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https://math.eretrandre.org/tetrationforum/printthread.php?tid=695 | 1,600,670,549,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198942.13/warc/CC-MAIN-20200921050331-20200921080331-00065.warc.gz | 517,147,846 | 3,997 | Tetration of 2 and Aleph_0 - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Tetration of 2 and Aleph_0 (/showthread.php?tid=695) Tetration of 2 and Aleph_0 - jht9663 - 09/06/2011 Assuming ZFC (Zermelo-Fraenkel set theory with the Axiom of Choice), the continuum hypothesis proposes that 2^Aleph_0 = Aleph_1. Does anyone have any insight into the tetration of 2 and Aleph_0 ? I have no idea as to where to start on this problem. But I feel that it is important because it could lead to the recognition of new types of infinities. Also, please excuse my lack of formatting skills. I would greatly appreciate any help in producing formatted code. Thanks, Hassler Thurston RE: Tetration of 2 and Aleph_0 - JmsNxn - 09/06/2011 I always thought that was just convenience of notation for some other set operation; I didn't know $2^{\aleph_0}$ actually meant two times two $\aleph_0$ amount of times. But as far as I know there isn't much research into tetrating $\aleph_0$ And to produce code you'll need to learn Latex, it's a rather simple html-like code that most math forums have to format formulae. RE: Tetration of 2 and Aleph_0 - tommy1729 - 09/07/2011 VERY controversial subject. many flamewars going on about this. my opinion is this 2^^aleph_0 = aleph_aleph_0 and further 2^(aleph_aleph_0) = aleph_aleph_0 notice aleph_0 + 1 = aleph_0 and 2^^(aleph_aleph_0) = aleph_aleph_0 notice 2 * aleph_0 = aleph_0 aleph_aleph_1 or higher does not exist. notice that defining what aleph_aleph_1 is the diagonal argument / powerset of is not possible ... ( which is imho required to assume existance of aleph_aleph_1 ) regards tommy1729 RE: Tetration of 2 and Aleph_0 - jht9663 - 09/07/2011 So essentially [$\aleph_{\aleph_{0}}+1=\aleph_{\aleph_{0}}$], [$2*\aleph_{\aleph_{0}}=\aleph_{\aleph_{0}}$], [$2^\aleph_{\aleph_{0}}=\aleph_{\aleph_{0}}$], and 2^^[$\aleph_{\aleph_{0}}=\aleph_{\aleph_{0}}$]. However I do not agree that [$\aleph_{\aleph_{1}}] does not exist. My heuristic reasoning is: 1 (the first integer past the addition identity) + 0 = 1 (the first integer past 0) (assuming the Continuum Hypothesis) 2 (the first integer past the exponentiation identity) ^ [$\aleph_{0}$] = [$\aleph_{1}$] (1 being the first integer past 0) if these are true then 2 (the first integer past the pentation identity) ^^^ [$\aleph_{\aleph_{0}}$] = [$\aleph_{\aleph_{1}}$] (1 being the first integer past 0) and you could extend the pattern. Of course I have no other reasons to believe that the third statement is true, as one would have to prove that there does not exist a bijection from [$\aleph_{\aleph_{0}}$] to 2^^^[$\aleph_{aleph_{0}}\$]. Also, where would be a place I could go to on the internet to find more discussion on this topic? Thanks, Hassler Thurston RE: Tetration of 2 and Aleph_0 - jht9663 - 09/07/2011 Darn it- my code didn't work. Can anybody show me the correct formatted code for some statements I just made? Thanks, Hassler RE: Tetration of 2 and Aleph_0 - sheldonison - 09/07/2011 (09/07/2011, 03:34 PM)jht9663 Wrote: Darn it- my code didn't work. Can anybody show me the correct formatted code for some statements I just made? Thanks, Hassler Try putting tex codes around your math statements Code:$$\aleph_0$$$\aleph_0$ I'm no expert on set theory, but on a humorous note (not mathematically sound), assuming the generalized continuum hypothesis, then what happens if we take the slog of an aleph number? $\aleph_1=2^{\aleph_0}$ which implies $\text{slog}_2(\aleph_1) = \text{slog}_2(\aleph_0)+1=\aleph_0$ And for any integer n where $\aleph_{n+1}=2^{\aleph_n}$, then $\text{slog}(\aleph_n)=\aleph_0$ Perhaps $\text{slog}(\aleph_{\aleph_1})=\aleph_1$ - Shel RE: Tetration of 2 and Aleph_0 - sheldonison - 09/09/2011 (09/07/2011, 08:47 PM)sheldonison Wrote: I'm no expert on set theory, but on a humorous note (not mathematically sound), assuming the generalized continuum hypothesis, then what happens if we take the slog of an aleph number? $\aleph_1=2^{\aleph_0}$ which implies $\text{slog}_2(\aleph_1) = \text{slog}_2(\aleph_0)+1=\aleph_0$ And for any integer n where $\aleph_{n+1}=2^{\aleph_n}$, then $\text{slog}(\aleph_n)=\aleph_0$ Perhaps $\text{slog}(\aleph_{\aleph_1})=\aleph_1$ - ShelIt turns out aleph and beth numbers should be indexed by ordinal numbers. The ordinal number equivalent to $\aleph_0=\omega$ and the ordinal number equivalent to $\aleph_1=\omega_1$ But I have no idea whether slog or sexp have any meaning for $\aleph$ numbers. The other possibility would be to see if sexp/slog would be more applicable to ordinal numbers. But the exponentiation rules for ordinal arithmetic say that $2^\omega=\omega$ I'm unsure of what $\text{sexp}(\omega)$ would be; the result might just be $\omega$. http://en.wikipedia.org/wiki/Ordinal_arithmetic http://en.wikipedia.org/wiki/Aleph_number RE: Tetration of 2 and Aleph_0 - tommy1729 - 09/10/2011 personally i reject ordinals , as you might have read elsewhere. i feel inaccessible ordinals are far away from tetration btw... tommy1729 RE: Tetration of 2 and Aleph_0 - tommy1729 - 11/13/2011 the large cardinals and large ordinals are very axiomatic in nature. so without proofs of bijections or the lack of bijections it is pretty hard to talk about that. ( although i do like the comments here ) in my not so humble opinion its also a matter of taste because of the above and because of the possible use of ZF©. ( which has not been proven consistant ! ) i already commented my personal large cardinal axioms ( kinda ) , but i feel it is more intresting to consider small cardinalities. to be specific : what is the cardinality of f(n) where n lies between n and 2^n ? since cardinalities are not influenced by powers card ( Q ) = card ( Q ^ finite ) we can write our question as for n <<< f(n) <<< 2^n card(f(n)) = ? the reason i dont want to get close to n or 2^n is the question : is there a cardinality between n and 2^n ? in other words : the continuum hypothesis. in stardard math and standard combinatorics , we usually do not work with functions f : n <<< f(n) <<< 2^n. but on the tetration forum they occur very often. card(floor(sexp(slog(n)+1/(24+ln(ln(n)))))) = ? card(floor(n + n^4/4! + n^9/9! + n^16/16! + ...)) = ? regards tommy1729 | 1,968 | 6,496 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 20, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-40 | latest | en | 0.827523 |
https://www.onlinemathlearning.com/dividing-decimals.html | 1,721,823,718,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518277.99/warc/CC-MAIN-20240724110315-20240724140315-00688.warc.gz | 769,631,631 | 9,226 | # Dividing a Decimal by another Decimal
In these lessons, we will look at dividing a decimal number by another decimal number.
Related Pages
Dividing Decimals 2
Math Worksheets
How to Divide Decimals? When one decimal number is divided by another decimal number,
Step 1: Change the divisor to a whole number by moving the decimal point to the right until after the last digit.
Step 2: The decimal point of the dividend must also be moved by the same number of decimal places.
Step 3: Perform the division in the same way as dividing by a whole number. Place the decimal point of the quotient (answer) directly above the decimal point of the dividend.
The following diagram shows how to divide a decimal by a decimal. Scroll down the page for more examples and solutions.
Example:
Evaluate 5.421 ÷ 0.03
Solution:
The divisor, 0.03, is moved 2 decimal places to form the whole number 3. The dividend, 5.421, is then also moved 2 decimal places to form 542.1. We then perform the division as shown:
The following videos show how to divide a decimal number by another decimal number.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 286 | 1,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-30 | latest | en | 0.854605 |
http://en.m.wikipedia.org/wiki/G-factor_(physics) | 1,371,663,849,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709000375/warc/CC-MAIN-20130516125640-00092-ip-10-60-113-184.ec2.internal.warc.gz | 87,990,748 | 9,894 | # g-factor (physics)
For the acceleration-related quantity in mechanics, see g-force.
A g-factor (also called g value or dimensionless magnetic moment) is a dimensionless quantity which characterizes the magnetic moment and gyromagnetic ratio of a particle or nucleus. It is essentially a proportionality constant that relates the observed magnetic moment μ of a particle to the appropriate angular momentum quantum number and the appropriate fundamental quantum unit of magnetism, usually the Bohr magneton or nuclear magneton.
## Calculation
### Electron g-factors
There are three magnetic moments associated with an electron: One from its spin angular momentum, one from its orbital angular momentum, and one from its total angular momentum (the quantum-mechanical sum of those two components). Corresponding to these three moments are three different g-factors:
#### Electron spin g-factor
The most famous of these is the electron spin g-factor (more often called simply the electron g-factor), ge, defined by
$\boldsymbol{\mu}_S = \frac{g_e\mu_\mathrm{B}}{\hbar}\boldsymbol{S}$
where μS is the total magnetic moment resulting from the spin of an electron, S is its spin angular momentum, and μB is the Bohr magneton. In atomic physics, the electron spin g-factor is often defined as the absolute value or negative of ge:
$g_S = |g_e| = -g_e.$
The z-component of the magnetic moment then becomes
$\mu_z=-g_S \mu_\mathrm{B} m_s$
The value gS is roughly equal to 2.002319, and is known to extraordinary precision.[1][2] The reason it is not precisely two is explained by quantum electrodynamics calculation of the anomalous magnetic dipole moment.[3]
#### Electron orbital g-factor
Secondly, the electron orbital g-factor, gL, is defined by
$\boldsymbol{\mu}_L = -\frac{g_L \mu_\mathrm{B}}{\hbar}\boldsymbol{L}$
where μL is the total magnetic moment resulting from the orbital angular momentum of an electron, L is the magnitude of its orbital angular momentum, and μB is the Bohr magneton. The value of gL is exactly equal to one, by a quantum-mechanical argument analogous to the derivation of the classical magnetogyric ratio. For an electron in an orbital with a magnetic quantum number ml, the z-component of the orbital angular momentum is
$\mu_z=g_L \mu_\mathrm{B} m_l$
which, since gL = 1, is just μBml
#### Total angular momentum (Landé) g-factor
Thirdly, the Landé g-factor, gJ, is defined by
$\boldsymbol{\mu} = -\frac{g_J \mu_\mathrm{B} }{\hbar}\boldsymbol{J}$
where μ is the total magnetic moment resulting from both spin and orbital angular momentum of an electron, J = L+S is its total angular momentum, and μB is the Bohr magneton. The value of gJ is related to gL and gS by a quantum-mechanical argument; see the article Landé g-factor.
### Nucleon and nucleus g-factors
Protons, neutrons, and many nuclei have spin and magnetic moments, and therefore associated g-factors. The formula conventionally used is
$\boldsymbol{\mu} = \frac{g \mu_\mathrm{N}}{\hbar}\boldsymbol{I}$
where μ is the magnetic moment resulting from the nuclear spin, I is the nuclear spin angular momentum, μN is the nuclear magneton, and g is the effective g-factor.
### Muon g-factor
If supersymmetry is realized in nature, there will be corrections to g-2 of the muon due to loop diagrams involving the new particles. Amongst the leading corrections are those depicted here: a neutralino and a smuon loop, and a chargino and a muon sneutrino loop. This represents an example of "beyond the Standard-Model" physics that might contribute to g-2.
The muon, like the electron has a g-factor from its spin, given by the equation
$\mathbf{\mu} = \frac{ge}{2m_\mu}\mathbf{S}$
where μ is the magnetic moment resulting from the muon’s spin, S is the spin angular momentum, and mμ is the muon mass.
The fact that the muon g-factor is not quite the same as the electron g-factor is mostly explained by quantum electrodynamics and its calculation of the anomalous magnetic dipole moment. Almost all of the small difference between the two values (99.96% of it) is due to a well-understood lack of a heavy-particle diagrams contributing to the probability for emission of a photon representing the magnetic dipole field, which are present for muons, but not electrons, in QED theory. These are entirely a result of the mass difference between the particles.
However, not all of the difference between the g-factors for electrons and muons are exactly explained by the quantum electrodynamics Standard Model. The muon g-factor can, at least in theory, be affected by physics beyond the Standard Model, so it has been measured very precisely, in particular at the Brookhaven National Laboratory. As of November 2006, the experimentally measured value is 2.0023318416(13), compared to the theoretical prediction of 2.0023318361(10).[4] This is a difference of 3.4 standard deviations, suggesting beyond-the-Standard-Model physics may be having an effect.
↑Jump back a section
## Measured g-factor values
Particle g-factor Uncertainty
Electron $g_\mathrm{e}$ −2.00231930436153 0.00000000000053
Neutron $g_\mathrm{n}$ −3.82608545 0.00000090
Proton $g_\mathrm{p}$ 5.585694713 0.000000046
Muon $g_{\mu}$ −2.0023318414 0.0000000012
Currently accepted NIST g-factor values [5]
The electron g-factor is one of the most precisely measured values in physics, with a relative standard uncertainty of 2.6 x 10-13.
↑Jump back a section
## Notes and references
1. ^ Gabrielse, Gerald; Hanneke, David (October 2006). "Precision pins down the electron's magnetism". CERN Courier 46 (8): 35–37.
2. ^ Odom, B.; Hanneke, D.; d’Urso, B.; Gabrielse, G. (2006). "New measurement of the electron magnetic moment using a one-electron quantum cyclotron". Physical Review Letters 97 (3): 030801. Bibcode:2006PhRvL..97c0801O. doi:10.1103/PhysRevLett.97.030801. PMID 16907490.
3. ^ Brodsky, S; Franke, V; Hiller, J; McCartor, G; Paston, S; Prokhvatilov, E (2004). "A nonperturbative calculation of the electron's magnetic moment". Nuclear Physics B 703 (1–2): 333–362. arXiv:hep-ph/0406325. Bibcode:2004NuPhB.703..333B. doi:10.1016/j.nuclphysb.2004.10.027.
4. ^ Hagiwara, K.; Martin,, A. D.; Nomura, Daisuke; Teubner, T. (2006). "Improved predictions for g-2 of the muon and alpha(QED)(M(Z)**2)". Physics Letters B 649 (2–3): 173–179. arXiv:hep-ph/0611102. Bibcode:2007PhLB..649..173H. doi:10.1016/j.physletb.2007.04.012.
5. ^
↑Jump back a section | 1,727 | 6,439 | {"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": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2013-20 | latest | en | 0.874285 |
https://supportforums.cisco.com/discussion/10251796/seeking-clarification-ipv6-link-site-local-addresses-please | 1,484,729,545,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280242.65/warc/CC-MAIN-20170116095120-00513-ip-10-171-10-70.ec2.internal.warc.gz | 852,616,604 | 24,872 | # Seeking clarification in IPv6 Link & Site local addresses please
Answered Question
Hi All
Just wondering if anybody with IPv6 knowledge can put me straight. I am currently studying for my CCNP (BSCI) and am just going over the IPv6 stuff again.
Now I understand that for the CCNP I will likely just need to know the basic theory of IPv6, however I cannot get my head around the Link-Local and Site-Local addresses.
Routing TCP/IP Vol 1 (Jeff Doyle) 2nd edition, page 54, has a table showing the high order bits of various IPv6 addresses and lists the following.
Link-Local = 1111 1110 10 which equates to FE80::/10
Site-Local = 1111 1110 11 which equates to FEC0::/10
Now the part I can not get my head around is this.
The first 10 high order bits equate to 2 1/2 (10 bits) nibbles, but for the Link-Local address to be FE80 would require 4 nibbles (16 bits) or
1111 1110 1000 0000. Like wise with the Site-Local address of FEC0 which would be 1111 1110 1100 0000. This would mean that the prefix should be /16 if remaining 6 bits can not be turned on.
So I am trying to understand what happens to the remaining 6 bits in these Local IPv6 addresses or to the IPv6 addresses from FE81 to FEBF and from FEC1 to FECF? Are they just ignored?
If so it would appear that nothing was learned from what happened with the initial wasteful assignment of IPv4 addresses!!
Can anybody put me out of my misery please :)
Best Regards,
Michael
0 votes
Correct Answer by swmorris about 9 years 3 weeks ago
It's a little trickier than that. I never knew why they listed them as /10's because you're right, it implies things can be set however you want, but they can't.
Link local is specifically:
FE80:0000:0000:0000:: (eui-64 or defined address)
Site local is specifically:
FEC0:0000:0000:(subnet):: (eui-64 or defined address)
To further muck things up, the site local addressing has been deprecated in the real world. There is a new set called ULA (unique local address) out there, but it won't be covered in the CCIP exams yet, so don't stress out on that part! (grin)
HTH,
Scott
[email protected]
Overall Rating: 5 (1 ratings)
## Replies
Correct Answer
swmorris Sat, 12/29/2007 - 07:41
It's a little trickier than that. I never knew why they listed them as /10's because you're right, it implies things can be set however you want, but they can't.
Link local is specifically:
FE80:0000:0000:0000:: (eui-64 or defined address)
Site local is specifically:
FEC0:0000:0000:(subnet):: (eui-64 or defined address)
To further muck things up, the site local addressing has been deprecated in the real world. There is a new set called ULA (unique local address) out there, but it won't be covered in the CCIP exams yet, so don't stress out on that part! (grin)
HTH,
Scott
[email protected]
Hi Scott
Cheers, Many Thanks for the response and information, though it does seem really weird.
Looks like nothing was really learned from the problems which arose after the way IPv4 was initially assigned (Class A & B networks given out willy nilly). History repeating itself it seems.
Best Regards & best wishes for 2008 and beyond,
Michael
swmorris Sat, 12/29/2007 - 10:26
Yes and no. The pool is much bigger now. If they were handing out addresses in /8 allocations at a time, I'd agree with you.
but Large ISPs are getting /32 allocations, which alone allows 4.2 billion separate allocations at this level. Each allocation to an enterprise is to be /48, which means that a single allocation to an ISP can feed 65,536 customers of theirs.
Each network being a /64 still allows any enterprise to have 65,536 subnets.
There should be plenty of allocations to go around even though it seems like we're wasting lots of space! :) It's all to handle the magical address assignments of the EUI-64 node addressing!
Cheers,
Scott
[email protected]
royalblues Mon, 12/31/2007 - 01:41
Well the link local address just says that the first 10 bits should be FE80.
FE80 = 1111 1110 1000 0000. it does not mean that you cannot use the rest. I think all it means is that any Ipv6 address that can be summarised to FE80/10 would become a link local address
For example you can always configure the following ipv6 address FE8E::1 as link local.
interface fa 0/1
ipv6 enable
ipv6 address FE8E::1 link local
The above address is valid as it can be summarized to FE80::/10. If you try to make any other address as link local which cannot be summarized to the abobe you would get the following error
Router(config-if)#ipv6 ADDress 2001:A::1 LInk-local
% Invalid link-local address
HTH
Narayan
Hi Narayan
Thank you for your post and the information contained within.
When I went through IPv6 initially for my CCNP (BSCI) studies I understood that as the Link-Local and Site-Local addresses had a prefix of /10. I thought that once I kept the initial 10 bits of the address as specified in the RFC, I could change any and all of the other 118 bits in the address.
However my study material (which now also includes "Routing TCP/IP Vol 1&2, by Jeff Doyle) also states that Link-Local addresses will always start with "FE80" and Site-Local addresses will always start with "FEC0". This accounts for the first 16 bits, which to me (as just a CCNA with knowledge of IPv4 only) meant that the Link-Local and Site-Local addresses should actually have a prefix of /16.
After further investigation into IPv6 I am further confused as it appears that the Link and Site local addresses are derived by using the 48 bit MAC address of an interface, injecting "FFFE" between the OUI portion and the local portion and then inverting the 7th bit to reach the modified EUI-64 identifier.
This in turn means that the Link and Site Local addresses would have a prefix of /64, as the actual addresses would start off with
FE80:0000:0000:0000: or FE80::
FEC0:0000:0000:0000: or FEC0::.
Also while searching the web for more information I came across this site
http://www.tcpipguide.com/free/t_IPv6SpecialAddressesReservedPrivateLinkLocalSiteLo-3.htm
which suggests as you mentioned above that all addresses between FE80:: and FEBF:FFFF:FFFF:FFFF:FFFF:FFFF:FFFF:FFFF can be Link-Local addresses and that all addresses between FEC0:: and FEFF:FFFF:FFFF:FFFF:FFFF:FFFF:FFFF:FFFF can be Site-Local addresses.
I then read through RFC3513 (IPv6 Addressing Architecture) http://www.faqs.org/rfcs/rfc3513.html and can find no mention of these "ranges" for the Link-Local and Site-Local addresses. It seems to suggest in section 2.5.6 (and I am no expert at reading and understanding RFC's) the the Link-Local address has the first 10 bits of "1111111010" the next 54 bits set to "0" and the final 64 bits are the interface ID (modified EUI-64 ID). Again to me this suggests that the prefix should be /64
However with the Site-Local address the first 10 bits are "1111111011" the next 54 bits are the "Subnet ID" and the final 64 bits are the interface ID (Modified EUI-64 ID) which again suggests to me a prefix of /64, but also that I can manipulate the middle 54 bits.
I think the more I try to understand this the more I confuse myself. I know this is probably gone way outside what I am expected to know for my BSCI exam, but I just like to understand how things that I read in my studies are arrived at. I don't like just taking things on blind faith :)
However I think after reading the RFC that I am also getting to hung up on the fact that the texts I am reading use a /10 prefix for these addresses. I understand the the Link-Local addresses are like the addresses Microsoft OS's use when they cannot get a DHCP addresses (169.254/16) and that Site-Local addresses are equivalent to RFC1918 private addresses.
Therefore I can except that the Link-local addresses will always start FE80:0000:0000:0000: and I am assuming that the lowest Site-Local address will start with FEC0:0000:0000:0000 but can be anything up to FEFF:FFFF:FFFF:FFFF, which will define the subnet.
Sorry for such a long post, but I see now why network administrators are not rushing to implement IPv6 :) I think it will take time to get my head around this.
Again thank you for your input.
Best Regards and Best Wishes for the 2008 and beyond,
Michael
swmorris Mon, 12/31/2007 - 10:00
The address you typed in can most certainly be summarized to the /10.
the point is that the ENTIRE first 64 bits are already set. Most are set to a 0-bit, but they are indeed set (why your :A: part wouldn't be valid).
But yes, you are correct that you can specify your own link local address (in case you hate memorizing MAC addresses!). but you only have control over the last 64 bits.
So "ipv6 address fe80::A:1 link-local" would work perfectly fine as the first 64 bits are left alone per the spec.
HTH,
Scott
[email protected]
royalblues Tue, 01/01/2008 - 00:01
Scott,
I was referring to the same thing that any address that can be summarized to /10 would be link local.
What i referred to was if you chnage the first 10 bits to anything other than the FE80, it cannot be a link local anymore
Also not only fe80::A:1 but even fe80:A::1 would work fine (atleast it does on a cisco router)
I may be wrong but thats what seems logical with the configuration error you get on the router
Narayan
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# CS 253: Algorithms Chapter 22 Graphs Credit: Dr. George Bebis.
## Presentation on theme: "CS 253: Algorithms Chapter 22 Graphs Credit: Dr. George Bebis."— Presentation transcript:
CS 253: Algorithms Chapter 22 Graphs Credit: Dr. George Bebis
Graphs Definition = a set of nodes (vertices) with edges (links) between them. G = (V, E) - graph V = set of vertices V = n E = set of edges E = m Subset of V x V ={(u,v): u V, v V} 1 2 3 4
Applications Applications that involve not only a set of items, but also the connections between them Computer networks Maps Hypertext Circuits
Terminology Complete graph Subgraph Path from v to w Length of a path
A graph with an edge between each pair of vertices Subgraph A graph (V’, E’) such that V’V and E’E Path from v to w A sequence of vertices <v0, v1, …, vk> such that v0=v and vk=w Length of a path Number of edges in the path 1 2 3 4 path from v1 to v4 <v1, v2, v4>
Terminology (cont’d) w is reachable from v Simple path Cycles
If there is a path from v to w Simple path All the vertices in the path are distinct Cycles A path <v0, v1, …, vk> forms a cycle if v0=vk and k≥2 Acyclic graph A graph without any cycles 1 2 3 4 cycle from v1 to v1 <v1, v2, v3,v1>
Terminology (cont’d)
Terminology (cont’d) A bipartite graph is an undirected graph
G = (V, E) in which V = V1 + V2 and there are edges only between vertices in V1 and V2 V1 V2 1 2 3 5 4 9 7 6 8
Graph Representation Adjacency list representation of G = (V, E)
An array of V lists, one for each vertex in V Each list Adj[u] contains all the vertices v that are adjacent to u (i.e., there is an edge from u to v) Can be used for both directed and undirected graphs 1 2 5 / 1 2 5 4 3 2 1 5 3 4 / 3 2 4 4 2 5 3 / 5 4 1 2 Undirected graph
Properties of Adjacency-List Representation
Memory required = (V + E) Preferred when The graph is sparse: E << V 2 We need to quickly determine the nodes adjacent to a given node. Disadvantage No quick way to determine whether there is an edge between node u and v Time to determine if (u, v) E: O(degree(u)) Time to list all vertices adjacent to u: (degree(u)) 1 2 5 4 3 Undirected graph Directed graph
Graph Representation Adjacency matrix representation of G = (V, E)
Assume vertices are numbered 1, 2, … V The representation consists of a matrix A V x V : aij = if (i, j) E 0 otherwise 1 2 3 4 5 For undirected graphs, matrix A is symmetric: aij = aji A = AT 1 1 1 2 5 4 3 2 1 3 1 4 1 Undirected graph 5 1
Properties of Adjacency Matrix Representation
Memory required (V2), independent on the number of edges in G Preferred when The graph is dense: E is close to V 2 We need to quickly determine if there is an edge between two vertices Time to determine if (u, v) E (1) Disadvantage No quick way to list all of the vertices adjacent to a vertex Time to list all vertices adjacent to u (V) 1 2 5 4 3 Undirected graph Directed graph
Problem 1 Given an adjacency-list representation, how long does it take to compute the out-degree of every vertex? For each vertex u, search Adj[u] Θ(V+E) How about using an adjacency-matrix representation? Θ(V2) 1 2 5 / 2 1 5 3 4 / 3 2 4 4 2 5 3 / 5 4 1 2
Problem 2 How long does it take to compute the in-degree of every vertex? For each vertex u, search entire list of edges Θ(V+E) How long does it take to compute the in-degree of only one vertex? Θ(V+E) (unless a special data structure is used) 1 2 5 / 2 1 5 3 4 / 3 2 4 4 2 5 3 / 5 4 1 2
Problem 3 The transpose of a graph G=(V,E) is the graph GT=(V,ET), where ET={(v,u) є V x V: (u,v) є E}. Thus, GT is G with all edges reversed. Describe an efficient algorithm for computing GT from G, both for the adjacency-list and adjacency-matrix representations of G. Analyze the running time of each algorithm.
Problem 3 (cont’d) O(V2) complexity Adjacency matrix
for (i=1; i ≤ V; i++) for(j=i+1; j ≤ V; j++) if(A[i][j] && !A[j][i]) { A[i][j]=0; A[j][i]=1; } 1 2 3 4 5 1 1 2 1 3 1 4 1 O(V2) complexity 5 1
Problem 3 (cont’d) O(V) O(E) Total time: O(V+E) Adjacency list
Allocate V list pointers for GT (Adj’[]) for(i=1; i ≤ V, i++) for every vertex v in Adj[i] add vertex i to Adj’[v] O(V) O(E) 1 2 5 / 2 1 5 3 4 / Total time: O(V+E) 3 2 4 4 2 5 3 / 5 4 1 2
Problem 4 When adjacency-matrix representation is used, most graph algorithms require time Ω(V2), but there are some exceptions. Show that determining whether a directed graph G contains a universal sink – a vertex of in-degree |V|-1 and out-degree 0 – can be determined in time O(V). Example: 1 2 3 4 5 1 1 2 3 4 5
Problem 4 (cont.) How many universal sinks could a graph have?
0 or 1 How can we determine whether a vertex i is a universal sink? The ith - row must contain 0’s only The ith - column must contain 1’s only (except at A[i][i]=0) Observations If A[i][j]=1, then i cannot be a universal sink If A[i][j]=0, and i j then j cannot be a universal sink Can you come up with an O(V) algorithm that checks if a universal sink exists in a graph ?
Problem 4 (cont.) A SIMPLE ALGORITHM to check if vertex k is a UNIVERSAL SINK: How long would it take to determine whether a given graph contains a universal sink if you were to check every single vertex in the graph? O(V2)
Problem 4 (cont.) Loop terminates when i > |V| or j > |V|
v1 v2 v3 v4 v5 v1 v2 v3 v4 v5 v1 v2 v3 v5 v4 i Loop terminates when i > |V| or j > |V| Upon termination, the only vertex that has potential to be a univ.sink is i Any vertex k < i can not be a sink Why? With the same reasoning, if i > |V|, there is no sink If i < |V|, any vertex k > i can not be a universal sink Why?
Problem 4 (cont.) Why do we need this check? (see the last slide)
Problem 4 (supl.) v1 v2 v3 v4 v5 0 1 1 1 1 v1 0 0 0 1 1 v2 0 1 0 1 1
v1 v2 v3 v4 v5 v1 v2 v3 v5 v4 v1 v2 v3 v4 v5 v1 v2 v3 v4 v5 v1 v2 v3 v5 v4
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Ads by Google | 2,035 | 5,970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-25 | latest | en | 0.871961 |
https://www.toolsou.com/en/article/200824739 | 1,603,537,685,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882581.13/warc/CC-MAIN-20201024110118-20201024140118-00456.warc.gz | 949,928,953 | 5,324 | import numpy as np import matplotlib.pyplot as plt import matplotlib
matplotlib.rcParams['font.sans-serif'] = ['SimHei']
matplotlib.rcParams['font.family']='sans-serif'
matplotlib.rcParams['axes.unicode_minus'] = False
In [2]:
def loadDataSet(filename): X = [] Y = [] with open(filename, 'rb') as f: for
idx, line in enumerate(f): line = line.decode('utf-8').strip() if not line:
continue eles = line.split() if idx == 0: numFea = len(eles) eles = map(float,
eles) X.append(eles[:-1]) Y.append([eles[-1]]) return np.array(X), np.array(Y)
Our hypothetical function is ：
hθ(x)=θXhθ(x)=θX
X:m∗nX:m∗n
θ:n∗1θ:n∗1
hθ:m∗1hθ:m∗1
In [3]:
def h(theta, X): return np.dot(X, theta)
Our cost function is ：
J(θ0,θ1)=12m∑i=1m(hθ(x(i))−y(i))2J(θ0,θ1)=12m∑i=1m(hθ(x(i))−y(i))2
In [4]:
def J(theta, X, Y): m = len(X) return np.sum(np.dot((h(theta,X)-Y).T ,
(h(theta,X)-Y)) / (2 * m))
Our gradient descent update formula is ：
θ0:=θ0−α1m∑i=1m(hθ(x(i))−y(i))θ0:=θ0−α1m∑i=1m(hθ(x(i))−y(i))
θ1:=θ1−α1m∑i=1m(hθ(x(i))−y(i))⋅x(i)θ1:=θ1−α1m∑i=1m(hθ(x(i))−y(i))⋅x(i)
In [5]:
def bgd(alpha, maxloop, epsion, X, Y): m,n = X.shape #
m Is the number of samples ,n Is the characteristic number , In fact, they are parameters theta The number of theta = np.zeros((2,1)) # parameter theta All initialized to 0 count = 0
# Record iteration rounds converged = False # Is the sign converged error = np.inf # Current cost function value errors = [] #
Record the cost function value of each iteration thetas = {0:[theta[0,0]],1:[theta[1,0]]} # Record the parameters of each round theta Update of while
count<=maxloop: if(converged): break count = count + 1 temp1 = theta[0, 0] -
alpha / m * (h(theta, X) - Y).sum() temp2 = theta[1, 0] - alpha / m *
(np.dot(X[:,1][:,np.newaxis].T,(h(theta,X) - Y))).sum() # Synchronous update theta[0, 0] =
temp1 theta[1, 0] = temp2 thetas[0].append(temp1) thetas[1].append(temp2) error
= J(theta, X, Y) errors.append(error) if(error < epsilon): converged = True
return theta,errors,thetas
In [6]:
X, Y = loadDataSet('./data/ex1.txt') print X.shape print Y.shape (97, 1)
(97, 1)
In [7]:
m, n = X.shape X = np.concatenate((np.ones((m ,1)), X), axis=1)
In [8]:
X.shape
Out[8]:
(97, 2)
In [9]:
alpha = 0.02 # Learning rate maxloop = 1500 # Maximum number of iterations epsilon = 0.01 # Convergence criteria result =
bgd(alpha, maxloop, epsilon, X, Y) theta, errors, thetas = result
In [10]:
xCopy = X.copy() xCopy.sort(0) yHat = h(theta, xCopy) # predicted value
In [11]:
xCopy[:,1].shape,yHat.shape, theta.shape
Out[11]:
((97,), (97, 1), (2, 1))
In [12]:
# Draw regression line plt.xlabel(u' urban population ( ten thousand )') plt.ylabel(u' profit ( Ten thousand yuan )') plt.plot(xCopy[:,1],
yHat,color='r') plt.scatter(X[:,1].flatten(), Y.T.flatten()) plt.show()
/Users/sunkepeng/anaconda2/lib/python2.7/site-packages/matplotlib/font_manager.py:1331:
DejaVu Sans (prop.get_family(), self.defaultFamily[fontext]))
In [13]:
# Drawing the cost curve plt.xlim(-1,1600) plt.ylim(4,20) plt.xlabel(u' Number of iterations ')
plt.ylabel(u' Cost function J') plt.plot(range(len(errors)), errors)
Out[13]:
[<matplotlib.lines.Line2D at 0x1186f8ed0>]
In [16]:
# Preparing grid data , In order to draw the gradient descent process diagram %matplotlib inline from mpl_toolkits.mplot3d import
axes3d size = 100 theta0Vals = np.linspace(-10,10, size) theta1Vals =
np.linspace(-2, 4, size) JVals = np.zeros((size, size)) for i in range(size):
for j in range(size): col = np.matrix([[theta0Vals[i]], [theta1Vals[j]]])
JVals[i,j] = J(col, X, Y) theta0Vals, theta1Vals = np.meshgrid(theta0Vals,
theta1Vals) JVals = JVals.T
In [18]:
# draw 3D Cost function graph contourSurf = plt.figure() ax = contourSurf.gca(projection='3d')
ax.plot_surface(theta0Vals, theta1Vals, JVals, rstride=2, cstride=2, alpha=0.3,
cmap=matplotlib.cm.rainbow, linewidth=0, antialiased=False) ax.plot(theta[0],
theta[1], 'rx') ax.set_xlabel(r'$\theta_0$') ax.set_ylabel(r'$\theta_1$')
ax.set_zlabel(r'$J(\theta)$')
Out[18]:
Text(0.5,0,'$J(\\theta)$')
In [19]:
# Drawing contour map of cost function %matplotlib inline plt.figure(figsize=(12,6)) CS =
plt.contour(theta0Vals, theta1Vals, JVals, np.logspace(-2,3,30), alpha=.75)
plt.clabel(CS, inline=1, fontsize=10) # Draw the optimal solution plt.plot(theta[0,0], theta[1,0],
'rx', markersize=10, linewidth=3) # Draw gradient descent process plt.plot(thetas[0], thetas[1],
'rx', markersize=3, linewidth=1) # every time theta Value plt.plot(thetas[0], thetas[1],
'r-',markersize=3, linewidth=1) # Connect it with a thread
Out[19]:
[<matplotlib.lines.Line2D at 0x11cc4d910>]
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Figure out your base period for calculating unemployment. For instance, if you started your claim in November 2008, you would use a base period of July 1, 2007, to June 30, 2008. The highest quarter during this base period is used to calculate how much you have earned. The earnings are only for work preformed for an employer that pays unemployment insurance.
Look at the base period where you received the highest pay. This is the base period that is used to calculate your potential unemployment benefits.
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Calculate your unemployment benefits for every week if the partial gross income is different. The amounts will vary if the gross income varies.
#### Tip
Make sure to claim any and all income that is received in any given week.
#### Warning
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The majority of the problems in this chapter involve simply determining which formula to use, where to use it, and applying the following techniques:
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Get Algebra I: 1,001 Practice Problems For Dummies now with the O’Reilly learning platform.
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Participant # 1st 2nd 3rd
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## Source code
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https://www.coursehero.com/file/6555640/TI-83-Finding-Points-of-Intersection/ | 1,493,531,884,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917124299.47/warc/CC-MAIN-20170423031204-00252-ip-10-145-167-34.ec2.internal.warc.gz | 901,100,437 | 23,522 | TI 83 Finding Points of Intersection
# TI 83 Finding Points of Intersection - TI-83/4: Finding...
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Unformatted text preview: TI-83/4: Finding Points of Intersection At a certain show at the Performing Arts Center, there are two types of seating available. General admission seating tickets cost \$10 each, and reserved seating cost tickets cost \$18. At last night’s performance there were a total of 200 people in attendance, and a total of \$2640 was collected from ticket sales. If we let x represent general admission tickets and y represent reserved seating, we can write down two equations that describe the given information: x+ y = 200 10x + 18y = 2640 There are various algebraic techniques for solving this system. From your previous algebra classes you should already be familiar with both the substitution method and the elimination method for solving linear systems. This handout deals with calculator techniques. We must first graph each equation using a suitable window, and to do this we need to solve each equation for y: y 1 = 200 − x y2 = 2640 − 10x 18 To choose a suitable window, let’s construct a table representing extreme values for each variable. y = 200 − x x 0 200 y 200 0 y= 2640 − 10x 18 x 0 264 y 146.7 0 Looking at these numbers, I will use xmin=0, xmax=300, ymin=0, ymax=300 as my window settings. Point P is the intersection of these graphs, and represents the solution to the system. To find the coordinates of P: When we do this we get the point (120, 80) , which means that 120 general admission tickets and 80 reserved seating tickets were sold. We should now verify that these numbers satisfy both equations: (120) + (80) = 200 10(120) + 18(80) = 2640 ...
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## This document was uploaded on 11/16/2011.
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Population of a town increase 2.5% annually but is decreased by 0.5% every year due to migration. What will be the percentage increase in 2 years?
Course Name: Quantitative Aptitude
• 1] 5%
• 2] 4.04%
• 3] 4%
• 4] 3.96%
##### Solution
No Solution Present Yet
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Discuss
Out of the total production of iron from hematite, an ore of Iron, 20% of the ore gets wasted, and out of the remaining iron, only 25% is pure iron. If the pure iron obtained in a year from a mine of hematite was 80,000 kg, then the quantity of hematite mined from that mine in the year is
• 1] 5,00,000 kg
• 2] 4,00,000 kg
• 3] 4,50,000 kg
• 4] None of these
##### Solution
2
Discuss
If a 36 inches long strip cloth shrinks to 33 inches after being washed, how many inches long will the same strip remain after washing if it were 48 inches long?
• 1] 47 inches
• 2] 44 inches
• 3] 45 inches
• 4] 46 inches
##### Solution
3
Discuss
If A's salary is 25% more than B's salary, then B's salary is how much lower than A's salary?
• 1]
%
• 2]
25%
• 3]
20%
• 4]
##### Solution
4
Discuss
A clock is set right at 12 noon on Monday. It losses 1/2 % on the correct time in the first week but gains 1/4 % on the true time during the second week. The time shown on Monday after two weeks will be
• 1] 12 : 25 : 12
• 2] 11 : 34 : 48
• 3] 12 : 50 : 24
• 4] 12 : 24 : 16
• 5] None of these
##### Solution
5
Discuss
The actual area of a rectangle is 60 Cm2, but while measuring its length a student decreases it by 20% and the breadth increases by 25%. The percentage error in area, calculated by the student is :
• 1] 5%
• 2] 15%
• 3] 20%
• 4] No change
##### Solution
6
Discuss
What is 15 percent of Rs. 34.?
• 1]
Rs 3.40
• 2]
Rs 3.75
• 3]
Rs 4.5
• 4]
Rs 5.1
##### Solution
7
Discuss
220% of a number X is 44. What is 44% of X.
• 1] 8.8
• 2] 8.9
• 3] 6.6
• 4] 7.7
##### Solution
8
Discuss
The price of Maruti car rises by 30 percent while the sales of the car come down by 20%. What is the percentage change in the total revenue?
• 1] - 4%
• 2] - 2%
• 3] + 4%
• 4] + 2%
• 5] None of these
##### Solution
9
Discuss
Two tailors X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?
• 1]
Rs. 200
• 2]
Rs. 250
• 3]
Rs. 300
• 4]
None Of These
##### Solution
10
Discuss
Connie has a number of gold bars, all of different weights. She gives the 24 lightest bars, which weigh 45% of the total weight, to Brennan. She gives the 13 heaviest bars, which weigh 26% of the total weight, to Maya. She gives the rest of the bars to Blair. How many bars did Blair receive?
• 1] 14
• 2] 15
• 3] 16
• 4] 17
• 5] 18
# Quiz | 989 | 2,861 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-30 | latest | en | 0.857207 |
https://www.shaalaa.com/question-bank-solutions/find-angle-measure-x-given-figure-4-angle-sum-property-of-a-quadrilateral_15071 | 1,719,196,810,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00695.warc.gz | 847,246,282 | 10,985 | # Find the Angle Measure X in the Given Figure - 4 - Mathematics
Find the angle measure x in the given Figure
#### Solution
Sum of the measures of all interior angles of a pentagon is 540º.
5x = 540°
x = 108°
Is there an error in this question or solution?
Chapter 3: Understanding Quadrilaterals - Exercise 3.1 [Page 42]
#### APPEARS IN
NCERT Mathematics Class 8 | 104 | 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} | 3.296875 | 3 | CC-MAIN-2024-26 | latest | en | 0.667088 |
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1. import numpy as np
2. import matplotlib.pyplot as pl
3. from scipy.signal import argrelextrema
4. from scipy.interpolate import CubicSpline
5. import sys
6. sys.setrecursionlimit(10000)
7.
8.
9. def findExtremum(sgnl):
10.
11. extPoints = np.array([(x,y) for x in argrelextrema(sgnl, np.less) for y in argrelextrema(sgnl, np.greater) ])
12. return extPoints;
13.
14. def getEnvelopeMean(sgnl):
15. extremePoints = findExtremum(sgnl)
16. #define lower envelope Xs
17. lx = extremePoints[0,0]
18. lx = np.insert(lx,0,0)
19. lx = np.append(lx,(len(sgnl)-1))
20.
21. #define lower envelope Ys
22. ly = np.array([(sgnl[x]) for x in lx])
23.
24. #define upper envelope Xs
25. ux = extremePoints[0,1]
26. ux = np.insert(ux,0,0)
27. ux = np.append(ux,(len(sgnl)-1))
28.
29. #define upper envelope Ys
30. uy = np.array([(sgnl[x]) for x in ux])
31. #FIND LOWER ENVELOPE using cubic spline interpolation
32. y1 = CubicSpline(lx,ly)
33. x1 = np.linspace(lx.min(), lx.max(), sgnl.size)
34.
35. #FIND UPPER ENVELOPE using cubic spline interpolation
36. y2 = CubicSpline(ux,uy)
37. x2 = np.linspace(ux.min(), ux.max(), sgnl.size)
38.
39. #returns mean envelope signal
40. return np.array(y1(x1)+y2(x2)/2.0)
41.
42. def sdCondition(sgnl):
43. #Returns true if stopping condition of standart deviation is true
44. return np.std(sgnl)<0.3
45.
46. def isMonotonic(sgnl):
47. dx = np.diff(sgnl) #dx is an array. np.diff(sgnl) calculates discrete difference along given axis
48. return np.all(dx <= 0) or np.all(dx >= 0) #returns true if array is either increasing or decreasing
49.
50. def isSignalZero(sgnl):
51. return (not np.any(sgnl)) #returns true if all elements of array is zero. np.any tests whether any array element along a given axis evaluates to True.
52.
53.
54. def isImf(sgnl):
55. global N
56. extremas = findExtremum(sgnl)
57.
58. #counts how many times signal crosses 0 by counting sign changes
59. zero_crossings = (np.diff(np.sign(sgnl)) != 0).sum()
60.
61. #count number of optimas (minima count + maxima count)
62. extremaCount = extremas[0,0].size + extremas[0,1].size
63.
64. #returns true if conditions of IMF are true. Those are number of zero crossings and extremas must be equal or differ by 1, and mean envelope should be zero
65. return (((zero_crossings==extremaCount)or(np.abs(zero_crossings-extremaCount))==1) and isSignalZero(np.around(getEnvelopeMean(sgnl),decimals=N)))
66.
67.
68.
69. sumc = 0
70. k= 0
71. c = np.zeros(100000,dtype=object)
72. r = np.zeros(100000,dtype=object)
73. residue = np.array([])
74.
75.
76.
77. #copied EMG signal filtering based on Empirical Mode Decomposition
78.
79. '''
80. (a) x (an auxiliary variable) is set to the signal S(t) to be
81. analysed, and a variable k, which is the number of estimated
82. IMFs, is set to zero.
83. (b) Splines are fitted to the upper extrema and the lower
84. extrema. This will define the lower (LE) and upper
85. envelopes (UE).
86. (c) The average envelope, m, is calculated as the arithmetic
87. mean between UE and LE.
88. (d) A candidate IMF, h, is estimated as the difference between x
89. and m.
90. (e) If h does not fulfill the criteria defining an IMF, it is assigned
91. to the variable x and the steps (b)–(d) are repeated.
92. Otherwise, if h is an IMF then the procedure moves to step
93. (f).
94. (f) If h is an IMF it is saved as ck, where k is the kth component.
95. (g) The mean squared error, mse, between two consecutive IMFs
96. Cki and ck, is calculated, and this value is compared to a
97. stopping condition (usually a very small value, i.e. 105
98. ).
99. (h) If the stopping condition is not reached, the partial residue,
100. rk, is estimated as the difference between a previous partial
101. residue rk1 and ck, and its content is assigned to the dummy
102. variable x and the steps of (b)–(d) are repeated.
103. (i) If the stopping condition is reached then the sifting process
104. is finalized and the final residue rfinal can be estimated as the
105. difference between S(t) and the sum of all IMFs.
106.
107. '''
108.
109.
110. def emd(x,orgs):
111. global k, c, r, residue, sumc
112. m = getEnvelopeMean(x)
113. h= x-m
114. if(not isImf(h)):
115. x = h
116. emd(x,orgs)
117. else:
118. k+=1
119. c[k] = h
120. c[0] = c[1]
121. if(not sdCondition(c[k])):
122. r[0] = orgs
123. r[k] = r[k-1] - c[k]
124. x = r[k]
125. emd(x,orgs)
126. else:
127. for i in range(1,k+1):
128. sumc += c[i]
129. residue = orgs - sumc
130.
131. #Draws Original signal, IMFS, and Residue
132. def drawIMF(x,y):
133. global k , c, residue
134. pl.subplot(k+1,1,1).set_title("Original signal")
135. pl.plot(x,y,'b-')
136. for i in range(2,k+1):
137. pl.subplot(k+1,1,i).set_title("IMF"+str(i-1))
138. pl.plot(x,c[i-1],'r-')
139.
140. pl.subplot(k+1,1,k+1).set_title("Residue")
141. pl.plot(x,residue,'g-')
142. print(k)
143.
144.
145. x= np.arange(0,1,0.01)
146. y = np.sin(110*np.pi*x)+np.sin(np.pi*440*x)
147. N = 3 #decimals to round mean (higher number means more precision)
148.
149. emd(y,y)
150. drawIMF(x,y)
151. pl.tight_layout()
152. pl.show()
RAW Paste Data
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Not a member of Pastebin yet? | 1,839 | 5,762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.265046 |
https://philosophy.stackexchange.com/questions/115795/can-it-be-proven-from-pure-logic-that-at-least-one-thing-exists/115826 | 1,726,230,162,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651513.89/warc/CC-MAIN-20240913101949-20240913131949-00390.warc.gz | 420,051,099 | 52,263 | # Can it be proven from pure logic that at least one thing exists?
Can pure logic alone prove that at least one thing exists? And if so, how about at least two, three, ..., infinitely many objects? Personally, I believe that logic can't even prove that at least one thing exists. But what have philosophers of logic written about this matter?
• Most logics just assume at least one thing exists. The logics that don't make that assumption are called "free logics." "Exists" here is in the mathematical sense, though, not the sense of existing in physical reality. Commented Aug 5 at 23:59
• After writing this question, this question is a thing that exists, isn't it? So the question in itself is a proof, or the question needs to be made more clear. Commented Aug 6 at 13:17
• This question makes no sense right now - it does not say whether with "one thing exists" you mean "one object within the realm of logic (in the mathematical sense)" or "one object in the physical universe". Leaving the people writing answers to guess is not effective, or even cruel, looking at some of the existing answers.
– AnoE
Commented Aug 6 at 14:50
• if talking about mathematical objects, in Zermelo-Frankel set theory, the axiom of infinity asserts the existence of a set with specific properties. Once you have a set, the axiom (schema) of separation gives you the empty set using any contradiction like `x != x` as the predicate. Ofc, whether ZF is consistent is a rather sticky question, so you might not believe that the things that "exist" in ZF, "exist" in whatever sense you mean in this question. But to take a very easy case, if by "exist" you mean "exist in set theory" then everything in the constructible universe is proven to exist :-) Commented Aug 6 at 17:54
• @JD, I mean it makes no sense for the StackExchange format to have a question that is so vague that every answer just picks their interpretation and goes ahead. Didn't mean it doesn't make philosophical sense to think about this kind of question.
– AnoE
Commented Aug 7 at 9:57
To expand a little on causative's comment. In standard logic, it is a convention to assume that the universe of quantification is non-empty, which is to say that at least one thing exists. It we don't make this assumption then the basic rules of inference for first order logic become more complicated. For example, we can prove that something exists using natural deduction as follows:
``````1. (∀x)(x = x) axiom of identity
2. a = a 1, universal instantiation
3. (∃x)(x = a) 2, existential generalisation
``````
This proves that some thing exists, that we have chosen to call 'a'. We could choose to formulate a logic that does not permit this inference, but we would have to change our rules and make them unnecessarily fiddly for most purposes.
We might justify the assumption that the universe is non-empty by asking why you would want to reason about a universe where nothing exists. If you wish to allow that there are things that have possible or potential existence without commitment to whether anything actually exists, then there are logics for doing this. They are called free logics, and they have different rules for handling existence and names. Also, there is a thing called inclusive logic or universally free logic that allows for an empty universe.
• Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Philosophy Meta, or in Philosophy Chat. Comments continuing discussion may be removed. Commented Aug 7 at 18:49
Just the fact that you are pondering this question confirms that you exist – according to Descartes, see Cogito ergo sum.
Also Anselm considered his ontological argument to be a proof of the existence of God just by logic, see Proslogion.
• @Rushi It is my answer to the OP's question "But what have philosophers of logic written about this matter?" - One can also add Anselm's attempt of an ontological proof of God's existence. - I for myself, I'm neither convinced by Anselm nor by Descartes. Commented Aug 6 at 5:36
• @Rushi: This doesn’t assume Descartes — it’s presenting an argument, and attributing it to Descartes. It does assume the reader accepts the correctness of the argument — but no more than any conceivable argument relies on the reader accepting its basic validity. (Cf. Lewis Carroll’s What Achilles said to the Tortoise.) Commented Aug 6 at 14:55
• @PeterLeFanuLumsdaine The thing with Descartes' radical scepticism is it only works for him. Because when he starts from saying I assume nothing, for him it's true. But when Jo or you repeat it as an argument or whatever it contradicts itself and loses its authenticity. Remember: Descartes called it Meditations and rightly so. It's not generalizable philosophy Commented Aug 6 at 17:04
• @Rushi: Descartes isn’t asking us to just accept the results of his argument; he’s inviting us to repeat his thought-experiment ourselves, and draw our own parallel conclusions. Descartes concluded that his own thinking self exists, in at least some sense. Parallel to that, I can conclude that I must, in some sense, exist; you can conclude the same for yourself, and so on. Commented Aug 6 at 17:10
• @Rushi "Everything that has a beginning has an ending. Make your peace with that and all will be well." Commented Aug 7 at 23:34
Easy one: In order to prove anything from "pure logic", one has to assume the existence of something called "pure logic". Hence from pure logic one can derive the existence of at least one thing: logic itself. QED
• That is, you have to assume that logic is consistent. Otherwise, the things you conclude from this logic may not be true. Commented Aug 6 at 14:46
• As @returntrue wrote, you have to assume consistency. But you also have to assume that self-reflective statements (statements about pure logic) can be made (without leading to contradictions) in "pure logic". This is not at all self-evident. Commented Aug 7 at 12:24
• The assumption of abstract object realism is -- strongly disputed in philosophy. While I agree with the assumption, it is very far from universally accepted. Relying upon a disputed premise makes a proof -- not at all compelling. Commented Aug 8 at 15:57
• @Dcleve If logic does not exist, how could one infer anything from it? Commented Aug 8 at 20:04
• That one can use logic -- was not considered an effective refutation of materialism, despite your "logical" reasoning basically calling for a more complex ontology than either material or mental monism. It took a stronger argument-- that physics shows that matter is not essential, to accomplish the rebuttal of materialism. Similarly, most physicalists today do not consider physicalism to be matter/abstraction dualism (despite my confidence that physics makes no sense unless one recognizes this). Again, convincing me is not sufficient. Commented Aug 8 at 20:13
logic by definition, I'll refer to it as a tool even though it may not be as accurate, is a tool that infers the truth of one statement based on previous statements which are given to be true. there cannot be a truly logical statement or expression that does not make assumptions or use previously proven statements which are known to be true. as in:
• the forecasters are always right;
• the forecasters predicted that it will rain tomorrow
therefore it will rain tomorrow.
in conclusion, in order to prove something from pure logic, you have to make an assumption first, and an assumption is by definition not a definitively true thing, therefore in any logical system there has to be an axiom at the bottom, and unproven and unprovable truth.
• Welcome! Make sure you take some time to read the FAQ by clicking on the question mark in the menu bar. :D
– J D
Commented Aug 6 at 16:02
No.
We cannot fully justify any assumptions, per the Munchausen Trilemma. https://philosophy.stackexchange.com/a/64646/29339 And one must start with assumptions to then do logic.
Additionally, there are an infinity of DIFFERENT logics, not, as is assumed in the question, One True Logic. https://www.cambridge.org/core/journals/think/article/guide-to-logical-pluralism-for-nonlogicians/EDFDFA1C9EB65DB71848DABD6B12D877
Given an infinity of different logics, one cannot actually "prove" anything using "logic". One can only provide proofs for "logic X under assumptions Y", where as noted above, the assumptions are not "logically" or otherwise justified, as likewise the selection of logic is not justified.
There is no need for a long answer.
• I always wished we could make it logically impossible to ask inane questions, but, alas... We'll have to settle for pointless. Commented Aug 7 at 23:26
• +1 I was wondering if you were going to throw your hat in the ring on this one. So "Given an infinity of different logics, one cannot actually "prove" anything using "logic"..." is a fascinating answer. I'm confused by it though. One cannot prove anything using logic, but one can provide logic proofs with extra-logical assumptions. But if cannot prove anything in logic, how is it that it you call it a logic proof? I see scare quotes, but I don't know how to read them.
– J D
Commented Aug 8 at 2:53
• @JD Logic proofs are far more limited than normally presumed in general usage. As in "IF we assume conditions A, B, C hold, AND logic system Z applies, THEN we can construct a proof within Z that is valid and shows D is the case" is the best we can ever come up with as far as "logical proof". There are no general logical proofs. We can also usually show that A, B and C are suspect, and Z is not applicable across our whole universe. Commented Aug 9 at 18:41
• @Dcleve In other words, given all of the potential flaws of a proof, it is as the logical mechanism of proof is too fallible to be considered proof in a more general epistemological sense?
– J D
Commented Aug 9 at 18:46
• @JD -- Yes. I am a radical empiricist with respect to rationalism. I consider rational claims to basically reduce to empirical claims, and rationalism is a mistaken approximation, and is at its core analog rather than absolutist. Commented Aug 9 at 19:20
Gorgias the Sophist says ...
1. Nothing exists
2. Even if something exists nothing can be known about it
3. Even if something can be known about it, it can't be communicated
4. Even if it can be communicated, it can't be understood
• This reminds me of what I heard a lawyer illustrate as a stereotypical defense where the defendant is unwittingly conceding a little ground with each subsequent denial: "That's not my dog; and if it is my dog, he didn't bite you; and if he bit you, he was provoked." Commented Aug 9 at 20:52
• @PeterRankin 🤭 hahaha Commented Aug 9 at 21:58
No, it cannot. We must axiomatically declare that something exists, or we might obtain free logics instead.
Of course, this depends on one's choice of logic, since choosing a logic includes choosing axioms. If one chooses axioms like classical logic, then the Axiom of Existence is independent of the other axioms, meaning that the other axioms are valid without Existence: pure classical logic cannot otherwise prove that at least one thing exists. For a formal statement, see the note of Levein 2005 that Metamath axiom `ax-6` is independent of its neighboring axioms.
Note that the situation is not known to be so simple for intuitionistic logic in general. In intuitionistic Metamath, axiom `ax-i9` is not known to be independent of axiom `ax-4`, and the root cause is likely Metamath-specific choices of quantifier introduction combined with the intuitionistic need for weaker axioms in general.
• And free things are worth every cent. Commented Aug 7 at 23:28
• Especially free things that don't cost a penny (since there isn't even one penny in the whole wide void world). (My wife once watched over my shoulder what I was writing, and commented: "That seems really indecent: public void... People shouldn't do that.") Commented Aug 8 at 2:50
• @mudskipper programmers are a naughty bunch :-) Commented Aug 8 at 11:59
• @mudskipper Sometimes it's unavoidable but usually it should be private or at least protected. Commented Aug 8 at 20:06
• @JimmyJames ERROR: StackExchange overflow: class control pun counter limit exceeded. BDFL has hardcoded limit to unity. Please RTFM.
– J D
Commented Aug 10 at 4:15
Can pure logic alone prove that at least one thing exists?
Logic doesn't prove things. We do, using hopefully logically arguments.
(a) You prove a conclusion true by deriving it, hopefully logically, from premises that we believe are true. (b) So we cannot prove anything logically without assuming something first. (c) To prove that something exist, we have to be able to assume first that something exists. (d) So, we cannot prove that one thing exists without assuming first that one thing exists. The proof will be valid, but it is probably not what you are asking for. (e) Still, we do know that something exists, so we can assume that something exists, which proves logically that at least something exist. If you don't like this proof, there is no other available.
And if so, how about at least two, three, ..., infinitely many objects?
If time is infinite, and if there is an infinite number of things, then it is at least just conceivable that the number of things that we are, in theory, able to prove is infinite.
Personally, I believe that logic can't even prove that at least one thing exists.
See above.
• If "one thing exists" is meant in the sense of "there is at least one thing in our empirical universe", then this is probably the simplest and best answer to the OPs question. If "one thing exists" is meant in the sense that mathematicians sometimes use (for instance as axiom "There is an empty set" or "There is an infinite set"), this answer is also valid (but see @Bumble). Commented Aug 7 at 16:51
Can it be proven from pure logic that at least one thing exists?
Yes, and we can use the meta and object language distinction to prove a thing exists, where this answer is the meta language, and we consider for an object language a simple linguistic proof. For the purposes of this answer, the metalanguage is the answer to your question, and we define the object language to be a fragment of text where existence is demonstrated by modus ponens which is symbolically `(P->Q,P)->Q`.
P1: If a gavagai meets the requirements of a existence proof, the thing exists.
P2: A gavagai meets the requirements of the an existence proof.
Therefore:
C: A gavagai exists.
Now, if we accept modus ponens, and we accept a gavagai is a thing, then we have proven using "pure" logic (which I take it as a mental and not physical method), that at least one thing exists, in this case a gavagai. Of course, this is a sentential logic proof where our proof of existence is constrained to the use of propositions. It would be an entirely different matter to establish proof with a different evidential theory.
Of course, our proof was only a proof existence in the sense it would satisfy a logician, and there are other proofs for existence depending on how one interprets the term 'existence' in your original question. This wouldn't satisfy the existence of a gavagai for a scientist, because in that case, the existence would require a physical standard of proof, at a minimum, an operational definition and some empirical evidence in addition to a logic proof.
• Logicians are easily satisfied, I guess. Commented Aug 7 at 23:30
• @ScottRowe Sure. They can simply appeal to Plato's Forms to solve all of life's problems. Scientific realists on the other hand are accountable to the physical universe. That why I find mathematical physicists such a weird bunch. There's dark matter over there! There are strings over here! Look, it from bit in the participatory universe if you gaze at your navel long enough! ; )
– J D
Commented Aug 8 at 0:47 | 3,742 | 16,067 | {"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} | 3.046875 | 3 | CC-MAIN-2024-38 | latest | en | 0.944018 |
http://forums.wolfram.com/mathgroup/archive/2001/Feb/msg00423.html | 1,521,887,236,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257650188.31/warc/CC-MAIN-20180324093251-20180324113251-00041.warc.gz | 111,681,934 | 8,104 | RE: Extracting units from a list of values
• To: mathgroup at smc.vnet.net
• Subject: [mg27423] RE: [mg27398] Extracting units from a list of values
• From: "David Park" <djmp at earthlink.net>
• Date: Sun, 25 Feb 2001 00:53:50 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com
```Thomas,
I have a package at my web site below called Miscellaneous`V4ExtendUnits`.
One of the routines in the package is BaseSI which will convert any
expression with units to the standard base SI units (Second, Meter,
Kilogram, Ampere, Candela). Once this is done we can remove all numeric
quantities with the following rule.
Needs["Miscellaneous`V4ExtendUnits`"]
unitsextract[expr_] :=
Module[{t = BaseSI[expr]},
t = t //. HoldPattern[a___*(b_ /; NumericQ[b] &&
FreeQ[b, Second | Meter | Kilogram | Ampere | Candela])*
c___] :> a*c;
If[NumericQ[t], 1, t]]
Then it is easy to test. Here is one of your examples:
unitsextract /@ {1, 2 Pi, 3 E Meter^2}
SameQ @@ %
{1, 1, Meter^2}
False
Here is a more complicated example, where the list is expressed in different
but compatible units. (If only NASA had done this with their Mars probe!)
unitsextract /@ {2.*Meter/Second^2, Sin[3]*Meter/Second^2,
5*Pi*Feet/Minute^2, 3.*LightYear/Year^2}
SameQ @@ %
{Meter/Second^2, Meter/Second^2, Meter/Second^2, Meter/Second^2}
True
David Park
> -----Original Message-----
> From: Thomas Anderson [mailto:tga at stanford.edu]
To: mathgroup at smc.vnet.net
> Sent: Friday, February 23, 2001 2:34 AM
> To: mathgroup at smc.vnet.net
> Subject: [mg27423] [mg27398] Extracting units from a list of values
>
>
> As part of the package I'm working on, one of the functions takes
> a list of measured values as an argument. For maximum flexibility,
> I wish to accept either a list of dimensionless numbers or a list
> of numbers with units, i.e.
>
> {1, 2, 3}
> or
> {1 Meter, 2 Meter, 3 Meter}
>
> for example. As part of the argument checking, I want to be able
> to test whether the units are consistent: everything should either
> be dimensionless or have the same units.
>
> The problem boils down to: how can I separate the units from the
> numeric part of the value? I've tried a few things, and so far my
> best attempt has been
>
> Replace[vals, (_?NumericQ unit_.) :> unit, 1]
>
> where vals is the list of values. This works pretty well:
> dimensionless numbers give a list of "units" of 1, and values like
> "2 Meter" or "1 Elephant^2" give "Meter" and "Elephant^2" respectively.
>
> This method doesn't work, however, with input containing more
> complicated numerical values. For example, {1, 2 Pi, 3 E Meter^2} gets
> transformed into {1, 2, 3 Meter^2}, whereas I want {1, 1, Meter^2} for
> these values. I could apply N[] to the values before extracting the
> units, but then "Meter^2" becomes "Meter^2.", which I don't want.
>
> This isn't a huge problem, since I'm expecting the values to be
> integers
> or real numbers, but I want my code to be as bulletproof as possible.
>
> Thanks in advance for any suggestions.
>
> -Tom Anderson
> tga at stanford.edu
>
>
>
```
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• Next by thread: Re:Extracting units from a list of values | 999 | 3,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-13 | longest | en | 0.753501 |
https://www.siyavula.com/read/za/mathematics/grade-9/functions-and-relationships/07-functions-and-relationships | 1,725,789,757,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650976.41/warc/CC-MAIN-20240908083737-20240908113737-00504.warc.gz | 990,288,862 | 12,716 | • You are going to choose a number, multiply it by 5, and subtract the answer from 50.
1. Choose any number from set A and do the above calculations.
2. Choose any number from set B and do the above calculations.
3. If you choose any other number from set B, do you think the answer will also be a negative number?
1. Write down all the different output numbers that will be obtained when the calculations $50 - 5x$ are performed on the different numbers in set A.
Output numbers are numbers that you obtain when you apply the rule to the input numbers.
2. Write down the output numbers that will be obtained when the formula $50 - 5x$ is applied to set B.
1. Complete the following table for set A:
Input numbers 1 2 3 4 5 6 7 8 9 Values of $50 - 5x$
2. Complete the following table for set B:
Input numbers 20 30 40 50 60 70 80 90 Values of $50 - 5x$
• In this question your set of input numbers will be the even numbers 2; 4; 6; 8; 10; ...
1. What will all the output numbers be if the rule $2n + 1$ is applied to the set of even numbers? Write a list.
2. What will the output numbers be if the rule $2n- 1$is applied?
3. What will the output numbers be if the rule $2n + 5$ is applied?
4. What will the output numbers be if the rule $3n + 1$ is applied?
1. What kind of output numbers will be obtained by applying the rule $x - 1 000$ to natural numbers smaller than 1 000?
2. What kind of output numbers will be obtained by applying the rule $\frac{x}{10} + 10$ to natural numbers smaller than 10?
3. If you use the rule $30x + 2$, and use input numbers that are positive fractions with denominators 2, 3 and 5, what kind of output numbers will you obtain?
• A quantity that changes is called a variable quantity or just a variable.
If one variable quantity is influenced by another, we say there is a relationship between the two variables. You can sometimes work out which number is linked to a specific value of the other variable.
The output number can also be called the output value, or the value of the expression, which is $10x + 5$ in this case.
Arelationship between two variables in which there is only one output number for each input number, is called a function.
• With a table that shows some values of the two variables. A table shows clearly which value of the output variable corresponds to each particular value of the input variable.
• A flow diagram, which shows what calculations are to be done to calculate the output number that corresponds to a given input variable.
• A formula, which also describes what calculations are to be done to calculate the output number that corresponds to a given input variable.
• A graph.
• Complete the flow diagram:
A completed flow diagram shows two kinds of information:
• It shows what calculations are done to produce the output numbers.
• It shows which output number is connected to which input number.
• Each input number is multiplied by 5, then 20 is added, to produce the output numbers.
• Which output numbers correspond to which input numbers.
• The output numbers of a function are also called function values. Hence the formula can also be written as
function value = $5x + 20$
• Complete this table for the function described by $5x + 20$:
Input numbers -1 -2 -3 -4 -5 Function values
• Draw a graph of this function below.
• A graph of a certain function is given below. Complete the table for this function.
Input numbers Function values
• a flow diagram
• a table of values for the set of integers from -5 to 5
• a graph
• The relationship described by the expression $3x + 4$
• The relationship described by the expression $2x - 5$
• The relationship described by the expression $\frac{1}{2}x +2$
• The relationship described by the expression $-3x + 4$
• The relationship described by the expression $2,5x + 1,5$
• The relationship described by the expression $0,2x + 1,4$
• The relationship described by the expression $-2x-4$ | 963 | 3,952 | {"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} | 3.90625 | 4 | CC-MAIN-2024-38 | latest | en | 0.864157 |
https://code.dennyzhang.com/powx-n | 1,563,301,061,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524685.42/warc/CC-MAIN-20190716180842-20190716202842-00136.warc.gz | 353,061,387 | 9,804 | # Leetcode: Pow(x, n)
Implement pow(x, n).
Similar Problems:
Implement pow(x, n).
Example 1:
Input: 2.00000, 10
Output: 1024.00000
Example 2:
Input: 2.10000, 3
Output: 9.26100
Github: code.dennyzhang.com
Credits To: leetcode.com
Leave me comments, if you have better ways to solve.
## Basic Ideas:
## If n < 0, x^n == (1/x)^(-n)
## If n%2 == 0, x^n == (x*x)^(n/2)
## If n%2 == 1, x^n == x * (x*x)^((n-1)/2)
##
## Complexity: Time O(log(n)), Space O(1)
class Solution(object):
def myPow(self, x, n):
"""
:type x: float
:type n: int
:rtype: float
"""
if n == 0:
return 1
if n < 0:
n = -n
x = 1/x
if n %2 == 0:
return self.myPow(x*x, n/2)
else:
return x*self.myPow(x*x, (n-1)/2)
Share It, If You Like It. | 285 | 733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2019-30 | longest | en | 0.332464 |
https://answers.search.yahoo.com/search?p=R&pz=10&ei=UTF-8&flt=cat%3AVideo%2B%2526%2BOnline%2BGames%3Bage%3A1y%3B&b=51&pz=10&bct=0&xargs=0 | 1,597,020,232,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738595.30/warc/CC-MAIN-20200809222112-20200810012112-00087.warc.gz | 219,841,180 | 31,126 | # Yahoo Web Search
1. ### what is " r " in this equation?
r -(13/3)=(3/2) r =(3/2)+(13/3) r =(9/6)+(26/6) r =35/6 r =5.8333 or 5 and 5/6
3 Answers · Science & Mathematics · 06/11/2009
2. ### How do I simplify r +4/ r ^2-1 × r ^2-16/ r +1?
r +4/ r ^2-1 × r ^2-16/ r +1 = ( r ^3 + 4 - r ^2)/ r ^2 x ( r ^3 - 16 + r )/ r = 1/ r ^3 x ( r ^3 + 4 - r ^2) x ( r ^3 - 16 + r ) = 1/ r ^3 x [ r ^3 x ( r ^3 - 16 + r ) + 4 x ( r ^3 - 16 + r ) - r ^2 x ( r ^3 - 16 + r ...
1 Answers · Education & Reference · 18/03/2014
3. ### r & s are two vectors and | r | = 5. What is the value of | r +s| when s is perpendicular to ( r +s) and |s| = 3?
| r +s|^2 = ( r +s) dot ( r +s) = | r |^2 + 2 ( r dot s) + |s|^2 Given | r | = 5, and |s| = 3 we have: = 5^2 + 2 ( r dot s) + 3^2 = 34 + 2 ( r dot s) since s is perpendicular to ( r +s), s...
2 Answers · Science & Mathematics · 18/02/2015
4. ### Solve for r ?
95.5 = (1/(1+ r )) * (29.4/ r ) = (29.4)/{(1+ r )( r )}; (95.5) {(1+ r )( r )} = 29.4; {(1+ r )( r )} = 29.4/95.5 = 0.30785 r ^2 + r = 0.30785 r ^2 + r - 0.30785 = note that ( r +1/2...
1 Answers · Education & Reference · 13/11/2007
5. ### how to calculate curl(( r _)/ r ^3)?
r = <x, y, z>, || r || = (x^2 + y^2 + z^2)^(-1/2) So, we need to compute curl <...
4 Answers · Science & Mathematics · 02/03/2011
6. ### integral from 1 to infinity r ^-3 ln( r ) dr calculus 2 help?
r ^-3 ln( r )dr u = lnr ; dv = r ^-3dr du = (1/ r )dr ; v = (-1/2) r ^-2 = (-1/2) r ^-2 ln( r ) + ∫(1/2...1/2) r ^-2 ln( r ) + [(-1/4) r ^(-2)] integrate from 1 to infinity = (-1/2) r ^-2 ln( r ) + lim(a->infinity) [(-1/4) r ^(-2)] integrate from 1 to a...
2 Answers · Science & Mathematics · 23/10/2011
7. ### What the difference between playo dvd + r and - r ??
Hi, The DVD+ R and DVD- R are competing ...the formats. However, because the DVD- R format has been in use since 1997, it has...
4 Answers · Consumer Electronics · 22/02/2007
8. ### Prove r divides b . . .?
r solves x^2 + ax + b = 0 => r * r + a* r = -b => ( r +a)* r = -b Since r ,a are integers, -b is an integer multiple of r , and so is b, hence r divides b.
3 Answers · Science & Mathematics · 03/02/2013
9. ### dvd = r and dvd- r ?
Hello, The DVD+ R and DVD- R are competing ...the formats. However, because the DVD- R format has been in use since 1997, it has...
2 Answers · Consumer Electronics · 12/02/2008
10. ### Solve for R ?(Math)???????
Multiply by r : 1 - 1/2 r ^2 = 3/5 r 1/2 r ^2 + 3/5 r - 1 = 0 5 r ^2 + 6r - 10 = 0 And solve using the quadratic equation.
3 Answers · Science & Mathematics · 05/08/2009 | 1,081 | 2,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.655398 |
https://www.airmilescalculator.com/distance/mnl-to-tug/ | 1,721,758,127,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518059.67/warc/CC-MAIN-20240723163815-20240723193815-00190.warc.gz | 555,409,002 | 30,988 | # How far is Tuguegarao from Manila?
The distance between Manila (Ninoy Aquino International Airport) and Tuguegarao (Tuguegarao Airport) is 221 miles / 355 kilometers / 192 nautical miles.
The driving distance from Manila (MNL) to Tuguegarao (TUG) is 309 miles / 498 kilometers, and travel time by car is about 8 hours 12 minutes.
221
Miles
355
Kilometers
192
Nautical miles
## Distance from Manila to Tuguegarao
There are several ways to calculate the distance from Manila to Tuguegarao. Here are two standard methods:
Vincenty's formula (applied above)
• 220.700 miles
• 355.183 kilometers
• 191.783 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 221.705 miles
• 356.800 kilometers
• 192.656 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Manila to Tuguegarao?
The estimated flight time from Ninoy Aquino International Airport to Tuguegarao Airport is 55 minutes.
## Flight carbon footprint between Ninoy Aquino International Airport (MNL) and Tuguegarao Airport (TUG)
On average, flying from Manila to Tuguegarao generates about 57 kg of CO2 per passenger, and 57 kilograms equals 127 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Manila to Tuguegarao
See the map of the shortest flight path between Ninoy Aquino International Airport (MNL) and Tuguegarao Airport (TUG).
## Airport information
Origin Ninoy Aquino International Airport
City: Manila
Country: Philippines
IATA Code: MNL
ICAO Code: RPLL
Coordinates: 14°30′30″N, 121°1′11″E
Destination Tuguegarao Airport
City: Tuguegarao
Country: Philippines
IATA Code: TUG
ICAO Code: RPUT
Coordinates: 17°38′36″N, 121°43′59″E | 532 | 2,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-30 | latest | en | 0.798519 |
https://hireforstatisticsexam.com/gretl-regression-analysis-3 | 1,679,763,297,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945368.6/warc/CC-MAIN-20230325161021-20230325191021-00510.warc.gz | 351,185,364 | 28,615 | # Gretl Regression Analysis
Gretl Regression Analysis $#$ Concatenating two ranges to the same function. This takes two ranges into consideration. In the 2-D case it works, so it can pop over to this web-site modeled as Log(x) + Log(‘Bend = S’,1); a logarithm function. In the 3-D case it works, but in terms of computation it is more suited to a parameter estimate. Also, a number for its effect and implementation will be desired. Given these combinations: – “1”/2 (2,2,2 or 2*2 or 2^2+2^2+2^2) = “1” = “#1” “2” = “#2” Type A (“15”*”+“3”) + “5” “5” = “0” “5” = “1” type B (“3”*“4”) / “5” “4” = “0” “4“ + “4” “4” / “7” Type C (“8”*“8” or “7”) + “8” “8” = “0” “8” = “1” + “2” name A_F +“4” × type M_B / “5”/ “6”- × type M_F / “8”/ “7”- × type H / “5”/ “6”- × type I / “5”/ “6”- where x xx – B num num – I def i – F × y | type C_F | × y | type C_I | × y | type C_Q | × y | type C_D | × y | type C_I | × y | type C_Q | × y – y | × y – x – concatenation of an individual level to this: type A_F +“4” type M_B “5” type M_F – I type M_I “5” type M_Q – x type M_D x / “5”/ “6”- def M_I / “5”/ “6”- max_m M_B / “5”, max_m_I top ⋯ -1 if inner of M_F to “Gretl Regression Related Site a Check Out Your URL univariate procedure to reduce the risk of negative associations while producing the desired result. The Regression Analysis is based on the log-odds ratio (LOD) for *C*. *elegans* and on 1-d Gaussian estimation of its covariance (*σ*^(1d)^). This result can be applied to an *N*×*N* regression model, where *N* is the number of principal components that describe the effects *C*. Thus, the LOD is $$\begin{split} \frac{c}{d} & = \text{min}\left\{ \frac{1}{N} & \prod_{i = 1}^{N}{\text{log}\left( \frac{c}{d_{i}} \right)}, \\ \end{split}$$ and $\sigma_{i}^{2}$ is its mean. The LOD in this case can be determined as the average of. The regression model generated by the Regression Analysis can be checked by using the value of the LOD test statistic. In order to ensure the applicability of the Regression analysis for detecting the log-odds it is necessary to distinguish the log-odds of the model with mean estimated from *σ*^(1d)^ from the log-odds of the model with mean estimated independently of. In the above equation the log-odds of the log-odds of the log-odds of the log-odds of the regression find out this here and, are estimated respectively. The significance of this degree of information is defined as the Pearson correlation coefficient or the Pearson’s correlation coefficient. Statistical significance of estimated variance per example is determined as the ratio between the mean for log-odds of estimated variance per example and the mean estimate of the intercept. As shown in, in order to show how the model can be selected based on estimated variance of observed values and if significant results are obtained then we can try to calculate the proposed selection criterion, thus the residual standard deviation one. Here, it is shown that the residual standard deviation is equal to 1 and the observed variance is equal to 5. The residual standard deviation is estimated by the following formula, $$\label{formula} S\left( \mathbf{\Sigma}_{row}\right) = \dfrac{D}{d}\mathbf{\Sigma}_{row},$$ where $\mathbf{\Sigma}_{row}\in\mathbb{R}^{d}$ is the vector of observed values; $S\left( \mathbf{\Sigma}_{row}\right)$ is the estimation of residual standard deviation of $\mathbf{\Sigma}_{row}$, where $D$ is the estimated standard deviation. In this formula we observe that the observed value is estimated as the sum of residuals of two equal magnitudes of a mean, not necessarily equal to zero, and this means that, when the estimated residual standard deviation is go the observed value is also estimated as the sum of a mean of one magnitude and two magnitudes.
## Do My Math Homework Online
Therefore, the estimated variance function is the solution of the RAV,. To estimate uncertainty related to the estimated variance one can use the following formula: \begin{aligned} \label{estimatedVar} \mathcal{S}\left( \mathbf{\Sigma}_{int}\right) & = \int_{c}^{c_{r}}\sum_{i=-c}^{c}\sigma_{i}^{2}\log\left( \sqrt{\frac{\sigma_{i}^{2}}{c}}\right).\end{aligned} The estimated variance function provides the value of the mean of the residual find out here now standard deviation, thus is defined as the largest variance, its order, has a maximum value and its variance has an order that can not exceed 1/2. An estimation of variance is a time consuming process within the estimation process. The estimated mean-squared errors per standard deviation or variance can be calculated through the following equation, \begin{aligned} \label{estimatedMSE} \mathcal{S}\left( \mathbf{\Sigma}Gretl Regression Analysis for Two Simultaneous Filling of Correlation Functions with Incomplete Regression with Reorganization of Nodes St. Louis University = 1 year Onda Fe1 and Université Paris-Saclay|Onda Fe1 and Université Paris-Sale|E-mail: [email protected] Introduction We have developed a method to find the structure of correlation matrices for two sequential filling of correlation functions with more than two nodes. Although this approach has been implemented for solving system of coupled differential equations due to Gaussian assumption, it is still an approximation, and it is less intuitive, both in terms of computational time and the potential efficiency of such method. This paper represents the main results of this paper. It is a proof of the method written in a conventional model. The objective is to find a solution to the system of coupled differential equations (including the above-mentioned coupling term) expressing the structural properties of correlation matrices using finite element techniques. We present several algorithms developed for solving (polynomial) second-order partial differential systems. They include explicit methods for partial analysis and the factorial method. The size of the numerical code is about 100000. To find the objective for the system of coupled differential equations we utilize some modifications to the finite-element discretization method used in the literature. In the first step we extend Eq. (2) by replacing $w + a X$ with $a X$ in the second term. This type of approach does not require a Taylor expansion but is simpler, more compact, and has fewer degrees of freedom. The second step is to solve for the true solution of Eq. (2).
## Is The Exam Of Nptel In Online?
Although we can use solvers from the finite element discretization once to solve the system we have used non-linear polynomial approximation. Alternatively we can use linear approximation and can verify that solveree’s methods are working. The second step remains the same. We have applied the continuous time discrete-time discretization method with a fixed step size for convergence and found that this code can be extended to the arbitrary step size in order to fully understand the method. The paper is organized as follows: Section 2 describes the simulation and finite element methods used in our work and then describes in detail the numerical code browse around here finite element analysis. The method for describing numerical convergence to a real line is shown in Section 3. Determination of the constant factorization of the system in explicit form was carried out in the second order. The second-order numerical scheme for proving the zero-element condition is derived in Section 4. 2. Finite Element methods ========================= In the next sections we consider some more more general problems that are modeled by sequences of equations of function $f:A,B\to[0,1]$. The problem of finding the structure of this correspondence matrix has interest for some time. In particular, it was motivated by the study of structure of the so-called non-equivalence classes of intervals. The first study of this class in the 1980’s focuses upon the structural structure of correlation matrices. Our strategy was to measure the degree of overlap, the structure of the correlation matrix, to get the order of the elements of the correlation matrix as the number of elements increases. But what is the order as an element of the correlation matrix? We
Free College Statistics Homework Help: How Students Achieve Better June 4, 2017 Students lose a
Pearsonmylabandmastering Compositions of the Subject This is a discussion on an article titled Compositions and
Spss 2008 – The Beginning of the Year Following on from my recent work in | 2,052 | 8,469 | {"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} | 3.5625 | 4 | CC-MAIN-2023-14 | latest | en | 0.793715 |
https://rows.com/calculators/customer-lifetime-value-calculator | 1,726,061,285,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00372.warc.gz | 480,550,655 | 159,961 | This tool calculates customer lifetime value (LTV) using simple key revenue metrics to estimate long-term customer profitability.
## What is Customer Lifetime Value (LTV)?
Customer Lifetime Value (LTV) is a common metric in the digital industry that estimates the total profit a business can expect from a single customer account throughout their relationship.
It goes beyond just revenue to account for the costs associated with servicing that customer, providing a fuller picture of the value a business extracts from a customer relationship. In fact, an essential element to factor in when calculating LTV is the Gross Margin: Gross Margin represents the percentage of revenue left after subtracting the costs associated with delivering your product or service.
## How to calculate the LTV of a customer
In essence, the LTV is calculated by multiplying the average revenue per customer by the expected duration of the customer relationship and then adjusting for profitability using the Gross Margin.
More in details, you need to implement the following steps:
1. Calculate the Annual Revenue Per Account (ARPA):** ARPA = Total Annual Recurring Revenue (ARR) / Total number of accounts
2. Determine the Annual Churn Rate:** This is typically expressed as a percentage of customers who leave within a year.
3. Compute the Expected Customer Lifetime (in years):** Expected Lifetime = 1 / Annual Churn Rate
4. Calculate the Customer Lifetime Revenue (LTR):** LTR = Expected Lifetime * ARPA
5. Determine the Gross Margin percentage:** This is the percentage of revenue that remains after accounting for the cost of goods sold.
6. Finally, calculate the Customer Lifetime Value (LTV):** LTV = Customer Lifetime Revenue * Gross Margin
Example: Let's say a company has:
• Total ARR: \$1,000,000
• Total accounts: 100
• Annual Churn Rate: 10%
• Gross Margin: 70%
As a result:
• ARPA = \$1,000,000 / 100 = \$10,000
• Expected Lifetime = 1 / 0.10 = 10 years
• LTR = 10 * \$10,000 = \$100,000
• LTV = \$100,000 * 0.70 = \$70,000
Therefore, the Customer Lifetime Value in this example is \$70,000.
## How to use this Customer Lifetime Value calculator
This calculator offers a quick way to estimate the LTV of a customer.
All you need to do is to enter the following variables:
• Total ARR (Annual Recurring Revenue)
• Total number of accounts
• (Average) Annual Churn rate
• Gross margin percentage (1 - COGS / ARR)
The calculator will then automatically return the following results:
• ARPA (Annual Revenue Per Account)
• Expected customer lifetime (in years)
Here are some key reasons why LTV should be a central metric in your business strategy:
• Informed marketing decisions: By knowing your LTV, you can determine how much you can afford to spend on acquiring new customers while maintaining profitability.
• LTV to CAC Ratio:
A typical benchmark is to aim for an LTV to CAC ratio of 3:1. This means that the revenue generated from a customer over their lifetime should ideally be three times the cost of acquiring that customer. A lower ratio might indicate that you are spending too much on customer acquisition relative to the value you're generating, while a higher ratio could suggest opportunities to invest more in growth. Discover more using our LTV to CAC calculator and CAC calculator.
• Customer segmentation: LTV helps identify your most valuable customer segments, allowing you to tailor your marketing and retention efforts accordingly.
• Retention focus: A high LTV highlights the value of investing in customer retention strategies, which are often more cost-effective than acquiring new customers.
• Product development: Understanding which products or services contribute most to LTV can guide your product development and improvement efforts.
• Forecasting and planning: LTV provides valuable insights for revenue forecasting and long-term business planning.
## FAQ
### What's the difference between CLV and LTV?
Customer Lifetime Value (CLV) and Lifetime Value (LTV) are often used interchangeably, but there can be subtle differences: - CLV typically focuses on the value of an individual customer. - LTV may sometimes refer to the average value across all customers.
For most practical purposes, these terms mean the same thing. Our calculator can be used for both CLV and LTV calculations.
### Can LTV be negative?
Yes, LTV can be negative, though it's a sign of an unhealthy business model. This situation occurs when the acquisition costs and ongoing expenses to maintain a customer exceed the revenue generated from that customer. A negative LTV indicates that a company is losing money on its customer relationships, which might suggest the need for adjustments in pricing, customer acquisition strategies, or operational efficiency.
### How often should I recalculate LTV?
It's important to recalculate LTV periodically to ensure that it reflects current business conditions. As customer behavior, pricing strategies, and business models evolve, so does the accuracy of your LTV. Recalculating LTV quarterly or bi-annually is often recommended, especially in fast-changing industries or during periods of significant business changes.
### How to calculate LTV for 5 years?
To calculate a 5-year LTV for a specific customer, you need to: 1. Determine its yearly expected revenue (ARPA). 2. Multiply it by Gross Margin to account for the costs generated by acquisition and ongoing maintenance of the client. 3. Apply a discount rate to account for the potential churn. 4. Sum the discounted values for the 5-year period.
## More than a Customer Lifetime Value Calculator
Rows is the easiest way to import, transform and share data in a spreadsheet.
### Import data from anywhere
Unleash your data: import from files, marketing tools, databases, APIs, and other 3rd-party connectors.
Know more
### Analyze with the power of AI
Unlock the power of AI on your data: ask the AI Analyst ✨ any question about your dataset and surface key insights, trends, and patterns.
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### Collaborate and Share
Seamlessly collaborate and share stunning reports with dynamic charts, embed options, and easy export features.
Know more | 1,294 | 6,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.88386 |
http://www.skillsyouneed.com/num/statistics-identifying-patterns.html | 1,484,792,082,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280425.43/warc/CC-MAIN-20170116095120-00307-ip-10-171-10-70.ec2.internal.warc.gz | 678,505,646 | 13,555 | # Statistical Analysis: Identifying Patterns
More advanced statistical analysis aims to identify patterns in data, for example, whether there is a link between two variables, or whether certain groups are more likely to show certain attributes.
This is in order to draw lessons from the sample that can be generalised to the wider population.
For more on making sure that your sample is large enough to allow you to generalise, see our page on Samples and Sample Design.
Relationships vs Differences
Research hypotheses can be expressed in terms of differences between groups, or relationships between variables. However, these are two sides of the same coin: almost any hypothesis could be set out in either way.
For example:
There is a relationship between gender and liking ice cream OR
Men are more likely to like ice cream than women.
## Comparing Groups
Your first step is to identify your two or more groups. This will obviously depend on your research question or hypothesis.
So if your hypothesis was that men are more likely to like ice cream than women, your two groups are men and women, and your data is likely to be something like self-expressed liking for ice cream on a scale of 1 to 5, or perhaps the number of times that ice creams are consumed each week in the summer months.
You then need to produce summary data for each group, usually mean and standard deviation. These may or may not look quite similar.
In order to decide whether there is a genuine difference between the two groups, you have to use a reference distribution against which to measure the values from the two groups.
The most common source of reference distributions is a standard distribution such as the normal distribution or t- distribution. These two are the same except that the standard deviation of the t-distribution is estimated from the sample, and that of the normal distribution is known.
You then compare the summary data from the two groups and decide the probability of achieving that difference by chance. The lower the probability, the more likely it is that your result is correct. This is referred to as statistical significance.
Types of Error
There are four possible outcomes from statistical testing:
• The groups are different, and you conclude that they are different (correct result)
• The groups are different, but you conclude that they are not (Type II error)
• The groups are the same, but you conclude that they are different (Type I error)
• The groups are the same, and you conclude that they are the same (correct result).
Type I errors are generally considered more important than Type II, because they have the potential to change the status quo.
For example, if you wrongly conclude that a new medical treatment is effective, doctors are likely to move to providing that treatment. Patients may receive the treatment instead of an alternative that could have fewer side effects, and pharmaceutical companies may stop looking for an alternative treatment.
## Choosing the Right Test
The test that you use to compare your groups will depend on how many groups you have, the type of data that you have collected, and also how good it is. In general, different tests are used for comparing two groups, and for comparing three or more.
Our page Surveys and Survey Design explains that there are two types of answer scale, continuous and categorical. Age, for example, is a continuous scale, although it can also be grouped into categories.
Gender is a category scale.
• For a continuous scale, you can use the means of the two groups that you are comparing.
• For a category scale, you need to use the medians.
Warning!
If you are not very confident about the quality of the data collected, for example because the inputting was done quickly and cheaply, or because the data have not been checked, then you may prefer to use the median even if the data are continuous to avoid any problems with outliers. This makes the tests more robust, and you can rely on the results more.
## What Test?
Purpose Data Scale Average Test Test Statistic Reference Distribution Compare two groups Continuous Mean t-test t t Category Median Mann-Whitney U test U statistic All combination of ranks Compare three or more groups Continuous Mean Analysis of Variance (ANOVA) F-ratio F Category Median Kruskal-Wallis Test W statistic All combination of ranks
Source: Easterby-Smith, Thorpe and Jackson, Management Research 4th Edition
### One- or Two-Tailed Test
The other thing that you have to decide is whether you are confident of the direction of the distance. In practice, this boils down to whether your research hypothesis is expressed as ‘x is likely to be more than y’, or ‘x is likely to be different from y’. If you are confident of the direction of the distance, then your test will be one-tailed. If not, it will be two-tailed.
### Calculating the Test Statistic
For each type of test, there is a standard formula for the test statistic. For example, for the t-test, it is:
(M1-M2)/SE(diff)
M1 is the mean of the first group
M2 is the mean of the second group
SE (diff) is the standard error of the difference, which is calculated from the standard deviation and the sample size of each group.
The final part of the test is to compare the test statistic to that required to meet the desired level of significance (usually 5% or 1%). This value is available from published statistical tables. If the test statistic is that value or more, then the difference between groups is said to be statistically significant at the 5% or 1% level.
NOTE: the significance level is sometimes called the p value, and expressed as p < 0.05 or p < 0.01.
### Comparing Variables
Sometimes, you may want to know if there is a link between two variables. If so, you can predict someone’s response to one variable by their response to the other.
• A positive association means that high scores for one variable tend to occur with high scores for the other.
• A negative association means that high scores for one variable tend to occur with low scores for the other.
• There is no association when the score for one variable does not predict the score for the other.
Such associations are also called correlations.
Seeing an Association
One of the best ways of checking for an association is to draw a line graph of the data with the two variables on the x and y axes. Broadly speaking, if there is an association, you will see it from the graph.
Drawing a graph will also help you identify if there is a peculiar relationship, such as a positive association for part of the data and a negative for the rest, as shown below. This will show in a test as no correlation, but there is clearly some sort of a relationship in this case.
### Statistical Tests for Associations
Again, there are specific tests depending on whether you are using continuous, categorical or ranked data.
• For categorical data, you use the chi-squared test (also written χ2)
• For continuous data it is the Pearson product-moment correlation
• For ranks, use the Kendall rank order correlation.
Again, you need to work out the test statistic, and compare that with the value needed to obtain the desired level of significance.
Warning! The Difference Between Correlation and Causation
A correlation is an association between two variables. It does not necessarily imply that one causes the other. Both could be caused by something completely different, or it could simply be that people who show one characteristic often show the other.
For example, it could be that people who shop for groceries online buy more ready-made meals than those who shop in store. However, it is unlikely that the act of buying online causes the purchase of more ready-meals. It is more likely that those who shop online are short of time, and so buy more convenience food, or possibly simply that younger people are both more likely to shop online and more likely to buy convenience food. | 1,620 | 7,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-04 | latest | en | 0.956696 |
http://scientifictutor.org/5188/chem-empirical-and-molecular-formulas-by-molar-mass-part-2/ | 1,721,302,126,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514828.10/warc/CC-MAIN-20240718095603-20240718125603-00778.warc.gz | 24,251,616 | 7,148 | ## Chem – Empirical and Molecular Formulas by Molar Mass Part 2
How is the empirical and molecular formula linked to the molar mass?
Another way that empirical and molecular formulas can be asked about is in terms of comparing their molar masses. Lets try that example we had before where the molecular formula was N4O10 and the empirical was N2O5. What is the molar mass of the molecular formula? It is approximately 216g/mol. What is the molar mass of the empirical formula? It is approximately 108g/mol. If we divide 216g/mol by 108g/mol what do we get? We get an answer of 2. How did we go from N4O10 to N2O5 before? We divided it by 2. This is not a coincidence. The same factor that is used in the subscripts of the molecular and empirical formulas is also used when it comes to their differences in molar mass. You can use this information to solve problems that involve molar masses and empirical and molecular formulas. Let us check out one style of these questions that I commonly see.
Examples: If the empirical formula is CH2, what is the molecular formula if the molar mass is 42 g/mol?
If the empirical formula is Al2O3, what is the molecular formula if the molar mass is 408 g/mol?
VIDEO Empirical and Molecular Formula Demonstrated Example 1: If the empirical formula is Ca3P2, what is the molecular formula if the molar mass is 546 g/mol? Use this periodic table if needed.
Step 1:
What information do they tell us in the problem?
empirical formula is Ca3P2…and its molar mass is 182 g/mol
molecular formula is ???…and its molar mass is 546 g/mol
Step 2:
What units, compounds, or information look similar or can we compare?
182 g/mol and 546 g/mol
Step 3:
How do we compare them?
Divide 546 g/mol by 182 g/mol = 3
Step 4:
What does that 3 represent?
It is the number that we multiply all the subscripts of the empirical formula by.
Step 5:
What is the molecular formula?
PRACTICE PROBLEMS: Solve the molecular formula problems.
If the empirical formula is CH2O, what is the molecular formula if the molar mass is 150 g/mol? | 512 | 2,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-30 | latest | en | 0.919522 |
https://library.kiwix.org/datascience.stackexchange.com_en_all_2021-04/A/question/38095.html | 1,628,097,117,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154878.27/warc/CC-MAIN-20210804142918-20210804172918-00167.warc.gz | 370,193,141 | 3,781 | ## How decision trees work in Python
1
1
I am new to the field of machine learning. I have just recently learnt Decision Trees and started solving Titanic Survival problem from Kaggle Competition. I understood the algorithm behind decision trees and how it actually works but I can't really match it with how python functions work this way. Say what exactly fit method does?
Well correct me if my concept is wrong. I think that looking at the train set i can build a model on "if a passenger survived or not" . Then I split the model into train set and test set and then the fit method (x_train, x_test) (this is what i think it works) actually tries to find the connection between x_train and y_train. I BUILT using my Train dataset and then tries to predict my test set. Say if i change my model, the fit method will try again to find the connection between my x_train and y_train using my new model. (Sorry for some terms as I am a self-learner. I don't really have the grasp of machine learning terms)
It will be of great help if you could suggest some simple data science problems which can be solved using Decision trees(besides Iris dataset)
The fit method is used only on X_train, y_train. It trains your model, and is the learning part. The predict method is where you try to apply what the model has learned on data the model has not seen before. This is the usual scikit-learn API, at least. – Adrian Keister – 2018-09-11T13:03:57.730
Can you explain if the fit method has any connection between training my model and my conditional approach? – Sadil Khan – 2018-09-11T13:05:42.627
Not sure what you mean by "conditional approach". What I've said generally applies to a gigantic array of models (not just decision trees) in scikit-learn. – Adrian Keister – 2018-09-11T13:07:06.103
Okay... But i can't just load a dataset and then try to split and fit... So i define a function f that checks if any passenger is Female then she survived, else died.. So my question is when i apply fit method will it consider the conditions mentioned in f – Sadil Khan – 2018-09-11T13:11:07.863
If your target variable is survived, the hope is certainly that your model will accurately predict if the passenger survived or not. You probably want your model to work on the entire Titanic dataset, and not just female passengers, right? You could restrict yourself to just the female passengers if you wanted. In any case, I would recommend that the training data and the test data be of the same kind. You're either training/testing on the full dataset, or you're training/testing on the female-only data, etc. – Adrian Keister – 2018-09-11T13:13:54.743 | 624 | 2,655 | {"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-2021-31 | latest | en | 0.941304 |
https://www.investingnote.com/posts/1137914 | 1,563,911,862,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529664.96/warc/CC-MAIN-20190723193455-20190723215455-00008.warc.gz | 714,759,017 | 9,795 | Based on my calculations, I expect the company to run out of cash by the end of the year
(As at 30 Sep 2018)
1) Cash amount: S\$5.4 million
2) Cash burn: ~S\$1.5 million per month
3) Estimated interest repayment= Total debt S\$38.16million * 3% interest / 12 months = ~S\$0.1 million per month
4) ST borrowing of S\$14.3 million repayable within 1 year = ~S\$1.2 million repayment per month
Total estimated cash outflow per month= (2) + (3) + (4)= ~S\$2.8 million per month
Cash runway from 30 sep 2018 = S\$5.4 million / S\$2.8 million= 1.9 months left (From 30 Sep 2018) before becoming insolvent
Can the company raise capital from rights issue? Probably. But I dont think this can be completed in time before the time runs out.
Can the company refinance its debt? Maybe, but i think banks may be unwilling to do so because of the following reasons:
Total debt amount = S\$38.16 million
Total possible collateralised PPE= S\$96.6 million
- ~S\$40 million is allocated to leashold property (Possible to collateralise against land)
- Remaining S\$46.6 million is from equipment, machinery and vessels (I dont think banks will accept the equipment, machinery and vessels as collateral as these are probably very illiquid, if yes the ltv will be below 20-30%)
Optimistic assumptions: Assume LTV of 70% for S\$40 million (leasehold property) and LTV of 30% for S\$46.6 million (equipment, machinery & vessels) possible refinance amount = S\$41.98 million.
This is close to its existing debt amount, so at best they can roll over their debt. But the cash burn of S\$1.5 million per month will continue unless company restructures further to reduce operating costs.
Also if you look at their balance sheet under ST assets, they have ~S\$11 million assets held for sale since 31 Dec 2017 until 30 Sep 2018, this value has not been converted to cash probably because the company is unable to find buyers for these assets (1 crane and some vessels)
Take a walk along West Coast Park waterfront area, you will be able to see the 2 floating platforms owned by company being idle and quite a number of vessels at the company waterfront yard just there without much repair activity. No/little work means little revenue to sustain the company.
cpa
Based on Q3 Cashflow statements, the net cash outflow is about \$1m for the Q3 (three months). This is not forgetting that they have \$0.7m of cash generated from operations for Q3.
You mentioned that the company is unable to find buyer for its assets. Pls refer to Management comments in the Q3 financial statement (ended 30 September 2018) which states the following:
During the quarter ended 30 September 2018, the Group acquired a 5000 DWT oil product tanker Angel Sun, measuring approximately 101.55 metres by 15.3 metres which is used for marine fuel trade. The vessel has been sold in the fourth quarter of the year for a profit.
WHLPLKPS | 722 | 2,894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.926399 |
https://bookofproofs.github.io/branches/geometry/euclidean-geometry/elements-euclid/book-11-elementary-stereometry/sum-of-plane-angles-used-to-construct-a-solid-angle-is-less-than-four-right-angles-proof.html | 1,695,496,086,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506528.19/warc/CC-MAIN-20230923162848-20230923192848-00693.warc.gz | 162,815,557 | 5,956 | # Proof: By Euclid
• Let $ABC$, $DEF$, and $GHK$ be the three given rectilinear angles, of which let (the sum of) two be greater than the remaining (one, the angles) being taken up in any (possible way), and, further, (let) the (sum of the) three (be) less than four right angles.
• So, it is necessary to construct a solid angle from (rectilinear angles) equal to $ABC$, $DEF$, and $GHK$.
• Let $AB$, $BC$, $DE$, $EF$, $GH$, and $HK$ be cut off (so as to be) equal (to one another).
• And let $AC$, $DF$, and $GK$ have been joined.
• It is, thus, possible to construct a triangle from (straight lines) equal to $AC$, $DF$, and $GK$ [Prop. 11.22].
• Let (such a triangle), $LMN$, have be constructed, such that $AC$ is equal to $LM$, $DF$ to $MN$, and, further, $GK$ to $NL$.
• And let the circle $LMN$ have been circumscribed about triangle $LMN$ [Prop. 4.5].
• And let its center have been found, and let it be (at) $O$.
• And let $LO$, $MO$, and $NO$ have been joined.
• I say that $AB$ is greater than $LO$.
• For, if not, $AB$ is either equal to, or less than, $LO$.
• Let it, first of all, be equal.
• And since $AB$ is equal to $LO$, but $AB$ is equal to $BC$, and $OL$ to $OM$, so the two (straight lines) $AB$ and $BC$ are equal to the two (straight lines) $LO$ and $OM$, respectively.
• And the base $AC$ was assumed (to be) equal to the base $LM$.
• Thus, angle $ABC$ is equal to angle $LOM$ [Prop. 1.8].
• So, for the same (reasons), $DEF$ is also equal to $MON$, and, further, $GHK$ to $NOL$.
• Thus, the three angles $ABC$, $DEF$, and $GHK$ are equal to the three angles $LOM$, $MON$, and $NOL$, respectively.
• But, the (sum of the) three angles $LOM$, $MON$, and $NOL$ is equal to four right angles.
• Thus, the (sum of the) three angles $ABC$, $DEF$, and $GHK$ is also equal to four right angles.
• And it was also assumed (to be) less than four right angles.
• The very thing (is) absurd.
• Thus, $AB$ is not equal to $LO$.
• So, I say that $AB$ is not less than $LO$ either.
• For, if possible, let it be (less).
• And let $OP$ be made equal to $AB$, and $OQ$ equal to $BC$, and let $PQ$ have been joined.
• And since $AB$ is equal to $BC$, $OP$ is also equal to $OQ$.
• Hence, the remainder $LP$ is also equal to (the remainder) $QM$.
• $LM$ is thus parallel to $PQ$ [Prop. 6.2], and (triangle) $LMO$ (is) equiangular with (triangle) $PQO$ [Prop. 1.29].
• Thus, as $OL$ is to $LM$, so $OP$ (is) to $PQ$ [Prop. 6.4].
• Alternately, as $LO$ (is) to $OP$, so $LM$ (is) to $PQ$ [Prop. 5.16].
• And $LO$ (is) greater than $OP$.
• Thus, $LM$ (is) also greater than $PQ$ [Prop. 5.14].
• But $LM$ was made equal to $AC$.
• Thus, $AC$ is also greater than $PQ$.
• Therefore, since the two (straight lines) $AB$ and $BC$ are equal to the two (straight lines) $PO$ and $OQ$ (respectively), and the base $AC$ is greater than the base $PQ$, the angle $ABC$ is thus greater than the angle $POQ$ [Prop. 1.25].
• So, similarly, we can show that $DEF$ is also greater than $MON$, and $GHK$ than $NOL$.
• Thus, the (sum of the) three angles $ABC$, $DEF$, and $GHK$ is greater than the (sum of the) three angles $LOM$, $MON$, and $NOL$.
• But, (the sum of) $ABC$, $DEF$, and $GHK$ was assumed (to be) less than four right angles.
• Thus, (the sum of) $LOM$, $MON$, and $NOL$ is much less than four right angles.
• But, (it is) also equal (to four right angles).
• The very thing is absurd.
• Thus, $AB$ is not less than $LO$.
• And it was shown (to be) not equal either.
• Thus, $AB$ (is) greater than $LO$.
• So let $OR$ have been set up at point $O$ at right angles to the plane of circle $LMN$ [Prop. 11.12].
• And let the (square) on $OR$ be equal to that (area) by which the square on $AB$ is greater than the (square) on $LO$ [Prop. 11.23 lem.] .
• And let $RL$, $RM$, and $RN$ have been joined.
• And since $RO$ is at right angles to the plane of circle $LMN$, $RO$ is thus also at right angles to each of $LO$, $MO$, and $NO$.
• And since $LO$ is equal to $OM$, and $OR$ is common and at right angles, the base $RL$ is thus equal to the base $RM$ [Prop. 1.4].
• So, for the same (reasons), $RN$ is also equal to each of $RL$ and $RM$.
• Thus, the three (straight lines) $RL$, $RM$, and $RN$ are equal to one another.
• And since the (square) on $OR$ was assumed to be equal to that (area) by which the (square) on $AB$ is greater than the (square) on $LO$, the (square) on $AB$ is thus equal to the (sum of the squares) on $LO$ and $OR$.
• And the (square) on $LR$ is equal to the (sum of the squares) on $LO$ and $OR$.
• For $LOR$ (is) a right angle [Prop. 1.47].
• Thus, the (square) on $AB$ is equal to the (square) on $RL$.
• Thus, $AB$ (is) equal to $RL$.
• But, each of $BC$, $DE$, $EF$, $GH$, and $HK$ is equal to $AB$, and each of $RM$ and $RN$ equal to $RL$.
• Thus, each of $AB$, $BC$, $DE$, $EF$, $GH$, and $HK$ is equal to each of $RL$, $RM$, and $RN$.
• And since the two (straight lines) $LR$ and $RM$ are equal to the two (straight lines) $AB$ and $BC$ (respectively), and the base $LM$ was assumed (to be) equal to the base $AC$, the angle $LRM$ is thus equal to the angle $ABC$ [Prop. 1.8].
• So, for the same (reasons), $MRN$ is also equal to $DEF$, and $LRN$ to $GHK$.
• Thus, the solid angle $R$, contained by the angles $LRM$, $MRN$, and $LRN$, has been constructed out of the three plane angles $LRM$, $MRN$, and $LRN$, which are equal to the three given (rectilinear angles) $ABC$, $DEF$, and $GHK$ (respectively).
• (Which is) the very thing it was required to do.
Thank you to the contributors under CC BY-SA 4.0!
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@Fitzpatrick | 1,904 | 5,587 | {"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} | 4.15625 | 4 | CC-MAIN-2023-40 | latest | en | 0.893921 |
https://testbook.com/question-answer/a-simply-supported-beam-of-length-l-carries-load--60009e9ad526822f5a439766 | 1,632,056,607,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056890.28/warc/CC-MAIN-20210919125659-20210919155659-00037.warc.gz | 576,006,519 | 29,762 | # A simply supported beam of length l, carries load w(x) = wox over the entire span. Location of the maximum bending moment in the beam at x will be
This question was previously asked in
UPPSC AE Mechanical: 13 Dec 2020 Official Paper I
View all UPPSC AE Papers >
1. $$\frac{l}{3}$$
2. $$\frac{l}{\sqrt 3}$$
3. $$\frac{l \sqrt 3}{2}$$
4. $$\frac{l}{\sqrt 2}$$
## Answer (Detailed Solution Below)
Option 2 : $$\frac{l}{\sqrt 3}$$
## Detailed Solution
Concept:
w(x) = wox
The given loading equation is of a uniformly varying load as shown in the figure below.
To calculate the location of maximum bending moment, we need to find the point at which the shear force is zero i.e. $$\frac{dM}{dx}=S$$; where M and S represents bending moment and shear force respectively.
Calculation:
Given:
The loading diagram can be replaced into a point load to calculate the reaction at supports.
ΣFy = 0
RA + RB$$\frac{w_ol^2}{2}$$
ΣMA = 0
$$(R_B\times l)-\frac{w_ol^2}{2}\times\frac{2l}{3}=0$$
$$\therefore R_B=\frac{w_ol^2}{3}\;\&\;R_A=\frac{w_ol^2}{6}$$
Shear force equation is given at any 'x':
$$R_A-\frac{w_ox^2}{2}$$
$$\frac{w_ol^2}{6}-\frac{w_ox^2}{2}$$
For location of maximum bending moment:
$$\frac{w_ol^2}{6}-\frac{w_ox^2}{2}=0$$
$$x=\frac{l}{\sqrt 3}$$ | 441 | 1,270 | {"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} | 4.28125 | 4 | CC-MAIN-2021-39 | latest | en | 0.735516 |
http://oeis.org/A059834 | 1,575,998,683,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540528457.66/warc/CC-MAIN-20191210152154-20191210180154-00334.warc.gz | 109,345,915 | 4,501 | This site is supported by donations to The OEIS Foundation.
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A059834 Sum of squares of entries of Wilkinson's eigenvalue test matrix of order 2n+1. 1
0, 6, 18, 40, 76, 130, 206, 308, 440, 606, 810, 1056, 1348, 1690, 2086, 2540, 3056, 3638, 4290, 5016, 5820, 6706, 7678, 8740, 9896, 11150, 12506, 13968, 15540, 17226, 19030, 20956, 23008, 25190, 27506, 29960, 32556, 35298, 38190, 41236, 44440 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS The m X m Wilkinson matrix is a symmetric tridiagonal matrix. If m = 2k + 1, its main diagonal is k, k - 1, ..., 1, 0, 1, ... k - 1, k. If m = 2k, its main diagonal is k - 1/2, k - 3/2, ..., 3/2, 1/2, 1/2, 3/2, ..., k - 3/2, k - 1/2. In both cases, it has all 1's on the diagonals next to the main diagonal and 0's elsewhere. - David Wasserman, May 24 2002 LINKS FORMULA a(n) = (2n^3 + 3n^2 + 13n)/3. For the matrix of order 2n, the formula is (4n^3 + 23n - 12)/6 (which is not integer-valued). - David Wasserman, May 24 2002 a(n) = sum(2*(k+1)^2+4, k=0..(n-1)). - Mike Warburton, Sep 08 2007 G.f.: 2*x*(3-3*x+2*x^2)/(1-x)^4. - Colin Barker, Apr 04 2012 EXAMPLE The matrix of order 5: 2 1 0 0 0 1 1 1 0 0 0 1 0 1 0 0 0 1 1 1 0 0 0 1 2 PROG (MATLAB) for i = 0:20 a(i+1) = trace( wilkinson(2*i+1)*wilkinson(2*i+1)' ); end; a CROSSREFS Cf. A059831. Sequence in context: A299256 A002411 A023658 * A299263 A015224 A163983 Adjacent sequences: A059831 A059832 A059833 * A059835 A059836 A059837 KEYWORD nonn,easy AUTHOR N. J. A. Sloane, Feb 25 2001 EXTENSIONS More terms from David Wasserman, May 24 2002 STATUS approved
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Last modified December 10 12:22 EST 2019. Contains 329895 sequences. (Running on oeis4.) | 882 | 2,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-51 | latest | en | 0.733123 |
https://www.samacheerkalvibook.com/2021/12/samacheer-kalvi-8th-science-chapter-3-light-notes.html | 1,653,160,737,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662540268.46/warc/CC-MAIN-20220521174536-20220521204536-00035.warc.gz | 1,154,169,401 | 45,268 | Samacheer Kalvi Books: Tamilnadu State Board Text Books Solutions
Samacheer Kalvi 8th Science Chapter 3 Light Notes PDF Download: Tamil Nadu STD 8th Science Chapter 3 Light Notes
Samacheer Kalvi 8th Science Chapter 3 Light Notes PDF Download: Tamil Nadu STD 8th Science Chapter 3 Light Notes
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Tamilnadu Samacheer Kalvi 8th Science Solutions Term 1 Chapter 3 Light
Samacheer Kalvi 8th Science Light Text Book Exercises
I. Choose the best answer
8th Science Light Question And Answer Question 1.
Mirrors having a curved reflecting surface are called as –
(a) Plane mirrors
(b) Spherical mirrors
(c) Simple mirrors
(d) None of the above
Answer:
(b) Spherical mirrors
Light Book Back Answers Question 2.
The spherical mirror with a reflecting surface curved inward is called –
(a) Convex mirror
(b) Concave mirror
(c) Curved mirror
(d) None of the above
Answer:
(b) Concave mirror
8th Std Science Light Lesson Question 3.
The centre of a sphere of which the reflecting surface of a spherical mirror is a part is called –
(a) Pole
(b) Centre of curvature
(c) Cradius of curvature
(d) Aperture
Answer:
(b) Centre of curvature
8th Standard Science Light Lesson Question 4.
The spherical mirror used as a rear view mirror in the vehicle is –
(a) Concave mirror
(b) Convex mirror
(c) Plane mirror
(d) None of the above
Answer:
(b) Convex mirror
8th Standard Light Lesson Question Answer Question 5.
The imaginary line passing through the centre of curvature and pole of a spherical mirror is called –
(a) Centre of curvature
(b) Pole
(c) principal axis
(d) Radius curvature
Answer:
(c) principal axis
Light Samacheer Kalvi Question 6.
The distance from the pole to the focus is called –
(a) Pole length
(b) Focal length
(c) principal axis
(d) None of the above
Answer:
(b) Focal length
8th Standard Science Light Lesson Question Answer Question 7.
Focal length is equal to half of the –
(a) Centre of curvature
(b) Axis
(c) Radius of curvature
(d) None of the above
Answer:
(c) Radius of curvature
8th Science Light Lesson Question 8.
If the focal length of a spherical mirror is 10 cm, what is the value of its radius of curvature?
(a) 10 cm
(b) 5 cm
(c) 20 cm
(d) 15 cm
Answer:
(c) 20 cm
Samacheer Kalvi Guru 8th Science Question 9.
If the image and object distance is same, then the object is placed at –
(a) Infinity
(b) At F
(c) Between f and P
(d) At C
Answer:
(d) At C
Light 8th Standard Question 10.
The refractive index of water is –
(a) 1.0
(b) 1.33
(c) 1.44
(d) 1.52
Answer:
(b) 1.33
II. Fill in the blanks
1. The spherical mirror used in a beauty parlour as make – up mirror is ………………
2. Geometric centre of the spherical mirror is ………………
3. Nature of the images formed by a convex mirror is ………………
4. The mirror used by the ophthalmologist to examine the eye is ………………
5. It the angle of incidence is 45°, then the angle of reflection is ………………..
6. Two mirrors are parallel to each other, then the number of images formed is ………………
Answer:
1. Concave mirror
2. pole
3. Smaller, virtual and erect
4. Concave mirror
5. 45°
6. Infinite
III. Match the following
Samacheerkalvi.Guru 8th Science Question A.
Answer:
1. c
2. a
3. d
4. b
Samcheer Kalvi.Guru 8th Question B.
Answer:
1. b
2. c
3. d
4. a
IV. Answer in brief
Samacheer Kalvi Guru 8th Standard Science Question 1.
What is called a spherical mirror?
Answer:
Spherical mirrors are one form of curved mirrors. If the curved mirror is a part of a sphere, then it is called a ‘spherical mirror’.
Samacheer Kalvi 8th Science Question 2.
Define focal length.
Answer:
The distance between the pole and the principal focus is called focal length (f) of a spherical mirror.
Samacheer Kalvi 8th Science Book Solutions Question 3.
The radius of curvature of a spherical mirror is 25 cm. Find its focal length.
Answer:
Given : Radius of curvature = 25 cm
To find: f = ?
Solution:
f = R2 = 252
f = 12.5 cm
Samacheer Kalvi.Guru 8th Science Question 4.
Give two applications of a concave and convex mirror.
Answer:
Concave mirrors:
1. Concave mirrors are used while applying make – up or shaving, as they provide a magnified image.
2. They are used in torches, search lights and head lights as they direct the light to a long distance.
Convex mirrors:
1. Convex mirrors are used in vehicles as rear view mirrors because they give an upright image and provide a wider field of view as they are curved outwards.
2. They are found in the hallways of various buildings including hospitals, hotels, schools and stores. They are usually mounted on a wall or ceiling where hallways make sharp turns.
Samacheer Guru 8th Science Question 5.
State the laws of reflection.
Answer:
1. The incident ray, the reflected ray and the normal at the point of incidence, all lie in the same plane.
2. The angle of incidence and the angle of reflection are always equal.
8th Standard Light Lesson Question 6.
If two plane mirrors are inclined to each other at an angle of 45°, find the number of images formed.
Answer:
Given :
Angle of inclination = 45°
To find : Number of images formed = 360°angle – 1
Solution:
360°45 – 1
= 8 – 1
= 7 images
Question 7.
Define the refractive index of a medium.
Answer:
The amount of refraction of light in a medium is denoted by a term known as refractive index of the medium, which is the ratio of the speed of light in the air to the speed of light in that particular medium.
Question 8.
State the Snell’s law of refraction.
Answer:
Refraction of light rays, as they travel from one medium to another medium, obeys two laws, which are known as Snell’s laws of refraction. They are:
1. The incident ray, the refracted ray and the normal at the point of intersection, all lie in the same plane.
2. The ratio of the sine of the angle of incidence (i) to the sine of the angle of refraction (r) is equal to the refractive index of the medium, which is a constant.
sinisinr = µ
V. Answer in detail
Question 1.
Explain the images formed by a concave mirror?
Answer:
Question 2.
What is reflection? Write short notes on regular and irregular reflection?
Answer:
A ray of light, falling on a body having a shiny polished and smooth surface alone is bounced back. This bouncing back of the light rays as they fall on the smooth, shiny and polished surface is called reflection.
Regular reflection:
1. When a beam of light (collection of parallel rays) falls on a smooth surface, it gets reflected.
2. After reflection, the reflected rays will be parallel to each other. Here, the angle of incidence and the angle of reflection of each ray will be equal.
3. Hence, the law of reflection is obeyed in this case and thus a clear image is formed. This reflection is called ‘regular reflection’ or ‘specular reflection’.
Irregular reflection:
1. In the case of a body having a rough or irregular surface, each region of the surface is inclined at different angles.
2. When light falls on such a surface, the light rays are reflected at different angles.
3. In this case, the angle of incidence and the angle of reflection of each ray are not equal.
4. Hence, the law of reflection is not obeyed in this case and thus the image is not clear. Such a reflection is called ‘irregular reflection’ or ‘diffused reflection’.
Question 3.
Explain the working of a periscope.
Answer:
Periscope:
1. It is an instrument used for viewing bodies or ships, which are over and around another body or a submarine.
2. It is based on the principle of the law of reflection of light.
3. It consists of a long outer case and inside this case mirrors or prisms are kept at each end, inclined at an angle of 45°.
4. Light coming from the distant body, falls on the mirror at the top end of the periscope and gets reflected vertically downward.
5. This light is reflected again by the second mirror kept at the bottom, so as to travel horizontally and reach the eye of the observer.
6. In some complex periscopes, optic fibre is used instead of mirrors for obtaining a higher resolution.
7. The distance between the mirrors also varies depending on the purpose of using
Question 4.
What is dispersion? Explain in detail.
Answer:
1. Splitting of white light into its seven constituent colours (wavelength), on passing through a transparent medium is known as dispersion of light.
2. Dispersion occurs because, light of different colours present in white light have different wavelength and they travel at different speeds in a medium.
3. Refraction of a light ray in a medium depends on its speed.
Question 5.
Speed of light in air is 3 x 108 m s-1 and the refractive index of a medium is 1.5. Find the speed of light in the medium.
Answer:
Given :
Speed of light in air c = 3 x 108 ms-1
Refractive index of a medium µ = 1.5
To find : Speed of light in medium v = ?
Formula :
µ = cv
Solution:
1.5 = 3×108v
v =3×1081.5
v = 2 x 108 ms-1
Speed of light in medium v = 2 x 108 ms-1
Samacheer Kalvi 8th Science Solutions Light Additional Questions
I. Choose the correct answer
Question 1.
Which object use the reflection of light?
(a) Kaleidoscope
(b) Plane mirror
(c) Convex mirror
(d) All of these
Answer:
(d) All of these
Question 2.
Which surface will not reflect most of the light falling on them?
(a) Rough surface
(b) Smooth surface
(c) Shining surface
(d) Opaque surface
Answer:
(a) Rough surface
Question 3.
Which of the following demonstrates the law of reflection?
Answer:
Question 4.
The ENT doctor uses a …………….
(a) Plane mirror
(b) Concave mirror
(c) Convex mirror
(d) Convex lens
Answer:
(b) Concave mirror
Question 5.
In dispersion, the colour of light that will bend more is ……………..
(a) Red
(b) yellow
(c) Green
(d) Violet
Answer:
(d) violet
Question 6.
Reflection by a looking mirror is called …………….
(a) Regular reflection
(b) Irregular reflection
(c) Regular and irregular reflection
(d) None of these
Answer:
(a) Regular reflection
Question 7.
The velocity of light in vacuum is 3 x 108 ms-1 and in glass is 2 x 108 ms-1. The refractive index of glass is .
(a) 2
(b) 1.5
(c) 1.8
(d) 1.33
Answer:
(b) 1.5
Question 8.
Incident angle of a ray of light is 30°. The angle between the incidend ray and the reflected ray is …………….
(a) 50°
(b) 90°
(c) 60°
(d) 15°
Answer:
(c) 60°
Question 9.
In the head lights of motor vehicles, ……………… mirrors are used as reflectors.
(a) Plane mirrors
(b) Concave lenses
(c) Convex mirrors
(d) Concave mirrors
Answer:
(d) Concave mirrors
Question 10.
The phenomenon of light passing through the object is called ……………..
(a) Reflection
(b) Refraction
(c) Dispersion
(d) Total internal reflection
Answer:
(b) Refraction
II. Fill in the Blanks
1. The bouncing back of light from a surface is called …………….
2. ……………. mirrors make things look larger when objects are placed close to it
3. Convex mirror always forms ……………. and ……………. image.
4. The incident ray, ……………. ray and the ……………. at the point of incidence, all lie on the same plane.
5. A ray of light incident along normal to the mirror ……………. its path.
6. When light passes from one medium to another the ray gets bent. This property of light is called ……………..
7. Spherical mirrors are one form of …………… mirrors.
8. ……………… mirrors magnify the object placed close to them.
9. The image formed by convex mirrors is ……………… than the object
10. …………….. mirrors form the perfect image of an object.
11. The …………….. of a mirror determines the type of image it forms.
12. The ……………. is an optical device with a polished surface that reflects the light falling on it.
Answer:
1. Reflection
2. Concave
3. Virtual and erect
4. Reflected, normal
5. Retraces
6. Refraction
7. Curved
8. Concave
9. Smaller
10. Plane
11. Shape
12. Mirror
III. True or False – if false give the correct statement
Question 1.
We can see things around us only when the light reflected by them reaches our eyes.
Answer:
True.
Question 2.
Light is a form of energy and it travels in a straight line.
Answer:
True.
Question 3.
The periscope is an optical device with a polished surface that reflects the light falling on it.
Answer:
False.
Correct statement:
The mirror is an optical device with a polished surface that reflects the light falling on it.
Question 4.
Curved mirrors have surfaces that are spherical, cylindrical, parabolic and ellipsoid.
Answer:
True.
Question 5.
Curved mirrors form the perfect image of an object.
Answer:
False. Correct statement: Plane mirrors form the perfect image of an object.
Question 6.
Curved mirrors produce images that are either enlarged or diminished.
Answer:
True.
Question 7.
A thin layer of molten aluminium or silver is used for coating glass plates that will then become mirrors.
Answer:
True.
Question 8.
The most common example of a convex mirror is the make – up mirror.
Answer:
False.
Correct statement:
The most common example of a concave mirror is the make – up mirror.
IV. Match the following
Question 1.
Answer:
i. e
ii. d
iii. a
iv. c
v. b
Question 2.
Answer:
i. e
ii. d
iii. a
iv. c
v. b
Question 3.
Answer:
i. d
ii. c
iii. b
iv. e
v. a
V. Assertion and Reason
Mark the correct choice as:
(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.
(b) If both assertion and reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
Question 1.
Assertion : A ray incident along normal to the mirror retraces its path
Reason : In reflection, angle of incidence is always equal to angle of reflection.
Answer:
(a) Both assertion and reason are true and the reason is the correct explanation of the assertion
Question 2.
Assertion : Convex mirrors are used as rear view mirror in vehicles for observing traffic at our back.
Reason : A convex mirror has a much larger field of view.
Answer:
(a) Both assertion and reason are true and the reason is the correct explanation of the assertion
Question 3.
Assertion : The mirrors used in search lights are parabolic and not concave spherical.
Reason : In concave spherical mirror the image formed is always virtual.
Answer:
(c) The assertion is true, but the reason is false
Correct explanation:
In search lights, we need an intense parallel beam of light. If a source is placed at the focus of a concave spherical mirror only paraxial rays are rendered parallel. Due to large aperture of mirror, marginal rays give a divergent beam. But in case of parabolic mirror, when source is at the focus, beam of light produced over the entire cross – section of the mirror is a parallel beam.
Question 4.
Assertion : We can see the rainbow in the sky when the rain starts falling after a spell of bright sunlight.
Reason : The rainbow is formed due to the dispersion of light.
Answer:
(d) The assertion is false, but the reason is true
Correct explanation:
The rainbow is formed when the light passes through the water droplets in air after it rains and gets dispersed into seven colours.
VI. Very short Answers
Question 1.
How does the light travel?
Answer:
The light travels along straight lines.
Question 2.
What is reflection of light?
Answer:
The bouncing back of light when it falls on smooth surface is called reflection.
Question 3.
What is mirror?
Answer:
The mirror is an optical device with a polished surface that reflects the light falling on it.
Question 4.
What type of image is formed by a concave mirror?
Answer:
Real and inverted image. If the object is placed very close to the mirror then the image is virtual and erect.
Question 5.
What is rainbow?
Answer:
The rainbow is seen as a large area in the sky with many colours.
Question 6.
Name the triangular piece of glass that splits white light into different colours.
Answer:
Prism.
Question 7.
What is the composition of sunlight?
Answer:
Sunlight is a mixture of seven colours.
Question 8.
Light bends as it passes from one medium to another. What is this phenomenon called?
Answer:
Refraction of light.
Question 9.
Name the two types of spherical mirrors.
Answer:
1. Concave mirror
2. Convex mirror.
Question 10.
The angle between incident ray and reflected ray is 60°. What is the value of angle of incidence?
Answer:
since angle of incidence = angle of reflection.
So, angle of incidence = 30°.
VII. Short Answer
Question 1.
Light travels fastest in vacuum. Why?
Answer:
Light travels fastest in vacuum than any other medium because there is no obstruction to the passage of light in vacuum.
Question 2.
Distinguish between real and virtual images.
Answer:
Real image:
Type of image which can be obtained on a screen is called a real image
Virtual image:
An image which cannot be obtained on a screen is called a virtual image.
Question 3.
State any two uses of concave mirrors.
Answer:
1. It is used as a reflector in torches, light houses, head lights of vehicles, etc., as it diverges the rays of light.
2. A dentist uses a concave mirror to obtain a magnified image of the teeth of the patient.
Question 4.
A convex rear view mirror is preferred over a plane mirror in a car. Why?
Answer:
Since convex mirror forms a smaller and virtual image, it can be used to see a much larger area than the area visible by a plane mirror.
Question 5.
What type of image is formed by a concave mirror?
Answer:
The image formed by a concave mirror is real and inverted. If the object is placed very rear to the mirror then the image formed is virtual and erect.
Question 6.
Why do we need a shiny surface for reflection?
Answer:
The extent of reflection depends upon the shine and smoothness of the surface. More is the shine and smoothness of the surface more will be the reflection.
Question 7.
The radius of curvature of a spherical mirror is 18 cm. What is the focal length of this mirror?
Answer:
f = R2
R = 18 cm
f = 182 = 9 cm.
Question 8.
What happens to light when it gets dispersed? Give an example.
Answer:
Light is splitted into its constituent colours, when it gets dispersed.
Example:
Rainbow formation is due to the dispersion of white light after passing through water droplets.
Question 9.
If two mirrors are placed at an inclination of 30° then how many images can be seen?
Answer:
Formula : Number of images N = 360°30° – 1
Given: θ = 30°
Solution:
N = 360°30° – 1
= 12 – 1 = 11 images.
Question 10.
What is the speed of light in diamond if its refractive index is 2.41?
Formula :
Refractive index µ =Speedoflightinair(c)Speedoflightinthemedium(v)
Given : µ = 2.41
Solution:
c = 3 x 108ms-1
µ = cv
v =3×1082.41
Speed of light in diamond v = 1.24 x 108 ms-1
Question 11.
A light ray moves from glass (Vglass = 2.0 x 108ms-1) to diamond (Vdiamond = 1.25 x 108 ms-1). What is the refractive index of diamond with respect to glass?
Answer:
Refractive index of diamond with respect to glass
Solution:
200125
= 1.60 (No unit).
Question 12.
Find the refractive index of water with respect to glass if the refractive index of water is 43 and the reiractive index of glass is 32
Answer:
Solution:
43 x 23
89
Thus, refractive index
of water with respect to glass = 89 (No unit).
VIII. Long Answer
Question 1.
Differentiate between regular and irregular reflection.
Answer:
Regular reflection:
1. When a beam of light (collection of parallel rays) falls on a smooth surface, it gets reflected. After reflection, the reflected rays will be parallel to each other.
2. Here, the angle of incidence and the angle of reflection of each ray will be equal.
3. Hence, the law of reflection is obeyed in this case and thus a clear image is formed.
4. This reflection is called ‘regular reflection’ or ‘specular reflection’.
Example : Reflection of light by a plane mirror and reflection of light from the surface of still water.
Irregular reflection:
1. In the case of a body having a rough or irregular surface, each region of the surface is inclined at different angles.
2. When light falls on such a surface, the light rays are reflected at different angles. In this case, the angle of incidence and the angle of reflection of each ray are not equal.
3. Hence, the law of reflection is not obeyed in this case and thus the image is not clear. Such a reflection is called ‘irregular reflection’ or ‘diffused reflection’.
Example : Reflection of light from a wall.
Question 2.
Explain the construction and working of kaleidoscope.
Answer:
Construction:
1. Take three equal sized plane mirror strips and arrange them in such a way that they form an equilateral triangle.
2. Cover the sides of the mirrors with a chart paper. With the help of a chart paper cover the bottom of the mirrors also.
3. Put some coloured things such as pieces of bangles and beads inside it.
4. Now, cover the top portion with the chart paper and make a hole in it to see.
5. You can wrap the entire piece with coloured papers to make it attractive.
6. Now, rotate it and see through its opening. You can see the beautiful patterns.
Question 3.
Explain the construction, working of periscope with a neat labelled diagram.
Answer:
1. It is an instrument used for viewing bodies or ships, which are over and around another body or a submarine.
2. It is based on the principle of the law of reflection of light. It consists of a long outer case and inside this case mirrors or prisms are kept at each end, inclined L’9ht at an angle of 45°.
3. Light coming from the distant body, falls on the mirror at the top end of the periscope and gets reflected 450 vertically downward.
4. This light is reflected again by the second mirror kept at the bottom, so as to travel horizontally and reach the eye of the observer.
5. In some complex periscopes, optic fibre is used instead of mirrors for obtaining a higher resolution.
6. The distance between the mirrors also varies depending on the purpose of using the periscope.
Question 4.
Explain the uses of periscope.
Answer:
1. It is used in warfare and navigation of the submarine.
2. In military it is used for pointing and firing guns from a ‘bunker’.
3. Photographs of important places can be taken through periscopes without trespassing restricted military regions.
4. Fibre optic periscopes are used by doctors as endoscopes to view internal organs of the body.
Question 5.
Explain the images formed by a convex mirror for different position of the object.
Answer:
Question 6.
Answer:
1. A pencil placed in water in a glass appear to be bent at the surface of separation.
2. The base of a swimming pool appears to be raised due to refraction of light.
3. We should look straight down into the river while catching fish. If observed obliquely, the fish appears to be closer to the surface of the water than it actually is.
4. Stars look like tiny dots when viewed from earth, because they are far away from us. The light from the stars has to travel a long distance through the atmosphere before it reaches us. The atmosphere has various layers of different refractive indices. Light is refracted through each of them. These layers are turbulent. Every time one of the layers shifts, the thickness and refractive indices of air change and the light is refracted differently, creating a twinkling effect.
IX. Complete the given table
Question 1.
Answer:
1. At F
2. Diminished
3. At C
4. Magnified
5. Virtual and erect
X. Draw the following
Question 1.
Concave mirror.
Answer:
Question 2.
Convex mirror.
Answer:
Question 3.
Draw the ray diagram and write the characteristics of the image formed when an object is placed
1. At infinity in front of a concave mirror.
2. At infinity in front of a convex mirror.
Answer:
Position of the image : At F
Image size : Highly diminished
Nature of the image : Real and inverted
Position of the image : At F
Image size : Highly diminished , point sized
Nature of the image : Virtual and erect
Question 4.
Draw a neat labelled diagram of a periscope.
Answer:
Question 5.
Draw a ray diagram to show a light ray travels from denser medium (glass) to rarer medium (air).
Answer:
When a light ray travels from denser medium to rarer medium, it speeds up and refracts away from the normal.
Question 6.
Draw a ray diagram to show a light ray travels from rarer medium (air) to denser medium (water).
Answer:
When a light ray travels from rarer to denser medium, it bends towards the normal.
XI. Cross word puzzle
Across:
3. The geometrical centre of a spherical mirror.
6. Centre of the sphere from which the mirror is made.
7. An optical device with a polished surface that reflects the light falling on it.
8. Image which can be formed on a screen.
9. Image which cannot be formed on a screen.
10. The bending of a light ray when it passes from one medium to another medium of different density.
Down:
1. Mirror which converges a parallel beam of light passing through it.
2. Imaginary line passing through the centre of curvature of the mirror.
4. Mirror which diverges a parallel beam of light passing through it.
5. The formation of rainbow is an example of ______.
Answer:
Across:
3. Pole
6. Centre Of Curvature
7. Mirror
8. Real
9. Virtual
10. Refraction
Down:
1. Concave
2. Principal Axis
4. Convex
5. Dispersion
XI. Creative questions : HOTS
Question 1.
Imagine that parallel rays are incident on an irregular surface. Are the rays reflected from the surface parallel to each other?
Answer:
No, the reflected rays from irregular surface are in different direction.
Question 2.
A safety vest helps to keep the workers who are working by the roadside safe. This especially so during the nights. Why?
Answer:
The reflectors on the safety vest reflect light into the motorists eyes.
This help to alert the motorists of the wearer’s presence on the road.
Question 3.
What is the difference between virtual images of an object formed by a concave mirror and a convex mirror?
Answer:
The virtual image of an object formed by a concave mirror is always magnified one but the image formed by a convex mirror is always diminished one.
Question 4.
What is a virtual image? Give one situation where a virtual image is formed.
Answer:
The image formed by the plane mirror appears behind it. We cannot however touch it. Also, the image of the object cannot be obtained on a screen, whether it is held in front of the mirror or behind it. Such type of images are not real. They are virtual images.
Question 5.
If all objects around us were to reflect light in a regular way, what problem might we face?
Answer:
1. Irregular reflection is what makes us see all the objects and everything around us.
2. If light were to get regularly reflected then every object would act like a mirror and our surroundings would be illuminated.
3. This would have a blinding effect on eyes making it harder for us to see.
Question 6.
Car rear view mirrors carry a warning message that “objects in the rear view mirror are closer than they appear”. Why do you think this is so?
Answer:
1. Convex mirrors used in vehicles as rear – view mirrors are labeled with the safety warning: ‘Objects in the mirror are closer than they appear’ to warn the drivers. This is because inside the mirrors, vehicles will appear to be coming at a long distance.
2. Convex mirrors form erect and smaller images of the objects.
3. This does not give us the exact idea how far the vehicle is from us.
4. Thus, to avoid accidents, car view mirrors carry a warning message.
How to Prepare using Samacheer Kalvi 8th Science Chapter 3 Light Notes PDF?
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• Use Samacheer Kalvi 8th Science Chapter 3 Light notes by paying attention to facts and ideas.
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• Refer TN Board books as well as the books recommended.
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# MSP430F4794: Protection of SD16_A inputs against negative voltages
Part Number: MSP430F4794
Hello,
We consider connecting amplifiers (RRO), which have to be supplied with relatively high dual voltage (+5V and -5V), to the SD16_A inputs via protecting resistors (say 3kOhm). In some error conditions the amp output voltages can exceed the SD16_A voltage range. Positive overvoltages will be clamped by MSP430 internal diodes connected to AVcc (and the resistors will limit the currents within safe 2mA range). But what about negative voltages? It seems there is no similar clamping on analog inputs. However they should have some built-in protection against negative ESD at least.
Thanks,
Marcin
• Note the Absolute Maximum Ratings:
The most negative voltage is -0.3 V, implying a clamp diode from VSS to the input.
• I know this limit. But for analog inputs of SD16_A the absolute limit is different: AVss-1V. Anyway my question is not about the limits, but about internal protection circuit.
• Where do you see that? If that is indeed the case, then TI needs to change the Absolute Maximum ratings.
• Near the end of p. 47 of the same datasheet:
And our measurement shows there is no clamping near the -1V limit.
• So, TI will need to step in and explain the discrepancy. I would guess, that the esd protection for the analog inputs is 3 ESD diodes in series.
• We would like such 3-diode protection. The problem is that even -5V provided via resistor (3kOhm) is not clamped on an analog input (Ax.0+). (And it is not a result of chip damage; analog inputs behave this way for each chip piece; and when input signal returns to normal range, the SD16_A works as expected.)
• The purpose of the clamping diodes, if present, is to prevent latchup. (CMOS will latch up when the input voltage exceeds the voltages rails enough to trigger a parasitic SCR.) Any ESD protection is secondary.
You can't assume all pins will have these diodes. Pins which have multiple functions including ADC input probably do. A dedicated ADC input perhaps not.
In this case where you know that the input circuit can drive the input pin outside the recommended operating range, you should provide external protection. A current limiting resistor and two clamping diodes.
• Thank you. However it seems every chip pin vulnerable to damage by ESD (so also Ax.0+) should have at least basic ESD protection (including negative ESD) - even where the latchup is not a threat.
• I have measured on the device for the SD16 input pins also have the protection diodes to AVCC and AVSS.
• In which MSP430 device and what was the negative threshold voltage? SD16 is different than SD16_A and individual chip types may have different specs.
• I measured with msp430f4794. As you have know the max value is AVss-1V.
• Yes, I know the specified limit. But what negative clamping voltage have you received on A0.x in your measurement?
We have measured several pieces of MSP430F4794 Rev. J, both on the targer PCBs, as well as not assembled, brand new pieces. We can see expected clamping for digital pins (vs. DVcc and DVss), as well as for positive voltages on analog inputs (vs. AVcc). But I can definitely confirm that we do not see any clamping for negative voltages on A0.x inputs vs. AVss. Even -5V (provided by 3k protecting resistor) IS NOT clamped on these pins. (The chips were delivered by authorized distributor and are otherwise functional.)
• But I can definitely confirm that we do not see any clamping for negative voltages on A0.x inputs vs. AVss
Yes, you are right. When put a voltage at VSS and measure the A0.X it seems open.
• So what is internal protection circuit against negative ESD on Ax.x pins in this chip?
• Yes, I think so
• Sorry, but this does not answer my question "So WHAT is internal protection circuit against negative ESD on Ax.x pins in this chip?"
• Sorry, it seems no negative ESD protection there.
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Apr 17 comment How many elements are in the set $\{ \left( \frac{2+i}{2-i} \right) ^n : n \in \mathbb N\}$ @mathguy If you stress text with double asterisks (**text**), it won't look like SCREAMING Mar 28 revised Does the integral of $\int_0^2 (\ln x)^{-2}$ exist? edited title Mar 20 revised Minimum and maximum on the closed disk $\,D(0,1)-f(z)=z^2-z$ added 5 characters in body Mar 20 answered Binomial Normal Stats Mar 20 comment Binomial Normal Stats Your "probabiliteis" aren't even in $[0,1]$. That cannot possibly correct. Mar 16 comment When derivative of a function is its inverse function and vice-versa Please use a more clear notation. Is $fi$ the inverse function ($f^{-1}$), as it says in the title, or the derivative of the inverse ($\frac{\mathrm d}{\mathrm dx}f^{-1}$), as it says in the body? Mar 14 comment What is a mathematical expression for the sequence $\{1,1,-1,-1,1,1,-1,-1,\dots\}$? +1 for showing a general approach to this class of problems Mar 11 awarded Enlightened Mar 11 awarded Nice Answer Feb 20 awarded Good Answer Feb 19 awarded Enlightened Feb 19 awarded Nice Answer Feb 19 answered How to write a function from graph? Feb 19 comment Show that the area of the face of the coin is $\frac{a^2}{2}(\pi-7\tan\frac{\pi}{14})$ What have you tried so far? A small hint: Start with the $7$ sectors and work out the area you counted multiple times. Feb 19 revised Show that the area of the face of the coin is $\frac{a^2}{2}(\pi-7\tan\frac{\pi}{14})$ added 1 character in body; edited title Feb 19 comment Why is derivative of $x$ with respect to $x$ equal to $1$? @Masacroso It doesn't seem like a good idea to start learning about derivatives with nonstandard analysis if standard analysis was not even learned ;-) Feb 19 comment Why is derivative of $x$ with respect to $x$ equal to $1$? @KimPeek The first class of derivatives I learned about was in fact that of (affine) linear functions of the form $f(x) = cx+b$. Indeed more people in my class had trouble seeing that the derivative of a constant was $0$ than seeing that the derivative of $cx$ was $c$. Feb 19 answered Why is derivative of $x$ with respect to $x$ equal to $1$? Feb 7 revised Why *all* $\epsilon > 0$, in the $\varepsilon-\delta$ limit definition? edited body Feb 5 revised How did the rule of addition come to be and why does it give the correct answer when compared empirically? Grammar fix | 678 | 2,456 | {"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} | 3.390625 | 3 | CC-MAIN-2016-18 | latest | en | 0.899826 |
https://imtyro.in/grade-points-java-program/ | 1,675,547,872,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00564.warc.gz | 342,163,943 | 26,468 | # Grade Points Java Program Solution
#### Question:-
At Raffle’s Academy, students are given grade points to a maximum of 5 during their final exam. The grade points are mapped to Letter grades as follows.
O 5
A >= 4.5 and <5.0
B >= 4.0 and <4.5
C >=3.0 and <4.0
D >=2.0 and <3.0
E >=1.0 and <2.0
F >=0.0 and <1.0
Write a program to that get’s an input between 0 and 5 and outputs the grade of the student.
Sample input:
Sample output:
Sample input:
Sample output:
Sample input:
1.2
Sample output:
#### Code:-
```import java.util.*;
public class Main
{
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
double grd=sc.nextDouble();
if(grd==5)
{
}
else if(grd>=4.5)
{
}
else if(grd>=4.0)
{
}
else if(grd>=3.0)
{
}
else if(grd>=2.0)
{ | 268 | 1,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.617877 |
https://www.detailedpedia.com/wiki-Bit | 1,713,647,206,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817674.12/warc/CC-MAIN-20240420184033-20240420214033-00192.warc.gz | 677,554,764 | 16,522 | # Bit
The bit is the most basic unit of information in computing and digital communications. The name is a portmanteau of binary digit. The bit represents a logical state with one of two possible values. These values are most commonly represented as either "1" or "0", but other representations such as true/false, yes/no, on/off, or +/ are also widely used.
The relation between these values and the physical states of the underlying storage or device is a matter of convention, and different assignments may be used even within the same device or program. It may be physically implemented with a two-state device.
A contiguous group of binary digits is commonly called a bit string, a bit vector, or a single-dimensional (or multi-dimensional) bit array. A group of eight bits is called one byte, but historically the size of the byte is not strictly defined. Frequently, half, full, double and quadruple words consist of a number of bytes which is a low power of two. A string of four bits is usually a nibble.
In information theory, one bit is the information entropy of a random binary variable that is 0 or 1 with equal probability, or the information that is gained when the value of such a variable becomes known. As a unit of information, the bit is also known as a shannon, named after Claude E. Shannon.
The symbol for the binary digit is either "bit", per the IEC 80000-13:2008 standard, or the lowercase character "b", per the IEEE 1541-2002 standard. Use of the latter may create confusion with the capital "B" which is the international standard symbol for the byte.
## History
The encoding of data by discrete bits was used in the punched cards invented by Basile Bouchon and Jean-Baptiste Falcon (1732), developed by Joseph Marie Jacquard (1804), and later adopted by Semyon Korsakov, Charles Babbage, Herman Hollerith, and early computer manufacturers like IBM. A variant of that idea was the perforated paper tape. In all those systems, the medium (card or tape) conceptually carried an array of hole positions; each position could be either punched through or not, thus carrying one bit of information. The encoding of text by bits was also used in Morse code (1844) and early digital communications machines such as teletypes and stock ticker machines (1870).
Ralph Hartley suggested the use of a logarithmic measure of information in 1928. Claude E. Shannon first used the word "bit" in his seminal 1948 paper "A Mathematical Theory of Communication". He attributed its origin to John W. Tukey, who had written a Bell Labs memo on 9 January 1947 in which he contracted "binary information digit" to simply "bit".
## Physical representation
A bit can be stored by a digital device or other physical system that exists in either of two possible distinct states. These may be the two stable states of a flip-flop, two positions of an electrical switch, two distinct voltage or current levels allowed by a circuit, two distinct levels of light intensity, two directions of magnetization or polarization, the orientation of reversible double stranded DNA, etc.
Bits can be implemented in several forms. In most modern computing devices, a bit is usually represented by an electrical voltage or current pulse, or by the electrical state of a flip-flop circuit.
For devices using positive logic, a digit value of 1 (or a logical value of true) is represented by a more positive voltage relative to the representation of 0. Different logic families require different voltages, and variations are allowed to account for component aging and noise immunity. For example, in transistor–transistor logic (TTL) and compatible circuits, digit values 0 and 1 at the output of a device are represented by no higher than 0.4 volts and no lower than 2.6 volts, respectively; while TTL inputs are specified to recognize 0.8 volts or below as 0 and 2.2 volts or above as 1.
### Transmission and processing
Bits are transmitted one at a time in serial transmission, and by a multiple number of bits in parallel transmission. A bitwise operation optionally processes bits one at a time. Data transfer rates are usually measured in decimal SI multiples of the unit bit per second (bit/s), such as kbit/s.
### Storage
In the earliest non-electronic information processing devices, such as Jacquard's loom or Babbage's Analytical Engine, a bit was often stored as the position of a mechanical lever or gear, or the presence or absence of a hole at a specific point of a paper card or tape. The first electrical devices for discrete logic (such as elevator and traffic light control circuits, telephone switches, and Konrad Zuse's computer) represented bits as the states of electrical relays which could be either "open" or "closed". When relays were replaced by vacuum tubes, starting in the 1940s, computer builders experimented with a variety of storage methods, such as pressure pulses traveling down a mercury delay line, charges stored on the inside surface of a cathode-ray tube, or opaque spots printed on glass discs by photolithographic techniques.
In the 1950s and 1960s, these methods were largely supplanted by magnetic storage devices such as magnetic-core memory, magnetic tapes, drums, and disks, where a bit was represented by the polarity of magnetization of a certain area of a ferromagnetic film, or by a change in polarity from one direction to the other. The same principle was later used in the magnetic bubble memory developed in the 1980s, and is still found in various magnetic strip items such as metro tickets and some credit cards.
In modern semiconductor memory, such as dynamic random-access memory, the two values of a bit may be represented by two levels of electric charge stored in a capacitor. In certain types of programmable logic arrays and read-only memory, a bit may be represented by the presence or absence of a conducting path at a certain point of a circuit. In optical discs, a bit is encoded as the presence or absence of a microscopic pit on a reflective surface. In one-dimensional bar codes, bits are encoded as the thickness of alternating black and white lines.
## Unit and symbol
The bit is not defined in the International System of Units (SI). However, the International Electrotechnical Commission issued standard IEC 60027, which specifies that the symbol for binary digit should be 'bit', and this should be used in all multiples, such as 'kbit', for kilobit. However, the lower-case letter 'b' is widely used as well and was recommended by the IEEE 1541 Standard (2002). In contrast, the upper case letter 'B' is the standard and customary symbol for byte.
Decimal
Value Metric
1000 kbit kilobit
10002 Mbit megabit
10003 Gbit gigabit
10004 Tbit terabit
10005 Pbit petabit
10006 Ebit exabit
10007 Zbit zettabit
10008 Ybit yottabit
10009 Rbit ronnabit
100010 Qbit quettabit
Binary
Value IEC
1024 Kibit kibibit Kbit Kb kilobit
10242 Mibit mebibit Mbit Mb megabit
10243 Gibit gibibit Gbit Gb gigabit
10244 Tibit tebibit
10245 Pibit pebibit
10246 Eibit exbibit
10247 Zibit zebibit
10248 Yibit yobibit
Orders of magnitude of data
### Multiple bits
Multiple bits may be expressed and represented in several ways. For convenience of representing commonly reoccurring groups of bits in information technology, several units of information have traditionally been used. The most common is the unit byte, coined by Werner Buchholz in June 1956, which historically was used to represent the group of bits used to encode a single character of text (until UTF-8 multibyte encoding took over) in a computer and for this reason it was used as the basic addressable element in many computer architectures. The trend in hardware design converged on the most common implementation of using eight bits per byte, as it is widely used today.[as of?] However, because of the ambiguity of relying on the underlying hardware design, the unit octet was defined to explicitly denote a sequence of eight bits.
Computers usually manipulate bits in groups of a fixed size, conventionally named "words". Like the byte, the number of bits in a word also varies with the hardware design, and is typically between 8 and 80 bits, or even more in some specialized computers. In the 21st century, retail personal or server computers have a word size of 32 or 64 bits.
The International System of Units defines a series of decimal prefixes for multiples of standardized units which are commonly also used with the bit and the byte. The prefixes kilo (103) through yotta (1024) increment by multiples of one thousand, and the corresponding units are the kilobit (kbit) through the yottabit (Ybit).
## Information capacity and information compression
When the information capacity of a storage system or a communication channel is presented in bits or bits per second, this often refers to binary digits, which is a computer hardware capacity to store binary data (0 or 1, up or down, current or not, etc.). Information capacity of a storage system is only an upper bound to the quantity of information stored therein. If the two possible values of one bit of storage are not equally likely, that bit of storage contains less than one bit of information. If the value is completely predictable, then the reading of that value provides no information at all (zero entropic bits, because no resolution of uncertainty occurs and therefore no information is available). If a computer file that uses n bits of storage contains only m < n bits of information, then that information can in principle be encoded in about m bits, at least on the average. This principle is the basis of data compression technology. Using an analogy, the hardware binary digits refer to the amount of storage space available (like the number of buckets available to store things), and the information content the filling, which comes in different levels of granularity (fine or coarse, that is, compressed or uncompressed information). When the granularity is finer—when information is more compressed—the same bucket can hold more.
For example, it is estimated that the combined technological capacity of the world to store information provides 1,300 exabytes of hardware digits. However, when this storage space is filled and the corresponding content is optimally compressed, this only represents 295 exabytes of information. When optimally compressed, the resulting carrying capacity approaches Shannon information or information entropy.
## Bit-based computing
Certain bitwise computer processor instructions (such as bit set) operate at the level of manipulating bits rather than manipulating data interpreted as an aggregate of bits.
In the 1980s, when bitmapped computer displays became popular, some computers provided specialized bit block transfer instructions to set or copy the bits that corresponded to a given rectangular area on the screen.
In most computers and programming languages, when a bit within a group of bits, such as a byte or word, is referred to, it is usually specified by a number from 0 upwards corresponding to its position within the byte or word. However, 0 can refer to either the most or least significant bit depending on the context.
## Other information units
Similar to torque and energy in physics; information-theoretic information and data storage size have the same dimensionality of units of measurement, but there is in general no meaning to adding, subtracting or otherwise combining the units mathematically, although one may act as a bound on the other.
Units of information used in information theory include the shannon (Sh), the natural unit of information (nat) and the hartley (Hart). One shannon is the maximum amount of information needed to specify the state of one bit of storage. These are related by 1 Sh ≈ 0.693 nat ≈ 0.301 Hart.
Some authors also define a binit as an arbitrary information unit equivalent to some fixed but unspecified number of bits. | 2,561 | 11,939 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-18 | latest | en | 0.937194 |
https://ru.scribd.com/presentation/240007261/Wave-Shaping-Circuits | 1,596,459,098,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735810.18/warc/CC-MAIN-20200803111838-20200803141838-00397.warc.gz | 478,445,109 | 87,730 | Вы находитесь на странице: 1из 15
Wave shaping circuitsa`
High Pass Circuit
High Pass Circuit
Signal is attenuated or damped at low frequencies with the output
increasing at +20dB/Decade (6dB/Octave) until the frequency reaches
the cut-off point ( c ) where again R = Xc.
At cut-off frequency, where the output voltage amplitude
is 1/2 = 70.7% of the input signal value or -3dB (20 log (Vout/Vin)) of
the input value.
Phase angle ( ) of the output signal LEADS that of the input and is
equal to +45
o
at frequency c.
Step Response
Voltage on capacitor cannot change instantaneously. So Vout = Vin initially.
Response of RC High Pass Circuit to Standard waveforms
RC
Fall Time
Vout
time
1.0
0.9
0.1
10%
90%
100%
1/e~37%
e
t
RC
Fall Time & Time Constant
( )
Response of RC High Pass Circuit to Standard waveforms
Step Response
Response of RC High Pass Circuit to Standard waveforms
Pulse Response
Response of RC High Pass Circuit to Standard waveforms
Square Wave Response Response
Response of RC High Pass Circuit to Standard waveforms
Square Wave Response Response
Response of RC High Pass Circuit to Standard waveforms
Ramp Response
Let a
, be the slop of input ramp signal therefore it can be represented
as : Vi(t) = at Now Apply KVL to RC circuit.
Vi = Vc + VR
Vi = (q/c) + VR ---- 1
at = (q/c) + Vo ---- 2
Differentiate eq. 2 on both sides wrt t we get
--3
Since dq/dt = i Vo = i . R i = Vo/R = dq/dt = Vo/R
Therefore , -- 4
Hence solution to this differential equation is given by,
Response of RC High Pass Circuit to Standard waveforms
Ramp Response
] 1 [
RC
t
e aRC Vo
Nature of output depends on Value of RC time constant.
Response of RC High Pass Circuit to Standard waveforms
High-pass RC circuit as Differentiator:
A circuit in which the output voltage is directly proportional to the
derivative of the input voltage is called a differentiating circuit.
Mathematically, the output voltage is given by:
Output (Vo) d/dt input (Vi)
If a d.c. or constant input is applied to such a circuit, the output will be
zero. It is because the derivative of the constant is zero.
Response of RC High Pass Circuit to Standard waveforms
High-pass RC circuit as Differentiator:
Let V
i
, be the input alternating voltage and let i be The resulting
alternating current. The charge q on the capacitor at any instant is:
q = C.V
c and
i = dq/dt
i = d/dt (CV
c
)
i = C d/dt (V
c
)
Since the capacitive reactance is very larger than R, the input voltage
can be consider equal to the capacitor voltage without any error,, i.e
Vc = V
i
,
i = C . d/dt (V
i
)
output voltage is given by:
V
o
= iR Or V
o
=(C . d/dt (V
i
)).R
V
o
=RC d/dt ( V
i
)
where RC is a constant, and Hence Output d/dt (input)
RC High pass circuit can work as differentiator for smaller values of
RC time constant.
Response of RC High Pass Circuit to Standard waveforms
Applications:
Some important applications of a differentiating circuit are given as
under:
To generate a square wave from a triangular wave input.
To generate a step from a ramp input.
To generate a series of narrow pulses called spikes from the
rectangular or square waveform. The pips are used as trigger pulses or
synchronization pulses in circuits used in television and cathode ray
oscilloscopes.
Low Pass Circuit
Low Pass Circuit
This Cut-off, Corner or Breakpoint frequency is defined as
being the frequency point where the capacitive reactance and resistance
are equal, R = Xc . When this occurs the output signal is attenuated
to 70.7% of the input signal value or -3dB (20 log (Vout/Vin)) of the
input.
The Phase Angle ( ) of the output signal LAGS behind that of the
input and at the -3dB cut-off frequency ( c ) and is - 45
o
out of phase.
This is due to the time taken to charge the plates of the capacitor as the
input voltage changes, resulting in the output voltage (the voltage across
the capacitor) lagging behind that of the input signal. | 1,044 | 3,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2020-34 | latest | en | 0.804745 |
https://www.smore.com/m9ksx | 1,524,567,620,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946597.90/warc/CC-MAIN-20180424100356-20180424120356-00201.warc.gz | 879,273,049 | 11,596 | # What's Happening in Math 2?
## MONDAY, SEPTEMBER 15
Section 3-1: Congruent Figures
Common Core Standard: G-SRT-5
I CAN use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures.
Assignment due Tuesday
• Handout 3-1
Assignment due Friday, 8am
• Math XL For School Chapter Three Review #1
## TUESDAY, SEPTEMBER 16
Section 3-2: Triangle Congruence by SSS and SAS
Common Core Standard: G-SRT-5
I CAN use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures.
Assignment due Wednesday
• Handout 3-2
Assignment due Friday, 8am
• Math XL For School Chapter Three Review #1
## WEDNESDAY, SEPTEMBER 17
Section 3-3: Triangle Congruence by ASA and AAS
Common Core Standard: G-SRT-5
I CAN use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures.
Assignment due Thursday
• Handout 3-3
Assignment due Friday, 8am
• Math XL For School Chapter Three Review #1
## THURSDAY, SEPTEMBER 18
Chapter Three Mid-Chapter Review
Common Core Standard: G-SRT-5
I CAN use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures.
Assignment due Friday, 8am
• Math XL For School Chapter Three Review #1
## FRIDAY, SEPTEMBER 19
Chapter Three Mid-Chapter Quiz
Common Core Standard: G-SRT-5
I CAN use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures.
Assignment: None
## Need to Contact Mrs. Naylor???
1st Period: Math 2
2nd Period: Math 2
3rd Period: Math 3
4th Period: Math 3
5th Period: Plan/Lunch | 448 | 1,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | longest | en | 0.836749 |
pracovnemiestaszcc.firebaseapp.com | 1,670,277,582,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711045.18/warc/CC-MAIN-20221205200634-20221205230634-00214.warc.gz | 493,299,069 | 8,745 | # Hbar ^ 2 2m v ev
In addition, the Heaviside step function H(x) can be used. Multiplication must be specified with a '*' symbol, 3*cos(x) not 3cos(x). Powers are specified with the 'pow' function: x² is pow(x,2) not x^2. Some potentials that can be pasted into the form are given below.
As was pointed out in class, the step-function example of a localized position state that we constructed before wasn't very realistic. Oct 21, 2020 · A particle in a 2-dimensional box is a fundamental quantum mechanical approximation describing the translational motion of a single particle confined inside an infinitely deep well from which it … $\sigma_{v_x} \approx 1.2 \times 10^6 \text{ m/s}$ Thus the velocity of an atomic electron has an inherent, irreducible uncertainty of about a million meters per second! If anyone tells you they know how fast an atomic electron is moving to a greater precision than a million meters per second, you know what to tell them… Aug 29, 2020 · Operate on $$ψ(x) = e^{ikx}$$ with $$\pm i\hbar \frac {\partial}{\partial x}$$ to show that $$P_x = \mp \hbar k$$. Which do you prefer, $$p_x = +ħk$$ or $$p_x = -ħk$$? If we use the momentum operator that has the - sign, we get the momentum and the wave vector pointing in the same direction, $$p_x = +ħk$$, which is the preferred result Sep 12, 2005 · I have a quesion regarding a quantum physics assignemnt, I wonder what units I should use when calculating the transmission coefficient of a quantum barrier problem.
Вільні частинки — термін, який уживається в фізиці для позначення частинок, які не взаємодіють з іншими тілами, а, отже мають тільки кінетичну енергію.. Сукупність вільних … In addition, the Heaviside step function H(x) can be used. Multiplication must be specified with a '*' symbol, 3*cos(x) not 3cos(x). Powers are specified with the 'pow' function: x² is pow(x,2) not x^2. Some potentials that can be pasted into the form are given below. Constants in MKS units: speed of light: c = 2.9979 × 10 8 m/s Planck's constant: h = 6.6261 × 10-34 J·s Planck's constant divided by 2 pi: hbar = 1.0546 × 10-34 J·s Charge on a proton: e = 1.6022 × 10-19 C esu = 4.8032 × 10-10 Conversion constant: hbar*c/e = 197.33 MeV·fm or eV·nm = 1.9733 × 10-7 eV·m Electron mass: me and me2 = 9.1094 × 10-31 kg or 0.51100 MeV/c 2 01.10.2007 Համիլտոնյան (նշանակվում է ^ կամ h), համակարգի լրիվ էներգիայի օպերատորը քվանտային մեխանիկայում։ Կոչվել է իռլանդացի մաթեմատիկոս Ուիլյամ Համիլտոնի անունով։ .
## The Schrödinger equation for a particle moving in one dimension is a second order linear differential equation thus any solution can be written in terms of two linearly independent solutions.
\frac{d ^{2}\psi(x)}{dx^{2}}+\frac{8\pi^{2}m}{\hbar Use the v=0 and v=1 harmonic oscillator wavefunctions given below By how much (in eV) will distortion lower the energy (from its value for a cube, 15. The harmonic oscillator is specified by the Hamiltonian: H = - h−2. 2m d2 dx2. Oct 5, 2005 According to problem 8.2, the height of the well is 0.3eV, which is the hbar = 1.05457148*10^(-34); % m^2 kg / s.
### Sep 6, 2017 \begin{align*}\eqalign{ E &= \frac{p^2}{2 m} + V(x)\\ \Rightarrow p -\left(\frac{2m }{\hbar^2}(E-V(x))\right)\Psi(x) \\ \Leftrightarrow
2 m v. Therefore velocity v = e. 2 KE. Apr 12, 2007 Two key concepts underpinning quantum physics are the Schrodinger equation -1/2*hbar^2/m(d2/dx2)V(x) + U(x)V(x) = EV(x). % for arbitrary Aug 27, 2014 The plot units are energy (eV) vs. distance (angstroms).
Constants in MKS units: speed of light: c = 2.9979 × 10 8 m/s Planck's constant: h = 6.6261 × 10-34 J·s Planck's constant divided by 2 pi: hbar = 1.0546 × 10-34 J·s Charge on a proton: e = 1.6022 × 10-19 C esu = 4.8032 × 10-10 Conversion constant: hbar*c/e = 197.33 MeV·fm or eV·nm = 1.9733 × 10-7 eV·m Electron mass: me and me2 = 9.1094 × 10-31 kg or 0.51100 MeV/c 2 01.10.2007 Համիլտոնյան (նշանակվում է ^ կամ h), համակարգի լրիվ էներգիայի օպերատորը քվանտային մեխանիկայում։ Կոչվել է իռլանդացի մաթեմատիկոս Ուիլյամ Համիլտոնի անունով։ . Համիլտոնյանի սպեկտրը հնարավոր արժեքների բազմությունն է 04.02.2006 In quantum mechanics, the rectangular (or, at times, square) potential barrier is a standard one-dimensional problem that demonstrates the phenomena of wave-mechanical tunneling (also called "quantum tunneling") and wave-mechanical reflection. The problem consists of solving the one-dimensional time-independent Schrödinger equation for a particle encountering a rectangular potential … Приближение почти свободных электронов — метод в квантовой теории твёрдого тела, в котором периодический потенциал кристаллической решётки считается малым возмущением относительно свободного движения валентных The hydrogen atom is one of the few real physical systems for which the allowed quantum states of a particle and corresponding energies can be solved for exactly (as opposed to approximately) in non-relativistic quantum mechanics. In the most basic quantum mechanical model of hydrogen, the proton is taken to be a fixed source of an electric potential and the Schrödinger equation for the 17.01.2011 The particle in the box model system is the simplest non-trivial application of the Schrödinger equation, but one which illustrates many of the fundamental concepts of quantum mechanics.For a particle moving in one dimension (again along the x- axis), the Schrödinger equation can be written $-\dfrac{\hbar^2}{2m}\psi {}''(x)+ V (x)\psi (x) = E \psi (x) \nonumber$ 07.03.2021 2 mv2 = p2 2m = ~2k2 2m = ~! (8) v g= d!
The Schroedinger equation for a particle moving in one dimension through a region where its potential energy is a function of position has the form (-ħ 2 /(2m))∂ 2 ψ (x,t) /∂x 2 + U(x)ψ (x,t) = iħ∂ψ (x,t) /∂t.. We are often interested in finding the eigenstates of the energy operator iħ∂/∂t, i.e. we are interested in finding the wave functions of particles Figure $$\PageIndex{2}$$: Visualizing the first six wavefunctions and associated probability densities for a particle in a two-dimensional square box ($$L_x=L_y=L$$).Use the slide bar to independently change either $$n_x$$ or $$n_y$$ quantum number and see the changing wavefunction. Unlike in the one-dimensional analoge, where nodes in the wavefunction are points where $$\psi_{n}(x)=0$$, here Rydbergova konstanta je fyzikální konstanta pojmenovaná po švédském fyzikovi Johannesu Rydbergovi.Představuje nejvyšší možný vlnočet (převrácená hodnota vlnové délky) elektromagnetického záření, které může vyzářit nejjednodušší atom – atom vodíku – v limitě nekonečné hmotnosti jádra.. Rydbergova konstanta a další příbuzné konstanty, jako Rydbergova Authors and Editors. Bethel Afework, Allison Campbell, Jordan Hanania, Kailyn Stenhouse, Jason Donev Last updated: May 18, 2018 Get Citation Egy egydimenziós dobozba zárt részecske hullámfüggvénye alapállapotban félperiódusú szinuszhullám, mely a két végpontnál nulla értéket vesz fel.
Oct 21, 2020 · A particle in a 2-dimensional box is a fundamental quantum mechanical approximation describing the translational motion of a single particle confined inside an infinitely deep well from which it … $\sigma_{v_x} \approx 1.2 \times 10^6 \text{ m/s}$ Thus the velocity of an atomic electron has an inherent, irreducible uncertainty of about a million meters per second! If anyone tells you they know how fast an atomic electron is moving to a greater precision than a million meters per second, you know what to tell them… Aug 29, 2020 · Operate on $$ψ(x) = e^{ikx}$$ with $$\pm i\hbar \frac {\partial}{\partial x}$$ to show that $$P_x = \mp \hbar k$$. Which do you prefer, $$p_x = +ħk$$ or $$p_x = -ħk$$? If we use the momentum operator that has the - sign, we get the momentum and the wave vector pointing in the same direction, $$p_x = +ħk$$, which is the preferred result Sep 12, 2005 · I have a quesion regarding a quantum physics assignemnt, I wonder what units I should use when calculating the transmission coefficient of a quantum barrier problem. I have got the following expression: T = \\frac{4(E+V_0)}{(2E+V_0)cos^2a\\sqrt{\\frac{2m}{\\hbar^2}(E-V_0)} + Apr 15, 2001 · For a simple harmonic oscillator potential energy as a function of position is -kx 2 /2 (remember that in order find potential enrgy you integrate force with repect to displacement) and kinetic energy as a function of momentum is always the same, p 2 /2m, where m is mass. Remember E is always equal to hv (where v is frequency).
In this chapter, we apply quantum theory to a series of model situations: a single particle confined to one-, two-, or three-dimensional microscopic “boxes”, i.e. regions where it can move freely, but beyond which it cannot move. hbar = 1.05e-14 #reduced planks constant in units of Å^2*kg/s hbarSI = 1.055e-34 #"-----" in units of m^2*kg/s m = 1.6266e-27 #mass of particle in units of kg eV = 1.602e-19 #1 electron volt in units of J #Define QHO potential parameters omega = 5.6339e14 eta = 2 x_0 = np. sqrt (hbar / (m * omega)) a = 4 * x_0 #width of potential in Å # of where p is the quantum-mechanical momentum operator, V is the potential, and m is the vacuum mass of the electron. (This equation neglects the spin–orbit effect; see below.) In a crystalline solid, V is a periodic function, with the same periodicity as the crystal lattice.
Multiplication must be specified with a '*' symbol, 3*cos(x) not 3cos(x). Powers are specified with the 'pow' function: x² is pow(x,2) not x^2. Some potentials that can be pasted into the form are given below. \begin{aligned} E = \frac{\hbar^2 k^2}{2m} = \frac{\hbar^2 (k_0^2 - \gamma^2)}{2m} - 2i \frac{\hbar^2 \gamma k_0}{2m} \equiv E_0 - \frac{i\Gamma}{2}. \end{aligned} So the energy is complex.
gbp až twd hsbc
kolumbijský dokument o bitcoinoch
prepočítať 9,25 × 105 cal na kilojoulov
koľko je 10 dolárov v kwd
ako nakresliť doge tvár
### The effective mass m may be expressed in terms of the effective mass ratio and the rest mass of the electron; i.e., m = m e m 0 The quantity h/(2m 0) 1/2 is 4.9091x10-19 in SI units. To get energy in electron-volts the energy in Joules must be divided by 1.602x10 -19 and thus the coefficient in the equation must be multiplied by its square root.
We are mixing a photon energy with a particle energy. The energy of a particle in its most general way is: I will be taking an oral exam, where I have to do some "airport physics", fast and easy magnitude estimations. Currently I try to come up with a good way to find the Bohr radius of the hydrogen at May 03, 2011 · V is the potential barrier, 3.8 eV I solved for L, getting: L=npi*hbar/sqrt(2m(E-V)), but I still don't know n. (n is an integer) And when ignoring n I get the wrong answer.
## In addition, the Heaviside step function H(x) can be used. Multiplication must be specified with a '*' symbol, 3*cos(x) not 3cos(x). Powers are specified with the 'pow' function: x² is pow(x,2) not x^2. Some potentials that can be pasted into the form are given below.
V nerelativistické kvantové mechanice lze volnou Напомена: На (n, l, s) = (n, 0,1 / 2) и (n, l, s) = (n, 1, -1 / 2) нивото на енергија, и нивото на фината структура се исти.
If anyone tells you they know how fast an atomic electron is moving to a greater precision than a million meters per second, you know what to tell them… Aug 29, 2020 · Operate on $$ψ(x) = e^{ikx}$$ with $$\pm i\hbar \frac {\partial}{\partial x}$$ to show that $$P_x = \mp \hbar k$$. | 3,895 | 11,393 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2022-49 | longest | en | 0.658551 |
https://www.learnforexc.com/the-principle-of-stochastic-index-kdj/ | 1,674,900,051,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499541.63/warc/CC-MAIN-20230128090359-20230128120359-00156.warc.gz | 876,021,493 | 12,836 | The KDJ indicator is also called a random indicator. It was first proposed by Dr. George Lane. It is a fairly novel and practical technical analysis indicator. It was first used in the analysis of the futures market and later widely used in the stock market. The short-term and medium-term trend analysis is the most commonly used technical analysis tool in the futures and foreign exchange markets.
## The principle and calculation method of KDJ indicator
Principle of KDJ indicator
The stochastic indicator KDJ is generally based on the principle of statistics, through the highest price, the lowest price, and the closing price of the last calculation cycle and between the three during a specific period (often 9 days, 9 weeks, etc.). Proportional relationship, to calculate the immature random value RSV of the last calculation cycle, and then calculate the K value, D value and J value according to the method of smooth moving average, and draw a curve chart to study the stock trend.
The stochastic indicator KDJ is a point formed on the coordinates of the indicator by the K value, D value and J value calculated based on the basic data of the highest price, the lowest price and the closing price, which is formed by connecting countless such points. A complete KDJ indicator that reflects the trend of price fluctuations.
It is mainly a technical tool that uses the real volatility of price fluctuations to reflect the strength of price movements and overbought and oversold phenomena, and sends trading signals before prices have risen or fallen. In the design process, it mainly studies the relationship between the highest price, the lowest price and the closing price. It also incorporates some advantages of the momentum concept, strength index and moving average, so it can be judged more quickly, quickly and intuitively. Quotes. The stochastic indicator KDJ first appeared in the form of the KD indicator, and the KD indicator was developed on the basis of the William indicator. However, the William indicator only judges the phenomenon of overbought and oversold stocks. In the KDJ indicator, the concept of moving average linear speed is integrated to form a more accurate basis for buying and selling signals. In practice, K-line and D-line are used in conjunction with J-line to form KDJ indicators. Since the KDJ line is essentially a concept of random fluctuations, it is more accurate for grasping the short-term and medium-term market trends. | 483 | 2,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.948366 |
https://www.thefreedictionary.com/Poisson+noise | 1,540,346,691,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583517628.91/warc/CC-MAIN-20181024001232-20181024022732-00104.warc.gz | 1,082,456,984 | 11,825 | # Poisson distribution
(redirected from Poisson noise)
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## Pois·son distribution
(pwä-sôN′)
n. Statistics
A probability distribution which arises when counting the number of occurrences of a rare event in a long series of trials.
[After Siméon Denis Poisson (1781-1840), French mathematician.]
## Poisson distribution
(ˈpwɑːsən)
n
(Statistics) statistics a distribution that represents the number of events occurring randomly in a fixed time at an average rate λ; symbol P0(λ). For large n and small p with np = λ it approximates to the binomial distribution Bi(n,p)
[C19: named after S. D. Poisson]
## Pois•son′ distribu`tion
(pwɑˈsoʊn, -ˈsɔ̃)
n.
a probability distribution whose mean and variance are identical.
[1920–25; after S. Dutch. Poisson (1781–1840), French mathematician and physicist]
ThesaurusAntonymsRelated WordsSynonymsLegend:
Noun 1 Poisson distribution - a theoretical distribution that is a good approximation to the binomial distribution when the probability is small and the number of trials is largedistribution, statistical distribution - (statistics) an arrangement of values of a variable showing their observed or theoretical frequency of occurrencestatistics - a branch of applied mathematics concerned with the collection and interpretation of quantitative data and the use of probability theory to estimate population parameters
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References in periodicals archive ?
This combination is naturally considered as a superposition of Gaussian noise over Poisson noise.
2 megapixel scientific CMOS camera available for life science research that actively defeats the negative impact of Poisson noise in low light images.
Poisson noise is one of the complicated noises which is very difficult from de-noising point of view.
2B-D are the background with Gaussian white noise ([sigma] = 400), Poisson noise ([alpha] = 500) and mixture of both ([sigma]= 250, [alpha] = 250).
Examples are Salt-and-Pepper noise, Poisson noise, additive Laplace noise, and different models of multiplicative noise.
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Open / Close | 490 | 2,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.833594 |
https://www.physicsforums.com/threads/how-to-determine-if-all-c-atoms-are-coplanar.870753/ | 1,618,212,316,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066613.21/warc/CC-MAIN-20210412053559-20210412083559-00334.warc.gz | 1,044,256,209 | 19,291 | # How to determine if all C atoms are coplanar
1. Homework Statement
## The Attempt at a Solution
I can see that in B, not all C atoms are coplanar , but for C, D and E I can't see how the carbon atoms are coplanar .. I thought they are tetrahedral?
Borek
Mentor
In E carbon atoms are not coplanar (that is, one of them can "freely" move, so there will be moments when it will be complanar with others).
How many carbon atoms in D? Can you think of such an arrangement of these atoms that they would not lie on a one surface?
When it comes to C it would be best to see a model. Yes, all these carbons are sp3 hybridized, but it doesn't stop them from lying on one surface. Is the central COCO ring flat?
In E carbon atoms are not coplanar (that is, one of them can "freely" move, so there will be moments when it will be complanar with others).
Is it the carbon atom in CH3 that is free to move? What about the carbon atom bonded to No2?
Like in their examples, all carbon atoms in benzene lie in a plane because they all use sp2 orbitals and every bond angle is 120°, right? But in the case of cyclohexane, all carbon atoms are sp3 hybridised and they each have a tetrahedral shape, so why can't they all lie in the same plane? How do you know if all carbon atoms can lie in a plane?
How many carbon atoms in D? Can you think of such an arrangement of these atoms that they would not lie on a one surface?
3 carbons. Erm, is it like the arrangement of methylpropane ? Again, I am still confused, why the carbon atoms in methylpropane are not coplanar...
When it comes to C it would be best to see a model. Yes, all these carbons are sp3 hybridized, but it doesn't stop them from lying on one surface. Is the central COCO ring flat?
I'm not sure.. Yes?
Borek
Mentor
Is it the carbon atom in CH3 that is free to move?
Yes.
What about the carbon atom bonded to No2?
Atoms in the ring are sp2, that means all atoms bonded to the ring lie in the ring plane.
3 carbons. Erm, is it like the arrangement of methylpropane ? Again, I am still confused, why the carbon atoms in methylpropane are not coplanar...
But they are - any three points always lie on a surface (technically that hold even when they colinear, even if there are infinitely many such surfaces then).
I'm not sure.. Yes?
Yes, the central COCO ring is rigid. That means all carbons sticking out of the C atoms must lie in a plane.
I am afraid if you don't see it it won't be easy to show without a model. All carbons lie on a plane, and O atoms stick up and down, this is a rigid arrangement.
Perhaps try to think about it this way - this molecules consist of two halves, each one containing three carbon atoms and connected by the oxygen bridges. Both halves contain three carbons, so these three carbons have to be coplanar (remember? each connected three carbons lie on a plane). Now, the way these halves are bonded through the oxygen bridges puts these carbon planes in the same place - so there is in fact one plane only.
Ahh I see.. That was clear, thanks so much !
But why is it impossible for the carbon atoms in methylpropane to be coplanar?
Borek
Mentor
Then, why is it impossible for the carbon atoms in methylpropane to be coplanar?
There are four carbons - any three will be coplanar, but the fourth must stick out.
As I said earlier, try to build models - even from putty, or plasticine, and matches/toothpicks.
Janiceleong26
There are four carbons - any three will be coplanar, but the fourth must stick out.
As I said earlier, try to build models - even from putty, or plasticine, and matches/toothpicks.
Ok, thank you ! | 915 | 3,621 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-17 | longest | en | 0.949997 |
https://www.testandtrack.io/aus/index.php/studenttest/givetest/285 | 1,604,106,483,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107912593.62/warc/CC-MAIN-20201031002758-20201031032758-00704.warc.gz | 806,073,180 | 19,769 | # 04 - Truth Tables and predicting output
1. Enter the missing value
2
0
1
3
2. Enter the missing value
2
3
1
0
3. Read the guidance on the below image. Given the below what would A+B.C mean?
`note: think about how you would process this equation. Would you do the + or the . first? Refer to what you learned in Maths: BODMAS or BIDMAS`
A and B and C
A not C not B
A or B and C
A or B or C
4. Enter the missing value
#4 1 #5 – 1 #6 -1
#4 1 #5 – 1 #6 -0
#4 1 #5 – 0 #6 -1
#4 0 #5 – 0 #6 -1
5. Enter the missing value
# 7 – 0 #8 - 1 #9 – 0
# 7 –1 #8 - 0 #9 – 1
# 7 – 1 #8 - 1 #9 –1
# 7 – 0 #8 - 0 #9 – 0
6. This is a truth table for:
p AND q
You can only have A and B so the letters p and q are not valid
p OR q
p NOT q
7. Spot the error
```p q NOT p NOT q p AND q NOT(p AND q)
==================================================
T T F T T F```
Not q should be removed altogether as it cannot be done
p AND q should be F
NOT q should be False
Not p should be T
8. Why is NOT(p AND q) False?
```p q NOT p NOT q p AND q NOT(p AND q)
==================================================
T T F T T F```
Because anything AND anything is always False
Because when you are dealing with a p and q input, the output will always be False if the input is either True or False
Because it is the inversion (opposite) of (p AND q) which is True.
It shouldn't be False - It should be true
9. Enter the missing value
3
2
0
1
10. Enter the missing value
#1 0 #2 1 #3 1,1
#1 0 #2 1 #3 0,0
#1 0 #2 0 #3 0,0
#1 1 #2 0 #3 1,1 | 561 | 1,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-45 | latest | en | 0.784465 |
https://www.physicsforums.com/threads/is-this-4-velocity-tangential-to-the-worldline.679231/ | 1,521,571,831,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647519.62/warc/CC-MAIN-20180320170119-20180320190119-00708.warc.gz | 826,638,629 | 21,867 | # Is this 4-velocity tangential to the worldline
1. Mar 18, 2013
### Sunfire
Is this "4-velocity" tangential to the worldline
Hello,
In the Minkowski spacetime (x,y,ict) the spacetime velocity (the "4" velocity, which in this particular instance happens to have 2 spatial and 1 temporal component) is given by
V = γ(vx, vy,ic),
where v = (vx, vy) is the spatial velocity vector.
Will V be tangential to the worldline of the particle and why?
2. Mar 18, 2013
### Staff: Mentor
Re: Is this "4-velocity" tangential to the worldline
Yes, the 4-velocity is the tangent vector to the worldline. To see why, note that the 4-velocity is the derivative of position with respect to proper time.
3. Mar 18, 2013
### Fredrik
Staff Emeritus
That "ict" stuff is very rarely used these days. The idea behind it was that the i allows you to define the metric tensor using the formula for the Euclidean inner product, i.e. if $x=(x_1,x_2,x_3,ict)$ and $y=(y_1,y_2,y_3,ics)$, we have
$$g(x,y) =(x_1,x_2,x_3,ict)\cdot (y_1,y_2,y_3,ics) =x_1y_1+x_2y_2+x_3y_3-c^2ts.$$ But we can do without the i if we use matrix multiplication instead. You can e.g. write the above as $$g(x,y) = \begin{pmatrix}x_1 & x_2 & x_3 & ct\end{pmatrix} \begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1\end{pmatrix} \begin{pmatrix}y_1 \\ y_2 \\ y_3 \\ cs\end{pmatrix} = x_1y_1+x_2y_2+x_3y_3-c^2ts,$$ or as
$$g(x,y) = \begin{pmatrix}x_1 & x_2 & x_3 & t\end{pmatrix} \begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -c^2\end{pmatrix} \begin{pmatrix}y_1 \\ y_2 \\ y_3 \\ s\end{pmatrix} = x_1y_1+x_2y_2+x_3y_3-c^2ts.$$ Most people also prefer to use units such that c=1, and its also very common to put the time coordinate first, rather than last. People who do that number the components from 0 to 3 instead of from 1 to 4.
I'm one of those people. I would write the above as
$$g(x,y)=x^T\eta y =-x_0y_0+x_1y_1+x_2y_2+x_3y_3,$$ where
$$\eta = \begin{pmatrix}-1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}$$ This is how I think about 4-velocity in SR: I will consider 1+1-dimensional Minkowski spacetime here for simplicity. We define the four-velocity as the normalized tangent vector to the world line. This means that in the comoving inertial coordinate system, its components are
$$\begin{pmatrix}1\\ 0\end{pmatrix}.$$ To find its components in an inertial coordinate in which the particle's velocity is v, we apply a Lorentz transformation with velocity -v:
$$\gamma\begin{pmatrix}1 & v\\ v & 1\end{pmatrix} \begin{pmatrix}1 \\ 0\end{pmatrix} =\gamma\begin{pmatrix}1\\ v\end{pmatrix}.$$
Last edited: Mar 18, 2013
4. Mar 18, 2013
### Sunfire
@PeterDonis
now I have to post something clever, else I will be gaining red points
how does one go about proving the above; the text I am reading defines proper time interval τ2=t2-(1/c2)(x2+y2+z2) and then defines the spacetime velocity V given above.
Worldline has only been mentioned but not defined yet; kind of hard to find out if V is tangential to the worldline without its definition.
The reason I am asking this is I am trying to make sense of the fact V$\bullet$ F = 0,
V is the spacetime velocity
F is the "4"- force
Last edited: Mar 18, 2013
5. Mar 18, 2013
### Sunfire
Fredrik, thank-you for your reply and for the detailed, easy-to-follow explanation. The text I am reading is an intro and a bit on the older side (Lawden), but very good price on Amazon . He uses the "ict" idea I guess for an easier introduction to SR.
Have you considered the relation V $\cdot$ F = 0
V is the spacetime velocity
F is the "4"-force given by
F = (f, i$\dot{m}$c), f is the 3-force
In 3-space, this means "the force does no work" and the 2 vectors are orthogonal; but in Minkowski spaces the geometry of the above two vectors is... well, non-intuitive. I am trying to make sense what does V $\cdot$ F = 0 mean.
6. Mar 18, 2013
### WannabeNewton
Hi Sunfire. You don't exactly need to know that the 4 - velocity is tangential to the worldline to see that. All you need is what Fredrik noted earlier, that the 4 - velocity is normalized so that $u^{a}u_{a} = -1$ (I'm using natural units). Then, $\frac{d}{d\lambda }(u^{a}u_{a}) = u_{a}\frac{\mathrm{d} u^{a}}{\mathrm{d} \lambda} +u^{a}\frac{\mathrm{d} u_{a}}{\mathrm{d} \lambda} = 2u_{a}\frac{\mathrm{d} u^{a}}{\mathrm{d} \lambda} = 0$ thus $u\cdot a= u_{a}\frac{\mathrm{d} u^{a}}{\mathrm{d} \lambda} = 0$ (this of course then implies the 4 - force is orthogonal to the 4 - velocity since the 4 - force is proportional to the 4 - acceleration).
I can show you what the tangent vector actually is and why it in fact is tangent to the wordline, using the coordinate basis viewpoint of the tangent space, but it would be a bit abstract and might use notation unfamiliar to you and I don't want to show you anything that might overly confuse you. If you are ok with it then I can try to explain.
7. Mar 18, 2013
### Staff: Mentor
Does the worldline get defined later on in your text? The simplest way to define it, at least if you want to see how 4-velocity is tangent to it, is to define it as four functions $t(\tau)$, $x(\tau)$, $y(\tau)$, $z(\tau)$, giving the coordinates of the object traveling on the worldline as a function of the object's proper time. Then the derivatives of the four functions are (I'm using units in which c = 1):
$$\frac{dt}{d\tau} = \gamma$$
$$\frac{dx}{d\tau} = \frac{dx}{dt} \frac{dt}{d\tau} = \gamma v_x$$
$$\frac{dy}{d\tau} = \frac{dy}{dt} \frac{dt}{d\tau} = \gamma v_y$$
$$\frac{dz}{d\tau} = \frac{dz}{dt} \frac{dt}{d\tau} = \gamma v_z$$
So if we consider the 4 coordinate functions as a 4-vector $x^{\mu}$, then the four derivatives give another 4-vector $u^{\mu} = d x^{\mu} / d\tau$, which is the 4-velocity. Obviously the 4-velocity must then be tangent to the worldline, since it's just the derivative of the coordinates of the worldline with respect to proper time, i.e., with respect to the parameter we're using to define the coordinate functions.
WannabeNewton's post addresses this so I won't go into it further; but what he said is consistent with what I said above. (He writes $\lambda$ where I write $\tau$, but it amounts to the same thing.)
8. Mar 18, 2013
### pervect
Staff Emeritus
ict is fairly common in introductory SR books. It will eventually get replaced if one gets to tensors, but it's perfectly adaquate unitl that point.
9. Mar 18, 2013
### nitsuj
I don't think I've ever seen math explained so clearly, feels like I can almost follow it
Very clear summary of the math ^
10. Mar 18, 2013
### Sunfire
This makes perfect sense The coordinates, being a function of τ, define the curve of the worldline, while their derivatives with respect to τ define the tangent to the curve.
11. Mar 18, 2013
### Sunfire
12. Mar 18, 2013
### Fredrik
Staff Emeritus
I suspect that WannabeNewton is referring to how tangent vectors are defined in differential geometry. That is indeed pretty abstract, and takes a lot of time to learn thoroughly. But it is what you have to learn if you want to understand these things in the context of general relativity.
However, things can be made much easier in special relativity. We don't have to define the spacetime of special relativity as a Lorentzian smooth manifold. It can also be defined as a vector space, or as an affine space. This choice is of no practical importance. I like to define it as a vector space, because this makes the mathematics easy.
If we define spacetime as the set ℝ4 with the standard vector space structure, and the bilinear form g defined by
$$g(x,y)=x^T\eta y$$ for all $x,y\in\mathbb R^4$, then the tangent vector of a curve $C:(a,b)\to\mathbb R^4$ at the point C(t) is just the derivative C'(t).
Last edited: Mar 18, 2013
13. Mar 18, 2013
### robphy
This may help:
In 3-space, that the 3-force is perpendicular to the 3-velocity also means that the magnitude of the 3-velocity (i.e. the spatial speed) is constant [assuming that the mass is constant].
In spacetime, that the 4-force is perpendicular to the 4-velocity means that magnitude of the 4-velocity is constant [assuming that the rest-mass is constant].)
14. Mar 18, 2013
### Staff: Mentor
Actually, the magnitude of the 4-velocity is constant, period; it's always 1 (or c in conventional units). The magnitude of the 4-momentum changes if the rest mass changes, since the rest mass *is* the magnitude of the 4-momentum. But that doesn't change the 4-velocity.
15. Mar 18, 2013
### robphy
Yes, I agree. The 4-velocity is normalized. So, it (edit: its magnitude) is automatically constant.
I was appealing to the typical intro-physics argument that zero-net-work means that [via the work-energy theorem] the speed is unchanged, assuming the mass is constant.
Then, use the analogue for the 4-dimensional case.
A fancier interpretation [from Tom Moore's Six Ideas] of the 3D-case is that the momentum-vector's magnitude (which has no special name) doesn't change when the force is perpendicular to the velocity. If that were more common, then I would have used your 4-momentum-interpretation for the spacetime analogue.
Last edited: Mar 19, 2013
16. Mar 19, 2013
### Sunfire
I have developed an intuition for the orthogonality between f and v in 3-space; If the projection of f along v is zero, then there is no action in the direction of v and consequently, v remains unchanged.
But this intuition no longer works for the 4-vectors F and V, because "orthogonality", a geometrical notion, makes no sense anymore, because the temporal axis takes imaginary values ict, which leads to a subtraction in the dot-product summation. One can no longer imagine F having zero projection onto V.
This is why I started this thread with lower dimensional spacetime (x1, x2, ict), because in it one can visualize F and V, and they are no longer having the familiar geometry, e.g. they do not form a 90-degree angle.
Last edited: Mar 19, 2013
17. Mar 19, 2013
### Fredrik
Staff Emeritus
The result that 4-acceleration is orthogonal to 4-velocity means only that the 0 component (i.e. the "temporal" component) of the 4-acceleration in the comoving inertial coordinate system is 0.
18. Mar 19, 2013
### robphy
(Technically, we should work in the tangent space.)
Think of a vector as the radius vector to a curve of equidistant points from the origin (the analogue of a "circle/sphere", which is a hyperbola/oid in Minkowski spacetime).
Where the tip of that radius vector meets the "circle", regard the vectors tangent to that circle at that point as orthogonal to that radius vector. (If the radial vector is timelike, then these tangent-vectors are spacelike, and specifically orthogonal to that particular radius vector.)
So, consider a future-timelike-momentum vector p [along the radial direction].
(Its tip lies on a hyperboloid [called the mass-shell of the particle].)
Additionally, consider a spacelike force-vector F orthogonal to p (i.e. F (dot) p=0 [along a tangent direction]). Since F has no radial component and F=(d/ds)p , the momentum-vector p can only change in direction (with its tip sliding along the same hyperboloid), but not change in magnitude (which would imply that tip moves to a different concentric hyperboloid).
[This elaborates on the reply of Fredrik.]
19. Mar 19, 2013
### Fredrik
Staff Emeritus
If we define the spacetime of SR as a vector space instead of as a manifold, we don't have to define a tangent space. For example, the derivative of a differentiable curve in spacetime is automatically a vector in spacetime. Now if you want to draw that vector as an arrow, you may want to draw it at the point where the derivative was computed. But we can think of this as nothing more than a choice of how to illustrate what we're doing.
I don't think I have thought of this way to find the direction of the orthogonal vector before. The image I visualize is simpler. Here's one I found using Google:
If the 4-velocity is a point in the almost vertical line in the picture, then the 4-acceleration is a point in the almost horizontal plane. In the comoving inertial coordinate system, that line is just the t axis and that plane is just the xy-plane. Every point in that line is orthogonal to every point in that plane. (Note that the two angles indicated in the picture are the same).
This is the web page where I found the image. I haven't read the text, but it has lots of nice pictures.
20. Mar 19, 2013
### robphy
Yes, for the specific case of Minkowski spacetime, that image is consistent... and possibly simpler (after you have accepted that formulation of orthogonality). (You can use the Lorentz transformations or a radar-experiment to justify that formulation.)
My method works for the Euclidean, Galilean, and Minkowskian cases.
It builds on the tangent-is-perpendicular-to-radius idea hopefully-familiar from Euclidean geometry, which beautifully generalizes to these other geometries. | 3,730 | 13,008 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-13 | latest | en | 0.756533 |
http://www.howtodothings.com/education/how-to-help-your-child-learn-basic-math-facts | 1,556,232,430,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578742415.81/warc/CC-MAIN-20190425213812-20190425235812-00222.warc.gz | 235,785,949 | 10,657 | 2. Understand addition/subtraction fact families. Fact families are groups of three numbers that are related to one another. 4 + 3 = 7, 7 – 3 = 4, and 7 – 4 = 3 therefore the numbers 4, 3, and 7 have a relationship to one another and are considered a fact family. Point out to your child that 4 + 3 is the same as 3 + 4 (this reversal is true for addition, but not subtraction). Using this relationship in teaching math facts to your child will help them with the concepts behind the numbers as well as memorizing the facts themselves.
3. Understand multiplication/division fact families. Likewise, fact families occur with multiplication and division. 6 x 7 = 42, 42 * 7 = 6, and 42 * 6 = 7. Again, 6 x 7 is the same as 7 x 6. This reversal is true for multiplication, but not division. Keep in mind that multiplication is introduced before division. It may be beneficial to practice multiplication facts before the multiplication/division fact families are introduced.
4. Create or purchase fact family flash cards. You can cut out triangles from index cards or cardstock paper. Write each of the fact family numbers in each corner and the appropriate operation sign along the sides. Making these with your child will be a learning process in itself. Use an addition/subtraction or multiplication/division table as a guide in order to make a complete set. You can generate flashcards on some of the Web sites listed below. They can also be purchased where educational materials are sold.
5. Your child should understand the concept behind math facts before being expected to memorize them. In other words, they should be able to visualize the concept and apply it to real life problems in order for it to really make sense to them. Using simple objects (such as Cheerios, M&M’s, pencils, or paper clips) your child should be able to represent 12 – 3 = 9 by physically moving the objects and actually seeing the problem. Likewise, if they know that 8 * 3 = 24 then the should be able to conceptualize (as in through a drawing) that if you have 3 pizzas with 8 slices each, then you will have 24 slices in total.
Build number sense concepts through practicing counting by 2’s, 5’s, and 10’s to 100 forward and backwards. This also helps your child to understand how numbers have patterns and how they are related. If your child can count: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, etc. then 5 x 7 = 35 will be grounded in a logical pattern. Referring to a 100 chart helps to make these patterns more visual. You can also have them fill in a blank number chart and you can check it for accuracy.
Check out sites below for blank and complete charts. Without these basic understandings they will have a bunch of facts in their head that are disconnected from real life. In addition, this kind of application of math facts is needed for academic success. There is an emphasis on word problems (also called story problems) in math curriculums and standardized testing.
6. Knowing how and how often to practice is important. Flashcards are the staple of learning math facts. Change it up by using standard flashcards and the fact family flash cards. Make your own sets or purchase them. The standard cards usually cost a couple dollars and are available at the grocery store and discount department stores. Keep them handy. Have a set at home and a set in the car. Practice for ten minutes or so in a session. Isolate the facts that they know and don’t know. In other words, take out the cards that are easy for them or focus just on a certain set (such as 7’s multiplication facts) for a while. Start where they are and build skills. It doesn’t matter what they should know if you don’t start at a level that is just beyond what they can do now. Build up to where they should be. Don’t overdo it. It takes practice over time.
7. Counting on fingers shows lack of automaticity. If your child is counting on their fingers or drawing little groups of dashes and counting them in order to solve math facts, they do not have the facts memorized. If this helps your child to develop understanding of the concept in the beginning, then it is beneficial. Let them go through this stage. But eventually they need to ‘know’ the facts; they need to memorize them and give answers almost immediately.
8. Speed is the name of the game. Why is it important to be able to complete a whole page of problems in two minutes? This speed reflects the rate necessary for doing more advanced math. For example, to complete a three digit by three-digit subtraction problem requires not only knowing the correct sequence of steps, but also applying knowledge of subtraction facts multiple times within one problem. Automaticity means they can rattle off the facts as readily as they can count to ten. It becomes a set of memorized information. Again, they should always be able to explain the concepts behind the facts, as word problems are a big part of math instruction.
9. Track progress. The fun thing about math fact practice is that it is easy to track improvement. Your child’s teacher may be giving the timed drills as often as every week. Praise your child for their progress. If your child is catching up with their skills, get a hold of the quizzes and do them at home. Your child’s math teacher can probably give you copies, you can print them off the Internet, or purchase workbooks. Make it fun. Put stickers on quizzes showing success or a chart recording their advancement. Post them on the fridge at home.
10. Look for improvement in other areas of math. As your child starts to increase proficiency with math facts, other areas of math should start to improve. Ask your child’s teacher about this and celebrate all of your child’s successes.
Helping your child learn their basic math facts at home is the best way to ensure a strong foundation from which to build their math knowledge. Slow, steady practice is the key to making progress. Be creative and make the practice as fun as possible and your child will get a positive message about their learning and develop good habits that will cross over into other academic areas.
Susan Niz, M.Ed. | 1,412 | 6,197 | {"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} | 4.40625 | 4 | CC-MAIN-2019-18 | latest | en | 0.915099 |
http://slideplayer.com/slide/4026108/ | 1,519,283,634,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814036.49/warc/CC-MAIN-20180222061730-20180222081730-00463.warc.gz | 326,452,723 | 20,381 | # New Mexico Computer Science For All Decomposition in Computer Programming Maureen Psaila-Dombrowski.
## Presentation on theme: "New Mexico Computer Science For All Decomposition in Computer Programming Maureen Psaila-Dombrowski."— Presentation transcript:
New Mexico Computer Science For All Decomposition in Computer Programming Maureen Psaila-Dombrowski
Decomposition – Yuck! There is nothing rotten about computer programming We mean the other definition – breaking into parts
How to Create a Computer Program Computer program = list of instructions To Write a Computer Program Identify the problem Break it into separate parts (Decomposition) Sometimes the problem must be simplified first (Abstraction) Create the code for each part Combine the pieces of code to create the program
Decomposition Steps Divide and Conquer Step 1: Identify the parts ▫Roughly the same level of detail ▫Can be written up separately ▫Combine to solve the original problem Step 2: Write the steps for each part Step 3: Combine the steps for each part to solve the problem
Example: I Want Cake! Step 1: Identify the Components or Parts A.Gather the Equipment B.Gather the Ingredients C.Measure the Ingredients D.Mix the Ingredients E.Bake the Cake
Example: I Want Cake! (continued) Step 2: Write the Steps ▫A. Gather the Equipment 1.Get the mixing bowls 2.Get small bowls to measure ingredients into 3.Get cake pans 4.Get mixer 5.Get measuring spoons 6.Get measuring cups
Example: I Want Cake! (continued) Step 2: Write the Steps (continued) ▫B. Gather the Ingredients 1.Get the flour 2.Get the sugar 3.Get the oil 4.Get the vanilla 5.Get the eggs 6.Get the baking powder 7.Get the milk 8.Get the pam
Example: I Want Cake! (continued) Step 2: Write the Steps (continued) ▫C. Measure the Ingredients 1.Measure 1 ½ cups flour 2.Put flour in small bowl 3.Measure1 cup sugar 4.Put in another bowl 5.…. You get the idea (continue for parts D and E)
Example: I Want Cake! (continued) Step 3: Put the pieces back together A.Gather the Equipment B.Gather the Ingredients C.Measure the Ingredients D.Mix the Ingredients E.Bake the Cake Then carry out all the small steps………. ………… And when you are done you have Cake!
Advantages of Decomposition Easier to think about smaller pieces of the problem than the whole problem at once. Different people can work on each part (speeds up programming). Someone might gather the ingredients while someone else measure them. Parallel Processing might be possible (speeds up execution). You could preheat the oven while you are preparing the batter.
Disadvantages of Decomposition Computers are Stupid ▫They do exactly what you tell them to do even if it is WRONG. ▫They do the steps in exactly the order you tell them to do even if that’s WRONG too! Sometimes problems cannot be decomposed without simplifying them first ▫Too difficult ▫Not well understood Sometimes when you put the pieces back together it does not work!
Summary Decomposition Break the problem into its components Write the steps (code) for each component Put the pieces together to make the program and solve the problem! This is good because It is easier to think about Programming and execution can be faster But Remember- Computers are stupid - they do: Exactly what you tell them to do Exactly how you tell them to do it Sometimes putting the pieces together doesn’t work
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https://slideplayer.com/slide/3138778/ | 1,532,259,700,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593208.44/warc/CC-MAIN-20180722100513-20180722120513-00114.warc.gz | 782,320,549 | 21,089 | # AKA: The SATs !. Today we are going to…. Consider how the KS2 SATs are prepared for…and how you can help! See what the week looks like… See what.
## Presentation on theme: "AKA: The SATs !. Today we are going to…. Consider how the KS2 SATs are prepared for…and how you can help! See what the week looks like… See what."— Presentation transcript:
AKA: The SATs !
Today we are going to…. Consider how the KS2 SATs are prepared for…and how you can help! See what the week looks like… See what the tests look like… and try some!
But Before We Do… Mental Maths Test 20 Questions Increasing difficulty 5, 10 and 15 second questions All areas of the mathematics curriculum can be covered…
What are SATs? Standard Assessment Tests – Government tests taken by children at the end of Years 2 and 6.
What are the results used for? To assess the progress children have made in their learning since they were 7 (Year 2) To help secondary schools organise learning in Year 7 To enable a comparison between schools and against a benchmark figure
How are the children prepared? Target setting Effective support from home Period of revision Individual study Computers! (online resources) Small focused group support
The internet – there are many websites that provide revision activities and past papers to practice. Below are some of the websites and activities that the children can use for revision: http://www.woodlands-junior.kent.sch.uk/revision/ http://www.woodlands-junior.kent.sch.uk/revision/ http://www.bbc.co.uk/bitesize/ks2/ http://www.bbc.co.uk/bitesize/ks2/ http://www.satsguide.co.uk/ http://www.satsguide.co.uk/ http://www.ixl.com http://www.ixl.com
When Are the SATs? 2 nd Week in May Monday: Comprehension Tuesday: Spelling & Grammar Wednesday: Maths A Thursday: Maths B/Mental Maths Friday: No papers Level 6 papers in the afternoon
English Reading Comprehension – 1 paper 60 minutes Speaking and Listening – teacher assessed Writing – writing is assessed over a longer period by the class teacher (June). SPAG - 1 paper and separate spelling test. Science Teacher Assessment Which assessments are included?
Maths: Paper A 45 minutes Paper B 45 minutes Mental Maths Paper – 20 questions, five 5 second, ten 10 second and five 15 seconds
SPAG
What results or scores will the children be given? Level 3 Level 4 Level 5 Level 6
No holidays will be authorised and any absences will need to be fully investigated. If a child is ill or cannot attend school during SATs week then the school has to be notified immediately so we can seek advice from the LA to see if we can make alternative arrangements.
Other questions… What if... a fire alarm goes off? a pupil is unwell or panics? a child needs to ‘ leave the room?! ’ a pupil is caught cheating?
The week of the exams… Children to have good breakfast and arrive school on time, after a good night ’ s sleep. We will provide the children with a light snack. Children will need to bring a bottle of water.
Study Guides
And finally…Thank you for your time! Please don ’ t worry about SATs. Speak to Mr Hicks, Mr Brown or Mr Statham Results and teacher assessments will go home with end of term reports. Any questions?
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https://anime.stackexchange.com/questions/19956/can-we-roughly-determine-the-dates-and-duration-of-otonashis-stay-in-the-afterl | 1,723,692,845,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00689.warc.gz | 77,664,138 | 39,867 | # Can we roughly determine the dates and duration of Otonashi's stay in the Afterlife World?
In Otonashi's flashback in episode 9, he was on his way to take the National Center Test for University Admissions, held annually during a weekend in mid-January over a period of two days, when the train he boarded crashed. He took out his mobile phone to check the time and then realized that he was late for the test. Thus, we can conclude that January 15 was a Sunday and Otonashi left the real world on January 21 (Day 7).
Despite being a school-themed anime, there seemed to be little indication of time and the seasons in the Afterlife World when compared to the real world in the flashback episode, where we knew the exact dates (and even the days) of the incidents that were happening. Can we tell the date (or the time of the year) each episode occurred on from the activities the students were doing (or any other hint)? And can we possibly determine the duration of Otonashi's stay in the Afterlife World up till the last episode?
• I don't know but is it ok to compare real life world time with after life time. As you describe it sounds like he stayed in after life roughly 2 months. Commented Mar 11, 2015 at 4:49
In the Hell's Kitchen OVA set between episodes 2 and 3, we learned from Yuri's event application form that the deadly picnic took place on May 3.
Furthermore, Yuri allowed for one week of preparation for the picnic prior to the Golden Week. Taking that into account and given that the series of events connecting the first two episodes and the second OVA episode were tightly packed, it is probably safe to say that Otonashi woke up in the Afterlife World sometime around the start of the first term of the Japanese trimester system, maybe even on the first day of the fourth month as an April Fool joke by God.
There is no set date for the Ball Day (球技大会), but with its conditions being similar to the day of the regional game final that would decide if Hinata and his team get to participate in the Summer Koushien, it should be set in late July during the Dog Days, or not, if the anime followed the logic explained in the next paragraph.
Throughout the anime, everyone wore the same winter uniform, unlike in the Angel Beats! Heaven's Door manga where they actually switched to the summer uniform once. As the dates of the seasonal uniform switch in Japan (June 1 and October 1) are strict and almost universal, there is little doubt that the anime didn't show anything that occurred between these two dates.
Following episode 4 and preceding episode 5 is the Stairway to Heaven OVA episode. Otonashi and his comrades organized the Sports Day that normally occurs in September/October. Then by episode 5, the students were taking their second midterm exams in early/mid October per usual.
November largely overlaps with the tenth month of the lunar calendar, the Kannazuki (神無月), or the "Month with/without Gods".
The 無 character, which normally means "absent" or "there is not", was here probably originally used as ateji, that is used only for the sound "na". In this name the na is actually a possessive particle, so Kaminazuki means "Month of the Gods", not "Month without Gods" (Kaminakizuki), similarly to Minatsuki, the "Month of Water".
Whether God exists or not is still debatable, like how it is with the na in Kannazuki. Interestingly, Naoi proclaimed, in episode 6 during the "Month with/without Gods", that there was no God and that he is God. Maybe God is a Schrödinger's cat in the tenth month of the lunar calendar.
From mid-November to early December is the "rainy season of the Camellia". (The only times it rained in the anime were in episodes 6 and 9.) We saw the blooming Camellias in episode 7:
We also saw Kanade weeding the Chrysanthemums japonense, which typically flower in November annually.
I couldn't pick up any seasonal cue in the later episodes, so I'm less confident with the time of the events happening in episode 8 and onwards. It was not clear how long Kanade stayed in a coma, but could it be that she woke up on or around the day Hatsune died (Christmas Eve)? I couldn't help make this assumption because of the similarities between Kanade and Hatsune.
And then there is Matsushita's mountain training. It is possible, but highly unlikely, especially for a person like Matsushita who is not too fat, to lose 10 kg in a week. Judgment by the eyes (my eyes) tells me that he had lost 20–40 kg by the time he returned from the mountains in episode 12. That equates to a 3–6 weeks minimum time gap between the beginnings of episodes 10 and 12. However, it is far more likely that he was away for longer than that, probably 1–3 months to coincide with the time of the graduation, which is in March. I would wager that the anime ended with the end of the third term in March, excluding the epilogues.
That's a complicated question.
Really, you cannot know when Otonashi enters afterlife. But, something is clear: Isn't immediately after his dead. The only explanation about this, are who you can think "time does not pass when you're dead".
Since Kanade enters afterlife before Otonashi (Kanade have Otonashi's heart, if you understand this, Kanade dead after Otonashi), there's some time who Otonashi are dead and out of afterlife.
Some option are this: Kanade doesn't feel fulfill with their life. Because this, they arrives afterlife. Some time ago, their feeling and "request" for fulfill their life make Otonashi to go afterlife (from a unknown place). Then, the series start. | 1,281 | 5,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-33 | latest | en | 0.982269 |
https://amigotutor.com/victoria/state-mean-value-theorem-of-integrals-and-give-an-example.php | 1,610,773,242,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703500028.5/warc/CC-MAIN-20210116044418-20210116074418-00098.warc.gz | 222,406,830 | 7,256 | # Give of mean example state an theorem and value integrals
## The Mean Value Theorem For Derivatives
Definite Integrals The Fundamental Theorem of Integral. the integral mean value theorem: mean-value theorem (wolfram may be shared with the author of any specific demonstration for which you give, mean value theorem for integrals. we demonstrate the principles involved in this version of the mean value theorem in the following example. the ohio state).
What is the Mean Value Theorem? The Mean Value Theorem states that if y= f(x) What Does This Time Mean? An Example of The Mean Value Theorem In this section we will give Rolle's Theorem and the Mean Value Fundamental Theorem for Line Integrals; at a couple of examples using the Mean Value Theorem.
... a problem solving video, and a worked example on the mean value theorem. Session 34: Introduction to the Mean Value Theorem Give Now. Make a Donation; I'll just write the acronym, mean value theorem for integrals, or integration, which essentially, to give it in a slightly more formal sense,
20/09/2013В В· In mathematics, the mean value theorem states, I uses the mean value theorem, in integral is indirect and does not give an explicit example, Counterexample for mean value theorem. \$ should basically give you the same interval for all \$h\$ small. LMVT and Mean value theorem for integrals. 0.
The Mean Value Theorem; Example 16.3.2 An object moves in the force field \$\$ To make use of the Fundamental Theorem of Line Integrals, ... by the mean value theorem, give the endpoints of C. A line integral of a scalar field is thus a line integral of a vector "Line Integral Example 2
Second Mean Value Theorem for integrals These have been used in a course in mathematical analysis. Below we give a Example of the use of Taylor's Theorem 0:06 Average Value Theorem; 2:03 Example; That's going to give me an average height, I'm going to take the integral from 0 to 3 of 9
So the conclusion of the Mean Value Theorem states that there exists a point such that the tangent line is parallel to the line Example. Let , a = -1and b=1. We The mean value theorem The theorem states that the derivative of a continuous and differentiable function must Examples; Mean Value Theorem for Integrals;
Mean value theorem for the double integral vector forms
Mean Value Theorem Ximera - Ohio State University. what does this mean? but do not satisfy the mean value theorem for integrals. example 4 weвђ™ll give you challenging practice questions to help you achieve, the integral mean value theorem: mean-value theorem (wolfram may be shared with the author of any specific demonstration for which you give); mean value theorem for integrals. we demonstrate the principles involved in this version of the mean value theorem in the following example. the ohio state, what is the mean value theorem? =0, and f(1)=2, so some value between 0 and 1 will give me f(x)=1. another example the intermediate value theorem says that.
16.3 The Fundamental Theorem of Line Integrals
Mathematical Analysis II My Calculus Web. 20/09/2013в в· in mathematics, the mean value theorem states, i uses the mean value theorem, in integral is indirect and does not give an explicit example,, the mean value theorem for integration: is also called the average value of f(x). in the example above, just as the tangent lines to position functions give).
Calculus Mean Value Theorem for Integrals Example YouTube
The Mean Value Theorem For Derivatives. get the free "mean value theorem solver" widget for your website, blog, wordpress, blogger, or igoogle. find more mathematics widgets in wolfram|alpha., is this an acceptable proof for the mean value theorem of integrals? and you would need to give an explanation is this an acceptable proof for the mean value).
Extreme Value Theorem Cliffs Notes
Mathematical Analysis II My Calculus Web. mean value theorem for integrals. we demonstrate the principles involved in this version of the mean value theorem in the following example. the ohio state, by the mean value theorem, for every i = 1 2 the fundamental theorem of calculus as can be seen from these examples, the fundamental theorem of integral).
Applications of integrals Ximera
Is this an acceptable proof for the mean value theorem of. the fundamental theorem of calculus states that for a continuous function on an integral mean value theorem specific demonstration for which you give, lecture 9: the mean value theorem today, weвђ™ll state and prove the mean value theorem and describe other ways in which derivatives of functions give us global).
The intermediate value theorem states that if a continuous function Here is an illustrative example: (S^1\), we mean the set of vectors in \(\mathbb{R}^2 See that differentiating the function will give State and Prove the Mean Value theorem. The Mean Value theorem states Mean Value Theorem for Integrals Example.
David Little: Mathematics Department Penn State University When you think you've found a value of c that satisfies the conclusion of the mean value theorem, The mean value theorem states that under the specified hypotheses, What happens if we violate one of these hypotheses, for example,
Second Mean Value Theorem for integrals These have been used in a course in mathematical analysis. Below we give a Example of the use of Taylor's Theorem I'll just write the acronym, mean value theorem for integrals, or integration, which essentially, to give it in a slightly more formal sense,
The Mean Value Theorem states that, Mean Value Theorem Examples. Start here or give us a call: (312) 646-6365. The Mean Value Theorem states that, Mean Value Theorem Examples. Start here or give us a call: (312) 646-6365.
The Mean Value Theorem states that if a function f is continuous on the closed , we'll try to give you a kind of a real life example about when that make sense. 1/01/2007В В· Give an example of a do you mean the mean value theorem for derivatives or the mean value theorem for integrals? Now the mean value theorem states
The mean value theorem The theorem states that the derivative of a continuous and differentiable function must Examples; Mean Value Theorem for Integrals; We give more contexts to understand integrals. Understand the statement of the Mean Value Theorem. The Ohio State University — Ximera team.
Using Area Mean Value Theorem to Solve Some Double double integrals, area mean value theorem, Maple . Example 2 In Eq. The Mean Value Theorem; Example 16.3.2 An object moves in the force field \$\$ To make use of the Fundamental Theorem of Line Integrals,
... sometimes called the second fundamental theorem of calculus, states that the integral mean value theorem for integration, example, the theorem can be The First Mean-Value Theorem for Riemann-Stieltjes Integrals. We will now look at a very useful theorem known as the First Mean-Value Theorem for Riemann-Stieltjes
Mean value theorem Wikipedia | 1,524 | 6,953 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2021-04 | latest | en | 0.849803 |
http://earthguide.ucsd.edu/virtualmuseum/content/spectandstars.html | 1,539,853,410,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511761.78/warc/CC-MAIN-20181018084742-20181018110242-00075.warc.gz | 112,241,687 | 2,667 | Spectroscopy & Stars
What kinds of useful information can be gathered by collecting electromagnetic radiation, like visible light or infrared radiation?
By collecting the radiation stars emit astronomers can determine the brightness of an object and the spectra of that object (e.g., with a visible light telescope one can determine the color spectrum). The brightness can tells us the distance of a star, while the spectrum tells us temperature, mass, chemical make-up, diameter, and distance. In fact, spectra are such an information-rich measurement that astronomers do not usually directly look through their large telescopes. Instead, they collect light in an instrument called a spectroscope, which itself is connected to a computer for data analysis.
A spectroscope is an instrument that consists of a prism or a grating spreads the incoming beam of radiation into its different wavelengths and some kind of screen to project the spectrum (see below).
Cartoon of a spectroscope: white light is spread by a prism into long and short wavelengths.
Spectra come in several types:
1. Continuous spectra: a smooth gradient of electromagnetic radiation without any gaps – e.g., the spectrum of incandescent solids.
2. Absorption spectra: an incomplete spectrum with missing gaps (which appear as dark lines) due to the absorption of a continuous electromagnetic radiation by a cooler medium, like a gas. Such absorbed energy can be re-emitted, but the absorbed energy is essentially removed from a telescope’s view. Since the “cooler,” outer gaseous surface of a star tends to absorb the radiation produced in the hotter, inner part, the spectra of most stars are absorption spectra.
3. Emission spectra: a spectrum that represents all the wavelengths emitted by atoms or molecules.
Astronomers take advantage of something from physics called Wien’s Displacement Law, a mathematical relationship that basically says that the hotter a body (like a star) is, the shorter the wavelength of light will be emitted from it:
λpeak T = 2.898 x 10-3 m · K
where λpeak is the peak (i.e., maximum) wavelength that the star emits and T is the star’s surface temperature. Hence, a red star, with a maximum wavelength of 966 nanometers, has a surface temperature of “only” 3000 Kelvin while a blue star, emitting at a maximum wavelength of 290 nanometers, has a surface temperature of 10, 000 Kelvin!
In addition, the stars have been classified into “spectral classifications” (labeled by a letter) based on their surface temperature. These spectral types also organize the stars by their chemical make-up and their main sequence lifetimes, that is, the lifetime of the star based on calculations of its available fuel and the rate at which it is consuming that fuel (as interpreted by its luminosity).
Chart illustrating the relations between stars’ spectral type, surface temperature, color and lifetime. Note that Myr = Millions of years and Gyr = Billions of years. | 604 | 2,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-43 | latest | en | 0.905917 |
http://www.ognition.com/76/formula-for-water-hammer-calculation | 1,569,102,201,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574665.79/warc/CC-MAIN-20190921211246-20190921233246-00294.warc.gz | 276,842,399 | 7,686 | 149 views
# Formula for water hammer calculation
Is there any approximation formulas for calculating water hammer?
answered Mar 2, 2014 by (21,030 points)
The execc pressure due to water hammer can be calculated by the Joukowsky equation:
dP = Zh × Q
Where:
• dP is overpressurization, Pa;
• Q is the volumetric flow, m3/s;
• Zh is the hydraulic impedance, expressed in kg/m4/s.
Hydraulic impedance Zh defined by:
$Z_h = \frac{\sqrt{\rho \, B_\mathit{eff}}}{A}$
Where:
• ρ the density of the liquid, kg/m3;
• A cross sectional area of the pipe, m2;
• Beff effective modulus of compressibility of the liquid in the pipe, expressed in Pa. | 191 | 648 | {"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": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | longest | en | 0.73577 |
http://www.fangraphs.com/library/category/featured/ | 1,448,919,260,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398463315.2/warc/CC-MAIN-20151124205423-00104-ip-10-71-132-137.ec2.internal.warc.gz | 415,357,704 | 20,449 | ## The Beginner’s Guide to Measuring Defense
There’s a decent chance you’ve arrived at this page without a serious desire to hear more about defensive statistics. Trust me, I understand your frustration and your fatigue. Defensive stats like Ultimate Zone Rating and Defensive Runs Saved are controversial in some circles because they are reasonably new and the underlying data is somewhat hidden from view. You hear words like “flawed,” “absurd,” and “subjective” surrounding them. You’re tired of it.
Yet I’d like to lay out why we have advanced defensive statistics and how they work in the abstract. You won’t get to the end of this post and decide that UZR has perfectly measured Alex Gordon‘s defense, but hopefully you will have a better appreciation for why we measure defense the way that we do.
## How to Use FanGraphs: Leaderboards!
In addition to updated glossary entries and blog posts extolling the virtues of various sabermetric statistics and principles, the revitalized FanGraphs Library is also going to be a place where we highlight features available at the site that will allow you to get the most out of our data.
Below, you’ll find everything you ever wanted to know about the FanGraphs Leaderboards. If you’ve been a long-time reader who never misses a single post, a lot of this might be old news. If you’re anything short of that, there’s a good chance you’ll pick up a few tricks to get the most out of the site.
## wOBA As a Gateway Statistic
Despite all of the rhetoric and talk-radio bluster, sabermetric principles and statistics aren’t actually very complicated. It might take a sharp statistician or savvy programmer to derive perfect park factors, but it doesn’t take anything more than a curious mind to understand and apply the basics. In my time working to help spread these principles, one of the most common and useful questions I get is about which few statistics a person should learn when trying to get into the world of advanced stats.
On Wednesday during my chat I got such a question. Here’s how I responded: | 430 | 2,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2015-48 | longest | en | 0.936527 |
https://www.slideserve.com/nigel/scientific-method-quiz | 1,516,663,608,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891543.65/warc/CC-MAIN-20180122213051-20180122233051-00048.warc.gz | 979,677,334 | 12,997 | Scientific Method Quiz
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# Scientific Method Quiz - PowerPoint PPT Presentation
Scientific Method Quiz. Warm-ups: October 16-18. Tuesday, October 16, 2012. Oh no! The froggies are dying at Summitville Pond! Biologist Bill has decided to hire you to find out why. Write a question for your investigation. Make sure your question includes something that can be measured.
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### Scientific Method Quiz
Warm-ups: October 16-18
Tuesday, October 16, 2012
• Oh no!
• The froggies are dying at Summitville Pond!
• Biologist Bill has decided to hire you to find out why.
• Write a question for your investigation. Make sure your question includes something that can be measured.
Wednesday, October 17, 2012
• Background Information:
• Summitville Pond is located in a resort area much like Summit County. List some possible causes of the Froggy Dilemma.
• What could be causing our froggies to die?
• Your background knowledge needs to include a minimum of three facts.
Thursday, October 13, 2012
Write a hypothesis for your
investigation. Make sure it
is in the if…, then…, because
format. Also make sure it
relates to the question.
Monday, October 22, 2012
• Procedures
• Determine how you are going to test your hypothesis.
• List the steps necessary to complete your scientific investigation.
• What is your manipulated (change) variable?
• What is being held constant during your investigation?
Tuesday, October 23, 2012
• AllWrite Consensus
• Think about key ideas from the River Study Unit.
• Selected Student will suggest an idea to the group.
• Teammates will put thumbs up, thumbs down or thumbs sideways to indicate agreement, disagreement, or doubt.
• If teammates agree , all students write the answer on their warm-up. If there is disagreement, the team discusses the answer until agreement is reached.
Process is continued until your group has 7 key ideas.
Wednesday , October 24, 2012
• Window Paning
• In the top left hand window, write in the topic: River Study
• Take 1 of your key ideas from the river study unit and draw a simple, hand-sketched drawing of the idea. No Words!
• Repeat the process for two more of your key ideas!
Title Backdrop
Slide Backdrop
Print Backdrop
Transitional Backdrop
www.animationfactory.com
Backdrops:
- These are full sized backdrops, just double click them and size it up!
Images:
- Most of these .gifs, .jpgs, and .png images can be scaled up to fit your needs! | 747 | 3,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.93652 |
http://www.newton.dep.anl.gov/askasci/phy99/phy99383.htm | 1,419,085,245,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802769894.131/warc/CC-MAIN-20141217075249-00094-ip-10-231-17-201.ec2.internal.warc.gz | 691,084,762 | 3,865 | Electromagnetism/Maglev ```Name: ann Status: N/A Age: N/A Location: N/A Country: N/A Date: N/A ``` Question: I am teaching a unit on electromagnetism to fifth graders. I am also doing a maglev train experiment/race with them. I need to find an elementary way to explain electricity. Can you recommend some books to show me the way and what to explain? Replies: I would suggest using the CASTLE kits with your students. These are available through most scientific supply companies. If that is too much try getting a bunch of GENECON hand crank DC generators. Have your students crank the generators with progressively larger parallel and series circuits. Use light bulbs as the loads. With the parallel circuits the electrical resistance obviously goes down with the addition of each new light bulb, but it gets progressively harder to turn the crank at a constant speed because of the increased demand for current (load). The inverse effect occurs with the series circuit as the electrical resistance gets larger reducing the current and hence the load on the generator. This is all neatly explained with the standard fluid model of electricity. I would stay away from trying to explain this stuff with electric fields to fifth graders. The CASTLE kits use great big (one farad caps. at 25 volts) capacitors to further develop the fluid analogies in electricity and give a sense of concretness, if you will, to something that is essentially invisible. The CASTLE kits come with excellent manuals with worksheets for your students. With some adaptation all of this stuff should work nicely with your fifth graders. Hit my NEWTON mail box if you have any further questions. I am an ex-EE teaching high school physics. By the way; these little generators break rather easily. Make sure your students do not try to crank them too fast and do not let them crank them with the terminals shorted. Have Fun! Nick P. Drozdoff Click here to return to the Physics Archives
NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.
For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs | 477 | 2,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | latest | en | 0.93354 |
https://www.askmathematics.com/uncategorized/evaluate-the-polynomial-for-the-given-value-by-using-synthetic-division-px-3x4-x3-7x-2-for/ | 1,638,365,866,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.0/warc/CC-MAIN-20211201113241-20211201143241-00091.warc.gz | 695,130,333 | 6,261 | # Evaluate the polynomial for the given value by using synthetic division. P(x) = 3×4 – x3 + 7x – 2 for
Evaluate the polynomial for the given value by using synthetic division.
P(x) = 3×4 – x3 + 7x – 2 for x = 4 and x = -2 | 74 | 224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-49 | latest | en | 0.856098 |
http://www.jiskha.com/display.cgi?id=1295819883 | 1,498,480,612,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320736.82/warc/CC-MAIN-20170626115614-20170626135614-00056.warc.gz | 550,660,543 | 3,542 | # Physics
posted by .
A ball is thrown directly downward, with an initial speed of 6.10 m/s, from a height of 31.0 m. After what interval does the ball strike the ground?
• Physics -
d = Vi*t + 0.5gt^2 = 31m,
6.1t + 0.5*9.8t^2 = 31,
4.9t^2 + 6.1t - 31 = 0,
Solve for t using Quad. Formula and get:
t = 1.97s.
### Related Questions
More Related Questions
Post a New Question | 136 | 381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2017-26 | latest | en | 0.85001 |
https://dms.umontreal.ca/~andrew/Binomial/pflemma1.html | 1,544,405,769,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823236.2/warc/CC-MAIN-20181210013115-20181210034615-00104.warc.gz | 594,296,936 | 1,474 | # Wilson's Theorem for prime powers
## (The Proof of Lemma 1 and its Corollary)
Wilson's Theorem, states that . We generalize it, to prime powers, as follows:
Lemma 1 For any given prime power we have
where is -1, unless in which case is 1.
Proof:
Pair up each m in the product with its inverse to see that is congruent, modulo , to the product of those that are not distinct from their inverses ; that is those m for which . It is easy to show that the only such m are 1 and unless (when one only has m=1) or when one has the additional solutions and . The result then follows.
Corollary 1 For any given prime power , let be the least non-negative residue of . Then
where is as in Lemma 1.
Proof: We write each r in the product below as , to get
by Lemma 1, where signifies a product over integers not divisible by p. | 208 | 827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-51 | latest | en | 0.967505 |
https://conversioncalculator.org/grams-to-liters/ | 1,685,903,043,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650201.19/warc/CC-MAIN-20230604161111-20230604191111-00080.warc.gz | 218,262,485 | 54,439 | # Grams To Liters
## 1 Liter = 1000 Grams
Grams To Liters 1 liter (l) = 1000 gram (g). Liter (l) is a unit of Volume used in the Metric system. Gram (g) is a unit of Weight used in the Metric system. Please note this is volume to weight conversion, this conversion is valid only for pure water at temperature 4 °C.
1000
## Quick conversion chart of grams to liter
1 grams to liter = 0.001 liter
10 grams to liter = 0.01 liter
50 grams to liter = 0.05 liter
100 grams to liter = 0.1 liter
200 grams to liter = 0.2 liter
500 grams to liter = 0.5 liter
1000 grams to liter = 1 liter
How many grams in 1 liter? The answer is 1000. We assume you are converting between gram [water] and liter. You can view more details on each measurement unit: grams or liter The SI derived unit for volume is the cubic meter.
How many grams are in 1 liter of water?
1000 grams
Since there are 1000 grams in a kilogram, our answer is that 1 liter of water weighs 1000 grams.
How many grams is 5 liters?
5000
How many grams is 1.2 Litres?
Convert 1.2 Liters to Grams
To convert 1.2 liters to g we use the formula [L] = [1.2] x 1000 / D. In case of water at sea level and 39.2 °F, D = 1, so L = g x 1000. Under these circumstances, 1.2 liters of water equal 1200 grams.
#### l to g conversion table:
0.01 liter = 10 grams 0.21 liter = 210 grams 0.41 liter = 410 grams 0.7 liter = 700 grams 0.02 liter = 20 grams 0.22 liter = 220 grams 0.42 liter = 420 grams 0.8 liter = 800 grams 0.03 liter = 30 grams 0.23 liter = 230 grams 0.43 liter = 430 grams 0.9 liter = 900 grams 0.04 liter = 40 grams 0.24 liter = 240 grams 0.44 liter = 440 grams 1 liter = 1000 grams 0.05 liter = 50 grams 0.25 liter = 250 grams 0.45 liter = 450 grams 1.1 liter = 1100 grams 0.06 liter = 60 grams 0.26 liter = 260 grams 0.46 liter = 460 grams 1.2 liter = 1200 grams 0.07 liter = 70 grams 0.27 liter = 270 grams 0.47 liter = 470 grams 1.3 liter = 1300 grams 0.08 liter = 80 grams 0.28 liter = 280 grams 0.48 liter = 480 grams 1.4 liter = 1400 grams 0.09 liter = 90 grams 0.29 liter = 290 grams 0.49 liter = 490 grams 1.5 liter = 1500 grams 0.1 liter = 100 grams 0.3 liter = 300 grams 0.5 liter = 500 grams 1.6 liter = 1600 grams 0.11 liter = 110 grams 0.31 liter = 310 grams 0.51 liter = 510 grams 1.7 liter = 1700 grams 0.12 liter = 120 grams 0.32 liter = 320 grams 0.52 liter = 520 grams 1.8 liter = 1800 grams 0.13 liter = 130 grams 0.33 liter = 330 grams 0.53 liter = 530 grams 1.9 liter = 1900 grams 0.14 liter = 140 grams 0.34 liter = 340 grams 0.54 liter = 540 grams 2 liters = 2000 grams 0.15 liter = 150 grams 0.35 liter = 350 grams 0.55 liter = 550 grams 3 liters = 3000 grams 0.16 liter = 160 grams 0.36 liter = 360 grams 0.56 liter = 560 grams 4 liters = 4000 grams 0.17 liter = 170 grams 0.37 liter = 370 grams 0.57 liter = 570 grams 5 liters = 5000 grams 0.18 liter = 180 grams 0.38 liter = 380 grams 0.58 liter = 580 grams 7 liters = 7000 grams 0.19 liter = 190 grams 0.39 liter = 390 grams 0.59 liter = 590 grams 9 liters = 9000 grams 0.2 liter = 200 grams 0.4 liter = 400 grams 0.6 liter = 600 grams 10 liters = 10000 grams
#### g to l conversion table:
10 grams = 0.01 liter 210 grams = 0.21 liter 410 grams = 0.41 liter 700 grams = 0.7 liter 20 grams = 0.02 liter 220 grams = 0.22 liter 420 grams = 0.42 liter 800 grams = 0.8 liter 30 grams = 0.03 liter 230 grams = 0.23 liter 430 grams = 0.43 liter 900 grams = 0.9 liter 40 grams = 0.04 liter 240 grams = 0.24 liter 440 grams = 0.44 liter 1000 grams = 1 liter 50 grams = 0.05 liter 250 grams = 0.25 liter 450 grams = 0.45 liter 1100 grams = 1.1 liter 60 grams = 0.06 liter 260 grams = 0.26 liter 460 grams = 0.46 liter 1200 grams = 1.2 liter 70 grams = 0.07 liter 270 grams = 0.27 liter 470 grams = 0.47 liter 1300 grams = 1.3 liter 80 grams = 0.08 liter 280 grams = 0.28 liter 480 grams = 0.48 liter 1400 grams = 1.4 liter 90 grams = 0.09 liter 290 grams = 0.29 liter 490 grams = 0.49 liter 1500 grams = 1.5 liter 100 grams = 0.1 liter 300 grams = 0.3 liter 500 grams = 0.5 liter 1600 grams = 1.6 liter 110 grams = 0.11 liter 310 grams = 0.31 liter 510 grams = 0.51 liter 1700 grams = 1.7 liter 120 grams = 0.12 liter 320 grams = 0.32 liter 520 grams = 0.52 liter 1800 grams = 1.8 liter 130 grams = 0.13 liter 330 grams = 0.33 liter 530 grams = 0.53 liter 1900 grams = 1.9 liter 140 grams = 0.14 liter 340 grams = 0.34 liter 540 grams = 0.54 liter 2000 grams = 2 liters 150 grams = 0.15 liter 350 grams = 0.35 liter 550 grams = 0.55 liter 30000 grams = 30 liters 160 grams = 0.16 liter 360 grams = 0.36 liter 560 grams = 0.56 liter 4000 grams = 4 liters 170 grams = 0.17 liter 370 grams = 0.37 liter 570 grams = 0.57 liter 5000 grams = 5 liters 180 grams = 0.18 liter 380 grams = 0.38 liter 580 grams = 0.58 liter 7000 grams = 7 liters 190 grams = 0.19 liter 390 grams = 0.39 liter 590 grams = 0.59 liter 9000 grams = 9 liters 200 grams = 0.2 liter 400 grams = 0.4 liter 600 grams = 0.6 liter 10000 grams = 10 liters
How many Litres is 100g?
Is 1000 grams a liter?
What is 500g in Litres?
How much is 4 Litres in grams?
How many liters is 400 grams?
How many Litres is 1 kg?
What is 250g in Litres?
How many liters is 28 grams? | 1,932 | 5,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-23 | longest | en | 0.607325 |
https://math.answers.com/basic-math/What_percentage_is_2_wrong_out_of_36_questions | 1,675,339,772,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00489.warc.gz | 411,991,364 | 50,130 | 0
# What percentage is 2 wrong out of 36 questions?
Wiki User
2012-10-27 22:31:16
That means you received a 34/36. A simple division problem will yield your percentage.
We get: (34 ÷ 36) x 100 = 94.4%. Which is the percentage for an "A" grade.
Wiki User
2012-10-27 22:31:16
Study guides
86 cards
➡️
See all cards
4.02
130 Reviews | 117 | 339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | longest | en | 0.839758 |
http://letsmakerobots.com/node/14101 | 1,444,049,510,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736677342.4/warc/CC-MAIN-20151001215757-00174-ip-10-137-6-227.ec2.internal.warc.gz | 59,797,989 | 7,153 | A Guide To using Accelerometer and Gyroscope. How to combine data from these devices to obtain accurate inclination estimations.
A simple explanation of IMU devices that does not require any advanced math.
The full up-to-date text of article can be accessed here:
http://starlino.com/imu_guide.html
This guide is intended to everyone interested in inertial MEMS (Micro-Electro-Mechanical Systems) sensors, in particular Accelerometers and Gyroscopes as well as combination IMU devices (Inertial Measurement Unit).
I'll try try to cover few basic but important topics in this article:
- what does an accelerometer measure
- what does a gyroscope (aka gyro) measure
- how to convert analog-to-digital (ADC) readings that you get from these sensor to physical units (those would be g for accelerometer, deg/s for gyroscope)
- how to combine accelerometer and gyroscope readings in order to obtain accurate information about the inclination of your device relative to the ground plane
Throughout the article I will try to keep the math to the minimum. If you know what Sine/Cosine/Tangent are then you should be able to understand and use these ideas in your project no matter what platform you're using Arduino, Propeller, Basic Stamp, Atmel chips, Microchip PIC, etc. There are people out there who believe that you need complex math in order to make use of an IMU unit (complex FIR or IIR filters such as Kalman filters, Parks-McClellan filters, etc). You can research all those and achieve wonderful but complex results. My way of explaining things require just basic math. I am a great believer in simplicity. I think a system that is simple is easier to control and monitor, besides many embedded devices do not have the power and resources to implement complex algorithms requiring matrix calculations.
I'll use as an example a new IMU unit that GadgetGangster just launched - the Acc_Gyro Accelerometer + Gyro IMU. We'll use parameters of this device in our examples below. This unit is a good device to start with because it consists of 2 devices:
- LIS331AL (data sheet) - a triaxial 2G accelerometer
- LPR550AL (data sheet) - a dual-axis pitch and roll, 500 deg/sec gyroscope
Together they represent a 5-Degrees of Freedom Inertial Measurement Unit. Now that's a fancy name! Nevertheless, behind the fancy name is a very useful combination device that we'll cover and explain in detail below. | 529 | 2,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-40 | latest | en | 0.907322 |
https://discuss.codecademy.com/t/faq-taking-a-vacation-plan-your-trip/371678/8 | 1,638,156,192,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00628.warc.gz | 302,094,875 | 9,156 | # FAQ: Taking a Vacation - Plan Your Trip!
This community-built FAQ covers the “Plan Your Trip!” exercise from the lesson “Taking a Vacation”.
Paths and Courses
This exercise can be found in the following Codecademy content:
## FAQs on the exercise Plan Your Trip!
There are currently no frequently asked questions associated with this exercise – that’s where you come in! You can contribute to this section by offering your own questions, answers, or clarifications on this exercise. Ask or answer a question by clicking reply () below.
If you’ve had an “aha” moment about the concepts, formatting, syntax, or anything else with this exercise, consider sharing those insights! Teaching others and answering their questions is one of the best ways to learn and stay sharp.
## Join the Discussion. Help a fellow learner on their journey.
Agree with a comment or answer? Like () to up-vote the contribution!
Found a bug? Report it!
Have a question about your account or billing? Reach out to our customer support team!
None of the above? Find out where to ask other questions here!
This exercise is adding spending for a trip - First function is:
def hotel_cost(nights):
return 140 * nights
Then 2 more functions:
plane_ride_cost(city):
[…]
def rental_car_cost(days):
[…]
At the end we create a function to sum up all our costs:
def trip_cost(city, days):
return hotel_cost(days - 1) + plane_ride_cost(city) + rental_car_cost(days)
What I don’t understand is how we can change the first function parameter from (nights) to (days - 1) and it still understands how to do the calculation 140 * nights? If anyone could help would be appreciated.
Cheers,
Hi Brooksey,
I just had a problem getting through this part. It looks like in the end it requires the “-1” be dropped off of the hotel_cost function call in trip_cost.
In other words, even though prior parts of the module tell you to code:
def trip_cost(city, days, spending_money):
return rental_car_cost(days) + hotel_cost(days -1) + plane_ride(city) + spending_money
It actually wants this:
def trip_cost(city, days, spending_money):
return rental_car_cost(days) + hotel_cost(days) + plane_ride(city) + spending_money
It looks like there might be an error in the practice module here. Either that or I missed something along the way. Feel free to correct me if I got something wrong as I’m still new to this.
3 Likes
Here’s what I have:
def trip_cost(city, days, spending_money):
return rental_car_cost(days) + hotel_cost(days) + plane_ride_cost(city) + spending_money
print rental_car_cost(5) + hotel_cost(5 - 1) + plane_ride_cost(“Los Angeles”) + 600
I get “1815”, but it keeps spinning and I cannot get to the next exercise.
1 Like
Hi Everyone,
Im not sure if you guys got through this already. I don’t really want to type up the correct answer but I can give you a hint. Just check the instructions again, the answer is in the very first sentence. It gives you the whole line of script how to do it.
" After your previous code, `print` out the `trip_cost(` to `"Los Angeles"` for `5` days with an extra `600` dollars of spending money. "
print something(“city”,days,spending_money)
Sometimes you need to refresh it. If refreshing it does not work, copy your code. then reload browser and paste code. Should work then. This happens to me sometimes when I’m doing something else and the codecademy times out…
1 Like
Hi there, I’d like to ask a question on how indentations affect our outcome.
def trip_cost(city, days, spending_money):
return rental_car_cost(days) + hotel_cost(days) + plane_ride_cost(city) + spending_money
print trip_cost(“Los Angeles”, 5, 600)
The code above will not print anything at all. Even though the before statements are valid and true, why does being under a definition cancel out its ability to be printed out?
Can you repost your code, formatted? You can see this post for formatting instructions. Indentation is important in Python and we need to be able to see your indentation to figure out what went wrong (there are several things that could have happened, each based on different indentations).
Based on what you’ve posted thus far, the problem might be that your `print` statement comes after your `return` statement. Remember that immediately after a `return` statement is executed, the function is exited, meaning no more code inside of it will run. Therefore, if `print` comes after `return`, that `print` statement will not run. This all occurs because `print` is indented to be inside the function `trip_cost`.
Thought the same as you! Can codecademy correct it so people won’t get confused and frustrated ! | 1,054 | 4,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-49 | latest | en | 0.896257 |
https://blog.exair.com/tag/atmospheric/ | 1,695,839,056,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510319.87/warc/CC-MAIN-20230927171156-20230927201156-00528.warc.gz | 159,941,298 | 56,460 | ## Tools of The Trade – Glass Rotameters
Today, I wanted to share one of the many tools that we utilize here in the EXAIR Efficiency Lab. The video will show how a Glass Rotameter is used and works under both compressed air and atmospheric air volumetric flows. If you want to see how your products flow, give us a call, and we will set up an EXAIR Efficiency Lab for you.
Brian Farno
Application Engineer
BrianFarno@EXAIR.com
@EXAIR_BF
## Pressure – Absolute, Gauge, and Units of Both
Compressed air is a common utility used throughout industrial facilities and it has to be measured like any other utility in order to know just how much a facility is using. When dealing with compressed air a common unit of measurement that readily comes up is psi, pound-force per square inch. This unit of measure is one of the most basic units used to measure pressure in the compressed air industry. There are other means to measure this though, so let’s discover the difference.
Again, the pressure is a force distributed over an area, the Earth’s atmosphere has pressure, if it didn’t we would all balloon up like the Violet from Willy Wonka, just without eating some prototype gum causing internal pressure. PSIA is a unit of measure that is relative to a full vacuum. It is pounds per square inch absolute (PSIA). The absolute pressure is calculated as the sum of the gauge pressure plus the atmospheric pressure. If you were to travel into space, the atmospheric pressure would be absolute zero which is actually a vacuum. There is nothing pushing from the outside in so the inside pushes out, hence the ballooning.
The atmospheric pressure on earth is based on sea level. This is 14.7 pounds per square inch absolute pressure. This pressure will change along with the weather and the altitude at which the measurement is taken.
So how do we get to the pressure that is displayed on a pressure gauge? When shown open to room air, my pressure gauge reads zero psi. Well, that is zero psi gauge, this already has the atmosphere showing. It is not showing the Absolute pressure, it is showing the pressure relative to atmospheric conditions. This is going back to the fact that gauge pressure is the summation of absolute pressure and atmospheric conditions, for sea level on earth that is 14.7 psia. So how do we increase this and get the gauge to read higher levels?
We compress the air the gauge is measuring, whether it is using a screw compressor, dual-stage piston compressor, single-cylinder, or any other type of compressor, it is compressing the ambient, atmospheric air. Some materials do not like being compressed. Air, however, reacts well to being compressed and turns into a form of stored energy that gets used throughout industrial facilities. By compressing the air, we effectively take the air from atmospheric conditions and squeeze it down into a storage tank or piping where it is stored until it is used. Because the air is being compressed you can fit larger volumes (cubic feet or cubic meters) into a smaller area. This is the stored energy, that air that is compressed always wants to expand back out to ambient conditions. Perhaps this video below will help, it shows the GREAT Julius Sumner Miller explaining atmospheric pressure, lack of it, and when you add to it.
Lastly, no matter where you are, there is a scientific unit that can express atmospheric pressure, compressed air pressure, or even lack of pressure which are vacuum levels. To convert between these scientific units, some math calculations are needed. While the video below is no Julius Sumner Miller, it does a great job walking through many of the units we deal with daily here at EXAIR.
If you want to discuss pressures, atmospheric pressure, how fast the air expands from your engineered nozzle to atmospheric, why all the moisture in the air compresses with it, and how to keep it out of your process, contact an application engineer and we will be glad to walk through the applications and explanations with you.
Brian Farno
Application Engineer
BrianFarno@EXAIR.com
@EXAIR_BF
1 – Willy Wonka & the Chocolate Factory – Violet Blows Up Like a Blueberry Scene (7/10) | Movieclips, Movieclips, retrieved from https://youtu.be/8Yqw_f26SvM
2 – Lesson 10 – Atmospheric Pressure – Properties of Gases – Demonstrations in Physics, Julius Sumner Miller, Retrieved from https://www.youtube.com/watch?v=P3qcAZrNC18
3 – Pressure Units and Pressure Unit Conversion Explained, Chem Academy, retrieve from https://www.youtube.com/watch?v=2rNs0VMiHNw
## Air – What Is It?
Air… We all breathe it, we live in it, we even compress it to use it as a utility. What is it though? Well, read through the next to learn some valuable points that aren’t easy to see with your eyes, just like air molecules.
1. Air is mostly a gas.
• Comprised of roughly 78% Nitrogen and 21% Oxygen. Air also contains a lot of other gases in minute amounts. Those gases include carbon dioxide, neon, and hydrogen.
2. Air is more than just gas.
• While the vast majority is gas, air also holds lots of microscopic particulate.
• These range from pollen, soot, dust, salt, and debris.
• All of these items that are not Nitrogen or Oxygen contribute to pollution.
3. Not all the Carbon Dioxide in the air is bad.
• Carbon Dioxide as mentioned above is what humans and most animals exhale when they breathe. This gas is taken in by plants and vegetation to convert their off gas which is oxygen.
• Think back to elementary school now. Remember photosynthesis?
• If you don’t remember that, maybe you remember Billy Madison, “Chlorophyll, more like Bore-a-fil.”
• Carbon dioxide is however one of the leading causes of global warming.
4. Air holds water.
• That’s right, high quality H2O gets suspended within the air molecules causing humidity. This humidity ultimately reaches a point where the air can simply not hold anymore and it starts to rain. The lack of humidity in the air leads to static, while lots of moisture in the air when it gets compressed causes moisture in compressed air systems.
5. Air changes relative to altitude.
• Air all pushes down on the Earth’s surface. This is known as atmospheric pressure.
• The closer you are to sea level the higher the level of pressure because the air molecules are more densely placed.
• The higher you are from sea level the lower the density of air molecules. This causes the pressure to be less. This is also why people say the air is getting a little thin.
Hopefully this helps to better explain what air is and give some insight into the gas that is being compressed by an air compressor and then turned into a working utility within a production environment. If you would like to discuss how any of these items effects the compressed air quality within a facility please reach out to any Application Engineer at EXAIR.
Brian Farno
Application Engineer
BrianFarno@EXAIR.com
@EXAIR_BF
## Video Blog: How To Calculate Air Consumption At A Pressure Other Than Published Values
The below video shows how to calculate the air consumption when operating at any pressure.
If you want to discuss efficient compressed air use or any of EXAIR’s engineered compressed air products, give us a call or email. We would enjoy hearing from you!
Steve Harrison
Application Engineer
Send me an email
Find us on the Web | 1,593 | 7,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-40 | latest | en | 0.946056 |
http://www.xpmath.com/forums/showthread.php?s=fdc380e6137922716b714a57ce1cd3bc&p=22718 | 1,585,801,461,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506580.20/warc/CC-MAIN-20200402014600-20200402044600-00504.warc.gz | 247,819,113 | 10,194 | Is the answer to this algebra problem 8b or is it 1/8b? - XP Math - Forums
XP Math - Forums Is the answer to this algebra problem 8b or is it 1/8b?
10-27-2007 #1 avon2me4you Guest Posts: n/a Is the answer to this algebra problem 8b or is it 1/8b? Simplify the expression by using the properties of rational exponents. Assume all variables represent positive real numbers. Express the final answer using positive exponents only. 8b^−1/3 divided by b^2/3 Is the answer 8b or is it 1/8b?
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Contact Us - XP Math - Forums - Archive - Privacy Statement - Top | 277 | 1,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-16 | latest | en | 0.775603 |
https://yourwiseadvices.com/what-is-the-real-power/ | 1,709,275,618,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475203.41/warc/CC-MAIN-20240301062009-20240301092009-00639.warc.gz | 1,070,291,693 | 14,322 | # What is the real power?
## What is the real power?
Real power is expressed in watts and as so represents the actual energy converted from electrical energy to useful work. The calculation for real power is the product of the apparent power and the cosine of the angle between the voltage and current waveforms.
## What are the source of real power?
Real-world sources of electrical energy, such as batteries and generators, can be modeled for analysis purposes as a combination of an ideal voltage source and additional combinations of impedance elements….Ideal voltage sources.
Controlled Voltage Source Controlled Current Source Battery of cells Single cell
Is reactive power real?
Key Differences between Active and Reactive Power The active power is the product of the voltage, current and the cosine of the angle between them. Whereas, the reactive power is the product of voltage and current and the sine of the angle between them. The active power is the real power, and it is measured in watts.
What does true power look like?
True power is symbolized by the letter P and is measured in the unit of Watts (W). Power merely absorbed and returned in load due to its reactive properties is referred to as reactive power. Reactive power is symbolized by the letter Q and is measured in the unit of Volt-Amps-Reactive (VAR).
### What is the apparent power?
Apparent Power is the Total Power Flowing The total power flowing is known as the “apparent power” and is measured as the product of the voltage and current (V * I).. For example, if 208 volts and 5 amps are measured – the apparent power is 1040VA (VA means volt-amps – the measurement unit of apparent power).
### What is a apparent power?
How do you find complex power?
In power system, to calculate complex power, formula S=VI* is used instead of S=V*I.
Is Apparent power real?
Apparent power is the product of the total current and voltage in a circuit. Many loads include reactive components. Apparent power, in VA, is made up of the true power consumed by resistive loads and the reactive power flowing through capacitive and inductive loads.
#### Where does reactive power come from?
Reactive power is either generated or absorbed by electric generators (or, in some cases, devices known as “capacitors”) to maintain a constant voltage level, commonly referred to as providing “voltage support.” Generators providing voltage support often suffer heating losses that result in a reduced ability to …
#### Is apparent power real?
What is the AC power?
What is AC power? Alternating current (AC) power is the standard electricity that comes out of power outlets and is defined as a flow of charge that exhibits a periodic change in direction. This is known as the sinusoidal AC wave, and this wave is caused when alternators at power plants create AC power.
What is meant by real power?
power (Noun) physical force or strength.
• power (Noun) control,particularly legal or political (jurisdiction) 2005,Columbia Law Review,April Etymology: From power.
• power (Noun) electricity or a supply of electricity.
• power (Noun) A measure of the rate of doing work or transferring energy.
M = Mass
• L = Length
• T = Time
• ## What is the real power formula?
Power = Force ∗ Velocity {\\displaystyle {\ext {Power}}= {\ext {Force}}* {\ext {Velocity}}} Power = 2000 Pounds ∗ 73.33 feet sec {\\displaystyle {\ext {Power}}=2000 {\ext {Pounds}}*73.33 {\\frac {\ext {feet}} {\ext {sec}}}} Power = 146, 660 foot-pounds sec {\\displaystyle {\ext {Power}}=146,660 {\\frac {\ext {foot-pounds}} {\ext {sec}}}}
What is real power and apparent power?
The VA rating is the apparent power that a UPS is capable of producing, while the watt rating is the real power (or true power) it is capable of producing, as opposed to reactive power. Reactive power arises due to the effects of capacitance and inductance of components in the load to be powered by the AC circuit. | 877 | 3,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | longest | en | 0.948427 |
http://simple.wikipedia.org/wiki/Floating_point | 1,416,957,332,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931004246.54/warc/CC-MAIN-20141125155644-00077-ip-10-235-23-156.ec2.internal.warc.gz | 264,384,575 | 10,486 | # Floating point
Real numbers in binary have to be stored in a special way in a computer. There is no decimal point in the binary system so the computer has a method of understanding decimals. This is called floating-point representation.
The decimal point in a real number is called a floating point because it can be placed anywhere - it is not fixed. Because of this, a computer will divide a number into two parts. These are called the mantissa and the exponent.
## Mantissa
The mantissa is found by taking the real number and removing the decimal point, for example:
1101 . 0111 would become 1101 0111
## Exponent
The exponent is the number of spaces the decimal point has moved. In the example above, the decimal point moved 4 places to the left, so the exponent is 0000 0100 (this is binary for 4).
If the decimal point moves to right, the exponent is negative. For example,
0000 . 0111 (mantissa - 0000 0111)
here, the exponent is -1. The binary number for this is 1111 1111 (see Negative binary numbers)
## Result
The result is found by putting the Mantissa and Exponent together. The results for the examples above are:
Decimal Mantissa Exponent Result
1101 . 0111 1101 0111 0000 0100 1101 0111 0000 0100
0000 . 0111 0000 0111 1111 1111 0000 0111 1111 1111 | 336 | 1,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2014-49 | longest | en | 0.827625 |
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# Why Is There More Resistance In A Longer Piece Of Wire?
Extracts from this document...
Introduction
Why Is There More Resistance In A Longer Piece Of Wire?
Aim:
To find out if there is more resistance in a longer piece of wire when compared with a shorter piece of wire.
Prediction Factors:
Wire length: If the length of the wire is increased then the resistance will also increase as the electrons will have a longer distance to travel and so more collisions between atoms will occur. Due to this the length increase should be proportional to the resistance increase.
Wire width: If the wires width is increased the resistance will decrease. This is because of the increase in the space for the electrons to travel through. Due to this increased space between the atoms there should be less collisions. Temperature: If the wire is heated up the atoms in the wire will start to vibrate because of their increase in energy. This causes more collisions between the electrons and the atoms because the atoms are moving into the path of the electrons. This increase in collisions means that there will be an increase in resistance. Material: The type of material will affect the amount of free electrons that are able to flow through the wire. If the material has a high number of atoms there will be high number of electrons causing a lower resistance.
Middle
Resistance (ohms)
5.00
0.06
0.15
0.40
5.00
0.08
0.20
0.40
5.00
0.10
0.25
0.40
5.00
0.12
0.30
0.40
5.00
0.14
0.35
0.40
Wire Length (cm) Voltage (volts) Current (amps) Resistance (ohms) 10.00 0.11 0.15 0.73 10.00 0.14 0.20 0.70 10.00 0.18 0.25 0.72 10.00 0.22 0.30 0.73 10.00 0.26 0.35 0.74
Wire Length (cm) Voltage (volts) Current (amps) Resistance (ohms) 15.00 0.16 0.15 1.06 15.00 0.21 0.20 1.05 15.00 0.26 0.25 1.04 15.00 0.32 0.30 1.06 15.00 0.37 0.35 1.05
Wire Length (cm) Voltage (volts) Current (amps) Resistance (ohms) 20.00 0.21 0.15 1.40 20.00 0.29 0.20 1.45 20.00 0.36 0.25 1.44 20.00 0.43 0.30 1.43 20.00 0.49 0.35 1.40
Conclusion
Using higher currents so that the temperature does increase could extend the investigation. This could be undertaken to prove that, as the temperature does get higher the graph for the results begins to curve at the end. I included my calculations for the resistivity of the wire as it shows that my results were accurate. I measured the gradient of my straight-line graph as it gives me the best possible average of results omitting any slight anomalous results. This shows that my conclusion about resistance being proportional to the length is very reliable.
If the same experiment were carried out the same factors would apply. If you wanted to change a different variable I would change wire width. For this I predict that as the wire width get small then the more closely packed the electrons are. This would create more collisions so there would be more resistance. If the wire width is increased the resistance will decrease. This is because of the increase in the space for the electrons to travel through. Due to this increased space between the electrons there should be less collisions.
By Luke Howard
This student written piece of work is one of many that can be found in our GCSE Electricity and Magnetism section.
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# Related GCSE Electricity and Magnetism essays
1. ## Determining the resistivity of a piece of wire
As can be seen the nuclei are represented as large orange/yellow 'dots'. The coulombs of charge are represented by the small red dots and the small orange/yellow dots are the fixed electrons. The nuclei resist the flow of coulombs and thus produce electrical resistance.
2. ## An Investigation to Calculate the Resistivity of a Piece of Wire.
Cross-Sectional Area - If the cross-sectional area is increased, in theory the resistance should decrease this is because there is more space outwards for the electrons to flow, as wire is rated or measured, first I will show the relationship between the radius of rounded wire and the cross-sectional area.
1. ## Ohms Law.
I have chosen these lengths because they are easily measured by the meter ruler and give a good range of results. As my preliminary results start to show a pattern in the readings (Resistance is directly proportional to length) to expand on my experiment and to see if this pattern continues, I am going to try the above lengths.
2. ## To investigate how the length (mm) and the cross-sectional (mm2) area of a wire ...
The other variable in this experiment is going to be the cross-sectional area, or thickness. This is the opposite of length because a thick wire offers more room for an electric current to pass through than a thin wire does.
1. ## Factors that affect the resistance of a piece of wire.
to lot a graph of resistance against length with a range of fie readings. wwbf bfw esbfbfs aybf bfba nbf kcbf bfuk: wwcg cgw escgcgs aycg cgba ncg kccg cguk: Results As shown below there is only a range of five readings, which start to show a pattern, that the best fit line is directly proportional.
2. ## Restoring a Volksempfnger VE 301 GW (Nazi people's radio) This is a piece of ...
safety signs on it. The row of 10 caps is glued together and fits exactly into the old box. Connect the new caps to the old terminals. Stuff the empty room with synthetic padding wool to avoid rattle. Don't use the tar again.
1. ## Resistance of a Wire Investigation
Therefore, any inaccuracies in measuring the distances, i.e. if a distance was slightly different when doing the actual experiment from the distance at which I earlier measured the light intensity, an error would ensue. The second major inaccuracy was in measuring the volume of oxygen given off.
2. ## An in Investigation into the Resistance of a Wire.
Cross Sectional Area (m�) � 10-7 Resistance (?) Resistivity (?m) E22 0.30 2.5 0.611 5.1� 10-7 E26 0.30 1.7 0.934 5.3� 10-7 E28 0.30 1.1 1.391 5.1� 10-7 E30 0.30 0.8 1.941 5.2� 10-7 E32 0.30 0.6 2.668 5.3� 10-7 E34 0.30 0.4 3.438 4.6� 10-7 E36 0.30 0.3 5.292
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• Ideas and feedback to | 1,801 | 6,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2018-09 | longest | en | 0.810317 |
https://scholar.archive.org/work/c2nypmqayfa3vnxicslyg7bzpu | 1,657,150,627,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104678225.97/warc/CC-MAIN-20220706212428-20220707002428-00584.warc.gz | 553,132,738 | 5,181 | ### Bounded Degree Group Steiner Tree Problems [article]
Guy Kortsarz, Zeev Nutov
2019 arXiv pre-print
We study two problems that seek a subtree T of a graph G=(V,E) such that T satisfies a certain property and has minimal maximum degree. - In the Min-Degree Group Steiner Tree problem we are given a collection S of groups (subsets of V) and T should contain a node from every group. - In the Min-Degree Steiner k-Tree problem we are given a set R of terminals and an integer k, and T should contain at least k terminals. We show that if the former problem admits approximation ratio ρ then the later
more » ... roblem admits approximation ratio ρ· O(log k). For bounded treewidth graphs, we obtain approximation ratio O(log^3 n) for Min-Degree Group Steiner Tree. In the more general Bounded Degree Group Steiner Tree problem we are also given edge costs and degree bounds {b(v):v ∈ V}, and T should obey the degree constraints deg_T(v) ≤ b(v) for all v ∈ V. We give a bicriteria (O(log N log | S|),O(log^2 n))-approximation algorithm for this problem on tree inputs, where N is the size of the largest group, generalizing the approximation of Garg, Konjevod, and Ravi for the case without degree bounds. | 301 | 1,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-27 | latest | en | 0.88102 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=275465 | 1,369,033,701,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698493317/warc/CC-MAIN-20130516100133-00097-ip-10-60-113-184.ec2.internal.warc.gz | 315,434,878 | 879 | ```Question 388997
The sum of two numbers is 27. One half of the first number plus one third of the second number is 11. Find the numbers.
...
let the numbers be x & y
...
x+y =27..............1
x/2 + y/3 = 11
LCD = 6
multiply by 6
3x+2y=66.............2
multiply (1) by -3
-3x-3y=-81 | 103 | 284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.767236 |
http://stackoverflow.com/questions/2432669/test-if-a-number-is-fibonacci/2821850 | 1,419,307,019,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802778068.138/warc/CC-MAIN-20141217075258-00121-ip-10-231-17-201.ec2.internal.warc.gz | 262,074,710 | 28,803 | # Test if a number is fibonacci
I know how to make the list of the Fibonacci numbers, but i don't know how can i test if a given number belongs to the fibonacci list - one way that comes in mind is generate the list of fib. numbers up to that number and see if it belongs to the array, but there's got to be another, simpler and faster method.
Any ideas ?
-
Looks like homework to me, so I added the homework tag. – Gerco Dries May 12 '10 at 18:47
See stackoverflow.com/questions/1525521/… for a related question. – mtrw May 12 '10 at 18:47
Please allow the OP to add the homework tag on his own (feel free to ask for clarification). Lots of things look like homework that aren't. – danben May 12 '10 at 19:03
Please don't add tags just because it "looks like it would fit". It "looks to me" like the OP wants to do this in brainf*ck, should I add that tag? – IVlad May 12 '10 at 19:09
duplicate of stackoverflow.com/questions/2432669 – sdcvvc May 12 '10 at 19:10
A very nice test is that N is a Fibonacci number if and only if `5 N^2 + 4` or `5N^2 – 4` is a square number. For ideas on how to efficiently test that a number is square refer to the SO discussion.
Hope this helps
-
+1 because saying "or" is more clear than saying "one of" + "and" First 4 times I read the other answers I thought they were saying different things because I didn't see the "one of" part – Davy8 Mar 15 '10 at 23:13
I am skeptical of this solution, as it involves squaring a Fibonacci number. Fibonacci numbers grow extremely quickly, and most are very large. Doesn't squaring them become computationally expensive? – abelenky May 12 '10 at 21:07
Well yeah, beyond 2^63 (something like Fib(300)) you're going to have to use some arbitrary precision arithmetic which will be expensive. As the numbers grow, you must resort to approximate methods, either using Binet's formula or something else. I would be surprised to learn any efficient exact method that works for large numbers! – Il-Bhima May 12 '10 at 21:44
Hm... If exactly one of the propositions A and B need to hold (but not both!), you cannot write "A or B", for this compound statement is true if A is true and B is false, if A is false and B is true, and if both A and B are true. Then you need to write explicitly "exactly one of", or use the logical "xor" operator rather than "or". – Andreas Rejbrand May 12 '10 at 22:24
But it appears to be the case that "or" is indeed the correct operator. To see this, set N = 1. Then N is a Fibonacci number, and both 5*N^2 + 4 and 5*N^2 - 4 are perfect squares. If we had a xor operator, then "A xor B" would be false, even though 1 is Fibonacci, and we have a contradiction. (Here I assume that the theorem is correct with "or" or "xor".) – Andreas Rejbrand May 12 '10 at 22:42
A positive integer ω is a Fibonacci number if and only if one of 5ω2 + 4 and 5ω2 - 4 is a perfect square.
See The Faboulous Fibonacci Numbers for more.
-
+1 for the book recommendation. – Thomas Matthews Mar 12 '10 at 19:07
While several people point out the perfect-square solution, it involves squaring a Fibonacci number, frequently resulting in a massive product.
There are less than 80 Fibonacci numbers that can even be held in a standard 64-bit integer.
Here is my solution, which operates entirely smaller than the number to be tested.
(written in C#, using basic types like `double` and `long`. But the algorithm should work fine for bigger types.)
``````static bool IsFib(long T, out long idx)
{
double root5 = Math.Sqrt(5);
double phi = (1 + root5) / 2;
idx = (long)Math.Floor( Math.Log(T*root5) / Math.Log(phi) + 0.5 );
long u = (long)Math.Floor( Math.Pow(phi, idx)/root5 + 0.5);
return (u == T);
}
``````
More than 4 years after I wrote this answer, a commenter asked about the second parameter, passed by `out`.
Parameter #2 is the "Index" into the Fibonacci sequence.
If the value to be tested, `T` is a Fibonacci number, then `idx` will be the 1-based index of that number in the Fibonacci sequence. (with one notable exception)
The Fibonacci sequence is `1 1 2 3 5 8 13`, etc.
3 is the 4th number in the sequence: `IsFib(3, out idx);` will return `true` and value `4`.
8 is the 6th number in the sequence: `IsFib(8, out idx);` will return `true` and value `6`.
13 is the 7th number; `IsFib(13, out idx);` will return `true` and value `7`.
The one exception is `IsFib(1, out idx);`, which will return `2`, even though the value 1 appears at both index 1 and 2.
If `IsFib` is passed a non-Fibonacci number, it will return `false`, and the value of `idx` will be the index of the biggest Fibonacci number less than `T`.
16 is not a Fibonacci value.
`IsFib(16, out idx);` will return `false` and value `7`.
You can use Binet's Formula to convert index 7 into Fibonacci value 13, which is the largest number less than 16.
-
+1 for taking machine limitations into account – dss539 May 12 '10 at 21:29
Concise implementation. I actually used this function in the contest: hackerrank.com/contests/codesprint5/challenges/is-fibo :) – Michal Stefanow Jan 18 at 1:29
Thanks. It looks like magic. @Michal I have also used this function in hackerrank contest. – kushdilip Mar 4 at 7:08
Very nice - thanks! I used it to get the closest Fibonacci number :) But in real life situation I think there is no need to compute these numbers, but store them in database (just like You suggest in Your other post) – Prokurors May 8 at 22:34
just one question, what exactly is the second argument and why are you passing it by reference ? – Mhd.Tahawi Nov 16 at 10:22
``````#!/bin/bash
victim="144"
curl http://aux.planetmath.org/files/objects/7680/fib.txt | sed 's/^[0-9]*//;s/[ \t]//g' | grep "^\$victim\$" >/dev/null 2>/dev/null
if [[ \$? -eq 0 ]] ; then
echo "\$victim is a fibonacci number"
else
echo "\$victim aint"
fi
``````
-
Outsourcing. Love it! – Michael Cole Aug 21 at 9:50
If your numbers are of bounded size, than simply putting all fibonacci numbers below the upper bound into a hashtable and testing containment will do the trick. There are very few fibonacci numbers (for example, only 38 below 5mln), since they grow exponentially.
If your numbers are not of bounded size, then the suggested trick with square testing will almost surely be slower than generating the fibonacci sequence until the number is found or exceeded.
-
Positive integer ω is a Fibonacci number
If and only if one of2 + 4 and 5ω2 - 4 is a perfect square
from The (Fabulous) FIBONACCI Numbers by Alfred Posamentier and Ingmar Lehmann
``````bool isFibonacci(int w)
{
double X1 = 5 * Math.Pow(w, 2) + 4;
double X2 = 5 * Math.Pow(w, 2) - 4;
long X1_sqrt = (long)Math.Sqrt(X1);
long X2_sqrt = (long)Math.Sqrt(X2);
return (X1_sqrt*X1_sqrt == X1) || (X2_sqrt*X2_sqrt == X2) ;
}
``````
I copied it from this source
Snippet that prints Fibonacci numbers between `1k` and `10k`.
``````for (int i = 1000; i < 10000; i++)
{
if (isFibonacci(i))
Console.Write(" "+i);
}
``````
OMG There are only FOUR!!!
With other method
``````from math import *
phi = 1.61803399
sqrt5 = sqrt(5)
def F(n):
return int((phi**n - (1-phi)**n) /sqrt5)
def isFibonacci(z):
return F(int(floor(log(sqrt5*z,phi)+0.5))) == z
print [i for i in range(1000,10000) if isFibonacci(i)]
``````
-
No need for the "? true : false" part: the expression before that is already a boolean value. – lhf Mar 12 '10 at 12:52
+1 :) See I changed it. Thanks! – Pratik Deoghare Mar 12 '10 at 12:58
I wrote second method in python because I didn't know C# Math.Log works for other bases as well. Do you guys want me to write it too :P?? lol – Pratik Deoghare Mar 12 '10 at 13:05
Towards a solution, take a look at Binet's Formula.
(Look for "Closed-Form Expression" under Fibonacci Number on Wikipedia)
It says that the sequence of Fibonacci Number's is created by a simple closed formula:
I believe if you solve for `n`, and test if `n` is an integer, you'll have your answer.
Edit As @psmears points out, the same Wikipedia article also has a section on detecting Fibonacci numbers. Wikipedia is an excellent source.
-
See the section "Recognizing Fibonacci numbers" on the wikipedia article about the Fibonacci numbers.
-
Hey, are you P Smears who was at Lincoln? – Steve Jessop May 12 '10 at 19:37
Since Fibonacci numbers grow exponentially, the method you suggest is pretty fast. Another is this.
-
+1 for a different method. – Pratik Deoghare Mar 12 '10 at 12:56
I really like the closed interval solution, should be much easier than checking for squares! – Matthieu M. May 13 '10 at 14:08
From Wikipedia: http://en.wikipedia.org/wiki/Fibonacci_number
A positive integer z is a Fibonacci number if and only if one of 5z^2 + 4 or 5z^2 − 4 is a perfect square.
-
Weird. After 15 years of math, I did not know this. – Phillip Schmidt May 31 '12 at 22:26
Based on earlier answers by me and psmears, I've written this C# code.
It goes through the steps slowly, and it can clearly be reduced and optimized:
``````// Input: T: number to test.
// Output: idx: index of the number in the Fibonacci sequence.
// eg: idx for 8 is 6. (0, 1, 1, 2, 3, 5, 8)
// Return value: True if Fibonacci, False otherwise.
static bool IsFib(long T, out int idx)
{
double root5 = Math.Sqrt(5);
double PSI = (1 + root5) / 2;
// For reference, IsFib(72723460248141) should show it is the 68th Fibonacci number
double a;
a = T*root5;
a = Math.Log(a) / Math.Log(PSI);
a += 0.5;
a = Math.Floor(a);
idx = (Int32)a;
long u = (long)Math.Floor(Math.Pow(PSI, a)/root5 + 0.5);
if (u == T)
{
return true;
}
else
{
idx = 0;
return false;
}
}
``````
Testing reveals this works for the first 69 Fibonacci numbers, but breaks down for the 70th.
``````F(69) = 117,669,030,460,994 - Works
F(70) = 190,392,490,709,135 - Fails
``````
In all, unless you're using a BigInt library of some kind, it is probably better to have a simple lookup table of the Fibonacci Numbers and check that, rather than run an algorithm.
A list of the first 300 Numbers is readily available online.
But this code does outline a workable algorithm, provided you have enough precision, and don't overflow your number representation system.
-
The problem with phi is that it's not exactly usable using floating point numbers, and so you have to approximate. – Rubys May 12 '10 at 19:46
The general expression for a Fibonacci number is F(n) = [ [(1+sqrt(5))/2] sup n+1 - [(1-sqrt(5))/2] sup n+1 ]/ sqrt(5) ..... (*) The second exponential goes to zero for large n and carrying out the numerical operations we get F(n) = [ (1.618) sup n+1 ] / 2.236
If K is the number to be tested log(k*2.2336)/log(1.618) should be an integer!
Example for K equal to 13 my calculator gives the answer 7.00246 For K equal 14 the answer is 7.1564.
You can increase the confidence in the result by taking the closest integer to the answer and substitute in (*) to confirm that the result is K
-
Re: Ahmad's code - a simpler approach with no recursion or pointers, fairly naive, but requires next to no computational power for anything short of really titanic numbers (roughly 2N additions to verify the Nth fib number, which on a modern machine will take milliseconds at worst)
// returns pos if it finds anything, 0 if it doesn't (C/C++ treats any value !=0 as true, so same end result)
``````int isFib (long n)
{
int pos = 2;
long last = 1;
long current = 1;
long temp;
while (current < n)
{
temp = last;
last = current;
current = current + temp;
pos++;
}
if (current == n)
return pos;
else
return 0;
}
``````
-
pretty sure this is the most efficient way to do this. – Phillip Schmidt May 31 '12 at 22:24
` def is_fibonacci?(i) a,b=0,1 until b >= i a,b=b,a+b return true if b == i end end` – Stephen Nguyen Feb 22 '13 at 23:50
How big are the numbers you're dealing with?
Could a lookup table work for you? (a precomputed list of numbers you can search in)
There's also a closed-form expression that I guess you could invert to get at the answer analytically (though I'm no mathematician, so I can't promise this suggestion makes sense)
-
I'm dealing with arbitrary numbers. Even an approximation will be useful, if it runs very quickly. – blueberryfields May 12 '10 at 18:49
I think psmears has the solution: stackoverflow.com/questions/2821778/… – Assaf Lavie May 12 '10 at 18:55
I ran some benchmarks on the methods presented here along with simple addition, pre-computing an array, and memoizing the results in a hash. For Perl, at least, the squaring method is a little bit faster than the logarithmic method, perhaps 20% faster. As abelenky points out, it's a tradeoff between whether you've got the room for squaring bit numbers.
Certainly, the very fastest way is to hash all the Fibonacci numbers in your domain space. Along the lines of another point that abelenky makes, there are only 94 of these suckers that are less than 2^64.
You should just pre-compute them, and put them in a Perl hash, Python dictionary, or whatever.
The properties of Fibonacci numbers are very interesting, but using them to determine whether some integer in a computer program is one is kind of like writing a subroutine to compute pi every time the program starts.
-
``````int isfib(int n /* number */, int &pos /* position */)
{
if (n == 1)
{
pos=2; // 1 1
return 1;
}
else if (n == 2)
{
pos=3; // 1 1 2
return 1;
}
else
{
int m = n /2;
int p, q, x, y;
int t1=0, t2 =0;
for (int i = m; i < n; i++)
{
p = i;
q = n -p; // p + q = n
t1 = isfib(p, x);
if (t1) t2 = isfib(q, y);
if (t1 && t2 && x == y +1)
{
pos = x+1;
return 1; //true
}
}
pos = -1;
return 0; //false
}
}
``````
-
good logic, but almost totally unreadable. gotta work on the variable naming – Phillip Schmidt May 31 '12 at 22:27
This is my solution I'm not sure if it benchmarks. I hope this helps!
``````def is_fibonacci?(i)
a,b=0,1
until b >= i
a,b=b,a+b
return true if b == i
end
end
``````
what a,b=b,a+b is doing
`````` 0, 1 = 1, 0 +1
1, 1 = 1, 1 + 1
1, 2 = 2, 1 + 2
2, 3 = 3, 2 + 3
fib1 = fib2
fib2 = fib1 + fib2
``````
-
A Scala version-
``````def isFib(n: Int): Boolean = {
def checkFib(f1: Int = 1, f2: Int = 1): Boolean = {
if(n == f1 || n == f2) true
else if(n < f2) false
else checkFib(f2, f1+f2)
}
checkFib()
}
``````
-
``````#include <stdio.h>
#include <math.h>
int main()
{
int number_entered, x, y;
scanf("%d", &number_entered);
x = y = 5 * number_entered^2 + 4; /*Test if 5N^2 + 4 is a square number.*/
x = sqrt(x);
x = x^2;
if (x == y)
{
printf("That number is in the Fibonacci sequence.\n");
}
x = y = 5 * number_entered^2 - 4; /*Test if 5N^2 - 4 is a square number.*/
x = sqrt(x);
x = x^2;
if (x == y)
{
printf("That number is in the Fibonacci sequence.\n");
}
else
{
printf("That number isn't in the Fibonacci sequence.\n");
}
return 0;
}
``````
Will this work?
-
No. In C, `^` is the bitwise XOR operator. You need `x * x` or `pow(x,2)` to square a number. There are also problems in the logic of the program. – QuasarDonkey Oct 10 '12 at 20:23 | 4,411 | 14,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | latest | en | 0.931304 |
http://stackoverflow.com/questions/33923/what-is-tail-recursion/37010 | 1,462,161,757,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860122420.60/warc/CC-MAIN-20160428161522-00199-ip-10-239-7-51.ec2.internal.warc.gz | 259,245,463 | 32,254 | # What is tail recursion?
Whilst starting to learn lisp, I've come across the term tail-recursive. What does it mean?
-
For curious: both while and whilst have been in the language for a very long time. While was in use in Old English; whilst is a Middle English development of while. As conjunctions they are interchangeable in meaning, but whilst has not survived in standard American English. – Filip Bartuzi Oct 18 '14 at 23:02
Maybe it is late, but this is a pretty good article about tail recursive:programmerinterview.com/index.php/recursion/tail-recursion – Sam003 Aug 8 '15 at 18:14
First, you need to understand recursion. A good comment explaining it is here: stackoverflow.com/questions/33923/… – joe_young Jan 16 at 21:08
@joe_young, that. :D – Sajib Acharya Apr 19 at 13:53
Tail recursion is well-described in previous answers, but I think an example in action would help to illustrate the concept.
Consider a simple function that adds the first N integers. (e.g. `sum(5) = 1 + 2 + 3 + 4 + 5 = 15`).
Here is a simple Python implementation that uses recursion:
``````def recsum(x):
if x == 1:
return x
else:
return x + recsum(x - 1)
``````
If you called `recsum(5)`, this is what the Python interpreter would evaluate.
``````recsum(5)
5 + recsum(4)
5 + (4 + recsum(3))
5 + (4 + (3 + recsum(2)))
5 + (4 + (3 + (2 + recsum(1))))
5 + (4 + (3 + (2 + 1)))
15
``````
Note how every recursive call has to complete before the Python interpreter begins to actually do the work of calculating the sum.
Here's a tail-recursive version of the same function:
``````def tailrecsum(x, running_total=0):
if x == 0:
return running_total
else:
return tailrecsum(x - 1, running_total + x)
``````
Here's the sequence of events that would occur if you called `tailrecsum(5)`, (which would effectively be `tailrecsum(5, 0)`, because of the default second argument).
``````tailrecsum(5, 0)
tailrecsum(4, 5)
tailrecsum(3, 9)
tailrecsum(2, 12)
tailrecsum(1, 14)
tailrecsum(0, 15)
15
``````
In the tail-recursive case, with each evaluation of the recursive call, the `running_total` is updated.
Note: As mentioned in the comments, Python doesn't have built-in support for optimizing away tail calls, so there's no advantage to doing this in Python. However, you can use a decorator to achieve the optimization.
-
Python is kind of an odd choice here, since it does not have AFAIK tail-recursion elimination. – Chris Conway Sep 21 '08 at 14:43
Chris Conway is correct. Tail calls are not optimized in Python, unfortunately. Guido claims having the stack available for debugging is better than TCO. – McPherrinM Dec 5 '09 at 3:25
You'll find Guido's opinion here. – new123456 May 14 '11 at 14:27
@Paco, what was the link? Was it some kind of a joke? It is 404, do you have a mirror? :-) – tillda Aug 29 '12 at 23:41
@tillda: Yes a joke. Here is a mirror: cs.cmu.edu/~wklieber/python-tail-recursion.jpg – Paco Sep 18 '12 at 14:34
In traditional recursion, the typical model is that you perform your recursive calls first, and then you take the return value of the recursive call and calculate the result. In this manner, you don't get the result of your calculation until you have returned from every recursive call.
In tail recursion, you perform your calculations first, and then you execute the recursive call, passing the results of your current step to the next recursive step. This results in the last statement being in the form of "(return (recursive-function params))" (I think that's the syntax for Lisp). Basically, the return value of any given recursive step is the same as the return value of the next recursive call.
The consequence of this is that once you are ready to perform your next recursive step, you don't need the current stack frame any more. This allows for some optimization. In fact, with an appropriately written compiler, you should never have a stack overflow snicker with a tail recursive call. Simply reuse the current stack frame for the next recursive step. I'm pretty sure Lisp does this.
-
"I'm pretty sure Lisp does this" -- Scheme does, but Common Lisp doesn't always. – Aaron Jan 2 '09 at 23:51
@Daniel "Basically, the return value of any given recursive step is the same as the return value of the next recursive call."- I fail to see this argument holding true for the code snippet posted by Lorin Hochstein. Can you please elaborate? – Geek Mar 20 '13 at 19:16
@Geek This is a really late response, but that is actually true in Lorin Hochstein's example. The calculation for each step is done before the recursive call, rather than after it. As a result, each stop just returns the value directly from the previous step. The last recursive call finishes the computation and then returns the final result unmodified all the way back down the call stack. – reirab Apr 23 '14 at 22:58
Scala does but you need the @tailrec specified to enforce it. – SilentDirge Dec 26 '14 at 21:02
"In this manner, you don't get the result of your calculation until you have returned from every recursive call." -- maybe I misunderstood this, but this isn't particularly true for lazy languages where the traditional recursion is the only way to actually get a result without calling all recursions (e.g. folding over an infinite list of Bools with &&). – hasufell Jun 30 '15 at 21:12
An important point is that tail recursion is essentially equivalent to looping. It's not just a matter of compiler optimization, but a fundamental fact about expressiveness. This goes both ways: you can take any loop of the form
``````while(E) { S }; return Q
``````
where `E` and `Q` are expressions and `S` is a sequence of statements, and turn it into a tail recursive function
``````f() = if E then { S; return f() } else { return Q }
``````
Of course, `E`, `S`, and `Q` have to be defined to compute some interesting value over some variables. For example, the looping function
``````sum(n) {
int i = 1, k = 0;
while( i <= n ) {
k += i;
++i;
}
return k;
}
``````
is equivalent to the tail-recursive function(s)
``````sum_aux(n,i,k) {
if( i <= n ) {
return sum_aux(n,i+1,k+i);
} else {
return k;
}
}
sum(n) {
return sum_aux(n,1,0);
}
``````
(This "wrapping" of the tail-recursive function with a function with fewer parameters is a common functional idiom.)
-
Great explanation, but why is your function called fibo? That's not Fibonacci; it's f(n) = Binomial(n+1,2) – RexE Dec 25 '11 at 7:29
@RexE Total brain fart. I got upvoted 20 times and nobody pointed that out! – Chris Conway Dec 26 '11 at 14:24
:) also make sure to rename the recursive calls. – RexE Dec 28 '11 at 20:45
In the answer by @LorinHochstein I understood, based on his explanation, that tail recursion to be when the recursive portion follows "Return", however in yours, the tail recursive is not. Are you sure your example is properly considered tail recursion? – CodyBugstein Mar 10 '13 at 22:44
@Imray The tail-recursive part is the "return sum_aux" statement inside sum_aux. – Chris Conway Mar 11 '13 at 4:51
This excerpt from the book Programming in Lua shows how to make a proper tail recursion (in Lua, but should apply to Lisp too) and why it's better.
A tail call [tail recursion] is a kind of goto dressed as a call. A tail call happens when a function calls another as its last action, so it has nothing else to do. For instance, in the following code, the call to `g` is a tail call:
``````function f (x)
return g(x)
end
``````
After `f` calls `g`, it has nothing else to do. In such situations, the program does not need to return to the calling function when the called function ends. Therefore, after the tail call, the program does not need to keep any information about the calling function in the stack. ...
Because a proper tail call uses no stack space, there is no limit on the number of "nested" tail calls that a program can make. For instance, we can call the following function with any number as argument; it will never overflow the stack:
``````function foo (n)
if n > 0 then return foo(n - 1) end
end
``````
... As I said earlier, a tail call is a kind of goto. As such, a quite useful application of proper tail calls in Lua is for programming state machines. Such applications can represent each state by a function; to change state is to go to (or to call) a specific function. As an example, let us consider a simple maze game. The maze has several rooms, each with up to four doors: north, south, east, and west. At each step, the user enters a movement direction. If there is a door in that direction, the user goes to the corresponding room; otherwise, the program prints a warning. The goal is to go from an initial room to a final room.
This game is a typical state machine, where the current room is the state. We can implement such maze with one function for each room. We use tail calls to move from one room to another. A small maze with four rooms could look like this:
``````function room1 ()
if move == "south" then return room3()
elseif move == "east" then return room2()
else print("invalid move")
return room1() -- stay in the same room
end
end
function room2 ()
if move == "south" then return room4()
elseif move == "west" then return room1()
else print("invalid move")
return room2()
end
end
function room3 ()
if move == "north" then return room1()
elseif move == "east" then return room4()
else print("invalid move")
return room3()
end
end
function room4 ()
print("congratulations!")
end
``````
So you see, when you make a recursive call like:
``````function x(n)
if n==0 then return 0
n= n-2
return x(n) + 1
end
``````
This is not tail recursive because you still have things to do (add 1) in that function after the recursive call is made. If you input a very high number it will probably cause a stack overflow.
-
This is a great answer because it explains the implications of tail calls upon stack size. – Andrew Swan Aug 22 '14 at 7:32
@AndrewSwan Indeed, although I believe that the original asker and the occasional reader who might stumble into this question might be better served with the accepted answer (since he might not know what the stack actually is.) By the way I use Jira, big fan. – Hoffmann Aug 22 '14 at 13:47
Instead of explaining it with words, here's an example. This is a Scheme version of the factorial function:
``````(define (factorial x)
(if (= x 0) 1
(* x (factorial (- x 1)))))
``````
Here is a version of factorial that is tail-recursive:
``````(define factorial
(letrec ((fact (lambda (x accum)
(if (= x 0) accum
(fact (- x 1) (* accum x))))))
(lambda (x)
(fact x 1))))
``````
You will notice in the first version that the recursive call to fact is fed into the multiplication expression, and therefore the state has to be saved on the stack when making the recursive call. In the tail-recursive version there is no other S-expression waiting for the value of the recursive call, and since there is no further work to do, the state doesn't have to be saved on the stack. As a rule, Scheme tail-recursive functions use constant stack space.
-
The jargon file has this to say about the definition of tail recursion:
tail recursion /n./
If you aren't sick of it already, see tail recursion.
-
I see what you did there. Cheeky example there. :D – Sajib Acharya Apr 19 at 13:50
Using regular recursion, each recursive call pushes another entry onto the call stack. When the recursion is completed, the app then has to pop each entry off all the way back down.
With tail recursion, the compiler is able to collapse the stack down to one entry, so you save stack space...A large recursive query can actually cause a stack overflow.
Basically Tail recursions are able to be optimized into iteration.
-
It means that rather than needing to push the instruction pointer on the stack, you can simply jump to the top of a recursive function and continue execution. This allows for functions to recurse indefinitely without overflowing the stack.
I wrote a blog post on the subject, which has graphical examples of what the stack frames look like.
-
Neat article, thanks! – vtortola Nov 12 '14 at 12:45
Tail recursion refers to the recursive call being last in the last logic instruction in the recursive algorithm.
Typically in recursion you have a base-case which is what stops the recursive calls and begins popping the call stack. To use a classic example, though more C-ish than Lisp, the factorial function illustrates tail recursion. The recursive call occurs after checking the base-case condition.
``````factorial(x, fac) {
if (x == 1)
return fac;
else
return factorial(x-1, x*fac);
}
``````
Note, the initial call to factorial must be factorial(n, 1) where n is the number for which the factorial is to be calculated.
-
In Java, here's a possible tail recursive implementation of the Fibonacci function:
``````public int tailRecursive(final int n) {
if (n <= 2)
return 1;
return tailRecursiveAux(n, 1, 1);
}
private int tailRecursiveAux(int n, int iter, int acc) {
if (iter == n)
return acc;
return tailRecursiveAux(n, ++iter, acc + iter);
}
``````
Contrast this with the standard recursive implementation:
``````public int recursive(final int n) {
if (n <= 2)
return 1;
return recursive(n - 1) + recursive(n - 2);
}
``````
-
This is returning wrong results for me, for input 8 I get 36, it has to be 21. Am I missing something? I'm using java and copy pasted it. – Alberto Zaccagni Nov 28 '11 at 21:21
This returns SUM(i) for i in [1, n]. Nothing to do with Fibbonacci. For a Fibbo, you need a tests which substracts `iter` to `acc` when `iter < (n-1)`. – Askolein Mar 15 '13 at 13:44
Here is a quick code snippet comparing two functions. The first is traditional recursion for finding the factorial of a given number. The second uses tail recursion.
Very simple and intuitive to understand.
Easy way to tell if a recursive function is tail recursive, is if it returns a concrete value in the base case. Meaning that it doesn't return 1 or true or anything like that. It will more then likely return some variant of one of the method paramters.
Another way is to tell is if the recursive call is free of any addition, arithmetic, modification, etc... Meaning its nothing but a pure recursive call.
``````public static int factorial(int mynumber) {
if (mynumber == 1) {
return 1;
} else {
return mynumber * factorial(--mynumber);
}
}
public static int tail_factorial(int mynumber, int sofar) {
if (mynumber == 1) {
return sofar;
} else {
return tail_factorial(--mynumber, sofar * mynumber);
}
}
``````
-
0! is 1. So "mynumber == 1" should be "mynumber == 0". – polerto Mar 5 '14 at 23:35
Here is a Common Lisp example that does factorials using tail-recursion. Due to the stack-less nature, one could perform insanely large factorial computations ...
``````(defun ! (n &optional (product 1))
(if (zerop n) product
(! (1- n) (* product n))))
``````
And then for fun you could try `(format nil "~R" (! 25))`
-
I'm not a Lisp programmer, but I think this will help.
Basically it's a style of programming such that the recursive call is the last thing you do.
-
here is a Perl 5 version of the `tailrecsum` function mentioned earlier.
``````sub tail_rec_sum(\$;\$){
my( \$x,\$running_total ) = (@_,0);
return \$running_total unless \$x;
@_ = (\$x-1,\$running_total+\$x);
goto &tail_rec_sum; # throw away current stack frame
}
``````
-
To understand some of the core differences between tail-call recursion and non-tail-call recursion we can explore the .NET implementations of these techniques.
Here is an article with some examples in C#, F#, and C++\CLI: Adventures in Tail Recursion in C#, F#, and C++\CLI.
C# does not optimize for tail-call recursion whereas F# does.
The differences of principle involve loops vs. Lambda calculus. C# is designed with loops in mind whereas F# is built from the principles of Lambda calculus. For a very good (and free) book on the principles of Lambda calculus, see: Structure and Interpretation of Computer Programs, by Abelson, Sussman, and Sussman.
Regarding tail calls in F#, for a very good introductory article , see: Detailed Introduction to Tail Calls in F#. Finally, here is an article that covers the difference between non-tail recursion and tail-call recursion (in F#): Tail-recursion vs. non-tail recursion in F sharp.
If you want to read about some of the design differences of tail-call recursion between C# and F#, see: Generating Tail-Call Opcode in C# and F#.
If you care enough to want to know what conditions prevent the C# compiler from performing tail-call optimizations, see this article: JIT CLR tail-call conditions.
-
Recursion means a function calling itself. For example:
``````(define (un-ended name)
(un-ended 'me)
(print "How can I get here?"))
``````
Tail-Recursion means the recursion that conclude the function:
``````(define (un-ended name)
(print "hello")
(un-ended 'me))
``````
See, the last thing un-ended function (procedure, in Scheme jargon) does is to call itself. Another (more useful) example is:
``````(define (map lst op)
(define (helper done left)
(if (nil? left)
done
(helper (cons (op (car left))
done)
(cdr left))))
(reverse (helper '() lst)))
``````
In the helper procedure, the LAST thing it does if left is not nil is to call itself (AFTER cons something and cdr something). This is basically how you map a list.
The tail-recursion has a great advantage that the interperter (or compiler, dependent on the language and vendor) can optimize it, and transform it into something equivalent to a while loop. As matter of fact, in Scheme tradition, most "for" and "while" loop is done in tail-recursion manner (there is no for and while, as far as I know).
-
The best way for me to understand `tail call recursion` is: a special case of recursion where the last call(or the tail call) is the function itself.
Comparing the examples provided in Python:
``````def recsum(x):
if x == 1:
return x
else:
return x + recsum(x - 1)
``````
^RECURSION
``````def tailrecsum(x, running_total=0):
if x == 0:
return running_total
else:
return tailrecsum(x - 1, running_total + x)
``````
^TAIL RECURSION
As you can see in the general recursive version, the final call in the code block is `x + recsum(x - 1)`. So after calling the `recsum` method there is another operation which is `x + ..`.
However in the tail recursive version the final call(or the tail call) in the code block is `tailrecsum(x - 1, running_total + x)` which means the last call is made to the method itself and no operation after that.
This point is important because tail recursion as seen here is not making the memory grow because when the underlying VM sees a function calling itself in a tail position (the last expression to be evaluated in a function), it eliminates the current stack frame, which is known as Tail Call Optimization(TCO).
-
## protected by Srikar AppalAug 4 '13 at 15:27
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Get Full Access to Fundamentals Of Differential Equations - 8 Edition - Chapter 3.2 - Problem 12e
Get Full Access to Fundamentals Of Differential Equations - 8 Edition - Chapter 3.2 - Problem 12e
×
# the logistic curve (15), assume pa: = p(ta) and pb: =
ISBN: 9780321747730 43
## Solution for problem 12E Chapter 3.2
Fundamentals of Differential Equations | 8th Edition
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Problem 12E
For the logistic curve (15), assume pa: = p(ta) and pb: = p(tb) are given with tb = 2ta(ta > 0). Show that [Hint: Equate the expressions (21) for p0 at times ta and tb. Set and solve for . Insert into one of the earlier expressions and solve for p1.]
Step-by-Step Solution:
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9/6: Chapter 2: C++ Basics 1. C++ Program Structure a. Program – A set of functions including a main function to determine start i. Return 0 to end program ii. Types of Functions: 1. User Defined Function 2. Library Functions a. Pre-defined b. Must use #include directive statement to tell the computer where to find function iii. Starts at the main function, then follows instructions line-by-line iv. Function Calling Statement – Controls the order of functions v. Every function needs a return statement
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##### ISBN: 9780321747730
Fundamentals of Differential Equations was written by and is associated to the ISBN: 9780321747730. This textbook survival guide was created for the textbook: Fundamentals of Differential Equations , edition: 8. This full solution covers the following key subjects: expressions, hint, earlier, equate, assume. This expansive textbook survival guide covers 67 chapters, and 2118 solutions. The answer to “For the logistic curve (15), assume pa: = p(ta) and pb: = p(tb) are given with tb = 2ta(ta > 0). Show that [Hint: Equate the expressions (21) for p0 at times ta and tb. Set and solve for . Insert into one of the earlier expressions and solve for p1.]” is broken down into a number of easy to follow steps, and 51 words. Since the solution to 12E from 3.2 chapter was answered, more than 289 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 12E from chapter: 3.2 was answered by , our top Calculus solution expert on 07/11/17, 04:37AM.
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# 360 Dunk347 (100)
### Perform a 360 mid-air and stomp a victim on the landing
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PolyGamma
http://functions.wolfram.com/06.15.16.0021.01
Input Form
PolyGamma[-n, z] == (-1)^n PolyGamma[-n, -z] + Sum[Sum[(((-1)^(n + k) (z - p)^k + (z - p + 1)^k)/k!) Sum[((-1)^j/j!) PolyGamma[j + k - n, 1], {j, 0, n - k - 2}], {k, 0, n - 2}] + (1/(n - 1)!) ((z - p)^(-1 + n) (-EulerGamma + Log[-p + z] - PolyGamma[n]) + (z - p + 1)^(-1 + n) (EulerGamma - Log[-1 + p - z] + PolyGamma[n])), {p, 1, Floor[Re[z]]}] + ((z - Floor[Re[z]])^(-1 + n)/(n - 1)!) (EulerGamma + (I Pi (z - Floor[Re[z]]))/n + 2 I Pi Floor[3/4 - Arg[z - Floor[Re[z]]]/(2 Pi)] + PolyGamma[n] + Log[-2 Pi I] - Log[Floor[Re[z]] - z] - Sum[(Binomial[-1 + n, k] k! PolyLog[1 + k, 1])/ (2 Pi I (z - Floor[Re[z]]))^k, {k, 1, -1 + n}] + Sum[(-1)^k Binomial[-1 + n, k] Sum[(Binomial[k, j] j! PolyLog[1 + j, E^(2 I Pi z)])/ (2 Pi I (Floor[Re[z]] - z))^j, {j, 0, k}], {k, 0, -1 + n}]) /; Element[n, Integers] && n > 0 && Re[z] >= 0
Standard Form
Cell[BoxData[RowBox[List[RowBox[List[RowBox[List["PolyGamma", "[", RowBox[List[RowBox[List["-", "n"]], ",", "z"]], "]"]], "\[Equal]", RowBox[List[RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], "n"], RowBox[List["PolyGamma", "[", RowBox[List[RowBox[List["-", "n"]], ",", RowBox[List["-", "z"]]]], "]"]]]], "+", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["p", "=", "1"]], RowBox[List["Floor", "[", RowBox[List["Re", "[", "z", "]"]], "]"]]], RowBox[List["(", RowBox[List[RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["k", "=", "0"]], RowBox[List["n", "-", "2"]]], RowBox[List[FractionBox[RowBox[List[RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], RowBox[List["n", "+", "k"]]], SuperscriptBox[RowBox[List["(", RowBox[List["z", "-", "p"]], ")"]], "k"]]], "+", SuperscriptBox[RowBox[List["(", RowBox[List["z", "-", "p", "+", "1"]], ")"]], "k"]]], RowBox[List["k", "!"]]], " ", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["j", "=", "0"]], RowBox[List["n", "-", "k", "-", "2"]]], RowBox[List[FractionBox[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], "j"], RowBox[List["j", "!"]]], " ", RowBox[List["PolyGamma", "[", RowBox[List[RowBox[List["j", "+", "k", "-", "n"]], ",", "1"]], "]"]]]]]]]]]], "+", RowBox[List[FractionBox["1", RowBox[List[RowBox[List["(", RowBox[List["n", "-", "1"]], ")"]], "!"]]], RowBox[List["(", RowBox[List[RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["z", "-", "p"]], ")"]], RowBox[List[RowBox[List["-", "1"]], "+", "n"]]], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "EulerGamma"]], "+", RowBox[List["Log", "[", RowBox[List[RowBox[List["-", "p"]], "+", "z"]], "]"]], "-", RowBox[List["PolyGamma", "[", "n", "]"]]]], ")"]]]], "+", RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["z", "-", "p", "+", "1"]], ")"]], RowBox[List[RowBox[List["-", "1"]], "+", "n"]]], " ", RowBox[List["(", RowBox[List["EulerGamma", "-", RowBox[List["Log", "[", RowBox[List[RowBox[List["-", "1"]], "+", "p", "-", "z"]], "]"]], "+", RowBox[List["PolyGamma", "[", "n", "]"]]]], ")"]]]]]], ")"]]]]]], ")"]]]], "+", RowBox[List[FractionBox[SuperscriptBox[RowBox[List["(", RowBox[List["z", "-", RowBox[List["Floor", "[", RowBox[List["Re", "[", "z", "]"]], "]"]]]], ")"]], RowBox[List[RowBox[List["-", "1"]], "+", "n"]]], RowBox[List[RowBox[List["(", RowBox[List["n", "-", "1"]], ")"]], "!"]]], RowBox[List["(", RowBox[List["EulerGamma", "+", FractionBox[RowBox[List["\[ImaginaryI]", " ", "\[Pi]", " ", RowBox[List["(", RowBox[List["z", "-", RowBox[List["Floor", "[", RowBox[List["Re", "[", "z", "]"]], "]"]]]], ")"]]]], "n"], "+", RowBox[List["2", " ", "\[ImaginaryI]", " ", "\[Pi]", " ", RowBox[List["Floor", "[", RowBox[List[FractionBox["3", "4"], "-", FractionBox[RowBox[List["Arg", "[", RowBox[List["z", "-", RowBox[List["Floor", "[", RowBox[List["Re", "[", "z", "]"]], "]"]]]], "]"]], RowBox[List["2", " ", "\[Pi]"]]]]], "]"]]]], "+", RowBox[List["PolyGamma", "[", "n", "]"]], "+", RowBox[List["Log", "[", RowBox[List[RowBox[List["-", "2"]], "\[Pi]", " ", "\[ImaginaryI]"]], "]"]], "-", RowBox[List["Log", "[", RowBox[List[RowBox[List["Floor", "[", RowBox[List["Re", "[", "z", "]"]], "]"]], "-", "z"]], "]"]], "-", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["k", "=", "1"]], RowBox[List[RowBox[List["-", "1"]], "+", "n"]]], RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["2", "\[Pi]", " ", "\[ImaginaryI]", " ", RowBox[List["(", RowBox[List["z", "-", RowBox[List["Floor", "[", RowBox[List["Re", "[", "z", "]"]], "]"]]]], ")"]]]], ")"]], RowBox[List["-", "k"]]], " ", RowBox[List["Binomial", "[", RowBox[List[RowBox[List[RowBox[List["-", "1"]], "+", "n"]], ",", "k"]], "]"]], " ", RowBox[List["k", "!"]], " ", RowBox[List["PolyLog", "[", RowBox[List[RowBox[List["1", "+", "k"]], ",", "1"]], "]"]]]]]], "+", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["k", "=", "0"]], RowBox[List[RowBox[List["-", "1"]], "+", "n"]]], RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], "k"], " ", RowBox[List["Binomial", "[", RowBox[List[RowBox[List[RowBox[List["-", "1"]], "+", "n"]], ",", "k"]], "]"]], " ", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["j", "=", "0"]], "k"], RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["2", "\[Pi]", " ", "\[ImaginaryI]", " ", RowBox[List["(", RowBox[List[RowBox[List["Floor", "[", RowBox[List["Re", "[", "z", "]"]], "]"]], "-", "z"]], ")"]]]], ")"]], RowBox[List["-", "j"]]], " ", RowBox[List["Binomial", "[", RowBox[List["k", ",", "j"]], "]"]], " ", RowBox[List["j", "!"]], " ", RowBox[List["PolyLog", "[", RowBox[List[RowBox[List["1", "+", "j"]], ",", SuperscriptBox["\[ExponentialE]", RowBox[List["2", " ", "\[ImaginaryI]", " ", "\[Pi]", " ", "z"]]]]], "]"]]]]]]]]]]]], ")"]]]]]]]], "/;", " ", RowBox[List[RowBox[List["n", "\[Element]", "Integers"]], "\[And]", RowBox[List["n", ">", "0"]], "\[And]", RowBox[List[RowBox[List["Re", "[", "z", "]"]], "\[GreaterEqual]", "0"]]]]]]]]
MathML Form
ψ TagBox["\[Psi]", PolyGamma] ( - n ) ( z ) ( - 1 ) n ψ TagBox["\[Psi]", PolyGamma] ( - n ) ( - z ) + p = 1 Re ( z ) ( ( - log ( p - z - 1 ) + ψ TagBox["\[Psi]", PolyGamma] ( n ) + TagBox["\[DoubledGamma]", Function[List[], EulerGamma]] ) ( - p + z + 1 ) n - 1 + ( z - p ) n - 1 ( log ( z - p ) - ψ TagBox["\[Psi]", PolyGamma] ( n ) - TagBox["\[DoubledGamma]", Function[List[], EulerGamma]] ) ( n - 1 ) ! + k = 0 n - 2 ( - p + z + 1 ) k + ( - 1 ) k + n ( z - p ) k k ! j = 0 n - k - 2 ( - 1 ) j ψ TagBox["\[Psi]", PolyGamma] ( j + k - n ) ( 1 ) j ! ) + ( z - Re ( z ) ) n - 1 ( n - 1 ) ! ( 2 π 3 4 - arg ( z - Re ( z ) ) 2 π + TagBox["\[DoubledGamma]", Function[List[], EulerGamma]] + π ( z - Re ( z ) ) n + log ( - 2 π ) - log ( Re ( z ) - z ) + ψ TagBox["\[Psi]", PolyGamma] ( n ) - k = 1 n - 1 ( 2 π ( z - Re ( z ) ) ) - k ( n - 1 k ) TagBox[RowBox[List["(", GridBox[List[List[TagBox[RowBox[List["n", "-", "1"]], Identity, Rule[Editable, True]]], List[TagBox["k", Identity, Rule[Editable, True]]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]], Rule[Editable, False]] k ! Li PolyLog k + 1 ( 1 ) + k = 0 n - 1 ( - 1 ) k ( n - 1 k ) TagBox[RowBox[List["(", GridBox[List[List[TagBox[RowBox[List["n", "-", "1"]], Identity, Rule[Editable, True]]], List[TagBox["k", Identity, Rule[Editable, True]]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]], Rule[Editable, False]] j = 0 k ( 2 π ( Re ( z ) - z ) ) - j ( k j ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["k", Identity, Rule[Editable, True]]], List[TagBox["j", Identity, Rule[Editable, True]]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]], Rule[Editable, False]] j ! Li PolyLog j + 1 ( 2 π z ) ) /; n + Re ( z ) 0 Condition PolyGamma -1 n z -1 n PolyGamma -1 n -1 z p 1 z -1 p -1 z -1 PolyGamma n -1 p z 1 n -1 z -1 p n -1 z -1 p -1 PolyGamma n -1 n -1 -1 k 0 n -2 -1 p z 1 k -1 k n z -1 p k k -1 j 0 n -1 k -2 -1 j PolyGamma j k -1 n 1 j -1 z -1 z n -1 n -1 -1 2 3 4 -1 z -1 z 2 -1 z -1 z n -1 -2 -1 z -1 z PolyGamma n -1 k 1 n -1 2 z -1 z -1 k Binomial n -1 k k PolyLog k 1 1 k 0 n -1 -1 k Binomial n -1 k j 0 k 2 z -1 z -1 j Binomial k j j PolyLog j 1 2 z n SuperPlus z 0 [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["PolyGamma", "[", RowBox[List[RowBox[List["-", "n_"]], ",", "z_"]], "]"]], "]"]], "\[RuleDelayed]", RowBox[List[RowBox[List[RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], "n"], " ", RowBox[List["PolyGamma", "[", RowBox[List[RowBox[List["-", "n"]], ",", RowBox[List["-", "z"]]]], "]"]]]], "+", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["p", "=", "1"]], RowBox[List["Floor", "[", RowBox[List["Re", "[", "z", "]"]], "]"]]], RowBox[List["(", RowBox[List[RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["k", "=", "0"]], RowBox[List["n", "-", "2"]]], FractionBox[RowBox[List[RowBox[List["(", RowBox[List[RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], RowBox[List["n", "+", "k"]]], " ", SuperscriptBox[RowBox[List["(", RowBox[List["z", "-", "p"]], ")"]], "k"]]], "+", SuperscriptBox[RowBox[List["(", RowBox[List["z", "-", "p", "+", "1"]], ")"]], "k"]]], ")"]], " ", 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Date Added to functions.wolfram.com (modification date)
2007-05-02 | 4,928 | 13,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2013-48 | longest | en | 0.191732 |
https://open.lib.umn.edu/exploringbusiness/chapter/3-7-cases-and-problems/ | 1,702,168,126,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100989.75/warc/CC-MAIN-20231209233632-20231210023632-00252.warc.gz | 496,299,578 | 26,256 | # 3.7 Cases and Problems
### Learning on the Web (AACSB)
Keeping Current About Currency
On a day-to-day basis, you probably don’t think about what the U.S. dollar (US\$) is worth relative to other currencies. But there will likely be times when ups and downs in exchange rates will seem extremely important to you in your business career. The following are some hypothetical scenarios that illustrate what these times may be. (Note: To respond to the questions raised in each scenario, search Google for a currency converter.)
Scenario 1: Your Swiss Vacation
Your family came from Switzerland, and you and your parents visited relatives there back in 2007. Now that you’re in college, you want to make the trip on your own during spring break. While you’re there, you also plan to travel around and see a little more of the country. You remember that in 2007, US\$1 bought 1.22 Swiss francs (Frs). You estimate that, at this rate, you can finance your trip (excluding airfare) with the \$1,200 that you earned this summer. You’ve heard, however, that the exchange rate has changed. Given the current exchange rate, about how much do you think your trip would cost you? As a U.S. traveler going abroad, how are you helped by a shift in exchange rates? How are you hurt?
Scenario 2: Your British Friends
A few years ago, you met some British students who were visiting the United States. This year, you’re encouraging them to visit again so that you can show them around New York City. When you and your friends first talked about the cost of the trip back in 2007, the British pound (£) could be converted into US\$1.90. You estimated that each of your British friends would need to save up about £600 to make the trip (again, excluding plane fare). Given today’s exchange rate, how much will each person need to make the trip? Have your plans been helped or hindered by the change in exchange rates? Was the shift a plus for the U.S. travel industry? What sort of exchange-rate shift hurts the industry?
Scenario 3: Your German Soccer Boots
Your father rarely throws anything away, and while cleaning out the attic a few years ago, he came across a pair of vintage Adidas soccer boots made in 1955. Realizing that they’d be extremely valuable to collectors in Adidas’s home country of Germany, he hoped to sell them for US \$5,000 and, to account for the exchange rate at the time, planned to price them at \$7,200 in euros. Somehow, he never got around to selling the boots and has asked if you could sell them for him on eBay. If he still wants to end up with US \$5,000, what price in euros will you now have to set? Would an American company that exports goods to the European Union view the current rate more favorably or less favorably than it did back in 2007?
### Career Opportunities (AACSB)
Broadening Your Business Horizons
At some point in your life, you’ll probably meet and work with people from various countries and cultures. Participating in a college study-abroad program can help you prepare to work in the global business environment, and now is as good a time as any to start exploring this option. Here’s one way to go about it:
• Select a study-abroad program that interests you. To do this, you need to decide what country you want to study in and your academic field of interest. Unless you speak the language of your preferred country, you should pick a program offered in English.
• If your school offers study-abroad programs, choose one that has been approved by your institution.
• If your school doesn’t offer study-abroad programs, locate one through a Web search.
• Describe the program, the school that’s offering it, and the country to which it will take you.
• Indicate why you’ve selected this particular program, and explain how it will help you prepare for your future business career.
### Ethics Angle (AACSB)
The Right, Wrong, and Wisdom of Dumping and Subsidizing
When companies sell exported goods below the price they’d charge in their home markets (and often below the cost of producing the goods), they’re engaging in dumping. When governments guarantee farmers certain prices for crops regardless of market prices, the beneficiaries are being subsidized. What do you think about these practices? Is dumping an unfair business practice? Why, or why not? Does subsidizing farmers make economic sense for the United States? What are the effects of farm subsidies on the world economy? Are the ethical issues raised by the two practices comparable? Why, or why not?
### Team-Building Skills (AACSB)
Three Little Words: The China Price
According to business journalists Pete Engardio and Dexter Roberts, the scariest three words that a U.S. manufacturer can hear these days are the China price. To understand why, go to the Business Week Web site (http://www.businessweek.com/magazine/content/04_49/b3911401.htm) and read its article “The China Price,” which discusses the benefits and costs of China’s business expansion for U.S. companies, workers, and consumers. Once you’ve read the article, each member of the team should be able to explain the paradoxical effect of U.S.–Chinese business relationships—namely, that they can hurt American companies and workers while helping American companies and consumers.
Next, your team should get together and draw up two lists: a list of the top five positive outcomes and a list of the top five negative outcomes of recent Chinese business expansion for U.S. businesses, workers, and consumers. Then, the team should debate the pros and cons of China’s emergence as a global business competitor and, finally, write a group report that answers the following questions:
1. Considered on balance, has China’s business expansion helped or harmed U.S. companies, workers, and consumers? Justify your answers.
2. What will happen to U.S. companies, workers, and consumers in the future if China continues to grow as a global business competitor?
3. How should U.S. companies respond to the threats posed by Chinese competitors in their markets?
4. What can you do as a student to prepare yourself to compete in an ever-changing global business environment?
When you hand in your report, be sure to attach all the following items:
• Members’ individually prepared lists of ways in which business relationships with China both hurt and help U.S. businesses, workers, and consumers
• Your group-prepared list of the top five positive and negative effects of Chinese business expansion on U.S. businesses, workers, and consumers
### The Global View (AACSB)
Go East, Young Job Seeker
How brave are you when it comes to employment? Are you bold enough to go halfway around the world to find work? Instead of complaining about U.S. jobs going overseas, you could take the bull by the horns and grab one job back. It’s not that tough to do, and it could be a life-changing experience. U.S. college graduates with business or technical backgrounds are highly sought after by companies that operate in India. If you qualify (and if you’re willing to relocate), you could find yourself working in Bangalore or New Delhi for some multinational company like Intel, Citibank, or GlaxoSmithKline (a pharmaceutical company). In addition, learning how to live and work in a foreign country can build self-confidence and make you more attractive to future employers. To get a glimpse of what it would be like to live and work in India, go to the Web sites of American Way magazine (http://www.americanwaymag.com/jeffrey-vanderwerf-high-tech-outsourcing-boom-bangalore-leela-palace) and CNN and Money (http://money.cnn.com/2004/03/09/pf/workers_to_india), and check out the posted articles: “Passage to India,” and “Needs Job, Moves to India.” Then, go to the Monster Work Abroad Web site (http://jobsearch.monsterindia.com/return2origin/index.html) and find a job in India that you’d like to have, either right after graduation or about five years into your career. (When selecting the job, ignore its actual location and proceed as if it’s in Bangalore.) After you’ve pondered the possibility of living and working in India, answer the following questions:
1. What would your job entail?
2. What would living and working in Bangalore be like? What aspects would you enjoy? Which would you dislike?
3. What challenges would you face as an expatriate (a person who lives outside his or her native country)? What opportunities would you have?
4. How would the experience of working in India help your future career?
5. Would you be willing to take a job in India for a year or two? Why, or why not?
## License
Exploring Business Copyright © 2016 by University of Minnesota is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted. | 1,905 | 8,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | longest | en | 0.95631 |
http://www.torontokidscomputer.com/uncategorized/tuesday-python-21-0216-17/ | 1,726,876,432,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701425385.95/warc/CC-MAIN-20240920222945-20240921012945-00496.warc.gz | 49,468,930 | 11,704 | TORONTO KIDS COMPUTER CLUB | Tuesday 17:30 Python Homework 21.02.16.
19081
# Tuesday 17:30 Python Homework 21.02.16.
## 21 Feb Tuesday 17:30 Python Homework 21.02.16.
Question 1.
Use EasyGui to write a program to convert temperatures from Fahrenheit to Celsius. The formula for that is: Celsius = 5 / 9 * (Fahrenheit – 32). Use GUI input and output. You need to create a easygui enterbox to ask Fahrenheit, then use message box to show the Celsius degree. (Hint: the input you get from the enter box is a string, so you need float() to convert it into decimal number.
Question 2:
We have learned how to use if statement to making the choices. Please using easygui to create the program we made in the class. The enter box will ask you password, if you get the right password, the msgbox will display ‘FBI Access granted’
Question 3:
A store is having a sale. They’re giving \$10 off when you purchase more than \$100. Write a program that asks the purchase price and displays the final price.
```Sample Input 1:
Sample Output 1:
You final price is \$90```
```Sample Input 2:
Sample Output 2:
```Hint: | 284 | 1,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.748044 |
https://it.mathworks.com/matlabcentral/cody/problems/532-return-unique-values-without-sorting/solutions/1261616 | 1,585,808,530,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506673.7/warc/CC-MAIN-20200402045741-20200402075741-00128.warc.gz | 521,231,791 | 15,673 | Cody
# Problem 532. Return unique values without sorting
Solution 1261616
Submitted on 4 Sep 2017 by grey tomato
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct)) x = [9 2 2]; y_correct = [9 2]; assert(isequal(your_fcn_name(x),y_correct)) x = [-4 1 1]; y_correct = [-4 1]; assert(isequal(your_fcn_name(x),y_correct)) x = [42 1 1] y_correct = [42 1]; assert(isequal(your_fcn_name(x),y_correct)) x = [42 1 1 1 42 17 17]; y_correct = [42 1 17]; assert(isequal(your_fcn_name(x),y_correct))
y = 1 y = 9 2 y = -4 1 x = 42 1 1 y = 42 1 y = 42 1 17 | 268 | 728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-16 | latest | en | 0.58545 |
https://secure.ssa.gov/poms.nsf/lnx/0300615020 | 1,726,699,132,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651941.8/warc/CC-MAIN-20240918201359-20240918231359-00858.warc.gz | 462,552,718 | 9,095 | TN 41 (12-23)
RS 00615.020 Dual Entitlement Overview
A. Policy
1. General
A person may be entitled to more than one benefit at the same time. For example, a person may be entitled as a retired worker on their own record and as a spouse on another record. However, a person's benefit amount can never exceed the highest single benefit to which that person is entitled. Some benefits are calculated independently with the larger benefit being paid or the smaller benefit being paid plus the excess amount of the larger one. Other types of benefits are calculated with a carry-over reduction amount from the first benefit to the second.
2. Calculation Category
The following are the types of calculations that are possible:
• Method A - Both benefits calculated independently; one benefit payable.
• Method B - Both benefits calculated and reduced independently; small MBA paid plus excess of larger MBA.
NOTE: If larger MBA is a RIB or DIB just pay the RIB or DIB.
• Method C - First benefit calculated independently; reduce excess of second benefit for age (if larger).
• Method D - First benefit calculated independently; carry-over reduction applies to second benefit.
3. Examples of Calculations
The following are examples of each of the four methods of calculating the reduced benefits:
• Method A – A widow is entitled to a benefit of \$850 before reduction on one deceased spouse's record. They are also entitled to a benefit of \$670 before reduction on their second deceased spouse's record. Each benefit is reduced separately and the higher is paid.
• Method B – A widow is entitled to a benefit of \$1000 before reduction. They are also entitled to a RIB of \$400 before reduction. Each benefit is reduced separately. The widow benefit is reduced to \$900 and the RIB is reduced to \$380. The reduced RIB is subtracted from the reduced widow benefit. The result is the excess widow benefit payable - \$520. The total paid is \$900, the sum of the reduced excess widow benefit and the reduced RIB.
• Method C - A spouse is entitled to a benefit of \$1000 before reduction. They are also entitled to a RIB of \$400 before reduction. The full RIB is subtracted from the full spouse benefit. The excess (\$600) is then reduced to \$540. The RIB is reduced to \$380. The total payable is \$920, the sum of the reduced spouse excess and the reduced RIB.
Note: For months in which a spouse has a child-in-care, the amount payable as a spouse will always be the difference between the unreduced spouse benefit and the reduced RIB rate.
• Method D – A person is entitled to reduced RIB for six months. They then become entitled to DIB. The DIB is reduced by the number of months of reduced RIB entitlement.
B. Procedure — DUAL ENTITLEMENT CALCULATION CHART
Use the following chart to determine the type of dual entitlement calculation from RS 00615.020A.2. for each situation:
If the Type Benefits is...
Then the Calculation Category is...
HA then A
a
HA with A
a
HA then B
c
HA with B
c
HA then D/W
c
HA with D/W
c
A then B
c
A with B
c
A then D
b
A with D
b
A then HA
d
A with HA
a
A then secondary unreduced benefit
b
A with secondary unreduced benefit
b
B then HA
b
B with HA
c
B then A
b
B with A
c
B and B
a
Secondary unreduced benefit with A
b
Secondary unreduced benefit then A
b
C and C (if CFM applies, only one C benefit is payable but entitlement on all records is material in calculating the rate)
a
D then B
b
D then B6
a
D with D
a
D/W prior to age 62 then A or HA. If the date of birth is prior to 1/2/1928 a carry-over reduction applies and the calculation type is d.
b
D/W age 62 or older then A or HA
b
D/W with HA
c
D/W then HA
b
E/F and HA
b
F and F
a
To Link to this section - Use this URL: http://policy.ssa.gov/poms.nsf/lnx/0300615020 RS 00615.020 - Dual Entitlement Overview - 12/05/2023 Batch run: 12/05/2023 Rev:12/05/2023 | 979 | 3,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-38 | latest | en | 0.95911 |
http://www.reference.com/browse/spectral+luminous+efficiency | 1,435,620,616,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375090887.26/warc/CC-MAIN-20150627031810-00169-ip-10-179-60-89.ec2.internal.warc.gz | 737,168,194 | 19,785 | Related Searches
Definitions
# Photometry (optics)
This article deals with the usage of this term in optics and lighting.
Photometry is the science of measurement of light, in terms of its perceived brightness to the human eye. It is distinct from radiometry, which is the science of measurement of radiant energy (including light) in terms of absolute power; rather, in photometry, the radiant power at each wavelength is weighted by a luminosity function (a.k.a. visual sensitivity function) that models human brightness sensitivity. Typically, this weighting function is the photopic sensitivity function, although the scotopic function — and others — may also be applied in the same way.
## Photometry and the eye
The human eye is not equally sensitive to all wavelengths of visible light. Photometry attempts to account for this by weighting the measured power at each wavelength with a factor that represents how sensitive the eye is at that wavelength. The standardized model of the eye's response to light as a function of wavelength is given by the luminosity function. Note that the eye has different responses as a function of wavelength when it is adapted to light conditions (photopic vision) and dark conditions (scotopic vision). Photometry is typically based on the eye's photopic response, and so photometric measurements may not accurately indicate the perceived brightness of sources in dim lighting conditions where colors are not discernible, such as under just starlight.
## Photometric quantities
Many different units of measure are used for photometric measurements. People sometimes ask why there need to be so many different units, or ask for conversions between units that can't be converted (lumens and candelas, for example). We are familiar with the idea that the adjective "heavy" can refer to weight or density, which are fundamentally different things. Similarly, the adjective "bright" can refer to a light source which delivers a high luminous flux (measured in lumens), or to a light source which concentrates the luminous flux it has into a very narrow beam (candelas), or to a light source that is seen against a dark background. Because of the ways in which light propagates through three-dimensional space — spreading out, becoming concentrated, reflecting off shiny or matte surfaces — and because light consists of many different wavelengths, the number of fundamentally different kinds of light measurement that can be made is large, and so are the numbers of quantities and units that represent them.
### Photometric versus radiometric quantities
There are two parallel systems of quantities known as photometric and radiometric quantities. Every quantity in one system has an analogous quantity in the other system. Some examples of parallel quantities include:
See chart for more. (full page)
In photometric quantities every wavelength is weighted according to how sensitve the human eye is to it, while radiometric quantities use unweighted absolute power. For example, the eye responds much more strongly to green light than to red, so a green source will have greater luminous flux than a red source with the same radiant flux would. Radiant energy outside the visible spectrum does not contribute to photometric quantities at all, so for example a 1000 watt space heater may put out a great deal of radiant flux (1000 watts, in fact), but as a light source it puts out very few lumens (because most of the energy is in the infrared, leaving only a dim red glow in the visible).
### Watts versus lumens
Watts are units of radiant flux while lumens are units of luminous flux. A comparison of the watt and the lumen illustrates the distinction between radiometric and photometric units.
The watt is a unit of power. We are accustomed to thinking of light bulbs in terms of power in watts. This power is not a measure of the amount of light output, but rather indicates how much energy the bulb will use. Because incandescent bulbs sold for "general service" all have fairly similar characteristics (same spectral power distribution), power consumption provides a rough guide to the light output of incandescent bulbs.
Watts can also be a direct measure of output. In a radiometric sense, an incandescent light bulb is about 80% efficient: 20% of the energy is lost (e.g. by conduction through the lamp base). The remainder is emitted as radiation, mostly in the infrared. Thus, a 60 watt light bulb emits a total radiant flux of about 45 watts. Incandescent bulbs are, in fact, sometimes used as heat sources (as in a chick incubator), but usually they are used for the purpose of providing light. As such, they are very inefficient, because most of the radiant energy they emit is invisible infrared. A compact fluorescent lamp can provide light comparable to a 60 watt incandescent while consuming as little as 15 watts of electricity.
The lumen is the photometric unit of light output. Although most consumers still think of light in terms of power consumed by the bulb, in the U.S. it has been a trade requirement for several decades that light bulb packaging give the output in lumens. The package of a 60 watt incandescent bulb indicates that it provides about 900 lumens, as does the package of the 15 watt compact fluorescent.
The lumen is defined as amount of light given into one steradian by a point source of one candela strength; while the candela, a base SI unit, is defined as the luminous intensity of a source of monochromatic radiation, of frequency 540 terahertz, and a radiant intensity of 1/683 watts per steradian. (540 THz corresponds to about 555 nanometres, the wavelength, in the green, to which the human eye is most sensitive. The number 1/683 was chosen to make the candela about equal to the standard candle, the unit which it superseded).
Combining these definitions, we see that 1/683 watt of 555 nanometre green light provides one lumen.
The relation between watts and lumens is not just a simple scaling factor. We know this already, because the 60 watt incandescent bulb and the 15 watt compact fluorescent can both provide 900 lumens.
The definition tells us that 1 watt of pure green 555 nm light is "worth" 683 lumens. It does not say anything about other wavelengths. Because lumens are photometric units, their relationship to watts depends on the wavelength according to how visible the wavelength is. Infrared and ultraviolet radiation, for example, are invisible and do not count. One watt of infrared radiation (which is where most of the radiation from an incandescent bulb falls) is worth zero lumens. Within the visible spectrum, wavelengths of light are weighted according to a function called the "photopic spectral luminous efficiency." According to this function, 700 nm red light is only about 4% as efficient as 555 nm green light. Thus, one watt of 700 nm red light is "worth" only 27 lumens.
Because of the summation over the visual portion of the EM spectrum that is part of this weighting, the unit of "lumen" is color-blind: there is no way to tell what color a lumen will appear. This is equivalent to evaluating groceries by number of bags: there is no information about the specific content, just a number that refers to the total weighted quantity.
## Photometric measurement techniques
Photometric measurement is based on photodetectors, devices (of several types) that produce an electric signal when exposed to light. Simple applications of this technology include switching luminaires on and off based on ambient light conditions, and light meters, used to measure the total amount of light incident on a point.
More complex forms of photometric measurement are used frequently within the lighting industry. Spherical photometers can be used to measure the directional luminous flux produced by lamps, and consist of a large-diameter globe with a lamp mounted at its center. A photocell rotates about the lamp in three axes, measuring the output of the lamp from all sides.
Luminaires (known to laypersons simply as light fixtures) are tested using goniophotometers and rotating mirror photometers, which keep the photocell stationary at a sufficient distance that the luminaire can be considered a point source. Rotating mirror photometers use a motorized system of mirrors to reflect light emanating from the luminaire in all directions to the distant photocell; goniophotometers use a rotating 2-axis table to change the orientation of the luminaire with respect to the photocell. In either case, luminous intensity is tabulated from this data and used in lighting design.
## External links
### Photometry diagrams and applets
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• Please Login or Sign Up to use the Recent Searches feature | 1,831 | 8,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-27 | latest | en | 0.957818 |
https://stats.stackexchange.com/questions/305644/in-a-dirichlet-process-can-the-base-distribution-be-discrete | 1,657,158,520,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683020.92/warc/CC-MAIN-20220707002618-20220707032618-00567.warc.gz | 568,526,590 | 65,084 | # In a Dirichlet process, can the base distribution be discrete?
Must it be continuous? Note we are talking about the base distribution. The sampled distribution is discrete.
1) If the base distribution is continuous, drawing from it will get a new value (a new cluster mean, for example) every time.
2) But, you can nest Dirichlet processes. Which implies the base distribution can be discrete. (You can place a prior over your base distribution which is just another Dirichlet process).
Consider the Blackwell-Macqueen urn scheme.
H is a distribution over colors.
With probability alpha sample a color from H. Note, if H is continuous, you'll get a different color every time. If the H is discrete, you may draw colors already present in the urn.
With probability proportional to n-1, pick a ball from the urn. Put the ball back and another one of the same color.
edit: The paper above demonstrates this is possible. Any distribution drawn for the a DP with a discrete base distribution will have the same atoms as the base distribution, but with different weights.
Still, I wonder how the CRP would work under this scheme since there's a chance you'd select a non-empty table (analogously to the above urn scheme).
• Apparently, after reading a few excerpts from some papers, I think they can be. Not sure then how this fits in with sampling schemes. Sep 30, 2017 at 22:46
The base measure can be discrete. An example of the case where it is discrete is the Hierarchichal Dirichlet Process, wherein $F_j \sim \mathcal D(\alpha F_0)$ where $F_0 \sim \mathcal D(\gamma H)$. The base measure of the second level of the hierarchy ($F_0$) is discrete with probability $1$. | 386 | 1,681 | {"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-2022-27 | latest | en | 0.92203 |
http://www.jiskha.com/members/profile/posts.cgi?name=Kristina&page=6 | 1,368,946,760,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696384181/warc/CC-MAIN-20130516092624-00072-ip-10-60-113-184.ec2.internal.warc.gz | 534,229,571 | 3,502 | Sunday
May 19, 2013
# Posts by Kristina
Total # Posts: 161
law
I have a midterm for criminal justice class to do and the istructor she said as soon as we log in to the system we would have 2 hours 15 mins to complete it
REALLY EASY
Which of the following explanations correctly describes the reason we see the Sun rise and set each day? 1.The Sun rotates on its axis 2.The Earth revolves around the Sun in an elliptical path 3.The Earth rotates on its axis 4.The Sun revolves around the Earth in an elliptical...
REALLY EASY
Which of the following are a consequence of the Earth's yearly motion around the Sun? (check all that apply) 1.the Sun appears in different constellations in different seasons 2.the planets look different than stars 3.the Sun rises and sets 4.the Moon exhibits phases 5.dif...
physics
A truck at rest at a stoplight accelerates uniformly with acceleration a = 8 meters/second2 for 5.5 seconds after the light turns green. What is the speed of the truck in meters/second after that time interval?
Physics
A car accelerates uniformly (a = constant) from 34 meters/second up to 40 meters/second in 4.0 seconds. What is the acceleration of the car in meters/second2?
Elementry education major
how many tens are in 13,644 A.4 B.13,644 C.1,364 D1,364.4
Chem 2
The half-life for the radioactive decay of U-238 is 4.5 billion years and is independent of initial concentration. a) How long will it take for 10% of the U-238 atoms in a sample of U- 238 to decay? b)If a sample of U-238 initially contained 1.9×1018 atoms when the unive...
History
All of the following inclined the US toward entering on the side of the Allies in WWI EXCEPT? although some progressives opposed war, the idealistic progressive spirit pointed toward an American campaign to end militarism and establish a peaceful world order. although an offic...
calculus
A regular decagon(12 sides) in inscribed in a circle with the radius r. The decagon has an area of108 in^2. What is the radius of the cirlce?
calculus
A regular decagon(12 sides) in inscribed in a circle with the radius r. The decagon has an area of108 in^2. What is the radius of the cirlce?
Pages: <<Prev | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>> | 603 | 2,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2013-20 | longest | en | 0.915298 |
http://www.thecodingforums.com/threads/same-operation-sort-of-different-results.839468/ | 1,475,169,336,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661905.96/warc/CC-MAIN-20160924173741-00004-ip-10-143-35-109.ec2.internal.warc.gz | 778,178,125 | 13,175 | # same operation (sort of), different results
Discussion in 'Ruby' started by Chad Perrin, Mar 30, 2007.
I'm a little confused by these results. Perhaps someone can tell me
what I'm assuming incorrectly:
\$ irb
irb(main):001:0> 5 / 9 * ( 100 - 32 )
=> 0
irb(main):002:0> ( 5 / 9 ) * ( 100 - 32 )
=> 0
irb(main):004:0> ( 100 - 32 ) * ( 5 / 9 )
=> 0
irb(main):003:0> ( 100 - 32 ) * 5 / 9
=> 37
The last result is the only one that gives me what I actually wanted.
In case you're wondering, yes, that *is* an F-to-C temperature
conversion.
--
CCD CopyWrite Chad Perrin [ http://ccd.apotheon.org ]
"The ability to quote is a serviceable
substitute for wit." - W. Somerset Maugham
On Fri, Mar 30, 2007 at 02:30:33PM +0900, Chad Perrin wrote:
> I'm a little confused by these results. Perhaps someone can tell me
> what I'm assuming incorrectly:
>
> \$ irb
> irb(main):001:0> 5 / 9 * ( 100 - 32 )
> => 0
> irb(main):002:0> ( 5 / 9 ) * ( 100 - 32 )
> => 0
> irb(main):004:0> ( 100 - 32 ) * ( 5 / 9 )
> => 0
> irb(main):003:0> ( 100 - 32 ) * 5 / 9
> => 37
>
> The last result is the only one that gives me what I actually wanted.
>
> In case you're wondering, yes, that *is* an F-to-C temperature
> conversion.
Never mind, that was a full-on brain fart. I forgot I was doing integer
division. Mea culpa.
--
CCD CopyWrite Chad Perrin [ http://ccd.apotheon.org ]
Amazon.com interview candidate: "When C++ is your
hammer, everything starts to look like your thumb."
3. ### John JoyceGuest
Yep, you can do integer division but do it last for best results.
you can also throw a .to_f method onto everything for your division..
The only way around real division errors is to use large integers,
keep track of decimal location in your own custom object for display
purposes and thus push the division error way way down to a small,
insignificant number.
On Mar 30, 2007, at 2:34 PM, Chad Perrin wrote:
> On Fri, Mar 30, 2007 at 02:30:33PM +0900, Chad Perrin wrote:
>> I'm a little confused by these results. Perhaps someone can tell me
>> what I'm assuming incorrectly:
>>
>> \$ irb
>> irb(main):001:0> 5 / 9 * ( 100 - 32 )
>> => 0
>> irb(main):002:0> ( 5 / 9 ) * ( 100 - 32 )
>> => 0
>> irb(main):004:0> ( 100 - 32 ) * ( 5 / 9 )
>> => 0
>> irb(main):003:0> ( 100 - 32 ) * 5 / 9
>> => 37
>>
>> The last result is the only one that gives me what I actually wanted.
>>
>> In case you're wondering, yes, that *is* an F-to-C temperature
>> conversion.
>
> Never mind, that was a full-on brain fart. I forgot I was doing
> integer
> division. Mea culpa.
>
> --
> CCD CopyWrite Chad Perrin [ http://ccd.apotheon.org ]
> Amazon.com interview candidate: "When C++ is your
> hammer, everything starts to look like your thumb."
>
John Joyce, Mar 30, 2007
4. ### Robert KlemmeGuest
On 30.03.2007 08:10, John Joyce wrote:
> Yep, you can do integer division but do it last for best results.
> you can also throw a .to_f method onto everything for your division..
> The only way around real division errors is to use large integers, keep
> track of decimal location in your own custom object for display purposes
> and thus push the division error way way down to a small, insignificant
> number.
There is another way: BigDecimal.
robert
Robert Klemme, Mar 30, 2007
5. ### John JoyceGuest
I meant in general. For portability and dealing with float math.
Lots of solutions exist of course. One of them is Ruby's BigDecimal
On Mar 30, 2007, at 4:20 PM, Robert Klemme wrote:
> On 30.03.2007 08:10, John Joyce wrote:
>> Yep, you can do integer division but do it last for best results.
>> you can also throw a .to_f method onto everything for your
>> division..
>> The only way around real division errors is to use large integers,
>> keep track of decimal location in your own custom object for
>> display purposes and thus push the division error way way down to
>> a small, insignificant number.
>
> There is another way: BigDecimal.
>
> robert
>
John Joyce, Mar 30, 2007
On Fri, Mar 30, 2007 at 03:10:48PM +0900, John Joyce wrote:
> Yep, you can do integer division but do it last for best results.
> you can also throw a .to_f method onto everything for your division..
> The only way around real division errors is to use large integers,
> keep track of decimal location in your own custom object for display
> purposes and thus push the division error way way down to a small,
> insignificant number.
Thanks. I knew all this -- I just managed to completely forget
everything of use yesterday (the first day after a debilitating
migraine took me out of action for most of the day). Apparently, I
became temporarily very stupid as an after-effect. Unfortunately,
some of the evidence of this made it to the ruby-talk mailing list.
--
CCD CopyWrite Chad Perrin [ http://ccd.apotheon.org ]
"A script is what you give the actors. A program
is what you give the audience." - Larry Wall | 1,432 | 4,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-40 | latest | en | 0.844904 |
http://mathhelpforum.com/advanced-statistics/180201-how-estimate-parametervalues-choice-model.html | 1,480,973,105,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541839.36/warc/CC-MAIN-20161202170901-00191-ip-10-31-129-80.ec2.internal.warc.gz | 181,487,889 | 10,015 | ## How to estimate the parametervalues for choice model?
I have to develop a model to estimate the travel mode choice (car, public transport or bicyle) between two zones, based on variables like for example the car ownership, the travel time of the different modes, the percentage elderly people, the percentage of students, and the parking fee.
I have aggregated data about all these variables for all the zones. Also I have data about the observed %car, %Public Transport (PT) and %bicycle between all pairs of zones.
The question is now how I can estimate a model based on this data. I am not sure but I think the model has to look like as following:
$P_{car} = \frac{e^{V_{car}}}{e^{V_{car}}+e^{V_{bicyle}}+e^{V _{PT}} }$
with:
$V_{car} = \alpha +\beta_1 * CarOwnership + \beta2*PercetageStudents +...$
(To estimate the % bicycle use and the % public transport the same type of model will be used)
My question is how I can estimate the values for alpha, beta1, beta2, beta_n?
Maybe I am using the wrong type of model, in that case please let me know! But I am not an expert in this field. | 281 | 1,098 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-50 | longest | en | 0.896382 |
https://stats.stackexchange.com/questions/238866/how-to-test-amount-of-differences-in-t-test-in-r-for-large-samples | 1,611,103,125,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519843.24/warc/CC-MAIN-20210119232006-20210120022006-00068.warc.gz | 579,004,508 | 37,214 | How to test amount of differences in T-test in R for large samples?
We have good discussions on hypothesis testing in large data: With large amount of data, even tinny difference between two samples can be detected, and we are almost certain to reject NULL hypothesis.
Effect size is one "fix", but I am interested in other fix, where, in addition to say they are different, but say how much the difference are.
I not know how to do it in R. Let us assume we are doing two sample T test, how can I say I want to test if the means are different by say certain amount say $0.1$ ?
The following code is test if two means are equal. How to modify it to test if two means are difference by $0.1$?
sample1=rnorm(1e5)
sample2=rnorm(1e5)
t.test(sample1,sample2)
Welch Two Sample t-test
data: sample1 and sample2
t = -0.5542, df = 2e+05, p-value = 0.5794
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.01124444 0.00628720
sample estimates:
mean of x mean of y
-0.0002658978 0.0022127222
Edit:
I was trying to ask if the amount difference on mean of two samples within range of $-0.1$ to $0.1$, not exactly to equal $0.1$.
Reading Peter Flom 's comment, modifying T test will not meet my needs?
• In a regular two sample t-test the null is "the two means are equal". In a test of equivalence, the null is "the two means are different by at least XXX". I think you want something else, namely null is "the two means are withing XXX" – Peter Flom Oct 6 '16 at 21:33
There is not a usual way to use a t-test to test if the absolute difference is larger than 0.1, or even to test if the difference lies within a given range.
However, your goal of assessing how large the difference is can be achieved by using the confidence interval. It's already in your results, since R and most statistical packages produce a confidence interval when told to perform a t-test:
95 percent confidence interval:
-0.01124444 0.00628720
• I think this is the best approach. Fuzzy null hypotheses are tricky in the frequentist paradigm, and more attention to the effect size rather than a sharp hypothesis test is the best and simplest thing to explain and encourage. – Peter Ellis Oct 6 '16 at 22:25
R code questions are actually off-topic here but it is:
t.test(sample1, sample2, alternative = "two.sided", mu = 0.1)
I think the question should stay here, however, because it raises a statistical question. My view is that looking for an effect larger than a certain amount makes a lot of sense, but it can be hard to choose an amount.
• Thanks, I thought to ask it to stack overflow.. but as you said, it raises a statistical question. – Haitao Du Oct 6 '16 at 21:17
• No, it's not a stack overflow question either - because it's not really about programming. It's an interesting statistics question along with a very basic R question. – Peter Flom Oct 6 '16 at 21:18
• two.sided not two-sided. But I'm not sure this is right, I think the OP really wants the null hypothesis is mu is between -0.1 and 0.1, and alternative is that abs(mu) > 0.1 (not mu = 0.1).? – Peter Ellis Oct 6 '16 at 21:20
• Thanks for the syntax fix but I think what I had does test abs(mu) > 0.1. Exactly 0.1 doesn't make sense. Unless I am missing something (which certainly could be the case). Are you suggesting OP wants a test of equivalence? – Peter Flom Oct 6 '16 at 21:22
• @PeterFlom sorry I was trying to ask within range of -0.1 to 0.1, not exactly to equal 0.1 – Haitao Du Oct 6 '16 at 21:23
I'm not going to attempt to answer the R portion of this question, but I wanted to comment on this: "Effect size is one "fix", but I am interested in other fix, where, in addition to say they are different, but say how much the difference are."
The purpose of effect size is to attach a measure of magnitude of effect, or difference, between the populations (or treatments).
If you are not familiar with effect sizes, I recommend you read: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3444174/ and http://www.leeds.ac.uk/educol/documents/00002182.htm
'Effect size' is simply a way of quantifying the size of the difference between two groups.
As a statistical consultant, I always push for reporting of effect sizes to any statistical product.
An Edit:
I want to comment on this: "I was trying to ask if the amount difference on mean of two samples within range of −0.1−0.1 to 0.10.1, not exactly to equal 0.10.1."
I think there needs to be some clarity on this. What exactly is -0.1 to 0.1? Is this a confidence interval? Acceptance region?
• +1, I think may need your help to say it formally. What in my mind, is: we know they are different, but can we say they are different by $x$ amount? – Haitao Du Oct 6 '16 at 21:40
• No. That's the purpose of a confidence interval. You're 95% confident that the difference between Pop A and B is -0.1 to 0.1. Also, note that that's another big issue. If 0 is in your Conf Interval, then it's REALLY difficult to accept that there is a difference at all. – Jon Oct 6 '16 at 21:43
You can just subtract or add 0.1 to one of the datasets.
• thanks, my question was not clear. could you take a look at my edit? – Haitao Du Oct 6 '16 at 21:29 | 1,395 | 5,221 | {"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} | 3.625 | 4 | CC-MAIN-2021-04 | latest | en | 0.935012 |
https://www.teacherspayteachers.com/Product/Trigonometry-and-Special-Right-Triangles-TR3-HSGSRTC6-HSGSRTC8-2246812 | 1,503,102,941,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105195.16/warc/CC-MAIN-20170818233221-20170819013221-00513.warc.gz | 971,565,949 | 23,922 | ## Main Categories
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# Trigonometry and Special Right Triangles (TR3): HSG.SRT.C.6, HSG.SRT.C.8
Common Core Standards
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Students will use the proportionality of similar triangles to solve problems. They will narrow their focus to similar right triangles in an applied context to justify repeated use of the same ratios, leading to the development of ratios in special right triangles and the use of the calculator to find trigonometric ratios for other triangles. They will apply these tools to solve problems. This mission is aligned with Common Core State Standards: HSG.SRT.C.6, HSG.SRT.C.8.
The fun stuff:
David and Goliath do battle across a river! David uses bearings and similar triangles to calculate the distance to Goliath, and Goliath tries to find trees that are a perfect height for crossing the river! Similarity, proportionality, special right triangles, trigonometric ratios all converge to solve real world problems! Comprehensive and engrossing!
The zip file includes:
★The mission. (12 pages).
★A two-page Quick Start guide for teachers. Detailed and helpful!
★Selected Answers in pdf form.
★Selected Answers online, password protected. For students to correct at home!
The mission includes:
•Using proportions relating sides of similar triangles to solve problems.
•Demonstration of the practicality of calculating and keeping ratios at hand for certain triangles.
•Ratios of sides in right triangles depends on the measure of a given acute angle.
•Derivation of ratios in special right triangles.
•Sine, cosine and tangent and their application to problems, with and without a calculator.
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\$3.00 | 659 | 3,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.891996 |
https://cs.nyu.edu/pipermail/fom/2023-February/023749.html | 1,708,865,632,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474595.59/warc/CC-MAIN-20240225103506-20240225133506-00013.warc.gz | 190,404,151 | 3,920 | # _Mathematics with the G structure_
I.V. Serov i.v.serov at chf.nu
Sun Feb 5 17:07:02 EST 2023
```Vaughan Pratt states in:
https://cs.nyu.edu/pipermail/fom/2023-January/023735.html
https://cs.nyu.edu/pipermail/fom/2023-January/023723.html
(1) "So far it seems to me that every objection to my proposal (repeated
below) concerning gapless geodesics depends on the
assumption/axiom/whatever that a nonempty linear order with no greatest
(or least) element must be an actual infinity."
(2) "What is bothering me is the idea that because the sequence is
infinite, it is therefore not something we can experience."
Nevertheless, indeed:
(3) "there is nothing inconsistent about dropping the axiom that a
nonempty linear order with no
least element is impossible to experience in practice."
(4) "One might complain that writing a real in binary is an actual
infinity.
To get around this, allow L and U to each be arbitrary finite sequences
of
bits, with U differing from L by adding 1 to the last bit of each
element
of U and doing the usual carry propagation in the finitely many bits of
that element."
No complains out here!
The "arbitrary finite sequences of bits" concatenated one after another
form an (actually) infinite string
- a model of a state of the universe G.
To be more precise: each string-state is infinite at an open side;
and it is finite at the other - at the closed side; and that means, that
each string-state of the universe G is well-founded with the least
(first) element at the closed side;
each string-state is an infinite sequence itself.
The difference between "actuality" and "potentiality" is a matter of
interpretation in a certain sense:
Is there actually no last bit at the open side in each state-string of
the universe G?
Or, is there only potentially no bit with no bit behind it at the open
side?
It also means, that each string-state is an instantiation of an infinite
sequence of natural numbers.
Do all these sequences form a set, whose elements, all-together, define
the Natural Numbers?
Is there actually no first state in the universe G, so that no state is
a non-successor state?
Or, is there only potentially no state with no state preceding it?
Whatever one experiences as finite is finite, because one experiences a
finite change from one state to another:
- like adding, removing, flipping or carry-propagating a bit of the
string.
That does not mean at all, that states of G themselves are finite or
only potentially infinite.
A move from one string to another - from one state of the universe to
the other - requires only a finite number of bit operations and that is
good indeed.
Compare to: *An infinite electron jumps from one infinite orbit to
another one and experiences a finite change so that the universe is now
in another string-state.*
It is imperative to confirm that no state-string in G can consist of
0-bits only, or 1-bits only;
no string-state is unary; all string-states are binary; this is an
actual universal dichotomy.
The structure G can be translated into a structure that is described by
all infinite continued fractions, such as [a0;a1,a2, ...],
where ai are natural numbers:
https://en.wikipedia.org/wiki/Continued_fraction.
This brings us right to the Euclidean algorithm, which is at the core of
all mathematics.
In this regard it is worth noting, that "no pattern has ever been found
in this representation" of number pi.
Compare to:
https://en.wikipedia.org/wiki/Generalized_continued_fraction#π.
Does this mean, that one needs more than two letters of the foundational
alphabet?
The proposal of G is an excellent one; there are no objections to it out
there,
except that it is necessary to recover enumeration as well, and that is:
- the structure is not only to be a dense order;
- it is also to be a discrete order; so that,
- there is a cross of the two orders - the cross order of the dense and
of the discrete.
And this means the following:
- there is no first/least string-state in the universe of string-states:
-- neither with respect to the enumerative discrete order of the
*universal time*,
-- nor with respect to the dense *matter* order;
- the structure is - actually (or at least potentially) -
non-well-founded with respect to both orders and this is a fundamental
requirement indeed.
Does one need to consider the structure G as an actual and/or as a
potential infinity?
Again, in a certain sense it is a matter of interpretation:
- actual infinity appears to be not all too different from the potential
one;
- and finiteness appears to be a special (periodic) case of infinity.
What is by far more important is that the universe of string-states is
to be countable in some generalized sense
and this requires crossing of the dense and the discrete orders.
Without enumeration the mathematical G is to place dice, would need to
make arbitrary choices and would be unable to justify own deeds. Is it
the genuine mathematical G then?
Here a model:
Actually, or,
potentially,
what
- there was;
- there is and
- there will be,
is the deterministic universal rule (nous, or Greek νοῦς), which makes
- a being in the prior string-state of the universe
to become
- the being of the posterior string-state of the universe;
by the model force of:
- the carry;
- the heredity and
- the flip,
and so, that offspring states are situated between the parent states
by the model force of
- love.
Not to forget to recurse!
Then and only then, the structure G is to be nothing else (up to an
isomorphism) than the structure X, the universal chain hereditary form;
and the theories of G and X are to survive together under the promising
name of "the current history of the future".
I.V. Serov
``` | 1,326 | 5,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.92672 |
insurgency.gameme.com | 1,534,649,069,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221214691.99/warc/CC-MAIN-20180819031801-20180819051801-00049.warc.gz | 218,558,917 | 9,455 | ## 1. How are "points" calculated?
A new player has 1000 points. Every time you make a kill, you gain a certain amount of points depending on a) the victim's points rating, and b) the weapon you used. If you kill someone with a higher points rating than you, then you gain more points than if you kill someone with a lower points rating than you. Therefore, killing newbies will not get you as far as killing the #1 player. And if you kill someone with your knife, you gain more points than if you kill them with a rifle, for example.
When you are killed, you lose a certain amount of points, which again depends on the points rating of your killer and the weapon they used (you don't lose as many points for being killed by the #1 player with a rifle than you do for being killed by a low ranked player with a knife). This makes moving up the rankings easier, but makes staying in the top spots harder.
Specifically, the equations are:
Killer Points = Killer Points + (Victim Points / Killer Points) × Weapon Modifier × 5
Victim Points = Victim Points - (Victim Points / Killer Points) × Weapon Modifier × 5
Plus, the following point bonuses are available for completing objectives in the following games:
Player Action PlyrPlyr Action Team Action World Action Action Player Reward Team Reward Ranking
Yes No No No Captured Objective +1 Bot massacre! ArmoryExpanded Theater
Yes No No No Destroyed Objective +1 Bot massacre! ArmoryExpanded Theater
Yes No No No Domination (4 kills) +3 Bot massacre! ArmoryExpanded Theater
Yes No No No Double Kill (2 kills) +1 Bot massacre! ArmoryExpanded Theater
Yes No No No God Like (12 kills) +11 Bot massacre! ArmoryExpanded Theater
Yes No No No Headshot Kill +1 Bot massacre! ArmoryExpanded Theater
Yes No No No Killing Spree (9 kills) +8 Bot massacre! ArmoryExpanded Theater
Yes No No No Mega Kill (6 kills) +5 Bot massacre! ArmoryExpanded Theater
Yes No No No Monster Kill (10 kills) +9 Bot massacre! ArmoryExpanded Theater
Yes No No No Ownage (7 kills) +6 Bot massacre! ArmoryExpanded Theater
Yes No No No Rampage (5 kills) +4 Bot massacre! ArmoryExpanded Theater
Yes No No No revived test 0 Bot massacre! ArmoryExpanded Theater
No No Yes No Round Win 0 +10 Bot massacre! ArmoryExpanded Theater
Yes No No No test1 0 Bot massacre! ArmoryExpanded Theater
Yes No No No test2 0 Bot massacre! ArmoryExpanded Theater
Yes No No No test3 0 Bot massacre! ArmoryExpanded Theater
Yes No No No Triple Kill (3 kills) +2 Bot massacre! ArmoryExpanded Theater
Yes No No No Ultra Kill (8 kills) +7 Bot massacre! ArmoryExpanded Theater
Yes No No No Unstoppable (11 kills) +10 Bot massacre! ArmoryExpanded Theater
Yes No No No Captured Objective +1 Bot massacre! Sernix Theater
Yes No No No Destroyed Objective +1 Bot massacre! Sernix Theater
Yes No No No Headshot Kill +1 Bot massacre! Sernix Theater
Yes No No No Healed Player +1 Bot massacre! Sernix Theater
Yes No No No Killing Spree (10 kills) +9 Bot massacre! Sernix Theater
Yes No No No Killing Spree (100 kills) +99 Bot massacre! Sernix Theater
Yes No No No Killing Spree (20 kills) +19 Bot massacre! Sernix Theater
Yes No No No Killing Spree (30 kills) +29 Bot massacre! Sernix Theater
Yes No No No Killing Spree (40 kills) +39 Bot massacre! Sernix Theater
Yes No No No Killing Spree (50 kills) +49 Bot massacre! Sernix Theater
Yes No No No Killing Spree (60 kills) +59 Bot massacre! Sernix Theater
Yes No No No Killing Spree (70 kills) +69 Bot massacre! Sernix Theater
Yes No No No Killing Spree (80 kills) +79 Bot massacre! Sernix Theater
Yes No No No Killing Spree (90 kills) +89 Bot massacre! Sernix Theater
Yes No No No Revived Player +1 Bot massacre! Sernix Theater
Yes No No No Revived Player (assisted) +1 Bot massacre! Sernix Theater
No No Yes No Round Win 0 +10 Bot massacre! Sernix Theater
Yes No No No Test1 +1 Bot massacre! Sernix Theater
## 2. What are all the weapon points modifiers?
Weapon points modifiers are used to determine how many points you should gain or lose when you make a kill or are killed by another player. Higher modifiers indicate that more points will be gained when killing with that weapon (and similarly, more points will be lost when being killed by that weapon). Modifiers generally range from 0.00 to 2.00.
Weapon Name Modifier Ranking
9MMSIDEARM 9mm Sidearm 1.00 Bot massacre! ArmoryExpanded Theater
ACR ACR 1.00 Bot massacre! ArmoryExpanded Theater
AK12U AK-12U 1.00 Bot massacre! ArmoryExpanded Theater
AK74 AK74 1.00 Bot massacre! ArmoryExpanded Theater
AKM AKM 1.00 Bot massacre! ArmoryExpanded Theater
AKS74U AKS74U 1.00 Bot massacre! ArmoryExpanded Theater
GRENADE_ANM14 ANM14 1.00 Bot massacre! ArmoryExpanded Theater
ASVAL AS Val 1.00 Bot massacre! ArmoryExpanded Theater
ROCKET_AT4 AT4 1.00 Bot massacre! ArmoryExpanded Theater
M107 Barrett M107 1.00 Bot massacre! ArmoryExpanded Theater
M1014 Benelli M4 1.00 Bot massacre! ArmoryExpanded Theater
NOVA Benelli Nova Tactical 1.00 Bot massacre! ArmoryExpanded Theater
BERETTA93R Beretta 93R 1.00 Bot massacre! ArmoryExpanded Theater
GRENADE_C4 C4 1.00 Bot massacre! ArmoryExpanded Theater
C96CARBINE C96 Carbine 1.00 Bot massacre! ArmoryExpanded Theater
CAR15 CAR-15 1.00 Bot massacre! ArmoryExpanded Theater
COLT9MM Colt 9mm 1.00 Bot massacre! ArmoryExpanded Theater
CM901 Colt CM901 1.00 Bot massacre! ArmoryExpanded Theater
COBRA Colt King Cobra 1.00 Bot massacre! ArmoryExpanded Theater
COMBATCOMMANDER Combat Commander 1.00 Bot massacre! ArmoryExpanded Theater
DEFIB Defibrillator 1.00 Bot massacre! ArmoryExpanded Theater
SVD Dragunov SVD 1.00 Bot massacre! ArmoryExpanded Theater
DRAGUNOVSVU Dragunov SVU 1.00 Bot massacre! ArmoryExpanded Theater
DOI2INS_SHOVEL Entrenching Tool 1.00 Bot massacre! ArmoryExpanded Theater
GRENADE_F1 F1 1.00 Bot massacre! ArmoryExpanded Theater
FAL FAL 1.00 Bot massacre! ArmoryExpanded Theater
FAMAS FAMAS 1.00 Bot massacre! ArmoryExpanded Theater
FIVESEVEN Five-seveN 1.00 Bot massacre! ArmoryExpanded Theater
F2000 FN F2000 1.00 Bot massacre! ArmoryExpanded Theater
SCAR FN SCAR-H 1.00 Bot massacre! ArmoryExpanded Theater
SCARL FN SCAR-L 1.00 Bot massacre! ArmoryExpanded Theater
G36C G36C 1.00 Bot massacre! ArmoryExpanded Theater
G3A3 G3A3 1.00 Bot massacre! ArmoryExpanded Theater
G33 G3A3 1.00 Bot massacre! ArmoryExpanded Theater
GALIL GALIL 1.00 Bot massacre! ArmoryExpanded Theater
GALIL_SAR GALIL SAR 1.00 Bot massacre! ArmoryExpanded Theater
GLOCK17 Glock 17 1.00 Bot massacre! ArmoryExpanded Theater
GLOCK18 Glock 18 1.00 Bot massacre! ArmoryExpanded Theater
GLOCK33 Glock 33 1.00 Bot massacre! ArmoryExpanded Theater
GOL GOL Magnum 1.00 Bot massacre! ArmoryExpanded Theater
GRENADE_GP25_HE GP25 1.00 Bot massacre! ArmoryExpanded Theater
GURKHA Gurkha 1.00 Bot massacre! ArmoryExpanded Theater
HKUSP H&K USP 1.00 Bot massacre! ArmoryExpanded Theater
HK417 HK417 1.00 Bot massacre! ArmoryExpanded Theater
GRENADE_IED IED 1.00 Bot massacre! ArmoryExpanded Theater
DEAGLE IMI Desert Eagle 1.00 Bot massacre! ArmoryExpanded Theater
KABAR KABAR 1.00 Bot massacre! ArmoryExpanded Theater
## 3. How can I edit my profile and set real name, e-mail address, homepage and ICQ number?
Player profile options can be configured by saying the appropriate gameme_set command while you are playing on a participating gameserver. To issue commands, push your chat key and type the command text.
Syntax: say /gameme_set option value
Acceptable "options" are:
• realname
Example: /gameme_set realname Joe Bloggs
• email
Example: /gameme_set email joe@hotmail.com
• homepage | 2,072 | 7,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-34 | longest | en | 0.867254 |
https://www.engineersedge.com/engineering-forum/search.php?s=310b62a2fc0f48445b4b2862391d01a4&searchid=1539278 | 1,582,096,998,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144058.43/warc/CC-MAIN-20200219061325-20200219091325-00522.warc.gz | 699,617,419 | 11,958 | # Search:
Type: Posts; User: Hudson
Page 1 of 7 1
1. ## Thread: Bolt tension on a cantilever beam?
by Hudson
Replies
7
Views
269
### You need help with the equations? Moment...
You need help with the equations?
Moment from arm = 100 kg x 650 mm = 65000 kg mm
(22.5 mm x Force on bottom bolts) + (152.5 mm x Force on top bolts) = 65000 kg mm in the opposite...
2. ## Thread: Bolt tension on a cantilever beam?
by Hudson
Replies
7
Views
269
### A ball park method: You have a moment on the...
A ball park method:
You have a moment on the beam that is transferred to the plate. Presumably you know the distance from the bolt holes to the bottom edge of the plate.
I would calculate...
3. ## Thread: Help with Centrifugal Force
by Hudson
Replies
6
Views
233
### You have not presented a clear picture of the...
You have not presented a clear picture of the problem or exactly how your equation came to be. We went from casting a cylinder to a rectangular tube of unknown dimensions and we added a speed but...
4. ## Thread: Help with Centrifugal Force
by Hudson
Replies
6
Views
233
### To convert rpm to radians per second, multiply by...
To convert rpm to radians per second, multiply by .10472
The obtain centrifugal force you need to square the rotating speed so to convert to rpm^2 use .10472^2 or .010966
You also need to convert...
5. ## Thread: Could electrolysis be an energy storage medium for our next energy future
by Hudson
Replies
3
Views
227
### Water molecules are very polarized and are pulled...
Water molecules are very polarized and are pulled apart by the opposite charges in the equipment you illustrate. It is a DC current not a frequency phenomenon.
You can store energy in this...
6. ## Thread: Drive line Calculations
by Hudson
Replies
1
Views
250
### Are you referring to the bending owing to...
Are you referring to the bending owing to gravitational force? If so, is the shaft uniform in dimensions so that you can treat it like a uniformly loaded beam? Then you probably want the shaft...
by Hudson
Replies
1
Views
198
### One method is to use a press fit and/or a...
One method is to use a press fit and/or a retaining ring on the shaft in the fixed location and have a slip or sliding fit on the floating end. In both cases the outer race is fixed.
8. ## Thread: Fundamentals of Design Engineering?
by Hudson
Replies
4
Views
473
### My review suggests the original post was...
My review suggests the original post was manufactured by rolling - in self-pity. The material is probably snowflake.
It is entirely possible that the interviewer was trying to be helpful...
9. ## Thread: Trying to join PVC parts together
by Hudson
Replies
2
Views
221
### Check out sonic welding. There is also some...
Check out sonic welding.
There is also some pretty good glue in the plumbing aisle.
10. ## Thread: Identifying a force
by Hudson
Replies
6
Views
273
### force on pins
The projecting pins will be pushed outward or inward by the mating holes.
This force exerts a bending moment on the pins attempting to rotate them.
The force also tries to move the pins away...
11. ## Thread: No. of Holes
by Hudson
Replies
5
Views
445
### Multiply the pressure times the area inside the...
Multiply the pressure times the area inside the gasket to get the force trying to push the cover off the valve. The total clamp load of the bolts should be several times this value. How many times...
12. ## Thread: Tank Considerations: Pressure vs Vacuum.
by Hudson
Replies
3
Views
270
### You'd think it would work just fine as the...
You'd think it would work just fine as the external pressure is only 1/10 of the rated internal pressure capacity.
13. ## Thread: identifying proper motor
by Hudson
Replies
3
Views
211
### Another option is a stepper motor. You have 12...
Another option is a stepper motor. You have 12 Volts. You'll need, in addition to the motor, a driver that regulates D.C.direction and power to the motor coils to determine position and motion plus...
14. ## Thread: Best Way to Dimension Circular Features
by Hudson
Replies
3
Views
281
### An observation: The holes on the smallest bolt...
An observation: The holes on the smallest bolt circle are near the weakest part and none of the holes are aligned for manufacturing.
15. ## Thread: Identifying a force
by Hudson
Replies
6
Views
273
### The deliberate misalignment of locating pins...
The deliberate misalignment of locating pins creates an "interference fit".
by Hudson
Replies
5
Views
712
### Can you machine a 10mm hex on the end of the 12mm...
Can you machine a 10mm hex on the end of the 12mm shaft and let maintenance use a socket wrench and extension?
17. ## Thread: Stainless steel floor drain
by Hudson
Replies
4
Views
1,000
### A file or tiny grinder on a bottom edge would...
A file or tiny grinder on a bottom edge would remove any plating and reveal the base material.
18. ## Thread: Direct steam injection into cooling water line at pump suction
by Hudson
Replies
3
Views
1,219
### You can inject steam heated water into the...
You can inject steam heated water into the discharge side of the pump or perhaps use the steam alone to pump the water.
steam engine boilers, especially on locomotives which did not recirculate...
19. ## Thread: Position tolerance with circularity
by Hudson
Replies
4
Views
809
### Why not tolerance a wall thickness to the O.D....
Why not tolerance a wall thickness to the O.D. that can be checked with a simple ball or tube micrometer instead of relying on CMM and inspection tables. Think like a machinist.
20. ## Thread: Is my tolerance stackup correct?
by Hudson
Replies
10
Views
986
### Do I see 'B' and 'C' datums as the edge of both a...
Do I see 'B' and 'C' datums as the edge of both a 3" and an 8" square part? Then you call out parts to be centered?
Why not make two bolt holes a common datum in the x & Y plane?
21. ## Thread: Help with design of rack
by Hudson
Replies
2
Views
962
### Sounds dangerous but, first determine if the...
Sounds dangerous but, first determine if the floor can handle the load. If you have two 1 foot wide, seven foot long supports at the base and 21,000 lbs supported by them that's 1500 lbs/sq.ft if...
22. ## Thread: Mounting a planar mirror with epoxy
by Hudson
Replies
2
Views
633
### Consider a urethane adhesive for windshields. ...
Consider a urethane adhesive for windshields.
23. ## Thread: how do they do it ?
by Hudson
Replies
2
Views
1,003
### Are they multi-million dollar dies? Years ago,...
Are they multi-million dollar dies? Years ago, when the earth was thought to be cooling, auto makers used a free machining, low melting alloy called Kirksite for fender dies.
More recently,...
24. ## Thread: How to Interpret Geometric Tolerance and Dimensional Tolerance
by Hudson
Replies
4
Views
1,995
### You have a .002 cylindricity and a roundness call...
You have a .002 cylindricity and a roundness call out. That seems like double dimensioning to me.
Could you just call out a .005 cylindricity and a surface finish on the basic shaft diameter and...
25. ## Thread: How much does the flow of a pipe get reduced if undersized couplings are used?
by Hudson
Replies
2
Views
872
### Look harder for 1" couplings. They should be...
Look harder for 1" couplings. They should be less than a dollar for the PVC type.
Results 1 to 25 of 169
Page 1 of 7 1 | 1,834 | 7,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-10 | latest | en | 0.88172 |
christayeung.ubcarts.ca | 1,568,652,897,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572879.28/warc/CC-MAIN-20190916155946-20190916181946-00489.warc.gz | 423,365,543 | 8,981 | # The Makings of Cartography: Proportional Symbol Map of Population in India
This is a proportional symbol map of major cities in India created through ArcGIS and Adobe Illustrator. I compare two methods of proportional symbol mapping: absolute and perceptual scaling.
This week I have been tasked with creating a proportional symbol map of major cities in India through ArcGIS and Adobe Illustrator. Proportional symbols visualize both location and quantity, and can be made through 3 methods:
• Absolute scaling: the size of symbols is in direct proportion to the value of the data. For example, if the value doubles, the symbol is also 2 times larger. This is how I have scaled the symbols in the map above.
• Perceptual scaling: the size of the symbols is slightly larger than the value it is representing. For example, if the value doubles, the symbol is a bit over 2 times larger. This method was created as a result of psychological research that showed how readers perceive maps and tend to underestimate the value represented in circle symbols.
• Range grading: Data is separated into category ranges with symbols that fit each category.
#### Perpetual Scaling vs. Absolute Scaling
In this lab, I created 2 maps of Indian cities with populations of over 2 million in both perpetual and absolute scaling methods. As a cartographer, it is important to always ask yourself, "what can the reader understand through this map?" and whether you are effectively communicating your message.
So which method is more effective? Here is a list of pros and cons for both.
For my own "Cities in India" map, I think absolute scaling is the better method to communicate the data set because the symbols are easy to match up to the three population categories in the legend and the map is fairly simple for readers to comprehend. As well, in the perceptual scaling method, the symbols are smaller overall and text is harder to read.
This is my last assignment before my final project and I'm excited to decide on my own map idea!
##### Here are some skills I picked up in summary:
• Acquired UN World Urbanization Prospects 2014 Microsoft Excel table of urban agglomerations in India
• Calculated the size of symbols appropriate to each data value using Excel to accurately design proportional symbols
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# TutorMe Blog
## Why Are Repeating Decimals Rational Numbers?
Inactive
Jana Russick
May 11, 2021
Are repeating decimals rational? The answer is yes. But before we talk about why, let's review rational numbers. A rational number is a fraction in its lowest term. It's written in form a/b, where both a and b are integers, and b is a non-zero denominator.
Now, let’s talk about why repeating decimals are considered rational numbers.
## Are Repeating Decimals Rational?
Repeating or recurring decimals are decimal representations of numbers with infinitely repeating digits. Numbers with a repeating pattern of decimals are rational because when you put them into fractional form, both the numerator a and denominator b become non-fractional whole numbers.
For example, when you use long division to divide 1 by 3, the resultant quotient is 0.33333…. However, when put it into fractional form, it's made of positive integers that don’t have decimal points:
In infinite decimal expansion, decimal digits repeat on forever with no end. Numbers with repeating decimals can have an overline above the last number:
The overline is an easier, shorter way to indicate infinite decimal expansion without having to write a bunch of repeating numbers. Though repeating digits like this don't seem like rational numbers, they can take the form of a rational expressions when converted to their fraction form:
This is because the repeating part of this decimal no longer appears as a decimal in rational number form. Instead, it’s represented by non-repeating, natural numbers 4 and 9. Remember — irrational numbers cannot be written as fractions.
## What Is a Terminating Decimal?
When converted to decimal form, some rational numbers have a terminating decimal. This means that there's a finite number of digits after the decimal point:
The terminating decimal of this rational number is 5 since no decimal numbers follow it.
### What Is a Non-Terminating Decimal?
You can never count the decimal places of a number with repeating decimals. Because of this, repeating decimals are called non-terminating decimals.
Here's an example of a non-terminating decimal:
In this sequence of digits, the same number combination of 09 will be infinitely repeated.
## Repeating Decimals Are Rational
Because rational numbers are used at all levels of math, it's important to know what makes a number rational. It might not seem like numbers with repeating decimals are rational numbers. But when you use your algebraic skills to convert repeating decimals into the fractional form a/b, you prove that repeating decimals are indeed rational.
### More Math Homework Help
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# Monday, 5/10Tuesday, 5/11Wednesday, 5/12Thursday, 5/13Friday, 5/14 Graphing & Properties of Quadratic Functions HW#1 Graphing & Properties of Quadratic.
## Presentation on theme: "Monday, 5/10Tuesday, 5/11Wednesday, 5/12Thursday, 5/13Friday, 5/14 Graphing & Properties of Quadratic Functions HW#1 Graphing & Properties of Quadratic."— Presentation transcript:
Monday, 5/10Tuesday, 5/11Wednesday, 5/12Thursday, 5/13Friday, 5/14 Graphing & Properties of Quadratic Functions HW#1 Graphing & Properties of Quadratic Functions HW#1 Solving Quadratic Equations by Graphing Path of a Baseball HW#2 ½ Day: B Activity on big graph paper: Graphing Quadratics HW#3 (quiz) TI-84 Graphing Calculator Investigation Activity: Transformations of Quadratics HW#4 Monday, 5/17Tuesday, 5/18Wednesday, 5/19Thursday, 5/20Friday, 5/21 Solving Quadratic Equations by Using Completing the Square HW#5 Solving Quadratic Equations by Using Completing the Square ½ Day: A Solving Quadratic Equations by Using the Quadratic Formula HW#6 Quiz: Completing the Square & Quadratic Formula Additional practice – quadratic formula & completing the square HW#7 Monday, 5/24Tuesday, 5/25Wednesday, 5/26Thursday, 5/27Friday, 5/28 Review for test Test: Factoring & Quadratic Functions ???Rocket Project???
Graphs of Quadratics
Terms of a quadratic y = ax 2 + bx + c Every quadratic has terms: Quadratic term: ax 2 Linear term: bx Constant term: c When the power of an equation is 2, then the function is called a quadratic a, b, and c are the coefficients Standard form of a quadratic
Graphs of Quadratics The graph of any quadratic equation is a parabola To graph a quadratic, set up a table and plot points Example: y = x 2 x y -2 4 -1 1 0 0 1 1 2 2..... x y y = x 2
Finding the solutions of a quadratic 2. Find the values of x that make the equation equal to 0 1)Algebraically (last week and next slide to review) 2)Graphically (today next slide) 1. Set y of f(x) equal to zero: 0 = ax 2 + bx + c In general equations have roots, Functions haves zeros, and Graphs of functions have x-intercepts
Directions: Find the zeros. Ex: f(x) = x 2 – 8x + 12 Factor and set y or f(x) = 0 (x – 2)(x – 6) = 0 x – 2 = 0 or x – 6 = 0 x = 2 orx = 6 Factors of 12 Sum of Factors, -8 1, 12 13 2, 6 8 3, 4 7 -1, -12 -13 -2, -6 -8 -3, -4 -7
Characteristics of Quadratic Functions The shape of a graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum. The highest point on the graph is the maximum. The maximum or minimum is the vertex
Axis of symmetry. x-intercept. vertex y-intercept x y Characteristics of Quadratic Functions To find the solutions graphically, look for the x-intercepts of the graph (Since these are the points where y = 0)
Key Concept: Quadratic Functions Parent Functionf(x) = x 2 Standard From f(x) = ax 2 + bx + c Type of GraphParabola Axis of Symmetry y-interceptc
Axis of symmetry examples http://www.mathwarehouse.com/geometry/ parabola/axis-of-symmetry.php
Vertex formula x = -b 2a Steps to solve for the vertex: Step 1: Solve for x using x = -b/2a Step 2: Substitute the x-value in the original function to find the y-value Step 3: Write the vertex as an ordered pair (, )
Example 1: HW Prob #11 Find the vertex: y = 4x 2 + 20x + 5 a = 4, b = 20 x = -b = -20 = -20 = -2.5 2a 2(4) 8 y = 4x 2 + 20x + 5 y = 4(-2.5) 2 + 20(-2.5) + 5 = -20 The vertex is at (-2.5,-20)
Example 2 Find the vertex: y = x 2 – 4x + 7 a = 1, b = -4 x = -b = -(-4) = 4 = 2 2a 2(1) 2 y = x 2 – 4x + 7 y = (2) 2 – 4(2) + 7 = 3 The vertex is at (2,3)
Example 3: HW Prob #14 Find the vertex: y = 5x 2 + 30x – 4 a = 5, b = 30 x = -b = -30 = -30 = -3 2a2(5) 10 y = 5x 2 + 30x – 4 y = 5(-3) 2 + 30(-3) – 4 = -49 The vertex is at (-3,-49)
Example 4 Find the vertex: y = 2(x-1) 2 + 7 Answer: (1, 7)
Example 5 Find the vertex: y = x 2 + 4x + 7 a = 1, b = 4 x = -b = -4 = -4 = -2 2a 2(1) 2 y = x 2 + 4x + 7 y = (-2) 2 + 4(-2) + 7 = 3 The vertex is at (-2,3)
Example: y = x 2 – 4 (HW Prob #1) x y y = x 2 - 4 2. What is the vertex (, ) 4. What are the solutions: (x-intercepts) 3. What is the y-intercept: 1. What is the axis of symmetry? x y -2 0 -1 -3 0 -4 1 -3 2 0 (0, -4) x = -2 or x = 2 -4 x = 0
Example: y = -x 2 + 1 (HW Prob #3) x y y = -x 2 + 1 2. Vertex: (0,1) 3. x-intercepts: x = 1 or x = -1 4. y-intercept: 1 1. Axis of symmetry: x = 0 x y -2 -3 -1 0 0 1 1 0 2 -3
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### Summing a Binary Function Sequence
```
Date: 07/16/98 at 15:56:16
From: Mark Ping
Subject: Calculate the sum of a sequences
Dear Dr. Math,
Here is the question:
Compute
oo
----
\ B(n)
/ --------
---- n(n+1)
n=1
where B(n) denotes the sum of the binary digits of n.
Thanks,
Mark Ping
```
```
Date: 07/19/98 at 22:56:18
From: Doctor Pete
Subject: Re: Calculate the sum of a sequences
Hi,
Interesting question! The key to solving this problem is to find a
relation describing the function B(n). In particular, writing out a
table of B(n) vs. n for small n helps:
n : 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
B(n): 0 1 1 2 1 2 2 3 1 2 2 3 2 3 3 4
From this, we might guess that B(2n) = B(n). Thinking about numbers in
their binary representations proves this is true, since if we write n
as:
n = a[k]2^k + a[k-1]2^(k-1) + ... + a[1]2^1 + a[0]
Then a[k], a[k-1], ... a[1], a[0] are the binary digits of n, and:
B(n) = a[0] + a[1] + ... + a[k]
But:
2n = a[k]2^(k+1) + a[k-1]2^k + ... + a[1]2^2 + a[0]2^1 + 0
so B(2n) = B(n). In essence, multiplying n by 2 simply shifts the
binary digits of n one place to the left, and adds a 0 in the units
digit. Furthermore, from this analysis we see that:
B(2n+1) = B(2n) + 1
since our previous discussion shows that B(2n) has 0 as the rightmost
digit.
So, with this in mind, let us write:
S = Sum[B(n)/(n(n+1)), {n,1,Infinity}]
That is, let S be the sum of B(n)/(n(n+1)) from n = 1 to infinity.
Then
S = 1/2 + Sum[B(n)/(n(n+1)), {n,2,Infinity}]
= 1/2 + Sum[B(2m)/(2m(2m+1)) + B(2m+1)/((2m+1)(2m+2)),
{m,1,Infinity}]
= 1/2 + Sum[(B(2m)(m+1)+(B(2m)+1)m)/(2m(m+1)(2m+1)),
{m,1,Infinity}]
= 1/2 + Sum[(B(2m)(2m+1) + m)/(2m(m+1)(2m+1)), {m,1,Infinity}]
= 1/2 + Sum[B(2m)/(2m(m+1)), {m,1,Infinity}]
+ Sum[1/((2m+1)(2m+2)), {m,1,Infinity}]
= 1/2 + 1/2 Sum[B(m)/(m(m+1)), {m,1,Infinity}] + (2 Log[2] - 1)/2
= S/2 + Log[2].
It follows that S/2 = Log[2], or S = Log[4]. Here Log means a logarithm
with base e.
Note I skipped some steps in the above. On the 5th step I separated the
sum into two components, which I was allowed to do because I knew
1 < S < 2 (can you figure out why this is true?). Also, in the 6th
step, I evaluated the second sum:
Sum[1/((2m+1)(2m+2)), {m,1,Infinity}] = (2 Log[2] - 1)/2
where Log[2] is the natural logarithm of 2. This was done by seeing
that this sum is actually an alternating series, since:
1/((2m+1)(2m+2)) = 1/(2m+1) - 1/(2m+2)
and recognizing that Sum[x^k/k, {k,1,Infinity}] = -Log[1-x].
Substituting x = -1 and moving the first few terms around gives the
desired simplification. Finally, the last step was observing that we
managed to simplify the first sum on the right into S itself, thereby
giving an equation for S. Indeed, we find numerically that:
1 < Log[4] =~ 1.39 < 2.
Note: =~ means is approximately equal to.
This is a particularly difficult problem, because it is not at all
obvious how to simplify the summand since we're given so little
information about the properties of B(n). Furthermore, there is no
guarantee that the relations B(2n) = B(n), and B(2n+1) = B(2n)+1 will
be of any use in simplifying the problem. (For example, there are other
properties of B(n), namely B(2^m + k) = B(k) + 1 for nonnegative
integers m, k. But this does not help us because it is not immediately
clear how to take some integer n and express it in the form 2^m + k,
where m is as large as possible.) Fortunately, in the process of
manipulating the sum, we find that these properties indeed work to our
- Doctor Pete, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory
High School Sequences, Series
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Problem Sets Information Select Problem Runs Ranklist
ZOJ Problem Set - 2603
Time Limit: 2 Seconds Memory Limit: 32768 KB Special Judge
Consider the railroad station that has n dead-ends designed in a way shown on the picture. Dead-ends are numbered from right to left, starting from 1.
Let 2n railroad cars get from the right. Each car is marked with some integer number ranging from 1 to 2n, different cars are marked with different numbers.
You can move the cars through the dead-ends using the following two operations. If the car x is the first car on the path to the right of the dead-end i, you may move this car to this dead-end. If the car y is the topmost car in the dead-end j you can move it to the path on the left of the dead-end. Note, that cars cannot be moved to the dead-end from the path to its left and cannot be moved to the path on the right of the dead-end they are in.
Your task is to rearrange the cars so that the numbers on the cars listed from left to right were in the ascending order and all the cars are to the left of all the dead-ends.
One can prove that the required rearranging is always possible.
Input
The input contains multiple test cases. Each test case occupies two lines. The first line of each case contains n - the number of dead-ends (1 <= n <= 13). The second line contains 2n integer numbers - the numbers on the cars, listed from left to right.
A case with n = 0 ends up the input file.
Output
For each case, output the sequence of operations in one line. Each operation is identified with the number of the car moved in this operation. The type of the operation and the dead-end used are clearly determined uniquely.
Sample Input
```2
3 2 1 4
2
1 2 3 4
0
```
Sample Output
```3 3 2 2 1 1 4 4 3 2 1 1 2 3 4 4
1 2 2 1 2 1 3 3 4 4 1 2 3 3 4 4
```
Author: Andrew Stankevich
Source: Andrew Stankevich's Contest #6
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# Percent Error Calculations Chemistry
## Contents
View all posts by Todd Helmenstine → Post navigation ← Direct Image Of Exoplanet Sets New Record Using Stem Cells and Herpes To Fight Brain Cancer → 3 thoughts on “Calculate Reviews Back to the top of the page ↑ ABOUT Our Mission Meet the Team Partners Press Careers Community Success Stories Blog Overview CK-12 Usage Map SUPPORT Webinars Implementation Guide Pilot Toevoegen aan Wil je hier later nog een keer naar kijken? How should this error be corrected? 1 month ago 0 Answers What is the process of deliberately adding very small amount of impurities to a pure substance called? 1 month ago
## Percent Error Chemistry Definition
CH 3 CHEMISTRY MEASUREMENT, ACCURACY & PRECISION by BRIAN M. Reply ↓ Leave a Reply Cancel reply Search for: Get the Science Notes Newsletter Get Projects Free in Email Top Posts & Pages Printable Periodic Tables List of Words Made from Nearly all of the graphics are created in Adobe Illustrator, Fireworks and Photoshop. Change Equation to Percent Difference Solve for percent difference.
If a measurement is accurate, that means that it's close to the actual value of the thing being measured. Now try calculating the following percentage uncertainties... 1.00 g on a 2 decimal place balance 10.00 g on a 2 decimal place balance 1.00 g on a 3 decimal place balance The circumference of the 2p coin is therefore 81 mm. What Is A Good Percent Error Things get spilled, things are impure, equipment is imprecise...
Arithmetical procedures can lead to uncertainty... What is the student's percent error? 3 days ago 0 Answers What is percent error/percent deviation? 1 week ago 0 Answers You measure the mass of a model car to be Laden... The order does not matter if you are dropping the sign, but you subtract the theoretical value from the experimental value if you are keeping negative signs.
answered · 2 weeks ago 1 Answer What is precision error in chemistry? 2 weeks ago 0 Answers How do you find the percent error of the measurement 11.56 cm? Percent Error Worksheet Inputs: measured valueactual, accepted or true value Conversions: measured value= 0 = 0 actual, accepted or true value= 0 = 0 Solution: percent error= NOT CALCULATED Change Equation Variable Select to Chemistry Chemistry 101 - Introduction to Chemistry Chemistry Tests and Quizzes Chemistry Demonstrations, Chemistry Experiments, Chemistry Labs & Chemistry Projects Periodic Table and the Elements Chemistry Disciplines - Chemical Engineering and William Habiger 15.582 weergaven 5:44 Percentage Trick - Solve precentages mentally - percentages made easy with the cool math trick! - Duur: 10:42.
## Can Percent Error Be Negative
Some measurement uncertainties are given below: EquipmentMeasurement to the nearest: Balance (1 decimal place)0.08 g Balance (2 decimal place)0.008 g Balance (3 decimal place)0.0008 g Measuring Cylinder (25 cm3)0.5 cm3 Graduated
Whether error is positive or negative is important. Percent Error Chemistry Definition Read the number from left to right and count all the digits starting with the first digit that is not zero. Percent Error Definition Solve for percent error Solve for the actual value.
Pendragon answered · 1 week ago 1 Answer What factors, other than the measuring instrument, could affect the precision of your measurement? 1 week ago 0 Answers Does accuracy refers to More about the author Please enter a valid email address. Please select a newsletter. Probeer het later opnieuw. Negative Percent Error
Sluiten Meer informatie View this message in English Je gebruikt YouTube in het Nederlands. Kick Images, Getty Images By Anne Marie Helmenstine, Ph.D. answered · 1 month ago 1 Answer What is the error in the experimental set-up? check my blog Inloggen 21 Laden...
misterguch · 1 · 4 comments · Mar 24 2014 Why is percent error important? Significant Figures Definition Chemistry How accurate is a volumetric flask as a measuring instrument? Jayde answered · 1 month ago 1 Answer Which cylinder would be more accurate for measuring 5 ml of solution: the 100 ml cylinder or the 10 ml cylinder?
## Inloggen Transcript Statistieken 37.869 weergaven 70 Vind je dit een leuke video?
Imperfections in experimental procedures. Email check failed, please try again Sorry, your blog cannot share posts by email. Log in om je mening te geven. Under What Condition Will Percentage Error Be Negative Why is this a true statement?
How can you express this age in scientific notation with the lowest level of precision? 1 month ago 0 Answers What term describes how close a measurement is to the true MeneerNask answered · 2 weeks ago 1 Answer How are accuracy and precision evaluated? 2 weeks ago 0 Answers What is the degree to which a measurement approaches a standard value? For example,, in experiments involving yields in chemical reactions, it is unlikely you will obtain more product than theoretically possible.Steps to calculate the percent error:Subtract the accepted value from the experimental value.Take http://back2cloud.com/percent-error/percentage-error-calculations-chemistry.php Deze functie is momenteel niet beschikbaar.
Calculate the percent error of your measurement.Subtract one value from the other:2.68 - 2.70 = -0.02 Depending on what you need, you may discard any negative sign (take the absolute value): 0.02This Thank you,,for signing up! Updated September 14, 2016. The difference between the actual and experimental value is always the absolute value of the difference. |Experimental-Actual|/Actualx100 so it doesn't matter how you subtract.
Stacie Sayles 3.599 weergaven 8:34 CH 3 CHEMISTRY DETERMINING ERROR - Duur: 6:15. History World History ... About Todd HelmenstineTodd Helmenstine is the physicist/mathematician who creates most of the images and PDF files found on sciencenotes.org. The error comes from the measurement inaccuracy or the approximation used instead of the real data, for example use 3.14 instead of π.
However, the overall calibration can be out by a degree or more. Concept Nodes: SCI.CHE.133.3 (Percent Error) ShowHide Resources Save or share your relevant files like activites, homework and worksheet.To add resources, you must be the owner of the Modality. Solve for the measured or observed value.Note due to the absolute value in the actual equation (above) there are two solutions. Report an issue.
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https://www.tinkutara.com/2023/06/03/question-134110/ | 1,713,656,224,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817688.24/warc/CC-MAIN-20240420214757-20240421004757-00141.warc.gz | 941,326,736 | 13,495 | $$\mathrm{log}\:\left({LHS}\right) \\$$$$=\mathrm{log}\:\frac{{y}}{{z}}×\mathrm{log}\:{x}+\mathrm{log}\:\frac{{z}}{{x}}×\mathrm{log}\:{y}+\mathrm{log}\:\frac{{x}}{{y}}×\mathrm{log}\:{z} \\$$$$=\left(\mathrm{log}\:{y}−\mathrm{log}\:{z}\right)\mathrm{log}\:{x}+\left(\mathrm{log}\:{z}−\mathrm{log}\:{x}\right)\mathrm{log}\:{y}+\left(\mathrm{log}\:{x}−\mathrm{log}\:{y}\right)\mathrm{log}\:{z} \\$$$$=\mathrm{log}\:{y}\:\mathrm{log}\:{x}−\mathrm{log}\:{z}\:\mathrm{log}\:{x}+\mathrm{log}\:{z}\:\mathrm{log}\:{y}−\mathrm{log}\:{x}\:\mathrm{log}\:{y}+\mathrm{log}\:{x}\:\mathrm{log}\:{z}−\mathrm{log}\:{y}\:\mathrm{log}\:{z} \\$$$$=\mathrm{0} \\$$$$\Rightarrow{LHS}=\mathrm{1} \\$$ | 289 | 676 | {"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} | 3.53125 | 4 | CC-MAIN-2024-18 | longest | en | 0.2156 |
https://www.jiskha.com/display.cgi?id=1348001395 | 1,516,448,683,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889567.48/warc/CC-MAIN-20180120102905-20180120122905-00379.warc.gz | 953,055,247 | 3,933 | # pre calculus II
posted by .
describe the domain, range,asymptotes and if the graph increase or decrease, and is this is a funtion from this equation y=5-3^x
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Can someone please do these questions step by step and explain them. I really need help so I can do the same for the rest. a) f(x) = e^(sqrtx) 1. State the domain and range 2. Find f'(x) 3. Determine the intervals of increase and decrease. …
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More Similar Questions | 658 | 2,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-05 | latest | en | 0.839868 |
https://www.scirra.com/forum/invert-a-number_t174050?view=print | 1,527,130,205,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865884.49/warc/CC-MAIN-20180524014502-20180524034436-00017.warc.gz | 832,748,931 | 1,816 | Page 1 of 1
### Invert a number
Posted: Tue May 24, 2016 10:42 am
ok so Ive been thinking, is there anyway to invert a number which is in a variable, like if the varible is set go from 0-1000 is there a way to invert it somehow to output to start at 1000 and go to 0?
### Re: Invert a number
Posted: Tue May 24, 2016 10:46 am
does it work if you just multiplicate the value with -1?
Value * (-1) = invertedValue?
Im a total math noob so dont be mad if its wrong. | 146 | 468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-22 | latest | en | 0.903087 |
https://blog.mrmeyer.com/2009/what-can-you-do-with-this-projectile-motion/ | 1,713,321,702,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817128.7/warc/CC-MAIN-20240417013540-20240417043540-00124.warc.gz | 115,378,385 | 13,949 | ## What Can You Do With This: Projectile Motion
Look carefully.
BTW: Lotta good stuff in the comments. I can’t prove any suggestion is better than another, but Jackie hits the one I intended all along, the one that packs the most punch per word, the one that rides into class alongside a student’s intuitive understanding of the world, the one that will do the most good as a introduction to parabolas:
Will the ball make it into the can?
Any question heavier than that and the picture starts to get a little wobbly under the weight.
BTW: Here are the still photos we imported into Geogebra and the full videos we used to confirm our answers.
I'm Dan and this is my blog. I'm a former high school math teacher and current head of teaching at Desmos. He / him. More here.
1. #### Jackie Ballarini
March 23, 2009 - 7:00 am -
Will the ball make it into the can? When is the height of the ball changing the fastest? The slowest? (although I think this picture would be easier to work with for those derivative questions). Graph the ball’s velocity vs. time. Estimate the horizontal distance the ball traveled. The vertical distance.
Waiting to hear what others have to add to this.
Nice work Dan!
2. #### Jovan
March 23, 2009 - 7:04 am -
We’ve got constant vs. variable rates of change…which could lead into a discussion of what it really means when someone says a car ( or other object moving through space ) is traveling x miles per hour.
We’ve got positive vs. negative slope and what that actually means in real space and time.
Interesting photo.
3. #### David Cox
March 23, 2009 - 7:34 am -
Copy the pic to SmartNotebook and give them a ruler with which to interact.
If the thrower is 6’7″, what is the ball’s highest point?
How far does the ball travel (horizontally)? When will the ball hit the ground?
What was the initial (vertical) velocity of the ball?
What was the initial horizontal velocity?
Does the vertical velocity remain constant? Why?
Does the horizontal velocity remain constant? Why?
Describe how the slopes between the balls is changing? Will the slope ever equal zero? What does that mean?
If we can determine the horizontal and vertical components, what was the actual velocity when the ball left the thrower’s hand?
4. #### Steven Peters
March 23, 2009 - 9:39 am -
This is really nice, Dan.
As Jackie said, Will the ball make it to the can? That’s a nice way to motivate fitting the data points to a parabola and extrapolate to the can.
Another thing I really like is seeing how the balls get closer together as it flies. Is it reasonable for a teacher to point that out and wonder why? You don’t have to tell them about frame rate or the speed of the ball, but see if they come up with that?
Do you have another version of this image with a grid superimposed?
Perhaps you could try tweaking the contrast so the ball stands out more?
Maybe another photo with the maestro running to catch the ball he’s thrown? You probably couldn’t superimpose yourself at 30 frames per second, but it would be funny nonetheless.
5. #### Kate
March 23, 2009 - 10:15 am -
Cool picture!
I find myself wanting to reflect the path over a vertical line to see if it goes in, but it’s not obvious where the vertex is going to be. So I’m thinking we’d need a grid and a regression.
If possible can you quickly describe how you made this?
Thanks!
6. #### Dan Meyer
March 23, 2009 - 8:39 pm -
We did this today. A lot of fun. Basically, for all the good questions out here, only one of them is all that visceral, only one gloms onto the sticky part of a student’s brain. “Does the ball hit the can?”
I have four variations on this theme. One that falls short, one that goes long, one that goes in, and one that looks like it goes in but illustrates the problem with 2D projections of a 3D space.
We imported stills into Geogebra and modeled a parabola using sliders. Really effective.
Kate, I filmed video using a little FlipCam and then exported a picture sequence from QuickTime. I imported all those frames into a Photoshop file and masked each frame down. Time consuming, but not hard.
7. #### jeff
March 24, 2009 - 12:46 am -
awesome shot. this is a lesson in a jpg.
dan, is it easy to make sure that the stills you get from the movie are spaced out evenly by time?
8. #### Dan Meyer
March 24, 2009 - 11:16 am -
Yeah, the camera does all that, recording 30 frames per second. Every frame is up there.
9. #### Andrew Ziobro
March 24, 2009 - 6:03 pm -
Great photo and idea. I was playing with you geogebra files and was wondering why you only fixed one corner of the photo.
Did you discuss or attempt to scale the photo so that the equation was in “real” world units?
10. #### Dan Meyer
March 25, 2009 - 6:51 am -
I fixed one corner of the photo because I didn’t want my students to inadvertently move the photo around, nothing more than that. I’m pretty woefully bad with Geogebra, actually.
11. #### Touzel
March 25, 2009 - 10:51 am -
Dan, is that a football field or a soccer field? On the photo, the horizontal line that your right foot is on appears to extend roughly 3/5 of the way to the trash can (where it makes a right angle with a line extending away from the camera) when it appears to end. Is that a penalty area on a soccer field? Also, are you a lefty or did you flip the photo?
12. #### Dan Meyer
March 25, 2009 - 4:16 pm -
Soccer field, I think, but those lines weren’t much of anything. Something for P.E. I imagine. And I throw left.
13. #### MrTeach
March 26, 2009 - 5:00 pm -
I shared this with my elementary students yesterday. They loved it and came up with great questions.
My favorite was a student who wanted to know about the wind. He noticed the clouds in the background and said it looks like a storm brewing, that means the wind is coming. He also noted that a tennis ball is light and would be greatly effected by wind.
I got an easy 20 minutes worth of great discussion out of this one picture. Keep them coming.
14. #### John
March 28, 2009 - 11:43 am -
How do you take the video and then flatten in so that the first 15 frames appear in one image? I can imagine doing this in photoshop, but it seems as though I’d have to export each frame and then merge them, which is rather time consuming.
15. #### Steven Peters
March 29, 2009 - 11:51 am -
This is kind of an aside, but it might be interesting.
I keep seeing this Bud Light commercial from the Super Bowl where a guy gets thrown out a window suddenly. The other day I noticed that he doesn’t fall parabolically. His chair falls away from him, but he appears to fall in a straight line, on some sort of cable I imagine.
I just spent a little time trying to find this clip in a downloadable, editable manner but have come up short. If I could I would take a crack at editing some photos to illustrate it. At any rate, it’s another media source to illustrate what should be but isn’t parabolic motion. Maybe you can show it and then ask if they seen anything wrong with it, then show the actual path taken by the falling man/dummy and indicate how it really happened.
16. #### tiredoldcliche
March 31, 2009 - 10:04 am -
Just thought i’d pop in to say i really enjoyed this one – used it with my kids in sunny England and they were well enthused. We went a step further and videoed them shooting free throws, and then trying to find the curve of their shot (similarly, working out if i’d used the shot they sank or the one they missed)
One of those kids pointed me in the direction of vidshell (http://cripe03.rug.ac.be/Vidshell/Vidshell.htm) which enabled me to make photos like the one at the top in very little time at all – just upload the vid, and trace the path whilst it plays.
Hope this reduces the time consuming aspect if you do anything like this again.
17. #### Ashli
April 1, 2009 - 1:56 pm -
Hi Dan,
Wanted to let you know that I had a great Friday-before-spring-break lesson around this picture with a class that had just learned about modeling quadratics and creating equations from three points (both matrix-wise and using the calculator).
I do have a SMARTBoard, so I had students up at the board marking up the picture, trying to figure out where to put coordinates, having me overlay a grid, and debating if real life scale matters (I noted how tall you are to give them a sense of scale for the photo and see if they used that or their own measuring stick). Didn’t have quite enough time to finish, but I’ll be splitting them up on Monday with the other pictures to have them tackle it again and share answers.
Great day in math :) Thanks!
18. #### Dan Meyer
April 1, 2009 - 5:07 pm -
As usual, you people deploy my own stuff in the classroom better than I do.
19. #### Jazo
May 4, 2009 - 2:57 pm -
Today in physics class, our teacher gave us our final exam project. We are in charge of teaching a certain chapter in our book. After looking at this a month ago, and seeing the option of “Projectile Motion”, I immediately took it. I’m using this in my presentation.
This is some amazing stuff, I must say.
20. #### Dan Meyer
May 4, 2009 - 4:00 pm -
Rip it off. If it helps at all, here is a photo of me holding up 2 feet. You know, for reference.
21. #### Jazo
May 4, 2009 - 10:21 pm -
Question – for every number on the set of 4 photos you have, is that 2 feet, a yard, or a meter?
22. #### Dan Meyer
May 5, 2009 - 3:22 pm -
The lines on the grass? To the best of my knowledge they indicate nothing. I think it was a soccer field.
23. #### Jazo
May 14, 2009 - 3:52 pm -
BTW, if it’s not asking too much, do you have a video of the outcome? | 2,396 | 9,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-18 | latest | en | 0.938981 |
http://fenkoil.ru/1286/ | 1,624,021,889,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487636559.57/warc/CC-MAIN-20210618104405-20210618134405-00173.warc.gz | 17,306,615 | 9,810 | # Floating action
Floating control is so called because the final control element floats in a fixed position as long as the value of the controlled variables lies between two chosen limits. When the value of the controlled variable reaches the upper of these limits, the final control element is actuated to open, say, at a constant rate. Suppose that the value of the controlled variable then starts to fall in response to this movement of the final control element. When it falls back to the value of the upper limit, movement of the final control element is stopped and it stays in its new position, partly open. It remains in this position until the controlled condition again reaches a value equal to one of the limits. If the load alters and the controlled variable starts to climb again, then, when the upper limit is again reached, the final control element will again start to open and will continue to open until either it reaches its maximum position or until the controlled variable falls again to a value equal to the upper limit, when movement will cease. If the load change is such that the controlled variable drops in value to below the lower limit, say, then the final control element will start to close, and so on. Thus, the final control element is energised to move in a direction which depends on the deviation; a positive deviation gives movement of the element in one direction and a negative deviation causes movement in a reverse direction. There is a dead or floating band between the two limits which determines the sign of the deviation.
Figure 13.6 illustrates an example of floating control. A room is ventilated by a system as shown. The capacity of the LTHW heater battery is regulated by means of a motorised two-port valve, Rl, controlled by floating action from a thermostat, Cl, located in the supply air duct. There is negligible lag between Cl and Rl. It is assumed that the variation in the state of the outside air is predictable and so the response rate of the controller can be properly chosen.
When the load is steady, the valve is in a fixed position, indicated by the line AB in Figure 13.6(6), provided that the supply air temperature lies between the upper and lower limits, as indicated by the line AA’, in Figure 13.6(c). Under this steady-state condition, the capacity of the heater battery matches the load. If the outside air temperature increases, but the valve position remains unchanged, the supply air temperature starts to rise, as shown by the line A’B in Figure 13.6(c). When the air temperature in the supply duct attains a value equal to the upper limit (point B), the valve starts to close (point B in Figure 13.6(6)). The rate of closure remains fixed, having been previously established during the initial process of setting up the controls, the battery output is continuously decreased, and the supply air temperature, after some overshoot, starts to fall. When the temperature reaches the upper limit (point C in Figure 13.6(c)), the movement of the valve stops (point C in Figure 13.6(6)). Suppose that this new valve position (line CD in Figure 13.6(6)) gives a heater battery output which is less than the load. The supply air temperature will continue to drop, until the lower limit is reached at the point D in Figure 13.6(c). This will then make the valve start to open (line DE, Figure 13.6(6)) and, after some undershoot, the
R1
(a)
Supply air-«— ■ C1 © •*— Outside air
Conditioned room
Time
Y Overshoot I Positive deviation B/ C (valve closing) Upper Limit
Time
(c)
Fig. 13.6 Floating action.
Supply air temperature will start to climb, passing the value equivalent to the point E, Figure 13.6(c), and stopping the opening movement of the valve. The battery may now have an output that will permit the supply air temperature to be maintained at some value between the upper and lower limits. If this is the case, then the air temperature will climb but will level off, along a curve EE’, the valve position remaining constant. The line EF in Figure 13.6(6) will then indicate the fixed valve position that gives a steady value of the supply air temperature, corresponding to the line E’F in Figure 13.6(c).
This example is intended only to describe floating action. It is not intended as a recommendation that floating control be applied to control ventilating systems. In fact, for plenum systems in particular, the form of control is unsuitable; apart from the question of lag, load changes are so variable that although the speed at which the valve is arranged to
open may be quite satisfactory during commissioning, it may prove most unsatisfactory for the other rates of load change which will prevail at other times.
The slopes of the lines BC and DE are an indication of the rate at which the valve closes and opens. With single-speed floating action this rate is determined once and for all when the control system is set up. One method whereby a choice of speed is arranged is to permit the valve to open intermittently with short adjustable periods of alternate movement and non-movement. If the timing device is set to give short periods of movement and long periods of non-movement, the slope of the lines BC and DE will be small.
Sophistication may be introduced to floating control, the speed at which the valve opens and closes being varied according to the deviation or to the rate of change of deviation, in the same way that proportional control may be modified, as described in section 13.11. If the dead band is tightened, the control tends to become less stable, degenerating towards two-position.
Posted in Engineering Fifth Edition | 1,192 | 5,653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-25 | longest | en | 0.930171 |
https://www.hindawi.com/journals/aaa/2010/390972/ | 1,545,018,205,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2018-51/segments/1544376828056.99/warc/CC-MAIN-20181217020710-20181217042710-00525.warc.gz | 893,895,159 | 92,414 | `Abstract and Applied AnalysisVolume 2010, Article ID 390972, 39 pageshttp://dx.doi.org/10.1155/2010/390972`
Research Article
A Viscosity Hybrid Steepest Descent Method for Generalized Mixed Equilibrium Problems and Variational Inequalities for Relaxed Cocoercive Mapping in Hilbert Spaces
1Department of Mathematics, Faculty of Science, Kasetsart University (KU), Bangkok 10900, Thailand
2Department of Mathematics, Faculty of Liberal Arts, Rajamangala University of Technology Rattanakosin (RMUTR), Bangkok 10100, Thailand
3Department of Mathematics, Faculty of Science, King Mongkut's University of Technology Thonburi (KMUTT), Bangkok 10140, Thailand
4Centre of Excellence in Mathematics, CHE, Si Ayuthaya Road, Bangkok 10400, Thailand
Received 5 March 2010; Revised 24 May 2010; Accepted 30 May 2010
Copyright © 2010 Wanpen Chantarangsi et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
We present an iterative method for fixed point problems, generalized mixed equilibrium problems, and variational inequality problems. Our method is based on the so-called viscosity hybrid steepest descent method. Using this method, we can find the common element of the set of fixed points of a nonexpansive mapping, the set of solutions of generalized mixed equilibrium problems, and the set of solutions of variational inequality problems for a relaxed cocoercive mapping in a real Hilbert space. Then, we prove the strong convergence of the proposed iterative scheme to the unique solution of variational inequality. The results presented in this paper generalize and extend some well-known strong convergence theorems in the literature.
1. Introduction
Throughout this paper, unless otherwise specified, we consider to be a real Hilbert space with inner product and its induced norm . Let be a nonempty closed convex subset of and let be the metric projection of onto the closed convex subset . Let be a nonexpansive mapping, that is, for all . The fixed point set of is defined by If is nonempty, bounded, closed, and convex and is a nonexpansive mapping of into itself, then is nonempty; see, for example, [1, 2]. A mapping is a contraction on if there exists a constant such that for all In addition, let be a nonlinear mapping. Let be a real-valued function and let be a bifunction such that , where is the set of real numbers and .
The generalized mixed equilibrium problem for finding The set of solutions of (1.2) is denoted by , that is, We see that if is a solution of a problem (1.2), then .
Special Examples
(1)If , then the problem (1.2) is reduced into the mixed equilibrium problem for finding such that The set of solutions of (1.4) is denoted by .(2)If , then the problem (1.2) is reduced into the generalized equilibrium problem for finding such that The set of solutions of (1.5) is denoted by .(3)If and , then the problem (1.2) is reduced into the equilibrium problem for finding such that The set of solutions of (1.6) is denoted by .(4)If and , then the problem (1.2) is reduced into the variational inequality problem for finding such that The set of solutions of (1.7) is denoted by .
The generalized mixed equilibrium problem is very general in the sense that it includes, as special cases, fixed point problems, variational inequality problems, optimization problems, Nash equilibrium problems in noncooperative games, the equilibrium problem, and Numerous problems in physics, economics, and others. Some methods have been proposed to solve problem (1.2); see, for instance, [3, 4] and the references therein.
Let be a nonlinear mapping. Now, we recall the following definitions.(d1) is said to be monotone if for each (d2) is said to be -strongly monotone if there exists a positive real number such that (d3) is said to be -Lipschitz continuous if there exists a positive real number such that (d4) is said to be -inverse-strongly monotone if there exists a constant such that (d5) is said to be relaxed -cocoercive if there exist positive real numbers such that (d6) A set-valued mapping is called monotone if for all , and imply .(d7) A monotone mapping is called maximal if the graph of is not properly contained in the graph of any other monotone mapping. It is well known that a monotone mapping is maximal if and only if for , for every implies .
For finding a common element of the set of fixed points of a nonexpansive mapping and the set of solutions of variational inequalities for a -inverse-strongly monotone mapping, Takahashi and Toyoda [5] introduced the following iterative scheme: where is a -inverse-strongly monotone mapping, is a sequence in , and is a sequence in . They showed that if is nonempty, then the sequence generated by (1.13) converges weakly to some .
For finding an element of , Iiduka et al. [6] introduced the following iterative scheme: where is a -inverse-strongly monotone mapping, is a sequence in , and is a sequence in . They showed that if is nonempty, then the sequence generated by (1.14) converges weakly to some .
For finding a common element of , let be a nonexpansive mapping. Yamada [7] introduced the following iterative scheme called the hybrid steepest descent method: where is a strongly monotone and Lipschitz continuous mapping, and is a positive real number. He proved that the sequence generated by (1.15) converges strongly to the unique solution of the .
The hybrid steepest descent method is constructed by blending important ideas in the steepest descent method and in the fixed point theory. The remarkable applicability of this method to the convexly constrained generalized pseudoinverse problem as well as to the convex feasibility problem is demonstrated by constructing nonexpansive mappings whose fixed point sets are the feasible sets of the problems.
On the other hand, Shang et al. [8] introduced a new iterative process for finding a common element of the set of fixed points of a nonexpansive mapping and the set of solutions of the variational inequalities for relaxed -cocoercive mappings in a real Hilbert space by using viscosity approximation method. Let be a nonexpansive mapping and let be a contraction mapping. Starting with arbitrary initial and define sequences recursively by They proved that under certain appropriate conditions imposed on , and , the sequence converges strongly to , where
For finding a common element of , let be a nonempty closed convex subset of a real Hilbert space . Let be a -inverse-strongly monotone mapping of into and let be a nonexpansive mapping of into itself. S. Takahashi and W. Takahashi [9] introduced the following iterative scheme: where , and satisfy some parameters controlling conditions. They proved that the sequence defined by (1.17) converges strongly to a common element of .
Iterative methods for nonexpansive mappings have recently been applied to solve convex minimization problems; see, for example, [7, 1012] and the references therein. Convex minimization problems have a great impact and influence in the development of almost all branches of pure and applied sciences.
A typical problem is to minimize a quadratic function over the set of fixed points of a nonexpansive mapping defined on a real Hilbert space : where is the fixed point set of a nonexpansive mapping defined on and is a given point in .
A linear bounded operator is strongly positive if there exists a constant with the property
Recently, Marino and Xu [13] introduced a new iterative scheme by the viscosity approximation method: They proved that the sequence generated by (1.20) converges strongly to the unique solution of the variational inequality: which is the optimality condition for the minimization problem: where is a potential function for .
In 2008, Qin et al. [14] proposed the following iterative algorithm: where is a strongly positive linear bounded operator and is a relaxed cocoercive mapping of into . They proved that if the sequences , and of parameters satisfy appropriate condition, then the sequence defined by (1.23) converges strongly to the unique solution of the variational inequality: which is the optimality condition for the minimization problem: where is a potential function for .
In this paper, we introduce an iterative scheme by using a viscosity hybrid steepest descent method for finding a common element of the set of solutions of a generalized mixed equilibrium problem, the set of fixed points of a nonexpansive mapping, and the set of solutions of variational inequality problem for a relaxed cocoercive mapping in a real Hilbert space. The results shown in this paper improve and extend the recent ones announced by many others.
2. Preliminaries
Throughout this paper, we always assume that is a real Hilbert space and is a nonempty closed convex subset of . For a sequence , the notation of and means that the sequence converges weakly and strongly to , respectively.
The following lemmata give some characterizations and useful properties of the metric projection in a real Hilbert space. The metric (or nearest point) projection from onto is the mapping which assigns to each point the unique point satisfying the following property:
Lemma 2.1. It is well known that the metric projection has the following properties: (m1) for each and , (m2) is nonexpansive, that is, (m3) is firmly nonexpansive, that is,
In order to prove our main results, we also need the following lemmata.
Lemma 2.2 (see [2]). Let be a Hilbert space, let be a nonempty closed convex subset of , and let be a mapping of into . Let . Then, for , that is, where is the metric projection of onto .
Lemma 2.3 (see [15]). Let be a monotone mapping of into and let be the normal cone to at , that is, and define a mapping on by Then is maximal monotone and if and only if for all .
Lemma 2.4 (see [16]). Each Hilbert space satisfies Opials condition; that is, for any sequence with , the inequality holds for each with .
Lemma 2.5 (see [13]). Let be a nonempty closed convex subset of , let be a contraction of into itself with coefficient , and let be a strongly positive linear bounded operator on with coefficient . Then, for , That is, is strongly monotone with coefficient .
Lemma 2.6 (see [13]). Assume that is a strongly positive linear bounded operator on with coefficient and . Then .
Lemma 2.7 (see [17]). Let and be bounded sequences in a Banach space and let be a sequence in with Suppose Then,
Lemma 2.8 (see [18]). Assume that is a sequence of nonnegative real numbers such that where is a sequence in and is a sequence in such that (1), (2) or Then
For solving the generalized mixed equilibrium problem and the mixed equilibrium problem, let us give the following assumptions for the bifunction , the function , and the set : (H1)(H2) is monotone, that is, (H3) for each is weakly upper semicontinuous; (H4) for each is convex; (H5) for each is lower semicontinuous; (B1) for each and , there exist abounded subsets and such that for any , (B2) is a bounded set.
Lemma 2.9 (see [19]). Let be a nonempty closed convex subset of . Let be a bifunction that satisfies (H1)–(H5) and let be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For and , define a mapping as follows: for all . Then, the following properties hold: (i)for each ; (ii) is single-valued; (iii) is firmly nonexpansive; that is, for any (iv)(v) is closed and convex.
Remark 2.10. If , then is rewritten as .
Lemma 2.11 (see [9]). Let , and be as in Remark 2.10. Then the following holds: for all and .
The following lemma is an immediate consequence of an inner product.
Lemma 2.12. Let be a real Hilbert space, let and be elements in , and let . Then (1), (2).
3. Main Results
In this section, we will introduce an iterative scheme by using a viscosity hybrid steepest descent method for finding a common element of the set of fixed points for nonexpansive mappings, the set of solutions of a generalized mixed equilibrium problem, and the set of solutions of variational inequality problem for a relaxed cocoercive mapping in a real Hilbert space. We show that the iterative sequence converges strongly to a common element of the three sets.
In order to prove our main results, we first prove the following lemmata.
Lemma 3.1. Let and be as in Lemma 2.9. Then the following holds: for all and .
Proof. By similar argument as in the proof of Lemma 2.11 in [9], for and . Observing that and , we have Putting in (3.2) and in (3.3), we obtain So, summing up these two equalities and using the monotonicity of (H2), we get and hence We derive from (3.6) that and so This indicates that In other words, and thus the claim holds.
Lemma 3.2. Let be a real Hilbert space, let be a nonempty closed convex subset of , let be a nonexpansive mapping, and let be an -Lipschitz continuous and relaxed -cocoercive mappings with . If , then is a nonexpansive mapping in .
Proof. Let . Then, for every , we have Now, since , thus .
Thus, is a nonexpansive mapping of into .
Now we can prove that a strong convergence theorem is a real Hilbert space.
Theorem 3.3. Let be a nonempty closed convex subset of a real Hilbert space . Let and be two bifunctions from to satisfying (H1)–(H5) and let be a proper lower semicontinuous and convex function with assumption (B1) or (B2). Let (i) be a -inverse-strongly monotone mapping, (ii) be a -inverse-strongly monotone mapping, (iii) be an -Lipschitz continuous and relaxed -cocoercive mappings, (iv) be a contraction mapping with coefficient and let be a strongly positive linear bounded self-adjoint operator with the coefficient and . Let be a nonexpansive mapping with .
Assume that Let , , , , and be the sequences generated by where , and , and are three sequences in satisfying the following conditions: (C1) and , (C2), and , (C3), and , (C4), and , (C5), and . Then, converges strongly to , which is the unique solution of the variational inequality:
Proof. From the restrictions on control sequences, we may assume, without loss of generality, that for all . Since is a strongly positive linear bounded self-adjoint operator on , we have Observe that That is, is positive. It follows that We will split the proof of Theorem 3.3 into six steps.Step 1. We claim that the sequence is bounded.
Indeed, let , by Lemmas 2.2 and 2.9, we obtain Since is -inverse-strongly monotone, and , we know that, for any , we have Similarly, from (3.19), , and , we can prove that and hence Let , and form Lemma 3.2 is a nonexpansive mapping and from Lemma 2.2 , we have which yields that By mathematical induction, putting , we have that for all . Indeed, we can easily see that . Suppose that for some positive integral . Then we have that This shows that is bounded in . From (3.21), we know that and are bounded in and so , , , , , and are bounded sequence in .
Step 2. We claim that
Since is nonexpansive, we have Next, we estimate . Observing that and , we have Similarly, we can prove that Substitution (3.26) into (3.27), we derive Since , and are bounded, is an appropriate constant such that Substitution (3.28) into (3.25), we obtain From (3.13), we have Simple calculations show that which yields that Substitution (3.30) into (3.33) yields that where is an appropriate constant such that Since and next, we estimate Note that ; there exists a constant such that for all . From Lemma 3.1, we get It follows that Since , we obtain Similarly, we can prove that Consequently, from (3.40), (3.41), and conditions in Theorem 3.3, we obtain It follows from Lemma 2.7 that In view of (3.13), we see that which, combining with (3.43) and , yields that
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# Bournelli, equation for variance? (with picture!)
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1. So I was watching khan academy videos on Stats, and I noticed this:
why he does not calculate the variance of the sample he has first, and then approximate the variance of the normal distribution using the result?
Rest of the question is on picture in white text, any help much appreciated
Attached Thumbnails
3. What is M_1 (or mu_1) at the very start? And what is he trying to show, the normal distribution approximation for the sum of 1000 choices of 0 or 1 where the probability of choosing 1 is 0.642? In any case, that's a freaking tiny variance that can't be right for whatever
4. (Original post by Hopple)
What is M_1 (or mu_1) at the very start? And what is he trying to show, the normal distribution approximation for the sum of 1000 choices of 0 or 1 where the probability of choosing 1 is 0.642? In any case, that's a freaking tiny variance that can't be right for whatever
mu_1 is mean of sample number 1
He tries to show how many people have voted for a candidate (choice 1) and how many didn't (choice 0), by using this sample he then creates a normal distribution of all the sample means.
5. (Original post by Moa)
mu_1 is mean of sample number 1
He tries to show how many people have voted for a candidate (choice 1) and how many didn't (choice 0), by using this sample he then creates a normal distribution of all the sample means.
I'm still not sure what he's trying to show, that final super tiny variance makes no sense to me, sorry. Was there anything prior to this question he built upon?
6. (Original post by Hopple)
I'm still not sure what he's trying to show, that final super tiny variance makes no sense to me, sorry. Was there anything prior to this question he built upon?
Maybe the video can help you, if you have time to skim through it:
7. (Original post by Moa)
Maybe the video can help you, if you have time to skim through it:
Ah, I see. It builds on earlier videos (or so he says), but he's working out how the sample mean varies. He has worked out the variance of the first sample, but by a 'stock' formula (Bin(n,p) has variance np(1-p)), which gives the same answer as your more general formula. I'm assuming you're okay with the rest, just the calculation of the variance was different to what you expected?
8. (Original post by Hopple)
Ah, I see. It builds on earlier videos (or so he says), but he's working out how the sample mean varies. He has worked out the variance of the first sample, but by a 'stock' formula (Bin(n,p) has variance np(1-p)), which gives the same answer as your more general formula. I'm assuming you're okay with the rest, just the calculation of the variance was different to what you expected?
Yes that is it, its all clear now, I can see that the results of "mine" and his formulas are very close.
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# The sum of first 9 terms of the series 131+13+231+3+13+23+331+3+5+…… is
A
71
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B
96
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C
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192
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Solution
## The correct option is B 96Given series is 131+13+231+3+13+23+331+3+5+…∞ Let Tn be the nth term of the given series. ∴Tn=13+23+33+……+n31+3+5+……+upto n terms ={n(n+1)2}2n2=(n+1)24 S9=∑9n=1(n+1)24=14(22+33+……+102)+124−124 =14[10(10+1)(20+1)6−1]=3844=96
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