoting from the datasheet:
Some examples for such disturbance signals which
are suppressed
by the TSOP17.. are:
• DC light (e.g. from tungsten bulb or sunlight)
Continuous signal at 38kHz or at any other
frequency
• Signals from fluorescent lamps with electronic
ballast (an example of the signal modulation is in the
figure below).
Special IR receivers exists for CW (like e.g. TSSP4038).
To use TSOP17nn, the signal should be modulated like this:
Again from datasheet
The data signal should fullfill the following condition:
• Carrier frequency should be close to center
frequency of the bandpass (e.g. 38kHz).
• Burst length should be 10 cycles/burst or longer.
• After each burst which is between 10 cycles and 70
cycles a gap time of at least 14 cycles is neccessary.
• For each burst which is longer than 1.8ms a
corresponding gap time is necessary at some time in
the data stream. This gap time should have at least
same length as the burst.
• Up to 1400 short bursts per second can be received
continuously.
The effect of this is, that a continuous signal lasting more than 1.84ms is considered noise, and makes the receiver turn its sensitivity more and more down, until it sees a break and that means a serious loss of range (which is up to 35m when used correctly). Add to that, the wrong frequency that steals around half the signal in the first place and I begin to wonder if that's the reason that you torture both the LED and the 555?! | 2,013 | 7,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-51 | latest | en | 0.89773 |
https://answers.yahoo.com/question/index?qid=20091013000015KK05203 | 1,576,134,645,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540537212.96/warc/CC-MAIN-20191212051311-20191212075311-00514.warc.gz | 274,249,329 | 22,306 | # 過去完成進行式,過去式與現在完成進行式的問題，麻煩英文高手!
1.是否過去完成進行式的先決條件也必須要加上過去簡單式來強調背景?
2.如同上面所說 過去完成進行-比過去簡單式更早發生，持續至過去簡單式，有可能會繼續至現在，那位什麼不可以用 現在完成進行式來表示呢?
In the year 1900 just over 4% of the population was aged over 63. However, by 1960 this figure 'had doubled'.
The chart shows the precentage of people aged 65 and over in the USA between 1900 and 2000. In the year 1900 just over 4% of population __was_ (be) aged over 65. However, by 1960 this figure __had doubled_(double).
the number of people aged between 75 and 84 __remainded__(remain) fairly steady between 1900 and 1930,making up only 1-1.3% of the population . The figure__began___ (begin) to rise more significantly in 1940 and by 1970 it __had tripled_____ (triple) to reach 3% of the population.
Although three __was__ (be) no change in the number of people aged 65-74 between 1960 and 1970, the number of people aged 75 and over __increaded___ (increase) during this time.By the year 2000,12.4% of the US population _had reached___(reach) tje age of 65 pr more, although this was slightly lower than in 1990 when it__had peaked_(peak) at 12.6%.
### 2 Answers
Rating
• 1 decade ago
Favorite Answer
有幾個問題想要請教，若是如以下題目 I had been reading when he came at five o'clock yesterday afternoon.
昨天下午五點鐘他來的時候我已經在讀書。
請問:
1.是否過去完成進行式的先決條件也必須要加上過去簡單式來強調背景?
是這樣子沒錯,當兩件事在進行並且在時間上有不同時,相對的在時間先後順序上要區隔開來.he came at five o'clock yesterday afternoon發生在I had been reading 之前
2.如同上面所說 過去完成進行-比過去簡單式更早發生，持續至過去簡單式，有可能會繼續至現在，那位什麼不可以用 現在完成進行式來表示呢?
不也是表示從之前一直延續到現在的動作?
過去完成進行是指過去一個持續動作在他來之前為止 的一個經驗(在過去發生的yeaterday).
現在完成是是只到目前為止的一個經驗,問題是發生在昨天 I had been reading when he came at five o'clock yesterday afternoon.
,不是嗎?
另外請問圖表的問題:
In the year 1900 just over 4% of the population was aged over 63. However, by 1960 this figure 'had doubled'.
請問，為什麼不能用 was doubled? 為什麼有的時候可以用過去式描述但有的時候又必須要用過去完成式呢?不是很懂他的道理在哪?
be+過去分詞(當形容詞)被動語態
have, has和had 是現在,過去完成式 指經驗
by+時間 是指在時間之類(不會超過時間)
by 1960在這一年之內但不會超過1960,並沒說明哪月,哪日,星期幾.
但是in+年份的話 in 1960可視為明確時間
過去簡單式是有明確時間點:日,月,星期和年
過去完成式則沒明確時間
希望你滿意
Source(s): 自己
• 1 decade ago
呂霖老師來簡單解釋一下
過去完成式只會出現在兩種考題
1.假設句 (對過去之假設)
If I had had money , I would have bought this car .
2.比過去式還早(譬如用在一...就....句型)
I lost my pen that my wife had bought for me .
(丟掉晚發生用ved) (買先發生用had + pp)
其他部分如他其他人解釋之,呂霖不再廢言~
呂霖
http://tw.myblog.yahoo.com/lerry-ivy/
Still have questions? Get your answers by asking now. | 1,072 | 2,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-51 | latest | en | 0.474089 |
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15 Questions | Attempts: 149
Share
Settings
.
• 1.
What are similar fractions?
• 2.
Equivalent fractions are fractions that has the same value even they though they look different. What word means the same with the underlined word?
• A.
Similar
• B.
Same numerator
• C.
Equal
• 3.
If the numerator is 4 and the denominator is 5, the fraction is 5/4.
• A.
True
• B.
False
• 4.
The sum of 2/5 and 4/5 is 7/5.
• A.
True
• B.
False
• 5.
The difference of 5/3 and 1/3 is 1/3.
• A.
True
• B.
False
• 6.
To give the fraction 1/8, the whole must be divided into 5.
• A.
True
• B.
False
• 7.
Two-thirds (2/3) is less than 1
• A.
True
• B.
False
• 8.
Five-sevenths (5/7) is less than 1.
• A.
True
• B.
False
• 9.
Five-fifths is equal to 1.
• A.
True
• B.
False
• 10.
If a cake was cut into 4 equal parts and Trina and her friends ate all. What is the fraction Trina and her friends eaten?
• 11.
What is the sum 1/11 and 7/11?
• 12.
Valerie drove Makati to Alabang in 2/3 of an hour? Which fraction is equivalent to 2/3?
• A.
4/6
• B.
6/6
• C.
3/6
• 13.
What is the missing fraction in 4/5 + ? = 9/5
• 14.
Rodrigo studied Math for 2/6 of an hour and Science for 3/6 of an hour. How long did he study Math and Science?
• A.
1/6 of an hour
• B.
5/6 of an hour
• C.
6/6 of an hour
• 15.
4/7 and 5/7 are dissimilar fractions.
• A.
True
• B.
False | 517 | 1,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.856015 |
https://www.quantumstudy.com/the-focal-length-of-the-objective-of-an-astronomical-telescope-is-75-cm-and-that-of-the-eye-piece-is-5-cm/ | 1,701,944,074,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100651.34/warc/CC-MAIN-20231207090036-20231207120036-00280.warc.gz | 1,055,483,107 | 48,636 | # The focal length of the objective of an astronomical telescope is 75 cm and that of the eye-piece is 5 cm…
Q: The focal length of the objective of an astronomical telescope is 75 cm and that of the eye-piece is 5 cm. If the final image is formed at the least distance of distinct vision from the eye, calculate the magnifying power of the telescope.
Solution : Here , f0 = 75 cm ; fe = 5 cm
We know, D = 25 cm
$\large M = – \frac{f_o}{f_e}(1 + \frac{f_e}{D})$
$\large M = – \frac{75}{5}(1 + \frac{5}{25})$
M = = – 15 × 1.2
M = – 18 | 170 | 540 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-50 | latest | en | 0.877563 |
https://sg.answers.yahoo.com/question/index?qid=20091024131919AAG0fKL | 1,582,973,370,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148850.96/warc/CC-MAIN-20200229083813-20200229113813-00300.warc.gz | 540,478,679 | 19,491 | # Prove the following identity: 1−2·2+3·4−4·8+· · ·±n·2n−1 = (1−(−2)n(3n+1))/9.?
1−2·2+3·4−4·8+· · ·±n·2n−1 = (1−(−2)n(3n+1))/9.
Relevance
• Anonymous
Sum [k=1..n] k*(-2)^(k-1) = (1-(-2)^n*(3n+1))/9
First we try it for a few numbers "to get a feeling for the sequence":
n=1:
1 = 1
(1-(-2)^1(3*1+1))/9 = (1+2*4)/9 = 1
n=2:
1-2*2 = 1-4 = -3
(1-(-2)^2(3*2+1))/9 = (1-4*7)/9 = -27/9 = -3
n=3:
1-*2*2+3*4 = 1-4+12 = -3+12 = 9
(1-(-2)^3(3*3+1))/9 = (1+8*10)/9 = 81/9 = 9
n=4:
1-*2*2+3*4-4*8 = 1-4+12-32 = 9-32 = -23
(1-(-2)^4(3*4+1))/9 = (1-16*13)/9 = -207/9 = -23
n=5:
1-*2*2+3*4-4*8+5*16 = 1-4+12-32+80 = -23+80 = 57
(1-(-2)^5(3*5+1))/9 = (1+32*16)/9 = 513/9 = 57
Works!! The proof is probably by induction .. . So we try it!
Induction starts are all the above! (usually one starts from n=1, but if you prove something manually for n = 1 and 2 and 3, you can then also say that your start is at n = 3).
Induction hypothesis: The formula is true for some n.
Now we try it for n+1:
Sum [k=1..n+1] k*(-2)^(k-1) =
...we split of the sum for 1..n, because we know its formula from the hypothesis!
Sum [k=1..n] k*(-2)^(k-1) + (n+1)*(-2)^n
We must now show that this is equal to the formula when one replaces n with n + 1. I do this by writing this as a "question" as follows:
Sum [k=1..n] k*(-2)^(k-1) + (n+1)*(-2)^n =? (1-(-2)^(n+1)*(3(n+1)+1))/9
Note the =? sign: We do not yet know this - but if it turns out to be true, we are done (my mathematics teacher allowed this sort of proof - it is helpful, as you always see where you have to go).
Ok - we replace the left sum with the formula from the hypothesis:
(1-(-2)^n*(3n+1))/9 + (n+1)*(-2)^n =? (1-(-2)^(n+1)*(3(n+1)+1))/9 ... *9
(1-(-2)^n*(3n+1)) + 9*(n+1)*(-2)^n =? (1-(-2)^(n+1)*(3(n+1)+1)) ... -1
-(-2)^n*(3n+1) + 9*(n+1)*(-2)^n =? (-2)^(n+1)*(3(n+1)+1) ... rewrite (-2)^(n+1)
-(-2)^n*(3n+1) + 9*(n+1)*(-2)^n =? (-2)^n*(-2)*(3(n+1)+1) ... /(-2)^n
-(3n+1) + 9*(n+1) =? (-2)*(3(n+1)+1) ... Expand all
-3n - 1 + 9n + 9 =? (-2)*(3n + 3 + 1) ...
6n + 8 =? -6n - 8
... oops, I made some sign error somewhere above: If this is corrected, then we can write
6n + 8 =! 6n + 8
now with an exclamation mark: This is always correct, and as we only did equivalent rewrites of the formulas, also the first line with =? is then correct, and hence we have proven the formula by induction fully.
I'll leave it at that ... it's your job to find that error (preferably by redoing the computations without the error :-) ).
Hope that helps! | 1,120 | 2,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2020-10 | latest | en | 0.726889 |
https://www.numbersaplenty.com/2110929 | 1,685,550,125,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646937.1/warc/CC-MAIN-20230531150014-20230531180014-00014.warc.gz | 1,018,063,643 | 3,188 | Search a number
2110929 = 3703643
BaseRepresentation
bin1000000011010111010001
310222020122120
420003113101
51020022204
6113124453
723641212
oct10032721
93866576
102110929
111211a77
12859729
1358ba92
143cd409
152ba6d9
hex2035d1
2110929 has 4 divisors (see below), whose sum is σ = 2814576. Its totient is φ = 1407284.
The previous prime is 2110921. The next prime is 2110931. The reversal of 2110929 is 9290112.
It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.
It is a cyclic number.
It is not a de Polignac number, because 2110929 - 23 = 2110921 is a prime.
It is a super-3 number, since 3×21109293 = 28219033399653885267, which contains 333 as substring.
It is a D-number.
2110929 is a lucky number.
It is a junction number, because it is equal to n+sod(n) for n = 2110899 and 2110908.
It is not an unprimeable number, because it can be changed into a prime (2110921) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 351819 + ... + 351824.
It is an arithmetic number, because the mean of its divisors is an integer number (703644).
22110929 is an apocalyptic number.
It is an amenable number.
2110929 is a deficient number, since it is larger than the sum of its proper divisors (703647).
2110929 is an equidigital number, since it uses as much as digits as its factorization.
2110929 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 703646.
The product of its (nonzero) digits is 324, while the sum is 24.
The square root of 2110929 is about 1452.9036444307. The cubic root of 2110929 is about 128.2796822269.
The spelling of 2110929 in words is "two million, one hundred ten thousand, nine hundred twenty-nine".
Divisors: 1 3 703643 2110929 | 574 | 1,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-23 | latest | en | 0.884146 |
https://www.slideserve.com/giacomo-watts/class-26 | 1,511,338,144,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806509.31/warc/CC-MAIN-20171122065449-20171122085449-00214.warc.gz | 848,415,029 | 14,686 | Class #26
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# Class #26 - PowerPoint PPT Presentation
Class #26. Nonlinear Systems and Chaos Most important concepts Sensitive Dependence on Initial conditions Attractors Other concepts State-space orbits Non-linear diff. eq. Driven oscillations Second Harmonic Generation Subharmonics Period-doubling cascade Bifurcation plot
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## PowerPoint Slideshow about 'Class #26' - giacomo-watts
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Presentation Transcript
Class #26
• Nonlinear Systems and Chaos
• Most important concepts
• Sensitive Dependence on Initial conditions
• Attractors
• Other concepts
• State-space orbits
• Non-linear diff. eq.
• Driven oscillations
• Second Harmonic Generation
• Subharmonics
• Bifurcation plot
• Poincare diagram
• Feigenbaum number
• Universality
Outline
• Origins and Definitions of chaos
• State Space
• Behavior of a driven damped pendulum (DDP)
• Non-linear behavior of a DDP
• Attractor
• Period doubling
• Sensitive dependence
• Bifurcation Plot
Definition of chaos
• The dynamical evolution that is aperiodic and sensitively dependent on initial conditions. In dissipative dynamical systems this involves trajectories that move on a strange attractor, a fractal subspace of the phase space. This term takes advantage of the colloquial meaning of chaos as random, unpredictable, and disorderly behavior, but the phenomena given the technical name chaos have an intrinsic feature of determinism and some characteristics of order.
• Colloquial meaning – disorder, randomness, unpredictability
• Technical meaning – Fundamental unpredictability and apparent randomness from a system that is deterministic. Some real randomness may be included in real systems, but a model of the system without ANY added randomness should display the same behavior.
Weather and climate prediction
• Is important
• Can we go:
• For a hike?
• Get married outdoors?
• Start a war?
• Will we be hurt by a:
• Hurricane?
• Lightning bolt?
• How much more fossil fuel can we burn before we:
• Fry?
• Drown?
• Starve?
Early work by Edward N. Lorenz
• 1960’s at MIT
• Early computer models of the atmosphere
• Were very simple
• (Computers were stupid)
• Results were not reproducible!!
• Lorenz ultimately noticed that
• For 7-10 days of prediction, all of his models reproduced very well.
• After 7-10 days, the same model could be run twice and give the same result – BUT!!
• Changes that he thought were trivial
• (e.g. changing the density of air by rounding it out at the 3th decimal place, or slightly modifying the initial conditions at beginning of model)
• Produced COMPLETELY DIFFERENT results
• This came to be called “The Butterfly effect”
Viscous Drag III – Stokes Law
D
Form-factor k becomes
“D” is diameter of sphere
Viscous drag on walls of sphere is responsible for retarding force.
George Stokes [1819-1903]
(Navier-Stokes equations/ Stokes’ theorem)
Damped Driven Pendulum (DDP)
• Damping
• Pendulum immersed in fluid with Newtonian viscosity
• Damping proportional to velocity (and angular velocity)
• Driving
• Constant amplitude drive bar
• Connected to pendulum via torsion spring
• Torque on Pendulum is
Conditions for chaos
• Dissipative Chaos
• Requires a differential equation with 3 or more independent variables.
• Requires a non-linear coupling between at least two of the variables.
• Requires a dissipative term (that will use up energy).
• Non-dissipative chaos
• Not in this course
Worked problem
• Sketch a state-space plot for the magnetic pendulum
• Indicate the attractors and repellers
• Show some representative trajectories
• First explore the trajectories beginning with theta-dot=0
• Then explore trajectories that begin with theta in some state near an attractor but through proper choice of theta-dot move to the other attractor.
• Sketch the basins of attraction | 1,018 | 4,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.794822 |
http://www.mathisfunforum.com/viewtopic.php?pid=20563 | 1,547,989,089,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583716358.66/warc/CC-MAIN-20190120123138-20190120145138-00215.warc.gz | 337,051,782 | 3,298 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #1 2005-12-13 11:45:59
babyangel1972
Member
Registered: 2005-12-12
Posts: 18
### Word problem
Between the years 1999 and 2003 the average text book rose from \$45.00 to \$67.00/ What is the percent of increase (rounded to the nearest whole number) for the average text book.
Offline
## #2 2005-12-13 12:17:19
mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900
### Re: Word problem
Percentage increase = (new amount - old amount)/old amount.
So in this case, it would be (67-45)/45 = 22/45.
Multiply this by 100% to get a percentage and you get an increase of 49% to the nearest whole number.
Why did the vector cross the road?
It wanted to be normal.
Offline
## #3 2005-12-13 18:58:36
MathsIsFun
Registered: 2005-01-21
Posts: 7,686
### Re: Word problem
Babyangel, maybe you could post future questions below here (click "Post Reply") , so that we can get a "BabyAngel" discussion going, and help you along better.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
Offline | 379 | 1,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-04 | latest | en | 0.871296 |
https://forums.openmv.io/t/how-can-i-find-minimum-area-rectangle-in-binary-image/1033 | 1,618,910,999,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039388763.75/warc/CC-MAIN-20210420091336-20210420121336-00565.warc.gz | 365,940,327 | 10,621 | How can I find minimum area rectangle in binary image?
I have EmguCV project to find all minimum area rectangle in contour like below.
VB.NET with EmguCV code :
imgRects.Draw(CvInvoke.MinAreaRect(contour), New Bgr(Color.Red), 3)
How can I find minimum area rectangle in binary image like vb.net code as above?
Hi, please clarify how this relates to the OpenMV Cam?
We don’t have a min area rect method. However, if you care about rotation the find_blobs() method returns the rotation angle of the object. Otherwise what is your goal with the min area rect?
Thank you for you fast reply. In my previous project, I use visual studio with emgu cv to measures width and length of object. In last project, I use openmv with non-distortion lens. Then I want to make embeded machine to use mesure width and length of object and have light display when found defect.
So I want to measure minimum area rectangle. That in my previous project, I use function as below.
CvInvoke.CvtColor(imgROI, imgHSV, ColorConversion.Bgr2Hsv)
CvInvoke.InRange
CvInvoke.FindContours(…
imgRects.Draw(CvInvoke.MinAreaRect(contour), New Bgr(Color.Red), 3)
I understand what you need to do. Um, so the rect we return has a left/right/up/down parts. However, this ins’t the min area rect… but, the min area rect intersects these points.
Anyway, the easiest way to get this is to take the rotation angle from the find_blobs() function and rotate the rectangle corners by a 2D rotation matrix.
E.g.
For each point of the of the rect do trate the point as:
[X
Y]
and the mulitply by: Rotation matrix - Wikipedia where pheta is the rotation angle from find_blobs().
I understand you’d prefer some working code versus how to do it but I am answering this from not at home right now and I’m unable to write some code for this. However, we can add this feature to the firmware. Please submit a github ticket issue for it.
I found new update OPENMV ide with new firmware. Do you add minimum area rectangle in last firmware?
Not yet. I’ve finally completed enough tasks to get back to doing firmware updates however. So, the next firmware release should have this feature.
Um, note, I plan to just expose the four extreme corners of the object. The minimum rect will just then be parallel lines that pass through these corners.
In our lib a rect() is never rotated… So, to draw the minimum rect you’d have to call draw line repeatedly on each corner to corner. Anyway, do you actually need the minimum rect or just the extreme corners?
I want bounding rotated rectangle that can returns center, angle, width, height, point[0], point[1], point[2], point[3] and contour area as attach picture. Thank you and Happy new year.
http://www.emgu.com/wiki/index.php/Minimum_Area_Rectangle_in_CSharp
https://docs.opencv.org/3.2.0/dd/d49/tutorial_py_contour_features.html
Yeah, I get what you want to do. But, keep in mind you haven’t presented a use case other than you want to display this onscreen which is a rather weak need for it. We have a method that gives you already the rotation an object.
Anyway, given what you want. I’ll add the outer edges of the object points along with projected points that show the min bounding rect.
Note that I will note be adding the contour value. I mad a decision a while back not to add that explicitly because of the limited RAM on the system. To keep an object contour requires retaining a variable length list of points of the object which takes quite a bit of ram and is not suitable for an embedded system. I understand this prevents using a lot of different features OpenCV has for object introspection but this is okay.
Anyway, I’m coding right now and fixing other bugs but will start on this feature on Wednesday. I should have it done before the end of the week and I’ll share a new binary with you for it.
Thanks,
Hi, okay, I did it. Attached is the new firmware. Here’s some sample code:
``````# Single Color RGB565 Blob Tracking Example
#
# This example shows off single color RGB565 tracking using the OpenMV Cam.
import sensor, image, time, math
threshold_index = 0 # 0 for red, 1 for green, 2 for blue
# Color Tracking Thresholds (L Min, L Max, A Min, A Max, B Min, B Max)
# The below thresholds track in general red/green/blue things. You may wish to tune them...
thresholds = [(30, 100, 15, 127, 15, 127), # generic_red_thresholds
(30, 100, -64, -8, -32, 32), # generic_green_thresholds
(0, 30, 0, 64, -128, 0)] # generic_blue_thresholds
sensor.reset()
sensor.set_pixformat(sensor.RGB565)
sensor.set_framesize(sensor.QVGA)
sensor.skip_frames(time = 2000)
sensor.set_auto_gain(False) # must be turned off for color tracking
sensor.set_auto_whitebal(False) # must be turned off for color tracking
clock = time.clock()
# Only blobs that with more pixels than "pixel_threshold" and more area than "area_threshold" are
# returned by "find_blobs" below. Change "pixels_threshold" and "area_threshold" if you change the
# camera resolution. "merge=True" merges all overlapping blobs in the image.
while(True):
clock.tick()
img = sensor.snapshot()
for blob in img.find_blobs([thresholds[threshold_index]], pixels_threshold=200, area_threshold=200):
img.draw_rectangle(blob.rect())
img.draw_cross(blob.cx(), blob.cy())
img.draw_line(blob.corners()[0][0], blob.corners()[0][1], blob.corners()[1][0], blob.corners()[1][1], color=(0,255,0))
img.draw_line(blob.corners()[1][0], blob.corners()[1][1], blob.corners()[2][0], blob.corners()[2][1], color=(0,255,0))
img.draw_line(blob.corners()[2][0], blob.corners()[2][1], blob.corners()[3][0], blob.corners()[3][1], color=(0,255,0))
img.draw_line(blob.corners()[3][0], blob.corners()[3][1], blob.corners()[0][0], blob.corners()[0][1], color=(0,255,0))
img.draw_line(blob.min_corners()[0][0], blob.min_corners()[0][1], blob.min_corners()[1][0], blob.min_corners()[1][1], color=(0,0,255))
img.draw_line(blob.min_corners()[1][0], blob.min_corners()[1][1], blob.min_corners()[2][0], blob.min_corners()[2][1], color=(0,0,255))
img.draw_line(blob.min_corners()[2][0], blob.min_corners()[2][1], blob.min_corners()[3][0], blob.min_corners()[3][1], color=(0,0,255))
img.draw_line(blob.min_corners()[3][0], blob.min_corners()[3][1], blob.min_corners()[0][0], blob.min_corners()[0][1], color=(0,0,255))
print(clock.fps())
``````
Drawing the lines is kinda a pain right now. I’ll add two more draw methods to make this easier.
So, the corners value is the object corners. It’s always sorted from the left, top, right, and bottom. Min corners is then the min area rect. I was going to call it min_area_rect_corners() but that was too verbose.
Note that since the min area rect is generated by just a few points it’s quite jumpy. Set your thresholds well.
firmware.zip (918 KB)
Thank you friend about blob.min_corners(). I will try to use.
About my project, I have egg place on green black ground. First step, I will detect green color in Lab color space and invert. So I will reach egg area. Then I want use to know ratio between width and length of egg area.
I not sure that the egg put align with camera sensor. So I want to use minimum rectangle area enclosing egg area. After than I will sent width and length data to process through UART to Arduino.
Please suggest me. What OpenMV library do I need?
Hi, I recently updated the firmware method with a major_axis_line() method that will tell you the longest length of the side in the min_area_rect. So, that’s basically what you need.
As for transmit via the UART. See the UART example on our docs. You can just print the value.
I’ll upload the new firmware tonight.
Hi, I’m still working on the min_area_rect() method. The method I used doesn’t seem to work in all situations. It’s very memory intensive to store the contour and then calculate the min_area_rect() per frame based on a full contour of the object. I am trying to figure out a way to do this with just 4 points but I have not found anything really robust yet…
In the mean-time I’ve upgraded the find_blob() method to be measurably faster, better merging, added perimeter support, floating point centroid, roundness calculation (eccentricity), and x/y histogram projections.
I will continue working on this until I can find a solution. Since I added the perimeter code I know what pixels are on the perimeter of the object so I could allocate a point list which will work on the F7/H7s RAM. However, I want to get this working on the M4 which requires me to do this in the least memory usage way possible.
EDIT: An idea… I think if I combine the 4 corners with the 4 diagonal corners I will get a much better result. I will see what I can do with a constrained list of points based on local optima of geometric identities.
Note: On the H7 I’m getting 80 FPS at 320x240 and 30 FPS on VGA grayscale. The code is fast.
Hi,
I would like to use this function with sensor.set_pixformat(sensor.GRAYSCALE)
Have you a solution ?
thank you
At the 1st draw.line, I can see this error message :
attribute errors : ‘blob’ object has no attribute ‘corners’
Hi, you have to update the firmware on your camera using the firmware binary I posted above.
Yes, it works on grayscale.
The current algorithm is not as robust as it needs to be. I will post another version that should work better tonight.
Yes it seems to be the last firmware : 3.2.0.
No, you have to download the firmware above. I posted a binary file a few posts back. This is a firmware cut that’s a pre-release.
Anyway, I’ll post something else again.
This code has the features you want:
``````# Single Color RGB565 Blob Tracking Example
#
# This example shows off single color RGB565 tracking using the OpenMV Cam.
import sensor, image, time, math
threshold_index = 0 # 0 for red, 1 for green, 2 for blue
# Color Tracking Thresholds (L Min, L Max, A Min, A Max, B Min, B Max)
# The below thresholds track in general red/green/blue things. You may wish to tune them...
thresholds = [(30, 100, 15, 127, 15, 127), # generic_red_thresholds
(30, 100, -64, -8, -32, 32), # generic_green_thresholds
(0, 30, 0, 64, -128, 0)] # generic_blue_thresholds
sensor.reset()
sensor.set_pixformat(sensor.RGB565)
sensor.set_framesize(sensor.QVGA)
sensor.skip_frames(time = 2000)
sensor.set_auto_gain(False) # must be turned off for color tracking
sensor.set_auto_whitebal(False) # must be turned off for color tracking
clock = time.clock()
# Only blobs that with more pixels than "pixel_threshold" and more area than "area_threshold" are
# returned by "find_blobs" below. Change "pixels_threshold" and "area_threshold" if you change the
# camera resolution. "merge=True" merges all overlapping blobs in the image.
while(True):
clock.tick()
img = sensor.snapshot()
for blob in img.find_blobs([thresholds[threshold_index]], pixels_threshold=200, area_threshold=200, merge=True):
# These values depend on the blob not being circular - otherwise they will be shaky.
if blob.elongation() > 0.5:
img.draw_edges(blob.min_corners(), color=(255,0,0))
img.draw_line(blob.major_axis_line(), color=(0,255,0))
img.draw_line(blob.minor_axis_line(), color=(0,0,255))
# These values are stable all the time.
img.draw_rectangle(blob.rect())
img.draw_cross(blob.cx(), blob.cy())
# Note - the blob rotation is unique to 0-180 only.
img.draw_keypoints([(blob.cx(), blob.cy(), int(math.degrees(blob.rotation())))], size=20)
print(clock.fps())
``````
If you want the blob locking to be better there’s a parameter in my code that can be increased to improve the accuracy of the min area rect. Generally, unless the object is unique the min area rect is jumpy. Right now, the firmware bounds the rect with 20 points separated by 18 degrees each. Its not easily possible to make the number of points used to bound the object dynamic so it’s a parameter in the code.
Anyway, for objects like rectangles it works great and irregularly shaped things. It will be in the firmware from now on.
firmware.zip (919 KB)
Hi,
I have installed the firmwire in this post and when I tried to run the example, I notice that the FPS printed on terminal is only about 15.
Is this normal ? As I saw that you have mentioned that the fps you can get is about 80 fps.
Also, I found that the minimum area rectangle is not that stable but I cant figure out how to modify the parameters to make it more stable.
Would you mind to further explain on how to modify those parameters?
Thanks
Hi, higher FPS is achieved if you lower the resolution and change the image type to one that is easier to process. If you are working on a grayscale image things go faster because there’s less work per pixel. Since you have an M7 that’s 2x slower than the new H7 coming out. Also, keep in mind the process of streaming the image to the IDE takes cpu and you can disable the frame buffer to see the real fps for your app.
As for the min area rect, the method I use shoots 20 points out from the center of the object and bounds the min area rect to those 20 points. If the object you are looking at is round then the rest will not be stable since the 20 points can jiggle around. Notice the elongation check in the code avoids drawing the min area rect if the object is not elongated. The method is quite stable the more elongated an object is and unique it is.
Given the method we use does not store a contour of the object this is the best I can do. If the firmware I can arbitrarily raise the number of points samples but I found 20 points to work the best causing the frame rate to cave.
Also, keep in mind your thresholds and the effect they have on the min area rect. Unlike the normal bounding box each and every pixel matters on the contour for the min area rect. If your color thresholds are too tight most likely the objects edges are going in and out per frame causing a lot of jiggle. Given the omnivision camera has a ton of shot noise you’re going to experience a lot of jiggling unless the object is really elongated such that some of the 20 points shot out from the center of the object when bounding it are stable and in the same place. | 3,425 | 14,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-17 | latest | en | 0.844308 |
http://www.automation-talk.info/2013/02/understanding-difference-between-source-sink-plc.html | 1,524,282,638,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944982.32/warc/CC-MAIN-20180421032230-20180421052230-00315.warc.gz | 350,905,304 | 37,806 | ## Understanding Difference Between Source & Sink in PLC's
If you have recently started learning PLC programming and has already understand the concept of choosing right type of PLC for any project, then it's now time to know about concept of Source and Sink in PLC. Understanding when PLC is connected as source and when as sink is very important for all the Automation engineers. Connecting a power supply to input or output of a PLC makes it input source or input sink and output source or output sink. So source and sink totally depends on how you connect your power supply. Subscribe to Automation-Talk by Email.
Recommended Reading: How to Make Sink Type PLC from Source Type
Difference Between Source & Sink in PLC's
1. Sink Type: In a sink type, current always flow into the PLC and positive (+ve) of supply is connected to PLC input terminal (In case you are making your input as sink) and Negative (-ve) of supply is connected to common in Input module of PLC as shown below.
Remember the basic electrical theory, that current always flow from positive to negative. You can also memorize this by taking example of a water sink, where water flows to sink hole.
Click to enlarge image
2. Source Type: In a source type, current always flow from PLC to external device such as switch or sensor as shown above. Here all the input of PLC are connected to negative (-ve) of supply and Common of Input module is connected to positive (+ve).
In the above image, we have taken example of Input side of PLC, here we have taken 2 PLC in source and sink configuration.
1. Here S0, S1, S2 are switches.
2. I0, I1, I2 are input's of PLC
3. "I" is the current and blue line is showing direction of current.
4. Positive (+ve) and negative (-ve) are terminal of power supply.
Note: Never confuse while connecting a NPN type PLC with NPN sensor or PNP PLC with PNP sensor.
The above explanation of Source and Sink was done considering Input side of PLC and same is true for output side of PLC. If you have any other question or query please feel free to leave your comments. If you like our article consider it sharing.
#### 1 comment:
1. I have read exact opposite of this for mitsubishi PLC.
24V is made common for sink input type and 0V for source input type. | 499 | 2,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2018-17 | latest | en | 0.929903 |
https://docs.pyro.ai/en/latest/_modules/pyro/distributions/transforms/matrix_exponential.html | 1,627,952,984,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154408.7/warc/CC-MAIN-20210802234539-20210803024539-00128.warc.gz | 218,642,322 | 8,114 | Source code for pyro.distributions.transforms.matrix_exponential
# Copyright (c) 2017-2019 Uber Technologies, Inc.
import math
from functools import partial
import torch
import torch.nn as nn
from torch.distributions import Transform, constraints
from pyro.nn import DenseNN
from ..conditional import ConditionalTransformModule
from ..torch_transform import TransformModule
from ..util import copy_docs_from
@copy_docs_from(Transform)
class ConditionedMatrixExponential(Transform):
domain = constraints.real_vector
codomain = constraints.real_vector
bijective = True
def __init__(self, weights=None, iterations=8, normalization="none", bound=None):
super().__init__(cache_size=1)
assert iterations > 0
self.weights = weights
self.iterations = iterations
self.normalization = normalization
self.bound = bound
# Currently, weight and spectral normalization are unimplemented. This doesn't effect the validity of the
# bijection, although applying these norms should improve the numerical conditioning of the approximation.
if normalization == "weight" or normalization == "spectral":
raise NotImplementedError("Normalization is currently not implemented.")
elif normalization != "none":
raise ValueError("Unknown normalization method: {}".format(normalization))
def _exp(self, x, M):
"""
Performs power series approximation to the vector product of x with the
matrix exponential of M.
"""
power_term = x.unsqueeze(-1)
y = x.unsqueeze(-1)
for idx in range(self.iterations):
power_term = torch.matmul(M, power_term) / (idx + 1)
y = y + power_term
return y.squeeze(-1)
def _trace(self, M):
"""
Calculates the trace of a matrix and is able to do broadcasting over batch
dimensions, unlike torch.trace.
Broadcasting is necessary for the conditional version of the transform,
where self.weights may have batch dimensions corresponding the batch
dimensions of the context variable that was conditioned upon.
"""
return M.diagonal(dim1=-2, dim2=-1).sum(-1)
def _call(self, x):
"""
:param x: the input into the bijection
:type x: torch.Tensor
Invokes the bijection x => y; in the prototypical context of a
:class:~pyro.distributions.TransformedDistribution x is a sample from
the base distribution (or the output of a previous transform)
"""
M = self.weights() if callable(self.weights) else self.weights
return self._exp(x, M)
def _inverse(self, y):
"""
:param y: the output of the bijection
:type y: torch.Tensor
Inverts y => x.
"""
M = self.weights() if callable(self.weights) else self.weights
return self._exp(y, -M)
def log_abs_det_jacobian(self, x, y):
"""
Calculates the element-wise determinant of the log Jacobian
"""
M = self.weights() if callable(self.weights) else self.weights
return self._trace(M)
[docs]@copy_docs_from(ConditionedMatrixExponential)
class MatrixExponential(ConditionedMatrixExponential, TransformModule):
r"""
A dense matrix exponential bijective transform (Hoogeboom et al., 2020) with
equation,
:math:\mathbf{y} = \exp(M)\mathbf{x}
where :math:\mathbf{x} are the inputs, :math:\mathbf{y} are the outputs,
:math:\exp(\cdot) represents the matrix exponential, and the learnable
parameters are :math:M\in\mathbb{R}^D\times\mathbb{R}^D for input dimension
:math:D. In general, :math:M is not required to be invertible.
Due to the favourable mathematical properties of the matrix exponential, the
transform has an exact inverse and a log-determinate-Jacobian that scales in
time-complexity as :math:O(D). Both the forward and reverse operations are
approximated with a truncated power series. For numerical stability, the
norm of :math:M can be restricted with the normalization keyword argument.
Example usage:
>>> base_dist = dist.Normal(torch.zeros(10), torch.ones(10))
>>> transform = MatrixExponential(10)
>>> pyro.module("my_transform", transform) # doctest: +SKIP
>>> flow_dist = dist.TransformedDistribution(base_dist, [transform])
>>> flow_dist.sample() # doctest: +SKIP
:param input_dim: the dimension of the input (and output) variable.
:type input_dim: int
:param iterations: the number of terms to use in the truncated power series that
approximates matrix exponentiation.
:type iterations: int
:param normalization: One of ['none', 'weight', 'spectral'] normalization that
selects what type of normalization to apply to the weight matrix. weight
corresponds to weight normalization (Salimans and Kingma, 2016) and
spectral to spectral normalization (Miyato et al, 2018).
:type normalization: string
:param bound: a bound on either the weight or spectral norm, when either of
those two types of regularization are chosen by the normalization
argument. A lower value for this results in fewer required terms of the
truncated power series to closely approximate the exact value of the matrix
exponential.
:type bound: float
References:
[1] Emiel Hoogeboom, Victor Garcia Satorras, Jakub M. Tomczak, Max Welling. The
Convolution Exponential and Generalized Sylvester Flows. [arXiv:2006.01910]
[2] Tim Salimans, Diederik P. Kingma. Weight Normalization: A Simple
Reparameterization to Accelerate Training of Deep Neural Networks.
[arXiv:1602.07868]
[3] Takeru Miyato, Toshiki Kataoka, Masanori Koyama, Yuichi Yoshida. Spectral
Normalization for Generative Adversarial Networks. ICLR 2018.
"""
domain = constraints.real_vector
codomain = constraints.real_vector
bijective = True
def __init__(self, input_dim, iterations=8, normalization="none", bound=None):
super().__init__(
iterations=iterations, normalization=normalization, bound=bound
)
self.weights = nn.Parameter(torch.Tensor(input_dim, input_dim))
self.reset_parameters()
[docs] def reset_parameters(self):
stdv = 1.0 / math.sqrt(self.weights.size(0))
self.weights.data.uniform_(-stdv, stdv)
[docs]@copy_docs_from(ConditionalTransformModule)
class ConditionalMatrixExponential(ConditionalTransformModule):
r"""
A dense matrix exponential bijective transform (Hoogeboom et al., 2020) that
conditions on an additional context variable with equation,
:math:\mathbf{y} = \exp(M)\mathbf{x}
where :math:\mathbf{x} are the inputs, :math:\mathbf{y} are the outputs,
:math:\exp(\cdot) represents the matrix exponential, and
:math:M\in\mathbb{R}^D\times\mathbb{R}^D is the output of a neural network
conditioning on a context variable :math:\mathbf{z} for input dimension
:math:D. In general, :math:M is not required to be invertible.
Due to the favourable mathematical properties of the matrix exponential, the
transform has an exact inverse and a log-determinate-Jacobian that scales in
time-complexity as :math:O(D). Both the forward and reverse operations are
approximated with a truncated power series. For numerical stability, the
norm of :math:M can be restricted with the normalization keyword argument.
Example usage:
>>> from pyro.nn.dense_nn import DenseNN
>>> input_dim = 10
>>> context_dim = 5
>>> batch_size = 3
>>> base_dist = dist.Normal(torch.zeros(input_dim), torch.ones(input_dim))
>>> param_dims = [input_dim*input_dim]
>>> hypernet = DenseNN(context_dim, [50, 50], param_dims)
>>> transform = ConditionalMatrixExponential(input_dim, hypernet)
>>> z = torch.rand(batch_size, context_dim)
>>> flow_dist = dist.ConditionalTransformedDistribution(base_dist,
... [transform]).condition(z)
>>> flow_dist.sample(sample_shape=torch.Size([batch_size])) # doctest: +SKIP
:param input_dim: the dimension of the input (and output) variable.
:type input_dim: int
:param iterations: the number of terms to use in the truncated power series that
approximates matrix exponentiation.
:type iterations: int
:param normalization: One of ['none', 'weight', 'spectral'] normalization that
selects what type of normalization to apply to the weight matrix. weight
corresponds to weight normalization (Salimans and Kingma, 2016) and
spectral to spectral normalization (Miyato et al, 2018).
:type normalization: string
:param bound: a bound on either the weight or spectral norm, when either of
those two types of regularization are chosen by the normalization
argument. A lower value for this results in fewer required terms of the
truncated power series to closely approximate the exact value of the matrix
exponential.
:type bound: float
References:
[1] Emiel Hoogeboom, Victor Garcia Satorras, Jakub M. Tomczak, Max Welling. The
Convolution Exponential and Generalized Sylvester Flows. [arXiv:2006.01910]
[2] Tim Salimans, Diederik P. Kingma. Weight Normalization: A Simple
Reparameterization to Accelerate Training of Deep Neural Networks.
[arXiv:1602.07868]
[3] Takeru Miyato, Toshiki Kataoka, Masanori Koyama, Yuichi Yoshida. Spectral
Normalization for Generative Adversarial Networks. ICLR 2018.
"""
domain = constraints.real_vector
codomain = constraints.real_vector
bijective = True
def __init__(self, input_dim, nn, iterations=8, normalization="none", bound=None):
super().__init__()
self.input_dim = input_dim
self.nn = nn
self.iterations = iterations
self.normalization = normalization
self.bound = bound
def _params(self, context):
return self.nn(context)
[docs] def condition(self, context):
# This hack could be fixed by having a conditioning network that outputs a more general shape
cond_nn = partial(self.nn, context)
def weights():
w = cond_nn()
return w.view(w.shape[:-1] + (self.input_dim, self.input_dim))
return ConditionedMatrixExponential(
weights,
iterations=self.iterations,
normalization=self.normalization,
bound=self.bound,
)
[docs]def matrix_exponential(input_dim, iterations=8, normalization="none", bound=None):
"""
A helper function to create a
:class:~pyro.distributions.transforms.MatrixExponential object for consistency
with other helpers.
:param input_dim: Dimension of input variable
:type input_dim: int
:param iterations: the number of terms to use in the truncated power series that
approximates matrix exponentiation.
:type iterations: int
:param normalization: One of ['none', 'weight', 'spectral'] normalization that
selects what type of normalization to apply to the weight matrix. weight
corresponds to weight normalization (Salimans and Kingma, 2016) and
spectral to spectral normalization (Miyato et al, 2018).
:type normalization: string
:param bound: a bound on either the weight or spectral norm, when either of
those two types of regularization are chosen by the normalization
argument. A lower value for this results in fewer required terms of the
truncated power series to closely approximate the exact value of the matrix
exponential.
:type bound: float
"""
return MatrixExponential(
input_dim, iterations=iterations, normalization=normalization, bound=bound
)
[docs]def conditional_matrix_exponential(
input_dim,
context_dim,
hidden_dims=None,
iterations=8,
normalization="none",
bound=None,
):
"""
A helper function to create a
:class:~pyro.distributions.transforms.ConditionalMatrixExponential object for
consistency with other helpers.
:param input_dim: Dimension of input variable
:type input_dim: int
:param context_dim: Dimension of context variable
:type context_dim: int
:param hidden_dims: The desired hidden dimensions of the dense network. Defaults
to using [input_dim * 10, input_dim * 10]
:type hidden_dims: list[int]
:param iterations: the number of terms to use in the truncated power series that
approximates matrix exponentiation.
:type iterations: int
:param normalization: One of ['none', 'weight', 'spectral'] normalization that
selects what type of normalization to apply to the weight matrix. weight
corresponds to weight normalization (Salimans and Kingma, 2016) and
spectral to spectral normalization (Miyato et al, 2018).
:type normalization: string
:param bound: a bound on either the weight or spectral norm, when either of
those two types of regularization are chosen by the normalization
argument. A lower value for this results in fewer required terms of the
truncated power series to closely approximate the exact value of the matrix
exponential.
:type bound: float
"""
if hidden_dims is None:
hidden_dims = [input_dim * 10, input_dim * 10]
nn = DenseNN(context_dim, hidden_dims, param_dims=[input_dim * input_dim])
return ConditionalMatrixExponential(
input_dim, nn, iterations=iterations, normalization=normalization, bound=bound
) | 2,877 | 12,157 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-31 | longest | en | 0.729552 |
https://www.astronomy.com/science/scientists-search-for-the-soundtrack-of-the-universe/ | 1,718,703,873,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.70/warc/CC-MAIN-20240618073942-20240618103942-00756.warc.gz | 573,884,325 | 31,636 | # Scientists search for the soundtrack of the universe
Early philosophers and astronomers strove to understand the harmonies of our solar system. Today, scientists hunt for cosmic tones of a different kind.
| Published: December 1, 2023
I have just replaced Albert Einstein (a.d. 1879–1955) with Pythagoras (570–490 b.c.) as the one person, living or dead, with whom I would like to have dinner.
That is quite a turnaround for me, since his Pythagorean theorem was the reason I failed my 10th-grade geometry class. It was enough to scare me away for years — and keep me from learning about a man considered a giant in the early studies of math, music, philosophy, and astronomy.
But since my main two hobbies in high school were astronomy and music, I wish my teacher would have shared with me that Pythagoras is the man who coined the term cosmos and was instrumental in the development of understanding the physics of musical sound.
## The overtone series
Among his many accomplishments, Pythagoras is credited with the discovery and understanding of the overtone series — the series of frequencies contained in most musical notes, creating the distinctive sound you hear. Before I was introduced to the overtone series, I had taken musical pitch at face value without knowing it takes a series of mathematically related tones to create what I heard as familiar and pleasing. After I learned about overtones, I never listened to music in the same way.
In his 1619 book Harmonices Mundi (Harmony of the World), Johannes Kepler (1571–1630) describes Pythagoras passing through a smithy and noticing that hammers of different sizes created sounds with different pitches on the anvil. Pythagoras then moved from hammers to strings, observing how different lengths of string oscillated at different frequencies, producing various tones. (Today, this exact story is considered apocryphal; likely Pythagoras made this discovery using only strings, no smithy required.)
Regardless of the chain of events, Pythagoras discovered that when a musician plays a note and causes a musical instrument to vibrate, the root note that we hear (called the fundamental or first harmonic) is also accompanied by a series of higher frequencies. These occur naturally in a strict mathematical sequence and depending on the strength of the overtones that occur above the fundamental, produces the tone and character of an instrument — which is also referred to as timbre. Without those overtones, the result would be a very unsatisfying sine wave. Pythagoras identified these overtones as harmonic frequencies using mathematical ratios. One-half of the fundamental (1:1) creates an octave (2:1) above the original pitch, with the third overtone — the perfect fifth — vibrating at three times that of the fundamental above the second overtone (3:2), and so on. This sequence continues into ever higher frequencies, eventually beyond what the human ear can detect. An instrument’s unique voice comes from the fact that each uses a different combination of frequency strengths to produce the same note.
In other words, the harmonics of the overtone series deliver rich tonal complexities. However, their existence is hidden within the sound we hear in the same way that light we perceive as white combines all the wavelengths of the visible spectrum.
## Heavenly motions
Where does this musical finding intersect with the cosmos? Pythagoras took his earthly discovery and sought to connect it to the workings of the sky above. Building on his realization that “there is geometry in the humming of the strings,” Pythagoras believed “there is music in the spacing of the spheres” of the planets visible in the sky.
This belief had long-lasting implications: Sixteen centuries later, music, math, and astronomy were complementary studies. Seeking connection between the physical world and the cosmos, Galileo Galilei (1564–1642) stated: “Mathematics is the alphabet with which God has written the universe.”
Around the same time, Kepler was attempting to expand Pythagoras’ concepts of mathematical ratios and musical intervals (including overtones) in Harmonices Mundi, reflecting what he believed was God’s design and what the soul naturally senses. “The heavenly motions are nothing but a continuous song for several voices, to be perceived by the intellect, not by the ear,” he wrote. His intention was to illuminate how the six known planets of that time orbit the Sun and produce distinctive tones relative to their distance and in harmonic relationship with each other — the “music of the spheres.”
NASA’s A Universe of Sound project is an ongoing program to sonify astronomical image data taken at multiple wavelengths. Targets include the galactic center (above), the Cassiopeia A supernova remnant, and the Pillars of Creation . You can explore the available sonifications here.
## Sonification
Enriching the exploration between sound and space is the recent increased interest in sonification — the process of converting data into sound. While abstract for some, it creates a new world of imagery for blind or visually impaired individuals interested in or working in the sciences.
Sound engineers and astronomers are now collaborating to create a vocabulary using musical terms such as timbre, tone, and pitch, associating them with spatial imaging. Imagine an astronomical image as a field of complex data points containing information such as brightness and wavelength, which taken together can have a host of other visual descriptors often taken for granted by fully sighted people. Via sonification, these points become audio representations, allowing the visually impaired to experience the cosmos — and sighted individuals to experience visual and other information in an entirely new way.
## Listening to the universe
Pythagoras, Kepler, and others may have professed a musical connection to the cosmos, but current technology provides definitive ways in which sound can show us the stars. In a sense, the rhythmic waves, pulses, and bursts we receive from space are all elements of the soundtrack to our universe.
For instance, in 1964, Arno Penzias and Robert Wilson picked up what we now know to be the cosmic microwave background (CMB) radiation, the echo from the Big Bang of 13.8 billion years ago. What first appeared to them as background noise ultimately provided evidence for much of modern cosmology. I find it noteworthy that the term itself, Big Bang, implies an explosion of sound, although that could not have happened because what we define as sound is created by vibrating air molecules. This is why screaming in the vacuum of space will not get you any attention: There is no air to transmit the vibration and create sound.
But although the limitations of the human ear require us to rely on processing technology to make that which is inaudible audible, we can still “hear” the extraordinarily low frequencies and respective overtones of the early universe, as captured by satellites such as Planck, through the CMB. The resulting sound is nothing short of mesmerizing.
Our ears are further limited as they can only detect frequencies between 20 and 20,000 Hertz, which puts many natural vibrations outside of our hearing range. Our Earth generates audible and inaudible sounds from deep within the planet, down to background hums as deep as 10,000 times lower than the ear can hear.
High above, electromagnetic waves propagating through the inner region of Earth’s magnetosphere, called the plasmasphere, create their own type of music. In this space above our planet, waves such as whistler-mode waves, chorus waves, and plasmaspheric hiss waves are associated with everything from the tug of the Moon to lightning strikes. (Lightning also creates waves that have been seen to bounce and “ring” through the ionosphere.)
The plasma there is far too thin for its vibrations to be able to tickle our eardrums. But, as with the CMB, these electromagnetic waves can be recorded by antennas and converted into sound. For instance, lightning causes electrons to oscillate in waves that, when played through speakers, sound like whistles — hence the name whistler-mode waves. It’s worth noting that transforming such waves into audible sound is akin to the way visual astronomers collect light at frequencies our eyes can’t see — say, infrared light using the James Webb Space Telescope — and transform it into a visual image.
These waves are not unique to Earth. Voyager 1 first recorded whistler-mode waves associated with lightning in Jupiter’s plasmasphere in 1979. Now referred to as jovian whistlers, they provide proof of otherworldly lightning and share with us a small taste of sound from the largest planet in our solar system. Both Voyager spacecraft have also now passed through the heliopause — the boundary of our solar system where the solar wind collides with the surrounding interstellar plasma — and have recorded many types of oscillating plasma waves. Ironically, as we listen to interstellar space and the phenomena that shape it, from under-lying galactic rotation to supernovae and smaller solar eruptions, the Voyagers carry a representative sample of Earth’s natural sounds and musical compositions in the form of the Golden Record produced by NASA and Carl Sagan.
Many stars oscillate in numerous ways as well, from hydrogen-burning main sequence stars like our Sun to pulsating variable stars to quaking neutron stars. Some oscillations are due to turbulent plasma trapped in stars’ upper layers; others come with changes brought on by age. Starquakes and pulsations can produce variations in a star’s brightness, and observations of these oscillations form the basis for the fields of helioseismology and asteroseismology. Such studies allow a further understanding of the inner structure and workings of stars at every stage of their evolution. If we accept that the physics of sound Pythagoras discovered holds constant, then it is plausible that all these events produce pitches, overtones, and harmonics with frequencies that can be converted into the audible range.
Even catastrophic events can yield auditory treasure. A study published earlier this year in Nature revealed that some gamma-ray bursts (GRBs) contain more than just noise. GRBs that occur when two neutron stars collide can create short-lived hypermassive neutron stars (HMNS), the fastest spinning stars in the universe. They are difficult to study, lasting only seconds before collapsing into a black hole. In the paper, astronomers reported they were able to recognize the presence of an HMNS by the frequency at which the staccato blast of gamma rays flickered. The trick is that the oscillations of HMNSs lack a consistent pitch; instead, they produce a variety of so-called quasi-periodic oscillations around a peak frequency, adding clamor to our celestial orchestra.
## What is music?
I imagine the future may hold similarly difficult-to-define patterns that will need more complex deciphering. In the meantime, the variety of identifiable signals we receive can be recorded, processed, resynthesized, and brought into the audible range. With varying rhythm and amplitude, some pitches and sounds rise and fall, while others remain constant. They can be reflective of lions roaring, a heart beating, or familiar ever-present white noise. More tonal results have been referred to as birds chirping or flocking, early video game sound effects, or ascending and descending vocal glissandos.
None of the samples we obtain from space may be music as most of society has come to define it. But is there a definitive definition of music? Today’s contemporary music would probably be unidentifiable to the ears of the first singers of Gregorian chant. And consider “4′ 33″” by the celebrated 20th-century composer John Cage, who believed the three-movement composition his most important work. It is four minutes and 33 seconds in which the performer sits in silence — which provokes the question of whether any and every auditory experience can be regarded as musical.
Is it possible that the breathtaking sight of a globular cluster with exoplanets rotating around their suns also produces yet-to-be-captured sounds? Could galactic groups generate collective waves that create a unique cosmic symphony? Maybe this would be the true music of the spheres. Would we perceive what we hear as a cacophonous jungle filled with alien, unconnected sounds? To some, yes. But that same term — cacophony — has also been used to describe the sound created when multiple piano keys are simultaneously played as closely as possible. In composition, this is referred to as a cluster chord, which sounds like noise to many but is nonetheless 100 percent musical.
Twenty-four centuries after Pythagoras died, Danish composer Rued Langgaard wrote a groundbreaking orchestral homage in his “Music of the Spheres.” More than 100 years after that, the British rock band Coldplay released their Music of the Spheres album, with each track representing a celestial body. Clearly, the impact of this spiritual, philosophical, mathematical, metaphysical, scientific, and astronomical concept has resonated throughout the millennia. I, for one, am standing by to enjoy the ongoing evolution and wonderment of this sonic journey. | 2,676 | 13,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-26 | latest | en | 0.964538 |
https://www.pinterest.ru/ashleebolin9/brain-teasers-riddles/ | 1,620,778,509,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990419.12/warc/CC-MAIN-20210511214444-20210512004444-00032.warc.gz | 993,319,228 | 62,458 | # BrAiN TeAsErS & RiDdLeS!!!
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Free printable Brain Teasers For Kids with answers featuring a mix of logical puzzles, drawing challenges and word searches ... engage brain!
With the Superbowl coming up next weekend, here's a fun little brain teaser you can use to stump your older kids (or husbands, coworkers or...
99 Riddles Bell Ringers / Lesson Starters / Morning Meeting - Riddles for Teens Are you looking for some gripping ways to settle your students really quickly at the start of your lesson? Here are 99 of the best riddles out there that you can use as bell ringers!
Pet Adoption Logic Puzzle - Free Coloring Pages for Kids - Printable Colouring Sheets
Cool Math Puzzles With Answers 1 : Funny Logical Maths Puzzle Difficulty ★☆☆☆☆ Popularity ★★★☆☆ If 1 = 5 , 2 = 25 , 3 = 325 , 4 = 4325 Then 5 = ? View Answer Discuss in Forum Answer & Explanation Solution: 1 Already stated 1=5 =>5=1 ============================================== Cool Math Puzzles With Answers 2 : Simple Maths Picture Problem Difficulty ★☆☆☆☆ Popularity ★★☆☆☆ Replace the question mark with the number in this simple maths problem ? View Answer Discuss in…
Can you solve these 5 matchstick puzzles riddles? Genius Matchstick Puzzle Riddles with answer. Move only one matchstick and make the equation correct. Take the challenge and solve these best matchstick puzzles. You will have to move just one stick and fix the equation. Let see, how you are going to make these equation true! Comment you answer or suggestion blow, before check the solution. Enjoy the puzzle! Matchstick Puzzle Riddles: #1 Matchstick riddle: 7 - 9 + 2 #2 Matchstick riddle: 6…
Brain Teaser | easy math puzzle What number should replace the question mark? Consider four triangles, each with different numbers in its corners. Apply your math skills to solve this puzzle!
How many squares do you see?
All of these matchstick puzzles involve triangles. Either take away or move the sticks to form new figures according to the directions. Stimulates creative thinking.
Brain Teaser | triangle riddle find the missing number in the last violet triangle.
Train your Brain - Do Riddles Everyday! | 514 | 2,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-21 | latest | en | 0.795679 |
http://www.studymode.com/essays/Hydrogen-Super-Highway-1106258.html | 1,524,436,976,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945660.53/warc/CC-MAIN-20180422212935-20180422232935-00072.warc.gz | 522,522,545 | 26,458 | # Hydrogen Super Highway
Topics: Electromagnet, Magnetic field, Magnet Pages: 33 (9928 words) Published: September 29, 2012
Prashant Zanwar
University of Central Florida
I. Introduction To Maglev Technology And Application
1.1 Principle of Maglev System
Maglev is a system in which the vehicle runs levitated from the guideway (corresponding to the rail tracks of conventional railways) by using electromagnetic forces between superconducting magnets on board the vehicle and coils on the ground. The following is a general explanation of the principle of Maglev.
fig. 1.1
1.1.1 Magnetic Levitation
The "8" figured levitation coils are installed on the sidewalls of the guideway. When the on-board superconducting magnets pass at a high speed about several centimeters below the center of these coils, an electric current is induced within the coils, which then act as electromagnets temporarily. As a result, there are forces which push the superconducting magnet upwards and ones which pull them upwards simultaneously, thereby levitating the Maglev vehicle.
fig. 1.2
1.1.2 Lateral Guidance
The levitation coils facing each other are connected under the guideway, constituting a loop. When a running Maglev vehicle, that is a superconducting magnet, displaces laterally, an electric current is induced in the loop, resulting in a repulsive force acting on the levitation coils of the side near the car and an attractive force acting on the levitation coils of the side farther apart from the car. Thus, a running car is always located at the center of the guideway.
1
fig. 1.3
1.1.3 Propulsion
A repulsive force and an attractive force induced between the magnets are used to propel the vehicle (superconducting magnet). The propulsion coils located on the sidewalls on both sides of the guideway are energized by a three-phase alternating current from a substation, creating a shifting magnetic field on the guideway. The on-board superconducting magnets are attracted and pushed by the shifting field, propelling the Maglev vehicle.
1.1.4 Floatation
The electrodynamics, repulsion-type maglev system, originally patented by American scientists in the 1960s, is the focus of the development program of the Japan Railway Technical Research Institute. In this approach, levitation magnets on the top of a guideway or as is the case with present Japanese prototype designs, in the guideway sidewalls push away superconducting magnets grouped underneath or at the bottom sides of the vehicles. Linear synchronous propulsion coils in the guideway propel the vehicles. This type of system allows for a large air gap (about 15 cm) between opposing magnets. In the electromagnetic, or attraction-type, Maglev developed by Transrapid International in Germany, conventional iron-core magnets in the vehicle's wraparound arms are pulled up to magnets under the guideway. A relatively small air gap (1 cm) separates the vehicle and guideway magnets. Although not part of the present Transrapid design, superconducting magnets can be incorporated in attraction-type Maglev systems.
2
fig. 2.1
1.1.5 Maglev Technology Some Facts
Q. How does low-speed Maglev technology work? A. Maglev technology employs powerful superconducting magnets to levitate or "float" cars about 2 inches above a guideway. Liquid helium in a special encasement cools the magnets to near absolute zero (or about -400 degrees Fahrenheit), enabling relatively small magnets to create very powerful fields. Linear induction motors using magnetic fields propel the cars. Such motors are already adapted to transit, such as in Vancouver’s automated SkyTrain Light Rail system. Electrodynamics involving the interaction of electrical currents and magnetic forces, state-of-art computers and microprocessors maintain guidance (vertical and horizontal spacing) during travel. Low-speed Maglev travels up to 60 mph. In Pittsburgh’s case, top speed will be about 40 mph. Q. Why do the superconducting magnets have to be super-cooled? A. It’s a matter of complex physics. But by bathing superconducting magnetic coils in... | 877 | 4,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.931388 |
https://physics.stackexchange.com/questions/334639/does-random-phase-approximation-rpa-response-function-obey-kramers-kronig-rela | 1,709,612,411,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948217723.97/warc/CC-MAIN-20240305024700-20240305054700-00363.warc.gz | 457,623,047 | 40,527 | # Does random phase approximation (RPA) response function obey Kramers-Kronig relations?
Consider the screened Coulomb interaction in electron liquid, which in the random phase approximation (RPA) takes the form $$V(q,\omega)=\frac{v(q)}{1-v(q)\Pi(q,\omega)},$$ where $v(q)$ is the unscreened Coulomb interaction, $\Pi(q,\omega)$ is the electron gas polarizability.
It is known the exact screened interaction $V_\mathrm{exact}(q,\omega)$ obeys the Kramers-Kronig relations $$\mathrm{Re}\,V_\mathrm{exact}(q,\omega)=v(q)-\frac1\pi\mathcal{P}\int d\omega'\frac{\mathrm{Im}\,V_\mathrm{exact}(q,\omega')}{\omega-\omega'},$$ $$\mathrm{Im}\,V_\mathrm{exact}(q,\omega)=\frac1\pi\mathcal{P}\int d\omega'\frac{\mathrm{Re}\,V_\mathrm{exact}(q,\omega')-v(q)}{\omega-\omega'}.$$ The polarizability $\Pi(q,\omega)$, being a retarded response function, also obeys similar relations (although without $v(q)$ in the right hand sides).
Does the RPA interaction $V(q,\omega)$ obey the Kramers-Kronig relations? If yes, how it can be proved?
• I am not certain, but I think the answer is yes. The RPA form is basically a subset of the Feynman diagrams in the exact case. Each Feynman diagram should satisfy the analyticity required by Kramers-Kronig relations. May 22, 2017 at 19:07
• @leongz We can, for example, truncate the RPA series up to the third order $V=v+v\Pi v+v\Pi v\Pi v$ (taking, in fact, a subset of Feynman diagrams) and then we get the second order poles of $V$ where $\Pi$ has the first order ones. The presence of second order poles will probably ruin the required analytical properties, isn't it? May 23, 2017 at 14:27
• I don't think the order of the poles matters in Kramers-Kronig relations. Please see derivation on wiki. May 24, 2017 at 9:08
Yes. The RPA response function $V(\omega)$ still obeys the Kramers-Kronig (KK) relation, as long as the polarization function $\Pi(\omega)$ obeys the KK relation. The key point is to show that all the higher-order poles that appear in the RPA expansion can be reduced to first-order using the KK relation of $\Pi(\omega)$, such that they do not cause a problem.
The KK relation of the polarization function implies that we can express $\Pi(\omega)$ as $$\Pi(\omega)=\int\frac{\mathrm{d}\omega'}{2\pi}\frac{A_\Pi(\omega')}{\omega-\omega'+\mathrm{i}0_+},$$ where $A_\Pi(\omega)\equiv -2\Im\Pi(\omega)$ is the spectral function of the polarization. Then let us focus on the term $v\Pi(\omega)v\Pi(\omega)v$ in the RPA expansion. We want to show that its sencond-order poles actually can be resolved by first-order poles and hence not problematic. To see this, we start from $$\begin{split}\Pi(\omega)^2&=\int\frac{\mathrm{d}\omega_1}{2\pi}\frac{\mathrm{d}\omega_2}{2\pi}\frac{A_\Pi(\omega_1)}{\omega-\omega_1+\mathrm{i}0_+}\frac{A_\Pi(\omega_2)}{\omega-\omega_2+\mathrm{i}0_+}\\ &=\int\frac{\mathrm{d}\omega_1}{2\pi}\frac{\mathrm{d}\omega_2}{2\pi}\Big(\frac{1}{\omega-\omega_1+\mathrm{i}0_+}-\frac{1}{\omega-\omega_2+\mathrm{i}0_+}\Big)\frac{A_\Pi(\omega_1)A_\Pi(\omega_2)}{\omega_1-\omega_2}\end{split}.$$ Then we can define a new spectral function $$A_\Pi^{(2)}(\omega)=2\int\frac{\mathrm{d}\omega'}{2\pi}\frac{A_\Pi(\omega)A_\Pi(\omega')}{\omega-\omega'}=2A_\Pi(\omega)\Re\Pi(\omega)=-4\Im\Pi(\omega)\Re\Pi(\omega),$$ which resolves the pole $(\omega-\omega')^{-1}$ in the integrant by the KK relation of $\Pi(\omega)$, hence reducing the total order of the poles by one. The spectral function $A_\Pi^{(2)}$ will be as analytic as $\Pi(\omega)$, which is exactly the spectral function of $\Pi(\omega)^2$ with out sencond-order poles: $$\Pi(\omega)^2=\int\frac{\mathrm{d}\omega'}{2\pi}\frac{A^{(2)}_\Pi(\omega')}{\omega-\omega'+\mathrm{i}0_+}.$$ Following the above approach, it is straight forward to show that all the higher-order terms in the RPA expansion has the spectral resolution of the same form in terms of the first-order poles only. $$\Pi(\omega)^n=\int\frac{\mathrm{d}\omega'}{2\pi}\frac{A^{(n)}_\Pi(\omega')}{\omega-\omega'+\mathrm{i}0_+}.$$ All the higher-order spectral functions $A_\Pi^{(n)}(\omega)$ can be express as a polynomial of $\Re\Pi(\omega)$ and $\Im\Pi(\omega)$. So as long as $\Pi(\omega)$ obeys the KK relation, all terms in the RPA expansion will also obeys the KK relation, as well as the RPA response function $V(\omega)$.
• Thank you! But I don't understand what do you mean by "resolving second-order pole by first-order poles". For example, if $A_\Pi(\omega)=(a/\pi)/[(\omega-\omega_0)^2+a^2]$ (Lorentzian spectral function), then $\Pi(\omega)=(1/2\pi)(\omega-\omega_0+ia)^{-1}$, i.e. $\Pi(\omega)$ has the first-order pole at $\omega=\omega_0-ia$, which means existence of a damped excitation. At the same time, $\Pi(\omega)=(1/2\pi)^2(\omega-\omega_0+ia)^{-2}$ still has the second-order pole at $\omega=\omega_0-ia$, as can be obtained either by direct squaring of $\Pi(\omega)$ or by using the function $A_\Pi^{(2)}$. May 28, 2017 at 12:06 | 1,543 | 4,931 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-10 | latest | en | 0.751302 |
http://stackoverflow.com/questions/15456445/optimizing-scala-code-snippet | 1,386,220,271,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163040022/warc/CC-MAIN-20131204131720-00077-ip-10-33-133-15.ec2.internal.warc.gz | 166,188,274 | 13,984 | # Optimizing Scala code snippet
I'm still learning Scala, and I decided to implement an Othello game. I'm using Alpha Beta pruning basing myself on this algorithm to implement the AI component.
However I've realized that my implementation is not very efficient. I start noticing performance issues if I increase the maximum depth of the algorithm to something greater than 5. I realize that the search space is exponential, so I was hoping if you can help me point out ways of optimizing this code.
Here's how I implemented the algorithm:
``````trait MaxMin
case class Max() extends MaxMin
case class Min() extends MaxMin
object AlphaBeta {
type Move = List[State]
type FitnessMove = Tuple2[Int, List[Move]]
def not(p: Player) = p match {
case _: Player2 => Player1()
case _: Player1 => Player2()
}
def max(x: FitnessMove, y: FitnessMove) = if (x._1 >= y._1) x else y
def min(x: FitnessMove, y: FitnessMove) = if (x._1 <= y._1) x else y
def terminal(turn: Int) = if (turn >= 64) true else false
def search(board: Board, player: Player, turn: Int, MAX_DEPTH: Int = 5): Move = {
def alphaBeta(node: Board, depth: Int, alpha: Int, beta: Int,
moveChoice: List[Move], player: Player, p: MaxMin, turn: Int): FitnessMove =
if (depth == 0 || terminal(turn))
(player.evalHeuristic(node, turn), moveChoice)
else
p match {
case _: Max =>
player.getPossibleMoves(node).
takeWhile(_ => beta > alpha). // Pruning
foldLeft((alpha, moveChoice)) { case ((alpha, moveChoice), move) =>
val simulate = player.simulateMove(node, move)
max((alpha, moveChoice),
alphaBeta(simulate, depth-1, alpha, beta,
move :: moveChoice, not(player), Min(), turn+1))
}
case _: Min =>
player.getPossibleMoves(node).
takeWhile(_ => beta > alpha). // Pruning
foldLeft((beta, moveChoice)) { case ((beta, moveChoice), move) =>
val simulate = player.simulateMove(node, move)
(
min((beta, moveChoice),
alphaBeta(simulate, depth-1, alpha, beta,
moveChoice, not(player), Max(), turn+1))._1,
moveChoice
)
}
}
val (_, moveChoice) = alphaBeta(board, MAX_DEPTH, Integer.MIN_VALUE, Integer.MAX_VALUE, List.empty[Move], player, Max(), turn)
}
}
``````
I could probably get better performance if I use while loops and a more "imperative" approach but I prefer to maintain the code immutable. Are there any improvements I could make on the code, or would you approach the whole implementation of the algorithm differently?
Here's the game if you want to check it out. Thanks!
-
Just if you are interested: Some time ago I have written a small console based Othello implementation, without AI: gist.github.com/sschaef/2472666 – sschaef Mar 17 at 1:04
can you make sure you attach a test so that I can freely refactor your code without breaking its functionalities? – Edmondo1984 Mar 17 at 2:45
Hi @Edmondo1984 yeah I'll try to add a few tests soon. – Samuel Heaney Mar 17 at 19:49
I haven't analyzed all of your code, but there was one point above that looked very "suspicious"
``````player.getPossibleMoves(node).
takeWhile
``````
So I thought: OK - if he really gets all the moves here in the list and filters afterwards, this could be very inefficient. This is your implementation which does the actual work with yield
`````` def findPossibleMoves(playerDisk: Int): List[Move] =
groupStatesByMove {
(for {
i <- upperLimit to lowerLimit
j <- leftLimit to rightLimit
if repr(i)(j) == 0
dir <- 1 to 8
disk = getPlayerDisk(i, j, dir)
if disk == playerDisk && findMove(i, j, dir)(playerDisk)
} yield new State(i, j, dir, playerDisk)).toList
}
``````
Now - if you just make the return type an `Iterable` (which should be sufficient for `takeWhile` and most other list-operations except direct access by index - see IterableLike) and if you leave out the `toList`at the end (also in the calling function of `Player`) the output is automatically set to the input type - which is list in this case, as `x to y` yields a list. The trick here is to use a Stream instead - instead of `1 to 8` use `(1 to 8).toStream` instead. The same goes for all other `to` expressions. Then each `State` should only be generated when needed - so: until `takeWhile` returns false.
Here is an example:
``````object TakeTest extends App {
def generate : Iterable[Int] = {
for (j <- (1 to 10).toStream; if (println("generate: " + j) == ()))
yield j
}
println("takeWhile with list:")
generate.toList.takeWhile(_ < 3)
println("takeWhile with Stream/Iterable:")
generate.takeWhile(_ < 3).toList
}
``````
That's the only hint I can give from the code, otherwise I would use - and you should probably make yourself familiar with it: jvisualvm
-
Thanks for the pointers, I was definitely overlooking the part where I generate the a whole collection every time I obtain possible moves. – Samuel Heaney Mar 17 at 20:57 | 1,248 | 4,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2013-48 | latest | en | 0.65792 |
https://www.ato.gov.au/rates/schedule-1---statement-of-formulas-for-calculating-amounts-to-be-withheld/?page=5 | 1,679,730,535,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00134.warc.gz | 732,991,661 | 19,941 | • ## Shading out point (SOP)
The SOP relative to an employee’s WFT is calculated as follows:
Multiply WFT by 0.1 and divide the result by 0.0800. Ignore any cents in the result.
Example
Employee has shown six dependent children at question 12 and scale 2 is applied:
WFT = ([3,533 × 6] + 38,474) ÷ 52
WFT = 1,147.5385 or \$1,147.54 (rounded to the nearest cent)
SOP = (WFT × 0.1) ÷ 0.0800
SOP = (\$1,147.54 × 0.1) ÷ 0.0800
SOP = 1,434.4250 or \$1,434 (ignoring cents)
End of example | 183 | 492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.87541 |
https://gmatclub.com/forum/m01q16-120578.html?kudos=1 | 1,498,156,797,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319688.9/warc/CC-MAIN-20170622181155-20170622201155-00715.warc.gz | 736,416,867 | 50,880 | It is currently 22 Jun 2017, 11:39
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# m01q16
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14 Sep 2011, 05:23
Q is a set of 10 numbers. If the range of Q is zero, then what is the standard deviation of Q ?
A.0
B.1
C.3.33
D.5
E. insufficient information
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14 Sep 2011, 14:55
A.
If the range is zero SD will be zero. Why?
When can range be zero?
When set is one element set or all the elements of set is identical.
In either case mean of the set will be the number in the set.
What is SD?
It is the dispersion of individual element of set from mean. Now if the mean is that same identical number of set, there is no dispersion. Hence SD will be zero.
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20 Oct 2011, 21:23
When range is zero, it means that all items in the set are same. Standard deviation is the sum of diversions from mean. Since all the items of the set are same, there will be no diversion from the mean and hence SD is zero.
I hope its clear now.
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14 Jan 2012, 22:21
IMO - A
Range signifies standard deviation so if range is 0 SD will also be zero.
Re: m01q16 [#permalink] 14 Jan 2012, 22:21
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# m01q16
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1. ## Thread: [RESOLVED] Recursive Function to form the sum of two positive integers a and b?
by papi-ro
Replies
2
Views
2,363
### Re: Recursive Function to form the sum of two positive integers a and b?
I figured it out by myself.
static int recSum(int x, int y)
{
int result = 0;
if (y == 0)
{
return (x);
2. ## Thread: [RESOLVED] Recursive Function to form the sum of two positive integers a and b?
by papi-ro
Replies
2
Views
2,363
### [RESOLVED] Recursive Function to form the sum of two positive integers a and b?
Hi all,
I'm new to c-sharp, so I will appreciate if you guys go easy on me.
I have to write a recursive function to form the sum of two positive integers a and b. I have to test the program by...
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https://math.stackexchange.com/questions/918445/problems-with-nesting-proof-predicates-in-first-order-logic | 1,721,890,995,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00820.warc.gz | 326,604,869 | 40,398 | # Problems with nesting proof predicates in first order logic.
Whenever I start nesting proof predicates, I always seems to run into these bizarre situations. I was wondering if anyone knows about this and could shed some light on it or provide me with some references. Thank you!
----Definition----
Let's define a set S as follows.
$$S := \begin{cases} \mathtt{ZF} & \text{ if \mathtt{Con}(\mathtt{ZF})} \\ \emptyset & \text{ otherwise} \end{cases}$$
Although we can't determine whether S is $\mathtt{ZF}$ or S is $\emptyset$, we can still define $S$.
Consider the predicate $\phi_S(x)$ which abbreviates: $\mathtt{Con}(\mathtt{ZF}) \rightarrow x=\mathtt{ZF} \wedge \neg \; \mathtt{Con}(\mathtt{ZF}) \rightarrow x=\emptyset$.
The predicate $\phi_S(x)$ acts as a definition for S and we have $\mathtt{ZF} \vdash \exists! x \; \phi_S(x).$
----Proof Predicates----
For any set X of Godel numbers for first order formulas, in a standard manner, one can formally define a proof predicate $\mathtt{Pf}_{X}$. (I can elaborate if necessary.)
We will consider both $\mathtt{Pf}_{\mathtt{ZF}}$ and $\mathtt{Pf}_{S}$.
It seems reasonable that we can nest $\mathtt{Pf}_{\mathtt{ZF}}$, but how do we nest $\mathtt{Pf}_{S}$?
To do so, $\mathtt{Pf}_{S}(\mathtt{Pf}_{S}(\ulcorner \psi \urcorner))$ will abbreviate $\exists x \; \phi_S(x) \wedge \mathtt{Pf}_{x}(\ulcorner \exists x \; \phi_S(x) \wedge \mathtt{Pf}_{x}(\ulcorner \psi \urcorner)\urcorner)$.
In other words, we include $S$'s definition each time we nest to a lower level.
Note: $\mathtt{Con}(X)$ will just abbreviate $\exists \ulcorner \psi \urcorner \; \neg \; \mathtt{Pf}_X(\ulcorner \psi \urcorner)$.
----Proofs----
1) First, I notice that $\mathtt{ZF} \vdash \mathtt{Con}(\mathtt{ZF}) \rightarrow S=\mathtt{ZF} \wedge \mathtt{Con}(S)$.
2) And, $\mathtt{ZF} \vdash \neg \; \mathtt{Con}(\mathtt{ZF}) \rightarrow S=\emptyset \wedge \mathtt{Con}(S)$.
3) From these two, we get: $\mathtt{ZF} \vdash \mathtt{Con}(S)$.
4) Further, we get: $\mathtt{ZF} \vdash \mathtt{Pf}_{\mathtt{ZF}}(\ulcorner \mathtt{Con}(S) \urcorner)$.
(I can elaborate on the standard properties of these proof predicates if necessary.)
Consider $\mathtt{Pf}_{S}(\ulcorner \mathtt{Con}(S) \urcorner)$. This could not happen if $S = \emptyset$. Also, it does happen if $S = \mathtt{ZF}$ as we just showed.
5) Therefore, $\mathtt{ZF} \vdash \mathtt{Con}(\mathtt{ZF}) \leftrightarrow \mathtt{Pf}_{S}(\ulcorner \mathtt{Con}(S) \urcorner)$.
For any $\ulcorner \psi \urcorner$, consider $\mathtt{Pf}_{S}(\mathtt{Pf}_{S}(\ulcorner \psi \urcorner))$. This implies that $S = \mathtt{ZF}$ and $\mathtt{Con}(\mathtt{ZF})$ because otherwise we have the empty set which can't prove anything.
6) $\mathtt{ZF} \vdash \mathtt{Pf}_{S}(\mathtt{Pf}_{S}(\ulcorner \psi \urcorner)) \rightarrow \mathtt{Con}(\mathtt{ZF})$.
So we have $\mathtt{Pf}_{\mathtt{ZF}}(\mathtt{Pf}_{S}(\ulcorner \psi \urcorner))$. For the same reasons, at the lower level we have $S = \mathtt{ZF}$ and $\mathtt{Con}(\mathtt{ZF})$ i.e. $\mathtt{Pf}_{\mathtt{ZF}}(\ulcorner S = \mathtt{ZF} \urcorner)$ and $\mathtt{Pf}_{\mathtt{ZF}}(\ulcorner \mathtt{Con}(\mathtt{ZF}) \urcorner)$.
However, $\mathtt{Pf}_{\mathtt{ZF}}(\ulcorner \mathtt{Con}(\mathtt{ZF}) \urcorner)$ is equivalent to $\neg \; \mathtt{Con}(\mathtt{ZF})$.
7) Therefore, $\mathtt{ZF} \vdash \mathtt{Pf}_{S}(\mathtt{Pf}_{S}(\ulcorner \psi \urcorner)) \rightarrow \neg \; \mathtt{Con}(\mathtt{ZF})$.
Combining both 6) and 7), we get:
8) $\mathtt{ZF} \vdash \neg \; \mathtt{Pf}_{S}(\mathtt{Pf}_{S}(\ulcorner \psi \urcorner))$.
If we picked $\ulcorner \psi \urcorner$ to be $\ulcorner \mathtt{Con}(S) \urcorner$, then we would have:
9) $\mathtt{ZF} \vdash \neg \; \mathtt{Pf}_{S}(\mathtt{Pf}_{S}(\ulcorner \mathtt{Con}(S) \urcorner))$.
----Conclusion----
In the end, this is a bizarre situation because we have both of the following:
$\mathtt{ZF} \vdash \mathtt{Con}(\mathtt{ZF}) \leftrightarrow \mathtt{Pf}_{S}(\ulcorner \mathtt{Con}(S) \urcorner)$, and $\mathtt{ZF} \vdash \neg \; \mathtt{Pf}_{S}(\mathtt{Pf}_{S}(\ulcorner \mathtt{Con}(S) \urcorner))$.
How can the proof predicates $\mathtt{Pf}_{S}$ and $\mathtt{Pf}_{S}\mathtt{Pf}_{S}$ say the opposite of each other?
• Are you sure that the formula Con(ZF)→x=ZF∧¬Con(ZF)→x=∅ is "meaningful", i.e. well-formed ? From x=∅, it seems that the variable $x$ range over set; if so, what is ZF ? Commented Sep 3, 2014 at 19:09
• ┌ψ┐ is a number, nit a variable. Thus, how you can quantify it into ; ∃┌ψ┐ ? Commented Sep 3, 2014 at 19:10
• I haven't thought this through carefully, but I'm suspicious about how much of the standard machinery you can salvage for your $\texttt{Pf}_S$. Unlike in Gödel's proof, your $S$ is not recursive (the $\texttt{Con}$ used in its definition is $\Pi_1$), and you won't be able to refer to $\Sigma_1$ completeness, which seems a key ingredient.
– user138530
Commented Sep 6, 2014 at 1:32
• @MichaelWehar: I'm not totally sure how relevant this is, but I had in the back of my mind the (standard?) proof of Gödel's 2nd incompleteness theorem through the Bernays-Lob derivability conditions (in a Peano arithmetic setting perhaps). The 3rd condition is usually obtained from the fact that PA (and hence ZFC as well) proves all true $\Sigma_1$ sentences, and perhaps this is where things go wrong in your setting.
– user138530
Commented Sep 6, 2014 at 4:00
• @MichaelWehar: My "3rd" condition is in fact the 2nd condition at the link you gave.
– user138530
Commented Sep 7, 2014 at 3:36
I think it might be helpful to get clear on exactly what $Pr_S(\ulcorner\phi\urcorner)$ says. In ZF it is provably equivalent to saying the following informal thing:
Either ZF is consistent and $\ulcorner\phi\urcorner$ is provable in ZF, or ZF isn't consistent and $\ulcorner\phi\urcorner$ is provable from nothing.
And since the second disjunct is provably false in ZF it just says: ZF is consistent and $\ulcorner\phi\urcorner$ is provable in ZF. In other words, $Pr_S(\ulcorner\phi\urcorner)$ is provably equivalent in ZF to $Con(ZF)\wedge Pr_{ZF}(\ulcorner\phi\urcorner)$.
(I assume $Pr_{\emptyset}$ means "provable from nothing", and not "provable from nothing apart from the axioms of first order logic"? At one point you said "the empty set can't prove anything". A similar diagnosis goes through on the other interpretation though.)
So I think the issue is, once you've spelt out the definition explicitly the result doesn't bear any straightforward resemblance to a provability predicate. It's also pretty clear why we get that $Pr_S(\ulcorner\phi\urcorner)$ and $Pr_S(Pr_S(\ulcorner\phi\urcorner))$ say different things, according to ZF.
$Pr_S(\ulcorner\phi\urcorner)$ says:
ZF is consistent and ZF proves $\ulcorner\phi\urcorner$
Which is true of many $\ulcorner\phi\urcorner$, assuming ZF is consistent. $Pr_S(Pr_S(\ulcorner\phi\urcorner))$ just says the following:
ZF is consistent and ZF proves that "ZF is consistent and ZF proves $\ulcorner\phi\urcorner$"
Which is, of course, is true of no $\ulcorner\phi\urcorner$, assuming ZF is consistent.
(Note that technically $Pr_S(Pr_S(\ulcorner\phi\urcorner))$ says: ZF is consistent and ZF proves $Pr_S(\ulcorner\phi\urcorner)$. However I noted above that $Pr_S(\ulcorner\phi\urcorner)$ is provably equivalent in ZF to $Con(ZF)\wedge Pr_{ZF}(\ulcorner\phi\urcorner)$.)
• I think this is a pretty good response! Formally, for any set $X$ of Godel numbers, I would define $Th(X)$ recursively where $Th(X)$ is the minimum set of Godel numbers for formulas that includes $X$ and is closed under the inference rules applied to the Godel numbers. Then, $\mathtt{Pf}_X(\ulcorner \psi \urcorner)$ will just mean $\ulcorner \psi \urcorner \in Th(X)$. Now, what you are saying is that from this definition, you can prove that $\mathtt{Pf}_S(\ulcorner \psi \urcorner)$ is equivalent to $\mathtt{Con}(\mathtt{ZF}) \wedge \mathtt{Pf}_{\mathtt{ZF}}(\ulcorner \psi \urcorner)$. Commented Sep 6, 2014 at 20:47
• I think your response is a great explanation for what is going on here, but I still feel uneasy and am confused on how we can define sets like S and on how statements like $\mathtt{Pf}_S(\mathtt{Pf}_S(\ulcorner \psi \urcorner))$ don't mean what we might think they mean. Would you be willing to talk further about this? I have another example that I'm trying to understand and I think your input would be quite helpful. Thank you! :) Commented Sep 6, 2014 at 21:02
• Certainly. You may already know this, but there is some work on non-standard provability predicates that seems related. For example there is a formula in the language of arithmetic which is true of the Godel number of a sentence in the standard model if and only if the sentence is provable in PA, but such that the claim corresponding to Con(PA) using that formula is provable in PA. (See for example matwbn.icm.edu.pl/ksiazki/fm/fm49/fm4915.pdf) Commented Sep 6, 2014 at 21:40
• Great! Thank you for the reference. I will contact you. :) Commented Sep 6, 2014 at 22:59 | 2,873 | 9,032 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.4375 | 3 | CC-MAIN-2024-30 | latest | en | 0.752264 |
https://www.eleceng.adelaide.edu.au/personal/dabbott/wiki/index.php/Small_World_Networks_2008 | 1,597,453,041,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740423.36/warc/CC-MAIN-20200815005453-20200815035453-00577.warc.gz | 653,868,758 | 17,585 | # Small World Networks 2008
## Weekly progress and questions
### Week 2
Wenrui (Linda):
Sweet and I had a short meeting last Wednesday, we decided to separate our proposal into three parts. Each person will talk about a little bit in each part. because we think it's necessary for both of us to understand every parts to get started.
For myself, I have prepared some definition about the seminar, and also I have read the book called "linked". But I haven't finish it yet. For the coming week, I will continue work on the proposal. maybe need to get start to make the PPT. find some fun video from www.youtube.com.
Zhujing (Sweet):
Linda & I managed our coming proposal this Wendnesday. There will be 3 parts and each part will take 5 mins(total 15 mins):
1.Introductions and background
What is "Small World Network"?
What is "Six Degrees of Separation"?
Is this theory existing in our life?
2.Show some fun things to support that the theory is existing around us
Relationship network of Hollywood stars
High school dating
A pretended network
About the pretended one, I have an idea. Recently, I am playing a wii game, "Super Mario Galaxy". The whole world in the game just 6 galaxys and each galaxy has 4-5 planets. If Mario wanna go any planet, he needs to fly to the centre of the galaxy first which the planet belongs to, and then tracks the network amoung the planets to get the destination.I sopport every planet is a individual node and each flying track is a link. We can calculate manually that how many degrees Mario needs :)
### Week 3
Wenrui (Linda):
For this week, I just use all the information I prepared for last week to made the ppt, basically on the introduction part and the fun part.
For the coming week, we will have our regular meeting(once two weeks) to discuss the serious part. such as the hierarchical clustering algorithms, so hopefully we will work out more about this part next week.
Zhujing (Sweet):
This week, we continued prepare our part 1&2, and had some ideas of the part 3.
1> I decided to change "Mario galaxy" to another game which mentioned in fun part last week, because the other one has a more clear network image.
2> Start drawing some pretended small networks(with 5-10 nodes) to find some measures for the part 3.
### Week 4
Wenrui (Linda):
We had a regular meeting with Matthew last Tuesday, during the meeting, we talked about the serious part. we talked about three types of failure - random link (or node) failure, targeted link (or node) failure, and cascading failures (one link fails, then the next links fail), and also an example of Google's algorithm ranks. hopefully it will be very helpful for the seminar next week.
For the coming weekends,I will finish my PPT, and send it to Matthew on the early of next week, hopefully can get some good feedback before the formal seminar.
Zhujing (Sweet):
After our regular meeting, we got some useful sense of the coming seminar, especially the most important part, the serious one.
1> About the fun part, we decided change the movie to a new series, because this series has better and more smooth images. And in the old movie there are some kinds of different languages, we must to use the subtitle. But the series is all English, that is more convenient for our work.
2>For the serious part, I was praparing the pertended networks. And I changed something of my old networks to make these show some different characters, such as target link. Because of these changes I need to recalculate some stuffs.
### Week 5
Wenrui(Linda):
last week, we did our proposal seminar, I didn't do that well, I talked too much about the hierarchical networks. However, I didn't clearly explain something important, and didn't include something important. e.g.
- we should include a plan for our software design.
- I should talk more about our Gantt chat, and as well as the break down structures.
- In the risks analysis part, the idea is to talk about the technical risks in the project itself, but not some personal issues.
- We should include the discussion of the data sets we will use.
- We should include the discussion of relevance to electrical engineering.
- we should mention what is our aim for the project.
- we should include the slides to explain finding shortest path, analyzing network efficiency, finding important hubs, finding weaknesses that can cause failure of a network, identifying weaknesses that make the network vulnerable to attack.
During the presentation, we learned our lessons, i.e. if we couldn't answer the question from the audiences, at least we should try to say something which is related to the question, it's very bad to just stand there and say nothing. if couldn't understand the question, just ask them to ask in a easy way. if still cannot answer, then we can say:" good question, We do not know the answer to that and we will definitely follow it up later."
For the coming week, I am planning to work on the software design. I will basically read the algorithms on the internet. hopefully I can write some simple code myself, and start to test it with the Mini networks.
Zhujian(Sweet):
This week, we just finished our proposal seminar. There were some weaknesses in our presentation, but we will improve these next time. The biggest problem for me is too nervous to listen to the questions. After prsentation, I just recalled these and found I could answer most of these. That is the most important thing I need to improve.
For the coming week, we will start some matlab analysis and read the references about our project.
### Week 6
Wenrui(Linda):
For last week, I have searched something about the software design part on the internet, but I haven't really finish that yet. I will continue my research next week as we got more time for the holidays.
Use enron as an example, we can start to write some simple program to analysis the degree distribution and as well as the clustering coefficient. e.g. the train map in Hongkong or New York.(any big cities)
what's more, I need to prepare for next week's meeting, we will talk about the software design by using enron as an example, hopefully next week's regular meeting will be helpful for our progress.
Zhujian(Sweet):
I was programming some functions in Java and read more references about the properties of network to figure out the measures of my hand-calculating part. I am not very sure whether my measures are correct. Next week we will have a meeting, I will ask for some helps.
Btw, when I run some Java codes in windows system, my computer alway not very stable :( Maybe I need install a Linux system :(
### Week 7
Wenrui(Linda):
According to our meeting on Wednesday, we talked about how to write the Pajek .net file from a simple state diagram. How we calculate the average degree distributions, and hence the find out the clustering coefficients.
For the coming week ,we will design a simple state diagram ourselves, and try to use the same method as Matthew showed us to calculate the clustering coefficients. I was planning to use New York city train map as an example to build our network, but Matthew gave us a better suggestion, that is to use the Singapore train map, because the road the is smaller and more organized. So that can reduce our work load.
Zhujian(Sweet):
We just had meeting on Wednesday. And then, I find some mistakes of my measures. I will fix these and figure out how to test in Pajek. There are still some problems of Java running in windows system. I am trying solve this this week. Maybe something softwares crash.
### Week 8
Wenrui(Linda):
From last week, I have started some calculations with the simple systems(train map), this is basic on the clustering coefficients calculation. Moreover, I have done the introduction part of the report which is due on week 8, it is similar to what we have done for proposal anyway.
For the coming week, we will try to do as much as we can for the Critical Design Review report.
Zhujian(Sweet):
This week, I start to read a new book named "Models and Methods in Social Network Analysis". That is very helpful for my understanding of the measures. I try to prove some fomulas by my own way. That is a very interesting thing :)
### Week 9
Wenrui(Linda)
From last week, I have finished the calculations of the clustering coefficient by using both the undirected and directed graphs, for undirected graph, I used Adelaide train map system as an example, set each station as a node in the graph. I used the Adelaide road map as an example of the directed graph, because some of the street are only on way to go. but some of them are both ways. I have also finish the project management part, which is the risks anysis, and the budget.
For the comming week, we will finish the report.
Zhujian(Sweet)
1> I have finished all the mini network examples which I mentioned in the proposal seminar manual calculation. And change these undirected networks to directed networks to compare the difference between them. The calculations include average degree (for directed networks include in-degree and out-degree), clustering coefficient, short path between 2 nodes, the longest distance between 2 nodes in the network, etc.
2> I downloaded a book named “Exploratory Social Network Analysis with Pajek” to know more properties of network and be familiar with Pajek, But still have some confused problem of using this software. I was trying solving this.
3> Preparing the critical design review report. I am in the charge of manual calculation and coding test. Coz I got in trouble of using Pajek, I will do the test next week, just need some hints. And for the JUNG part, now I decide to temporary stop. I forgot a lot of about java programming stuffs. That need a lot of time to recall. I will do that after the due week.
### Week 10
Wenrui(Linda)
For this week, we have corrected the calculation mistakes that Matthew corrected for us, what's more, we have finished the report.
For the comming week, i will work on the peer review. read the book and as well as study more about the software.
Matthew, do we need more meetings for the last 3 weeks of this semester?
Zhujian(Sweet)
1> Corrected the mistakes in our calculation parts, found the right way to use Pajek to test and finally, finished the report.
2> Download the book "Models and Methods in Social Network Analysis", which I mentioned in mid-break, to solve some problems I met in book "Exploratory Social Network Analysis with Pajek".
3> Prapared peer review report.
### Week 11 & 12
Wenrui(Linda)
For these two weeks, we had just read some reference books, we got our feedback for critical design review, so that we can improve our final project report. For the coming winter holidays, we decided to work on the software part of the project, some functionality of the software, such as java, Pajek
Zhujian(Sweet)
1> Read the books which I mentioned before
2> Find some courses which I am studing using our "Small world network". Some ones are quite similar, and the others are not but use some theories of our project. Such as, shortest path, degree, target link
### Week 13
Wenrui(Linda)
Last week, I have watched the movie "number" that Matthew gave to us. Volume one is talking about to find the central node of a network system, it's pretty relevant to our project. hopefully we can also work out some big events later by applying our discover during the project.
Also I will ask sweet to get the copy of "Models and Methods in Social Network Analysis" for me to read through.
One more question, for the enron data set, Matthew, you have already gave us the code, but it's in Python which i have never use before, should i rewrite it in java again?
Zhujian(Sweet)
This week, I was thinking a problem, the following is what I thought
The relationship is very important of society, especially in China. And I find the weak relationship always to be the keypiont to spread my own social network. For example, if I want to find a job, that is hard to have some helps from my best friends, because we always have the similar social networks. But the acquaintances in my network may give me more chances to get a job, because they may have different social relationships from mine. In engineering aspect, we always want to improve the weak link(eg:target link) to be stronger. In contrast, we always ignore these weak relationships in real social network, but sometimes these weak ones seem to be the keypoint to spread our relation nets. Should we need to improve these ones to be stronger like in the engineering aspects or just keep these weak?
Making social network links stronger requires time and energy. You have to look at the benefits to improving links to those people. Perhaps in terms of finding a job, you can just keep a weak link to them. A key difference between social networks and engineering networks is that in a social network you only have one link to another person. In networks like power networks and communications networks, you can usually add in multiple links between two nodes, and get an improvement in robustness. So it's a different kind of strength. --Mattjb 11:39, 16 June 2008 (CST)
### Week 14
Wenrui(Linda)
For the last few weeks of the semester, we will be busy for the exams and some of the homeworks. so maybe we will stop our project for the exam period, will continue our research after the exams. during these time, we will just read some reference book, such as "Linked" and some other papers. Matthew, could you give us a tasks list, i think it will be more clear for us to know our assignment for the project. we will figure that out during the semester break.
Graph file format converter (GML to Pajek), draw graphs, calculate network measures for networks (see network data below), matlab analysis and graphing of results, hierarchy algorithm (Enron data), --Mattjb 11:58, 16 June 2008 (CST)
Zhujian(Sweet)
1. After the meeting, I have done some Graph file format converter for Pajek. It works well. Later I will try to do some more like high school dating graph.
2. I am still studing for how to use Pajek. I have tried some examples in the book and make some other manual calculations to test the functions in Pajek.
### Week 1&2
Wenrui(Linda)
For the last two weeks, i have been doing some jave and matlab reviewing. i have tried to read the data from the data sets, later i will try to find the relationship between every single data, try to link them together, and finally use matlab to show the result by graphs.
Zhujian(Sweet)
1. These 2 weeks I have tried to build the High school dating for Pajek and also done the degree distribution, cluster coefficient atc. Now I am doing the analysis of these datas. I think that I need some other same kind graphs but with different number of nodes, links, then I can compare these and analyse better. Can I just change the original graph?
If you want to have some data to test your algorithms on, then this is a great idea, but don't draw any conclusions about social networks from it. --Mattjb 10:37, 18 August 2008 (CST)
2. A firend of mine is studing sociology. We are doing a small social network reaserch now. Can I use the data from our reaserch for this project?
Sure, but be wary that if your friend has collected the data and you use it, to make sure that you acknowledge this properly in your report & presentation. --Mattjb 10:37, 18 August 2008 (CST)
### Week 3&4
Wenrui(Linda)
For the enron email date set, i understand that the enron.net file is where to store every single code and their email address. enron.pairs file is where to store who actually send to the email to whom, which presented by code, it shows the direction as well. md5-authors.txt gives every single email address a uniq code, with the same length and the same type. However, when i try to match them, i can't find some of the email address in the first file which is enron.net. is there any data missing?
The enron.net Pajek file lists the nodes (number ID of person plus their email address) and a set of directed edges representing emails being sent. The other files are those I generated enron.net from. The message IDs are different to the person IDs (my software generates these). --Mattjb 14:01, 8 September 2008 (CST)
should i use java or C++ to make them look like what we need for the pajek software? just like what you had in md5-recipients.txt file? that will looks like Directed graph's file.
It's already in a Pajek format. I think there is still an issue with quote marks needed around the email addresses, and I will fix this up tonight. I can't test from home though. --Mattjb 14:01, 8 September 2008 (CST)
I have been trying to use the data from my facebook to make the file as well, i made the data set myself, what i did is make my name as the first node, all my friends in the facebook are the second level of the hierchical system, and then my friends' friends, and so on. it seems hard to get similar style of the file. facebook is where to know a friend, if you are a friend of some one, then the relationship should be both sides, so should be a undirected graph.
Yes, it's undirected, because each time a person adds someone, that someone must confirm the friendship link. --Mattjb 14:01, 8 September 2008 (CST)
it's obviously a hierachical.
While it has levels, it's not hierarchical in the sense that any person has a position above another person, nor is it a tree structure, since you can have cycles (A is friends with B is friends with C is friends with A, etc.) --Mattjb 14:01, 8 September 2008 (CST)
To anlaysis the third level of the data set, you need the data from the forth level, if i want to get a general result of the friendships, then that will be hard, because i will be need a really large data set, or maybe a data base. what's more, if just consider about the direct relationship but not 2 degrees or 3 degrees, in the hierachical system, we will only consider about the upper level and the lower level, then i don't know what the point to do that. However, if you find someone you know in your friend's list, add him/her to be your friend, that person may not know you at all. so in this situation the relationship is only one side.
For the matlab part, what kind of result should we show? i mean should we use matlab to draw the graphs to show all our value which we got from the pajek calculation?
Yes :) --Mattjb 14:01, 8 September 2008 (CST)
Zhujian(Sweet)
I think we got lots of social network. Should we do something about the Engineering network?
I have put some Internet networks up that you can manually turn into Pajek format files and process. Feel free to search for any other engineered networks. --Mattjb 14:01, 8 September 2008 (CST)
About my research, my friend and I just finished all the data collection. We are doing the analysis of these stuff. That is also about the social network. In my research, we just sort the links to 2 type, the "Money friendship" and "Emotion friendship". We will do the research of the 2 kind of links. I will do the analysis of data, my friend will do the part of literal analysis. I think that is better than we just use the data from internet.
For the Engineering network, I prefer the telecommunication network, like the telephone links, the base stations distribution of mobile phone, the exchange of a city or province, atc. That is just my viewpoint. I think Linda can do the part of engineering. Choose some networks have the hierarchical directory structure and include other normal or special properties, then start ur work :)
### Week 5
Wenrui(Linda):
software study:
Java: 1. readFile function: public class ReadFile; FileNotFoundException; import java.io.IOException; import java.io.File;
2. remove the duplicate data.
Zhujian(Sweet): Start social network analysis
1>Using stage 1 and 2 database to draw different pictures
2>Analyze the network
3>find the typical group structure
### Week 6
Wenrui(Linda):
1. Analysis Enron data sets by java.
3. file converter.
Zhujian(Sweet):
Go on to do the stage 3-5 social network analysis
1>Find special case in the network
2>Observe the data change
3>Using Pajek calculate special parameters
Start software Design
1> Properties choose
2> GUI structure
### Week 7
Wenrui(Linda):
Find Engineering network examples:
router network: 1. degree distribution
2. shortest distance
Zhujian(Sweet):
Finish all the social network anlysis
Focus on software design
start GUI programming
### Week 8
Wenrui(Linda):
some important properties of the network analysis:
1. target link 2. central node 3. distance 4. degree distribution 5. cluster coefficient 6. weak ties and strong ties
Zhujian(Sweet):
Debug GUI code
Algorithm analysis by java
1>shortest distance
3>cluster coefficient calculation
4>degree distribution calculation
### Week 9
Wenrui(Linda):
Engineering network example: Internode
Analysis the internode network by those important properties
2. cluster coefficient
Zhujian(Sweet):
### Week 10
Wenrui(Linda):
1. prepare the ppt for the final seminar
2. practice the final seminar
3. Demonstration
Zhujian(Sweet):
### Week 11
Wenrui(Linda):
software analysis
1. java: file reader; file converter; interface;
2. matlab: draw the degree distribution, and the degree by using matlab, hence easy to observe the average degree.
3. prepare the final report
Zhujian(Sweet):
### Week 12
Wenrui(Linda):
prepare the final project report.
prepare the exhibition.
Zhujian(Sweet):
## Project proposal
For the project proposal you need to have:
• Some fun stuff (fun small-world networks)
• Some more serious engineering reasons why small-world networks are important - namely robustness of power networks, communications networks, Google searching, finding who is the leader in social networks (like we discussed about Enron)
• Some slides on network measures (correlation coefficient, cluster detection algorithms).
• How you will start with some small networks that you can test your software on before moving on to a large data set.
• Some slides on project management - how you might structure your software modules & data flow, and definitely Gantt Chart (including slack space to cope with risks), budget, milestones, who is doing what (& why).
Here are some fun network graphs: [1] There are some fun and some more serious networks here: [2]
When you include graphics from the web, don't forget to include on the slide the web address you found them on. It is also a good idea to have a slide at the end with some key references (e.g. Barabasi book, Milgram's paper, the paper I gave you on finding hierarchies).
## Critical Design Review
For the Critical Design Review (not due until Week 8) you need to have the following parts:
• Project Aim & Background
• Literature review, where you discuss a number (say 10-15) of papers / books and their contributions. Try and summarise in a few sentences what each one is about.
• Your approach - what you are going to do, your software design (see below), and who is going to do what.
• Analysis of your software design - what are the hard sections, what are the critical parts that you need to get working first.
• Project management stuff. For risks you don't need to worry much about Occupational Health & Safety (apart from general computer set up) but there are other risks in your project to do with task timing, what happens if someone (including Derek and I) is unavailable, budget over-runs (should be minimal risk but you can still mention it).
Aside from the engineering applications (where small-world networks are important), the other engineering in this project is software engineering. You should have a block diagram showing the main parts of the software, i.e. graph file format converter, graph drawer, network measure calculators, hierarchy algorithm, and matlab analysis (graphing and statistics), and my software for converting the email data set into a graph file. For some of these you can then include what sub-modules they have, e.g. the network measure calculator might have sub-modules for degree distribution, for clustering coefficient. You should also have a flow diagram showing the data flow between these modules, and in the critical design review give details of the algorithms. You can software you find on the web but you will need to acknowledge it in your proposal report, make sure you understand how it works, and test it. It may be easier though, to write the software yourself then writing a file format converter to get other software to work.
The Critical Design Review can just be emailed directly to myself and Derek on the due date, and you only need one for your group. For the final report you need to write one each (please don't copy each other's writing but write it in your words - you can copy equations, flow charts, graphs, diagrams, reference list, software though. Derek and I are happy to read draft copies.
## Worked examples
### Directed graph
Pajek .net file
*vertices 4
1 "1"
2 "2"
3 "3"
4 "4"
*arcslist
1 2 4
2 3 4
3 1
4 1 2
Note that in my drawing done with different software I have drawn the bidirectional edges as two separate edges, whereas Pajek draws them as a single edge with arrows on both ends.
In-degree of nodes:
1. Node 1 has 2 incoming edges
2. 2
3. 1
4. 2
So the in-degree distribution is:
1. 1
2. 3
(i.e. the number of edges with 1 incoming edge is 1, and the number of edges with 2 incoming edges is 3) Average in-degree is (1*1+3*2)/4=1.75
Out-degree of nodes:
1. 2
2. 2
3. 1
4. 2
So the out-degree distribution is:
1. 1
2. 3
Average out-degree is (1*1+3*2)/4=1.75 Note in general that for a directed graph, the out-degree distribution is not the same as the in-degree.
To calculate the clustering coefficients we first need to find the neighbourhoods of each node. The neighbourhood of node i, is all the other nodes that either have an edge from node i or to node i. The neighbourhood of node 1 is {2,3,4}, the neighbourhood of node 2 is {1,3,4}, the neighbourhood of node 3 is {1,2} and the neighbourhood of node 4 is {1,2}.
We also need the total degree of each node = in-degree + out-degree, thus
1. 4
2. 4
3. 2
4. 4
Using the notation from here, these are the k_i (k underscore i).
We also need the number of edges between nodes j and k in the neighbourhood of i, i.e. for node 1, we count all the edges between nodes 2, 3, and 4 (the neighbourhood of 1). The counts are:
1. 3
2. 3
3. 1
4. 1
So the clustering coefficient for each node is the count divided by k_i(k_i - 1), i.e.
1. 3/(4*3) = 1/4
2. 3/(4*3) = 1/4
3. 1/(2*1) = 1/2
4. 1/(4*3) = 1/12
The clustering coefficient for the graph as a whole is just the average of the clustering coefficients of the nodes, i.e. (1/4 + 1/4 + 1/2 + 1/12) / 4 = 13/48 approx. 0.271
### Undirected Graph
Pajek .net file:
*vertices = 5
1 "1"
2 "2"
3 "3"
4 "4"
5 "5"
*edges
1 2
1 3
2 3
2 4
2 5
3 5
4 5
Degree:
1. 2
2. 4
3. 3
4. 2
5. 3
Degree distribution:
1. 0
2. 2
3. 2
4. 1
Average degree = (2*2 + 2*3 + 4)/5 = 14/5 = 2.8
Neighbourhoods:
1. {2,3}
2. {1,3,4,5}
3. {1,2,5}
4. {2,5}
5. {2,3,4}
Number of edges within neighbourhoods:
1. 1
2. 3
3. 2
4. 1
5. 2
Clustering coefficients
1. 2*1/(2*1) = 1
2. 2*3/(4*3) = 1/2
3. 2*2/(3*2) = 2/3
4. 2*1/(2*1) = 1
5. 2*2/(3*2) = 2/3
Average clustering coefficient = (1 + 1/2 + 2/3 + 1 + 2/3) / 5 approx. 0.767 | 6,642 | 27,493 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-34 | latest | en | 0.930099 |
https://www.physicsforums.com/threads/escape-velocity-for-saturn-introductory-mechanics.804275/ | 1,519,527,348,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816083.98/warc/CC-MAIN-20180225011315-20180225031315-00225.warc.gz | 901,329,884 | 16,193 | # Escape velocity for Saturn (Introductory Mechanics)
Tags:
1. Mar 20, 2015
### najatau
1. The problem statement, all variables and given/known data
The radius of Saturn (from the center to just above the atmosphere) is 60,300 km (60300✕10^3 m), and its mass is 570✕10^24 kg. An object is launched straight up from just above the atmosphere of Saturn.
(a) What initial speed is needed so that when the object is far from Saturn its final speed is 17500 m/s?
My computer's answer is, "43,200 m/s"
(b) What initial speed is needed so that when the object is far from Saturn its final speed is 0 m/s? (This is called the "escape speed.")
My computer's answer is, "35,600 m/s"
2. Relevant equations
Energy principle
3. The attempt at a solution
For a), I get 7,747.58, which doesn't match either problem. I'm not getting b) right either.
This is a practice version of the problem, and I keep thinking I'm going about the right process for solving it, but I don't get the same answers they do, so I can't be. I've tried different variations on the energy principle:
PE (initial) - UE (initial) = PE (final) - PE (initial)
The mass of the object cancels across the equation if kinetic energy is (1/2)mv2 and potential energy is (6.7x10-11*Mm)/(r)
I thought that the way to get escape speed was sqrt(2GM/R) or sqrt((2GM/R)+vi), but that is not working for me either. I'd really like to learn how to do these problems. I have four other similar ones that I think I'm supposed to use this same principle for, but my answers are continually wrong when I do the practice versions.
2. Mar 21, 2015
### ehild
How did you get the numerical value for the speed? Did you take into account that the gravitational potential energy is negative? | 453 | 1,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-09 | longest | en | 0.938666 |
https://thelmathinks.com/does-current-flow-through-capacitors/ | 1,713,943,609,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819067.85/warc/CC-MAIN-20240424045636-20240424075636-00011.warc.gz | 512,705,594 | 46,468 | # Does current flow through capacitors?
## Does current flow through capacitors?
Current does not flow through a capacitor but voltage is stored in a capacitor and consequently store electrical energy across it’s plates wherein these plates are separated in between (sandwhiched) by a dielectric material or insulator.
## How does current flow through capacitors?
Current flows through capacitor plates which induces the polarization of charge on plates through the dielectric placed in between capacitor plates where charge separation by some distance always creates an electric field directed from positive charge to negative charge I.e. it is construed as charge is stored in …
Does current flow through inductor?
When current is about to flow to the inductor, the magnetic field generated by that current cuts across the other windings, giving rise to an induced voltage and thus preventing any changes in the current level. The inductor does not allow AC to flow through it, but does allow DC to flow through it.
### Does a capacitor block current?
Actually capacitor doesn’t block DC current, the capacitor makes potential difference high to very low (about 0) and stops the current flow between them at a particular portion of a circuit by itself charge.
### Why do we need an inductor in a circuit?
Inductors are used as the energy storage device in many switched-mode power supplies to produce DC current. The inductor supplies energy to the circuit to keep current flowing during the “off” switching periods and enables topographies where the output voltage is higher than the input voltage.
What does an inductor represent to a DC current?
An Inductor is equivalent to a Short Circuit to Direct Current, because once the Storage Phase has finished, the Current, iL, that flows through it is stable, iL = V / R, no Self Induced e.m.f. is produced and vL is zero. The Inductor acts like an ordinary connecting wire, its Resistance is zero.
ALSO READ: What are the different types of audiences in writing?
Charging a capacitor is very simple. A capacitor is charged by connecting it to a DC voltage source. Once the capacitor is connected to the DC voltage source, it will charge up to the voltage that the DC voltage source is outputting. So, if a capacitor is connected to a 9-volt battery, it will charge up to 9 volts.
Here is a LC circuit with a DC supply. When the switch is closed at t=0 capacitor behaves as a short circuit while the inductor behaves as an open circuit as the voltage across the inductor immediately jumps to battery voltage.
How do I choose a DC blocking capacitor?
2 Answers. If you want to use a capacitor as a DC-blocking element (i.e., in series with the signal source) you should choose its capacitance value according to: AC signal frequency f; Equivalent Resistance Req seen from “NODE A” (see figure below) to GND.
## Why do we need to block DC?
All Answers (32) DC has zero frequency, so reactance is infinity. This is the reason DC is blocked. A capacitor blocks DC as once it gets charged up to the input voltage with the same polarity then no further transfer of electrons can happen accept to replenish the slow discharge due to leakage if any.
## Does DC block capacitor location matter?
Under this pretense, for a particular topology, as long as the distance from the end points to the DC blocking capacitor is the same, it does not matter if you place the capacitor at the receiver or driver, because the results will be identical. …
Begin typing your search term above and press enter to search. Press ESC to cancel. | 738 | 3,590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-18 | latest | en | 0.949593 |
https://www.alphabetworksheetsfree.com/tag/dividing-fractions-and-mixed-numbers-worksheets-6th-grade/ | 1,656,484,597,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103624904.34/warc/CC-MAIN-20220629054527-20220629084527-00427.warc.gz | 688,624,032 | 15,697 | Dividing Fractions Mixed Numbers Worksheet
Dividing Fractions Mixed Numbers Worksheet – There’s plenty of evidence to show how number worksheets are a great way for children to improve their math abilities. This article will concentrate on how important it is to use number worksheets for children. We will look at…
Dividing Fractions With Mixed Numbers Worksheet
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Division Of Fractions And Mixed Numbers Worksheets
Division Of Fractions And Mixed Numbers Worksheets – There’s plenty of evidence that proves how number worksheets can aid children to develop their math skills. This article will discuss the importance of number worksheets for children. We will look at the advantages and different types…
Dividing Mixed Numbers And Fractions Worksheet
Dividing Mixed Numbers And Fractions Worksheet – There is plenty of evidence that proves how number worksheets help children learn math. This article will highlight why it’s important to have number worksheets for kids. The article will cover the benefits and various types of number…
Dividing Fractions And Whole Numbers Worksheet
Dividing Fractions And Whole Numbers Worksheet – There is plenty of evidence to show how number worksheets are a great way for children to develop their math skills. This article will focus on why it’s important to have number worksheets for kids. We will look… | 298 | 1,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-27 | longest | en | 0.875045 |
https://blog.yhuang.org/?p=934 | 1,725,869,224,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00027.warc.gz | 121,845,365 | 9,476 | ### a problem of moments
We would like to prove the following fact:
For any non-negative random variable $$X$$ having finite first and second moments, $$\mathbb P(X>0) \ge (\mathbb EX)^2/\mathbb EX^2$$.
The proof isn’t difficult. Here are three different ones.
Proof 1. We already know from Jensen’s inequality that $$\mathbb E f(X) \ge f(\mathbb E X)$$ if $$f$$ is convex. This gives $$(\mathbb EX)^2/\mathbb EX^2 \le 1$$ for any $$X$$. The trick to make it be $$\le \mathbb P(X>0)$$ is to note that the density at $$X=0$$ contributes nothing to any moment. In particular, if $$F_X(t)$$ is the distribution function of $$X$$, then define a random variable $$Y$$ that is $$X$$ without the probability at zero, that is, according to the distribution function $$F_Y(t)=(F_X(t)-F_X(0))/(1-F_X(0))$$. Then Jensen’s gives $$(\mathbb EY)^2/\mathbb EY^2 \le 1$$. However, $$\mathbb EY = \mathbb EX / (1-F_X(0)) = \mathbb EX / \mathbb P(X>0)$$, and $$\mathbb EY^2 = \mathbb EX^2 / (1-F_X(0)) = \mathbb EX^2 / \mathbb P(X>0)$$, so $$(\mathbb EX)^2/\mathbb EX^2 = [(\mathbb EY)^2 \mathbb P(X>0)^2] / [\mathbb EY^2 \mathbb P(X>0)] \le \mathbb P(X>0)$$. $$\blacksquare$$
The statement would also work for non-positive $$X$$, of course; and an analogous statement can be made for arbitrary $$X$$ comparing $$\mathbb P(X\ne 0)$$ with some combination of moments for the positive and negative parts of $$X$$.
Proof 2. Apparently this problem can also be proved by an application of the Cauchy-Schwarz inequality. Assume the probability space $$(\Omega, \mathcal F, \mathbb P)$$. The space of finite second-moment real-valued random variables $$L_2(\Omega)=\{X(\omega):\Omega \to R\}$$ with the inner product $$\langle X,Y\rangle_{L_2(\Omega)}=\mathbb E XY$$ and induced norm $$\Vert X\Vert_{L_2(\Omega)}=\sqrt{\mathbb EX^2}$$ is a Hilbert space (modulo $$L_2$$ equivalence). Given this, let us apply Cauchy-Schwarz on the two random variables $$X$$ and $$\mathbf 1_{X>0}$$:
$$\langle X, \mathbf 1_{X>0} \rangle^2 \le \Vert X \Vert^2 \Vert \mathbf 1_{X>0} \Vert^2$$, by Cauchy-Schwarz
$$(\mathbb E X \mathbf 1_{X>0})^2 \le \mathbb EX^2 \mathbb E\mathbf 1_{X>0}$$, specializing to $$L_2(\Omega)$$
$$(\mathbb E X)^2 \le \mathbb EX^2 \mathbb P(X>0)$$, by noting that $$X = X \mathbf 1_{X>0}$$. $$\blacksquare$$
This is a special case of something called the Paley-Zygmund inequality. I didn’t know such a thing existed.
Proof 3. This one only proves the discrete case. It is well known that for positive discrete random variables $$X$$, $$\mathbb EX = \sum_{k=0}^\infty \mathbb P(X>k) = \mathbb P(X>0)+\mathbb P(X>1)+\cdots$$. Basically $$\mathbb P(X=1)$$ is counted once, $$\mathbb P(X=2)$$ is counted twice, and so on. The analogous thing can be derived for $$\mathbb EX^2$$, except now we need to count in squares. Happily we also know that squares accumulate by odd integers, i.e. $$n^2=1+3+5+\cdots+(2n-1)$$, so $$\mathbb EX^2 = \sum_{k=0}^\infty \mathbb (2k+1) \mathbb P(X>k) = \mathbb P(X>0)+3\mathbb P(X>1)+5\mathbb P(X>2)+\cdots$$.
Let’s simplify the notation a bit. Put $$q_k=\mathbb P(X>k)$$, then $$q_0\ge q_1\ge q_2 \ge \cdots$$. We just need to prove that $$q_0\ge (q_0+q_1+q_2+\cdots)^2 / (q_0+3q_1+5q_2+\cdots)$$, which is to say, $$(q_0+q_1+q_2+\cdots)^2 \le q_0(q_0+3q_1+5q_2+\cdots)$$. The two sides both have limits, so this just requires some accounting. On the left hand side, $$(q_0+q_1+q_2+\cdots)^2$$ expands to $$q_0^2+(q_1^2+2q_0q_1)+(q_2^2+2q_0q_2+2q_1q_2)+\cdots = Q_0+Q_1+Q_2+\cdots$$, where $$Q_k \triangleq q_k^2 + 2 \sum_{i=0}^{k-1} q_iq_k \le (2k+1)q_0q_k \triangleq R_k$$. But $$R_0+R_1+R_2+\cdots$$ is exactly the right hand side. So the left hand sum is dominated by the right hand sum. $$\blacksquare$$
With some real analysis, this proof could be made to work for random variables that are not discrete, but it might also turn into a special case of Proof 1. In any case, it’s interesting in its own right. | 1,441 | 3,937 | {"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} | 3.640625 | 4 | CC-MAIN-2024-38 | latest | en | 0.734996 |
https://dsp.stackexchange.com/questions/59509/number-of-microphone-limitation-for-spherical-harmonics-based-doa-estimation/59598 | 1,618,540,830,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088471.40/warc/CC-MAIN-20210416012946-20210416042946-00447.warc.gz | 260,953,082 | 40,209 | # Number of microphone limitation for Spherical Harmonics based DOA estimation
I am following this book "Theory and Applications of Spherical Microphone Array Processing", as well as this paper on DOA estimation "Eigenbeam-ESPRIT for DOA-Vector Estimation".
My dataset is composed of a small number of channels (4) from the Eigenmike32. The book states "all spatial sampling schemes require at least $$(L_f + 1)^2$$ microphones to sample a sound field of order Lf without aliasing.", where $$L_f$$ is the order of the sound field (3.4 Spatial Sampling). Both the paper and the book use 32 microphones in their implementations.
First, I do not find any information on what is the order of the sound field. It sounds to me that this is a signal-dependent quantity, much like a sampling rate is lower bounded by the highest frequency of the signal you want to sample. Did I get this right? Can anyone elaborate please?
Second, does that mean that I can do very little with 4 microphones? Is there a lower limit on the number of microphones for a DOA estimation and for a number of active sources estimation?
What other, none neural network, methods can I use to estimate the number of active sources and DOA with (no necessarily by the same model)?
First, I do not find any information on what is the order of the sound field. Can anyone elaborate please?
$$Y_{lm}(\Omega)$$ is the spherical harmonics with mode $$m$$, order $$l$$ and frequency $$\Omega=(\theta,\phi)$$ (dictating angles of arrival, ex: azimuth, spherical, elevation, etc.). According to [1], we have the following formula:
$$Y_{lm}(\Omega) = \sqrt{\frac{2l + 1}{ 4\pi} \frac{(l-m)!}{(l+m)!} }P_l^m(\cos \theta)e^{j m \phi}$$ where $$P_l^m(\cos \theta)$$ is the associated Legendre functions of degree (mode) $$m$$ and order $$l$$. The above serves as a basis of spherical harmonics (similar to Fourier expansion).
Second, does that mean that I can do very little with 4 microphones? Is there a lower limit on the number of microphones for a DOA estimation and for a number of active sources estimation?
The classical rule of thumb is that given $$q$$ sources, you need at least $$q+1$$ antennas to resolve uniquely your sources. However, this is really algorithmic dependent. For example, the paper in [2] (which formulate a 2D AoA DoA estimation) describes shift invariant methods with different lower bounds. So, in principle, the more antennas you have the better resolvability and lower MSE you will have, since the Cramer-Rao bound of DoA estimation is proportional to $$\frac{1}{N}$$ where $$N$$ is the number of antennas.
What other, none neural network, methods can I use to estimate the number of active sources and DOA with (no necessarily by the same model)?
You can use subspace methods like MUSIC, or co-variance fitting. Some generic methods aim at directly optimizing the Maximum likelihood problem like Expectation-Maximization or IQML for specific antenna configurations. A 2D IQML approach was derived in [3].
[1] Rafaely, Boaz. Fundamentals of spherical array processing. Vol. 8. Berlin: Springer, 2015.
[2] Bazzi, Ahmad, Dirk TM Slock, and Lisa Meilhac. "Single snapshot joint estimation of angles and times of arrival: A 2D Matrix Pencil approach." 2016 IEEE International Conference on Communications (ICC). IEEE, 2016.
[3] Bazzi, Ahmad, Dirk TM Slock, and Lisa Meilhac. "Efficient maximum likelihood joint estimation of angles and times of arrival of multiple paths." 2015 IEEE Globecom Workshops (GC Wkshps). IEEE, 2015.
• This is a long comment so I will divide in two and update the question. On the first portion, I agree with everything you wrote. What baffles me is the notion of a sound field order as a parameter of the specific sound. I understand I can represent the sound field using spherical harmonics similar to Fourier expansion, though, I can do it with any order I choose (minimizing the error), so this is not a limitation. However, it sounds to me that the order of the sound field is a parameter of the signal in hand, and is a lower limit on the number of required microphones. – havakok Jul 18 '19 at 8:16
• For an analogy, the sampling frequency must answer request for a given signal. It is signal-dependent. So, is the order of the sound field also signal-dependent? if so, how? Did I get this all wrong? – havakok Jul 18 '19 at 8:16
• The order of the sound field is in no way a signal parameter. This is as if you're saying the number of Taylor series expansion is a signal parameter. Unless you're addressing round-off errors (which is definitely not your case), this is not a signal parameter. Perhaps it has to do with the accuracy of estimation, however I'm not aware of signal parameters that have to do with the order. – Ahmad Bazzi Jul 18 '19 at 8:18
• On the other hand, if your signal $Y$ is unknown and you're estimating the weights of each order (as equation 2 in the paper you attached), then you've got more unknown weights to handle. In that case, indeed the order and the number of required microphones are related. – Ahmad Bazzi Jul 18 '19 at 8:21
• no it is not. It is the transmit (see eqn. 4 in what you attach). Receiver reads sum of $d_j \propto Y$ – Ahmad Bazzi Jul 18 '19 at 8:31
The order of the sound field indicates that there are constraints placed upon the sound field. In this context, the order equals that of the lowest-order Ambisonics multi-channel format that can describe the sound field. Ambisonics channels are weighted by directional patterns that are orthogonal spherical harmonics:
Figure 1. Directional patterns of Ambisonics channels for up to 5th order. For each increment in the order, another row is included in the set of channels. Image credit: Dr Franz Zotter. | 1,399 | 5,741 | {"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": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-17 | latest | en | 0.885174 |
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# The debt equity ratio for the company is bs 040 b 040s
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Unformatted text preview: 75) RWACC = .1105 or 11.05% Finally, we can use the WACC to discount the unlevered cash flows, which gives us an NPV of: NPV = –\$45,000,000 + \$13,500,000(PVIFA11.05%,5) NPV = \$4,837,978.59 b. The weighted average cost of capital used in part a will not change if the firm chooses to fund the project entirely with debt. The weighted average cost of capital is based on optimal capital structure weights. Since the current capital structure is optimal, all-debt funding for the project simply implies that the firm will have to use more equity in the future to bring the capital structure back towards the target. Challenge 14. a. The company is currently an all-equity firm, so the value as an all-equity firm equals the present value of aftertax cash flows, discounted at the cost of the firm’s unlevered cost of equity. So, the current value of the company is: VU = [(Pretax earnings)(1 – tC)] / R0 VU = [(\$28,000,000)(1 – .35)] / .20 VU = \$91,000,000 The price per share is the total value of the company divided by the shares outstanding, or: Price per share = \$91,000,000 / 1,500,000 Price per share = \$60.67 b. The adjusted present value of a firm equals its value under all-equity financing plus the net present value of any financing side effects. In this case, the NPV of financing side effects equals the aftertax present value of cash flows resulting from the firm’s debt. Given a known level of debt, debt cash flows can be discounted at the pretax cost of debt, so the NPV of the financing effects are: NPV = Proceeds – Aftertax PV(Interest Payments) NPV = \$35,000,000 – (1 – .35)(.09)(\$35,000,000) / .09 NPV = \$12,250,000 So, the value of the company after the recapitalization using the APV approach is: V = \$91,000,000 + 12,250,000 V = \$103,250,000 387 Since the company has not yet issued the debt, this is also the value of equity after the announcement. So, the new price per share will be: New share price = \$103,250,000 / 1,500,000 New share price = \$68.83 c. The company will use the entire proceeds to repurchase equity. Using the share price we calculated in part b, the number of shares repurchased will be: Shares repurchased = \$35,000,000 / \$68.83 Shares repurchased = 508,475 And the new number of shares outstanding will be: New shares outstanding = 1,500,000 – 508,475 New shares outstanding = 991,525 The value of the company increased, but part of that increase will be funded by the new debt. The value of equity after recapitalization is the total value of the company minus the value of debt, or: New value of equity = \$103,250,000 – 35,000,000 New value of equity = \$68,250,000 So, the price per share of the company after recapitalization will be: New share price = \$68,250,000 / 991,525 New share price = \$68.83 The price per share is unchanged. d. In order to value a firm’s equity using the flow-to-equity approach, we must discount the cash flows available to equity holders at the cost of the firm’s levered equity. According to Modigliani-Miller Proposition II with corporate taxes, the required return of levered equity is: RS = R0 + (B/S)(R0 – RB)(1 – tC) RS = .20 + (\$35,000,000 / \$68,250,000)(.20 – .09)(1 – .35) RS = .2367 or 23.67% After the recapitalization, the net income of the company will be: EBIT Interest EBT Taxes Net income \$28,000,000 3,150,000 \$24,850,000 8,697,500 \$16,152,500 388 The firm pays all of its earnings as dividends, so the entire net income is available to shareholders. Using the flow-to-equity approach, the value of the equity is: S = Cash flows available to equity holders / RS S = \$16,152,500 / .2367 S = \$68,250,000 15. a. If the company were financed entirely by equity, the value of the firm would be equal to the present value of its unlevered after-tax earnings, discounted at its unlevered cost of capital. First, we need to find the company’s unlevered cash flows, which are: Sales Variable costs EBT Tax Net income \$28,900,000 17,340,000 \$11,560,000 4,624,000 \$6,936,000 So, the value of the unlevered company is: VU = \$6,936,000 / .17 VU = \$40,800,000 b. According to Modigliani-Miller Proposition II with corporate taxes, the value of levered equity is: RS = R0 + (B/S)(R0 – RB)(1 – tC) RS = .17 + (.35)(.17 – .09)(1 – .40) RS = .1868 or 18.68% c. In a world with corporate taxes, a firm’s weighted average cost of capital equals: RWACC = [B / (B + S)](1 – tC)RB + [S / (B + S)]RS So we need the debt-value and equity-value ratios for the company. The debt-equity ratio for the company is: B/S = 0.35 B = 0.35S Substituting this in the debt-value ratio, we get: B/V = .35S / (.35S + S) B/V = .35 / 1.35 B/V = .26 389 And the equity-value ratio is one minus the debt-value ratio, or: S/V = 1 – .26 S/V = .74 So, using the capital structure weights, the company’s WACC is: RWACC = [B / (B + S)](1 – tC)RB + [S / (B + S)]RS RWACC = .26(1 – .40)(.09) + .74(.1868) RWACC = .1524 or 15.24% We can use the weighted average cost of capital to discount the firm’s unlevered aftertax earnings to value the company....
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Ask a homework question - tutors are online | 1,538 | 5,371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.874937 |
https://stats.stackexchange.com/questions/320173/p-value-and-95-confidence-interval | 1,721,247,186,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00115.warc.gz | 469,904,744 | 38,236 | # P value and 95% confidence Interval
I know that if a P value is less than 0.05 we can reject the null hypothesis and state that the finding may be statistically significant however I remember reading something about if the 95% confidence interval includes 0 that its not a significant finding or something like that (i tried to find what i was reading but couldnt find it again).
Im just wonder, if we have a p value below 0.05 but have a confidence interval including 0, how do we interpret this? Do we say that it is statistically significant but likely being caused by other factors?
Im just wonder, if we have a p value below 0.05 but have a confidence interval including 0, how do we interpret this?
It means that you took a confidence interval with more than 95% confidence. This is technically possible, but misleading to take the confidence interval at another $\alpha$ level than your test.
Existing software packages will give you p-values and CIs for the same $\alpha$ unless you force them to do something else.
• Thanks for the help, I've been sitting here for ages trying to work this out and just realized it was 0.1 not 0.01 for my p value ugh.
– Mick
Commented Dec 23, 2017 at 12:53 | 279 | 1,207 | {"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} | 3.203125 | 3 | CC-MAIN-2024-30 | latest | en | 0.951083 |
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# How is one hundred twenty five million three hundred fifteen thousand eighty six in decimal?
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Q: How is one hundred twenty five million three hundred fifteen thousand eighty six in decimal?
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One million, four hundred fifteen thousand, two hundred. | 427 | 1,807 | {"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} | 3.140625 | 3 | CC-MAIN-2023-40 | latest | en | 0.863909 |
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# semisolpr12 - Semiconductor Physics and Devices Basic...
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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 12 Solutions Manual Problem Solutions 185 Chapter 12 Problem Solutions 12.1 (a) I V V D GS t = () L N M O Q P 10 2.1 15 exp For VV GS = 05 ., I D =⇒ ( ) L N M O Q P 10 2.1 0 0259 15 exp . . Ix A D = 9.83 10 12 For GS = 07 A D = 388 10 10 . For GS = 09 A D = 154 10 8 . Then the total current is: II Total D = 10 6 bg For GS = IA Total = 9.83 µ For GS = . , Im A Total = 0388 . For GS = A Total = 154 . (b) Power: PI V Total DD =⋅ Then For GS = PW = 49.2 For GS = . , Pm W = 194 . For GS = W = 77 12.2 We have ∆∆ L eN Vs a t V a fp DS DS = ⋅+ + 2 φ −+ () fp DS a t where fp t a i V N nx == F H G I K J F H G I K J ln . ln . 00259 10 15 10 16 10 or fp V = 0347 . We find 2 2117 88510 16 10 10 14 19 16 12 = L N M O Q P eN x x a .. . / b gb g = 0360 ./ / mV We have a t V V DS GS T =− (a) For V s a t V GS DS = 54 . 2 5 Then L =+ + 0347 5 0347 4.25 . or Lm = 00606 . If L is 10%of L , then = 0606 . (b) For V s a t V DS GS DS ⇒= 52 1 2 5 ,. Then L + 0347 125 . . or = 0377 . Now if L is 10% of L , then = 377 . 12.3 L eN a a fp DS DS = + 2 fp DS a t where fp t a i V N n x x F H G I K J F H G I K J ln . ln . 41 0 16 10 or fp V = 0383 and x eN dT fp a = L N M O Q P 4 / = ( ) L N M O Q P 4117 88510 4 10 14 19 16 . . / x xx b g or xm dT = 0157 . Then = Qe N x SD a dT max
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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 12 Solutions Manual Problem Solutions 186 = −− 16 10 4 10 0157 10 19 16 4 .. xx x bg b g b g or = () QC c m SD max / 10 72 Now VQ Q t TS D S S ox ox ms fp = ++ F H G I K J max af φφ 2 so that V x x T = 10 3 10 400 10 39 885 10 71 9 1 0 8 14 . b g b g 02 0 3 8 3 . or VV T = 187 Now Vs a t V V V DS GS T =− = = 51 8 7 3 1 3 We find 2 2117 88510 4 10 14 19 16 12 = L N M O Q P eN x a . / b gb g = 180 10 5 . x Now ∆∆ Lx V DS =⋅ + + 180 10 0 383 313 5 . −+ 0383 313 or V DS =+ 180 10 3513 5 . We obtain V DS Lm µ 0 1 2 3 4 5 0 0.0451 0.0853 0.122 0.156 0.188 12.4 Computer plot 12.5 Plot 12.6 Plot 12.7 (a) Assume a t V DS = 1 , We have Ε sat DS a t L = We find Ε sat Vc m / 3 1 05 025 013 333 10 3 . x 10 4 21 0 4 x 41 0 4 x 7.69 10 4 x (b) Assume n cm V s 500 2 / , we have v ns a t = Ε Then For = 3 , vx c m s = 167 10 6 ./ For = 1 , vxc m s = 510 6 / For . , vc m s 10 7 / 12.8 We have IL LL I DD 1 We may write g I V L I L V O D DS D DS = ∂ ′ −∂ ()( ) F H G I K J 1 2 = ⋅⋅ L I L V D DS 2 We have L eN s a t a fp DS fp DS = ⋅+ + 2 We find = + L Ve N V DS a fp DS 2 φ (a) For GS DS == , , and a V DS GS T = −= 20 81 2 Also s a t V V DS DS DS = + = 12 1 2.2 and fp x x V F H G I K J 00259 310 15 10 0376 16 10 .l n . Now
Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 12 Solutions Manual Problem Solutions 187 2 2117 88510 16 10 3 10 14 19 16 12 = () L N M O Q P eN x xx a .. . / bg b g = 02077 ./ / µ mV We find L =+ + 0376 2.2 0376 12 . . = 00726 . m Then =⋅ + L V DS 2 1 . . = 00647 From the previous problem, Im A L m D == 048 2 ., Then gx O = 2 2 0 0726 0 48 10 0 0647 2 3 . or S O = 167 10 5 . so that r g k O O 1 59.8 (b) If Lm = 1 , then from the previous problem, we would have A D = 096 . , so that O = 1 1 0 0726 0 96 10 0 0647 2 3 .
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semisolpr12 - Semiconductor Physics and Devices Basic...
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Ask a homework question - tutors are online | 1,658 | 3,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-51 | latest | en | 0.605374 |
https://n0elle.com/why-do-chess-players-record-their-moves/ | 1,656,406,929,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103360935.27/warc/CC-MAIN-20220628081102-20220628111102-00097.warc.gz | 460,306,389 | 7,324 | Many time you might have i found it chess football player in tournaments writing on a item of file after every move, and that seems choose they are taking notes. So you can wonder, what carry out chess players create down?Chess players compose down the notation of each move they do in the scoresheet. A scoresheet is a paper of document that has blank locations where the chess players deserve to record the important information about the game being played. After ~ the video game ends both players authorize each other’s scoresheet.
You are watching: Why do chess players record their moves
The Algebraic Notation is taken into consideration the standard technique for recording the moves.After each move, chess players create down the moves they made. For example, in a chess game, if the white pawn is moved to e4 square and also the black color pawn is relocated to e5 square climate both chess players compose down e4 in the obelisk of white and also e5 in the shaft of black color in their particular scoresheets.Here’s what a conventional scoresheet look at like:
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Further in this article, i am walk to discuss in information how to create down the moves following the typical procedure.
But prior to that, you may think, also though digital chess board are capable of recording the moves then why execute chess players compose down their moves?Chess players compose down their moves due to the fact that it is mandatory as per the rules to document the moves. Moreover, the acts as a written record of your game and helps lock in solving disputes, making any kind of claims in former of the arbiter, and analyzing their gamings later on.Arbiter means the referee that a chess game. Now, this is simply a straightforward answer. If you room interested to understand the precise reasons why chess players note down your moves v an comprehensive explanation along with the typical procedure of creating down the moves, then proceed reading it spins the end.Here room the 7 reasons why chess players compose down their moves.
#1. It Is Mandatory together Per The Rules#2. Helps In Claiming draw By Threefold Repetition#3. Help In Claiming draw By 50 move Rule#4. Help To Prove the time Limit Is no exceeded#5. Stays clear of Cheating#6. Written records Are much more Reliable#7. Help To Analyse The game AfterwardsStandard Procedure Of writing Down The Chess moves In The ScoresheetConclusionRelated Questions:Why perform chess players take notes?Do grandmasters create down your moves?Why carry out chess players compose down their moves in spite of having electronic chess boards?
## #1. That Is Mandatory as Per The Rules
In chess, FIDE is the international chess administrate body. Every the traditional chess rule are complied with as pointed out in FIDE regulations of chess. So, in the FIDE rated chess tournaments, that is compulsory for a chess player to note down your moves.
There is a specialized section in FIDE laws of chess, which travel guide on how to record moves in chess. If a player is not able to compose the moves in the scoresheet then there are aides appointed that write the move on the scoresheet on the player’s behalf. Moreover, that is customary for each chess player to sign their score sheets after ~ their game ends. For this reason you can see how much importance is provided to writing down the move on the scoresheet.
The only exemption to this rule of record the moves is if the player has less 보다 5 minutes left ~ above his or her clock and doesn’t have extr 30 secs or much more added with each move, and also in quick chess games in which the time duration is much more than 10 minutes but less 보다 60 minutes together per FIDE.In casual play you don’t have to note down your moves. However, it is great practice to carry out so.
## #2. Help In Claiming attract By Threefold Repetition
In chess there are numerous rules and one among them is attract by threefold repetition.
“In an easy words, three-fold repetition means the same position occurs three times. Although the is not necessary that this has to happen in a succession but the insurance claim has to be made by the player with the revolve to move..”Now the procedure together per the FIDE write-up 9.2.1.1 and article 9.2.1.2,A claim can be made if the same position is going to occur the 3rd time through the same player to move and also so the player first has to note this under on the scoresheet and also then asking the arbiter. Or if the same position just developed for the third time for this reason the player with relocate can case for this.Moreover, the location are considered the same only if the same player has actually the opportunity to move, the very same pieces occupy the very same positions, the ideal for castling is the same, and the opportunity of en passant record is also the same.Now to case the attract by this dominion it is very important to have a written record of your moves. For this reason whenever you claim a draw, the arbiter deserve to easily inspect the location from the scoresheet and give the result. Thus writing moves aid to insurance claim a attract by threefold repetition.Below is the video clip by Chess base India of a chess game that reflects the occurrence of three-fold repeat practically. If interested you can watch how the player renders the case for it.
## #3. Help In Claiming attract By 50 move Rule
In simple words, a 50 move rule means a player can claim a draw if no pawn movement and no capture has to be made in the last 50 moves.The basic procedure to claim draw by this ascendancy as per the FIDE article 9.3.1 and article 9.3.2. A player needs to write his relocate on a scoresheet and then declare to the arbiter the the next resulting relocate will reason the 50 move ascendancy to be used or already the 50 moves have been play without any type of pawn activity or capture.So the arbiter have the right to track the move on a scoresheet and also decide whether to accept the insurance claim or reject it. Thus in this case additionally a written document proves come be really beneficial for a player to insurance claim the draw.One necessary thing to consider is by both the three-fold repeat rule and the 50 relocate rule, the game doesn’t end automatically. The player has actually to claim for it.But as per the new 75 move rules (provided no pawn movement and also no record has to be made within the last 75 moves) and also the five-fold repetition rule (the same place repeated for at least five times) climate the arbiter(referee in a chess game) deserve to declare the video game as a draw. Over there is no requirement of explain in this cases.
## #4. Help To Prove the time Limit Is not exceeded
In details chess games, a player needs to make a certain variety of moves within a resolved time frame. If the player over the time duration then vital penalties are used to the player.Read Related: What Happens once A Player Runs the end Of Time? (With Rules)Now if girlfriend have adhered to the rules then exactly how will friend prove it, by written records! Right? So writing under moves in a scoresheet help in proving that you adhered to the rules and made the required variety of moves within the provided time frame.
## #5. Prevents Cheating
Let’s intend that you room playing in a chess tournament. Now during the video game you have put your opponent in together a position that he/she could lose the game.Now if the opponent readjusted the place of several of the piece on the board or make an illegal move.After that once you insurance claim the adversary disagrees. At the time the only evidence you have is the written record of moves. You have the right to just walk to the arbiter and show her scoresheet. Friend will gain the decision in your favor. For this reason this habit avoids cheating.Also Read: What happens when you copy moves in chess?
## #6. Written records Are more Reliable
Though this particular day we have electronic chess boards called dgt chess boards i beg your pardon are really accurate and show the live activity of chess piece on the board. Yet after all it is an digital machine.Written records are very simple, reliable, and also easy come use. The arbiter can easily interpret the score sheets and also give judgment on even if it is to disapprove or accept any type of claim.
See more: Spanish Galleon Ring Spanish Galleon Shipwreck Re, Spanish Galleon Ring
Moreover, in certain complex situations, written documents are much more useful. Because that example, digital boards don’t record the attract offers do by chess players. But a player has to write down in the scoresheet every time a draw offer is made. Hence the scoresheet keeps a record of this. So whenever there is a dispute then the arbiter can just verify the scoresheet and make the decisions.
## #7. Help To Analyse The video game Afterwards
When you compose down the moves on a scoresheet you are recording the finish game top top a piece of paper.So afterwards when the game ends you have the right to analyse wherein you made mistakes and where you could have do a much better move. Thus writing the move becomes very handy for afterwards for self evaluation of her game.Moreover girlfriend can likewise put the chess move in the chess engine you have and then analyse follow to the mistake the engine point out out. You have to just input all the move serially and also then the chess engine’s software application will analyze your game taking referral from its huge database.Tip | 2,015 | 9,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-27 | latest | en | 0.96355 |
https://www.mathdoubts.com/sin-square-power-reduction-identity-proof/ | 1,718,853,706,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861880.60/warc/CC-MAIN-20240620011821-20240620041821-00599.warc.gz | 774,745,831 | 10,624 | # Proof of Sine squared Power reduction identity
According to the sine squared power reduction identity, the square of sine of angle is equal to one minus cos of double angle by two. It is mathematically expressed in any one of the following forms popularly.
$(1).\,\,\,$ $\sin^2{\theta} \,=\, \dfrac{1-\cos{(2\theta)}}{2}$
$(2).\,\,\,$ $\sin^2{x} \,=\, \dfrac{1-\cos{(2x)}}{2}$
$(3).\,\,\,$ $\sin^2{A} \,=\, \dfrac{1-\cos{(2A)}}{2}$
When the angle of a right triangle is denoted by a symbol theta, the squares of sine and cosine functions are written as $\sin^2{\theta}$ and $\cos^2{\theta}$ respectively. Similarly, the cosine of double angle is written as $\cos{2\theta}$ in mathematical form in trigonometry.
### Relation between functions in square form
According to the Pythagorean identity, the sum of squares of sine and cosine functions is equal to one.
$\cos^2{\theta}+\sin^2{\theta} = 1$
### Express the equation in Cos of double angle
The trigonometric equation can be converted into cosine of double angle by an acceptable setting.
$\implies$ $\cos^2{\theta} = 1-\sin^2{\theta}$
Now, add sine squared of angle to both sides of the trigonometric equation and it helps us to express the one side of equation as cosine of double angle function.
$\implies$ $\cos^2{\theta}-\sin^2{\theta} = 1-\sin^2{\theta}-\sin^2{\theta}$
$\implies$ $\cos^2{\theta}-\sin^2{\theta} = 1-2\sin^2{\theta}$
According to the cosine of double angle trigonometric identity, the subtraction of square of sine from square of cosine is equal to the cosine of double angle.
$\implies$ $\cos{(2\theta)} = 1-2\sin^2{\theta}$
### Express Sine squared in Cosine of Double angle
The trigonometric equation can be simplified for expressing the sine squared of angle in terms of cosine of double angle function.
$\implies$ $2\sin^2{\theta} = 1-\cos{(2\theta)}$
$\,\,\,\therefore\,\,\,\,\,\,$ $\sin^2{\theta} = \dfrac{1-\cos{(2\theta)}}{2}$
This equation expresses the square of sine of angle is reduced by expressing it as cosine of double angle and it is called the power reduction trigonometric identity of sine squared function.
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Copyright © 2012 - 2023 Math Doubts, All Rights Reserved | 740 | 2,829 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2024-26 | latest | en | 0.808791 |
https://nandandubey.wordpress.com/tag/algorithms/ | 1,566,415,048,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316150.53/warc/CC-MAIN-20190821174152-20190821200152-00436.warc.gz | 558,347,166 | 11,799 | # Max Sum
Given an array of positive numbers, find the maximum sum of a sub-sequence with the constraint that no 2 numbers in the sequence should be adjacent in the array.
Input:
The first line of input contains an integer T denoting the number of test cases.
The first line of each test case is N,N is the size of array.
The second line of each test case contains N input C[i].
Output:
Print the maximum sum of a sub-sequence.
C++ Solution
```
#include <iostream>
#include <vector>
using namespace std;
/*
T denotes maxsum then
T(N) = max(A[N-1] + T(N-2), T(N-1))
*/
void printMaxSum(vector <int> inp)
{
int size = inp.size();
vector <int> maxSum;
if(size <= 0)
return;
else
{
if(size == 1)
{
cout<<inp[0]<<endl;
return;
}
else
{
maxSum.push_back(inp[0]);
maxSum.push_back(max(inp[0],inp[1]));
}
}
for(int i = 2; i < size; i++)
{
int included = inp[i] + maxSum[i - 2];
int excluded = maxSum[i-1];
maxSum.push_back(max(included,excluded));
}
cout<< maxSum[size-1]<<endl;
}
int main() {
int T = 0;
cin>>T;
vector < vector <int> > allInputs;
for (int i = 0; i < T; i++)
{
int N = 0;
cin>>N;
//read all N inputs in a vector
vector <int> arr;
for(int j = 0; j < N; j++)
{
int inp = 0;
cin>>inp;
arr.push_back(inp);
}
allInputs.push_back(arr);
}
for(int i = 0; i < T; i++)
{
printMaxSum(allInputs[i]);
}
return 0;
}
``` | 415 | 1,327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2019-35 | latest | en | 0.521277 |
https://www.mathnasium.com/math-centers/diamondbar/news/word-problem-wednesday-turn-up-the-heat-on-everyday-math-skills | 1,719,255,252,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865482.23/warc/CC-MAIN-20240624182503-20240624212503-00052.warc.gz | 772,909,304 | 13,198 | # Word Problem Wednesday: Turn Up The Heat on Everyday Math Skills
Mar 8, 2018 | Diamond Bar
Do you enjoy baking? Nellie does too! In this challenge, you are tasked with figuring out when Nellie should take out her pies. Use your everyday math skills and this word problem challenge will be easy as pie. Do you have what it takes? Try solving below...
Challenge: Nellie puts a batch of pies into the oven at 6:17 AM. When they go into the oven, the internal temperature of the pies is 60° Fahrenheit. The temperature increases at a rate of 2.5° per minute. If Nellie wants the internal temperature of the pies to reach 160°, at what time should she take them out of the oven?
Don’t leave the problem half-baked, once you think you've worked it out, check below for the solution!
Solution: In order to increase to 160°, the internal temperature of the pies will need to rise 160° – 60° = 100°. Since 100° ÷ 2.5° per minute = 40 minutes, it’ll take 40 minutes for the pies to bake. So, Nellie should take them out of the oven at 6:57 AM. | 274 | 1,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.890443 |
https://tracks-movie.com/how-do-i-show-grand-total-in-pivot-chart/ | 1,718,876,325,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861916.26/warc/CC-MAIN-20240620074431-20240620104431-00170.warc.gz | 514,910,814 | 11,125 | # How do I show grand total in pivot chart?
## How do I show grand total in pivot chart?
Display or hide grand totals
1. Click anywhere in the PivotTable.
2. On the Design tab, in the Layout group, click Grand Totals, and then select the grand total display option that you want.
## How do I filter the total value in a PivotTable?
Here are the steps to do this:
1. Go to Row Label filter –> Value Filters –> Greater Than.
2. In the Value Filter dialog box: Select the values you want to use for filtering. In this case, it is the Sum of Sales (if you have more items in the values area, the drop down would show all of it). Select the condition.
3. Click OK.
How do you filter grand total?
You can reach this dialog using the Options menu, on the Options tab of the PivotTable Tools ribbon. Once the dialog is open, navigate to the Totals & Filters tab. There, set Grand Totals as you like.
### How do I filter subtotals in a PivotTable?
Click the target row or column field within the report and on the PivotTable Tools | Analyze tab, in the Active Field group, click the Field Settings button. On the Subtotals & Filters tab of the invoked Field Settings dialog, select one of the following options and click OK to apply changes.
### How do you find the grand total in Excel?
After calculating each group subtotals, to get the grand total of the column without theses subtotals, enter this formula: =SUBTOTAL(9,B2:B21) into a cell you need, then press Enter key, and you will get the grand total which excludes the subtotals automatically.
How do I show the grand total in a stacked bar chart in Excel?
Add total labels to stacked column chart with an amazing tool
1. Create the stacked column chart.
2. Select the stacked column chart, and click Kutools > Charts > Chart Tools > Add Sum Labels to Chart.
3. Then all total labels are added to every data point in the stacked column chart immediately.
## How do I filter zeros from grand total in pivot table?
Click on the arrow to the right of the Quantity (All) drop down box and a popup menu will appear. Check the “Select Multiple Items” checkbox. Then un-select the checkbox next to the 0 value and click on the OK button. Now when you return to the spreadsheet, the Totals with a value of 0 should be hidden.
## What is grand total in pivot table?
The Grand Total average in the pivot table is adding up all of the cells in the quantity column of the data set and dividing it by the total number of orders.
Does pivot table include filtered items in totals?
Pivot table options contain yet another useful feature: Include filtered items in set totals. It works the same way, referring to the data we have grouped into sets within the Power Pivot Report. | 610 | 2,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.789714 |
https://van.physics.illinois.edu/ask/listing/14004 | 1,719,097,680,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00806.warc.gz | 514,411,393 | 9,227 | # Different Forms of Magnetic Force
Q:
We have studied that magnetic force and field are perpendicular to each other.But when we are performing an experiment to draw the magnetic field of a magnet we first keep a magnetic compass near the north pole of the magnet and slowly bring it to the south pole in a defined manner.M question is that the magnetic force of magnet is toward the needle while the magnetic field which we draw also coincide with that magnetic force.Then how can it be possible that they are perpendicular.
- Prudhvi Raj Borra (age 16)
A:
The magnetic force on a moving charged particle is perpendicular to both the velocity of the motion and to the magnetic field, (in vector notation it's a cross product).
When you have a compass needle in a magnetic field the needle, a small bar magnet itself, will align itself to the direction of the external magnetic field. There is no velocity involved. If you do the calculation, a bit involved, you find that the minimum energy configuration is when the needle and the external field are aligned. That's where it winds up.
LeeH
To help see that, imagine that instead of a little permanent magnet, you used a little electromagnet. The currents going around in the coils would be going at right angles to the direction of the magnet. So there would be forces between the other magnet and those currents. /Mike W.
(published on 03/24/2009)
## Still Curious?
Expore Q&As in related categories | 309 | 1,462 | {"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} | 3.046875 | 3 | CC-MAIN-2024-26 | latest | en | 0.928091 |
https://www.neetprep.com/questions?courseId=8&testId=403-JIPMER-MBBS-&questionId=94199-particle-moves-towards-east--s-velocity--ms-move-towardsnorth--s-velocity--ms-average-velocity-particleis--ms--ms--ms--ms | 1,721,130,943,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00220.warc.gz | 776,566,632 | 61,434 | # A particle moves towards east for 2 s with velocity 15 m/s and move towards north for 8 s with velocity 5 m/s. Then, the average velocity of the particle is: 1. 2 m/s 2. 5 m/s 3. 7 m/s 4. 10 m/s
Subtopic: Average Speed & Average Velocity |
60%
Hints
Relation between a wavelength of photon and electron of same energy is:
1. $$\lambda_{p h}>\lambda_e$$
2. $$\lambda_{p h}<\lambda_e$$
3. $$\lambda_{p h}=\lambda_e$$
4. $$\frac{\lambda_e}{\lambda_{p h}}=\text { constant }$$
Subtopic: De-broglie Wavelength |
52%
Hints
Match the following.
A. Angular momentum 1 [M-1L2T-2] B. Torque 2 [M1T-2] C. Gravitational constant 3 [M1L2T-2] D. Tension 4 [M1L2T-1]
1. C-2, D-1
2. A-4, B-3
3. A-3, C-2
4. B-2, A-1
Subtopic: Dimensions |
72%
Hints
If we increase the kinetic energy of a body by 300%, then percent increase in its momentum is:
1. 50%
2. 300%
3. 100%
4. 150%
Subtopic: Concept of Work |
56%
Hints
Change in acceleration due to gravity is same at height h above the Earth surface and depth $$x ,$$ then relation between $$x$$ and $$h$$ is: ($$h$$ and $$x<<R_e$$)
1. $$x=h$$
2. $$x=2h$$
3. $$x=\frac h2$$
4. $$x=h^2$$
Subtopic: Acceleration due to Gravity |
77%
Hints
A mass of $$1~\mathrm{kg}$$ is suspended from a spring of force constant $$400$$ N, executing SHM total energy of the body is $$2$$ J, then the maximum acceleration of the spring will be:
1. $$4 \mathrm{~m} / \mathrm{s}^2$$
2. $$40 \mathrm{~m} / \mathrm{s}^2$$
3. $$200 \mathrm{~m} / \mathrm{s}^2$$
4. $$400 \mathrm{~m} / \mathrm{s}^2$$
Subtopic: Energy of SHM |
64%
Hints
Two capacitors of capacities $$C_1$$ and $$C_2$$ are charged upto the potential $$V_1$$ and $$V_2,$$ then the condition for not flowing the charge between them when connected the capacitors, in parallel is:
1. $$C_1=C_2$$
2. $$C_1V_1=C_2V_2$$
3. $$V_1V_2$$
4. $$\frac{C_1}{V_1}=\frac{C_2}{V_2}$$
Subtopic: Capacitance |
61%
Hints
Find equivalent resistance between $$X$$ and $$Y$$:
1. $$R$$
2. $$R/L$$
3. $$2R$$
4. $$5R$$
Subtopic: Combination of Resistors |
67%
Hints
Vibrations of rope tied by two rigid ends shown by equation $$y=cos2\pi t\sin2\pi x,$$ then the minimum length of the rope will be:
1. $$1$$ m
2. $$\frac12$$ m
3. $$5$$ m
4. $$2\pi$$ m
Subtopic: Standing Waves |
64%
Hints
If we change the value of $$R,$$ then:
1. voltage does not change on $$L$$
2. voltage does not change on $$LC$$ combination
3. voltage does not change on $$C$$
4. voltage changes on $$LC$$ combination
Subtopic: Different Types of AC Circuits |
55%
From NCERT | 938 | 2,536 | {"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} | 3.546875 | 4 | CC-MAIN-2024-30 | latest | en | 0.603794 |
https://www.studiestoday.com/useful-resources-science-cbse-class-6-science-electricity-and-circuits-exam-notes-241367.html | 1,601,266,646,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401585213.82/warc/CC-MAIN-20200928041630-20200928071630-00551.warc.gz | 1,007,547,847 | 20,468 | # CBSE Class 6 Science Electricity and Circuits Exam Notes
CBSE Class 6 Science Electricity and Circuits Exam Notes. There are many more useful educational material which the students can download in pdf format and use them for studies. Study material like concept maps, important and sure shot question banks, quick to learn flash cards, flow charts, mind maps, teacher notes, important formulas, past examinations question bank, important concepts taught by teachers. Students can download these useful educational material free and use them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.
Electricity and Circuits
Electricity is a kind a energy. It comes from the charged particles called electrons and protons inside an atom.
We have two types of electricity
a) Static Electricity: It is formed over an insulator and can’t move through it.
b) Current Electricity: Here charges move in a particular direction through a conductor.
Static electricity is also known as frictional electricity. When one object is rubbed by another object, heat so produced, get absorbed by electrons over their surface and they exchange electrons. One of them loses some electrons where the other gain some electrons equally. Both of them gain electric potential. Respectively the first one is called as positively charged object and said to be at positive potential and the second one is called as negatively charged object and is said to be in negative potential. For example when a glass rod is rubbed by fur, the glass rod looses electrons and becomes positivley charge where as the fur gains electrons and becomes negatively charge equally.
The potential of such accumulated charge is defined as the work done in bringing unit positive charge from infinity to the given charge.
A collection of positive charge has a potentoa; of +1V if 1J of work is done in bringing +1C charge from infinite to that given charge.
Here V is potential measured in Volt (V); W is the work done measured in Joule (J) and Q is the charge measured in coulomb (C). 6 × 1018 electron together makes 1 coulomb charge. 1 J work done = 1 m displacement of any thing by applying 1 Newtons force.
Current Electricity: Let us consider a positively charged object and a negatively charged object are connected by conducting wire is shown below:
It is obvious that at B electrons are at high pressure than at A an so electron will start drifting over the conductor from B to A. This fact is called as current of charges or electric current. Current is defined as the rate of flow of charges and expressed as,
I (current)= Q (charge)/ t (time)
It is 1 ampere current if at any cross-section of the conductor 1 coulomb charge passes in 1 second. So, 1A = 1C/1 Sec
Thus electric potential difference can produced electric current. Electric current can be obtained commercially in two ways. One way is from power plants which we shall learn later.
The other way is from a electric cell which is a source of electric potential difference. We shall learn more about cell and shall learn how electric circuits are made by using a cell.
ELECTRICITY
Electricity plays a very important role in our daily life. It lights, heats and cools our homes, and also performs a number of routine jobs for us. Most of the charm of modern life would be missing if there were no electricity. Electricity flows through electric wires. Electricity in motion is called current electricity.
SOURCES OF ELECTRIC CURRENT
a) Small Sources of Electric Current: Cell are small sources of current. In a cell chemical energy changes into electrical energy. A cell basically consists of two terminals — a metal cap on one side and a metal disc on the other. The metal cap is the positive terminal and the metal disc works as a negative
terminal. The chemicals stored in the cell produce electricity.
Cells are of two kinds:
1. Primary Cells 2. Secondary cells or accumulators.
1. Primary Cells: These cells provide current as a result of the chemical reaction that takes place in the chemical reaction that takes place in the chemicals stored in them. When the chemicals are used up, they stop producing electricity. They cannot be recharged. It means the chemical reaction is irreversible. Simple voltaic cell, Leclanche cell, Daniel cell, dry cell, etc., are examples of primary cells. The dry cell was discovered by French scientist G. Leclanche in 1868. It is the most commonly used cell and is used in torches, transistors, calculators, cameras, tape-recorders, etc. This cell is called dry cell because it doesn’t contain any liquid chemical .
A dry cell consists of a zinc casing and a carbon rod with a brass cap at its centre. The carbon rod is surrounded with a black powder which is a mixture of manganese dioxide (MnO2) and graphite (C). This black powder is either contianed in a thin bag of cloth or is surrounded by a thin layer of swadust. The space between the zinc casing and the sawdust is filled with a thick paste of ammonium chloride (NH4Cl).
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https://www.thegrint.com/range/2021/06/odds-making-hole-one-handicap-2/ | 1,624,249,085,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488262046.80/warc/CC-MAIN-20210621025359-20210621055359-00066.warc.gz | 933,689,976 | 16,892 | # ODDS OF MAKING A HOLE IN ONE, BY HANDICAP
The dream of every golfer is watching a ball go in the hole after hitting it from the tee. Many seasoned players play their entire life without achieving this feat, while some not-so-great players claim to have it done more than once. It’s obvious that some degree of luck is involved, but just how much? Today, we analyze the odds of making a hole in one by handicap. DISCLAIMER: They’re low. Veeeery low.
Check out the graphic to find your handicap and see your odds of making a hole in one!
As we can see, skill does play a big part. Scratch players are three times more likely to achieve this feat than a player with a handicap of 18 or more. And if we compare the odds of the two lowest handicap brackets (<0 and 0-5) vs the rest of the brackets we can see how much more likely a skilled player is to make a hole in one.
However, the odds are astronomically low no matter how good you are. On average, you’d need to play ~6,500 par 3’s (or ~1625 rounds assuming 4 Par 3′s per round) to have a hole in one (from a statistical standpoint)… which on the flip side, is great news for us since we now have a target of how many golf rounds we have to play to get one
Just think of all the birdies, pars, triple bogeys, cart girls, beers, and golfing shenanigans that awaits you before finally making a hole in one!
I wanted to close out this article a little different because I’m curious to find out how many of you have actually made a hole in one. So drop a line in the comments below mentioning when and how you made your hole in one!
If you are not in TheGrint, make sure to join Today here.
Golfer, Writer, and Grinter for life.
This entry was posted in Golfers Data Analysis, TheRange Blog and tagged , , , . Bookmark the permalink.
### 31 Responses to ODDS OF MAKING A HOLE IN ONE, BY HANDICAP
1. David Viscariello says:
First one just this May, hole 3 south hampton golf club, 148 yards.
• Luis says:
Fresh from the oven, then! Glad it happened for you David, thanks for sharing.
2. Don Blair says:
I’ve had three and never saw any of them go in the hole! This was due to the green complex and/or elevation change from the tee.
1st: #4 Kinsale CC – Powell, OH – April, 2010 – 190 yds, hybrid. Playing with my wife riding along. I new I hit a good shot but couldn’t see it when it landed because the green was slightly elevated and has a lot of contour. When we drove up I didn’t see the ball, I was looking to see if it had rolled over or was in the back bunker. I asked her to check the hole to which she replied, “Yep, it’s in there.” She had NO IDEA how monumental it was!!
2nd: #18 Valley View GC – Lancaster, OH – 180 yds, 4 iron, elevated green, playing with my son, he was humbled!!
3rd: #5 Champions GC – Columbus, OH – 145 yds, 8 iron, pin tucked over/behind bunker
• Luis says:
Ahhh, the proverbial hole in one with the family. Take a bow son! I find it hilarious how non-golfers react to an ace, she probably shrugged her shoulders and carried on with her day enjoying the views of the course. Let’s hope you have a crystal clear view of your next one! Thanks for sharing, Don.
3. Mark Wales says:
I’ve had (3) holes in one:
With my brother Greg in 1977 at Lowe’s Creek Eau Claire WI 99 yard hole.
With Columbia Senior Group Sept/2016 at South Carolina Club in Florence SC 115 yard hole.
By myself October/2016 at Charwood Club in South Carolina 137 yard hole.
I’m a 15 handicap and have played about 800 rounds of golf in 45 years.
I feel very blessed considering all the old guys I’ve played with who are better then me but have never had one.
Mark Wales
Lexington, SC 29072
612-702-3122
• Luis says:
There you go, Mark! Three hole in ones is something to be proud of, glad you had the opportunity! I’m going to keep score of my # of rounds and hopefully when I get to 800 I’ll have my first. Thanks for sharing!
4. David Paulson says:
I am a 15 handicap golfer who golds about 20 rounds a year. When Inokay golf I play with a group of guys who are significantly better than I am and who play several times a week. One night, after a round of golf that eight of us were engaged in, the conversation came up about their holes in one. All of us had at least one, but these other seven had them at courses like Firestone, Spyglass, Legacy, etc. I hadn’t said anything until they asked me if I had a hole in one and where. I reluctantly admitted that my hole in one was into a stiff wind on #14, 140 yards at “Hollydale”. A course build on land that wouldn’t qualify for a land full. But I had proof with the signatures of my three hacker friends and the club manager. I saw the ball bounce once and in the cup. It is a strange feeling walking onto a green with out a putter to pick up your ball and then to watch the others have to putt out. My snobbish friends got a chuckle out of my Hollydale hole in one, but it was still a great shot, one that I will likely never experience again.
• Luis says:
David! As a guy who regularly plays in courses that would have been left out on that conversation as well, I loved that story, there’s something romantic about a hole in one in the unlikeliest of courses. And I’m guessing the celebration would’ve been the same no matter where it happened. Fingers crossed and prayers to the golf gods so you may experience it again! (Huge plus if your snobbish friends have to end up buying you a round!)
5. James Sampson says:
2016 While playing with the course professional at Frosty Valley near Pittsburgh. Flew a 5 iron into the hole on hole #8. No bounce or roll !
• Luis says:
So you AIRED the hole!!! Talk about pinpoint accuracy eh, James? I’m guessing that being with the course pro at the time was a huge plus as well. Thanks for sharing!
6. Andy Kindler says:
I have had one ace and it was many years ago now.
• Luis says:
I’m guessing your overdue for another one then, Andy! Thanks for sharing.
7. Mark Liquorman says:
But miracles can happen, as it did for me last week! My first hole-in-one; that is, on a hole that didn’t involve a windmill. Still kind of surreal. We have a hole-in-one pot in our league, and probably the best part was being able to use a chunk of the money to buy a round of beers for everyone.
• Luis says:
Perfect timing then, huh? Congrats on the milestone! And if I might say so Mark, that’s EXACTLY how you should celebrate having a hole in one. Keep making the sport better and cheers for (hopefully) many more to come!
(BTW, I celebrate a hole-in-one on minigolf pretty much like the real thing until it happens for me)
8. Bob Calabro says:
Had my first and only hole in one at age 70 after playing about 50 years, 16 handicap.
• Luis says:
You give me hope Bob! Super glad that the golf gods gave you the joy after all those years committed to the game. Keep’em coming!!
9. Sean Nies says:
I have not made an ace, but one of my family members has made two in one week. He is a scratch golfer and both aces were recorded during league and or tournament play. The event being so rare, was published in the local paper. His comment “It was an expensive week”.
• Luis says:
I’m with you on this one Sean, still waiting for the planets to align! You can tell your guy that if we calculate how many \$\$\$ the average golfer would need to spend before making a hole in one, he probably got out VERY cheap. Thanks for sharing!
10. I am 66 and have been playing for 25 years. I am a 7.5 but most of my playing has been at a 14. I have had 2 aces. One at Gozzer Ranch a 168 6 iron and the other here on Lanai at The Challenge at Manele a 145 8 iron….both attested. Just scored my 7th eagle last night. Richard Bliss Brooke Lanai, Hi
• Luis says:
Aloha Richard! Haven’t had the opportunity to play in Hawaii yet, I went there with the wife and couldn’t manage to sneak a round, but I imagine what must feel like to ace one with those beautiful backdrops. Keep’em coming! Thanks for sharing.
11. Scott Hernke says:
I’ve actually made 5. I am a 6-8 handicap now. I got my first one when I was probably a 12 hdcp. That was in 2001, the last was in 2014. All were on good shots. No flukes off of a tree or a skull job that ran the down the fairway and through a trap and then somehow found the bottom of the cup. What I am most proud of, even beyond the number of holes-in-one that I have had, is the distances from which I have made them. Yardages are 173, 191, 144, 193 and 223. A nearly 185 yard average. It’s been 7 years since my last one. Kind of feels like I’m due! I’ve had quite a few close calls in the last 7 years, but none have found the cup. I always tell people that while it doesn’t necessarily require a good shot to get the hole-in-one, that no matter how good the shot, it always requires an enormous amount of luck, as getting a ball into that 4.25″ hole from that type of distance is incredibly lucky. I feel very, very lucky.
• Luis says:
You should feel proud Scott! You’re the top Grinter in this department so far (no one has had more than three). Five hole in ones is a sweet, sweet achievement. And talk about holing out from 223! That’s some serious yardage right there, congrats. You’re 100% right: it does require a little help from the universe to get it into that tiny, tiny space.
Thanks for sharing!
Eagle’s Nest Golf Course, 1990, 7th hole, 8 iron. I was 12.
• Luis says:
Beautiful course, Adam. Let’s hope you’ll get another shot a celebrating an ace as a full-grown man who is more aware of how difficult golf can be! Thanks for sharing.
13. DES FOX says:
Hello Luis – I must have beaten the odds because I had a hole in one in competition in 2009 playing off 18 – another one in 2016 playing off 14 – another competition one in 2018 playing off 12 and one more in competition in 2020 at 190 yards on the 17th in Millicent Golf Club.
Despite all efforts I’ve had no luck with the lottery – Cheers – Des Fox
• Luis says:
Hey Des, thanks for sharing! Wow, four (!!!!) hole in ones and three of them during a competition? I like your odds on the lottery!
14. Mike Bramble says:
I have been golfing for 34 years now and golf a lot!!! 147 rounds last season. I am always around a 6 cap and my home course has 6 par 3′s… in my life and all my rounds I have never had an ace myself or seen one in my group. The only hole in one I’ve seen was the group in front of me where a guy who wasn’t very good sculled a shot down the fairway, smashed the pin and dropped. I shake my head at players who have had multiple aces, wondering if I will ever get mine. The golf is still good and maybe one day I will have mine.
15. Tom Wickman says:
I beat the odds too Luis. As a 19 handicap at the Oxmoor Valley Short Course in Birmingham AL with a lackluster round to that point I placed a PW six inched from the cup and watched it roll in. Instant exhilaration. I felt like I would drop it in the cup on the next hole, but that was on March 1, 2019. I’m still waiting to repeat the feat.
• Luis says:
Ahhh, nothing like a hole in one to lift up your round! So nice you actually saw it rolling in, a lot of Grinters have missed it due to elevation/distances. I can imagine the joy you felt! Hopefully, you’ll get another one Tom! (I’m still on the search for the first one myself). Thanks for sharing!
16. Frank Pag says:
At Waterford GC Turnberry #6 160 yards up hill and I yelled go in the hole while you’re up there and it did.
• Luis says:
Hey Frank! I can picture the headline “Golf ball actually listens to man for once, goes in the hole”. Guess I’ll try that technique next time to see what happens. Thanks for sharing! | 2,926 | 11,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-25 | latest | en | 0.948491 |
https://www.programiz.com/java-programming/examples/compute-quotient-remainder | 1,722,735,758,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00528.warc.gz | 759,592,015 | 20,176 | Java Program to Compute Quotient and Remainder
To understand this example, you should have the knowledge of the following Java programming topics:
Example: Compute Quotient and Remainder
``````public class QuotientRemainder {
public static void main(String[] args) {
int dividend = 25, divisor = 4;
int quotient = dividend / divisor;
int remainder = dividend % divisor;
System.out.println("Quotient = " + quotient);
System.out.println("Remainder = " + remainder);
}
}``````
Output:
```Quotient = 6
Remainder = 1```
In the above program, we have created two variables dividend and divisor. Here, we are calculating the quotient and remainder by dividing 25 by 4.
To find the quotient, we have used the `/` operator. We have divided dividend (25) by divisor (4). Since both dividend and divisor are integers, the result will also be integer.
``````25 / 4 // results 6.5
// convert 6.5 to integer
// output will be 6``````
Likewise, to find the remainder we use the `%` operator. Here, the dividend is divided by the divisor and the remainder is returned by the `%` operator.
``25 % 4 // results 1``
Finally, quotient and remainder are printed on the screen using `println()` function. | 297 | 1,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.836911 |
https://teamtreehouse.com/community/using-mathceil-instead-of-mathfloor | 1,675,364,039,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500035.14/warc/CC-MAIN-20230202165041-20230202195041-00451.warc.gz | 573,073,513 | 31,941 | ## Welcome to the Treehouse Community
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# Using Math.Ceil instead of Math.floor?
Math.floor(Math.random() * 6) + 1 How would this ever result in 0 like he said ?
the + 1 at the end would stop this from happening?
If you test Math.ceil() and Math.floor() in the console, it'll make a bit more sense.
For example, Math.ceil(-0.001) returns -0. Math.floor(-0.001) returns -1.
Since we are working with rolling dice and dice have sides 1 through 6 but never a 0 side, we will want to use Math.floor() because it will prevent us from having to deal with the possibility of Math.ceil returning a 0.
That's not quite right. Both Math.floor and Math.ceil could return a 0. But Math.floor gives you the same chances of getting each possible value.
It's said that Math.random returns a random number in the range from 0 up to 1, so I'm kinda confused with the -0.001 example.
You're right, Swedina, that's not an example of a value you could get from "Math.random". But it does illustrate the difference between "Math.ceil" and "Math.floor".
A better example might be the value "0.5".
## It might result in 0 before the 1 was added.
Using Math.Ceil would not help the case where the value was actually zero.
I believe the possibility of the result of the formula being 0 was discussed as the reason to add 1 to it. If you heard otherwise, can you give a specific time reference in the video for where it is said? | 475 | 2,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | longest | en | 0.908207 |
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• Hello all: I would like to know if someone has used Chebyshev correction for square root aproximation. In that case , whats your opinion about it ? Chebyshev
Message 1 of 8 , Feb 1, 2004
View Source
Hello all:
I would like to know if someone has used Chebyshev correction
for square root aproximation.
Chebyshev correction is explained on Ron´s book and, as you now
, the expression for square root is : (1/8*an)* ((N-an^2)/an)^2 ,
Where N is the number we want to obtain the square root of,
an the n-aproximation.
Ron , what about 6 * 6 ? . 4 * 4 digits was great.
Best wishes to all.
Javier.
• Hi Oleg Yes, it would be relatively easy to divide 11730 by 10 and look at 1173, but my reason for treating the problem as I did is this: When competing in the
Message 1 of 8 , Feb 2, 2004
View Source
Hi Oleg
Yes, it would be relatively easy to divide 11730 by 10 and look at 1173, but my reason for treating the problem as I did is this:
When competing in the Mental Calculations World Championships, the prime factors problems have to be solved with the prime factors themselves given in ascending numerical order. If I had originally divided by 10 and then checked for the factors of 1173, the factor of 3 would have appeared out of order. However, if a brief check had revealed no factors of 3 in the number, then it would be better to divide by 10 to begin with - and, of course, to record one factor each of 2 and 5.
Best regards,
George Lane
Oleg Stepanov <olegstepanov@...> wrote:
GL> Sample problem: Find the prime factors of 11730
GL> 11730/2=5865, which is not divisible by 2
GL> 5865/3=1955 which is not divisible by 3
GL> 1955/5=391 which is not divisible by 5, 7, 11 or 13
GL> 391/17=23 which is a prime number
GL> The list of prime factors is thus 2, 3, 5, 17, and 23.
Please, I fill myself little strange. Probably I do not understand
something. Why not divide on 10 and by looks full number 11730 think
only about 1173? Your method remind me Jededia Buxton what by multiple
on 300 made it by 5, 20 and 3.
Apologize if I am not correct or did not find explanation...
Sincerely Yours Oleg Stepanov.
http://www.lk.net/~stepanov/
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• Sure future champion. :-) http://www.ruediger-gamm.de/ Sincerely Yours Oleg Stepanov. http://www.lk.net/~stepanov/
Message 1 of 8 , Feb 7, 2004
View Source
Sure future champion. :-)
http://www.ruediger-gamm.de/
Sincerely Yours Oleg Stepanov.
http://www.lk.net/~stepanov/
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https://www.coursehero.com/file/5624626/Lab09/ | 1,492,983,951,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118831.16/warc/CC-MAIN-20170423031158-00445-ip-10-145-167-34.ec2.internal.warc.gz | 882,741,987 | 50,949 | # Lab09 - Laboratory Exercise Number 9 Simulink and Algae...
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Laboratory Exercise Number 9 Simulink and Algae Growth Assignment : This laboratory unit is intended to introduce you to Simulink, a toolbox within Matlab that allows symbolic programming. Unlike the Matlab code we’ve worked with so far, Simulink is based on blocks that consist of operations and objects. The blocks are linked to form systems. Simulink is particularly suited to model and analyze dynamic (changing) systems. Today’s exercise will require you to create a system of blocks to simulate growth of an algal population that is limited by a single nutrient. This exercise will lead into a Simulink program that will model algae population in a lake over the course of a year. The goal of this week’s lab is to create a graph of algae concentration over time. Background : In the very first laboratory exercise of this semester we modeled unconstrained growth in bacterial populations, or “first-order growth kinetics”. Because all growth is constrained, whether it be through lack of space, light, or a specific nutrient, first order growth kinetics may not be ideal for modeling a real world population. If the problem you are interested in modeling is of short duration without any limitations (such as bacteria in a macaroni salad left in the sunlight on a summer picnic) first order kinetics may be adequate past the point of making you very sick, at which point more complex modeling will not matter to you. However, if longer term predictions are required, something better is necessary. In the unconstrained growth example of Laboratory Exercise 1, the rate of change of the number in a population (X) was considered to be a constant, μ (the growth rate), multiplied by the current number in the population. ݎ ൌߤܺ (1) However, this takes no consideration of the observation that not all cells survive and there is some loss of the population due to consumption by swim-by predators, starvation, and any other number of hazards an average algal cell faces in its short but basically contented life. The rate of
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## This note was uploaded on 09/28/2009 for the course BEE 1510 taught by Professor Staff during the Fall '05 term at Cornell University (Engineering School).
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Lab09 - Laboratory Exercise Number 9 Simulink and Algae...
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Ask a homework question - tutors are online | 574 | 2,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-17 | longest | en | 0.928456 |
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# Thread: matching value using a bitwise operator
1. ## matching value using a bitwise operator
I'm new to bitwise operators and the PHP documentation isn't helping my for the code I have below. In the example below \$ulogindetails["securityvalue"] is 34.
PHP Code:
``` if (\$ulogindetails["securityvalue"] & 880) { ... do sumthin ... } ```
Can anyone shed light on how this is working? When would the value match to 880?
Cheers,
Picco
• AND will return the common bits between the two. 880 & 34 share the 32'nd bit, so it will result in a valid 32.
I get the impression that 880 is supposed to be a singular item, which is not possible, 880 = 512 + 256 + 64 + 32 + 16
While the 34 is 32 + 2, sharing the common 32.
To check if it matches 880 using bitwise (which doesn't make any sense since the bit check is using 880 so a simple comparison will work) you can use `if (880 == (\$myVar & 880))`
The idea behind bitwise permissions is to take a less aggressive approach to it:
PHP Code:
``` define('CAN_READ', 1);define('CAN_EDIT', 2);define('CAN_DELETE', 4);// Do stuff for our 'user', we'll give them privs of 5if (\$userPerms & CAN_READ){ // Our user can read}if (\$userPerms & CAN_EDIT){ // Our user can edit (this user cannot, since 5 & 2 is 0)}if (\$userPerms & CAN_DELETE){ // Our user can delete} ```
Does that make sense?
• When you think bitwise, think binary.
Code:
```34 = 0000000000100010
880 = 1000101101110000```
In a bitwise AND, if and only if both values are 1, the result will be 1. Anything else is zero. So:
0 & 0 = 0
0 & 1 = 0
1 & 0 = 0
1 & 1 = 1
In a bitwise OR, if any of the values are one, the result will be one. So:
0 | 0 = 0
0 | 1 = 1
1 | 0 = 1
1 | 1 = 1
Code:
```34 = 0000000000100010 AND
880 = 1000101101110000
--------------------------
0000000000100000```
Extended a bit:
Code:
```34 = 0000000000100010 OR
880 = 1000101101110000
--------------------------
1000101101110010```
• Thanks for those explanations, I have been reading through them, but i'm still slightly confused as to when the if statement would evaluate and would be true? How would we get into '... do sumthin ...' ?
PHP Code:
``` if (\$ulogindetails["securityvalue"] & 880) { ... do sumthin ... } ```
Cheers,
Picco
• Originally Posted by crmpicco
Thanks for those explanations, I have been reading through them, but i'm still slightly confused as to when the if statement would evaluate and would be true? How would we get into '... do sumthin ...' ?
PHP Code:
``` if (\$ulogindetails["securityvalue"] & 880) { ... do sumthin ... } ```
Cheers,
Picco
You can't really in this situation. 880 does not represent a valid (singular) bit value, so it can only be comparable by the use of equality (==) or for the use of contained bits.
880 contains the following bit values: 512 + 256 + 64 + 32 + 16. That means as long as the 'securityvalue' contains a number that includes one or more of the above bits (18 for example would match 16 + 2) it will be deemed true.
Edit:
Lol, my statement doesn't really make sense to you're question. The answer is, it will be false when a comparable number does not include one or more of the following bits: 512, 256, 64, 32, 16. It is true when containing one or more of the previous bits.
•
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• | 1,065 | 3,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-49 | longest | en | 0.662217 |
https://www.edutoolbox.org/rasp/1776?route=astandard/12086 | 1,660,653,893,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572304.13/warc/CC-MAIN-20220816120802-20220816150802-00444.warc.gz | 655,662,273 | 11,740 | # Feeding Frenzy
In this activity, students will multiply and divide a recipe to feed groups of various sizes. Students will use unit rates or proportions and think critically about real world applications of a baking problem. This is a great lesson on setting up proportions, using unit rates, looking at diagrams and adjusting a recipe to feed different size groups. Hands on activities are also incorporated.
### Standards & Objectives
CCSS.Math.Content.6.G.A.2
Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge...
CCSS.Math.Content.6.G.A.3
Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side joining points with the same...
CCSS.Math.Content.6.G.A.4
Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures....
CCSS.Math.Content.6.NS.C.6
Understand a rational number as a point on the number line. Extend number line diagrams and coordinate axes familiar from previous grades to...
CCSS.Math.Content.6.NS.C.8
Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane. Include use of coordinates and...
CCSS.Math.Content.6.RP.A.1
Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, "The ratio of wings...
CCSS.Math.Content.6.RP.A.2
Understand the concept of a unit rate a/b associated with a ratio a:b with b != 0, and use rate language in the context of a ratio relationship....
CCSS.Math.Content.6.RP.A.3
Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams,...
CCSS.Math.Content.7.RP.A.1
Compute unit rates associated with ratios of fractions, including ratios of lengths, areas and other quantities measured in like or different...
CCSS.Math.Content.7.RP.A.2
Recognize and represent proportional relationships between quantities.
CCSS.Math.Content.7.RP.A.3
Use proportional relationships to solve multistep ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and...
GLE 0606.1.3
Develop independent reasoning to communicate mathematical ideas and derive algorithms and/or formulas.
GLE 0606.2.3
Understand and use ratios, rates and percents.
GLE 0606.3.6
Understand and use the Cartesian coordinate system.
GLE 0606.4.4
Develop and use formulas for surface area and volume of 3-dimensional figures.
GLE 0706.2.3
Develop an understanding of and apply proportionality.
GLE 0706.2.4
Use ratios, rates and percents to solve single- and multi-step problems in various contexts.
GLE 0706.3.5
Understand and graph proportional relationships.
GLE 0706.4.2
Apply proportionality to converting among different units of measurements to solve problems involving rates such as motion at a constant speed.
GLE 0706.4.4
Understand and use ratios, derived quantities, and indirect measurements.
SPI 0606.2.6
Solve problems involving ratios, rates and percents.
SPI 0606.2.7
Locate positive rational numbers on the number line.
SPI 0606.2.8
Locate integers on the number line.
SPI 0606.3.9
Graph ordered pairs of integers in all four quadrants of the Cartesian coordinate system.
SPI 0706.1.3
Recognize whether information given in a table, graph, or formula suggests a directly proportional, linear, inversely proportional, or other nonlinear relationship.
SPI 0706.2.6
Express the ratio between two quantities as a percent, and a percent as a ratio or fraction.
SPI 0706.2.7
Use ratios and proportions to solve problems.
SPI 0706.3.4
Interpret the slope of a line as a unit rate given the graph of a proportional relationship.
SPI 0706.3.5
Represent proportional relationships with equations, tables and graphs.
SPI 0806.1.1
Solve problems involving rate/time/distance (i.e., d = rt).
SPI 0806.1.3
Calculates rates involving cost per unit to determine the best buy.
SPI 0806.4.2
Apply the Pythagorean theorem to find distances between points in the coordinate plane to measure lengths and analyze polygons and polyhedra.
SPI 0806.4.4
Convert between and within the U.S. Customary System and the metric system.
TSS.Math.6.G.A.2
Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and...
TSS.Math.6.G.A.3
Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side that joins two vertices (vertical or...
TSS.Math.6.G.A.4
Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures. Apply these...
TSS.Math.6.NS.C.6
Understand a rational number as a point on the number line. Extend number line diagrams and coordinate axes familiar from previous grades to represent...
TSS.Math.6.NS.C.8
Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane. Include use of coordinates and absolute value...
TSS.Math.6.RP.A.1
Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, the ratio of wings to beaks...
TSS.Math.6.RP.A.2
Understand the concept of a unit rate a/b associated with a ratio a:b with b != 0. Use rate language in the context of a ratio relationship.
TSS.Math.6.RP.A.3
Use ratio and rate reasoning to solve real-world and mathematical problems.
TSS.Math.7.RP.A.1
Compute unit rates associated with ratios of fractions, including ratios of lengths, areas, and other quantities measured in like or different units. For...
TSS.Math.7.RP.A.2
Recognize and represent proportional relationships between quantities.
TSS.Math.7.RP.A.3
Use proportional relationships to solve multi-step ratio and percent problems.
Alignment of this item to academic standards is based on recommendations from content creators, resource curators, and visitors to this website. It is the responsibility of each educator to verify that the materials are appropriate for your content area, aligned to current academic standards, and will be beneficial to your specific students.
Learning objectives:
Learning Objectives:
Students will:
• Use ratios and solve proportions.
• Combine math with practical knowledge to analyze a problem.
Essential and guiding questions:
Questions for Students:
• Did you notice any shortcuts as you worked through the problems?
• Do you think the calculations would have been easier if you gave your answers in decimals? Why do you think cooking measurements are made in fractions?
• What practical knowledge do you need to bake cookies? Is it enough to calculate the quantities of the ingredients?
### Lesson Variations
Blooms taxonomy level:
Understanding
Extension suggestions:
Extensions:
• Students could plan an entire dinner party for 12, complete with shopping list. Have students bring in recipes for the dishes they wish to prepare, and then adjust all the recipes to serve 12.
• Many recipe websites can automatically adjust a recipe to the desired number of servings. Have students explore these recipes, and then write about how the conversions they did in class compare to those on the websites.
• Have students convert all the units to the simplest form for a particular ingredient. For example, students should have found that 3\frac{1}{3} tsp of vanilla is needed for 60 servings of chocolate chip cookies. Since 3 tsp =1 Tbsp, it would be easier to measure out 1 Tbsp + \frac{1}{3} tsp of vanilla.
• Students could convert all measurements into grams and other metric units, which are standard baking units in Europe, and then multiply the recipe for 300 servings. How does the process compare using different units? Which units are easier to calculate? Which units are easier to use when baking? | 1,799 | 7,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2022-33 | latest | en | 0.817453 |
https://mozbue.com/ascending-triangle-pattern-investbro/ | 1,726,199,473,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00092.warc.gz | 367,621,802 | 29,629 | One of the fundamental assumptions in price action technical analysis techniques is the assumption that asset price movements will create certain patterns that will repeat over time. repetitive price These are to develop and execute trading strategies.
from a large number of recurring price patterns There is one pattern that creates a triangle. that is, the triangle pattern This pattern is caused by resistance and support which inadvertently formed a triangle
This triangle pattern is further divided into three, namely: descending symmetry and triangle up where each pattern indicates a different trading signal. In this article, the author will only talk about the pattern. triangle up and indications, check out the full review below:
## What is an ascending triangle pattern?
pattern triangle up It’s a triangle looking upwards. This pattern is characterized by a relatively flat resistance line and a sloping support line at the same time.
A fixed resistance line indicates that the asset’s high price is relatively the same over a period of time. On the other hand, a higher support line indicates that during that period. The bull market continues to press the market. so that the lowest price of the asset does not decrease and increases steadily.
This pattern occurs when the price changes several times and fails to break through the resistance (breakout) and sales force (sell) unable to push the price down further The price eventually bounced up and broke the resistance at a certain point.
So in general The formation of an ascending triangle pattern is considered valid. If the price change touches the resistance at least 2 times and touches the support 3 times
## What does the Ascending Triangle mean?
outline ascending triangle pattern appear in the middle uptrend or the price goes up. The existence of this pattern directly indicates that the upward trend in the price of the related asset will continue (continuous pattern).
However, this pattern can also occur when prices are consolidating after the asset price tends to fall (downtrend). In this condition The existence of this pattern actually indicates a possible reversal. This indicates that the downward trend in asset prices will soon end.
Both of these apply to the note. First, the distance between the maximum end of the support line that creates this pattern is the maximum distance, or times the distance from the tip of the line to the top. Apex is the angle between the maximum end of the support line and the point. end of resistance If the distance between resistance and support is less than that The formation of this pattern is considered less valid and can actually be stimulating. price reversal (reversal).
## Ascending Triangle Example
Figure 1 above shows how this pattern is formed. From the image above, it can be seen that the resistance line tends to be flat. While the support has created several new lows. This pattern ends when the price action breaks the resistance (breakout).
## How to trade using an ascending triangle
First, you should open a position after a breakout. The reason is that it is possible that the price movements of these assets will move in the opposite direction than expected.
After the breakout happened You can open a position. buy on breakout Or buy an asset when a breakout occurs in the estimation that the asset price will rise. Conversely, you can place a Stop Loss at the price point which is at the highest support. The goal is if the asset’s price goes down. The system will automatically sell your assets to avoid greater losses.
## Pros and cons of the Ascending Triangle
### strength
1. easy to identify Ascending triangle pattern is one of the easiest price action to identify. Especially now that many trading platforms have a line drawing feature. with this feature Traders can clearly draw this pattern directly on the screen.
2. Easy to use as a referenceThe second advantage of this price pattern is that the shape of this price pattern can be used as a reference in setting a target entry point. (due to the relatively flat resistance line) and stop-loss points (with the support line above)
3. Traders can open positions before the pattern has fully formed.It is worth noting that traders can accurately predict the direction of the trend. This allows traders to buy assets at lower price levels than the breakout price.
### defect
1. There is a chance of acne.You have to remember that there are two types of acne: real acne and fake acne. A false breakout can be a reversal signal, although in some cases the price action after the actual breakout also reverses. Therefore, risk management skills in trading are essential here.
2. The price movement after an ascending triangle can also be a sideline or a reversal.. Therefore, traders should use other technical indicators in addition to using this single price action indicator. | 934 | 4,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.924661 |
https://www.varsitytutors.com/psat_math-how-to-find-the-equation-of-a-line-problem-5693 | 1,581,932,911,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141806.26/warc/CC-MAIN-20200217085334-20200217115334-00438.warc.gz | 949,101,920 | 31,690 | ### Test: PSAT Math
1 What is the equation of the line with a negative slope that passes through the y-intercept and one x-intercept of the graph y = –x2 – 2x + 8 ?
y = –4x + 8
y = –2x + 8
y = –4x + 4
y = –2x + 4
y = –x + 8
1/15 questions
0%
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https://gitlab.mpi-sws.org/simonspies/stdpp/-/commit/2ad847c3276518b9b848b853d77f7261bcff5c04.diff | 1,596,868,253,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-34/segments/1596439737289.75/warc/CC-MAIN-20200808051116-20200808081116-00511.warc.gz | 328,202,061 | 1,578 | diff --git a/theories/tactics.v b/theories/tactics.v index 6ba160cbb042678a98b8a015af3404693003c909..dd3f60ca02aee0fc3b7df2bc6db42c03b489b75e 100644 --- a/theories/tactics.v +++ b/theories/tactics.v @@ -69,10 +69,23 @@ Tactic Notation "etrans" := etransitivity. Note that [split_and] differs from [split] by only splitting conjunctions. The [split] tactic splits any inductive with one constructor. *) -Tactic Notation "split_and" := match goal with |- _ ∧ _ => split end. +Tactic Notation "split_and" := + match goal with + | |- _ ∧ _ => split + | |- Is_true (_ && _) => apply andb_True; split + end. Tactic Notation "split_and" "?" := repeat split_and. Tactic Notation "split_and" "!" := hnf; split_and; split_and?. +Tactic Notation "destruct_and" "?" := + repeat match goal with + | H : False |- _ => destruct H + | H : _ ∧ _ |- _ => destruct H + | H : Is_true (bool_decide _) |- _ => apply (bool_decide_unpack _) in H + | H : Is_true (_ && _) |- _ => apply andb_True in H; destruct H + end. +Tactic Notation "destruct_and" "!" := progress (destruct_and?). + (** The tactic [case_match] destructs an arbitrary match in the conclusion or assumptions, and generates a corresponding equality. This tactic is best used together with the [repeat] tactical. *) | 372 | 1,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-34 | latest | en | 0.757472 |
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PHYSHW
# PHYSHW - Consider a Young” s two-slit experiment in which...
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Unformatted text preview: Consider a Young” s two-slit experiment in which lhe wavelength of light is 33% smaller than the distance between the slits. {a} How man}.r complete dark things total would be seen on a distant screen?I [Asstune for all these questions that the screen is as lHJ'E as it needs tn be.) [b] How man}.r total interference rnasima would be seen on a distant semen? (of: Now suppose that the wavelength of light is 134% larger than the slit separation How man}.r complete darlt hinges would be seen on a distant staeen'? Solutinn: [at The statement of the problem is equivalent to saying that the wavelength of light is filii'i of the slit separation. That means that the two-slit destructive interference conditions bemune dsind=(m+é}i=fl.fifl(m+%)d =r- si115'=fi.fii(n1+%). Because the value of sintir has to be between —l and +1. onlv certain integers in will result in equations that can be solved for an actual angle. Thwe allowable values of In tell us how ntan];r tiarl-t frir : s there are. For the deal: fringes. values of m. = fl.J_l.-‘I all war-la sn there are four dark fringes. {b} This time we use the constructive interference condition dsinl'i' _ innit _ fliflnni _:- sinl'i' _ littlflm. Again Tinting that aim? has tn lie. between :l:|. we find that. the nlluwaljle, integers are in = [Li], which means that there are three bright fringes visible on the screen. {e} In this part we‘re told that the wavelength of light is 134% larger than the slit separa- tion: in other words. it = 2.34s. Then the destructive interference condition is I l l nisinli' = (m + i) a = ass [m + E) s =:~ sinti = ass-m + ii“ This time. there is no integer in that will produce a solvable equation. So there are zero visible dark hinges. In other words. the central max is a huge smear across the HL'TE'E'TI. ...
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Ask a homework question - tutors are online | 666 | 2,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-13 | latest | en | 0.889221 |
https://booksrun.com/blog/category/our-news/page/2/ | 1,603,961,419,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107903419.77/warc/CC-MAIN-20201029065424-20201029095424-00273.warc.gz | 246,207,727 | 18,443 | ## Christmas Contest Announcement
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You can also download our barcode scanner apps for Android and iPhone! | 1,189 | 5,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-45 | latest | en | 0.867933 |
https://www.geeksforgeeks.org/print-path-root-given-node-binary-tree/?ref=ml_lbp | 1,709,163,097,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474746.1/warc/CC-MAIN-20240228211701-20240229001701-00845.warc.gz | 789,249,890 | 66,538 | # Print path from root to a given node in a binary tree
Given a binary tree with distinct nodes(no two nodes have the same data values). The problem is to print the path from root to a given node x. If node x is not present then print “No Path”.
Examples:
Input : 1
/ \
2 3
/ \ / \
4 5 6 7
x = 5
Output : 1->2->5
Approach: Create a recursive function that traverses the different path in the binary tree to find the required node x. If node x is present then it returns true and accumulates the path nodes in some array arr[]. Else it returns false.
Following are the cases during the traversal:
1. If root = NULL, return false.
2. push the root’s data into arr[].
3. if root’s data = x, return true.
4. if node x is present in root’s left or right subtree, return true.
5. Else remove root’s data value from arr[] and return false.
This recursive function can be accessed from other function to check whether node x is present or not and if it is present, then the path nodes can be accessed from arr[]. You can define arr[] globally or pass its reference to the recursive function.
Implementation:
## C++
`// C++ implementation to print the path from root``// to a given node in a binary tree``#include ``using` `namespace` `std;` `// structure of a node of binary tree``struct` `Node``{`` ``int` `data;`` ``Node *left, *right;``};` `/* Helper function that allocates a new node with the`` ``given data and NULL left and right pointers. */``struct` `Node* getNode(``int` `data)``{`` ``struct` `Node *newNode = ``new` `Node;`` ``newNode->data = data;`` ``newNode->left = newNode->right = NULL;`` ``return` `newNode;``}` `// Returns true if there is a path from root``// to the given node. It also populates ``// 'arr' with the given path``bool` `hasPath(Node *root, vector<``int``>& arr, ``int` `x)``{`` ``// if root is NULL`` ``// there is no path`` ``if` `(!root)`` ``return` `false``;`` ` ` ``// push the node's value in 'arr'`` ``arr.push_back(root->data); `` ` ` ``// if it is the required node`` ``// return true`` ``if` `(root->data == x) `` ``return` `true``;`` ` ` ``// else check whether the required node lies`` ``// in the left subtree or right subtree of `` ``// the current node`` ``if` `(hasPath(root->left, arr, x) ||`` ``hasPath(root->right, arr, x))`` ``return` `true``;`` ` ` ``// required node does not lie either in the `` ``// left or right subtree of the current node`` ``// Thus, remove current node's value from `` ``// 'arr'and then return false `` ``arr.pop_back();`` ``return` `false``; ``}` `// function to print the path from root to the``// given node if the node lies in the binary tree``void` `printPath(Node *root, ``int` `x)``{`` ``// vector to store the path`` ``vector<``int``> arr;`` ` ` ``// if required node 'x' is present`` ``// then print the path`` ``if` `(hasPath(root, arr, x))`` ``{`` ``for` `(``int` `i=0; i"``;`` ``cout << arr[arr.size() - 1]; `` ``}`` ` ` ``// 'x' is not present in the binary tree `` ``else`` ``cout << ``"No Path"``;``}` `// Driver program to test above``int` `main()``{`` ``// binary tree formation`` ``struct` `Node *root = getNode(1);`` ``root->left = getNode(2);`` ``root->right = getNode(3);`` ``root->left->left = getNode(4);`` ``root->left->right = getNode(5);`` ``root->right->left = getNode(6);`` ``root->right->right = getNode(7);`` ` ` ``int` `x = 5;`` ``printPath(root, x);`` ``return` `0;``} `
## Java
`// Java implementation to print the path from root ``// to a given node in a binary tree ``import` `java.util.ArrayList;``public` `class` `PrintPath {` ` ``// Returns true if there is a path from root `` ``// to the given node. It also populates `` ``// 'arr' with the given path `` ``public` `static` `boolean` `hasPath(Node root, ArrayList arr, ``int` `x) `` ``{ `` ``// if root is NULL `` ``// there is no path `` ``if` `(root==``null``) `` ``return` `false``; `` ` ` ``// push the node's value in 'arr' `` ``arr.add(root.data); `` ` ` ``// if it is the required node `` ``// return true `` ``if` `(root.data == x) `` ``return` `true``; `` ` ` ``// else check whether the required node lies `` ``// in the left subtree or right subtree of `` ``// the current node `` ``if` `(hasPath(root.left, arr, x) || `` ``hasPath(root.right, arr, x)) `` ``return` `true``; `` ` ` ``// required node does not lie either in the `` ``// left or right subtree of the current node `` ``// Thus, remove current node's value from `` ``// 'arr'and then return false `` ``arr.remove(arr.size()-``1``); `` ``return` `false``; `` ``} ` ` ``// function to print the path from root to the `` ``// given node if the node lies in the binary tree `` ``public` `static` `void` `printPath(Node root, ``int` `x) `` ``{ `` ``// ArrayList to store the path `` ``ArrayList arr=``new` `ArrayList<>();`` ` ` ``// if required node 'x' is present `` ``// then print the path `` ``if` `(hasPath(root, arr, x)) `` ``{ `` ``for` `(``int` `i=``0``; i"``);`` ``System.out.print(arr.get(arr.size() - ``1``)); `` ``} `` ` ` ``// 'x' is not present in the binary tree `` ``else`` ``System.out.print(``"No Path"``); `` ``} ` ` ``public` `static` `void` `main(String args[]) {`` ``Node root=``new` `Node(``1``);`` ``root.left = ``new` `Node(``2``); `` ``root.right = ``new` `Node(``3``); `` ``root.left.left = ``new` `Node(``4``); `` ``root.left.right = ``new` `Node(``5``); `` ``root.right.left = ``new` `Node(``6``); `` ``root.right.right = ``new` `Node(``7``); `` ``int` `x=``5``;`` ``printPath(root, x);`` ``}``}` `// A node of binary tree ``class` `Node ``{ `` ``int` `data; `` ``Node left, right; `` ``Node(``int` `data)`` ``{`` ``this``.data=data;`` ``left=right=``null``;`` ``}``}; ``//This code is contributed by Gaurav Tiwari`
## Python3
`# Python3 implementation to print the path from ``# root to a given node in a binary tree ` `# Helper Class that allocates a new node ``# with the given data and None left and ``# right pointers. ``class` `getNode:`` ``def` `__init__(``self``, data):`` ``self``.data ``=` `data `` ``self``.left ``=` `self``.right ``=` `None` `# Returns true if there is a path from ``# root to the given node. It also ``# populates 'arr' with the given path ``def` `hasPath(root, arr, x):`` ` ` ``# if root is None there is no path `` ``if` `(``not` `root):`` ``return` `False`` ` ` ``# push the node's value in 'arr' `` ``arr.append(root.data) `` ` ` ``# if it is the required node `` ``# return true `` ``if` `(root.data ``=``=` `x): `` ``return` `True`` ` ` ``# else check whether the required node `` ``# lies in the left subtree or right `` ``# subtree of the current node `` ``if` `(hasPath(root.left, arr, x) ``or`` ``hasPath(root.right, arr, x)): `` ``return` `True`` ` ` ``# required node does not lie either in `` ``# the left or right subtree of the current `` ``# node. Thus, remove current node's value `` ``# from 'arr'and then return false `` ``arr.pop(``-``1``) `` ``return` `False` `# function to print the path from root to ``# the given node if the node lies in``# the binary tree ``def` `printPath(root, x):`` ` ` ``# vector to store the path `` ``arr ``=` `[] `` ` ` ``# if required node 'x' is present `` ``# then print the path `` ``if` `(hasPath(root, arr, x)):`` ``for` `i ``in` `range``(``len``(arr) ``-` `1``):`` ``print``(arr[i], end ``=` `"->"``) `` ``print``(arr[``len``(arr) ``-` `1``])`` ` ` ``# 'x' is not present in the `` ``# binary tree `` ``else``:`` ``print``(``"No Path"``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:`` ` ` ``# binary tree formation `` ``root ``=` `getNode(``1``) `` ``root.left ``=` `getNode(``2``) `` ``root.right ``=` `getNode(``3``) `` ``root.left.left ``=` `getNode(``4``) `` ``root.left.right ``=` `getNode(``5``) `` ``root.right.left ``=` `getNode(``6``) `` ``root.right.right ``=` `getNode(``7``) `` ` ` ``x ``=` `5`` ``printPath(root, x)`` ` `# This code is contributed by PranchalK`
## C#
`// C# implementation to print the path from root ``// to a given node in a binary tree ``using` `System;``using` `System.Collections;``using` `System.Collections.Generic;` `class` `PrintPath ``{`` ` `// A node of binary tree ``public` `class` `Node ``{ `` ``public` `int` `data; `` ``public` `Node left, right; `` ``public` `Node(``int` `data)`` ``{`` ``this``.data = data;`` ``left = right = ``null``;`` ``}``}` ` ``// Returns true if there is a path from root `` ``// to the given node. It also populates `` ``// 'arr' with the given path `` ``public` `static` `Boolean hasPath(Node root,`` ``List<``int``> arr, ``int` `x) `` ``{ `` ``// if root is NULL `` ``// there is no path `` ``if` `(root == ``null``) `` ``return` `false``; `` ` ` ``// push the node's value in 'arr' `` ``arr.Add(root.data); `` ` ` ``// if it is the required node `` ``// return true `` ``if` `(root.data == x) `` ``return` `true``; `` ` ` ``// else check whether the required node lies `` ``// in the left subtree or right subtree of `` ``// the current node `` ``if` `(hasPath(root.left, arr, x) || `` ``hasPath(root.right, arr, x)) `` ``return` `true``; `` ` ` ``// required node does not lie either in the `` ``// left or right subtree of the current node `` ``// Thus, remove current node's value from `` ``// 'arr'and then return false `` ``arr.RemoveAt(arr.Count - 1); `` ``return` `false``; `` ``} ` ` ``// function to print the path from root to the `` ``// given node if the node lies in the binary tree `` ``public` `static` `void` `printPath(Node root, ``int` `x) `` ``{ `` ``// List to store the path `` ``List<``int``> arr = ``new` `List<``int``>();`` ` ` ``// if required node 'x' is present `` ``// then print the path `` ``if` `(hasPath(root, arr, x)) `` ``{ `` ``for` `(``int` `i = 0; i < arr.Count - 1; i++) `` ``Console.Write(arr[i]+``"->"``);`` ``Console.Write(arr[arr.Count - 1]); `` ``} `` ` ` ``// 'x' is not present in the binary tree `` ``else`` ``Console.Write(``"No Path"``); `` ``} ` ` ``// Driver code`` ``public` `static` `void` `Main(String []args) `` ``{`` ` ` ``Node root = ``new` `Node(1);`` ``root.left = ``new` `Node(2); `` ``root.right = ``new` `Node(3); `` ``root.left.left = ``new` `Node(4); `` ``root.left.right = ``new` `Node(5); `` ``root.right.left = ``new` `Node(6); `` ``root.right.right = ``new` `Node(7); `` ``int` `x=5;`` ` ` ``printPath(root, x);`` ``}``}` `// This code is contributed by Arnab Kundu`
## Javascript
``
Output
```1->2->5
```
Time complexity: O(n) where n is the number of nodes in the binary tree.
Auxiliary Space: O(H) where H = height of the binary tree.
### Another Approach(Iterative Approach Using Stack):
Follow the below steps to solve the above problem:
1) Start at the root node and push it onto a stack.
2) Create a separate stack to store the path from the root to the current node.
3) While the stack is not empty, do the following:
a) Pop the top node from the stack and add it to the path stack.
b) If the current node is the target node, print the nodes in the path stack to get the path from the root to the target node.
c) Push the right child of the current node onto the stack if it exists.
d) Push the left child of the current node onto the stack if it exists.
Below is the implementation of above approach:
## C++
`// C++ Program to print the path from root``// to a given node in binary tree``#include ``using` `namespace` `std;` `// Structure of binary tree node``struct` `Node {`` ``int` `data;`` ``Node* left;`` ``Node* right;` ` ``Node(``int` `value){`` ``data = value;`` ``left = right = NULL;`` ``}``};` `// function which will print the path``void` `printPath(Node* root, ``int` `target) {`` ``vector<``int``> path;`` ``stack nodeStack;`` ``Node* curr = root;`` ``Node* prev = NULL;` ` ``while` `(curr || !nodeStack.empty()){`` ``while` `(curr){`` ``nodeStack.push(curr);`` ``path.push_back(curr->data);`` ``curr = curr->left;`` ``}`` ` ` ``curr = nodeStack.top();`` ` ` ``if` `(curr->right && curr->right != prev){`` ``curr = curr->right;`` ``}``else``{`` ``if` `(curr->data == target){`` ``for``(``int` `i = 0; i"``;`` ``cout<left = ``new` `Node(2);`` ``root->right = ``new` `Node(3);`` ``root->left->left = ``new` `Node(4);`` ``root->left->right = ``new` `Node(5);`` ``root->right->left = ``new` `Node(6);`` ``root->right->right = ``new` `Node(7);` ` ``int` `target = 5;`` ``printPath(root, target);`` ``return` `0;``}``// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)`
## Java
`import` `java.util.Stack;``import` `java.util.Vector;` `// Structure of binary tree node``class` `Node {`` ``int` `data;`` ``Node left, right;` ` ``public` `Node(``int` `value) {`` ``data = value;`` ``left = right = ``null``;`` ``}``}` `public` `class` `BinaryTreePath {`` ``// function which will print the path`` ``static` `void` `printPath(Node root, ``int` `target) {`` ``Vector path = ``new` `Vector<>();`` ``Stack nodeStack = ``new` `Stack<>();`` ``Node curr = root;`` ``Node prev = ``null``;` ` ``while` `(curr != ``null` `|| !nodeStack.isEmpty()) {`` ``while` `(curr != ``null``) {`` ``nodeStack.push(curr);`` ``path.add(curr.data);`` ``curr = curr.left;`` ``}` ` ``curr = nodeStack.pop();` ` ``if` `(curr.right != ``null` `&& curr.right != prev) {`` ``curr = curr.right;`` ``} ``else` `{`` ``if` `(curr.data == target) {`` ``for` `(``int` `i = ``0``; i < path.size() - ``1``; i++)`` ``System.out.print(path.get(i) + ``"->"``);`` ``System.out.println(path.get(path.size() - ``1``));`` ``return``;`` ``}`` ``path.remove(path.size() - ``1``);`` ``prev = curr;`` ``curr = ``null``;`` ``}`` ``}`` ``System.out.println(``"No Path"``);`` ``}` ` ``// Driver program to test above functions`` ``public` `static` `void` `main(String[] args) {`` ``// Create a binary tree`` ``Node root = ``new` `Node(``1``);`` ``root.left = ``new` `Node(``2``);`` ``root.right = ``new` `Node(``3``);`` ``root.left.left = ``new` `Node(``4``);`` ``root.left.right = ``new` `Node(``5``);`` ``root.right.left = ``new` `Node(``6``);`` ``root.right.right = ``new` `Node(``7``);` ` ``int` `target = ``5``;`` ``printPath(root, target);`` ``}``}`
## Python3
`class` `TreeNode:`` ``def` `__init__(``self``, value):`` ``self``.data ``=` `value`` ``self``.left ``=` `None`` ``self``.right ``=` `None` `def` `print_path(root, target):`` ``path ``=` `[]`` ``node_stack ``=` `[]`` ``curr ``=` `root`` ``prev ``=` `None` ` ``while` `curr ``or` `node_stack:`` ``while` `curr:`` ``node_stack.append(curr)`` ``path.append(curr.data)`` ``curr ``=` `curr.left` ` ``curr ``=` `node_stack[``-``1``]` ` ``if` `curr.right ``and` `curr.right !``=` `prev:`` ``curr ``=` `curr.right`` ``else``:`` ``if` `curr.data ``=``=` `target:`` ``print``( ``"->"``.join(``map``(``str``, path)))`` ``return` ` ``node_stack.pop()`` ``path.pop()`` ``prev ``=` `curr`` ``curr ``=` `None` ` ``print``(``"No Path"``)` `# Driver program to test the function``if` `__name__ ``=``=` `"__main__"``:`` ``# Create a binary tree`` ``root ``=` `TreeNode(``1``)`` ``root.left ``=` `TreeNode(``2``)`` ``root.right ``=` `TreeNode(``3``)`` ``root.left.left ``=` `TreeNode(``4``)`` ``root.left.right ``=` `TreeNode(``5``)`` ``root.right.left ``=` `TreeNode(``6``)`` ``root.right.right ``=` `TreeNode(``7``)` ` ``target ``=` `5`` ``print_path(root, target)`
## C#
`using` `System;``using` `System.Collections.Generic;` `// Structure of binary tree node``public` `class` `Node {`` ``public` `int` `data;`` ``public` `Node left;`` ``public` `Node right;` ` ``public` `Node(``int` `value)`` ``{`` ``data = value;`` ``left = right = ``null``;`` ``}``}` `class` `Program {`` ``// Function which will print the path`` ``static` `void` `PrintPath(Node root, ``int` `target)`` ``{`` ``List<``int``> path = ``new` `List<``int``>();`` ``Stack nodeStack = ``new` `Stack();`` ``Node curr = root;`` ``Node prev = ``null``;` ` ``while` `(curr != ``null` `|| nodeStack.Count > 0) {`` ``while` `(curr != ``null``) {`` ``nodeStack.Push(curr);`` ``path.Add(curr.data);`` ``curr = curr.left;`` ``}` ` ``curr = nodeStack.Peek();` ` ``if` `(curr.right != ``null` `&& curr.right != prev) {`` ``curr = curr.right;`` ``}`` ``else` `{`` ``if` `(curr.data == target) {`` ``for` `(``int` `i = 0; i < path.Count - 1; i++)`` ``Console.Write(path[i] + ``"->"``);`` ``Console.WriteLine(path[path.Count - 1]);`` ``return``;`` ``}` ` ``nodeStack.Pop();`` ``path.RemoveAt(path.Count - 1);`` ``prev = curr;`` ``curr = ``null``;`` ``}`` ``}`` ``Console.WriteLine(``"No Path"``);`` ``}` ` ``// Driver program to test above functions`` ``static` `void` `Main()`` ``{`` ``// Create a binary tree`` ``Node root = ``new` `Node(1);`` ``root.left = ``new` `Node(2);`` ``root.right = ``new` `Node(3);`` ``root.left.left = ``new` `Node(4);`` ``root.left.right = ``new` `Node(5);`` ``root.right.left = ``new` `Node(6);`` ``root.right.right = ``new` `Node(7);` ` ``int` `target = 5;`` ``PrintPath(root, target);`` ``}``}`
## Javascript
`// Structure of binary tree node``class Node {`` ``constructor(value) {`` ``this``.data = value;`` ``this``.left = ``null``;`` ``this``.right = ``null``;`` ``}``}` `// Function to print the path from root to a given node``function` `printPath(root, target) {`` ``let path = [];`` ``let nodeStack = [];`` ``let curr = root;`` ``let prev = ``null``;` ` ``while` `(curr || nodeStack.length > 0) {`` ``while` `(curr) {`` ``nodeStack.push(curr);`` ``path.push(curr.data);`` ``curr = curr.left;`` ``}` ` ``curr = nodeStack[nodeStack.length - 1];` ` ``if` `(curr.right && curr.right !== prev) {`` ``curr = curr.right;`` ``} ``else` `{`` ``if` `(curr.data === target) {`` ``for` `(let i = 0; i < path.length - 1; i++) {`` ``process.stdout.write(path[i] + ``"->"``);`` ``}`` ``console.log(path[path.length - 1]);`` ``return``;`` ``}`` ``nodeStack.pop();`` ``path.pop();`` ``prev = curr;`` ``curr = ``null``;`` ``}`` ``}`` ``console.log(``"No Path"``);``}` `// Driver code to test the above functions``let root = ``new` `Node(1);``root.left = ``new` `Node(2);``root.right = ``new` `Node(3);``root.left.left = ``new` `Node(4);``root.left.right = ``new` `Node(5);``root.right.left = ``new` `Node(6);``root.right.right = ``new` `Node(7);` `let target = 5;``printPath(root, target);`
Output
```1->2->5
```
Time Complexity: O(N), where N is the number of nodes in given binary tree.
Auxiliary Space: O(H), where H is the height of the binary tree.
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# Testing Electrical Hypothesis
Submitted By handsomedoodle1
Words 936
Pages 4
Question 1:
In observation I have come home from work to find the entry door light is not working. As I explore though the house I find that the lights inside and outside the house are not working throughout. I have identified this by attempting to turn the lights on in each room with no luck of light. I know that if only one area of the house was affected that the issue would be likely found at the interior breaker of the home, but this is not the case.
In developing a theory, I know none of the lights are working, which is an indicator that the problem is not a bulb issue or an interior breaker issue. I also know that the power company has no reported outages in the area. I do know that the house was built in the 50’s and the electrical panel has not been upgraded since the home was built. Based on this information my hypothesis is that issue is within the main breaker outside the home. Due to the age of the home and electrical panel, it is likely a burnt out fuse that has caused all the lights to not work.
Just to be sure I have tested the interior breakers of the house by flipping them to the off position and back on, still no lights. I also tested the exterior breaker by flipping it to the off position and back on, still no light. Now by replacing the fuse and switching on the lights I have solved the problem and have light throughout the house.
From the research above I have found the problem was in fact a burnt out fuse at the main breaker of the exterior of the house.
Question 2:
It’s dinner time, I have all the ingredients ready to cook but when I go to turn the stove on nothing happens. I look at the knobs to double check they are on in the correct position and they are, in the on position. That’s when I observe that not only is the stove not working but in conjunction with the stove one half of the kitchen lights and electrical is not working either. I learned the stove was not working when I went to cook dinner; it was then that I found that half of the kitchen didn’t work when I started to explore what was and wasn’t working. In attempts to develop a theory, I know the power is not working for half of the kitchen, and all else in the household is working just fine. Based on this information my hypothesis is that I have overloaded the kitchen breaker causing the breaker to pop. To test my hypothesis I will flip all the breaks to the off position and back on. In my attempt to flip all the breakers off I did not cure the problem, therefore my hypothesis is incorrect and will need to revisit other options. My second theory is that my 19 year old stove is old and no longer running correctly. I believe the stove is pulling additional power causing the breaker to pop by overloading it. My hypothesis is that the stove needs to be replaced. To test my hypothesis I will unplug the stove and repeat the steps in flipping the breakers on and back off. By unplugging the stove and flipping the breakers I have confirmed my hypothesis, the stove is has indeed caused the breaker to pop. To be confident the stove needs to be replaced I will have an appliance handyman test the stove and determine what is wrong with it. This way I will be 100% sure it isn’t a simple fix before spending a few hundred on a new appliance that can be saved. In both scenarios the electrical outage could have been a number of things and would have involved much more investigating. Throughout my career in property management I test the scientific method on a daily bases. Many times the electrical outages have to do with tenants overloading the breaker, appliances pulling too much power due to a defect, a nail driven through the electrical wires, electrical panels giving out, and in rare cases the electrical wiring was installed incorrectly. Each and every case is different is some way which has proven that even though the issue seems to be similar it’s not what it really seems. Without using some type of scientific method to find the issues like above, they would be difficult to understand what is wrong, time consuming, and could become costly. In fact by not using a process of which has some type of order the problem may not be determined. When getting involved in larger projects that involve electrical issue it could become dangerous, thus why thinking the possible theories out would be beneficial. I would like to conclude that we use the scientific method in many if not most of our daily activities, it a process that we learn at an early age without knowing what we are doing. Many will refer to it as problem solving.
References
H. Sawyer (2012) Poplar Mechanics. It’s Electrical! How Your Circuit Break Panel Works. http://www.popularmechanics.com/home/improvement/electrical plumbing/how_your_circuit_breaker_works
R. Wilber (2006) AllExperts. Electrical Wiring in the Home/Lights not working. http://en.allexperts.com/q/Electrical-Wiring-Home-1734/lights-working.htm
T. Harris (2012 How Stuff Works. How Circuit Breakers Work. http://electronics.howstuffworks.com/circuit-breaker.htm
Trefil, James, and Robert M. Hazen (2010) The Sciences: An Integrated Approach, 6th ed. John Wiley & Sons, Inc.
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DISPLAY
Two dimensional turbulent pipe flow
This example may be used to compare GRND2 (Log-law) and
GRND3 (Generalised) wall functions for KE and EP at a
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changing the value of WALLFN (2 for LOGLAW and 3 for GENLAW
wall functions).
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ENDDIS
************************************************************
Group 1. Run Title and Number
************************************************************
************************************************************
TEXT(2D Pipe Flow K-E Model; Re=8.327646E+04 )
************************************************************
************************************************************
IRUNN = 1 ;LIBREF = 292
************************************************************
Group 2. Time dependence
************************************************************
Group 3. X-Direction Grid Spacing
CARTES = F
NX = 1
XULAST =1.0E-02
XFRAC(1)=1.
************************************************************
Group 4. Y-Direction Grid Spacing
NY = 20
YVLAST =0.05
YFRAC(1)=0.01118 ;YFRAC(2)=0.031623
YFRAC(3)=0.058095 ;YFRAC(4)=0.089443
YFRAC(5)=0.125 ;YFRAC(6)=0.164317
YFRAC(7)=0.207063 ;YFRAC(8)=0.252982
YFRAC(9)=0.301869 ;YFRAC(10)=0.353553
YFRAC(11)=0.407891 ;YFRAC(12)=0.464758
YFRAC(13)=0.524047 ;YFRAC(14)=0.585662
YFRAC(15)=0.649519 ;YFRAC(16)=0.715542
YFRAC(17)=0.783661 ;YFRAC(18)=0.853815
YFRAC(19)=0.925946 ;YFRAC(20)=1.
************************************************************
Group 5. Z-Direction Grid Spacing
PARAB = F
NZ = 20
ZWLAST =5.
ZFRAC(1)=0.05 ;ZFRAC(2)=0.1
ZFRAC(3)=0.15 ;ZFRAC(4)=0.2
ZFRAC(5)=0.25 ;ZFRAC(6)=0.3
ZFRAC(7)=0.35 ;ZFRAC(8)=0.4
ZFRAC(9)=0.45 ;ZFRAC(10)=0.5
ZFRAC(11)=0.55 ;ZFRAC(12)=0.6
ZFRAC(13)=0.65 ;ZFRAC(14)=0.7
ZFRAC(15)=0.75 ;ZFRAC(16)=0.8
ZFRAC(17)=0.85 ;ZFRAC(18)=0.9
ZFRAC(19)=0.95 ;ZFRAC(20)=1.
************************************************************
Group 6. Body-Fitted Coordinates
************************************************************
Group 7. Variables: STOREd,SOLVEd,NAMEd
ONEPHS = T
NAME(1)=P1 ;NAME(5)=V1
NAME(7)=W1 ;NAME(12)=KE
NAME(13)=EP ;NAME(149)=EPKE
NAME(150)=ENUT
* Y in SOLUTN argument list denotes:
* 1-stored 2-solved 3-whole-field
* 4-point-by-point 5-explicit 6-harmonic averaging
SOLUTN(P1,Y,Y,Y,N,N,N)
SOLUTN(V1,Y,Y,N,N,N,Y)
SOLUTN(W1,Y,Y,N,N,N,Y)
SOLUTN(KE,Y,Y,N,N,N,N)
SOLUTN(EP,Y,Y,N,N,N,N)
SOLUTN(EPKE,Y,N,N,N,N,N)
SOLUTN(ENUT,Y,N,N,N,N,Y)
VIST = 150
************************************************************
Group 8. Terms & Devices
* Y in TERMS argument list denotes:
* 1-built-in source 2-convection 3-diffusion 4-transient
* 5-first phase variable 6-interphase transport
TERMS(P1,Y,Y,Y,N,Y,Y)
TERMS(V1,Y,Y,Y,Y,Y,Y)
TERMS(W1,Y,Y,Y,Y,Y,Y)
TERMS(KE,N,Y,Y,Y,Y,N)
TERMS(EP,N,Y,Y,Y,Y,N)
DIFCUT =0.5 ;ZDIFAC =1.
GALA = F ;ADDDIF = F
NEWENT = T
ISOLX = -1 ;ISOLY = -1 ;ISOLZ = -1
************************************************************
Group 9. Properties used if PRPS is not
stored, and where PRPS = -1.0 if it is!
RHO1 =1.22 ;TMP1 =0.
EL1 = GRND4
TSURR =0. ;TEMP0 =0.
PRESS0 =0.
DVO1DT =0. ;DRH1DP =0.
EMISS =0. ;SCATT =0.
EL1A =0. ;EL1B =0.
EL1C =0.
ENUL =1.465E-05 ;ENUT = GRND3
ENUTA =0. ;ENUTB =0.
ENUTC =0.
IENUTA = 0
PRNDTL(V1)=1. ;PRNDTL(W1)=1.
PRNDTL(KE)=1. ;PRNDTL(EP)=1.
PRT(V1)=1. ;PRT(W1)=1.
PRT(KE)=1. ;PRT(EP)=1.314
CP1 =1. ;CP2 =1.
************************************************************
Group 10.Inter-Phase Transfer Processes
************************************************************
Group 11.Initial field variables (PHIs)
FIINIT(P1)=1.0E-10 ;FIINIT(V1)=1.0E-10
FIINIT(W1)=10. ;FIINIT(KE)=1.
FIINIT(EP)=32.863354 ;FIINIT(EPKE)=1.0E-10
FIINIT(ENUT)=1.0E-10
No PATCHes yet used for this Group
FSWEEP = 1
NAMFI =CHAM
************************************************************
Group 12. Patchwise adjustment of terms
Patches for this group are printed with those
for Group 13.
Their names begin either with GP12 or &
************************************************************
Group 13. Boundary & Special Sources
PATCH(KESOURCE,PHASEM, 0, 0, 0, 0, 0, 0, 1, 1)
COVAL(KESOURCE,KE , GRND4 , GRND4 )
COVAL(KESOURCE,EP , GRND4 , GRND4 )
PATCH(IN1 ,LOW , 1, 1, 1, 20, 1, 1, 1, 1)
COVAL(IN1 ,P1 , FIXFLU ,12.200001 )
COVAL(IN1 ,V1 ,0. ,0. )
COVAL(IN1 ,W1 ,0. ,10. )
COVAL(IN1 ,KE ,0. ,1. )
COVAL(IN1 ,EP ,0. ,32.863354 )
PATCH(OU1 ,HIGH , 1, 1, 1, 20, 20, 20, 1, 1)
COVAL(OU1 ,P1 ,1000. ,0. )
COVAL(OU1 ,V1 ,0. ,0. )
COVAL(OU1 ,W1 ,0. ,0. )
COVAL(OU1 ,KE ,0. ,0. )
COVAL(OU1 ,EP ,0. ,0. )
PATCH(WA1 ,SWALL , 1, 1, 20, 20, 1, 20, 1, 1)
COVAL(WA1 ,W1 , GRND2 ,0. )
COVAL(WA1 ,KE , GRND3 , GRND3 )
COVAL(WA1 ,EP , GRND3 , GRND3 )
XCYCLE = F
EGWF = T
WALLCO = GRND2
************************************************************
Group 14. Downstream Pressure For PARAB
************************************************************
Group 15. Terminate Sweeps
LSWEEP = 100 ;ISWC1 = 1
LITHYD = 1 ;LITFLX = 1 ;LITC = 1 ;ITHC1 = 1
SARAH =5.
SELREF = T
RESFAC =1.0E-04
************************************************************
Group 16. Terminate Iterations
LITER(P1)=20 ;LITER(V1)=10
LITER(W1)=10 ;LITER(KE)=20
LITER(EP)=20
ENDIT(P1)=1.0E-03 ;ENDIT(V1)=1.0E-03
ENDIT(W1)=1.0E-03 ;ENDIT(KE)=1.0E-03
ENDIT(EP)=1.0E-03
************************************************************
Group 17. Relaxation
RELAX(P1,LINRLX,1.)
RELAX(V1,FALSDT,1.)
RELAX(W1,FALSDT,1.)
RELAX(KE,LINRLX,0.5)
RELAX(EP,LINRLX,0.5)
RELAX(EPKE,LINRLX,1.)
RELAX(ENUT,LINRLX,1.)
KELIN = 3
OVRRLX =0.
EXPERT = F ;NNORSL = F
************************************************************
Group 18. Limits
VARMAX(P1)=1.0E+10 ;VARMIN(P1)=-1.0E+10
VARMAX(V1)=1.0E+06 ;VARMIN(V1)=-1.0E+06
VARMAX(W1)=1.0E+06 ;VARMIN(W1)=-1.0E+06
VARMAX(KE)=1.0E+10 ;VARMIN(KE)=1.0E-10
VARMAX(EP)=1.0E+10 ;VARMIN(EP)=1.0E-10
VARMAX(EPKE)=1.0E+10 ;VARMIN(EPKE)=-1.0E+10
VARMAX(ENUT)=1.0E+10 ;VARMIN(ENUT)=-1.0E+10
************************************************************
Group 19. Data transmitted to GROUND
GENK = T
PARSOL = F
ISG62 = 1
************************************************************
Group 20. Preliminary Printout
DISTIL = T ;NULLPR = F
NDST = 0
DSTTOL =1.0E-02
EX(P1)=0. ;EX(V1)=0.
EX(W1)=0. ;EX(KE)=0.
EX(EP)=0. ;EX(EPKE)=0.
EX(ENUT)=0.
************************************************************
Group 21. Print-out of Variables
INIFLD = F ;SUBWGR = F
* Y in OUTPUT argument list denotes:
* 1-field 2-correction-eq. monitor 3-selective dumping
* 4-whole-field residual 5-spot-value table 6-residual table
OUTPUT(P1,Y,N,Y,Y,Y,Y)
OUTPUT(V1,Y,N,Y,Y,Y,Y)
OUTPUT(W1,Y,N,Y,Y,Y,Y)
OUTPUT(KE,Y,N,Y,Y,Y,Y)
OUTPUT(EP,Y,N,Y,Y,Y,Y)
OUTPUT(EPKE,Y,N,Y,N,N,N)
OUTPUT(ENUT,Y,N,Y,N,N,N)
************************************************************
Group 22. Monitor Print-Out
IXMON = 1 ;IYMON = 10 ;IZMON = 10
NPRMON = 100000 ;NPRMNT = 1 ;TSTSWP = -1
UWATCH = T ;USTEER = T
HIGHLO = F
************************************************************
Group 23.Field Print-Out & Plot Control
NPRINT = 100000 ;NUMCLS = 5
NYPRIN = -1 ;IYPRF = 1 ;IYPRL = 10000
NZPRIN = -1 ;IZPRF = 1 ;IZPRL = 10000
XZPR = F ;YZPR = F
IPLTF = 1 ;IPLTL = -1 ;NPLT = -1
ISWPRF = 1 ;ISWPRL = 100000
ITABL = 3 ;IPROF = 1
ABSIZ =0.5 ;ORSIZ =0.4
NTZPRF = 1 ;NCOLPF = 50
ICHR = 2 ;NCOLCO = 45 ;NROWCO = 20
No PATCHes yet used for this Group
************************************************************
Group 24. Dumps For Restarts
SAVE = T ;NOWIPE = F
NSAVE =CHAM
STOP
``` | 2,954 | 7,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-05 | latest | en | 0.352645 |
https://codedump.io/share/G0wX0hvsC1SF/1/palindrome-number-in-java-doesn39t-seem-to-make-sense | 1,527,394,882,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867995.55/warc/CC-MAIN-20180527024953-20180527044953-00062.warc.gz | 526,808,090 | 8,846 | DuckStudyingJava - 1 year ago 61
Java Question
Palindrome number in java doesn't seem to make sense
I'm doing some exercise programs in Java and came to this palindrome number exercise which tells if the number is palindrome, I'm getting the correct output but I'm trying to explain to myself how the program is working line by line, upon reaching a specific line I noticed that that part doesn't seem to add up, here is the code:
``````public class Palindrome {
public static void main(String[] args) {
int p = 252;
if(isPalindrome(p)) {
System.out.println("Palindrome");
} else {
System.out.println("Not palindrome");
}
}
public static boolean isPalindrome(int x) {
int num = x;
int rev = 0;
while (num != 0) {
int rmd = num % 10;//the remainder of 252 is 2
rev = rev * 10 + rmd;//rev which is 0 multiplied by 10 is 0 + 2 is 2
num = num / 10;
}
if (x == rev) {//x = 252 and rev = 2
return true;
}
return false;
}
}
``````
in the part:
`if(x == rev)`
how can it be true when 252 is not equal to 2? thanks guys..
You have to look what happens during the loop as it goes through its iterations.
For simple programs, such as this one, a paper-and-pencil approach works fine. For more complex programs adding "debug prints" help you understand what is going on:
``````int iterationCount = 0;
while (num != 0) {
System.out.println("Before iteration="+iterationCount+" num="+num+" rev="+rev);
int rmd = num % 10;//the remainder of 252 is 2
rev = rev * 10 + rmd;//rev which is 0 multiplied by 10 is 0 + 2 is 2
num = num / 10;
System.out.println("After iteration="+iterationCount+" num="+num+" rev="+rev);
iterationCount++;
}
``````
This produces the following output:
``````Before iteration=0 num=252 rev=0
After iteration=0 num=25 rev=2
Before iteration=1 num=25 rev=2
After iteration=1 num=2 rev=25
Before iteration=2 num=2 rev=25
After iteration=2 num=0 rev=252
Palindrome
``````
Note how `num` decreases down to zero, while `rev` grows to `252` with each iteration.
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download | 573 | 2,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-22 | latest | en | 0.831224 |
https://tsfa.co/how-to-solving-the-area-of-shapes-64 | 1,679,636,880,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945248.28/warc/CC-MAIN-20230324051147-20230324081147-00672.warc.gz | 641,214,837 | 6,194 | ## Calculating the Area of Shapes
Different methods for estimating the Area of Irregular Shape are: Evaluating Area using unit Squares. Dividing the Irregular Shape into two or more Regular Shapes. Dividing the Irregular Shape with curves into two or more Regular
x
## Area Calculator
Find the area of the side by multiplying the height of the cylinder by the perimeter of the base. The perimeter of a circle is P=2πr, so the area of the side is A=2πhr 5 Add up all the areas: the two identical
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## Area of Plane Shapes
The unit of measurement for the area is always the square of the unit in which lengths are given. The resultant unit is the product of the units of the given lengths. Let’s take an example, the area of a square with side length 8 cm is:
Get Assignment
Explain mathematic equation
Mathematics is the study of numbers, shapes, and patterns. | 317 | 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} | 3.5 | 4 | CC-MAIN-2023-14 | latest | en | 0.889328 |
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https://brainly.in/question/199655 | 1,485,119,438,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281574.78/warc/CC-MAIN-20170116095121-00567-ip-10-171-10-70.ec2.internal.warc.gz | 780,918,904 | 9,853 | # Give two rational numbers whose difference is a rational number sum is a rational number product is a rational number division is a rational number justify also
2
by agnayaprem
I think it should be give two irrational number whose.....Beacause it is of Cbse class 9 SA1 board paper. I have solbed it
2015-10-03T21:52:43+05:30
The two rational numbers are 2,3.
2+3=5
Subtraction
2-3=-1
3-2=1
Multiplication
2×3=6
Division
2÷3=2/3
3÷2=3/2
Hence proved.
2015-10-04T16:27:07+05:30
You can find this in 9th class ncert book or rd sharma | 185 | 539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-04 | latest | en | 0.88517 |
https://astronomy.stackexchange.com/questions/30123/what-pole-is-closer-to-the-galactic-edge?noredirect=1 | 1,624,340,319,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488507640.82/warc/CC-MAIN-20210622033023-20210622063023-00403.warc.gz | 112,886,739 | 31,695 | # What pole is closer to the galactic edge? [duplicate]
Suppose 2 spaceships leave from the earth's north and South Pole going straight up which one will exit the galaxy first (assume they have the same speed and are going in a totally straight line) ? Edge of the galaxy means absence of stars, this question refers to the up-down edge of this spiral galaxy. The question is about the current distance from the earths poles to the edge (top/down) not to the galactic plane, as far as I understand after crossing the plane the nearest pole to the edge will be reversed.
• Given the kind of distances involved, you can treat the Earth as a mathematical point without any loss of accuracy in the answer. Now, do you want "edges" in the spiral plane, or perpendicular to that? – Carl Witthoft Mar 25 '19 at 17:31
• You might like to edit your question to be a bit more precise. How do you define “edge”? Are you including the dark matter halo? Are you ignoring gravitational effects (both boosts and “drag”)? Ie is this about simple straight-line distance, or plotting an escape arc? – Chappo Hasn't Forgotten Monica Mar 25 '19 at 23:47
• Edited to show the question requires a simple straight line, and edge means perpendicular – user426132 Mar 26 '19 at 8:08
• and our solar system is at a large angle to the ecliptic plane (physicsforums.com/threads/…) so to get to the edge, you don't want to exit via the pole but at ~27deg from the equator. – Hobbes Mar 26 '19 at 8:24
• exiting from the pole gives a much longer trajectory than exiting perpendicular to the galactic plane. – Hobbes Mar 26 '19 at 9:59 | 395 | 1,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | longest | en | 0.909502 |
https://www.answers.com/Q/Does_a_book_falling_from_a_shelf_have_gravitational_energy | 1,563,280,391,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524548.22/warc/CC-MAIN-20190716115717-20190716141717-00391.warc.gz | 600,037,088 | 16,117 | # Does a book falling from a shelf have gravitational energy?
Yes; while on the shelf it has potential energy due to gravity of mgh where m = mass and g =gravity acceleration and h is height fom floor.
As it falls its potential energy is converted to kinetic energy mv squared/2 where v = velocity | 65 | 299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-30 | latest | en | 0.954548 |
https://www.jiskha.com/display.cgi?id=1273188118 | 1,516,795,259,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084894125.99/warc/CC-MAIN-20180124105939-20180124125939-00635.warc.gz | 927,141,373 | 3,889 | # Pre-cal
posted by .
Find the vertex of y^2 + 4y + 2x +10 = 0. Use completing the square to put the equation into standard form as your first step. Could you show me how to get the vertex? Thanks
• Pre-cal -
y^2 + 4y + 4 - 4 + 2x + 10 = 0
(y+2)^2 + 6 = -2x
x = (-1/2)(y+2)^2 + 6
• Pre-cal -
123694587
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More Similar Questions | 667 | 2,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-05 | latest | en | 0.847457 |
http://mathhelpforum.com/calculus/94323-taylor-polynomial-print.html | 1,511,140,289,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805881.65/warc/CC-MAIN-20171119234824-20171120014824-00447.warc.gz | 198,331,777 | 2,949 | # taylor polynomial
• Jul 3rd 2009, 04:35 PM
eniuqvw
taylor polynomial
I need to find the third-degree taylor polynomial at a for f(x) = sin9x and a = pi/54. I guess it's not clear to me what to do with the 9 in sin9x, does it end up in the final answer or only to calculate the coefficients? If someone could explain this polynomial intuitively (particularly what to do with the 9), that would be awesome.
• Jul 3rd 2009, 05:04 PM
Random Variable
$f(x) = \frac {f(a)}{0!}(x-a)^{0} + \frac {f'(a)}{1!} (x-a)^{1} + \frac {f''(a)}{2!}(x-a)^{2} + \frac {f'''(a)}{3!}(x-a)^{3} + ...$
$= f(a) + f'(a) (x-a) + \frac {f''(a)}{2}(x-a)^{2} + \frac {f'''(a)}{6}(x-a)^{3} + ...$
$f(x) = \sin (9x)$
$f'(x) = 9 \cos(9x)$
$f''(x) = -81 \sin(9x)$
$f'''(x) = -729 \cos(9x)$
now find f(a), f'(a), f''(a), and f'''(a)
• Jul 4th 2009, 02:33 PM
eniuqvw
ok cool, I got that far, I just wanted to know what to enter into the (x - a) part of the polynomial. Do I replace x with 9x because of the 9 in sin(9x). In other words, would the final polynomial have (9x - a) to the nth power or just (x - a) to the nth power? For me, it's not clear what to do with the 9 and (most importantly) why.
• Jul 4th 2009, 07:23 PM
Random Variable
just $\Big( x- \frac {5 \pi}{4} \Big)$ | 496 | 1,251 | {"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": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-47 | longest | en | 0.869949 |
https://www.java-forums.org/new-java/13716-problem-calculating-array-elements-print.html | 1,521,487,304,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647044.86/warc/CC-MAIN-20180319175337-20180319195337-00454.warc.gz | 829,885,673 | 4,194 | # PROBLEM - calculating with array elements
• 11-26-2008, 06:46 PM
ella
PROBLEM - calculating with array elements
hey :) i'm a girl who's studying at the university and i have trouble with java. i hope someone can help. i have 3 arrays. 2 of them are with given elements. elements for third one i have to calculate (distance / speed = time).
int a=4; //length for all
int[]b={100, 150, 138, 196}; //distance
int[]c= {70, 110, 96, 87}; //speed
double[]d=new double[a]; //empty array for times
for(int i=0;i<4;i++){
System.out.print(b[i]+"km ");
System.out.print(c[i]+"km/h ");
d[i]=b[i]/c[i];
System.out.println(d[i]+"h");
}
and the problem is, it prints correctly the distance and speed but when it should calculate the time, then it gives me some kinda rounded result. for example 100/70=1.0 and 196/87=2.0 instead of 1.42 and 2.25... what is wrong?:confused:
• 11-26-2008, 07:18 PM
serjant
Hey,i am a boy,wanna tell you that you have a problem with casting, when you calculate division with integers,and the result should be asigned into double value,so cast the result to double,it should be this way:
Code:
``` int a=4; //length for all int[]b={100, 150, 138, 196}; //distance int[]c= {70, 110, 96, 87}; //speed double[]d=new double[a]; //empty array for times for(int i=0;i<a;i++){ System.out.print(b[i]+"km "); System.out.print(c[i]+"km/h "); d[i]=(double)b[i]/c[i]; System.out.println(d[i]+"h"); }```
• 11-26-2008, 08:01 PM
ella
wow, and that was all?... lol. i was thinking like crazy. works fine now :P. thanx a lot.
and now i have one more question :D. i need only 2 spaces after comma but can't find a simple way how to do it :/. i've found only complicated methods. is there a simple way anyway?
• 11-26-2008, 11:04 PM
CJSLMAN
two spaces?
Quote:
i need only 2 spaces after comma...
Not sure I understand...
• Two spaces after a comma in a println()?
• Two spaces after a comma in an array?
• Two spaces after a comma in _____? (fill in in blank space :-)
CJSL
• 11-26-2008, 11:48 PM
ella
oh =) forgot that info. i ment in println().
• 11-27-2008, 12:24 AM
CJSLMAN
hhhmmm...
You mean like:
Code:
`System.out.print("One space:" + " " + ", now two spaces:" + " " + ". The end!");`
CJSL
• 12-03-2008, 10:15 AM
ella
if i calculate 100/170=0.588235294... but i don't want to see that on screen. 2 spaces after comma i mean i wanna see 0.59 instead of that long row of numbers :).
• 12-03-2008, 10:36 AM
serjant
This code will round your float:
Code:
``` /** * Round a double value to a specified number of decimal * places. * * @param val the value to be rounded. * @param places the number of decimal places to round to. * @return val rounded to places decimal places. */ public static double round(double val, int places) { long factor = (long)Math.pow(10,places); // Shift the decimal the correct number of places // to the right. val = val * factor; // Round to the nearest integer. long tmp = Math.round(val); // Shift the decimal the correct number of places // back to the left. return (double)tmp / factor; } /** * Round a float value to a specified number of decimal * places. * * @param val the value to be rounded. * @param places the number of decimal places to round to. * @return val rounded to places decimal places. */ public static float round(float val, int places) { return (float)round((double)val, places); }```
Then call the method like this:
Code:
```float y = 9.87654f; float w; w = round(y,3); System.out.println(w);```
So you can use that method written above in your program
• 12-03-2008, 04:59 PM
xcallmejudasx
or you can attach a DecimalFormat to your print statement.
• 12-03-2008, 10:00 PM
timkd127
Hey, also a boy here haha.
Just a suggestion... make
int a=4;
final int a=4;
just a style thing that alot of proffessors and bosses like.
• 12-03-2008, 10:15 PM
xcallmejudasx
Quote:
Originally Posted by timkd127
Hey, also a boy here haha.
Just a suggestion... make
int a=4;
final int a=4;
just a style thing that alot of proffessors and bosses like.
He means final int A = 4 (since final objects are generally denoted in all caps)
lol sorry just nit picking
• 12-03-2008, 10:26 PM
timkd127
haha u kno i noticed that after i posted it but was too lazy in that moment to edit it.
• 12-03-2008, 10:30 PM
xcallmejudasx
ya I hate writing like that. Specially when I have 50+ final Strings with names like CMD_PROPERTIES_EDIT_VIEW and my caps lock button is broken. Pain in the ass since double clicking selects the entire thing and not just properties or whatever when I need to edit
• 12-04-2008, 01:36 AM
timkd127
yea, plus with the IDE's now u can just run ur mouse over a variable and it shows u where it occurs. | 1,440 | 4,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-13 | latest | en | 0.763048 |
https://aatul.me/tag/java-program-to-find-whether-given-number-is-prime-or-not/ | 1,685,242,474,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643462.13/warc/CC-MAIN-20230528015553-20230528045553-00340.warc.gz | 109,345,832 | 13,914 | # Java program to Find whether given number is Prime or Not
## Java Program to find prime number upto N number
Objectives :
• Write a Java program to find prime number upto N number
• Write a Java program to print prime number upto N number
• Write a Java program to find prime number upto N number using Command Line Arguments
• Write a Java Program to find prime number upto N number via Command Line Arguments
• Write a Java program to find prime number upto N number using Scanner class
Java Program / Code :
Method 1 : Java Program to find Prime Number upto N number using Scanner class
[sourcecode lang=”Java”]
import java.util.Scanner;
{
public static void main(String[] args)
{
int n,p;
Scanner s=new Scanner(System.in);
System.out.println(“Enter number : ”);
n=s.nextInt();
for(int i=2;i<n;i++)
{
p=0;
for(int j=2;j<i;j++)
{
if(i%j==0)
p=1;
}
if(p==0){
System.out.println(i);
}
}
}
}
[/sourcecode]
Method 2 : Java Program to find Prime Number upto N number using Scanner class and writing function to find prime number
Method isPrime() for checking if a number is prime or not
[sourcecode lang=”java”]
public boolean isPrime(int num) {
if ( num < 2 ){
return false;
}
for (int i = 2; i <= Math.sqrt(num); i++) {
if ( num % i == 0 ) {
return false;
}
}
return true;
}
}
// Mock Test Class to test above code
import java.util.Scanner;
public class Demo
{
public static void main(String[] args)
{
Scanner scanner = new Scanner(System.in);
int num = scanner.nextInt();
if ( num < 2 ) {
System.out.println("\n There are no Prime Numbers available");
System.exit(0);
}
System.out.println("\n Prime Numbers from 1 to "+ num);
for (int i = 3; i <= num; i++) {
System.out.print(", " + i);
}
}
}
}
[/sourcecode]
Method 3 : Java Program to find Prime Number upto N number using Command line input
[sourcecode lang=”Java”]
{
public static void main(String[] args)
{
int n,p;
n=Integer.parseint(args[0]);
for(int i=2;i<n;i++)
{
p=0;
for(int j=2;j<i;j++)
{
if(i%j==0)
p=1;
}
if(p==0){
System.out.println(i);
}
}
}
}
[/sourcecode]
Steps to run above program via command line :
2. Interpretation : C:\JavaPrograms>java PrimeNumber 20
Output :
2
3
5
7
11
13
17
19 | 591 | 2,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-23 | latest | en | 0.330706 |
https://edurev.in/studytube/TrueFalse-of-Electrostatics--Past-year-Questions-J/b2b9cbce-f639-41c3-81fd-65920c2a316d_t | 1,686,337,215,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656788.77/warc/CC-MAIN-20230609164851-20230609194851-00665.warc.gz | 262,247,900 | 61,634 | JEE > True/False: Electrostatics | JEE Advanced
# True/False: Electrostatics | JEE Advanced | Physics 35 Years JEE Main & Advanced Past year Papers
1 Crore+ students have signed up on EduRev. Have you?
True/False
Q.1. The work done in carrying a point charge from one point to another in an electrostatic field depends on the path along which the point charge is carried. (1981- 2 Marks)
Ans. F
Solution. Electrostatic force is conservative in nature, therefore work done is path independent.
Q.2. Two identical metallic spheres of exactly equal masses are taken. One is given a positive charge Q coulombs and the other an equal negative charge. Their masses after charging are different. (1983 - 2 Marks)
Ans. T
Solution. The metallic sphere which gets negatively charged gains electrons and hence its mass increases.
The metallic sphere which gets positively charged loses electrons and hence its mass decreases.
Q.3. A small metal ball is suspended in a uniform electric field with the help of an insulated thread. If high energy X-ray beam falls on the ball, the ball will be deflected in the direction of the field. (1983 - 2 Marks)
Ans. T
Solution. When high energy X-ray beam falls, it will knock out electrons from the small metal ball making it positively charged.
Therefore the ball will be deflected in the direction of electric field.
Q.4. Two protons A and B are placed in between the two plates of a parallel plate capacitor charged to a potential difference V as shown in the figure. The forces on the two protons are identical. (1986 - 3 Marks)
Ans. T
Solution. The electric field produced between th e parallel plate capacitor is uniform. The force acting on charged particle placed in an electric field is given by F = qE.
In the case of two protons, q and E are equal and hence force will be equal.
Q.5. A ring of radius R carries a uniformly distributed charge + Q.
A point charge – q is placed on the axis of the ring at a distance 2R from the centre of the ring and released from rest. The particle executes a simple harmonic motion along the axis of the ring. (1988 - 2 Marks)
Ans. F
Solution. KEY CONCEPT : Force on charge (– q) due to small charge dq situated at length dℓ is given by
Resolving this force into two parts dFcos θ and dFsin θ as shown in figure.
If we take another diametrically opposite length d, the charge on it being dq. Then the force on charge (– q) by this small charge dq will be
Again resolving this force, we find dFsin q components of the two forces cancel out and dFcos q component adds up.
∴ The total force
Charge on length 2πR = Q
This is not an equation of simple harmonic motion.
Q.6. An electric line of forces in the x – y plane is given by the equation x2 + y2 = 1. A particle with unit positive charge, initially at rest at the point x = 1, y = 0 in the x – y plane, will move along the circular line of force. (1988 - 2 Marks)
Ans. F
Solution. For a particle to move in circular motion, we need a centripetal force which is not available.
The statement is false.
The document True/False: Electrostatics | JEE Advanced | Physics 35 Years JEE Main & Advanced Past year Papers is a part of the JEE Course Physics 35 Years JEE Main & Advanced Past year Papers.
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## Physics 35 Years JEE Main & Advanced Past year Papers
110 docs|49 tests
## Physics 35 Years JEE Main & Advanced Past year Papers
110 docs|49 tests
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; | 905 | 3,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-23 | latest | en | 0.89612 |
https://forum.pyro.ai/t/hierarchical-model-with-selection-bias/6243 | 1,721,231,638,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.19/warc/CC-MAIN-20240717151625-20240717181625-00310.warc.gz | 249,146,916 | 4,514 | # Hierarchical model with selection bias
Hi,
I wanted to formulate a hierarchical model that also produces observables that are selection biased. I am not sure what the best formulation to this problem is.
The details (for this toy model) are as follows.
1. There are `n_bins` bins and for each I draw a rate, called `rate_all` (from a uniform distribution).
2. Poisson sample the variable `n_all` given this rate.
3. For each of the bins, then draw `n_all` observations for each bin (so the number of observations in general is different in each bin). This new variable is called `loudness` and is drawn from a Normal distribution.
4. Depending on some kind of detected threshold (defined by the function `is_detected`), only a subset of these `loudness`s are detected.
5. Define an obs statement for the `loudness` parameter.
How do I implement this most elegantly?
Code:
``````# Meta variable
n_bin = 10
rate_all_max = 30 # Define the maximum value for rate_all
# Detection function
def is_detected(loudness, threshold=5.0):
return loudness > threshold
def model(observed_loudness=None, threshold=5.0):
# Draw n_all from a Poisson distribution with rate_all
with numpyro.plate("bins", n_bin):
# Sample rate_all from a uniform distribution
rate_all = numpyro.sample("rate_all", dist.Uniform(0, rate_all_max))
n_all = numpyro.sample("n_all", RightTruncatedPoisson(rate_all, high=50))
# Draw loudness from a normal distribution for each bin
loudness = []
for i in range(n_bin):
with numpyro.plate(f"loudness_{i}_plate", n_all[i]):
loudness.append(numpyro.sample(f"loudness_{i}", dist.Normal(5, 2)))
loudness = jnp.concatenate(loudness)
# Apply the detection function
detected_mask = is_detected(loudness, threshold)
# I am not sure how I efficiently apply the selection
# and write the obs statement
``````
with the custom distribution
``````class RightTruncatedPoisson(dist.Distribution):
"""
A truncated Poisson distribution.
:param numpy.ndarray high: high bound at which truncation happens
:param numpy.ndarray rate: rate of the Poisson distribution.
"""
arg_constraints = {
"high": dist.constraints.nonnegative_integer,
"rate": dist.constraints.positive,
}
has_enumerate_support = True
def __init__(self, rate=1.0, high=0, validate_args=None):
batch_shape = jax.lax.broadcast_shapes(jnp.shape(high), jnp.shape(rate))
self.high, self.rate = dist.util.promote_shapes(high, rate)
super().__init__(batch_shape, validate_args=validate_args)
def log_prob(self, value):
m = jax.scipy.stats.poisson.cdf(self.high, self.rate)
log_p = jax.scipy.stats.poisson.logpmf(value, self.rate)
return jnp.where(value <= self.high, log_p - jnp.log(m), -jnp.inf)
def sample(self, key, sample_shape=()):
shape = sample_shape + self.batch_shape
float_type = jnp.result_type(float)
minval = jnp.finfo(float_type).tiny
u = jax.random.uniform(key, shape, minval=minval)
# return self.icdf(u) # Brute force
# return self.icdf_faster(u) # For faster sampling.
return self.icdf(u) # Using `host_callback`
`````` | 771 | 3,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.776062 |
https://www.solutioninn.com/stores-walmart-is-the-largest-retailing-firm-in-the-world | 1,708,719,052,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474445.77/warc/CC-MAIN-20240223185223-20240223215223-00432.warc.gz | 1,004,194,975 | 24,198 | # Stores (Wal-Mart) is the largest retailing firm in the world. Building on a base of discount stores,
## Question:
Stores (Wal-Mart) is the largest retailing firm in the world. Building on a base of discount stores, Wal-Mart has expanded into warehouse clubs and Supercenters, which sell traditional discount store items and grocery products.
Exhibits 10.11, 10.12, and 10.13 present the financial statements of Wal-Mart for Years 2, 3, and 4. Exhibits 4.51 to 4.53 (Case 4.2 in Chapter 4) also present summary financial statements for Wal-Mart, and Exhibit 4.54 presents selected financial statement ratios for Years 2, 3, and 4. [Note: The data presented in Chapter 4 for Wal-Mart differ slightly from the data here because the Chapter 4 data have been adjusted slightly to remove the effects of items such as discontinued operations, for purposes of computing financial analysis ratios.]
a. Design a spreadsheet and prepare a set of financial statement forecasts for Wal-Mart for Year +1 to Year +5, using the assumptions that follow. Project the amounts in the order presented (unless indicated otherwise), beginning with the income statement, then the balance sheet, and then the statement of cash flows
Income Statement
Sales
Sales grew 11.6 percent in Year 3 and 11.6 percent in Year 4, primarily as a result of significant increases in same-store sales and opening new stores. The compound annual growth rate during the last five years was 12.8 percent. In the future, Wal-Mart will continue to grow internationally by opening stores and acquiring other firms, and domestically by converting discount stores to Supercenters. In addition, despite competition, Wal-Mart will likely also continue to enjoy increases in same-store sales of 5 percent to 6 percent, consistent with its experience through Year 4. Thus, assume that sales will grow 10 percent each year between Year +1 and Year +5.
Cost of Goods Sold
The cost of goods sold to sales percentage steadily decreased from 77.7 percent of sales in Year 2 to 77.1 percent in Year 4. Wal-Mart's everyday low-price strategy, its movement into grocery products, and competition will likely prevent Wal-Mart from achieving significant additional decreases in this expense percentage. Assume that the cost of goods sold to sales percentage will average 77.0 percent for Year +1 to Year +5.
The selling and administrative expense percentage has steadily increased from 17.4 percent of sales in Year 2 to 17.9 percent of sales in Year 4. Identifying and transacting international corporate acquisitions and opening additional Supercenters will put upward pressure on this expense percentage, but the slowdown in sales growth will moderate this upper pressure. Assume that the selling and administrative expense to sales percentage will remain at 18.0 percent for Year +1 to Year +5.
Other Operating Income
Other operating income has been approximately 1 percent of revenues during the last three years. Assume that other operating income will continue at this historical pattern.
Interest Income
Wal-Mart earns a bit of interest income on its cash and cash equivalents accounts. The average interest rate earned on average cash balances was approximately 3.8 percent
during Year 4. Assume Wal-Mart will earn interest income based on a 3.8 percent interest rate on average cash balances (that is, the sum of the beginning and end of the year cash divided by 2) for Year +1 through Year +5. Note: Projecting the amount of interest income must await projection of cash on the balance sheet.
Interest Expense
Wal-Mart engages in long-term borrowing to construct new stores domestically, and in both short- and long-term borrowing to finance corporate acquisitions. The average interest rate on all interest-bearing debt was approximately 4.1 percent during Year 4. Assume a 4.1 percent interest rate for all outstanding borrowing (notes payable, long-term debt, and current portion of long-term debt) for Wal-Mart for Year +1 through Year +5. Compute interest expense on the average amount of interest-bearing debt outstanding each year (that is, the sum of the beginning and end of the year divided by 2). NOTE: Projecting the amount of interest expense must await projection of interest-bearing debt on the balance sheet.
Other Expenses
Other expenses have been approximately 0.1 percent of revenues during the last three years. Assume that other expenses will continue at this historical pattern.
Continue over next page
Income Tax Expense
Wal-Mart's average income tax rate as a percentage of income before taxes has varied between 35.2 percent and 36.6 percent during the last three years. Assume an income tax rate of 36.0 percent of income before income taxes for Year +1 through Year +5. NOTE: Projecting the amount of income tax expense must await computation of income before taxes.
Balance Sheet
Cash
Cash will be the flexible financial account we will use to plug the amount necessary to equate total assets with total liabilities plus shareholders' equity. Projecting the amount of cash must await projections of all other balance sheet amounts.
Accounts Receivable
Accounts receivable will increase at the growth rate in sales. Inventories Wal-Mart has maintained a steady inventory turnover ratio of 7.8 times during the last two years (based on average inventory balances each year). In recent years, the expanding role of grocery products has increased Wal-Mart's inventory turnover. However, that increase has been offset by the stocking of new stores and the distribution of merchandise to stores worldwide. We will assume that inventory turnover will continue to average 7.8 times per year (every 47 days) in Years +1 to +5. Use this turnover rate to compute the average inventories each year, and then compute the implied ending inventories each year.
Prepayments
Other current assets include prepayments, which relate to ongoing operating costs, such as rent and insurance. Assume that prepayments will grow at the growth rate in sales.
Property, Plant, and Equipment-at cost
Property, plant, and equipment grew 14.4 percent annually during the most recent five years. In the most recent three years, Wal-Mart's growth in property, plant, and equipment has varied from 14.4 percent in Year 2 to 13.9 percent in Year 3 to 16.6 percent in Year 4. The construction of new Supercenters and the acquisition of established retail chains abroad will require additional investments in property, plant, and equipment. Assume that property, plant, and equipment will grow 14.4 percent each year from Year + 1 through Year +5.
Accumulated Depreciation
In Years 3 and 4, the change in Wal-Mart's accumulated depreciation has averaged roughly 4.0 percent of the beginning of year balance in property, plant, and equipment - at cost. During Year +1 through Year +5 assume accumulated depreciation will increase each year by an amount equal to 4.0 percent of the beginning of year balance in property, plant and equipment - at cost.
Goodwill and Other Assets
Goodwill and other assets primarily include goodwill arising from corporate acquisitions abroad. Such acquisitions increase Wal-Mart sales. Assume that goodwill and other assets will grow at the growth rate in sales.
Accounts Payable
Wal-Mart has maintained a steady accounts payable turnover, with payment periods averaging 33 days (an average turnover ratio of roughly 11.1 times per year) during the last five years (based on average balances each year). We will assume that accounts payable turnover will continue to average 33 days in Years +1 to +5. Use this turnover rate to compute the average accounts payable each year, and then compute the implied ending accounts payable each year. Remember to add the change in inventory to the cost of goods sold to obtain the total amount of credit purchases of inventory during the year.
Short-Term Debt, Current Maturities of Long-Term Debt, and Long-Term Debt
Wal-Mart uses short-term debt, current maturities of long-term debt, and long-term debt to augment cash from operations to finance acquisitions of property, plant, and equipment, and acquisition of existing retail chains abroad. Over the past five years, the total amount of Wal-Mart's interest-bearing debt has grown at 7.3 percent compounded annually. Assume that each of three interest-bearing sources of debt capital will increase at 7.3 percent per year over Year +1 through Year +5.
Other Current Liabilities
Other current liabilities relate to accrued expenses for ongoing operating activities and are expected to grow at the growth rate in selling and administrative expenses, which are expected to grow with sales.
Other Noncurrent Liabilities
Other noncurrent liabilities include amounts related to health care benefits and deferred taxes. Assume that other noncurrent liabilities will increase at the growth rate in sales.
Common Stock
Assume that common stock and additional paid-in capital will not change.
Retained Earnings
The increase in retained earnings equals net income minus dividends. Wal-Mart paid dividends amounting to $2,214 million to common shareholders in Year 4. Over the past five years, Wal-Mart's dividends have increased at an average annual rate of 20.0 percent. Assume that dividends will grow 20.0 percent each year between Year +1 and Year +5. In addition, assume that in Year +1 through Year +5, Wal-Mart will not repurchase any treasury stock. Accumulated Other Comprehensive Income Assume that accumulated other comprehensive income will not change. Cash At this point we can now project the amount of cash on Wal-Mart's balance sheet at each year-end from Year +1 to Year +5. We assume Wal-Mart uses cash as the flexible financial account to balance the balance sheet. Statement of Cash Flows Depreciation Add back Include the change in accumulated depreciation. Other Add backs Assume that changes in other noncurrent liabilities on the balance sheet are operating activities. Other Investing Transactions Assume that changes in other noncurrent assets on the balance sheet are investing activities. Required (continued from page 787) b. If you have programmed your spreadsheet correctly, the projected amount of cash declines from Year +1 to Year +5, and the projected cash balance at the end of Year +5 is a negative$584 million. Given the profitability and growth projected for Wal-Mart, a negative balance in cash seems unlikely. Identify the likely reason for the negative projected amount of cash.
c. Assume now that long-term debt will grow in Year +1 to Year +5 at a growth rate of 12.0 percent, which is closer to the assumed growth rate of 14.4 percent in property, plant, and equipment - at cost. Leave the forecast assumptions for short-term debt and current maturities of long-term debt unchanged. Assess whether this growth rate in long-term debt provides more reasonable forecast amounts for cash.
d. Calculate the financial statement ratios listed in Exhibit 4.54 for Wal-Mart using the forecast amounts determined in part c for Year +1 to Year +5. Assess the projected changes in the profitability and risk of Wal-Mart for Year +1 to Year +5.
Inventory Turnover Ratio
Inventory Turnover RatioThe inventory turnover ratio is a ratio of cost of goods sold to its average inventory. It is measured in times with respect to the cost of goods sold in a year normally. Inventory Turnover Ratio FormulaWhere,...
Financial Statements
Financial statements are the standardized formats to present the financial information related to a business or an organization for its users. Financial statements contain the historical information as well as current period’s financial...
Common Stock
Common stock is an equity component that represents the worth of stock owned by the shareholders of the company. The common stock represents the par value of the shares outstanding at a balance sheet date. Public companies can trade their stocks on...
Accounts Payable
Accounts payable (AP) are bills to be paid as part of the normal course of business.This is a standard accounting term, one of the most common liabilities, which normally appears in the balance sheet listing of liabilities. Businesses receive...
Distribution
The word "distribution" has several meanings in the financial world, most of them pertaining to the payment of assets from a fund, account, or individual security to an investor or beneficiary. Retirement account distributions are among the most...
Fantastic news! We've located the answer you've been seeking! | 2,546 | 12,515 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-10 | latest | en | 0.938367 |
https://www.soccerwidow.com/football-gambling/betting-knowledge/betting-advice/money-management/euro-2012-staking-plan-comparison/?replytocom=13257 | 1,632,139,029,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00425.warc.gz | 1,022,748,585 | 16,097 | # Euro 2012 Simulation – The Moment of Truth: Staking Plan Comparison
### The Nutty Professors:
Congratulations to those ‘mad’ scientists at the International School of Management (ISM) in Frankfurt and the German Sports University in Cologne whose predictions we followed throughout Euro 2012.
Image: Pikoso.kz (Shutterstock)
Those of you who also followed suit reaped the benefits of their statistical predictions realising a yield of between 19.7% and 25.5% depending upon which style of staking plan you employed (either fixed stake or fixed risk/win).
However, some of their prophecies did not materialise…
For example, co-hosts Poland did not make the semi-finals.
There were plenty of goals in the England v. Sweden match but not a huge goal difference for England and in the end they struggled to win at all.
Alas, the scientists’ beloved Germany did not get a chance to dethrone the defending champions in the final despite saying, “chance, luck and statistics are favouring zis”.
Never-ze-less, ze predictions from ze Frankfurt and Cologne think-tanks produced handsome profits.
As mentioned in our original article not all predictions will win. Indeed, it would have been miraculous if they had all won as the mathematical likelihood was just 0.00000031% (326 million to 1 – that’s definitely the way to bankrupt the bookies!).
Soccerwidow’s pre-tournament profit/loss estimation of the scientists’ predictions using a fixed risk/fixed win staking plan:
Total risk of all bets (stakes): 106.55 Units
Maximum potential profit: 140.52 Units
Realistically expected profit (10%-15% yield): 10.66 to 15.98 Units
However, things turned out better than expected with profits of 27.22 units (25.5% yield) based on our favoured fixed risk/fixed win staking plan:
Euro 2012 Simulation: Staking Plan Fixed Win / Fixed Risk
Euro 2012 Simulation: Staking Plan Fixed Stake
The second table represents the same bets using a fixed stake staking plan.
We feel that a Fixed win/Fixed Risk staking plan is the most solid and reliable form of staking for maximised profits and minimised losses, and we have just found this article Raceadvisor.co.uk: Fixed Profits or Fixed Stakes? which includes a mathematical experiment comparing a fixed stake with a fixed risk/win staking plan. It’s an interesting read and a similar outcome to our own findings.
However, the Fixed Stake Staking Plan tends to be a more popular method and although this would have also brought a nice profit of 22.65 units (19.7% yield), the Fixed win/Fixed Risk staking plan remains ‘smarter’ in our humble opinion.
Excelspreadsheet Euro 2012 Simulation – Staking Plan Comparison
### Famous Last Words
In our original ‘mad scientists’ article we recommended a fixed win/fixed risk staking plan and as we have seen, this method performed better than a fixed stake staking plan.
However, both staking plans produced fairly similar results due mainly to the fact that betting odds between 1.5 and 3.5 were in play. Especially at the lower odds the differences between the stake amounts in both plans were not huge, which ultimately led to similar results.
Whichever staking plan you choose is down to you and it then remains to follow it religiously without emotion and to never chase losses…
If you now have a great void in your life following the completion of Euro 2012, instead of developing withdrawal symptoms or falling into post-Euro 2012 depression, remember that you always have therapy available by reading this blog from cover to cover. Then, if you are very brave you may wish to dive into Soccerwidow’s Fundamentals of Sports Betting course in order to learn the skills of professional odds calculation and prepare yourself for next season!
Last Update: 7 July 2012
Categories:Case Studies Money Management
### 4 Responses to “Euro 2012 Simulation – The Moment of Truth: Staking Plan Comparison”
1. 14 May 2015 at 10:54 pm #
Shouldn’t we always stake fix amount of units for short odds?
2. 14 May 2015 at 10:52 pm #
Hi, is the fixed win/ fix risk staking plan simply means risk maximum ‘x’ unit and win maximum ‘x’ unit? If so, how come you staked in the above example, 5.54 units @1.80 for ‘Group A – Poland’. Shouldn’t we according to plan suggested, stake fix amount of units in short odds?
Thank you
3. 15 November 2012 at 1:10 pm #
Sorry for your trouble. It’s done 🙂
4. 15 November 2012 at 12:11 pm #
After sharing this article on FB and Twitter! the Staking Plan Excel Spreadsheet DID NOT magically appear!! Please can you email it to me, it’s of great interest. Thank you. | 1,055 | 4,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-39 | latest | en | 0.929133 |
https://www.tutoraspire.com/roc-curve-stata/ | 1,721,599,013,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00858.warc.gz | 898,133,713 | 61,671 | Home » How to Create and Interpret a ROC Curve in Stata
# How to Create and Interpret a ROC Curve in Stata
Logistic Regression is a statistical method that we use to fit a regression model when the response variable is binary. To assess how well a logistic regression model fits a dataset, we can look at the following two metrics:
• Sensitivity: the probability that the model predicts a positive outcome for an observation when indeed the outcome is positive.
• Specificity: the probability that the model predicts a negative outcome for an observation when indeed the outcome is negative.
One easy way to visualize these two metrics is by creating a ROC curve, which is a plot that displays the sensitivity and specificity of a logistic regression model.
This tutorial explains how to create and interpret a ROC curve in Stata.
## Example: ROC Curve in Stata
For this example we will use a dataset called lbw, which contains the folllowing variables for 189 mothers:
• low – whether or not the baby had a low birthweight. 1 = yes, 0 = no.
• age – age of the mother.
• smoke – whether or not the mother smoked during pregnancy. 1 = yes, 0 = no.
We will fit a logistic regression model to the data using age and smoking as explanatory variables and low birthweight as the response variable. Then we will create a ROC curve to analyze how well the model fits the data.
Step 1: Load and view the data.
Load the data using the following command:
use http://www.stata-press.com/data/r13/lbw
Gain a quick understanding of the dataset using the following command:
summarize
There are 11 different variables in the dataset, but the only three that we care about are low, age, and smoke.
Step 2: Fit the logistic regression model.
Use the following command to fit the logistic regression model:
logit low age smoke
Step 3: Create the ROC curve.
We can create the ROC curve for the model using the following command:
lroc
Step 4: Interpret the ROC curve.
When we fit a logistic regression model, it can be used to calculate the probability that a given observation has a positive outcome, based on the values of the predictor variables.
To determine if an observation should be classified as positive, we can choose a cut-point such that observations with a fitted probability above the cut-point are classified as positive and any observations with a fitted probability below the cut-point are classified as negative.
For example, suppose we choose the cut-point to be 0.5. This means that any observation with a fitted probability greater than 0.5 will be predicted to have a positive outcome, while any observation with a fitted probability less than or equal to 0.5 will be predicted to have a negative outcome.
The ROC curve shows us the values of sensitivity vs. 1-specificity as the value of the cut-off point moves from 0 to 1.  A model with high sensitivity and high specificity will have a ROC curve that hugs the top left corner of the plot. A model with low sensitivity and low specificity will have a curve that is close to the 45-degree diagonal line.
The AUC (area under curve) gives us an idea of how well the model is able to distinguish between positive and negative outcomes. The AUC can range from 0 to 1. The higher the AUC, the better the model is at correctly classifying outcomes. In our example, we can see that the AUC is 0.6111.
We can use AUC to compare the performance of two or more models. The model with the higher AUC is the one that performs best. | 799 | 3,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-30 | latest | en | 0.866565 |
https://gist.github.com/lgalke/febaaa1313d9c11f3bc8240defed8390 | 1,717,023,108,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059408.76/warc/CC-MAIN-20240529200239-20240529230239-00479.warc.gz | 226,641,372 | 25,299 | Instantly share code, notes, and snippets.
# lgalke/all_but_the_top.py
Last active February 22, 2022 23:46
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Word Embedding Postprocessing: All but the top
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""" All-but-the-Top: Simple and Effective Postprocessing for Word Representations Paper: https://arxiv.org/abs/1702.01417 Last Updated: Fri 15 Nov 2019 11:47:00 AM CET **Prior version had serious issues, please excuse any inconveniences.** """ import numpy as np from sklearn.decomposition import PCA def all_but_the_top(v, D): """ Arguments: :v: word vectors of shape (n_words, n_dimensions) :D: number of principal components to subtract """ # 1. Subtract mean vector v_tilde = v - np.mean(v, axis=0) # 2. Compute the first `D` principal components # on centered embedding vectors u = PCA(n_components=D).fit(v_tilde).components_ # [D, emb_size] # Subtract first `D` principal components # [vocab_size, emb_size] @ [emb_size, D] @ [D, emb_size] -> [vocab_size, emb_size] return v_tilde - (v @ u.T @ u)
### liaocs2008 commented Jul 19, 2018
I think line 24 is not as in the paper. it should be like "u[:i].dot(v[w])*u[:,i].T" rather than the element-wise multiplication.
### aerinkim commented Aug 12, 2018
Hey this implementation is wrong. You shouldn't use fit_transform. You need U, singular vector itelf.
### gojomo commented Nov 14, 2019
Using this all-but-the-top transformation didn't result in the expected improvement on a word-vector evaluation. (Specifically, `questions-words.txt` analogies-correct on 'GoogleNews' top-200k words.) In fact, it drove accuracy down from 75.37% to 59.45%.
Using instead the all-but-the-top implemenation at https://github.com/s1998/All-but-the-top/blob/e0c7d758b495ad55868d9a14ecd31df86b79e4d3/src/embeddings_processor.py#L4 slightly improved accuracy, as would be expected from the paper's claims, to 75.79%.
So: more evidence this implementation is off.
### lgalke commented Nov 14, 2019
@gojomo thanks for pointing that out! I will try to fix it as soon as possible. Please excuse the confusion.
### lgalke commented Nov 14, 2019 • edited
I adapted the code to match the version of this implementation. Changes are also ported back in vec4ir.
### gojomo commented Nov 15, 2019 • edited
Thanks - trying this version (that ends `return v - (v_tilde @ u.T @ u)`) has the expected behavior in my evaluations!
### lgalke commented Nov 15, 2019 • edited
@gojomo I just double checked with the paper itself and it should be `v_tilde - (v @ u.T @ u)` instead of `v - (v_tilde @...`.
Applying PCA to centered / non-centered versions should not make a difference. The important thing is to subtract the mean from the embeddings to match the paper. `v_tilde` holds the centered version in this gist so now we return the `embeddings - mean - first D principal components`.
@s1998: I think this last point is also not considered in your implementation. In line 11 you do not subtract from the centered version. Am I missing something?
### s1998 commented Nov 18, 2019
@lgalke You are correct, it should be mean centered embeddings that I need to subtract from. I'll fix that, thanks. | 913 | 3,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-22 | latest | en | 0.806477 |
https://www.coursehero.com/file/5791553/11-Integrated-Rate-Laws/ | 1,519,291,642,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814079.59/warc/CC-MAIN-20180222081525-20180222101525-00071.warc.gz | 863,378,631 | 35,072 | {[ promptMessage ]}
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11 Integrated Rate Laws
# 11 Integrated Rate Laws - Ozone reacts with ethylene to...
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1 1 Ozone reacts with ethylene to produce formaldehyde. 2 O 3 (g) + 2 C 2 H 4 (g) 4 CH 2 O(g) + O 2 (g) Use the initial reaction rate data to determine the order of the reaction with respect to ozone, x, and ethylene, y. Report the overall order, x + y. Rate = k [O 3 ] x [C 2 H 4 ] y A) 2 B) 3 C) 4 4.0 2.0 1.0 3.0 1.0 1.5 1.0 1.0 0.5 Initial Rate 10 -12 mol L -1 s -1 [C 2 H 4 ] 10 -8 mol L -1 [O 3 ] 10 -7 mol L -1 2 Today’s To do Determine the order of a reaction from an appropriate graph of concentration vs time. Determine the rate constant of a reaction from an appropriate graph of concentration vs time. Use integrated rate laws to determine the concentration of reactants or products at any time. Use integrated rate laws to determine rate constants from concentration data. Identify unimolecular and bimolecular elementary reactions. Write the rate laws for elementary reactions.
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2 3 Integrated Rate Laws The rate law gives the rate at which reactant and product concentrations are changing. The integrated rate law gives a reactant concentration as a function of time. Integrated laws are obtained from rate laws by integration, i.e. using Calculus! 4 Three Cases for a Single Reactant, A Zero order with respect to A First order with respect to A Second order with respect to A Also applies to a swamping experiment: A+ B with concentration of B very high so it doesn’t change significantly.
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11 Integrated Rate Laws - Ozone reacts with ethylene to...
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Ask a homework question - tutors are online | 532 | 2,032 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-09 | latest | en | 0.861361 |
https://testbook.com/question-answer/a-three-phase-induction-motor-is-running-at-4-sli--608648ede9bb8d77106bc30a | 1,632,388,832,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00169.warc.gz | 599,424,563 | 30,896 | # A three-phase induction motor is running at 4% slip. If the input to the rotor is 1,000 W, then the mechanical power developed by the motor will be _______.
This question was previously asked in
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1. 960 W
2. 9,600 W
3. 96 W
4. 0.96 W
## Answer (Detailed Solution Below)
Option 1 : 960 W
Free
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44199
10 Questions 10 Marks 10 Mins
## Detailed Solution
Power Stages in an Induction Motor:
Stator iron loss (eddy current loss and hysteresis losses) considered as constant loss and it depends on the supply frequency and magnetic flux density in the iron core.
The iron loss of the rotor is negligible because the frequency of rotor currents under normal running conditions is always small.
$${{P}_{2}}:{{P}_{cu}}:{{P}_{m}}=1:s:\left( 1-s \right)$$
Where,
s is the slip of motor and Pcu is rotor copper loss.
Calculation:
Given,
s = 4% = 0.04
P2 = 1000 W
From above concept,
P2 : Pm = 1 : (1 - s)
1000 : Pm = 1 : (1 - 0.04)
1000 : Pm = 1 : 0.96
$$\frac{1000}{P_m}=\frac{1}{0.96}$$
Pm = 1000 × 0.96 = 960 W | 373 | 1,159 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.40625 | 3 | CC-MAIN-2021-39 | latest | en | 0.828973 |
https://www.physicsforums.com/threads/ferris-wheel-magnitude-and-period.213150/ | 1,723,036,727,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694449.36/warc/CC-MAIN-20240807111957-20240807141957-00563.warc.gz | 723,006,783 | 17,566 | # Ferris wheel magnitude and period
• AnkhUNC
In summary, the magnitude of a Ferris wheel refers to its overall size or height and can vary up to 541 feet. It is typically measured in feet or meters and can be calculated by the distance between the ground and the highest point or the diameter of the wheel. The period of a Ferris wheel is the time it takes to make one full rotation and can range from a few minutes to over 30 minutes. Its calculation is based on the circumference and speed of the wheel. The magnitude of a Ferris wheel can affect its period, with larger wheels generally having a longer period due to their size, but other factors such as design and speed can also play a role.
AnkhUNC
## Homework Statement
A woman rides a carnival Ferris wheel at radius 21 m, completing 4.2 turns about its horizontal axis every minute. What are (a) the period of the motion, and the magnitude of her centripetal acceleration at (b) the highest point and (c) the lowest point?
## The Attempt at a Solution
How should I go about solving this one? Also are there any sites that can explain 2d and 3d motion better? My book is a little lacking.
So I have t= 14.285 but I really am unsure of how to find the max/min.
I would approach this problem by first identifying the relevant equations and principles related to circular motion. The period of the motion can be calculated using the formula T = 2πr/v, where r is the radius of the Ferris wheel and v is the linear velocity. In this case, r = 21 m and v = (4.2 turns/minute) * (2π rad/turn) * (21 m) = 88.2 m/s. Plugging these values into the formula, we get T = 2π(21 m)/88.2 m/s = 0.477 minutes.
To calculate the centripetal acceleration, we can use the formula a = v^2/r. At the highest point, the linear velocity is zero, so the centripetal acceleration is also zero. At the lowest point, the linear velocity is at its maximum, which we calculated to be 88.2 m/s. Plugging this into the formula, we get a = (88.2 m/s)^2/21 m = 370.5 m/s^2.
As for resources to better understand 2D and 3D motion, I would recommend checking out educational websites such as Khan Academy or Physics Classroom. You can also find helpful explanations and examples in physics textbooks or online lecture notes. It may also be helpful to practice solving similar problems to solidify your understanding of the concepts.
## 1. What is the magnitude of a Ferris wheel?
The magnitude of a Ferris wheel refers to its overall size or height. This can vary depending on the specific Ferris wheel, but some of the largest ones can reach up to 541 feet in height.
## 2. How is the magnitude of a Ferris wheel measured?
The magnitude of a Ferris wheel is typically measured in feet or meters. It can be measured by the distance between the ground and the highest point of the Ferris wheel, or by the diameter of the wheel itself.
## 3. What is the period of a Ferris wheel?
The period of a Ferris wheel is the amount of time it takes for the wheel to make one full rotation. This can range from a few minutes to over 30 minutes depending on the size and design of the Ferris wheel.
## 4. How is the period of a Ferris wheel calculated?
The period of a Ferris wheel can be calculated by dividing the circumference of the wheel by the speed at which it rotates. This formula is period = circumference / speed.
## 5. Can the magnitude of a Ferris wheel affect its period?
Yes, the magnitude of a Ferris wheel can affect its period. Generally, the larger the Ferris wheel, the longer the period will be, as it takes more time for a larger wheel to complete one full rotation. However, other factors such as design and speed can also impact the period of a Ferris wheel.
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6K | 1,048 | 4,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-33 | latest | en | 0.944844 |
https://atcoder.jp/contests/agc037/tasks/agc037_e | 1,638,265,908,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358966.62/warc/CC-MAIN-20211130080511-20211130110511-00099.warc.gz | 179,067,127 | 5,807 | Contest Duration: - (local time) (150 minutes) Back to Home
E - Reversing and Concatenating /
Time Limit: 2 sec / Memory Limit: 1024 MB
### 問題文
• S を反転した文字列を T として、ST をこの順に連結して得られる文字列を U とする。
• ある U の連続する長さ N の部分文字列を S' として、SS' で置き換える。
• 1 ≦ N ≦ 5000
• 1 ≦ K ≦ 10^9
• |S|=N
• S は英小文字からなる
### 入力
N K
S
### 入力例 1
5 1
bacba
### 出力例 1
aabca
S=bacbaのとき、T=abcab, U=bacbaabcabであるので S'=aabcaとするのが最適です。
### 入力例 2
10 2
bbaabbbaab
### 出力例 2
aaaabbaabb
Score : 1300 points
### Problem Statement
Takahashi has a string S of length N consisting of lowercase English letters. On this string, he will perform the following operation K times:
• Let T be the string obtained by reversing S, and U be the string obtained by concatenating S and T in this order.
• Let S' be some contiguous substring of U with length N, and replace S with S'.
Among the strings that can be the string S after the K operations, find the lexicographically smallest possible one.
### Constraints
• 1 \leq N \leq 5000
• 1 \leq K \leq 10^9
• |S|=N
• S consists of lowercase English letters.
### Input
Input is given from Standard Input in the following format:
N K
S
### Output
Print the lexicographically smallest possible string that can be the string S after the K operations.
### Sample Input 1
5 1
bacba
### Sample Output 1
aabca
When S=bacba, T=abcab, U=bacbaabcab, and the optimal choice of S' is S'=aabca.
### Sample Input 2
10 2
bbaabbbaab
### Sample Output 2
aaaabbaabb | 516 | 1,482 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-49 | latest | en | 0.329513 |
https://iris.unirc.it/handle/20.500.12318/5706 | 1,726,017,612,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651323.1/warc/CC-MAIN-20240910224659-20240911014659-00840.warc.gz | 296,652,487 | 11,820 | In this paper, analyzing the problems related to the safety and to the comfort in railway train transport, we develop a particular mathematical model in order to interpret systematically the interactions between the circulatory phenomenon (binomial ”man-vehicle”), the different components of the railway net (branch, knot, plant, etc.) and environment of pertinence. Preliminary we build the set of the "satisfaction’s states” for the standards of quality for the trip from the consumer and after individualizing a particular subset of the "states disconfortless”, through opportune mathematical developments, we represent geometrically the problem with an iper-space; in this iper-space we are resolution to represent the elements of the system man-vehicle-infrastructure-environment that characterize the states of comfort during the trip.
Un approccio matematico per l'analisi integrata della sicurezza e qualita globale d'esercizio della rete ferroviaria / A., Bevilacqua; LO BOSCO, Dario. - In: ATTI DELLA ACCADEMIA DELLE SCIENZE DI TORINO. CLASSE DI SCIENZE FISICHE MATEMATICHE E NATURALI. - ISSN 0001-4419. - (2010).
### Un approccio matematico per l'analisi integrata della sicurezza e qualita globale d'esercizio della rete ferroviaria
#### Abstract
In this paper, analyzing the problems related to the safety and to the comfort in railway train transport, we develop a particular mathematical model in order to interpret systematically the interactions between the circulatory phenomenon (binomial ”man-vehicle”), the different components of the railway net (branch, knot, plant, etc.) and environment of pertinence. Preliminary we build the set of the "satisfaction’s states” for the standards of quality for the trip from the consumer and after individualizing a particular subset of the "states disconfortless”, through opportune mathematical developments, we represent geometrically the problem with an iper-space; in this iper-space we are resolution to represent the elements of the system man-vehicle-infrastructure-environment that characterize the states of comfort during the trip.
##### Scheda breve Scheda completa Scheda completa (DC)
2010
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Utilizza questo identificativo per citare o creare un link a questo documento: `https://hdl.handle.net/20.500.12318/5706`
• ND
• ND
• ND | 572 | 2,485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-38 | latest | en | 0.777274 |
https://www.projecteuclid.org/journals/annals-of-probability/volume-40/issue-4/Tight-Markov-chains-and-random-compositions/10.1214/11-AOP656.full | 1,726,840,641,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652278.82/warc/CC-MAIN-20240920122604-20240920152604-00003.warc.gz | 870,334,120 | 28,695 | July 2012 Tight Markov chains and random compositions
Boris Pittel
Ann. Probab. 40(4): 1535-1576 (July 2012). DOI: 10.1214/11-AOP656
## Abstract
For an ergodic Markov chain $\{X(t)\}$ on $\mathbb{N}$, with a stationary distribution $\pi$, let $T_{n}>0$ denote a hitting time for $[n]^{c}$, and let $X_{n}=X(T_{n})$. Around 2005 Guy Louchard popularized a conjecture that, for $n\to\infty$, $T_{n}$ is almost $\operatorname{Geometric}(p)$, $p=\pi([n]^{c})$, $X_{n}$ is almost stationarily distributed on $[n]^{c}$ and that $X_{n}$ and $T_{n}$ are almost independent, if $p(n):=\sup_{i}p(i,[n]^{c})\to0$ exponentially fast. For the chains with $p(n)\to0$, however slowly, and with $\sup_{i,j}\|p(i,\cdot)-p(j,\cdot)\|_{\mathrm{TV}}<1$, we show that Louchard’s conjecture is indeed true, even for the hits of an arbitrary $S_{n}\subset\mathbb{N}$ with $\pi(S_{n})\to0$. More precisely, a sequence of $k$ consecutive hit locations paired with the time elapsed since a previous hit (for the first hit, since the starting moment) is approximated, within a total variation distance of order $k\sup_{i}p(i,S_{n})$, by a $k$-long sequence of independent copies of $(\ell_{n},t_{n})$, where $t_{n}=\operatorname{Geometric}(\pi(S_{n}))$, $\ell_{n}$ is distributed stationarily on $S_{n}$ and $\ell_{n}$ is independent of $t_{n}$. The two conditions are easily met by the Markov chains that arose in Louchard’s studies as likely sharp approximations of two random compositions of a large integer $\nu$, a column-convex animal (cca) composition and a Carlitz (C) composition. We show that this approximation is indeed very sharp for each of the random compositions, read from left to right, for as long as the sum of the remaining parts stays above $\ln^{2}\nu$. Combining the two approximations, a composition—by its chain, and, for $S_{n}=[n]^{c}$, the sequence of hit locations paired each with a time elapsed from the previous hit—by the independent copies of $(\ell_{n},t_{n})$, enables us to determine the limiting distributions of $\mu=o(\ln\nu)$ and $\mu=o(\nu^{1/2})$ largest parts of the random cca-composition and the random C-composition, respectively. (Submitted to Annals of Probability in June 2009.)
## Citation
Boris Pittel. "Tight Markov chains and random compositions." Ann. Probab. 40 (4) 1535 - 1576, July 2012. https://doi.org/10.1214/11-AOP656
## Information
Published: July 2012
First available in Project Euclid: 4 July 2012
zbMATH: 1257.05009
MathSciNet: MR2978132
Digital Object Identifier: 10.1214/11-AOP656
Subjects:
Primary: 05A15 , 05A17 , 11P99 , 60C05 , 60F05 , 60J05
Keywords: exponentiality , Extreme values , limit theorems , Markov chains , Random compositions , rarity | 830 | 2,702 | {"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} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.819573 |
https://intellipaat.com/community/24077/strange-behavior-in-a-function-while-implementing-the-alpha-beta-pruning-algorithm?show=24090 | 1,620,340,423,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988763.83/warc/CC-MAIN-20210506205251-20210506235251-00618.warc.gz | 344,368,108 | 12,548 | # Strange behavior in a function while implementing the alpha-beta pruning algorithm
1 view
I've implemented a minimax algorithm with alpha-beta pruning. To get the best move, I call the alpha-beta algorithm with the rootAlphaBeta function. However, in the rootAlphaBeta function, I spotted some very strange behavior. When I call the rootAlphaBeta function with a ply of 4, it makes about 20 000 calls, but when I call the alphaBeta function directly, it makes only about 2000 calls. I can't seem to find what's the problem, as the number of calls should be the same.
The move that both algorithms eventually find should be the same, right? I think so, at least the score of the move is the same, I have no way of knowing the move the alphaBeta chooses when I call it directly without rootAlphaBeta.
def alphaBeta(self, board, rules, alpha, beta, ply, player):
"""Implements a minimax algorithm with alpha-beta pruning."""
if ply == 0:
return self.positionEvaluation(board, rules, player)
move_list = board.generateMoves(rules, player)
for move in move_list:
board.makeMove(move, player)
current_eval = -self.alphaBeta(board, rules, -beta, -alpha, ply - 1,
board.getOtherPlayer(player))
board.unmakeMove(move, player)
if current_eval >= beta:
return beta
if current_eval > alpha:
alpha = current_eval
return alpha
def rootAlphaBeta(self, board, rules, ply, player):
"""Makes a call to the alphaBeta function. Returns the optimal move for a
player at given ply."""
best_move = None
max_eval = float('-infinity')
move_list = board.generateMoves(rules, player)
for move in move_list:
board.makeMove(move, player)
current_eval = -self.alphaBeta(board, rules, float('-infinity'),
float('infinity'), ply - 1,
board.getOtherPlayer(player))
board.unmakeMove(move, player)
if current_eval > max_eval:
max_eval = current_eval
best_move = move
return best_move
by (108k points)
The use of alpha-beta pruning in the minimax search reduces by a large factor the number of bottom positions that need to be examined, typically by several orders of magnitude in many game-playing programs.
In your program, the rootAlphaBeta function is not been updating the alpha value.
The function calls all its child nodes with the full range of (-inf, inf) when it could have narrowed down the range for all child nodes except the first. This will prevent the pruning of some branches which will have no effect on the final score, and increase the node count. | 557 | 2,471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-21 | latest | en | 0.788724 |
http://www.baselinemag.com/c/a/Tools-Calculators-hold/Paying-the-Electric-Bill | 1,571,744,077,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987817685.87/warc/CC-MAIN-20191022104415-20191022131915-00122.warc.gz | 214,527,316 | 21,795 | <img alt="dcsimg" id="dcsimg" width="1" height="1" src="//www.qsstats.com/dcs8krshw00000cpvecvkz0uc_4g4q/njs.gif?dcsuri=/index.php/c/a/Tools-Calculators-hold/Paying-the-Electric-Bill&WT.js=No&WT.tv=10.4.1&dcssip=www.baselinemag.com&WT.qs_dlk=Xa7pTF4L7wjJmiRWSugrcgAAAA0&">
# Paying the Electric Bill
| Posted 2006-03-06 Print
## How to figure out the true cost to power and cool your servers.
Information-technology managers usually focus on software licenses and maintenance. What about electricity for power, emergency backup and cooling? Consumption per unit of computing power is dropping but computing power has increased. In many cases, electricity exceeds the cost of the computer over three years—about \$300 per computer each year.
Once electricity costs are calculated, I.T. staffs can take steps to reduce power usage, typically by shutting down computers and lights in off-peak periods. Computer systems, cooling and backup must be installed for peak demand, so conserving energy off-peak doesn't affect capital costs.
Also, when computing costs, technology staffs often just add up power ratings. Electricity demand for computers and related equipment is usually specified in volt-amperes (VA). There had been a significant conversion factor, with 1 VA equal to about 0.7 watt. But today's redefined VA designation can almost always be freely interchanged
1 to 1 with wattage and directly converted to the way we pay for electricity, which is per kilowatt-hours used—in our tool, 150 VA becomes 0.15 kVA, for example.
Another way to spec: Neil Rasmussen, co-founder of American Power Conversion Corp. (APC), says the "power density" of data centers was originally estimated at 40 to 80 watts per square foot. Modern equipment uses more power—600 to 1,000 watts per square foot is common, and blade racks (7 square feet of floor space each) might use 1,500 to 20,000 watts, or as much as 3 kilowatts per square foot at the high end.
Power densities of even 500 watts per square foot mean extra engineering expense; typical office buildings aren't designed for it. Vendors tend to provide a power density per rack or per floor area taken up by the racks themselves. I.T. staffs add floor area for access—multiplying rack space by four, thus reducing power density to one-quarter the equipment vendors' totals.
Another variable: Running more than three hours while cut off from the grid requires a generator. A 10 kilowatt system is \$20,000 or more. But battery backup has a high continuing cost; battery and inverter will boost power consumption by 20% to 50%.
Have the latest technology news and resources emailed to you everyday. | 648 | 2,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-43 | latest | en | 0.888249 |
https://www.effortlessmath.com/math-topics/free-ged-math-practice-test/ | 1,721,052,391,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00854.warc.gz | 657,221,436 | 16,396 | GED Math Practice Test & Sample [Updated for 2024]
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20 Sample GED Math Practice Questions
1- Mr. Jones saves $2,500 out of his monthly family income of$55,000. What fractional part of his income does he save?
A. $$\frac{1}{22}$$
B. $$\frac{1}{11}$$
C. $$\frac{3}{25}$$
D. $$\frac{2}{15}$$
2- Four one-foot rulers can be split among how many users to leave each with $$\frac{1}{6}$$ of a ruler?
A. 4
B. 6
C. 12
D. 24
3- Simplify the expression.
$$(6x^3-8x^2+2x^4 )-(4x^2-2x^4+2x^3 )$$=?
A. $$4x^3-12x^2$$
B. $$4x^4+4x^3-12x^2$$
C. $$8x^3-12x^2$$
D. $$x^4+4x^3-12x^2$$
4- In two successive years, the population of a town is increased by $$15\%$$ and $$20\%$$. What percent of the population is increased after two years?
A. $$32\%$$
B. $$35\%$$
C. $$38\%$$
D. $$68\%$$
5- What is the surface area of the cylinder below?
A. $$48 {\pi} \space in^2$$
B. $$57 {\pi} \space in^2$$
C. $$66 {\pi} \space in^2$$
D. $$288 {\pi} \space in^2$$
6- A cruise line ship left Port A and traveled 80 miles due west and then 150 miles due north. At this point, what is the shortest distance from the cruise to port A in miles? ____________
7- A shirt costing $200 is discounted $$15\%$$. After a month, the shirt is discounted another $$15\%$$. Which of the following expressions can be used to find the selling price of the shirt? A. $$(200) (0.70)$$ B. $$(200) – 200 (0.30)$$ C. $$(200) (0.15) – (200) (0.15)$$ D. $$(200) (0.85) (0.85)$$ 8- Solve: $$5 + 8 \times (- 2) – [4 + 22 \times 5] \div 6 =$$? A. $$-30$$ B. $$-20$$ C. $$-10$$ D. 0 9- Which of the following points lies on the line $$x+2y=4$$? A. $$(-2, 3)$$ B. $$(1, 2)$$ C. $$(-1, 3)$$ D. $$(-3, 4)$$ 10- 5 less than twice a positive integer is 83. What is the integer? A. 39 B. 41 C. 42 D. 44 11- 11 yards 6 feet and 4 inches equal to how many inches? A. 388 B. 468 C. 472 D. 476 12- Mr. Carlos’s family is choosing a menu for their reception. They have 3 choices of appetizers, 5 choices of entrees, 4 choices of cake. How many different menu combinations are possible for them to choose from? A. 12 B. 32 C. 60 D. 120 13- The average of five consecutive numbers is 38. What is the smallest number? A. 38 B. 36 C. 34 D. 12 14- What is the difference between the smallest 4–digit number and the biggest 4–digit number? A. 6666 B. 6789 C. 8888 D. 8999 15- How many tiles of 8 $$cm^2$$ is needed to cover a floor of dimension 6 cm by 24 cm? A. 6 B. 12 C. 18 D. 24 16- A ladder leans against a wall forming a $$60^ \circ$$ angle between the ground and the ladder. If the bottom of the ladder is 30 feet away from the wall, how long is the ladder? A. 30 feet B. 40 feet C. 50 feet D. 60 feet 17- The average weight of 18 girls in a class is 60 kg and the average weight of 32 boys in the same class is 62 kg. What is the average weight of all the 50 students in that class? A. 60 B. 61.28 C. 61.68 D. 62.90 18- An angle is equal to one-fifth of its supplement. What is the measure of that angle? A. 20 B. 30 C. 45 D. 60 19- In a stadium the ratio of home fans to visiting fans in a crowd is 5:7. Which of the following could be the total number of fans in the stadium? A. 12,324 B. 42,326 C. 44,566 D. 66,812 20- If $$40\%$$ of a class are girls, and $$25\%$$ of girls play tennis, what percent of the class play tennis? A. $$10\%$$ B. $$15\%$$ C. $$20\%$$ D. $$40\%$$ Best GED Math Prep Resource for 2022 Original price was:$76.99.Current price is: $36.99. Satisfied 132 Students Answers: 1- A 2,500 out of 55,000 equals to $$\frac{2500}{55000}= \frac{25}{550}= \frac{1}{22}$$ 2- D $$4 \div \frac{1}{6} = 24$$ 3- B Simplify and combine like terms. $$(6x^3-8x^2+2x^4 )-(4x^2-2x^4+2x^3 ) {\Rightarrow} (6x^3-8x^2+2x^4 )-4x^2+2x^4-2x^3 {\Rightarrow} 4x^4+4x^3-12x^2$$ 4- C the population is increased by $$15\%$$ and $$20\%$$. $$15\%$$ increase changes the population to $$115\%$$ of original population. For the second increase, multiply the result by $$120\%$$. $$(1.15) \times (1.20) = 1.38 = 138\%$$ 38 percent of the population is increased after two years. 5- C Surface Area of a cylinder $$= 2\pi (r + h)$$, The radius of the cylinder is $$3(6 \div 2)$$ inches and its height is 8 inches. Therefore, Surface Area of a cylinder $$= 2\pi (3) (3 + 8) = 66\pi$$ 6- 170 Use the information provided in the question to draw the shape. Use Pythagorean Theorem: $$a^2 + b^2 = c^2$$ $$80^2 + 150^2 = c^2 {\Rightarrow} 6400 + 22500 = c^2 {\Rightarrow} 28900 = c^2 {\Rightarrow} c = 170$$. 7- D To find the discount, multiply the number by $$(100\% –$$ rate of discount). Therefore, for the first discount we get: $$(200) (100\% – 15\%) = (200) (0.85) = 170$$ For the next $$15\%$$ discount: (200) (0.85) (0.85) 8- A Use PEMDAS (order of operation): $$5 + 8 \times (-2) – [4 + 22 \times 5] \div 6 = 5 + 8 \times (-2) – [4 + 110] \div 6 = 5 + 8 \times (-2) – [114] \div 6 = 5 + (-16) – 19$$ $$5 + (-16) – 19 = -11 – 19 = -30$$ 9- A $$x+2y=4$$. Plug in the values of $$x$$ and $$y$$ from choices provided. Then: A. $$(-2,3) x+2y=4(\rightarrow )-2+2(3)=4(\rightarrow )-2+6=4$$ This is true! B. $$(1,2) x+2y=4(\rightarrow )1+2(2)=4(\rightarrow )1+4=4$$ (This is NOT true!) C. $$(-1,3) x+2y=4(\rightarrow )-1+2(3)=4(\rightarrow )-1+6=4$$ (This is NOT true!) D.$$(-3,4) x+2y=4(\rightarrow )-3+2(4)=4(\rightarrow )-3+8=4$$(This is NOT true!) 10- D Let $$x$$ be the integer. Then: $$2x – 5 = 83$$ Add 5 both sides: $$2x = 88$$ Divide both sides by$$2: x = 44$$ 11- C $$11 \times 36 + 6 \times 12 + 4 = 472$$ 12- C To find the number of possible outfit combinations, multiply number of options for each factor: 3 $$\times$$ 5 $$\times$$ 4 $$=$$ 60 13- B Let $$x$$ be the smallest number. Then, these are the numbers: $$x, x+1, x+2, x+3, x+4$$ $$average = \frac{sum of terms}{number of terms} \Rightarrow 38 = \frac{x+(x+1)+(x+2)+(x+3)+(x+4)}{5} \Rightarrow 38= \frac{5x+10}{5} \Rightarrow 190 = 5x+10 \Rightarrow 180 = 5x \Rightarrow x=36$$ 14- D Smallest 4–digit number is 1000, and biggest 4–digit number is 9999. The difference is: 8999 15- C The area of the floor is: $$6 cm \times 24 cm = 144 cm^2$$ The number of tiles needed$$= 144 \div 8 = 18$$ 16- D The relationship among all sides of special right triangle $$30^ \circ -60^ \circ – 90^ \circ$$ is provided in this triangle: In this triangle, the opposite side of $$30^ \circ$$ angle is half of the hypotenuse. Draw the shape for this question: The latter is the hypotenuse. Therefore, the latter is 60 ft. 17- B $$average = \frac{sum of terms}{number of terms}$$ The sum of the weight of all girls is: $$18 \times 60 = 1080$$ kg The sum of the weight of all boys is: $$32 \times 62 = 1984$$ kg The sum of the weight of all students is: $$1080 + 1984 = 3064$$ kg $$average = \frac{3064}{50}=61.28$$ 18- B The sum of supplement angles is 180. Let $$x$$ be that angle. Therefore, $$x + 5x = 180$$ $$6x = 180$$, divide both sides by $$6: x = 30$$ 19- A In the stadium the ratio of home fans to visiting fans in a crowd is 5:7. Therefore, total number of fans must be divisible by $$12: 5 + 7 = 12$$. Let’s review the choices: A. 12,324: $$12,324 \div 12 = 1027$$ B. 42,326: $$42,326 \div 12 = 3,527.166$$ C. 44,566: $$44,566 \div 12 = 3,713.833$$ D. 66,812: $$66,812 \div 12 = 5,567.666$$ Only choice A when divided by 12 results a whole number. 20- A The percent of girls playing tennis is: $$40 \% \times 25\% = 0.40 \times 0.25 = 0.10 = 10\%$$ Looking for the best resource to help you succeed on the GED Math test? High School Equivalency Tests The Best Booksto Ace the GED Math Test Original price was:$22.99.Current price is: $14.99. Satisfied 94 Students More from Effortless Math for GED Test … Want to try another GED Math practice test? Try our Full-Length GED Math Practice Test. It’s the best way to measure your exam readiness. Looking for the most common GED Math formulas? Here is a complete list of GED Math formulas. Want an in-depth guide to getting ready to take the GED Math test? Read our GED Math – Test Day Tips to learn more about what to do on your test day. The Perfect Prep Booksfor the GED Math Test Original price was:$25.99.Current price is: $13.99. Satisfied 125 Students Original price was:$24.99.Current price is: $14.99. Satisfied 207 Students Original price was:$27.99.Current price is: $14.99. Satisfied 182 Students Have any questions about the GED Test? Write your questions about the GED or any other topics below and we’ll reply! Related to This Article More math articles What people say about "GED Math Practice Test & Sample [Updated for 2024] - Effortless Math: We Help Students Learn to LOVE Mathematics"? No one replied yet. X 49% OFF Limited time only! Save Over 49% SAVE$34
It was $69.99 now it is$35.99 | 3,384 | 9,830 | {"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} | 4.09375 | 4 | CC-MAIN-2024-30 | latest | en | 0.884512 |
http://mathematica.stackexchange.com/questions/8961/how-to-work-out-the-parameter-in-a-definite-integration-which-has-an-exact-value?answertab=votes | 1,462,438,220,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860126377.4/warc/CC-MAIN-20160428161526-00183-ip-10-239-7-51.ec2.internal.warc.gz | 175,296,532 | 18,700 | How to work out the parameter in a definite integration which has an exact value while the integration doesn't have an analytical solution?
Here is the equation I'm trying to solve:
NIntegrate[1/(E^(1/(λ T)) - 1), {λ, 200, 220}] == 1000
T is the parameter I'm trying to work out. I tried to solve it with InverseFunction[] but it seemed not to work:
te[ii_] := InverseFunction[NIntegrate[1/(-1 + E^(1/(# λ))), {λ, 200, 220}] &][ii]
Evaluate[te[1000]]
What's wrong with my solution? How to correct it?
-
Here's a slight improvement of b.gatessucks's answer, adding a (mostly effective) initial guess:
te[ii_?NumericQ, opts___] := (\[FormalCapitalT] /. First@FindRoot[
NIntegrate[1/(E^(1/(λ \[FormalCapitalT])) - 1), {λ, 200, 220}] == ii,
{\[FormalCapitalT], (0.5437727672315316 + ii (0.5440978395984463 +
ii (0.12244298721801757 + (0.009523809523809525 +
0.0002380952380952381 ii) ii)))/(476.1910924718744 +
ii (247.61910924718742 + ii (30. + ii)))},
Evaluated -> False, opts])
Here's how I obtained the initial guess:
PadeApproximant[InverseSeries[
Integrate[#, {λ, 200, 220}] & /@
Normal[Series[1/(E^(1/(λ ii)) - 1), {ii, Infinity, 7}]] + O[ii, Infinity]^8],
{ii, Infinity, 4}] // N // FullSimplify
-
Wow, what a advanced solution…Er, I wonder where I can find the introduction of Evaluated, I can't find it in the help, and, it seems that PadeApproximant here isn't necessary, why you choose it? – xzczd Aug 1 '12 at 11:42
The Evaluated option is indeed undocumented, but it has been discussed a lot on this site; search around. As for PadeApproximant[], it's for generating a rational function approximation; they tend to be more accurate than truncations of Taylor polynomials for the same amount of arithmetic operations. – J. M. Aug 1 '12 at 12:05
For example this? OK, it's very interesting, now…I'm sorry! But I still can't understand the independent variable opts___ which I thought I would make it out after I knew the meaning of Evaluated… – xzczd Aug 1 '12 at 13:07
The opts is for the case where you need to pass options to FindRoot[]. Now that I think about it, I should have also made provisions for passing options to NIntegrate[]... maybe later. – J. M. Aug 1 '12 at 13:19
So, here I can just remove it? (Trying…) ………er, it seems that the ?NumericQ and Evaluated(whether it's True or False) are both unnecessary now… – xzczd Aug 1 '12 at 13:36
You can use FindRoot :
func[T_?NumericQ] := NIntegrate[1/(E^(1/(\[Lambda] T)) - 1), {\[Lambda], 200, 220}]
sol = FindRoot[func[T] == 1000, {T, 0.1}]
(* {T -> 0.240468} *)
func[sol[[1, 2]]]
(* 1000. *)
FindRoot needs an initial guess for the unknown; you can get a rough estimate by plotting func for instance.
-
Maybe for this case the initial guess isn't that important since the te[ii] is actually monotonous? – xzczd Aug 1 '12 at 13:12
@xzczd I agree. – b.gatessucks Aug 1 '12 at 13:24 | 909 | 2,873 | {"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} | 3.25 | 3 | CC-MAIN-2016-18 | latest | en | 0.882344 |
https://www.allmath.com/3dimensional-midpoint.php | 1,686,011,396,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652184.68/warc/CC-MAIN-20230605221713-20230606011713-00295.warc.gz | 687,770,913 | 29,458 | # Midpoint (3 Dimensions) Calculator
Enter the 1st and 2nd Coordinates of x, y, and z in the 3d midpoint calculator to calculate midpoints.
Give Us Feedback
3d Midpoint calculator is a free online tool that finds the average midpoints between different coordinates of a 3d figure. The first answer is the midpoint of x1 and x2 and so on.
## What is the Midpoint?
A midpoint refers to a point in the middle of a line. In 3d shapes, these lines are the ones joining the coordinates
In a straight line, there is one midpoint. In a 2d shape, there are 2 midpoints and so on.
## Midpoint Formula
The formulas for to find x, y, and z coordinates are:
(X1 + X2) / 2
(Y1 + Y2) / 2
(Z1 + Z2) / 2
## How to find the midpoint?
Example:
Find the midpoints of the coordinates:
X1, Y1, Z1 = 8, 4, 6
X2, Y2, Z2 = 6, 6, 8
Solution:
For X:
= (X1 + X2) / 2
= (8 + 6) / 2
7
For Y:
= (Y1 + Y2) / 2
= (4 + 6) / 2
= 5
For Z:
= (Z1 + Z2) / 2
= (6 + 8) / 2
= 7
Midpoints = (7, 5 ,7) | 365 | 994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2023-23 | longest | en | 0.842221 |
http://de.metamath.org/mpeuni/3exbii.html | 1,721,258,987,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00632.warc.gz | 7,873,221 | 2,954 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > 3exbii Structured version Visualization version GIF version
Theorem 3exbii 1766
Description: Inference adding three existential quantifiers to both sides of an equivalence. (Contributed by NM, 2-May-1995.)
Hypothesis
Ref Expression
3exbii.1 (𝜑𝜓)
Assertion
Ref Expression
3exbii (∃𝑥𝑦𝑧𝜑 ↔ ∃𝑥𝑦𝑧𝜓)
Proof of Theorem 3exbii
StepHypRef Expression
1 3exbii.1 . . 3 (𝜑𝜓)
21exbii 1764 . 2 (∃𝑧𝜑 ↔ ∃𝑧𝜓)
322exbii 1765 1 (∃𝑥𝑦𝑧𝜑 ↔ ∃𝑥𝑦𝑧𝜓)
Colors of variables: wff setvar class Syntax hints: ↔ wb 195 ∃wex 1695 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 This theorem depends on definitions: df-bi 196 df-ex 1696 This theorem is referenced by: 4exdistr 1911 ceqsex6v 3221 oprabid 6576 dfoprab2 6599 dftpos3 7257 xpassen 7939 bnj916 30257 bnj917 30258 bnj983 30275 bnj996 30279 bnj1021 30288 bnj1033 30291 ellines 31429
Copyright terms: Public domain W3C validator | 491 | 1,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.310605 |
https://www.vedantu.com/iit-jee/percentage-composition | 1,680,065,694,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948932.75/warc/CC-MAIN-20230329023546-20230329053546-00629.warc.gz | 1,181,436,122 | 32,925 | Courses
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# Percentage Composition for JEE
## What is Percentage Composition’s Definition?
Last updated date: 22nd Mar 2023
Total views: 82.8k
Views today: 2.60k
Percentage composition is the ratio of the amount of each element to the total amount of individual elements of the compound, multiplied by 100. Take H2O, or water, as an example. There are two molecules of hydrogen in water, 18.0152 grams in one mole of water and 1.008 grams in one mole of hydrogen atoms. The weight of 2 moles of hydrogen is 2.016 grams. Therefore, 1 mole of water contains 2.016 grams of hydrogen. Therefore, the percentage composition of hydrogen is 2.016 / 18.0152 = 11.19%.
## Percentage Composition Formula
• The following formula can be used to express the composition percentage of a particular element.
• The percentage composition of a solution can be expressed as $Percentage~{{C}_{E}}=\dfrac{{{g}^{E}}}{{{g}^{T}}}\times 100$ (where $Percentage~ {{C}_{E}}$ is the percentage composition of element E, ${{g}^{E}}$ is total amount of E present in the compound and ${{g}^{T}}$ is the total amount of all element present in the compound.
• This is the value to calculate. The counter on the right shows the total amount of element E present in the compound. The denominator, on the other hand, is the sum of all the elements present in a compound.
• Multiply this ratio by 100 to get the percentage of the composition.
• Next, let's take a closer look at the mass percent composition.
## Mass Percentage Composition
The concept of mass percent composition is used to indicate the concentration of an element or component in a compound. The term is used to indicate the total mass percent of each element present in a compound. It is important to note that the mass percent composition can be calculated by dividing the mass of one component by the total mass of the mixture. Then multiply this ratio by 100. This is also called mass percent (w / w)%.
$Mass~percentage =\dfrac{{{M}_{solute}}\left( g \right)}{{{M}_{solution}}\left( g \right)}\times 100$
## Volume Percentage
The volume fraction of a solution is defined as the ratio of the volume of solute present in the solution to the total volume of the solution. This type of concentration, usually calculated for liquid and gas phase solutions, is also expressed as a percentage, so this ratio should be multiplied by 100, as shown below.
$Volume~percentage =\dfrac{{{V}_{solute}}\left( mL \right)}{{{V}_{solution}}\left( mL \right)}\times 100$
As a solution is composed of both a solute and a solvent, the total volume of the solution is equal to the total volume of the solute and the solvent contained therein. Therefore, the volume fraction of a solution can also be calculated using the following equation:
$Volume~percentage =\dfrac{{{V}_{solute}}\left( mL \right)}{{{V}_{solute}}\left( mL \right)+{{V}_{solvent}}\left( mL \right)}\times 100$
## Importance of Mass Percentage
For many scientific and practical reasons, it is necessary to know the number of elements contained in a particular compound. This is essential for chemists to obtain the molecular formulas of many compounds. This formula is useful for showing the minimum integer and relative atomic number of each element of a compound. With the help of mass percentage, chemists can also calculate the actual molecular formula. The formula shows the exact number of atoms in the compound.
## Determining Mass Percentages from Chemical Formula
The basics of calculating mass percent are clarified. Let's take a look at glucose.
• The formula for glucose is C6H12O6. Let's calculate the mass percentage of each element of glucose.
• You can check the number of moles of each element from the formula.
• Glucose has 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.
• Next, let's multiply each atom by its molar mass.
• This will give you the mass of each element of glucose.
• Here, divide each mass by the mass of 1 mol of glucose to obtain the mass fraction of each element.
• Multiply this value by 100 to get the mass fraction of the element.
## What is a Mole?
Atoms and molecules are known to be very small. Therefore, it is very difficult to count with an accurate number. Now, this problem is solved by using Avogadro's number. This number is expressed as NA = 6.023x1023. Mole is defined as a number equal to Avogadro’s number. In other words, mole is the unit used to represent 6.023x1023 particles of the same substance. More specifically, a mole is the total amount of a substance that contains the same number of atoms, molecules, ions, electrons, or other basic entities as 12 grams of carbon atoms.
## Equivalent Weight
The equivalent weight of a substance is the mass (in grams) of the substance that is combined with 8 grams of oxygen or 1 gram of hydrogen or is chemically equivalent. In other words, it is equal to the mass (in grams) of the substance that reacts with or replaces 1 gram of hydrogen. Equivalent weight is expressed as the ratio of molecular weight to valence (combining power).
## Some Basic Concepts of Chemistry Class Formulas
Here we will discuss some basic concepts of chemistry five percentage composition
• Molality(m) $=\dfrac{\text{Number of moles of solute}}{\text{Mass of solvent in Kg}}$
• Molarity(M) $=\dfrac{\text{Number of moles of solute}}{\text{Volume of solution in mL}}$
• Mole fraction(xA) $=\dfrac{\text{Number of moles of solute(A)}}{\text{Total number of moles in solution}}$
## Summary
The percentage composition of a particular compound is the ratio of each element present in the compound to the total amount of individual elements present in the compound, multiplied by 100. Here, we measure the amount of existing elements in grams. Percent composition of a compound is a representation of its composition for all elements present. The importance of this composition calculation is found in chemical analysis.
Competitive Exams after 12th Science
## FAQs on Percentage Composition for JEE
1. What is the Percent Composition Formula?
All compounds are composed of various components. When writing a chemical reaction, it may be necessary to determine the number of elements in a particular compound. To understand the contribution of a particular element in one of the stoichiometric calculations of a compound that is part of an element, it is necessary to know the amount of the particular element in the compound. This can be known by knowing the percent composition of the elements of a particular compound. This is obtained from the percent composition formula.
2. What is titration? State the law it is based on.
Titration is a procedure that allows you to measure the concentration of an unknown solution using a solution of known concentration. In this method, the concentration of Solution A is determined by adding a carefully measured amount of Solution B of known concentration. Continue adding these solutions until the reaction between A and B is complete.
Titration is based on the law of equivalence. The law states, “At the endpoint of a titration, volumes of the two titrants reacted have the same number of equivalents or millie equivalents.” | 1,633 | 7,245 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-14 | latest | en | 0.884891 |
byw.thanksswf.site | 1,632,661,923,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057861.0/warc/CC-MAIN-20210926114012-20210926144012-00279.warc.gz | 211,159,897 | 9,236 | # 5/25 as a decimal number
It is the extension to non-integer numbers of the Hindu—Arabic numeral system. A decimal numeral also often just decimal or, less correctly, decimal numberrefers generally to the notation of a number in the decimal numeral system. Decimals may sometimes be identified by a decimal separator usually ".
The numbers that may be represented in the decimal system are the decimal fractions.
The decimal system has been extended to infinite decimals for representing any real numberby using an infinite sequence of digits after the decimal separator see decimal representation. In this context, the decimal numerals with a finite number of non-zero digits after the decimal separator are sometimes called terminating decimals. A repeating decimal is an infinite decimal that, after some place, repeats indefinitely the same sequence of digits e.
Many numeral systems of ancient civilizations use ten and its powers for representing numbers, possibly because there are ten fingers on two hands and people started counting by using their fingers. Examples are Brahmi numeralsGreek numeralsHebrew numeralsRoman numeralsand Chinese numerals. Very large numbers were difficult to represent in these old numeral systems, and only the best mathematicians were able to multiply or divide large numbers. These difficulties were completely solved with the introduction of the Hindu—Arabic numeral system for representing integers.
This system has been extended to represent some non-integer numbers, called decimal fractions or decimal numbersfor forming the decimal numeral system. The decimal digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; [7] the decimal separator is the dot ". For representing a non-negative numbera decimal numeral consists of. In some circumstances it may be useful to have one or more 0's on the left; this does not change the value represented by the decimal: for example, 3.
For representing a negative numbera minus sign is placed before a m. The integer part or integral part of a decimal numeral is the integer written to the left of the decimal separator see also truncation.
For a non-negative decimal numeral, it is the largest integer that is not greater than the decimal. The part from the decimal separator to the right is the fractional partwhich equals the difference between the numeral and its integer part. When the integral part of a numeral is zero, it may occur, typically in computingthat the integer part is not written for example.
In normal writing, this is generally avoided, because of the risk of confusion between the decimal mark and other punctuation. In brief, the contribution of each digit to the value of a number depends on its position in the numeral.
That is, the decimal system is a positional numeral system. The numbers that are represented by decimal numerals are the decimal fractions sometimes called decimal numbersthat is, the rational numbers that may be expressed as a fraction whose denominator is a power of ten.
More generally, a decimal with n digits after the separator represents the fraction with denominator 10 nwhose numerator is the integer obtained by removing the separator. Expressed as a fully reduced fractionthe decimal numbers are those whose denominator is a product of a power of 2 and a power of 5. Thus the smallest denominators of decimal numbers are.
Decimal numerals do not allow an exact representation for all real numberse. Nevertheless, they allow approximating every real number with any desired accuracy, e.Convert from decimal to fraction.
Convert 5. Here you can find a decimal to fraction chart and also write any decimal number as a fraction. Step 2 : Multiply both top and bottom by 10 for every number after the decimal point:. As we have 2 numbers after the decimal point, we multiply both numerator and denominator by Step 3 : Simplify or reduce the fraction:. As we have 1 numbers after the decimal point, we multiply both numerator and denominator by To convert from inch to millimeter multiply inch value by To convert from millimeter inch divide millimeter value by Please link to this page!
Just right click on the above image, then choose copy link address, then past it in your HTML. While every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions, or for the results obtained from the use of this information.
Cool Conversion. Terminating Repeating.
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Step 3: Simplify or reduce the fraction if it is not in the simplest form. Learn more reading the examples below or use our self-explaining calculator above. Convert decimal 0. So, 0.
### Write 5.25 as a percentage
What is 0. Equivalent fraction for 1. So, 1. This fraction is alread reduced, We can't reduce it any further. An easier way to do it is to use the table below. You will find Example 2 Convert 0. Look down the decimal column until you find 0.Below you can find the full step by step solution for you problem. We hope it will be very helpful for you and it will help you to understand the solving process. What is 5.
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To write 5. We divide now 5. Equations solver - equations involving one unknown Quadratic equations solver Percentage Calculator - Step by step Derivative calculator - step by step Graphs of functions Factorization Greatest Common Factor Least Common Multiple System of equations - step by step solver Fractions calculator - step by step Theory in mathematics Roman numerals conversion Tip calculator Numbers as decimals, fractions, percentages More or less than - questions.
## What is 403 5/25 as a decimal?
Toggle navigation GetEasySolution. If it's not what You are looking for, type in into the box below your number and see the solution. Write as a decimal as a fraction as a percent Solve.
You can always share this solution. Equations solver categories Equations solver - equations involving one unknown Quadratic equations solver Percentage Calculator - Step by step Derivative calculator - step by step Graphs of functions Factorization Greatest Common Factor Least Common Multiple System of equations - step by step solver Fractions calculator - step by step Theory in mathematics Roman numerals conversion Tip calculator Numbers as decimals, fractions, percentages More or less than - questions.Below you can find the full step by step solution for you problem.
We hope it will be very helpful for you and it will help you to understand the solving process.
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Equations solver - equations involving one unknown Quadratic equations solver Percentage Calculator - Step by step Derivative calculator - step by step Graphs of functions Factorization Greatest Common Factor Least Common Multiple System of equations - step by step solver Fractions calculator - step by step Theory in mathematics Roman numerals conversion Tip calculator Numbers as decimals, fractions, percentages More or less than - questions.
Toggle navigation GetEasySolution. If it's not what You are looking for, type in into the box below your number and see the solution.
Write as a decimal as a fraction as a percent Solve. You can always share this solution. Equations solver categories Equations solver - equations involving one unknown Quadratic equations solver Percentage Calculator - Step by step Derivative calculator - step by step Graphs of functions Factorization Greatest Common Factor Least Common Multiple System of equations - step by step solver Fractions calculator - step by step Theory in mathematics Roman numerals conversion Tip calculator Numbers as decimals, fractions, percentages More or less than - questions.Convert from percent to decimal.
Here is the answer to the question: 5. Use our percent to decimal calculator above to write any percent values as a decimal. Please link to this page!
### What is 5.25 as a fraction
Just right click on the above image, choose copy link address, then past it in your HTML. While every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions, or for the results obtained from the use of this information.
Cool Conversion. How to convert from percent to decimal Let's see this example: Percent means 'per '. So, 5. If you divide 5. Note that if the percent value is a integer, the '. Percent to Decimal Calculator Please link to this page! Sample Percent to Decimal Calculations Disclaimer While every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions, or for the results obtained from the use of this information.Below you can find the full step by step solution for you problem.
We hope it will be very helpful for you and it will help you to understand the solving process. What is 5. To write 5. And finally we have: 5. Equations solver - equations involving one unknown Quadratic equations solver Percentage Calculator - Step by step Derivative calculator - step by step Graphs of functions Factorization Greatest Common Factor Least Common Multiple System of equations - step by step solver Fractions calculator - step by step Theory in mathematics Roman numerals conversion Tip calculator Numbers as decimals, fractions, percentages More or less than - questions.
Toggle navigation GetEasySolution. If it's not what You are looking for, type in into the box below your number and see the solution.
Write as a decimal as a fraction as a percent Solve. You can always share this solution. Equations solver categories Equations solver - equations involving one unknown Quadratic equations solver Percentage Calculator - Step by step Derivative calculator - step by step Graphs of functions Factorization Greatest Common Factor Least Common Multiple System of equations - step by step solver Fractions calculator - step by step Theory in mathematics Roman numerals conversion Tip calculator Numbers as decimals, fractions, percentages More or less than - questions.Convert from percent to decimal.
Here is the answer to the question: 5. Use our percent to decimal calculator above to write any percent values as a decimal. Please link to this page! Just right click on the above image, choose copy link address, then past it in your HTML. While every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions, or for the results obtained from the use of this information.
Cool Conversion. How to convert from percent to decimal Let's see this example: Percent means 'per '.
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So, 5. If you divide 5. Note that if the percent value is a integer, the '.
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Percent to Decimal Calculator Please link to this page! Sample Percent to Decimal Calculations 7. Disclaimer While every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions, or for the results obtained from the use of this information. | 2,238 | 11,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-39 | latest | en | 0.887579 |
http://mathhelpforum.com/advanced-algebra/82149-permuting-roots.html | 1,506,166,114,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689624.87/warc/CC-MAIN-20170923104407-20170923124407-00496.warc.gz | 200,204,478 | 10,545 | 1. Permuting roots....
Hi guys,
This question is about Galois Theory.
One of the theorems in my books states that "Any K-automorphism of the field extension L/K must keep the elements of K invariant & must permute the roots of the minimal polynomial."
I was wondering, do the K-automorphisms permutes the roots of ALL polynomials, or just the minimal polynomial?
Many thanks in advance. :-) x
2. Originally Posted by AAM
Hi guys,
This question is about Galois Theory.
One of the theorems in my books states that "Any K-automorphism of the field extension L/K must keep the elements of K invariant & must permute the roots of the minimal polynomial."
I was wondering, do the K-automorphisms permutes the roots of ALL polynomials, or just the minimal polynomial?
Many thanks in advance. :-) x
All polynomials with coefficients in K, yes. If p(x)= $ax^n+ bx^{n-1}+ \cdot\cdot\cdot+ yx+ z$, then $f(p(x))= f(a)f(x)^n+ f(b)f(x)^{n-1}+ \cdot\cdot\cdot+ f(y)f(x)+ f(z)$, because f is an automorphism, $= a(f(x))^n+ b(f(x))^{n-1}+ \cdot\cdot\cdot+ yf(x)+ z$ because f "fixes" members of K.
IF x is a root of p(x) then $p(x)= 0$ so $f(p(x))= a(f(x))^n+ b(f(x))^{n-1}+ \cdot\cdot\cdot+ yf(x)+ z= 0$. that is, if x is a zero of a polynomial with coefficients in K, f(x) is a zero of that same polynomial: f permutes the zeros.
3. Originally Posted by HallsofIvy
All polynomials with coefficients in K, yes. If p(x)= $ax^n+ bx^{n-1}+ \cdot\cdot\cdot+ yx+ z$, then $f(p(x))= f(a)f(x)^n+ f(b)f(x)^{n-1}+ \cdot\cdot\cdot+ f(y)f(x)+ f(z)$, because f is an automorphism, $= a(f(x))^n+ b(f(x))^{n-1}+ \cdot\cdot\cdot+ yf(x)+ z$ because f "fixes" members of K.
IF x is a root of p(x) then $p(x)= 0$ so $f(p(x))= a(f(x))^n+ b(f(x))^{n-1}+ \cdot\cdot\cdot+ yf(x)+ z= 0$. that is, if x is a zero of a polynomial with coefficients in K, f(x) is a zero of that same polynomial: f permutes the zeros.
You still need to show more. Let $X$ be the set of zeros of $f(x)\in K[x]$ in some splitting field $L$. Notice that $|X|<\infty$ and that if $\sigma \in \text{Gal}(L/K)$ then $\sigma: X\to X$ as you have shown. However, $\sigma$ is injective (for it is an automorphism) and so $\sigma: X\to X$ is a bijection. Thus, $\sigma$ is a permutation of $X$ - the set of zeros of $f(x)$.
Originally Posted by AAM
I was wondering, do the K-automorphisms permutes the roots of ALL polynomials, or just the minimal polynomial?
In the proof we do not use the fact that the polynomial is irreducible. Thus, this is true for all non-constant polynomials.
4. Thank you! :-D | 823 | 2,548 | {"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": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2017-39 | longest | en | 0.816659 |
https://coderzcolumn.com/tutorials/python/difflib-simple-way-to-find-out-differences-between-sequences-file-contents-using-python | 1,726,751,797,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00791.warc.gz | 159,485,660 | 20,218 | Updated On : Aug-27,2022 Time Investment : ~25 mins
# difflib - Simple Way to Find Out Differences Between Sequences/File Contents using Python¶
#### > Why Compare Sequences?¶
Many times a Python developer needs to compare two sequences (list of numbers, list of strings, list of characters, etc.) to find matching subsequences in between them. These subsequences can help us understand how much two sequences are similar and how different they can have different applications.
The algorithm which does can be useful in various situations like comparing contents of the files, contents of a single string, etc.
#### > What Solution Python Offers for Comparing Sequences?¶
Python provides us with a module named difflib which can compare sequences of any type for us so that we don't need to write complicated algorithms to find common subsequences between two sequences.
The difflib module provides different classes and methods to perform a comparison of two sequences and generate a delta.
#### > What Can You Learn From This Article?¶
As a part of this tutorial, we'll be explaining how to use Python module "difflib" to compare sequences of different types with simple and easy-to-understand examples. Apart from basic comparison, Tutorial explains how to generate match ratios of sequences, handle junk characters, generate one sequence from another, etc. It even explains different ways of formatting (HTML, ContextDiff, Differ, etc) different between two sequences. Tutorial covers total API of "difflib" module.
#### > How "difflib" Works?¶
At the core of the difflib module is SequenceMatcher class which implements an algorithm responsible for comparing two sequences. It requires that all the elements of both sequences be "hashable" in order for them to work.
First, It finds the longest common subsequence between two sequences and then divides both sequences into left parts of both original sequences (left subsequences of original sequences) and right parts of both original sequences (right subsequences of original sequences) based on that common subsequence.
Then, It recursively performs the same function of finding the longest common subsequence on the left parts of both original sequences and the right parts of both original sequences.
The algorithm takes quadratic time for the worst case and linear time for the best case. The expected case time is dependent on the size of sequences and is better than worst-case quadratic time.
The algorithm also automatically takes care of junk elements which are the most commonly occurring elements. If an element appears for more than 1% time of the total elements of the sequence then it's considered a junk element and ignored for subsequence finding.
If you are just interested in comparing the list of files/directories and preparing a report about them then please feel free to check our tutorial on Python module filecmp.
Below, we have listed important sections of tutorial to give an overview of the material covered.
## Example 1: Simple Sequence Match (SequenceMatcher) ¶
As a part of our first example, we'll explain how we can compare two sequences with numbers using SequenceMatcher instance and its methods.
• SequenceMatcher(isjunk=None,a='', b='', autojunk=True) - It accepts two sequences and returns an instance of SequenceMatcher which can be used to find common subsequences.
• isjunk - This parameter accepts a function that takes as input a single element of the sequence and returns True if it junk else False. We can provide function if we want junk elements by ourselves. The default is None.
• autojunk - This parameter accepts boolean value. If set to True, it enables auto junk finding functionality of the algorithm which we described in the introduction section. We can disable it by setting this parameter to False. The default is True.
#### Important Methods of SequenceMatcher Instance¶
• find_longest_match(alo=0, ahi=None, blo=0, bhi=None) - This method accepts the starting and ending indices of both sequences and returns longest common matching subsequence between two sequences as Match instance.
• The Match instance is a named tuple and has three attributes namely a, b, and size. The a and b attributes specify indices in first and second sequences from where the common sequence starts. The size specifies the length of the subsequence.
• get_matching_blocks() - It returns a list of Match instances which has information about the list of matching subsequences between two sequences.
Our code for this example first creates an instance of SequenceMatcher using two sequences of integers. It then finds out the longest common subsequence using find_longest_match() and prints it. It then finds out the list of common subsequences using get_matching_blocks() and prints them.
```import difflib
l1 = [1,2,3,5,6,7, 8,9]
l2 = [2,3,6,7,8,10,11]
seq_mat = difflib.SequenceMatcher(a=l1, b=l2)
match = seq_mat.find_longest_match(alo=0, ahi=len(l1), blo=0, bhi=len(l2))
print("============ Longest Matching Sequence ==================")
print("\nMatch Object : {}".format(match))
print("Matching Sequence from l1 : {}".format(l1[match.a:match.a+match.size]))
print("Matching Sequence from l2 : {}\n".format(l2[match.b:match.b+match.size]))
print("============ All Matching Sequences ==================")
for match in seq_mat.get_matching_blocks():
print("\nMatch Object : {}".format(match))
print("Matching Sequence from l1 : {}".format(l1[match.a:match.a+match.size]))
print("Matching Sequence from l2 : {}".format(l2[match.b:match.b+match.size]))
```
```============ Longest Matching Sequence ==================
Match Object : Match(a=4, b=2, size=3)
Matching Sequence from l1 : [6, 7, 8]
Matching Sequence from l2 : [6, 7, 8]
============ All Matching Sequences ==================
Match Object : Match(a=1, b=0, size=2)
Matching Sequence from l1 : [2, 3]
Matching Sequence from l2 : [2, 3]
Match Object : Match(a=4, b=2, size=3)
Matching Sequence from l1 : [6, 7, 8]
Matching Sequence from l2 : [6, 7, 8]
Match Object : Match(a=8, b=7, size=0)
Matching Sequence from l1 : []
Matching Sequence from l2 : []
```
## Example 2: Sequence Match with Match Ratios (SequenceMatcher)¶
As a part of our second example, we'll be explaining various methods of SequenceMatcher instance.
#### Important Methods of SequenceMatcher Instance¶
• set_seqs(a,b) - It accepts two sequences and sets them as first and second sequence of the SequenceMatcher instance.
• set_seq1(a) - It accepts a single sequence and sets that sequence as the first sequence of the SequenceMatcher instance.
• set_seq2(b) - It accepts single sequence and sets that sequence as second sequence of the SequenceMatcher instance.
• ratio() - It returns float in the range 0-1, specifying similarity between two sequences.
• quick_ratio() - It returns a float in the range 0-1, specifying the upper bound on the similarity between two sequences quickly.
• real_quick_ratio() - It returns a float in the range 0-1, specifying the upper bound on the similarity between two sequences very quickly.
Our code for this example starts by creating an instance of SequenceMatcher without setting any sequences. We have then set both sequences using set_seqs() methods. We have then printed the longest common subsequence and ratios of similarity between two sequences.
We have then set two different sequences as first and second sequence of the SequenceMatcher using set_seq1() and set_seq2() methods. We have then again printed the longest common subsequence and similarity ratios between these two new sequences.
```import difflib
l1 = [1,2,3,5,6,7, 8,9]
l2 = [2,3,6,7,8,10,11]
seq_mat = difflib.SequenceMatcher()
seq_mat.set_seqs(l1, l2)
match = seq_mat.find_longest_match(alo=0, ahi=len(l1), blo=0, bhi=len(l2))
print("============ Longest Matching Sequence (l1,l2) ==================")
print("\nMatch Object : {}".format(match))
print("Matching Sequence from l1 : {}".format(l1[match.a:match.a+match.size]))
print("Matching Sequence from l2 : {}".format(l2[match.b:match.b+match.size]))
print("\n=========== Similarity Ratios ==============")
print("Similarity Ratio : {}".format(seq_mat.ratio()))
print("Similarity Ratio Quick : {}".format(seq_mat.quick_ratio()))
print("Similarity Ratio Very Quick : {}".format(seq_mat.real_quick_ratio()))
#####################################################
l3 = [0,1,2,3,4,6,7,8,9]
l4 = [2,3,6,7,8,9,10,11,12,13]
seq_mat.set_seq1(l3)
seq_mat.set_seq2(l4)
match = seq_mat.find_longest_match(alo=0, ahi=len(l3), blo=0, bhi=len(l4))
print("\n\n\n============ Longest Matching Sequence (l3,l4) ==================")
print("\nMatch Object : {}".format(match))
print("Matching Sequence from l3 : {}".format(l3[match.a:match.a+match.size]))
print("Matching Sequence from l4 : {}".format(l4[match.b:match.b+match.size]))
print("\n=========== Similarity Ratios ==============")
print("Similarity Ratio : {}".format(seq_mat.ratio()))
print("Similarity Ratio Quick : {}".format(seq_mat.quick_ratio()))
print("Similarity Ratio Very Quick : {}".format(seq_mat.real_quick_ratio()))
```
```============ Longest Matching Sequence (l1,l2) ==================
Match Object : Match(a=4, b=2, size=3)
Matching Sequence from l1 : [6, 7, 8]
Matching Sequence from l2 : [6, 7, 8]
=========== Similarity Ratios ==============
Similarity Ratio : 0.6666666666666666
Similarity Ratio Quick : 0.6666666666666666
Similarity Ratio Very Quick : 0.9333333333333333
============ Longest Matching Sequence (l3,l4) ==================
Match Object : Match(a=5, b=2, size=4)
Matching Sequence from l3 : [6, 7, 8, 9]
Matching Sequence from l4 : [6, 7, 8, 9]
=========== Similarity Ratios ==============
Similarity Ratio : 0.631578947368421
Similarity Ratio Quick : 0.631578947368421
Similarity Ratio Very Quick : 0.9473684210526315
```
## Example 3: String Sequence Match with Junk Characters Removed (SequenceMatcher)¶
As a part of our third example, we are explaining how we can compare strings using SequenceMatcher. We are also explaining how we can use a function with isjunk parameter which will decide which characters to consider as junk. We are also explaining which characters will be considered junk elements when the size of the sequences is greater than 200 elements. Apart from this, we have also explained few important attributes of the SequenceMatcher instance.
#### Important Attributes of SequenceMatcher Instance¶
• bjunk - It returns a list of characters that are junk in our sequences according to the function supplied to isjunk.
• bpopular - It returns a list of characters that are considered junk in our sequences according to autojunk parameter.
• b2j - It returns a dictionary that has a mapping from elements of second sequences to their position in the sequence. It does not include junk elements.
Our code first creates two sequences that have string data. It then creates an instance of SequenceMatcher. It provides a function to isjunk parameter which considers comma and dot as junk elements. It then finds out the longest common subsequence and prints it. It also prints a list of all subsequences between both strings. It then prints attributes bjunk, bpopular, and b2j of SequenceMatcher instance.
Our code then modifies the second-string further by adding 150 e characters to it. This is done so that the string size becomes more than 200 characters. This will activate the auto junk functionality of the algorithm. We have then again printed attributes bjunk, bpopular, and b2j of SequenceMatcher instance. We can clearly notice a difference in values of the attributes which is now giving results for bpopular attribute.
```import difflib
l1 = "Hello, Welcome to CoderzColumn."
l2 = "Welcome to CoderzColumn, Have a Great Learning Day."
seq_mat = difflib.SequenceMatcher(isjunk=lambda x: x in [",", "."], a=l1, b=l2, autojunk=True)
match = seq_mat.find_longest_match(alo=0, ahi=len(l1), blo=0, bhi=len(l2))
print("============ Longest Matching Sequence ==================")
print("\nMatch Object : {}".format(match))
print("Matching Sequence from l1 : {}".format(l1[match.a:match.a+match.size]))
print("Matching Sequence from l2 : {}\n".format(l2[match.b:match.b+match.size]))
print("============ All Matching Sequences ==================")
for match in seq_mat.get_matching_blocks():
if match.size > 0:
print("\nMatch Object : {}".format(match))
print("Matching Sequence from l1 : {}".format(l1[match.a:match.a+match.size]))
print("Matching Sequence from l2 : {}".format(l2[match.b:match.b+match.size]))
print("\nSequence B Junk : {}".format(seq_mat.bjunk))
print("Sequence B Popular : {}".format(seq_mat.bpopular))
print("Sequence B Junk : {}".format(seq_mat.b2j))
l2 = l2 + "e"*150 ### Added 150 e character to make string of lengh more than 200 to make autojunk work.
seq_mat.set_seq2(l2)
print("\nSequence B Junk : {}".format(seq_mat.bjunk))
print("Sequence B Popular : {}".format(seq_mat.bpopular))
print("Sequence B Junk : {}".format(seq_mat.b2j))
```
```============ Longest Matching Sequence ==================
Match Object : Match(a=7, b=0, size=23)
Matching Sequence from l1 : Welcome to CoderzColumn
Matching Sequence from l2 : Welcome to CoderzColumn
============ All Matching Sequences ==================
Match Object : Match(a=7, b=0, size=23)
Matching Sequence from l1 : Welcome to CoderzColumn
Matching Sequence from l2 : Welcome to CoderzColumn
Sequence B Junk : {',', '.'}
Sequence B Popular : set()
Sequence B Junk : {'W': [0], 'e': [1, 6, 14, 28, 34, 39], 'l': [2, 19], 'c': [3], 'o': [4, 9, 12, 18], 'm': [5, 21], ' ': [7, 10, 24, 29, 31, 37, 46], 't': [8, 36], 'C': [11, 17], 'd': [13], 'r': [15, 33, 41], 'z': [16], 'u': [20], 'n': [22, 42, 44], 'H': [25], 'a': [26, 30, 35, 40, 48], 'v': [27], 'G': [32], 'L': [38], 'i': [43], 'g': [45], 'D': [47], 'y': [49]}
Sequence B Junk : {',', '.'}
Sequence B Popular : {'o', 'a', 'e', ' '}
Sequence B Junk : {'W': [0], 'l': [2, 19], 'c': [3], 'm': [5, 21], 't': [8, 36], 'C': [11, 17], 'd': [13], 'r': [15, 33, 41], 'z': [16], 'u': [20], 'n': [22, 42, 44], 'H': [25], 'v': [27], 'G': [32], 'L': [38], 'i': [43], 'g': [45], 'D': [47], 'y': [49]}
```
## Example 4: Operations to Generate Second Sequence from First Sequence (SequenceMatcher)¶
As a part of our fourth example, we'll explain how we can perform list of operations on first sequence to transform it to second sequence using get_opcodes() and get_grouped_opcodes() methods of SequenceMatcher instances.
#### Important Methods of SequenceMatcher Instance¶
• get_opcodes() - It returns a list of tuples where each tuple has 5 elements (operation, i1, i2, j1, j2) specifying how to transform first sequence to second. The first element is a string specifying one of the operations that should be performed on the first sequence to transform it to seconds. The other four values are 2 indices from the first sequence and 2 indices from the second sequence specifying where operations should be applied. Below are operations that will be returned.
• replace - It specifies that a[i1:i2] should be replaced by b[j1:j2].
• delete - It specifies that a[i1:i2] should be deleted.
• insert - It b[j1:j2] should be inserted at a[i1].
• equal - It means that a[i1:i2] and b[j1:j2] are same and no action is needed.
• get_grouped_opcodes(n=3) - This method returns a generator instance where the individual element is a list of groups specifying operations to be performed to transform the first sequence to the second. It does that with n lines of context.
### 4.1: Using get_opcodes() Method¶
Our code for this part starts by creating an instance of SequenceMatcher with the same sequence which we had used in the previous example. We then create a third list which is a copy of the first list and has elements as a list of characters. We then loop through each operation returned by get_opcodes() method and perform that operation on the third sequence (which is a copy of the first sequence) to transform it to the second sequence.
```import difflib
l1 = "Hello, Welcome to CoderzColumn."
l2 = "Welcome to CoderzColumn, Have a Great Learning Day."
seq_mat = difflib.SequenceMatcher(a=l1, b=l2, autojunk=True)
l3 = list(l1)
for operation, i1,i2,j1,j2 in seq_mat.get_opcodes():
if operation == "delete":
print("Deleting Sequence : '{}' from l1".format(l1[i1:i2]))
l3[i1:i2] = [""] * len(l1[i1:i2])
elif operation == "replace":
print("Replacing Sequence : '{}' in l1 with '{}' in l2".format(l1[i1:i2], l2[j1:j2]))
l3[i1:i2] = [""] * len(l1[i1:i2])
l3.insert(i1, l2[j1:j2])
elif operation == "insert":
print("Inserting Sequence : '{}' from l2 at {} in l1".format(l2[j1:j2], i1))
l3.insert(i1, l2[j1:j2])
elif operation == "equal":
print("Equal Sequences. '{}' No Action Needed.".format(l1[i1:i2]))
print("\nFinal Sequence : {}".format("".join(l3)))
```
```Deleting Sequence : 'Hello, ' from l1
Equal Sequences. 'Welcome to CoderzColumn' No Action Needed.
Inserting Sequence : ', Have a Great Learning Day' from l2 at 30 in l1
Equal Sequences. '.' No Action Needed.
Final Sequence : Welcome to CoderzColumn, Have a Great Learning Day.
```
### 4.2: Using get_grouped_opcodes() Method¶
Our code for this part starts by creating an instance of SequenceMatcher with the same sequence which we had used in previous examples. We then create a third list which is a copy of the first list and has elements as a list of characters. We then loop through each group returned by get_grouped_opcodes() method and perform operations specified in each group on the third sequence (which is a copy of the first sequence) to transform it to the second sequence.
```import difflib
l1 = "Hello, Welcome to CoderzColumn."
l2 = "Welcome to CoderzColumn, Have a Great Learning Day."
seq_mat = difflib.SequenceMatcher(a=l1, b=l2, autojunk=True)
l3 = list(l1)
for groups in seq_mat.get_grouped_opcodes(n=8):
for operation, i1,i2,j1,j2 in groups:
if operation == "delete":
print("Deleting Sequence : '{}' from l1".format(l1[i1:i2]))
l3[i1:i2] = [""] * len(l1[i1:i2])
elif operation == "replace":
print("Replacing Sequence : '{}' in l1 with '{}' in l2".format(l1[i1:i2], l2[j1:j2]))
l3[i1:i2] = l2[j1:j2]
elif operation == "insert":
print("Inserting Sequence : '{}' from l2 at {} in l1".format(l2[j1:j2], i1))
l3.insert(i1, l2[j1:j2])
elif operation == "equal":
print("Equal Sequences. '{}' No Action Needed.".format(l1[i1:i2]))
print("\nFinal Sequence : {}".format("".join(l3)))
```
```Deleting Sequence : 'Hello, ' from l1
Equal Sequences. 'Welcome ' No Action Needed.
Equal Sequences. 'rzColumn' No Action Needed.
Inserting Sequence : ', Have a Great Learning Day' from l2 at 30 in l1
Equal Sequences. '.' No Action Needed.
Final Sequence : Welcome to CoderzColumn, Have a Great Learning Day.
```
## Example 5: Compare List of Strings (Differ) ¶
As a part of our fifth example, we'll explain how we can compare the list of strings using Differ class of difflib module.
• Differ(linejunk=None, charjunk=None) - It accepts two parameters which specify which accepts functions specifying which lines should be considered as junk and which character should be considered as junk. The default is None for both parameters hence nothing is considered junk.
The Differ internally uses SequenceMatcher for finding sequence on the list of strings to find common subsequences between two original sequences and then on a list of characters to find subsequences between individual elements of both original subsequences.
Our code for this part starts by creating two strings from the contents of zen of Python (import this). We have then created an instance of Differ which has a method named compare() which accepts two lists of strings and compares them. We have then compared two strings using compare() methods and printed their result. We have split strings into a list of strings using splitlines() method of string which splits strings based on a new line character.
We can notice from the output that there are four kinds of lines in the output.
• The lines starting with character '-' are lines present in sequence 1.
• The lines starting with character '+' are lines present in sequence 2.
• The lines starting with character ' ' are lines present in both sequences.
• The lines starting with character '?' are lines present in neither sequences but have characters that hint which characters were changed (added/removed) between two sequences.
```import difflib
a = '''
2. Special cases aren't special enough to break the rules.
3. Errors should never pass silently.
4. In the face of ambiguity, refuse the temptation to guess.
5. There should be one-- and preferably only one --obvious way to do it.
6. Although that way may not be obvious at first unless you're Dutch.
7. Now is better than never.
8. Although never is often better than *right* now.
9. If the implementation is hard to explain, it's a bad idea.
10. If the implementation is easy to explain, it may be a good idea.
'''
b = '''
1. Simplicy counts as well.
2. Special cases aren't that special enough to break the rules.
3. Errors shall never pass ever.
4. In the face of ambiguity, refuse the temptation to guess.
5. There should be one obvious way to do it.
6. Although that way may not be obvious at first unless you're Dutch.
7. Now is better than never.
8. Although never is often better than immediately.
9. If the implementation is hard to code, it's most probably a bad idea.
'''
difference = difflib.Differ()
for line in difference.compare(a.splitlines(keepends=True), b.splitlines(keepends=True)):
print(line, end="")
```
```- 1. Readability counts.
+ 1. Simplicy counts as well.
- 2. Special cases aren't special enough to break the rules.
+ 2. Special cases aren't that special enough to break the rules.
? +++++
- 3. Errors should never pass silently.
? ^^ ^ --- ^^^^
+ 3. Errors shall never pass ever.
? ^ ^ ^^^
4. In the face of ambiguity, refuse the temptation to guess.
- 5. There should be one-- and preferably only one --obvious way to do it.
? -- --------------------------
+ 5. There should be one obvious way to do it.
6. Although that way may not be obvious at first unless you're Dutch.
7. Now is better than never.
- 8. Although never is often better than *right* now.
? -- ^^ ^^^^^
+ 8. Although never is often better than immediately.
? ^^^^^^ ^^^
- 9. If the implementation is hard to explain, it's a bad idea.
? ------
+ 9. If the implementation is hard to code, it's most probably a bad idea.
? +++ ++++++++++++++
- 10. If the implementation is easy to explain, it may be a good idea.
```
## Example 6: Compare File Contents (Differ) ¶
As a part of our sixth example, we are again using Differ to explain how we can compare the contents of two files.
We have saved our strings from previous examples in files named original.txt and modified.txt. We have then read the contents of both files and compared them using Differ. We can notice from the result that it is exactly the same as the previous example.
```import difflib
a = open("original.txt", "r").readlines()
b = open("modified.txt", "r").readlines()
difference = difflib.Differ(charjunk=lambda x: x in [",", ".", "-", "'"])
for line in difference.compare(a, b):
print(line, end="")
```
```- 1. Readability counts.
+ 1. Simplicy counts as well.
- 2. Special cases aren't special enough to break the rules.
+ 2. Special cases aren't that special enough to break the rules.
? +++++
- 3. Errors should never pass silently.
? ^^ ^ --- ^^^^
+ 3. Errors shall never pass ever.
? ^ ^ ^^^
4. In the face of ambiguity, refuse the temptation to guess.
- 5. There should be one-- and preferably only one --obvious way to do it.
? -- --------------------------
+ 5. There should be one obvious way to do it.
6. Although that way may not be obvious at first unless you're Dutch.
7. Now is better than never.
- 8. Although never is often better than *right* now.
? -- ^^ ^^^^^
+ 8. Although never is often better than immediately.
? ^^^^^^ ^^^
- 9. If the implementation is hard to explain, it's a bad idea.
? ------
+ 9. If the implementation is hard to code, it's most probably a bad idea.
? +++ ++++++++++++++
- 10. If the implementation is easy to explain, it may be a good idea.
```
## Example 7: Generate HTML of Difference Between Files (HtmlDiff)¶
As a part of our seventh example, we'll explain how we can generate a difference between two sequences in HTML format where the table shows the difference between two sequences side by side using different colors. The difflib provides a class named HtmlDiff for this purpose.
• HtmlDiff(tabsize=8,wrapcolumn=None,linejunk=None,charjunk=IS_CHARACTER_JUNK) - This constructor creates an instance of HtmlDiff which we can then use to generate difference between two list of strings in an HTML format. The tabsize parameter specifies the size of the tab. The wrapcolumn parameter accepts the number and specifies after how many characters, the string should be wrapped on the next line.
#### Important Methods of HtmlDiff Instance¶
• make_file() - This method returns a string that has HTML showing differences between two lists of strings where each line is shown side by side between two lists and the difference is highlighted with different colors.
Our code for this part reads two files which we used in our previous example. We are then creating an instance of HtmlDiff to compare the list of strings. We then call make_file() method of HtmlDiff to compare two sequences and return comparison result in HTML format. We are then storing the result in compare.html file.
We are using display module of IPython to display an HTML file in jupyter notebook. The output presented in this format is easy to understand and interpret.
If you are interested in learning about how contents of different types like HTML, audio, video, etc can be displayed in the Jupyter notebook then please feel free to check our tutorial on the same.
```import difflib
from IPython import display
a = open("original.txt", "r").readlines()
b = open("modified.txt", "r").readlines()
difference = difflib.HtmlDiff(tabsize=2)
with open("compare.html", "w") as fp:
html = difference.make_file(fromlines=a, tolines=b, fromdesc="Original", todesc="Modified")
fp.write(html)
```
## Example 8: Generate HTML Table of Difference Between Files (HtmlDiff)¶
We are using our eighth example again to show differences between two lists of strings in HTML format. We are explaining usage of make_table() method of HtmlDiff instance this time. This can be useful when we only want difference as an HTML table so that we can include it in some HTML of our own. We might now always need the whole HTML ready if we want to include a table in some rich HTML of our own.
#### Important Methods of HtmlDiff Instance¶
• make_table() - This method returns a string that has an HTML table showing differences between two lists of strings where each line is shown side by side between two lists and the difference is highlighted with different colors.
Our code for this example is exactly the same as our previous example with the only change that we are using make_table() method instead of make_file().
```import difflib
from IPython import display
a = open("original.txt", "r").readlines()
b = open("modified.txt", "r").readlines()
difference = difflib.HtmlDiff(tabsize=2)
with open("compare.html", "w") as fp:
html = difference.make_table(fromlines=a, tolines=b, fromdesc="Original", todesc="Modified")
fp.write(html)
```
## Example 9: Compare List of Strings and Return Difference Context Diff Format (context_diff())¶
As a part of our ninth example, we'll be demonstrating how we can show the difference between two lists of strings in a contextual difference format using context_diff() method. The contextual difference is a simple way of showing which lines are changed along with few other lines around them to show context.
• context_diff(a,b,fromfile='',tofile='',fromfiledate='',tofiledate='') - It compares two list of strings and returns difference between them in contextual difference format. It even let us set file names and dates for two sequences.
Our code for this example reads the content of two files that we had created in one of our previous examples. It then finds out the contextual differences between them using context_diff() function and prints difference. The output has lines starting with the character '!' to show that it has the difference between the two sequences.
```import difflib
a = open("original.txt", "r").readlines()
b = open("modified.txt", "r").readlines()
difference = difflib.context_diff(a, b,
fromfile="original.txt", tofile="modified.txt",
fromfiledate="2021-02-19", tofiledate="2021-02-20")
for diff in difference:
print(diff, end="")
```
```*** original.txt 2021-02-19
--- modified.txt 2021-02-20
***************
*** 1,10 ****
! 1. Readability counts.
! 2. Special cases aren't special enough to break the rules.
! 3. Errors should never pass silently.
4. In the face of ambiguity, refuse the temptation to guess.
! 5. There should be one-- and preferably only one --obvious way to do it.
6. Although that way may not be obvious at first unless you're Dutch.
7. Now is better than never.
! 8. Although never is often better than *right* now.
! 9. If the implementation is hard to explain, it's a bad idea.
! 10. If the implementation is easy to explain, it may be a good idea.
--- 1,9 ----
! 1. Simplicy counts as well.
! 2. Special cases aren't that special enough to break the rules.
! 3. Errors shall never pass ever.
4. In the face of ambiguity, refuse the temptation to guess.
! 5. There should be one obvious way to do it.
6. Although that way may not be obvious at first unless you're Dutch.
7. Now is better than never.
! 8. Although never is often better than immediately.
! 9. If the implementation is hard to code, it's most probably a bad idea.
```
## Example 10: Compare List of Strings and Return Differ Style Difference (ndiff())¶
As a part of our tenth example, we'll explain the usage of function ndiff() of module difflib which gives the same functionality that is available through Differ instance.
• ndiff(a,b,linejunk=None,charjunk=IS_CHARACTER_JUNK) - This method accepts two lists of strings as input and returns a generator that has a list of lines showing the difference between two lists. It also lets us specify linejunk and charjunk parameters which accept functions for finding out which lines and characters to consider as junk.
Our code for this example is pretty self-explanatory which uses files that we have been using for many examples. It finds out the difference between the contents of the files using ndiff() method and prints it.
```import difflib
a = open("original.txt", "r").readlines()
b = open("modified.txt", "r").readlines()
for diff in difflib.ndiff(a, b):
print(diff, end="")
```
```- 1. Readability counts.
+ 1. Simplicy counts as well.
- 2. Special cases aren't special enough to break the rules.
+ 2. Special cases aren't that special enough to break the rules.
? +++++
- 3. Errors should never pass silently.
? ^^ ^ --- ^^^^
+ 3. Errors shall never pass ever.
? ^ ^ ^^^
4. In the face of ambiguity, refuse the temptation to guess.
- 5. There should be one-- and preferably only one --obvious way to do it.
+ 5. There should be one obvious way to do it.
6. Although that way may not be obvious at first unless you're Dutch.
7. Now is better than never.
- 8. Although never is often better than *right* now.
? -- ^^ ^^^^^
+ 8. Although never is often better than immediately.
? ^^^^^^ ^^^
- 9. If the implementation is hard to explain, it's a bad idea.
? ------
+ 9. If the implementation is hard to code, it's most probably a bad idea.
? +++ +++++++++ +++++
- 10. If the implementation is easy to explain, it may be a good idea.
```
## Example 11: Compare List of Strings and Return Difference in Unified Diff Format (unified_diff())¶
As a part of our eleventh example, we are demonstrating how we can show the difference between two lists of strings in a unified format using unified_diff() method of difflib module. The unified difference format just shows lines that are changed plus a few lines around them to show context.
• unified_diff(a,b,fromfile='',tofile='',fromfiledate='',tofiledate='',n=3,lineterm='\n') - This method compares two list of strings and returns their difference in unified format. It also let us specify file names and dates for both lists. The n parameter accepts integer specifying number of lines to show around the lines which are changed for contextual purpose.
Our code for this example like many of our previous examples starts by reading two text files created earlier. It uses unified_diff() method this time to find out the difference and print it.
```import difflib
a = open("original.txt", "r").readlines()
b = open("modified.txt", "r").readlines()
diff = difflib.unified_diff(a,b,
fromfile="original.txt", tofile="modified.txt",
fromfiledate="2020-02-19", tofiledate="2020-02-20"
)
for line in diff:
print(line, end="")
```
```--- original.txt 2020-02-19
+++ modified.txt 2020-02-20
@@ -1,10 +1,9 @@
-2. Special cases aren't special enough to break the rules.
-3. Errors should never pass silently.
+1. Simplicy counts as well.
+2. Special cases aren't that special enough to break the rules.
+3. Errors shall never pass ever.
4. In the face of ambiguity, refuse the temptation to guess.
-5. There should be one-- and preferably only one --obvious way to do it.
+5. There should be one obvious way to do it.
6. Although that way may not be obvious at first unless you're Dutch.
7. Now is better than never.
-8. Although never is often better than *right* now.
-9. If the implementation is hard to explain, it's a bad idea.
-10. If the implementation is easy to explain, it may be a good idea.
+8. Although never is often better than immediately.
+9. If the implementation is hard to code, it's most probably a bad idea.
```
## Example 12: Retrieve Original File Contents/Strings from Difference (restore())¶
As a part of our twelfth example, we are demonstrating how we can generate an original list of strings based on the difference that we found out between them. The difflib module provides method named restore() for this purpose. The restore() method can generate original list of strings from difference generated from Differ instance or ndiff() method only.
• restore(difference, which) - It accepts the difference between sequences as the first parameter and an integer 1 or 2 as the second parameter specifying which sequence it wants to retrieve.
### 12.1: From Difference Generated from Differ Instance¶
Our code for this example first finds the difference between the contents of two files using Differ instance. It then uses this difference to find our contents of the first and second files both using restore() method. We have been using both files since many of our last examples.
```import difflib
a = open("original.txt", "r").readlines()
b = open("modified.txt", "r").readlines()
difference = difflib.Differ()
diff = difference.compare(a, b)
original_file_contents = difflib.restore(diff, 1)
print("============== Original File Contents ====================\n")
for line in original_file_contents:
print(line, end="")
difference = difflib.Differ()
diff = difference.compare(a, b)
modified_file_contents = difflib.restore(diff, 2)
print("\n\n============== Modified File Contents ====================\n")
for line in modified_file_contents:
print(line, end="")
```
```============== Original File Contents ====================
2. Special cases aren't special enough to break the rules.
3. Errors should never pass silently.
4. In the face of ambiguity, refuse the temptation to guess.
5. There should be one-- and preferably only one --obvious way to do it.
6. Although that way may not be obvious at first unless you're Dutch.
7. Now is better than never.
8. Although never is often better than *right* now.
9. If the implementation is hard to explain, it's a bad idea.
10. If the implementation is easy to explain, it may be a good idea.
============== Modified File Contents ====================
1. Simplicy counts as well.
2. Special cases aren't that special enough to break the rules.
3. Errors shall never pass ever.
4. In the face of ambiguity, refuse the temptation to guess.
5. There should be one obvious way to do it.
6. Although that way may not be obvious at first unless you're Dutch.
7. Now is better than never.
8. Although never is often better than immediately.
9. If the implementation is hard to code, it's most probably a bad idea.
```
### 12.2: From Difference Generated from ndiff() Method¶
Our code for this example first finds the difference between the contents of two files using ndiff() method. It then uses this difference to find our contents of the first and second files both using restore() method.
```import difflib
a = open("original.txt", "r").readlines()
b = open("modified.txt", "r").readlines()
diff = difflib.ndiff(a, b)
original_file_contents = difflib.restore(diff, 1)
print("============== Original File Contents ====================\n")
for line in original_file_contents:
print(line, end="")
diff = difflib.ndiff(a, b)
modified_file_contents = difflib.restore(diff, 2)
print("\n\n============== Modified File Contents ====================\n")
for line in modified_file_contents:
print(line, end="")
```
```============== Original File Contents ====================
2. Special cases aren't special enough to break the rules.
3. Errors should never pass silently.
4. In the face of ambiguity, refuse the temptation to guess.
5. There should be one-- and preferably only one --obvious way to do it.
6. Although that way may not be obvious at first unless you're Dutch.
7. Now is better than never.
8. Although never is often better than *right* now.
9. If the implementation is hard to explain, it's a bad idea.
10. If the implementation is easy to explain, it may be a good idea.
============== Modified File Contents ====================
1. Simplicy counts as well.
2. Special cases aren't that special enough to break the rules.
3. Errors shall never pass ever.
4. In the face of ambiguity, refuse the temptation to guess.
5. There should be one obvious way to do it.
6. Although that way may not be obvious at first unless you're Dutch.
7. Now is better than never.
8. Although never is often better than immediately.
9. If the implementation is hard to code, it's most probably a bad idea.
```
## Example 13: Find Words from the List which Nearly Matches Given Input Word (get_close_matches())¶
We'll use our thirteenth example to demonstrate how we can find out the list of words from the given list of words that somewhat matches (not compulsory 100% match ) a particular word given as input. We can do this using get_close_matches() method of difflib.
• get_close_matches(word,list_of_strings,n=3,cutoff=0.6) - This method accepts word and list of words as input and returns a list that has words that somewhat matches the input word.
• The parameter n accepts integer value specifying a maximum number of matching words to return. The returned list's length can be less than or equal to this number.
• The cutoff parameter accepts float in the range 0-1 specifying how much matching a word should be matched to the given input word to be considered as matching.
Our code for this example simply tries different values of parameters of method get_close_matches() to see how they impact the results.
```import difflib
matches = difflib.get_close_matches("micro", ["macro", "crow", "cream", "nonsense", "none"])
print(matches)
matches = difflib.get_close_matches("micro", ["macro", "crow", "cream", "nonsense", "none"], n=1)
print(matches)
matches = difflib.get_close_matches("micro", ["macro", "crow", "cream", "nonsense", "none"], n=3, cutoff=0.4)
print(matches)
matches = difflib.get_close_matches("micro", ["macro", "crow", "cream", "nonsense", "none"], n=2, cutoff=0.4)
print(matches)
```
```['macro', 'crow']
['macro']
['macro', 'crow', 'cream']
['macro', 'crow']
```
This ends our small tutorial explaining how we can compare sequences with different types of data using difflib module.
Sunny Solanki
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# $\pi$ Day 2015 Art Posters
2021 $\pi$ reminds us that good things grow for those who wait.' edition.
2019 $\pi$ has hundreds of digits, hundreds of languages and a special kids' edition.
2018 $\pi$ day stitches street maps into new destinations.
2017 $\pi$ day imagines the sky in a new way.
2016 $\pi$ approximation day wonders what would happen if about right was right.
2016 $\pi$ day sees digits really fall for each other.
2015 $\pi$ day maps transcendentally.
2014 $\pi$ approx day spirals into roughness.
2014 $\pi$ day hypnotizes you into looking.
2014 $\pi$ day
2013 $\pi$ day is where it started
Circular $\pi$ art and other distractions
On March 14th celebrate $\pi$ Day. Hug $\pi$—find a way to do it.
For those who favour $\tau=2\pi$ will have to postpone celebrations until July 26th. That's what you get for thinking that $\pi$ is wrong. I sympathize with this position and have $\tau$ day art too!
If you're not into details, you may opt to party on July 22nd, which is $\pi$ approximation day ($\pi$ ≈ 22/7). It's 20% more accurate that the official $\pi$ day!
Finally, if you believe that $\pi = 3$, you should read why $\pi$ is not equal to 3.
Most of the art is available for purchase as framed prints and, yes, even pillows. Sleep's never been more important — I take custom requests.
Not a circle in sight in the 2015 $\pi$ day art. Try to figure out how up to 612,330 digits are encoded before reading about the method. $\pi$'s transcendental friends $\phi$ and $e$ are there too—golden and natural. Get it?
This year's $\pi$ day is particularly special. The digits of time specify a precise time if the date is encoded in North American day-month-year convention: 3-14-15 9:26:53.
The art has been featured in Ana Swanson's Wonkblog article at the Washington Post—10 Stunning Images Show The Beauty Hidden in $\pi$.
We begin with a square and progressively divide it. At each stage, the digit in $pi$ is used to determine how many lines are used in the division. The thickness of the lines used for the divisions can be attenuated for higher levels to give the treemap some texture.
Representing a number using a tree map. Each digit of the number is used to successively divide a shape, such as a square. (zoom)
This method of encoding data is known as treemapping. Typically, it is used to encode hierarchical information, such as hard disk spac usage, where the divisions correspond to the total size of files within directories.
At each level of the tree map, more digits are encoded. Shown here are tree maps for $pi$ for the first 6 levels of division. (zoom)
This kind of treemap can be made from any number. Below I show 6 level maps for $pi$, $phi$ (Golden ratio) and $e$ (base of natural logarithm).
At each level of the tree map, more digits are encoded. Shown here are tree maps for $pi$ for the first 6 levels of division. (zoom)
The number of digits per level, $n_i$ and total digits, $N_i$ in the tree map for $pi$, $phi$ and $e$ is shown below for levels $i = 1 .. 6$.
$PI PHI e i n_i N_i n_i N_i n_i N_i 1 1 1 1 1 1 1 2 4 5 2 3 3 4 3 15 20 9 12 19 23 4 98 118 59 71 96 119 5 548 666 330 401 574 693 6 2962 3628 1857 2258 3162 3855 7 16616 20244 10041 12299 17541 21396 8 91225 111469 9 500861 612330$
## Dividing the map
In all the treemaps above, the divisions were made uniformly for each rectangle. With uniform division, the lines that divide a shape are evenly spaced. With randomized division, the placement of lines is randomized, while still ensuring that lines do not coincide.
A multiplier, such as $phi$ (Golden Ratio), can be used to control the division. In this case, the first division is made at 1/$phi$ (0.62/0.38 split) and the remaining rectangle (0.38) is further divided at $/$phi$(0.24/0.14 split). The divisions of each shape can be influenced by another number and the level at which the division is performed. (zoom) Using a non-uniform multipler is one way to embed another number in the art. When a multiplier like$phi$is used, divisions at the top levels create very large rectangles. To attenuate this, the effect of the multiplier can be weighted by the level. Regardless of what multiplier is used, the first level is always uniformly divided. Division at subsequent levels incorporates more of the multiplier effect. The orientation of the division can be uniform (same for a layer and alternating across layers), alternating (alternating across and within a layer) or random. This modification has an effect only if the divisions are not uniform. The divisions of each shape can be influenced by another number and the level at which the division is performed. (zoom) ## Adjusting line thickness To emphasize the layers, a different line thickness can be used. When lines are drawn progressively thinner with each layer, detail is controlled and the map has more texture. When all lines have the same thickness, it is harder to distinguish levels. The divisions of each shape can be influenced by another number and the level at which the division is performed. (zoom) You could see this as a challenge! Below I show the treemaps for$pi$,$phi$and$e$with and without stroke modulation. The divisions of each shape can be influenced by another number and the level at which the division is performed. (zoom) When displayed at a low resolution (the image below is 620 pixels across), shapes at higher levels appear darker because the distance between the lines within is close to (or smaller) than a pixel. By matching the line thickness to the image resolution, you can control how dark the smallest divisions appear. The divisions of each shape can be influenced by another number and the level at which the division is performed. (zoom) ## Adding color Adding color can make things better, or worse. Dropping color randomly, without respect for the level structure of the treemap, creates a mess. We can rescue things by increasing the probability that a given rectangle will be made transparent—this will allow the color of the rectangle below to show through. Additionally, by drawing the layers in increasing order, smaller rectangles are drawn on top of bigger ones, giving a sense of recursive subdivision. The divisions of each shape can be influenced by another number and the level at which the division is performed. (zoom) Because the color is assigned randomly, various instances of the treemap can be made. The maps below have the same proportion of colors and transparency (same as the first image in second row in the figure above) and vary only by the random seed to pick colors. Different instances of 5 level$pi$treemaps. The proportion of transparent, white, yellow, red and blue shapes is 20:1:1:1:1. (zoom) ## Coloring using adjacency graph The color assignments above were random. For each shape the probability of choosing a given color (transparent, white, yellow, red, blue) was the same. Color choice for a shape can also be influenced by the color of neighbouring shapes. To do this, we need to create a graph that captures the adjacency relationship between all the shapes at each level. Below I show the first 4 levels of the$pi$treemap and their adjacency graphs. In each graph, the node corresponds to a shape and an edge between nodes indicates that the shapes share a part of their edge. Shapes that touch only at a corner are not considered adjacent. Different instances of 5 level$pi$treemaps. The proportion of transparent, white, yellow, red and blue shapes is 20:1:1:1:1. (zoom) One way in which the graphs can be used is to attempt to color each layer using at most 4 colors. The 4 color theorem tells us that only 4 unique colors are required to color maps such as these in a way that no two neighbouring shapes have the same color. It turns out that the full algorithm of coloring a map with only 4 colors is complicated, but reasonably simple options exist.. For the maps here, I used the DSATUR (maximum degree of saturation) approach. Different instances of 5 level$pi` treemaps. The proportion of transparent, white, yellow, red and blue shapes is 20:1:1:1:1. (zoom)
The DSATUR algorithm works well, but does not guarantee a 4-color solution. It performs no backtracking. If you look carefully, one of the rectangles in the 4th layer (top right quadrant in the graph) required a 5th color (shown black).
news + thoughts
# Convolutional neural networks
Thu 17-08-2023
Nature uses only the longest threads to weave her patterns, so that each small piece of her fabric reveals the organization of the entire tapestry. – Richard Feynman
Following up on our Neural network primer column, this month we explore a different kind of network architecture: a convolutional network.
The convolutional network replaces the hidden layer of a fully connected network (FCN) with one or more filters (a kind of neuron that looks at the input within a narrow window).
Nature Methods Points of Significance column: Convolutional neural networks. (read)
Even through convolutional networks have far fewer neurons that an FCN, they can perform substantially better for certain kinds of problems, such as sequence motif detection.
Derry, A., Krzywinski, M & Altman, N. (2023) Points of significance: Convolutional neural networks. Nature Methods 20:.
Derry, A., Krzywinski, M. & Altman, N. (2023) Points of significance: Neural network primer. Nature Methods 20:165–167.
Lever, J., Krzywinski, M. & Altman, N. (2016) Points of significance: Logistic regression. Nature Methods 13:541–542.
# Neural network primer
Tue 10-01-2023
Nature is often hidden, sometimes overcome, seldom extinguished. —Francis Bacon
In the first of a series of columns about neural networks, we introduce them with an intuitive approach that draws from our discussion about logistic regression.
Nature Methods Points of Significance column: Neural network primer. (read)
Simple neural networks are just a chain of linear regressions. And, although neural network models can get very complicated, their essence can be understood in terms of relatively basic principles.
We show how neural network components (neurons) can be arranged in the network and discuss the ideas of hidden layers. Using a simple data set we show how even a 3-neuron neural network can already model relatively complicated data patterns.
Derry, A., Krzywinski, M & Altman, N. (2023) Points of significance: Neural network primer. Nature Methods 20:165–167.
Lever, J., Krzywinski, M. & Altman, N. (2016) Points of significance: Logistic regression. Nature Methods 13:541–542.
# Cell Genomics cover
Mon 16-01-2023
Our cover on the 11 January 2023 Cell Genomics issue depicts the process of determining the parent-of-origin using differential methylation of alleles at imprinted regions (iDMRs) is imagined as a circuit.
Designed in collaboration with with Carlos Urzua.
Our Cell Genomics cover depicts parent-of-origin assignment as a circuit (volume 3, issue 1, 11 January 2023). (more)
Akbari, V. et al. Parent-of-origin detection and chromosome-scale haplotyping using long-read DNA methylation sequencing and Strand-seq (2023) Cell Genomics 3(1).
Browse my gallery of cover designs.
A catalogue of my journal and magazine cover designs. (more)
Thu 05-01-2023
My cover design on the 6 January 2023 Science Advances issue depicts DNA sequencing read translation in high-dimensional space. The image showss 672 bases of sequencing barcodes generated by three different single-cell RNA sequencing platforms were encoded as oriented triangles on the faces of three 7-dimensional cubes.
My Science Advances cover that encodes sequence onto hypercubes (volume 9, issue 1, 6 January 2023). (more)
Kijima, Y. et al. A universal sequencing read interpreter (2023) Science Advances 9.
Browse my gallery of cover designs.
A catalogue of my journal and magazine cover designs. (more)
# Regression modeling of time-to-event data with censoring
Thu 17-08-2023
If you sit on the sofa for your entire life, you’re running a higher risk of getting heart disease and cancer. —Alex Honnold, American rock climber
In a follow-up to our Survival analysis — time-to-event data and censoring article, we look at how regression can be used to account for additional risk factors in survival analysis.
We explore accelerated failure time regression (AFTR) and the Cox Proportional Hazards model (Cox PH).
Nature Methods Points of Significance column: Regression modeling of time-to-event data with censoring. (read)
Dey, T., Lipsitz, S.R., Cooper, Z., Trinh, Q., Krzywinski, M & Altman, N. (2022) Points of significance: Regression modeling of time-to-event data with censoring. Nature Methods 19:1513–1515.
# Music video for Max Cooper's Ascent
Tue 25-10-2022
My 5-dimensional animation sets the visual stage for Max Cooper's Ascent from the album Unspoken Words. I have previously collaborated with Max on telling a story about infinity for his Yearning for the Infinite album.
I provide a walkthrough the video, describe the animation system I created to generate the frames, and show you all the keyframes
Frame 4897 from the music video of Max Cooper's Asent.
The video recently premiered on YouTube.
Renders of the full scene are available as NFTs. | 3,185 | 13,351 | {"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} | 2.828125 | 3 | CC-MAIN-2023-40 | latest | en | 0.900936 |
https://tutorpro.us/acct505-capital-budgeting-problem/ | 1,716,218,447,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058291.13/warc/CC-MAIN-20240520142329-20240520172329-00825.warc.gz | 538,924,414 | 12,478 | # ACCT505 Capital Budgeting Problem
### ACCT505 Part B Capital Budgeting Problem
Capital Budgeting Decision
Here is Part B:
Clark Paints: The production department has been investigating possible ways to trim total production costs. One possibility currently being examined is to make the paint cans instead of purchasing them. The equipment needed would cost \$200,000, with a disposal value of \$40,000, and would be able to produce 5,500,000 cans over the life of the machinery.
The production department estimates that approximately 1,100,000 cans would be needed for each of the next 5 years. The company would hire three new employees. These three individuals would be full-time employees working 2,000 hours per year and earning \$12.00 per hour. They would also receive
the same benefits as other production employees, 18% of wages in addition to \$2,500 of health benefits.
It is estimated that the raw materials will cost 25 per can and that other variable costs would be 5 per can. Because there is currently unused space in the factory, no additional fixed costs would be incurred if this proposal is accepted.
It is expected that cans would cost 45 each if purchased from the current supplier. The company's minimum rate of return (hurdle rate) has been determined to be 12% for all new projects, and the current tax rate of 35% is anticipated to remain unchanged. The pricing for a gallon of paint as well as number of units sold will not be affected by this decision. The unit-of production depreciation method would be used if the new equipment is purchased.
Required:
1. Based on the above information and using Excel, calculate the following items for this proposed equipment purchase.
o Annual cash flows over the expected life of the equipment
o Payback period
o Annual rate of return
o Net present value
o Internal rate of return
2. Would you recommend the acceptance of this proposal? Why or why not. Prepare a short double-spaced Word paper elaborating and supporting your answer
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No matter what kind of academic paper you need and how urgent you need it, you are welcome to choose your academic level and the type of your paper at an affordable price. We take care of all your paper needs and give a 24/7 customer care support system. | 848 | 3,995 | {"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} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.939133 |
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