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https://www.mapleprimes.com/users/Pascal4QM/answers?page=1 | 1,660,588,505,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572198.93/warc/CC-MAIN-20220815175725-20220815205725-00441.warc.gz | 747,310,320 | 35,393 | 6 years, 18 days
## convert/exp...
Hi,
Maple result seems correct to me. The point is that the returned sum is a Dirac comb. Writting the result as a sum of exp or a sum of Dirac is a matter of choice depending of what you need. Using convert/exp might help:
>
>
>
>
>
(1)
>
(2)
>
(3)
Considering the Dirac comb
>
(4)
One has the relation
>
(5)
And finally
>
(6)
Note that the reciprocal conversion is not (yet) supported
>
(7)
>
>
## Possibly Physics:-Fundiff...
Hi,
For deriving an equation in Physics, Fundiff is often the good approach. This should give the result:
>
>
>
(1)
>
(2)
>
(3)
>
(4)
>
## Define quantum operators and use Commuta...
Hi,
In order to have * redefined, you need first to define quantum operators for instance. Then [a,b] has the same meaning whether Physics is loaded or not. I guess you'd like to use the commutator. For that purpose, just use "Commutator":
>
>
>
(1)
>
(2)
>
(3)
>
## anticommutativeprefix = {x, y}...
Hi,
In your case, you should rather use Setup(anticommutativeprefix = {x, y}), not "noncommutativeprefix". Then, it is unnecessary to specify "algebrarules={%AntiCommutator(x,x)=0, %AntiCommutator(y,y)=0}" as it will be automatically the case.
Finally, Maple's result:
AntiCommutator(x,y)=0
seems correct to me. As x and y anticommute, x*y + y*x = x*y - x*y = 0.
## Physics differentialoperators...
Hi,
A possible way is to define a formal differential operator within the Physics package. It can be define with an arbitrary function and trigged at will. For an exemple, have look at:
https://www.mapleprimes.com/posts/208710-Quantum-Commutation-Rules-Basics
## Component...
Hi,
Using "Component", it is possible to reconstruct a Vector:
V := Vector([seq(Component(q_, k), k = 1 .. 3)])
Component2Vector.mw
## Parentheses...
Try -3 - (-5) or 4*(-10), It should work as expected.
However, I agree that an expression like -3 - -5 should work without parentheses in Maple, mathematically the expression is correct, surprising. Entering -3-+-5 or 4*-10 in Matlab gives the expected result.
## Alias and Physics:-Latex...
Hi,
I think that using the new Physics:-Latex with alias, there is even no need for replacing.
>
>
(1)
>
(2)
>
x \left( t \right) ={\it \_C1}\,{{\rm e}^{t}}+{\it \_C2}\,{{\rm e}^{-{ \frac {t}{2}}}}\sin \left( {\frac {\sqrt {3}t}{2}} \right) +{\it \_C3} \,{{\rm e}^{-{\frac {t}{2}}}}\cos \left( {\frac {\sqrt {3}t}{2}} \right)
>
x \left(t \right) = c_{1} {\rm e}^{t}+c_{2} {\rm e}^{-\frac{t}{2}} \sin \left(\frac{\sqrt{3} t}{2}\right)+c_{3} {\rm e}^{-\frac{t}{2}} \cos \left(\frac{\sqrt{3} t}{2}\right)
>
Hi,
## Differential operator...
You might define a differential operator
>
>
(1)
>
(2)
>
(3)
>
(4)
>
(5)
>
(6)
>
>
## Setup(quantumoperators = V)...
Hi,
With Physics package, to get this feature, V should be declared as a quantum operator. Also, use Dagger instead of Transpose, or V_^*.
restart;
with(Physics);
with(Physics[Vectors]);
Setup(quantumoperators = V);
V_*Dagger(V_);
This question is related:
Hi,
L := [1, 2, 3, 4, 5];
L := [1, 2, 3, 4, 5]
15
## Restart...
Hi,
I have the same sometime. Just enter "restart" and this should load the last version, even if the message is the same. Otherwise, just close your worksheet and reopen it, it should work fine.
## Several ways...
Hi,
There are several ways. You can use "indets" to extract the variable of an expression, and then "select" according particular properties. For instance, assuming x and y have a t depedency, while A and B are independent of that parameter:
EE := (x(t)/A)^2 + (y(t)/B)^2 = 1;
2 2
x(t) y(t)
EE := ----- + ----- = 1
2 2
A B
FF := indets(EE);
FF := {A, B, t, x(t), y(t)}
select(u -> type(u, freeof(t)), FF);
{A, B}
## Close the Palette...
Hi,
In addition to Edgardo's recommendations, you could try to close the palettes (left Panel). There is a known issue at startup for Maple 2020 involving those palettes. Once Maple has started again, you can reopen the required palettes.
1 2 3 4 Page 1 of 4
| 1,237 | 4,107 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2022-33 | latest | en | 0.832402 |
http://www.typologycentral.com/forums/science-technology-and-future-tech/35445-math-question-3d-integrals-2.html | 1,513,199,944,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948530841.24/warc/CC-MAIN-20171213201654-20171213221654-00037.warc.gz | 486,586,516 | 12,067 | # Thread: Math question - 3d integrals
1. Originally Posted by ubiquitous1
I think I understand what you are saying, but triple integrals are still used to find a volume.
V=∫∫∫ dz dy dx
Where the limits of integration for z are 0 to 1-y, limits for y are 0 to 1 and finally limits of x are 0 to 2.
I miss Calculus
That's how we used to do it, too, although it's conceptually rather primitive.
2. How you approach it depends on what you are solving. For A(h)=pi*h^2 (ie. a form of cone), you wouldn't need to perform a triple integral to find the volume. If you work it from basics it is a triple integral though, because the area came from integration over r and theta, then setting r=h.
So triple integrals are used to find volume from scratch, but are not always a necessary step if you recognise how the cross-sectional area behaves. That was the formula onemoretime had, which may or may not be useful depending on what needs to be solved.
#### Posting Permissions
• You may not post new threads
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Single Sign On provided by vBSSO | 283 | 1,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2017-51 | latest | en | 0.93524 |
https://www.airmilescalculator.com/distance/pmc-to-zal/ | 1,708,489,633,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473370.18/warc/CC-MAIN-20240221034447-20240221064447-00026.warc.gz | 675,741,785 | 16,733 | # How far is Valdivia from Puerto Montt?
The distance between Puerto Montt (El Tepual Airport) and Valdivia (Pichoy Airport) is 123 miles / 199 kilometers / 107 nautical miles.
The driving distance from Puerto Montt (PMC) to Valdivia (ZAL) is 156 miles / 251 kilometers, and travel time by car is about 3 hours 2 minutes.
123
Miles
199
Kilometers
107
Nautical miles
## Distance from Puerto Montt to Valdivia
There are several ways to calculate the distance from Puerto Montt to Valdivia. Here are two standard methods:
Vincenty's formula (applied above)
• 123.435 miles
• 198.650 kilometers
• 107.262 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
• 123.602 miles
• 198.918 kilometers
• 107.407 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 Puerto Montt to Valdivia?
The estimated flight time from El Tepual Airport to Pichoy Airport is 44 minutes.
## Flight carbon footprint between El Tepual Airport (PMC) and Pichoy Airport (ZAL)
On average, flying from Puerto Montt to Valdivia generates about 43 kg of CO2 per passenger, and 43 kilograms equals 95 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Puerto Montt to Valdivia
See the map of the shortest flight path between El Tepual Airport (PMC) and Pichoy Airport (ZAL).
## Airport information
Origin El Tepual Airport
City: Puerto Montt
Country: Chile
IATA Code: PMC
ICAO Code: SCTE
Coordinates: 41°26′20″S, 73°5′38″W
Destination Pichoy Airport
City: Valdivia
Country: Chile
IATA Code: ZAL
ICAO Code: SCVD
Coordinates: 39°39′0″S, 73°5′9″W | 511 | 1,916 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.807894 |
https://testbook.com/question-answer/the-ratio-of-ages-of-two-persons-a-and-b-is-3nbsp--5ff83af33ed6cde622f66c16 | 1,639,058,194,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964364169.99/warc/CC-MAIN-20211209122503-20211209152503-00238.warc.gz | 619,020,414 | 30,553 | # The ratio of ages of two persons A and B is 3 ∶ 4 and that of ages of B and C is 4 ∶ 5. If the average of all three is 40 years. What is B's age?
This question was previously asked in
Official Paper 20: Held on 11th Nov 2020 Shift 2
View all UGC NET Papers >
1. 20 years
2. 30 years
3. 40 years
4. 50 years
Option 3 : 40 years
## Detailed Solution
Given:
Average age = 40 years
Formula used:
Average = Sum of all the terms/Total number of terms
Calculation:
A ∶ B = 3 ∶ 4
And, B ∶ C = 4 ∶ 5
Then, A ∶ B ∶ C = 3 ∶ 4 ∶ 5
Let the age of A, B and C be 3x, 4x and 5x
Sum of ages of all the persons = 3x + 4x + 5x = 12x
Average = 12x/3 = 4x
⇒ 4x = 40
⇒ x = 10
∴ The age of B is 40 | 292 | 697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-49 | latest | en | 0.897945 |
https://www.cit.cat/how-to-multiply-single-digits | 1,656,539,025,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00759.warc.gz | 761,985,235 | 14,818 | # How To Multiply Single Digits
How To Multiply Single Digits. These are all single digit multiplication facts and you should have those memorized. The multiplication of 1 digit number with the number starts from the unit’s place and goes on.
The calculator is recommended if students want to solve long. As we start counting the table of 3, we can easily get the answer. Similarly, to multiply a number by 100, put two zeros to the right of the number.
### Examples, Solutions, Videos, Stories, And Songs To Help Grade 3 Students Learn How To Multiply 2 Or More Digits By 1 Digit With And Without Regrouping.
For instance, the 'zero rule' tells you that when you multiply any number by 0, the. We can also say that the multiplication of digits from right to left. For example, 435 x 100 = 43500.
### The Tens Place Of Each Product Are Added To The Next Product.
As we start counting the table of 3, we can easily get the answer. It shows how to multiply single digit. 3 x 4 = 2.
### Now, We Multiply 3 X 6 = 18.
Multiplying three digits of place value. The calculator is recommended if students want to solve long. Make sure that you are able to prove that your product is correct to another person.
### Multiply That Digit By Single Digit Multiplicand.
So multiplication is nothing more than adding a series of multiplication facts. The video explains how to multiply a two digit number by single digit (e.g. Like our prior examples, multiplying three digit numbers by a single digit is an exercise in applying our multiplication facts individually to each place value.
### Setup The Numbers You Are Multiplying (The “Multiplicands”) Vertically.
To keep things simple, we’ll call this digit from the lower. Many of them are multiplied by columns, for my case is need to get the first two digit from a single column and multiply. These are all single digit multiplication facts and you should have those memorized. | 426 | 1,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2022-27 | latest | en | 0.88761 |
https://forums.kleientertainment.com/forums/topic/77553-metheus-puzzle-part-3-spoilers-updated-with-chest-answers/page/3/ | 1,686,413,426,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657720.82/warc/CC-MAIN-20230610131939-20230610161939-00368.warc.gz | 309,112,172 | 33,458 | # Metheus Puzzle Part 3 (Spoilers!!! Updated With Chest Answers).
## Recommended Posts
How come there is no clue?? everything up until now was in pairs... It's simple logic.
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So can anyone help me and my friends puzzle since I never did Mastermind or Code Breaker before
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@MrDeepDarkmind I tried to explain an efficient algorithm for solving part 2 here:
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The total number of glyphs is 32. A legitimate number for an alphabet, for example. Most of the glyphs on the mural appear in strings of 2-3. Hmmmmmmmmmmmm.
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ugh this puzzle only gets harder.
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27 minutes ago, Captain_Rage said:
The total number of glyphs is 32. A legitimate number for an alphabet, for example. Most of the glyphs on the mural appear in strings of 2-3. Hmmmmmmmmmmmm.
But how would that help us to put items into the chest? Would the symbols spell a word? Because if not, we'd be starting over with the chest theory.
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It seems that this alphabet is probably pictographic. This gives us a very limited number of words, even if they can be combined as modifiers. Alternatively, we have distinct symbol groups - The blocks, the black-code and the gold-code being the main distinct sets. The blocks seem numeric, and base-13 would seem in keeping with Klei logic. Special characters in this set for stack-size limits would also be in keeping. If these are distinct sets, the gold-code symbols could be syllabic (who's to say the ancients used the same sound set?) and the black-code pictographic. The Fuelweaver "speaks", but also has telepathic abilities (mind control), so there is little reason to suppose it has a common language.
Regarding the "staff", is there reason to think it could not be the Shadow Thurible? This would limit the chest solution to mature worlds where the Fuelweaver has been defeated (cheats aside), and it bears a fairly direct association. Klei puzzles are usually clear from the back-end, even if difficult to decipher immediately, and this seems like a reasonable leap.
I don't know that Klei would force those on unreliable connections (generally played as local games) to join wider servers in order to meet their puzzle-partners from earlier stages, without providing capacity to relink accounts. Alternatively, this might be a deliberate strategy in order to try to foster (force) players to expand into wider communities.
There are three potential cases for the victory condition and relating to the player count: Require multiple players, allow multiple players, prevent multiple players. Pairing may play a role here. We have no information on that. However, "prevent" would stop this puzzle becoming a stealing tool for those on public servers, while "require" would be in line with earlier puzzles.
These are my considerations on the topic for the moment - do with them as you will. I haven't been around for any previous puzzles, so this may-or-may-not be useful. The though also occurs that the wild theories these puzzles generate, appropriately mixed, would provide a rich ore to distil new lore from.
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7 minutes ago, Whisperr said:
ok this is as far as me and my friend got
we cant go any farther , any ideas ?
i already tried it , it didn't work ....
The point of the spreadsheet is to figure out the answer, it doesn't tell you it :P. None of us have figured it out yet.
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10 minutes ago, Cheetos said:
The point of the spreadsheet is to figure out the answer, it doesn't tell you it :P. None of us have figured it out yet.
gg klei
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My friend and I noticed parts of our mural filling in (we have completed part one a two and the chest has appeared in the world), symbols in the creatures, like the one with the carrots and gems, do we know what that is about yet?
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excuse me, I have a little issue, did you saw the new symbols on the obelisk?,wellI was trying but I can't scroll to the top of the obelisk, this seems as some kind of glitch I can scroll to the bottom but not the top, I see an incomplete picture, I can zoom out to see the full image but I the image is very small now, anyone had that issue?
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2 minutes ago, Rily said:
excuse me, I have a little issue, did you saw the new symbols on the obelisk?,wellI was trying but I can't scroll to the top of the obelisk, this seems as some kind of glitch I can scroll to the bottom but not the top, I see an incomplete picture, I can zoom out to see the full image but I the image is very small now, anyone had that issue?
I have the same issue, I use safari and I go to View - Zoom Out and it makes more of it visable. Should work in chrome too.
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3 minutes ago, Rily said:
excuse me, I have a little issue, did you saw the new symbols on the obelisk?,wellI was trying but I can't scroll to the top of the obelisk, this seems as some kind of glitch I can scroll to the bottom but not the top, I see an incomplete picture, I can zoom out to see the full image but I the image is very small now, anyone had that issue?
Change the number from 1 upon 4 to see the obelisk. Your welcome.
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Anyone having this same issue? i already completed the second part of the puzzle but now everytime i enter the site it shows like this, does your friend need to be online on steam for it to show complete or what's going on.
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Oh, symbols on the obelisk? that sounds interesting, can someone upload a picture?
I tried the link but I get an error.
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14 hours ago, Tumalu said:
From what other threads have figured, the most probable solution is:
You need the combination that results in 6 lights on the Metheus website puzzle. The set of 6 blocky things have notches to indicate how MANY for the stack. The set of 6 shadow letters translates on the mural to what kind of item to put in that slot.
Match the symbols to the correct items on the mural and we should have it solved.
The chest calls for your id, so yes, you need to solve the puzzle on the site and get your own personal solution with your partner. There most likely is no "one" solution that will work. It has to be YOUR solution.
Whether this turns out to be accurate or not, this is really helpful with logically grasping the second puzzle. Each item being a certain number of a certain item going into a certain chest slot.
Much easier to do that way.
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1 minute ago, Tumalu said:
Oh, symbols on the obelisk? that sounds interesting, can someone upload a picture?
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12 minutes ago, Canal_WP said:
Change the number from 1 upon 4 to see the obelisk. Your welcome.
thanks I was doing that, nothing will stop me to discover the end of the puzzle, but I would like to made it as it was intended, in this way the puzzle feels a bit flawed
edit:
maybe those symbols are the real number, I mean, one thulian shrimp thing have a symbol and "4" carrots and the other one with the other symbol have "3" gems
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Oooh, so the obelisk symbols are just the murals. I see, that makes sense they'd show up in a "normal" fashion eventually.
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14 minutes ago, Tumalu said:
Oooh, so the obelisk symbols are just the murals. I see, that makes sense they'd show up in a "normal" fashion eventually.
the murals have been on the obelisk sense we saw it but it looks like the murals are changing seems like white tiles and symbols have been added
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1 minute ago, yujinthegamer said:
the murals have been on the obelisk sense we saw it but it looks like the murals are changing seems like white tiles and symbols have been added
I don't see those?
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4 minutes ago, Zampano said:
I don't see those?
the symbols in the shrimp peps on the first panel and some of the shadows are white now or atleast from what i can tell im pretty sure that wasnt there b4
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To the video/painting symbols...
The painting symbols next to the Staff are (speculated to be) "Key/Staff/Caves".
Using the translation of those with the symbol we already knew, the video version shown next to the Moonbase is "Key/Staff/Tenebrae".
One interpretation of "Tenebrae" is "Darkness".
"Key/Staff/Caves" - The KeyStaff in Caves.
"Key/Staff/Darkness" - The KeyStaff in the Darkness.
Take the KeyStaff from the Caves, and put it in Darkness.
Alright.
Step back for a moment.
The items we put into the chest seemingly always include what has been interpreted as the Walking Cane, and some amount of Nightmare Fuel.
Of interesting note, the Cane anim files were recently updated, and accompanying cane_shadow_fx, which is a collection of shadows seemingly dripping off of some cane-like object, unseen in that file; ie it's an overlay.
I'm going to assume that what we get out of this will indeed be the in-game Staff people data-mined, acting as an alternate Cane 'skin'.
Back to the puzzle.
We probably get the staff from doing the chest puzzle right, which will likely involve that other topic that's figuring out the symbol meanings. They're getting Klei shadowedits, btw.
Anyway...we know that both the New Moon and Full Moon have power in DST. The Moonbase channels the "moon's" power.
'Take the KeyStaff from the Caves, and put it in Darkness.' could mean...
Wait for a New Moon (empty, black moon, already has some stuff to do with Shadow empowerment).
Put KeyStaff (what we probably get from the chest) into the Moonbase during that night.
If anyone manages to correctly solve the chest puzzle, please try this.
I'm not sure what it will 'unlock', raining shadows down on a shadow staff...but it should be interesting.
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1 hour ago, Pyr0mrcow said:
Whether this turns out to be accurate or not, this is really helpful with logically grasping the second puzzle. Each item being a certain number of a certain item going into a certain chest slot.
Much easier to do that way.
A very valid point. Plus, the Sacred Chest seems to have appeared in-game after compleing the 2nd phase on the website.
Hence, it is a super strong indication that it is the next step. | 2,403 | 10,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.937683 |
https://mathoverflow.net/questions/211381/eta-invariants-of-riemann-surface | 1,632,553,924,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057598.98/warc/CC-MAIN-20210925052020-20210925082020-00366.warc.gz | 431,999,854 | 30,306 | # $\eta$-invariants of Riemann Surface
I am curious about a concrete computation of $\eta$-invariants for Riemann surface, e.g. Torus.
Is there any nice review or notes talking about the computation? Or is it possible to express it as some topological data, analogous to 3d case?
A related post to this subject: $\eta$ invariants of Pin+ manifolds $\mathbb{RP}^{8k}$
• Kirby-Taylor ( www3.nd.edu/~taylor/old/papers/PSKT.pdf, see Lemma 3.6) describes how to compute an invariant for surfaces with Pin$^-$ structures, as the Gauss sum of a quadratic refinement on the first homology group. I think it is basically the $\eta$ invariant, in light of the relation to fermion SPT phases. Jul 13 '15 at 15:55
• Recently I got some more understanding on $\eta$ invariant, it can be computed directly with the knowledge of the spectrum of, e.g., Dirac operator. e.g, the torus certainly provides a symmetric spectrum for Dirac operator, and therefore has vanishing $\eta$ invariants. Some less trivial examples are real projective space at even dimension, which are unorientable. See the post cited and comments there. Dec 11 '15 at 2:10
• Isn't it defined in terms of the spectrum of Dirac operators? So if we know the spectrum, we can compute the invariant. Dec 11 '15 at 2:32
• @MengCheng yes, that is one direct way to compute that, for example you can compute $\eta$ for $\mathbb{RP}^{2l}$. Other way to compute that is by applying APS index theorem, $e.g.,$ constructing a higher dimensional manifold bounding the manifold you'd like to compute. Dec 11 '15 at 5:31
• @MengCheng maybe I didn't understand your comment? I guess you might ask whether it has be defined for Dirac operator? if so, the answer is no, and it could be defined for more general operators. Dec 11 '15 at 5:35 | 466 | 1,782 | {"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} | 2.640625 | 3 | CC-MAIN-2021-39 | latest | en | 0.89504 |
https://www.johndcook.com/blog/2021/07/12/spline-convergence-rate/ | 1,721,214,373,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.39/warc/CC-MAIN-20240717090242-20240717120242-00700.warc.gz | 726,406,886 | 13,092 | Rate of natural cubic spline convergence
Suppose you want to approximate a function with a polynomial by interpolating it at evenly spaced points. You might reasonably expect that the more points you use, the better the approximation will be. That might be true, but it might not. As explained here, for some functions the maximum approximation error actually increases as the number of points increases. This is called the Runge phenomenon.
What about cubic splines? Instead of fitting one high-degree polynomial, you fit a different cubic polynomial on each sub-interval, with the constraint that the pieces fit together smoothly. Specifically, the spline must be twice differentiable. This means that the first and second derivatives from both sides match at the end of each interval.
Say our function f is defined over an interval [a, b] and we break the interval into n sub-intervals. This means we interpolate f at n + 1 evenly spaced points. We have 4n degrees of freedom: four polynomial coefficients over each interval. Interpolation adds 2n constraints: the value of the polynomial is specified at the ends of each sub-interval. Matching first and second derivatives at each of the n − 1 interior nodes gives 2(n − 1) constrains. This is a total of 4n − 2 equations for 4n unknowns.
We get the two more equations we need by specifying something about the derivatives at a and b. A natural cubic spline sets the second derivatives equal to zero at a and b.
So if we interpolate f at n + 1 evenly spaced points using a natural cubic spline, does the splines converge uniformly to f as we increase n? Indeed they do. Nothing like Runge phenomenon can happen. If f is four times continuously differentiable over [ab], then convergence is uniform on every compact subinterval of (ab) and the rate of convergence is O(1/n4). If the second derivative of f is zero at a and b, then the rate of convergence is O(1/n4) over the whole interval [a, b].
Source: Kendall E. Atkinson. On the Order of Convergence of Natural Cubic Spline Interpolation. SIAM Journal on Numerical Analysis, Vol. 5, No. 1 (Mar., 1968), pp. 89-101.
Update: See this post for results for other kinds of cubic splines, i.e. cubic splines with different boundary conditions.
Experiment
We will look at sin(x) and exp(x) on the interval [0, 2π]. Note that the second derivatives of former at zero at both ends of the interval, but this is not true for the latter.
We will look at the error when interpolating our functions at 11 points and 21 points (i.e. 10 sub-intervals and 20 sub-intervals) with natural cubic splines. For the sine function we expect the error to go down by about a factor of 16 everywhere. For the exponential function, we expect the error to go down by about a factor of 16 in the interior of [0, 2π]. We’ll see what happens when we zoom in on the interval [2, 4].
We’re basing our expectations on a theorem about what happens as the number of nodes n goes to infinity, and yet we’re using fairly small values of n. So we shouldn’t expect too close of an agreement with theory, but we’ll see how close our prediction gets.
Computation
The following Python code will plot the error in the interpolation by natural cubic splines.
``` import numpy as np
from scipy.interpolate import CubicSpline
import matplotlib.pyplot as plt
for n in [11, 21]:
for f in [np.sin, np.exp]:
knots = np.linspace(0, 2*np.pi, n)
cs = CubicSpline(knots, f(knots))
x = np.linspace(0, 2*np.pi, 200)
plt.plot(x, f(x) - cs(x))
plt.xlabel("\$x\$")
plt.ylabel("Error")
plt.title(f"Interpolating {f.__name__} at {n} points")
plt.savefig(f"{f.__name__}_spline_error_{n}.png")
plt.close()
```
Notice an interesting feature of NumPy: we call the `__name__` method on `sin` and `exp` to get their names as strings to use in the plot title and in the graphic file.
Here’s what we get for the sine function.
First for 10 intervals:
Then for 20 intervals:
When we doubled the number of intervals, the maximum error went down by about 30x, better than the theoretical upper bound on error.
Here’s what we get for the exponential function.
First for 10 intervals:
Then for 20 intervals:
The error in approximating the exponential function by splines is much greater than the error in approximating the sine function. That’s because the former acts less like a polynomial than the latter. Polynomial approximation, and piecewise polynomial approximation, works better on things that behave like polynomials.
When we doubled the number of intervals, the maximum error went down by a factor of 12. We shouldn’t be surprised that the error to go down by a factor of less than 16 since the hypotheses for the theorem that gives a factor of 16 aren’t satisfied.
Here’s what we get when we zoom in on the interval [2, 4].
For 10 subintervals:
And for 20 subintervals:
The error is much smaller over the interior interval. And when we double the number of interpolation points, the error over [2, 4] goes down by about a factor of 3 just as it did for sine.
Note that in these last two plots, we’re still interpolating over the entire interval [0, 2π], but we’re looking at the error over [2, 4]. We’re zooming in on part of the error plot, not allocating our interpolation points specifically to the smaller interval. | 1,259 | 5,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-30 | latest | en | 0.882356 |
https://arzung.org/adding-fractions-with-unlike-denominators-worksheets-pdf/ | 1,621,280,932,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992440.69/warc/CC-MAIN-20210517180757-20210517210757-00479.warc.gz | 138,120,756 | 9,991 | # Adding fractions with unlike denominators worksheets pdf Latest
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Adding Fractions With Unlike Denominators Worksheets Pdf. Use this link to go to our Adding and Subtracting Fractions with Unlike Denominators page. Adding or Subtracting Fractions with Different Denominators Evaluate each expression. 6 3 6 2 6 6 same 3 5 2 2 5 2 14 6 -10 2 -10 -10 10 10 2 10 2 10 10 14 1-4 2 3 3 5 5 3 6 9 3 9 8 1-4 14 8 2. Our printable adding unlike fractions worksheets have vowed that no student in 4th grade and 5th grade shall suffer while adding fractions with different denominators.
Adding Fractions Color By Code Worksheets Fractions Worksheets Adding Fractions Fractions From pinterest.com
Our printable adding unlike fractions worksheets have vowed that no student in 4th grade and 5th grade shall suffer while adding fractions with different denominators. In these worksheets the child has to add together two fractions which have different denominators. Or choose for adding. Fractions mathematics math adding. You have to option to set a range for the denominators. These worksheets require students to find a common denominator so that the numerators may be combined and then the answer reduced to final form.
### Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6.
Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6. Or choose for adding. Adding and subtracting fractions Otros contenidos. Regardless of whether youre seeking to save money on curriculum or maybe your child demands extra work in fractions these worksheets can certainly help out making it simple for you concurrently. Adding fractions worksheet maker for like and unlike denominators. 1 3 4 2 5 2 1 9 5 3 13 7 3 2 4.
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Complete work with steps is shown for each problem on the answer keys. This is the first series of worksheets that deal with unlike denominators. Fractions Worksheet – Adding Fractions with Unlike Denominators Author. Worksheets for adding fractions with common denominators with unlike denominators as simple fractions and as mixed fractions. 3 bKBuptTa7 aSMo3fOtRw9aEroe6 hLvLOCRE k kAIl8lj SrXiEgxh7txsD Wr1ezsVerevNe1dk2 P SMvagdWe0 gw5it6hJ pIEnCfXibnsiWtUer XPPrCe0-dAcl1ggeZbwr0aVH Worksheet by Kuta Software LLC Answers to Adding and Subtracting Fractions with Unlike Denominators ID.
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Adding Fractions Worksheets Pdf. Or choose for adding. Click here for our other Fractions Worksheets. 1 3 4 2 5 2 1 9 5 3 13 7 3 2 4. Pdf adding fractions with unlike denominators worksheets pdf is available here for you to download.
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Regardless of whether youre seeking to save money on curriculum or maybe your child demands extra work in fractions these worksheets can certainly help out making it simple for you concurrently. Preview images of the first and second if there is one pages are shown. These worksheets are pdf files. 15 4 13 7 16 4 5 1 3 17 3 2 11 6 18 4 7 3 2 19 5 7 1 4 20 3 2 13 7. The worksheets in this series avoid mixed numbers to introduce the skills needed to add with different denominators.
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They should first multiply the numerator and denominator of one of the fractions so that both fractions share a common denominator and then do the addition. 1 1 48 35 2 2 3 15 8 4 31 12 5 8 15 6 52 21 7 19 15 8 25 28 9 5 6 10 10 21 11 7. Adding and subtracting fractions Other contents. The size of the PDF file is 27095 bytes. You have to option to set a range for the denominators.
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1 3 4 2 5 2 1 9 5 3 13 7 3 2 4. The worksheets in this series avoid mixed numbers to introduce the skills needed to add with different denominators. An indispensable bunch these pdfs incorporate adding proper fractions adding improper fractions adding. Fractions Worksheet – Adding Fractions with Unlike Denominators Author. Click on the following links to download adding fractions worksheets as pdf document.
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In these worksheets the child has to add together two fractions which have different denominators. Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6. Click here to return to the main worksheet index. These worksheets are pdf files. An indispensable bunch these pdfs incorporate adding proper fractions adding improper fractions adding.
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In these worksheets the child has to add together two fractions which have different denominators. Adding fractions with different denominators. Answers to Adding or Subtracting Fractions with Different Denominators 1 23 20 2 14 5 3 47 14 4 27 56 5 10 7 6 23 14 7. The worksheets in this series avoid mixed numbers to introduce the skills needed to add with different denominators. Click on the following links to download adding fractions worksheets as pdf document.
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The key is to stick to the steps of the operation to a tee. Answers to Adding or Subtracting Fractions with Different Denominators 1 23 20 2 14 5 3 47 14 4 27 56 5 10 7 6 23 14 7. 1 3 4 2 5 2 1 9 5 3 13 7 3 2 4. Printable Adding And Subtracting Fractions With Unlike Denominators Worksheets Pdf might be a life saver into a property school mommy. All fractions are proper fractions.
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You have to option to set a range for the denominators. 3 bKBuptTa7 aSMo3fOtRw9aEroe6 hLvLOCRE k kAIl8lj SrXiEgxh7txsD Wr1ezsVerevNe1dk2 P SMvagdWe0 gw5it6hJ pIEnCfXibnsiWtUer XPPrCe0-dAcl1ggeZbwr0aVH Worksheet by Kuta Software LLC Answers to Adding and Subtracting Fractions with Unlike Denominators ID. This is the first series of worksheets that deal with unlike denominators. Fractions with unlike denominators Aadir a mis cuadernos 2 Descargar archivo pdf Insertar en mi web o blog Aadir a Google Classroom Aadir a Microsoft Teams Compartir por Whatsapp. Complete work with steps is shown for each problem on the answer keys.
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Download any other Fraction worksheets on this website including pdf adding fractions with unlike denominators worksheets pdf. Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6. These math worksheets are pdf files. Answers to Adding or Subtracting Fractions with Different Denominators 1 23 20 2 14 5 3 47 14 4 27 56 5 10 7 6 23 14 7. Adding fractions with different denominators.
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This is the first series of worksheets that deal with unlike denominators. The size of the PDF file is 27095 bytes. Fractions Worksheet – Adding Fractions with Unlike Denominators Author. Adding fractions worksheet maker for like and unlike denominators. 3 bKBuptTa7 aSMo3fOtRw9aEroe6 hLvLOCRE k kAIl8lj SrXiEgxh7txsD Wr1ezsVerevNe1dk2 P SMvagdWe0 gw5it6hJ pIEnCfXibnsiWtUer XPPrCe0-dAcl1ggeZbwr0aVH Worksheet by Kuta Software LLC Answers to Adding and Subtracting Fractions with Unlike Denominators ID.
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3 bKBuptTa7 aSMo3fOtRw9aEroe6 hLvLOCRE k kAIl8lj SrXiEgxh7txsD Wr1ezsVerevNe1dk2 P SMvagdWe0 gw5it6hJ pIEnCfXibnsiWtUer XPPrCe0-dAcl1ggeZbwr0aVH Worksheet by Kuta Software LLC Answers to Adding and Subtracting Fractions with Unlike Denominators ID. If there are more versions of this worksheet the other versions will be available below the preview images. Worksheets given in this section will be much useful for the students who would like to practice problems on adding fractions. Regardless of whether youre seeking to save money on curriculum or maybe your child demands extra work in fractions these worksheets can certainly help out making it simple for you concurrently. Fractions with unlike denominators Add to my workbooks 2 Download file pdf Embed in my website or blog Add to Google Classroom Add to Microsoft Teams Share through Whatsapp.
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The worksheets in this series avoid mixed numbers to introduce the skills needed to add with different denominators. Adding and subtracting fractions Other contents. Adding Unlike Fractions Different Denominators Worksheets. 3 bKBuptTa7 aSMo3fOtRw9aEroe6 hLvLOCRE k kAIl8lj SrXiEgxh7txsD Wr1ezsVerevNe1dk2 P SMvagdWe0 gw5it6hJ pIEnCfXibnsiWtUer XPPrCe0-dAcl1ggeZbwr0aVH Worksheet by Kuta Software LLC Answers to Adding and Subtracting Fractions with Unlike Denominators ID. Preview images of the first and second if there is one pages are shown.
Source: pinterest.com | 2,406 | 9,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.882461 |
https://www.physicsforums.com/threads/calculating-the-thermodynamic-efficency-of-an-electrical-energy-power-station.629843/ | 1,674,856,077,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764495012.84/warc/CC-MAIN-20230127195946-20230127225946-00755.warc.gz | 922,890,592 | 17,769 | # Calculating the thermodynamic efficency of an electrical energy power station
higheye
I have to calculate the thermodynamic efficiency of a power station. I have to assume the cooling towers maintain 70 degrees C and hence the amount of electrical energy (in J) from one mole of methan. I've been given the enthalpy of combustion of Methane. I know the thermo efficiency is (sum of net useful power + and useful thermal outputs) / total fuel input...(am i right?) i just can't figure out how to get the electrical energy! I have figured out how many moles of water the combustion of methane heats, and figured out the energy needed to take that one mole to steam at 500 degrees C (a question earlier i think it has relevance) but that's all i got, i don't know how to calculate the electrical energy output from just this information! Is there something in the question missing, or am i missing something?? Any help is greatly appreciated as I've been up all night doing this thanks
voko
It is really difficult to understand your problem. Could you describe it exactly as it is given to you, with all the data, and show what you have already done?
higheye
This is the 4th question of the sheet, it directly asks;
Once the power station ia ready, calculate its thrmodynamic efficiency, assuming cooling towers maintain a temperatue of 70 degrees C and hence the amount of electrical energy (in J) from one mole of methane.
Thats the question as it is on paper.
The whole sheet is on a power station starting up from cold. Initally the boiler contains 23,000 kg of water at a temp of 10 degree C.
I was then asked to calculate the energy need to raise water to boiling point
The energy required to vaporise all water to steam
Energy required to raise temp of steam to 500 degrees C
It told me to add these 3 results then, (from which I got 119 GJ), and stated this is the power stations (rough estimate) amount of energy needed to prepare it for electrical generation.
My next question is to calculate the mass of methane that must be burnt to prepare the power station for generation. I calculated this, and now on the 5th question that I wrote at the begining.
Data I been given: Molar heat capacity of water - 75.29 J K-1 Mol-1
'' '' '' steam - 36.54 J K-1 Mol-1
Enthalpy of Vapourisation of water - 40.656 kJ Mol-1
Enthalpy of combustion if methane - -890 kJ Mol-1
Gibbs free energy for the combustion of methane - -845 kJ Mol-1
voko
I am lead to believe that 500 C is the "high" temperature, and 70 C is the "low" temperature. This is enough to determine the Carnot cycle efficiency.
higheye
Yeah i picked up on this difference in temperature but didn't know what to do with it and what the loss of 430 C ment, i'll research the carnot cycle now thanks
higheye
efficiency = work done by system / energy put in
am i right in calculating the work done is the energy used to transfer 890 kJ from the methane, heating up the water to boiling then vapourising then heating steam to 500 C?
Is the above equation the same as: 1 - (the absolute temperature of cold resevoir/the absolute temperature of hot resevoir)
voko
efficiency = work done by system / energy put in
am i right in calculating the work done is the energy used to transfer 890 kJ from the methane, heating up the water to boiling then vapourising then heating steam to 500 C?
No, this is the work done by the heat engine, resulting from the heat it consumes. In your case the work done is spinning the generator's shaft. You can't calculate this directly, rather, you compute the efficiency in some other way and since you know the input heat, you can then estimate the work done.
Is the above equation the same as: 1 - (the absolute temperature of cold resevoir/the absolute temperature of hot resevoir)
In the case of the Carnot cycle, yes. Note that a real power plant does not use the Carnot cycle, the Carnot cycle is only the the upper bound theoretical estimate.
higheye
ah so i can take the second equation like this: 1 - ( 70 / 500 ) = 0.86
then I can re-arrange the the top one: work done = 0.86 * 890 kJ?
this come out to work done = 765.4 kJ
Is this sensible? i really don't know?
voko
ah so i can take the second equation like this: 1 - ( 70 / 500 ) = 0.86
You need to use the absolute temperature.
then I can re-arrange the the top one: work done = 0.86 * 890 kJ?
this come out to work done = 765.4 kJ
Yes, that's the principle (but the numbers are wrong). Note that what you get is mechanical energy. There is also some loss of energy when it is converted to electrical energy, but the loss is negligible. I leave it to you to find out the efficiency of power plant electrical generators to prove that.
higheye
I re-calculated the mechanical energy and resulted in 356 kJ, I have been researching and to no avail have I been able to progress with working out how much electrical energy is produced. Maybe I am looking for the wrong thing? I been looking for equations or conversions for mechanical enrergy into electrical energy through generators. Am i at least on the right path? Maybe I need sleep
voko
I re-calculated the mechanical energy and resulted in 356 kJ
I have been researching and to no avail have I been able to progress with working out how much electrical energy is produced. Maybe I am looking for the wrong thing? I been looking for equations or conversions for mechanical enrergy into electrical energy through generators. Am i at least on the right path? Maybe I need sleep
You merely need to find some facts on the current electrical generators. For example: http://www.energy.siemens.com/hq/en/power-generation/generators/sgen-4000w.htm
higheye
I mistaskenly used the upper temperature as 575.15 K instead of 773.15 K. I have repremanded this and my new results is 498.4 kJ
higheye
1 - ( 343.15 / 773.15 )
voko
This result seems correct.
higheye
Is there a way to find out how much electrical energy can be generated form 498.4 kJ of mechanical? I am seriously out of practice with area of physics sorry
higheye
is the carnot efficiency, at 56%, not the answer?
why has it given me the 1 mole information and electrical energy information?
voko
The efficiency of good electrical generators is close to 100% (see the page I linked earlier). So you can assume 100% mechanical energy is converted into electrical. Now, given the energy content of one mole of the fuel, you can find out the resultant mechanical energy (via efficiency) and thus the electrical energy. | 1,582 | 6,505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-06 | latest | en | 0.954557 |
http://www.mathworks.com/matlabcentral/answers/87952-fp-ridge-orientation-code-only-returning-45-or-135-degrees-please-help?requestedDomain=www.mathworks.com&nocookie=true | 1,480,973,494,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541839.36/warc/CC-MAIN-20161202170901-00469-ip-10-31-129-80.ec2.internal.warc.gz | 575,752,880 | 13,911 | MATLAB Answers
FP ridge orientation code only returning 45 or 135 degrees. Please help
Asked by Sam
Sam (view profile)
on 23 Sep 2013
Hi all, I am trying calculate fp orientation using equation given the image (Jain 1995). I am only getting two values 2.3562 and 0.7875 rad(45 deg and 135 deg) in the output matrix. Please help me to know where i am going wrong.
clc;clear all;close all;
img1=imread('1_1.png');
img1=normalise(img1);
[m n]=size(img1);
[gx gy]=imgradientxy(img1);
for i=9:m-9
` for j=9:n-9`
` gxy=0;gxx=0;gyy=0;`
``` for i1=-8:+8
for j1=-8:+8
gxy=gxy+((gx(i+i1,j+j1)).*(gy(i+i1,j+j1)));
gxx=gxx+power((gx(i+i1,j+j1)),2);```
` gyy=gyy+power((gx(i+i1,j+j1)),2);`
``` end
end
Q(i,j)=pi/2+0.5*(atan2(double(2*gxy),double((gxx-gyy))));
end
end```
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1 Answer
Image Analyst (view profile)
Answer by Image Analyst
Image Analyst (view profile)
on 23 Sep 2013
Are gx and gy integers or doubles before you start multiplying them together. If things are being quantized, the first thing I look at is if it's doing integer math instead of floating point math. By the way, you can use ^ to raise to a power instead of calling the power() function.
Sam
Sam (view profile)
on 23 Sep 2013
gx,gy are double.. but looking for class of variables I found a typing error in gyy=gyy+power((gx(i+i1,j+j1)),2); (for gyy power arg should be gy). :|
just one problem remains...cant compute Q's for border regions
Q is 330x240
img1 is 338x248
is there a function where i can pad img1 with 8 cols(on right) and 8 rows on bottom.????
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Learn MATLAB today! | 585 | 1,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2016-50 | latest | en | 0.753209 |
https://documen.tv/question/there-are-25-students-in-the-class-36-of-the-students-ate-hotdogs-for-lunch-how-many-students-at-20808341-60/ | 1,627,481,542,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153729.44/warc/CC-MAIN-20210728123318-20210728153318-00399.warc.gz | 224,441,412 | 17,543 | ## There are 25 students in the class. 36% of the students ate hotdogs for lunch. How many students ate hotdogs?
Question
There are 25 students in the class. 36% of the students ate hotdogs for lunch. How many students ate hotdogs?
in progress 0
1 hour 2021-07-22T08:06:11+00:00 2 Answers 0 views 0
9
Step-by-step explanation:
25 times .36 is 9!
I hope this helps! | 115 | 373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.946262 |
http://tweleve.org/winchester-treasure/16913-gold-8.html | 1,566,202,366,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314696.33/warc/CC-MAIN-20190819073232-20190819095232-00002.warc.gz | 192,495,511 | 12,253 | 1. Junior Twelever Bronze
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Yes, I've been thinking about that bird. Trouble is, I've not been able to link it to a fish. I reckon they're both together somewhere ??
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Did you have a good holiday Winnie - or is it now raining for 40 days and 40 nights?
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Oooooooo. You cheeky piggy! You know it's not St. Swithun's day yet.
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Pitter Patter ; Pitter Patter - Winnie of Winchester
5. Getting the hang of it. Copper
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So, we need to be broad as well as focussed.
Could this mean we need to use a bit more lateral thinking, in which case as we have been working on maths, how about we need to change the letters round the border to numbers? We haven't used numbers at all yet although there are many in the text given by dates.
I'll let you know if I come up with anything.
Opal
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Yes, I reckon you could be right. I've been looking at the "offsets" on each page, as each sentence starts in a different place. Could these be translated into numbers?
You see, we have several things that we think we've found. We know we're looking for a keyword, and hence a code, and although we can make guesses as to what that keyword is, it's useless unless we know how to use it. I've been playing around with vigenere squares, and now I'm so desperate that I'm thinking about vigenere rectangles
So, it's back to the numbers for me.
7. Getting the hang of it. Copper
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Well, no luck so far with the numbers!
I think it could be many moons before I solve this one, having said that, the most prominent circle in all of nature is the SUN! This being the likely symbolic meaning of the circle number pi in ancient Egyptian Numerology.
In TWT we have a pub called 'The Eclipse' we are told it was called The Eclipse because 'The Sun' was once opposite. Winnie has spoken of the 'Sun' shining in Winchester, plus Monday being a holiday. The word Monday derives from MOON - day. So should we be thinking about Astrology or Astronomy?
Found this website that maybe of interest. www.recoveredscience.com/const123Pisunphimoon
There again...
8. Getting the hang of it. Copper
Join Date
Apr 2006
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If you can open the web address go to 'Constants'.
Nice find! It combines pies, opposites and Italy's sun.
10. Originally Posted by Hog
We know we're looking for a keyword, and hence a code
Using 2 lines of the poem there are three possible references to ciphers:
"Italy's sun, France's snow"
"And come to rest in this fair place"
Italy's sun could be read as Italy's son who was Caesar, hence Caesar shift (who knows which shift, maybe 6 or 20)
France's snow - snow in French is neige, with a lot of imagination it could refer to the Vigenere cipher
fair place could be talking about a Playfair cipher | 770 | 2,988 | {"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} | 2.6875 | 3 | CC-MAIN-2019-35 | longest | en | 0.97292 |
https://bestftxtgxr.netlify.app/gutierrex30536nuhi/calculate-rate-of-interest-online-348 | 1,718,703,954,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.70/warc/CC-MAIN-20240618073942-20240618103942-00545.warc.gz | 107,296,457 | 11,624 | ## Calculate rate of interest online
Calculates principal, principal plus interest, rate or time using the standard compound interest formula A = P(1 + r/n)^nt. Calculate compound interest on an investment or savings. Compound interest formulas to find principal, interest rates or final investment value including continuous compounding A = Pe^rt. Bankrate.com provides a FREE return on investment calculator and other ROI calculators to compare the impact of taxes on your investments. CD Calculator Calculate your earnings and more. Use this CD calculator to find out how much interest is earned on a certificate of deposit (CD). Just enter a few pieces of information and this CD
Calculate your housing loan payment with MoneySmart mortgage calculator. Check interest rates, how much you can borrow, loan tenure, etc. Guide to Simple Interest Rate formula, here we discuss its uses with practical examples In the case of Simple Interest, Interest is calculated on the Loan amount, 250+ Online Courses | 1000+ Hours | Verifiable Certificates | Lifetime Access Calculates a table of the future value and interest of periodic payments. Calculate rate for long term ins policy vs straight savings. [3] 2020/01/31 23:16. Check rates and calculate PPF amount online with PPF account calculator. You can get loan against PPF account. Calculate PPF Maturity Amount @ 7.90%. 15 Apr 2019 Today, several banks offer online FD calculators, which will accurately give you the maturity period and interest earned. Effortless and convenient.
## Your estimated annual interest rate. Interest rate variance range. Range of interest rates (above and below the rate set above) that you desire to see results for.
Calculate the monthly instalments, total interest payable and total payment required for your term loan. Type of loan. Term Loan Interest Rate. %. Interest Rate Simple interest calculator with step by step explanations. Calculate Principal, Interest Rate, Time or Interest. in the formula. Free online tool by Math Warehouse! A (final amount), P ( principal), r (interest rate) or T (how many years to compound) Auto Calculate ? HDFC's home loan calculator helps you calculate your Home Loan Emi with ease . With a low-interest rate and long repayment tenure, HDFC ensures a for a home loan online from the comfort of your living room easily with Online Home Financial Analyst Training. Get world-class financial training with CFI's online certified Calculate interest rate and effective cost of loan if you know Equated Monthly Instalment (EMI)
### It pays a fixed interest rate for a specified amount of time, giving an easy-to-determine rate of return and investment length. Normally, the longer that money is left in a CD, the higher the rate of interest received.
Calculate Principal Amount, solve for P P = A / (1 + rt) Calculate rate of interest in decimal, solve for r r = (1/t)(A/P - 1) Calculate rate of interest in percent How to Calculate Simple Interest, Principal, Rate, or Time When 3 Values Are Become a Virtual Assistant using all free online tools and the skills you already Interest calculation is like percentage calculation. What do the words capital, annual interest and interest rate mean? The capital is the amount of money you got. Interest rate is the amount charged by lenders to borrowers for the use of money, expressed as a percentage of the principal, or original amount borrowed; it can also be described alternatively as the cost to borrow money. For instance, an 8% interest rate for borrowing \$100 a year will obligate a person to pay \$108 at year end. Calculate the simple interest for the loan or principal amount of Rs. 5000 with the interest rate of 10% per annum and the time period of 5 years. P = 5000, R = 10% and T = 5 Years Applying the values in the formula, you will get the simple interest as 2500 by multiplying the loan amount (payment) with the interest rate and the time period. It will take 9 years for the \$1,000 to become \$2,000 at 8% interest. This formula works best for interest rates between 6 and 10%, but it should also work reasonably well for anything below 20%. Fixed vs. Floating Interest Rate. The interest rate of a loan or savings can be "fixed" or "floating".
### Financials institutions vary in terms of their compounding rate requency - daily, monthly, yearly, etc. Should you wish to work the interest due on a loan, you can use the loan calculator. Compound interest formula. Compound interest, or 'interest on interest', is calculated with the compound interest formula.
15 Apr 2019 Today, several banks offer online FD calculators, which will accurately give you the maturity period and interest earned. Effortless and convenient. Regular Deposit \$. Deposit and Interest Frequency. Monthly. Weekly, Fortnightly, Monthly, Annually. Interest Rate % p.a.. Savings Term year/s. Calculate Investigating the impact of interest rates on savings and borrowing. Simple interest is calculated annually using the interest rate. Simple interest is always Calculate Principal Amount, solve for P P = A / (1 + rt) Calculate rate of interest in decimal, solve for r r = (1/t)(A/P - 1) Calculate rate of interest in percent How to Calculate Simple Interest, Principal, Rate, or Time When 3 Values Are Become a Virtual Assistant using all free online tools and the skills you already Interest calculation is like percentage calculation. What do the words capital, annual interest and interest rate mean? The capital is the amount of money you got.
## To calculate the monthly cost of your car loan, use the free online EMI calculator. Enter the principal loan amount, loan tenure, interest rate and processing fee
5 Dec 2017 Generally, interest on student loans is calculated daily. Number of Days in a Year (used for calculations)*. 365 Annual Interest Rate (%)*. Our compound interest calculator shows you how compound interest can increase your savings. your savings interest; the difference between saving now and saving later; how to calculate compound interest Effective interest rate : 5.12%
Check rates and calculate PPF amount online with PPF account calculator. You can get loan against PPF account. Calculate PPF Maturity Amount @ 7.90%. 15 Apr 2019 Today, several banks offer online FD calculators, which will accurately give you the maturity period and interest earned. Effortless and convenient. Regular Deposit \$. Deposit and Interest Frequency. Monthly. Weekly, Fortnightly, Monthly, Annually. Interest Rate % p.a.. Savings Term year/s. Calculate Investigating the impact of interest rates on savings and borrowing. Simple interest is calculated annually using the interest rate. Simple interest is always Calculate Principal Amount, solve for P P = A / (1 + rt) Calculate rate of interest in decimal, solve for r r = (1/t)(A/P - 1) Calculate rate of interest in percent | 1,453 | 6,868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-26 | latest | en | 0.891216 |
https://everything2.com/title/Richard+Paradox | 1,539,925,974,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512323.79/warc/CC-MAIN-20181019041222-20181019062722-00149.warc.gz | 677,245,169 | 6,316 | This paradox, proposed by the French logician Jules Antoine Richard, is summed up (hrm) with the phrase,
"The smallest number that cannot be defined by a phrase in the English language containing fewer than twenty words."
The brute-force approach to defining such a number lies in assembling a set of all possible phrases, each 19 or less words in length, that define a number, and finding the smallest number not defined by a phrase in the set. The paradox is that the statement itself is 19 words long, and thus is a member of the set it implies. Like Jabberwocks and the square root of -1, Richard's Number must remain imaginary.
Ian Stewart, in the January 2001 Scientific American ("Mathematical Recreations" sidebar, pg 103), proposes further problems with list items that refer to other list items, since their inclusion in the set may be situational. Consider the phrases:
• "The number named in the next expression, if a number is named there, and zero if not." (17 words)
• "One plus the number named in the preceding expression." (9 words)
Taken together in the order shown, neither phrase defines a number, and so they don't seem to be valid set members. With the right "neighbors", however, they would be valid expressions. | 271 | 1,238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-43 | longest | en | 0.931252 |
https://www.proofwiki.org/wiki/Brouncker%27s_Formula | 1,680,218,921,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949506.62/warc/CC-MAIN-20230330225648-20230331015648-00714.warc.gz | 1,019,430,097 | 11,628 | # Brouncker's Formula
## Theorem
$\dfrac 4 \pi = 1 + \cfrac {1^2} {2 + \cfrac {3^2} {2 + \cfrac {5^2} {2 + \cfrac {7^2} {2 + \cfrac {9^2} {2 + \cdots} } } } }$
## Also presented as
This can also be presented as:
$\dfrac \pi 4 = \dfrac 1 {1 + \cfrac {1^2} {2 + \cfrac {3^2} {2 + \cfrac {5^2} {2 + \cfrac {7^2} {2 + \cfrac {9^2} {2 + \cdots} } } } } }$
## Source of Name
This entry was named for William Brouncker. | 197 | 419 | {"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} | 3.34375 | 3 | CC-MAIN-2023-14 | latest | en | 0.66517 |
https://www.sarthaks.com/805/find-the-distance?show=806 | 1,653,370,244,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662564830.55/warc/CC-MAIN-20220524045003-20220524075003-00014.warc.gz | 883,492,382 | 14,526 | # Find the distance:
1.0k views
Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is the distance from the plane on to the top of the larger sphere?
by (9.9k points)
Let the centers of your 3 spheres of radius 1 be C1,C2,C3. Let the center of the sphere of radius 2 be C4. These 4 points forms a tetrahedron with equilateral triangle base of length 2 ( C1C2=C2C3=C3C1=2) , and a side length of 3 (C4C1=C4C2=C4C3=3). Let O be the center of the equilateral base, then OC1=2/3√3
Hence, the distance from the plane to the top is 3+√69/3. | 197 | 599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.777851 |
http://math.stackexchange.com/questions/169126/is-using-the-symbol-with-the-associative-law-of-multiplication-invalid | 1,469,787,622,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830064.24/warc/CC-MAIN-20160723071030-00096-ip-10-185-27-174.ec2.internal.warc.gz | 161,982,133 | 18,145 | # Is using the - symbol with the Associative Law of multiplication invalid?
I was trying to prove that $-(x + y) = -x - y$ and as you can see in the image below, I took the liberty of using the $-$ symbol as a number and applying the associative law with it. Is it kosher in all rigorousness given the axioms professional mathematicians use?
-
If you're not assuming commutativity I feel like it should be $-(x+y) = -y -x$. – Cocopuffs Jul 10 '12 at 17:43
@Cocopuffs Commutativity is assumed in the reals. – Pedro Tamaroff Jul 10 '12 at 17:44
@PeterTamaroff Oh, I didn't see that. Then why not use it? Just $(x+y) + -x - y = (y + x) + -x - y = 0$ by associativity. – Cocopuffs Jul 10 '12 at 17:45
There's a problem in there. You're "pulling out" the $-$ under the "associative law". I guess you know that, given $x$, $(-1) \cdot x =-x$ is its additive inverse. So what you want to do is to use the distributive law as $-(x+y)=(-1)\cdot(x+y)=(-1)\cdot x+(-1)\cdot y$ – Pedro Tamaroff Jul 10 '12 at 17:48
$$(x+y)+(-x-y)=(y+x)+(-x-y)=y+(x-x)+(-y)=y+0+(-y)=0$$ So $(x+y)$ is the additive inverse of $(-x-y)$. Hence $-(x+y)=-x-y$
Well the LHS of your equation is just saying "the additive inverse of $x+y$".
So all you have to show is that the additive inverse of $x+y$ really is the RHS of the equation, i.e. $-x-y$, then by uniqueness of inverses in a group the two must be equal. | 449 | 1,380 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2016-30 | latest | en | 0.91532 |
https://fr.mathworks.com/matlabcentral/cody/problems/43079-check-if-two-matrices-are-permutations-of-each-other/solutions/1738190 | 1,579,305,222,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250591431.4/warc/CC-MAIN-20200117234621-20200118022621-00149.warc.gz | 474,115,226 | 15,739 | Cody
Problem 43079. Check if two matrices are permutations of each other
Solution 1738190
Submitted on 27 Feb 2019
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Fail
x = [1 2 3; 4 5 6; 7 8 9] y = [3 5 6; 7 1 2; 4 9 8] isPerm = true; assert(isequal(isPermute(x,y),isPerm))
x = 1 2 3 4 5 6 7 8 9 y = 3 5 6 7 1 2 4 9 8 a = 9 b = 9 isPerm = logical 1
Matrix dimensions must agree. Error in isPermute (line 9) if x==2:51 Error in Test1 (line 4) assert(isequal(isPermute(x,y),isPerm))
2 Fail
x = [1 2; 4 5; 7 8]; y = x'; isPerm = true; assert(isequal(isPermute(x,y),isPerm))
a = 6 b = 6 isPerm = logical 1
Matrix dimensions must agree. Error in isPermute (line 9) if x==2:51 Error in Test2 (line 4) assert(isequal(isPermute(x,y),isPerm))
3 Pass
x = 1:50; y = randperm(50); isPerm = true; assert(isequal(isPermute(x,y),isPerm))
a = 50 b = 50 isPerm = logical 1
4 Fail
x = 2:51; y = randperm(50); isPerm = false; assert(isequal(isPermute(x,y),isPerm))
a = 50 b = 50 isPerm = logical 1 y = logical 0
Assertion failed.
5 Fail
x = [1 2 3; 4 5 6; 7 8 9] y = [3 5; 7 1; 4 9] isPerm = false; assert(isequal(isPermute(x,y),isPerm))
x = 1 2 3 4 5 6 7 8 9 y = 3 5 7 1 4 9 a = 9 b = 6 isPerm = logical 0
Matrix dimensions must agree. Error in isPermute (line 9) if x==2:51 Error in Test5 (line 4) assert(isequal(isPermute(x,y),isPerm)) | 598 | 1,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.55288 |
http://psychology-papers.com/342q6c0p8rupp6fd/critical-thinking-and-problem-solving | 1,670,225,596,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711013.11/warc/CC-MAIN-20221205064509-20221205094509-00738.warc.gz | 37,712,557 | 39,514 | Assignment 1: Discussion—Prescriptive and Descriptive Arguments
Descriptions “describe”—they depict the “what is” of a statement. Prescriptions “prescribe”—they express the “what ought to be” of a statement. When approaching controversial or emotionally charged subjects, critical thinkers are mindful of the different roles that facts and values play in people’s judgments and the different roles they play in different kinds of judgments. This assignment will help deepen your understanding of those distinctions.
To see how an issue can be approached using a descriptive or prescriptive argument, review these examples. Descriptive and prescriptive arguments are considered in the lecture pages for Module 3.
Based on your understanding of descriptive and prescriptive arguments, respond to the following:
• Identify a topic of interest for which arguments of different perspectives can be created.
• Construct one original descriptive argument and one original prescriptive argument for the topic you select.
Support your arguments with scholarly references. Be sure to provide citations for your sources as well as citations for a premise you state to be a fact.
Write your initial response in a total of 200–300 words. Apply APA standards to citation of sources.
By Saturday, December 3, 2016, post your response to the appropriate Discussion Area. Through Wednesday, December 7, 2016, review at least two peers’ responses. Each response should be at least 75 words in length. Critically comment on how they have used evidence in different types of arguments. Be sure to address the following:
• Identify the supporting evidence for their arguments.
• Offer an assessment of the strength of the evidence provided in support of the argument. Include a rationale for your statements. You may offer a suggestion for improved supporting evidence.
Assignment 1 Grading Criteria Maximum Points Initial Discussion Response 16 Discussion Participation 16 Writing Craftsmanship and Ethical Scholarship 8 Total: 40
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During Reconstruction, Black Codes were enacted in several former states in the Confederacy. A Republican controlled Congress would later...
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\$65 | 798 | 3,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-49 | latest | en | 0.890021 |
https://minuteshours.com/113-4-hours-in-hours-and-minutes | 1,627,496,132,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153739.28/warc/CC-MAIN-20210728154442-20210728184442-00512.warc.gz | 415,487,555 | 5,012 | # 113.4 hours in hours and minutes
## Result
113.4 hours equals 113 hours and 24 minutes
You can also convert 113.4 hours to minutes.
## Converter
One hundred thirteen point four hours is equal to one hundred thirteen hours and twenty-four minutes. | 61 | 253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-31 | latest | en | 0.838983 |
http://mathdl.maa.org/mathDL/19/?pa=reviews&sa=viewTocFront&bookId=73449 | 1,369,244,032,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702127714/warc/CC-MAIN-20130516110207-00030-ip-10-60-113-184.ec2.internal.warc.gz | 175,263,341 | 10,306 | Search
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The Joy of x: A Guided Tour of Math, from One to Infinity
Steven Strogatz
Table of Contents
Preface
Part One: NUMBERS
1. From Fish to Infinity
An introduction to numbers, pointing out their upsides (they’re efficient) as well as their downsides (they’re ethereal)
2. Rock Groups
Treating numbers concretely—think rocks—can make calculations less baffling.
3. The Enemy of My Enemy
The disturbing concept of subtraction, and how we deal with the fact that negative numbers seem so ... negative
4. Commuting
When you buy jeans on sale, do you save more money if the clerk applies the discount after the tax, or before?
5. Division and Its Discontents
Helping Verizon grasp the difference between .002 dollars and .002 cents
6. Location, Location, Location
How the place-value system for writing numbers brought arithmetic to the masses
Part Two: RELATIONSHIPS
7. The Joy of x
Arithmetic becomes algebra when we begin working with unknowns and formulas.
8. Finding Your Roots
Complex numbers, a hybrid of the imaginary and the real, are the pinnacle of number systems.
9. My Tub Runneth Over
Turning peril to pleasure in word problems
10. Working Your Quads
The quadratic formula may never win any beauty contests, but the ideas behind it are ravishing.
11. Power Tools
In math, the function of functions is to transform.
Part Three: SHAPES
12. Square Dancing
Geometry, intuition, and the long road from Pythagoras to Einstein
13. Something from Nothing
Like any other creative act, constructing a proof begins with inspiration.
14. The Conic Conspiracy
The uncanny similarities between parabolas and ellipses suggest hidden forces at work.
15. Sine Qua Non
Sine waves everywhere, from Ferris wheels to zebra stripes
16. Take It to the Limit
Archimedes recognized the power of the infinite and in the process laid the groundwork for calculus.
Part Four: CHANGE
17. Change We Can Believe In
Differential calculus can show you the best path from A to B, and Michael Jordan’s dunks help explain why.
18. It Slices, It Dices
The lasting legacy of integral calculus is a Veg-O-Matic view of the universe.
19. All about e
How many people should you date before settling down? Your grandmother knows—and so does the number e.
20. Loves Me, Loves Me Not
Differential equations made sense of planetary motion. But the course of true love? Now that’s confusing.
21. Step Into the Light
A light beam is a pas de deux of electric and magnetic fields, and vector calculus is its choreographer.
Part Five: DATA
22. The New Normal
Bell curves are out. Fat tails are in.
23. Chances Are
The improbable thrills of probability theory
24. Untangling the Web
How Google solved the Zen riddle of Internet search using linear algebra
Part Six: FRONTIERS
25. The Loneliest Numbers
Prime numbers, solitary and inscrutable, space themselves apart in mysterious ways.
26. Group Think
Group theory, one of the most versatile parts of math, bridges art and science.
27. Twist and Shout
Playing with Möbius strips and music boxes, and a better way to cut a bagel
28. Think Globally
Differential geometry reveals the shortest route between two points on a globe or any other curved surface.
29. Analyze This!
Why calculus, once so smug and cocky, had to put itself on the couch
30. The Hilbert Hotel
An exploration of infinity as this book, not being infinite, comes to an end
Acknowledgments
Notes
Credits
Index | 813 | 3,482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.868584 |
https://www.esaral.com/q/if-sin-cos-x-prove-that-sin6-cos6-4-3-x2-1-24-35119 | 1,723,341,876,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640843545.63/warc/CC-MAIN-20240811013030-20240811043030-00823.warc.gz | 575,162,988 | 11,645 | # If sin θ + cos θ = x, prove that sin6θ+cos6θ=4−3 (x2−1)24.
Question:
If $\sin \theta+\cos \theta=x$, prove that $\sin ^{6} \theta+\cos ^{6} \theta=\frac{4-3\left(x^{2}-1\right)^{2}}{4}$.
Solution:
Given: $\sin \theta+\cos \theta=x$
Squaring the given equation, we have
$(\sin \theta+\cos \theta)^{2}=x^{2}$
$\Rightarrow \sin ^{2} \theta+2 \sin \theta \cos \theta+\cos ^{2} \theta=x^{2}$
$\Rightarrow\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+2 \sin \theta \cos \theta=x^{2}$
$\Rightarrow \quad 1+2 \sin \theta \cos \theta=x^{2}$
$\Rightarrow \quad 2 \sin \theta \cos \theta=x^{2}-1$
$\Rightarrow \quad \sin \theta \cos \theta=\frac{x^{2}-1}{2}$
Squaring the last equation, we have
$(\sin \theta \cos \theta)^{2}=\frac{\left(x^{2}-1\right)^{2}}{4}$
$\Rightarrow \sin ^{2} \theta \cos ^{2} \theta=\frac{\left(x^{2}-1\right)^{2}}{4}$
Therefore, we have
$\sin ^{6} \theta+\cos ^{6} \theta=\left(\sin ^{2} \theta\right)^{3}+\left(\cos ^{2} \theta\right)^{3}$
$=\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{3}-3 \sin ^{2} \theta \cos ^{2} \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$
$=(1)^{3}-3 \frac{\left(x^{2}-1\right)^{2}}{4}(1)$
$=1-3 \frac{\left(x^{2}-1\right)^{2}}{4}$
$=\frac{4-3\left(x^{2}-1\right)^{2}}{4}$
Hence proved. | 522 | 1,264 | {"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} | 4.4375 | 4 | CC-MAIN-2024-33 | latest | en | 0.324411 |
http://mail.scipy.org/pipermail/numpy-discussion/2011-March/055457.html | 1,406,304,772,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997894319.36/warc/CC-MAIN-20140722025814-00132-ip-10-33-131-23.ec2.internal.warc.gz | 228,724,588 | 1,799 | # [Numpy-discussion] Multiple Linear Regression
Angus McMorland amcmorl@gmail....
Wed Mar 16 06:47:44 CDT 2011
```On 16 March 2011 02:53, dileep kunjaai <dileepkunjaai@gmail.com> wrote:
> Dear sir,
> Can we do multiple linear regression(MLR) in python.... is there any
> inbuilt function for MLR
Yes, you can use np.linalg.lstsq [1] for this.
Here's a quick example:
import numpy as np
# model is y = b0.x0 + b1.x1 + b2.x2
b = np.array([3.,2.,1.])
noise = np.random.standard_normal(size=(10,3)) * 0.1
bn = b[None] + noise
x = np.random.standard_normal(size=(10,3))
y = np.sum(bn * x, axis=1)
be = np.linalg.lstsq(x,y)
and be[0] should be close to the original b (3,2,1.).
[1] http://docs.scipy.org/doc/numpy-1.4.x/reference/generated/numpy.linalg.lstsq.html
> --
> DILEEPKUMAR. R
> J R F, IIT DELHI
--
AJC McMorland
Post-doctoral research fellow
Neurobiology, University of Pittsburgh
``` | 312 | 897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2014-23 | longest | en | 0.589656 |
http://oeis.org/A281863 | 1,596,994,538,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738562.5/warc/CC-MAIN-20200809162458-20200809192458-00203.warc.gz | 80,353,983 | 4,320 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A281863 Alternating powers of 60 and 10 times powers of 60. 4
1, 10, 60, 600, 3600, 36000, 216000, 2160000, 12960000, 129600000, 777600000, 7776000000, 46656000000, 466560000000, 2799360000000, 27993600000000, 167961600000000, 1679616000000000, 10077696000000000, 100776960000000000, 604661760000000000 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS These numbers are the values for the positions in the sumerian (and babylonian) alternating sexagesimal - decimal system (used at least up to 10*60^2 = 36000, but here extended). For the numbers in this mixed base system see A055643. For the number of symbols needed for representing n see A131650. For the number of digits (including 0) of the representation of n see A282622. REFERENCES Georges Ifrah, Universalgeschichte der Zahlen, Campus Verlag, Frankfurt, New York, 2. Auflage, 1987, pp.210-221. Histoire Universelle des Chiffres, Paris, 1981. From one to zero, A universal history of numbers, Viking Penguin Inc., 1985. LINKS Colin Barker, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (0,60). FORMULA a(2*n) = 60^(n/2), a(2*n+1) = 10*60^((n-1)/2), n >= 0. From Colin Barker, Feb 21 2017: (Start) a(n) = 60*a(n-2) for n>1. G.f.: (1 + 10*x) / (1 - 60*x^2). (End) MATHEMATICA LinearRecurrence[{0, 60}, {1, 10}, 21] (* or *) a[0]=1; a[1]=10; a[n_]:=60*a[n-2]; Table[a[n], {n, 0, 20}] (* Indranil Ghosh, Feb 21 2017 *) PROG (PARI) Vec((1 + 10*x) / (1 - 60*x^2) + O(x^30)) \\ Colin Barker, Feb 21 2017 CROSSREFS Cf. A131650, A055643, A282622. Sequence in context: A002493 A054364 A004309 * A219368 A052664 A090373 Adjacent sequences: A281860 A281861 A281862 * A281864 A281865 A281866 KEYWORD nonn,easy AUTHOR Wolfdieter Lang, Feb 19 2017 STATUS approved
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Last modified August 9 12:53 EDT 2020. Contains 336323 sequences. (Running on oeis4.) | 784 | 2,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2020-34 | latest | en | 0.536919 |
https://www.perlmonks.org/?node_id=427672 | 1,631,986,752,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056548.77/warc/CC-MAIN-20210918154248-20210918184248-00209.warc.gz | 985,039,375 | 6,777 | Just another Perl shrine PerlMonks
### Re^4: JAPH randomness
by awwaiid (Friar)
on Feb 03, 2005 at 15:23 UTC ( #427672=note: print w/replies, xml ) Need Help??
in reply to Re^3: JAPH randomness
*BIG SMILE*
Welcome, my friend, to the world of pseudo-random number generators! One of the best things about computers is that they are (supposed to be) deterministic; so random number generators are based on neat mathematical equations which appear random to an observer. You start with a seed, and then each number generated is based on the number(s) which have already been generated, going back to the seed.
Now, not all of these pseudo-random number generator algorithms are as "random" as others (see Knuth's work for example). Not long ago I had a Commodore-128 on which I was doing some lottery simulations and statistics to see if I could predict the lottery. I got stunningly good results... but as it turns out it was a weakness in the built-in random-number generating algorithm, not a weakness in the lottery :)
In any case, one thing that most of the pseudo-random number generators have in common is the seed. You give it the same seed, you get the same sequence of random numbers. All sorts of fun follow that. If you don't specify the seed one is created for you, based off of timing-based things like the clock and the PIDs and things like that (see function srand)
Our pseudo-random number generating days are most certainly limited, however. Hardware-based random number generators are increasingly innexpensive, and I doubt it will take long before all devices which need a random-number (all desktops for example) have one built-in. These are based on random numbers generated from physics... sort of the opposite of the timing accuracy of your quartz watch. Some are even based on magical quantum physics stuff...
I love living in the future!
Replies are listed 'Best First'.
Re^5: JAPH randomness
by wolfger (Deacon) on Feb 03, 2005 at 15:34 UTC
I knew computer randomizers were pseudo-random, it just surprised me how pseudo it was, to be able to seed the generator and come up with the same number every single time. I thought that on a system running multiple processes, the length of time to execute commands would vary marginally to produce a set (3, 6?) of "random" numbers that reliably comes up.
Hopefully the days of truly random numbers on computers comes soon.
--
Linux, sci-fi, and Nat Torkington, all at Penguicon 3.0
perl -e 'print(map(chr,(0x4a,0x41,0x50,0x48,0xa)))'
OTOH, seeding the random number generator to allways get back the same results is very handy during testing.
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for(ref bless{},just'another'perl'hacker){s-:+-\$"-g&&print\$_.\$/}
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Notices? | 746 | 3,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.950418 |
https://www.ufoscience.org/what-is-the-relationship-between-effect-size-and-power/ | 1,696,118,655,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510730.6/warc/CC-MAIN-20230930213821-20231001003821-00224.warc.gz | 1,129,246,763 | 11,169 | ## What is the relationship between effect size and power?
Like statistical significance, statistical power depends upon effect size and sample size. If the effect size of the intervention is large, it is possible to detect such an effect in smaller sample numbers, whereas a smaller effect size would require larger sample sizes.
## Why does power increase with effect size?
As the sample size gets larger, the z value increases therefore we will more likely to reject the null hypothesis; less likely to fail to reject the null hypothesis, thus the power of the test increases.
Is Correlation an effect size?
The Pearson product-moment correlation coefficient is measured on a standard scale — it can only range between -1.0 and +1.0. As such, we can interpret the correlation coefficient as representing an effect size. It tells us the strength of the relationship between the two variables.
### How do you calculate the effect size between two groups?
Effect size measures the intensity of the relationship between two sets of variables or groups. It is calculated by dividing the difference between the means pertaining to two groups by standard deviation.
### Does effect size affect power?
The statistical power of a significance test depends on: • The sample size (n): when n increases, the power increases; • The significance level (α): when α increases, the power increases; • The effect size (explained below): when the effect size increases, the power increases.
What does it mean to have a large effect size?
Effect size tells you how meaningful the relationship between variables or the difference between groups is. It indicates the practical significance of a research outcome. A large effect size means that a research finding has practical significance, while a small effect size indicates limited practical applications.
## How do you calculate effect size and power?
Generally, effect size is calculated by taking the difference between the two groups (e.g., the mean of treatment group minus the mean of the control group) and dividing it by the standard deviation of one of the groups.
## How do I calculate effect size?
The effect size of the population can be known by dividing the two population mean differences by their standard deviation.
Is a large effect size good?
Effect size tells you how meaningful the relationship between variables or the difference between groups is. A large effect size means that a research finding has practical significance, while a small effect size indicates limited practical applications. | 475 | 2,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-40 | latest | en | 0.920274 |
https://mathematica.stackexchange.com/questions/269652/how-to-make-these-callout-labels-not-overlap | 1,713,296,707,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817106.73/warc/CC-MAIN-20240416191221-20240416221221-00804.warc.gz | 354,963,745 | 43,190 | # How to make these callout labels not overlap
This is my code:
list = {{1368, 1398}, {1399, 1402}, {1403, 1424}, {1424, 1425},
{1426, 1435}, {1436, 1449}, {1450, 1457}, {1457, 1464},
{1465, 1487}, {1488, 1505}, {1506, 1521}, {1522, 1566},
{1567, 1572}, {1573, 1620}, {1620, 1620}, {1621, 1627},
{1628, 1644}};
ListPlot[
Table[
Callout[{0, -First[i]}, First[i], Left, LabelVisibility -> All],
{i, list}
],
Axes->False
]
Notice that two points are very close to each other, which causes their labels to overlap. I really don't want to set the coordinates of a point separately. Is there an elegant way to automatically avoid overlap of their labels?
The question may seem a Convex quadratic minimization with linear constraints:
\begin{align}\mathrm{minimize}&\; \sum_\limits{i=1}^n{(x_i-a_i)^2}\\\mathrm{subject\;to}&\; x_{i+1}-x_i>h,\;i=1,2,\cdots,n-1\end{align}
there, $$a_i$$ is known data in your "list", $$x_i$$ is the appropriate position of label $$a_i$$.
It's easy to solve $$x_i$$ with the function "FindArgMin" in Mathematica.
But I can't find out a self-adaption font-size in the function "Style".
(*define function: join line*)
line[pLeft_, pRight_] :=
With[{meanX = Mean[{pLeft[[1]], pRight[[1]]}]},
BezierCurve[{pLeft, {meanX, pLeft[[2]]}, {meanX, pRight[[2]]},
pRight}]];
(*define function: show label*)
showLabel[yData_, label_, labelH_, labelS_, imageHeight_, left_ : 0,
right_ : 100] := Block[{
n = Length[yData],
y = Sort[yData],
vX, ySolve, minmax, scale, object, condition
},
(*solve appropriate position*)
vX = ToExpression["x" <> ToString[#] & /@ Range[n]];
object = Total[(vX - y)^2];
condition = And @@ (# > labelH + labelS & /@ Differences[vX]);
ySolve = FindArgMin[{object, condition}, vX];
(*sacle*)
minmax = MinMax[Flatten[{y, ySolve}]];
scale = imageHeight/Differences[minmax][[1]];
(*Show*)
Graphics[{
{(*right vertical line*)
Opacity[0.5], AbsoluteThickness[1], Gray,
Line[{{right, minmax[[1]]}, {right, minmax[[2]]}}]},
{(*join line*)
Opacity[0.5], AbsoluteThickness[1], Gray,
Table[line[{left, ySolve[[k]]}, {right, y[[k]]}], {k, n}]},
{(*right data point*)
RGBColor["#1d3557"], AbsolutePointSize[3],
Point[{right, #} & /@ y]},
{(*decorate point*)
White, AbsolutePointSize[4], Point[{left, #} & /@ ySolve]},
{(*label*)
RGBColor["#457b9d"],
Table[Inset[
FrameMargins -> Tiny, ImageSize -> {Automatic, scale labelH},
Background -> RGBColor["#f1faee"]], {left, ySolve[[k]]},
Right], {k, n}]}
},
ImageSize -> {Automatic, imageHeight},
PlotRange -> {Automatic, minmax + {-labelH, labelH}}]
];
(*Demo*)
y = Sort[RandomReal[{0, 1000}, 15]];
showLabel[y, y, 30, 5, 500, 0, 100]
The code and image for your data:
list = {{1368, 1398}, {1399, 1402}, {1403, 1424}, {1424, 1425}, {1426,
1435}, {1436, 1449}, {1450, 1457}, {1457, 1464}, {1465,
1487}, {1488, 1505}, {1506, 1521}, {1522, 1566}, {1567,
1572}, {1573, 1620}, {1620, 1620}, {1621, 1627}, {1628, 1644}};
y = Sort[list[[;; , 1]]];
showLabel[y, y, 12, 3, 400, 0, 30]
• This is very nice! Jun 25, 2022 at 10:15
The code below distributes the labels on either sides of the plot points, trying to ensure that labels for values that are very close end up on opposite sides. I also "cheated" a little by increasing the AspectRatio: this got me some more vertical space and improved the vertical separation between labels that were still too close even after the left-right separation.
threshold = 50;
values = list[[All, 1]];
left =
First[#, #] & /@
DeleteDuplicatesBy[Nearest[values, values, {2, threshold}], Sort];
right = Complement[values, left];
annotated =
Function[{list, position},
Callout[
{0, -#}, #,
position, LabelVisibility -> All
] & /@ list],
{{left, right}, {Left, Right}}
];
ListPlot[
annotated,
Axes -> False, PlotStyle -> Black,
AspectRatio -> 1.1
]
Here is an alternative presentation that retains all labels on the left of the data, but differentiates them using short vs. long leaders in the Callout:
short = left; long = right;
annotated2 =
Function[{list, length},
Callout[
{0, -#}, #,
Left, LabelVisibility -> All,
] & /@ list
],
{{short, long}, {10, 50}}
];
ListPlot[
annotated2,
Axes -> False, PlotStyle -> Black,
AspectRatio -> 2
]
• Hi, thanks for your answer. But all labels should be on left, which is important for me.
– yode
Jun 20, 2022 at 1:57
• @yode Sure would have been nice to know that beforehand! But if you want the labels on the same side, what other solution do you envision apart from just stretching the aspect ratio to make more space vertically? I guess you could also use my data splitting code to make the callout leaders longer for some of the labels, but keep them all on the same side. Would that be acceptable, if it can be done? Jun 20, 2022 at 2:22
• Yes, I can accept longer leaders.
– yode
Jun 20, 2022 at 2:44
• What is your values?
– yode
Jun 20, 2022 at 4:56
• @yode Added, sorry for the omission. It's just list[[All, 1]] Jun 20, 2022 at 5:08
In fact I think avoiding overlap is a function that the Callout itself needs to implement. I have make workaround:
box = {};
ListPlot[Table[dist = Min[
EuclideanDistance[#, {0, -First[i]} - {0.06, 0}] & /@ box];
If[dist < 7, pos = {0, -First[i]} - {0.06, 7 - dist},
pos = {0, -First[i]} - {0.06, 0}]; AppendTo[box, pos];
Callout[{0, -First[i]}, First[i], pos, LabelVisibility -> All], {i,
list}], Axes -> False, ImageSize -> 600]
Or use DynamicName,DynamicLocation and stretchText here make a custom Callout:
pts = Table[
DynamicName[Point[{0, -First[list[[i]]]}],
TemplateApply["point<*i*>"]], {i, Length[list]}];
textboxs = {};
labels = Table[
dist = Min[
RegionDistance[#, {0, -First[list[[i]]]} - {25, 0}] & /@
textboxs];
If[dist < 5 + 2, pos = {0, -First[list[[i]]]} - {25, 5 + 2 - dist},
pos = {0, -First[list[[i]]]} - {25, 0}];
AppendTo[textboxs, Rectangle[pos, pos + {10, 5}]];
DynamicName[
stretchText[ToString[First[list[[i]]]], pos - {0, 5/2}, {10, 5}],
TemplateApply["label<*i*>"]], {i, Length[list]}]; | 1,962 | 5,932 | {"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": 5, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-18 | latest | en | 0.616829 |
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May 24 comment RSA with probable primes "crept happens" $\: \mapsto \:$ "crept in happens" $\;\;\;\;$ May 24 comment RSA with probable primes residuals $\mapsto$ residual $\;$ May 24 comment Executing encrypted code? May 24 comment RSA with probable primes May 23 revised Does the One Time Pad rely on confusion or diffusion? fixed title's spelling May 23 revised How to prove that someone encrypted a specific (large) chunk of data fixed title's grammar May 23 comment Huffman encoding of hashes How about a base-256 encoding? $\:$ (aka, not encoded at all) $\;\;\;\;$ May 23 comment Huffman encoding of hashes They can be encoded with just 20 characters. $\;$ May 23 comment Huffman encoding of hashes No, but getting farther away from that requires more output for a given security level. $\hspace{1.41 in}$ May 23 revised Factorization of a number obtained by a modular multiplication operation can reveal factors of the used operands? fixed spelling and grammar May 22 comment Authenticated EC key exchange without a signing/signature scheme? @StevePeltz : $\:$ Yes. $\;\;\;\;$ May 22 revised Authenticated EC key exchange without a signing/signature scheme? changed PAKE link May 22 comment Hard-core predicates: should the adversary be given $1^n$? $B$ can just run $A$ with $n$ ranging from $0$ to $\:2\hspace{-0.03 in}\cdot \hspace{-0.03 in}\operatorname{length}(\hspace{.03 in}y)\:$,$\:$ since injective functions have negligible probability of shrinking random inputs by large amounts. $\;\;\;\;$ May 22 comment Hash length extension attack - SHA256 to 512 - impossible, correct? Of course, it would probably be better is the user can see a commitment to the secret instead of the SHA256 hash of the secret, and if the user-key is then HMACed with the secret instead of them being hashed with SHA512. $\;$ May 21 comment Would this method deliver a perfectly non-malleable encryption for at least two blocks? "The second block" of the ciphertext "includes the propagation ... intermediate block". $\:$ "The second block" of the plaintext is not affected by the IV. $\;\;\;\;$ May 21 comment Would this method deliver a perfectly non-malleable encryption for at least two blocks? The version you have at the moment (you might still be editing) is still malleable because changing $\hspace{.42 in}$ the IV would change block1 of the decryption result but not block2 of the decryption result. $\hspace{.92 in}$ May 21 comment Would this method deliver a perfectly non-malleable encryption for at least two blocks? Now that I'm actually looking at your decryption procedure, I see that your scheme is very malleable, $\hspace{.13 in}$ since xoring the IV with any string will xor the result of decryption with the same string. $\hspace{1.19 in}$ May 21 comment Would this method deliver a perfectly non-malleable encryption for at least two blocks? You should also note that if 128 is large, then there is a simple&standard way to do the no-ciphertext-expansion version of what you're trying for, although it's slightly less efficient than your third version. $\;$ May 21 comment Would this method deliver a perfectly non-malleable encryption for at least two blocks? It looks like your third version requires 4 encryptions per block, 2 with each key. $\;$ May 21 comment prepaid meters that rely on a disconnected system Why doesn't asymmetric "fit the architecture of one server sending to many clients"? $\hspace{1.29 in}$ | 829 | 3,520 | {"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} | 2.609375 | 3 | CC-MAIN-2016-07 | longest | en | 0.851164 |
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1 1. Mark the correct statement(s) 1.1 A theorem in Boolean algebra: a) Can easily be proved by e.g. logic induction b) Is a logical statement that is assumed to be true, c) Can be contradicted by another theorem in the same system d) Is the main building block of all computers, e) None of the above. 1.2 Sum of products in principle: a) Uses more OR gates than AND gates, b) Uses more AND gates than OR gates, c) Uses XOR gates in combination with, d) None of the above. 1.3 Multiplexer: a) Has usually more inputs than outputs, b) Has usually more outputs than inputs, c) Can be used to implement logic functions in a convenient way, d) Cannot be implemented using standard basic logic gates as NOR, XOR, NAND, e) None of the above. 1.4 An axiom: a) Can easily be proved by induction, b) Is a logical statement that is assumed to be true, c) Can be contradicted by another axiom in the same system d) Is the main building block of all computers, e) None of the above 1.5 Product of sums in principle: a) Uses more OR gates than AND gates, b) Uses more AND gates than OR gates, c) Uses XOR gates in combination with inverters, d) None of the above
2 1.6 Ripple-carry adder: a) Gives simultaneously the result of addition of all bits, b) Adds MSB first and LSB last when the carry is propagated, c) Adds LSB first and MSB last when the carry is propagated, d) Adds bits of binary numbers disregarding carry, e) None of the above 1.7 A Karnaugh map for a function of six (6) variables would have: a) 8 cells, b) 16 cells, c) 32 cells, d) 64 cells, e) 128 cells. 1.8 A priority encoder with 4 inputs has: a) 2 outputs, b) 3 outputs, c) 4 outputs, d) 16 outputs. 1.9 The product of maxterms f x, y, w M 0,1,4,7 b) f x, y, w m 2,3,5, 6, b) f x, y, w m 0,1,4, 7, c) f x, y, w m 2,3,5, 6, d) f x, y, w M 2,3,5,6 '. can be written as: 1.10 Which of the following is an axiom of Boolean algebra: a) x. x = x, b) = 1, c) x. x = 0, d) x + x = A multiplexer with 4 select inputs has: a) 1 output, b) 2 outputs, c) 4 outputs, d) 16 outputs.
3 1.12 Which of the following Boolean equations is in Product-Of-Sums (POS) form? a) F = (A B C) (A C ) (A B C) b) F = A B C + A (B + C) c) F = (A + B + C ) (A + C) (A + B + C) d) F = A B C + A C + A B C 2. Answer in your own words 2.1 What does DeMorgans theorem state? 2.2 What does Shannon s expansion theorem state? 2.3 What is a DECODER? 2.4 What is a MULTIPLEXER / DEMULTIPLEXER? 2.5 Analyze the CMOS circuit shown in the figure! Fill the truth table and state what logical function does it implement? Indicate pull-up and push-down networks! A B C i Out Write the expression for the function f(a,b) and make the truth-table of the circuit!
4 2.7 What are the outputs Y in the figures below?! 2.8 What is terribly wrong in the figure below?! 2.9 What is the output F in the figure below?! 2.10 What is a maxterm? 2.11 What is a minterm? 2.12 Determine by perfect induction whether the following is true or false A' + A B = A' + B
5 2.13 What is the simplest Boolean expression for the circuit shown below 2.14 Using DeMorgan's theorem and Boolean algebra determine whether the two circuits shown here are equivalent or not 3. Solve the following problems 3.1 Make a truth table of function F(x0, x1, x2, x3) that has output logic 1 when inputs form binary numbers divisible by three, and logic 0 otherwise. Construct the reduced circuit using K-map to optimize the expression for F(x0, x1, x2, x3)! 3.2 A minority circuit is one whose output is equal to 1 if the input variables have less 1 s than 0 s. Show the truth table for a 3-input version of this circuit. Write the expression for the reduced circuit using K-map to optimize F(x, y, z)! 3.3 Make a truth table of function F(x0, x1, x2, x3) that has output logic 1 when inputs form a prime binary number, and logic 0 otherwise. Construct the reduced circuit using K-map to optimize the expression for F(x0, x1, x2, x3)! 3.4 Using Karnaugh map, find a minimal sum-of-products expression for the following function: F(x,y,z)= (m 1 m 2 m 4 m 5 m 6 )
6 3.5 Using 2-to-1 MULTIPLEXER, implement the function from previous question. 3.6 Use Karnaugh Map to simplify the following Boolean function F A,B,C, = (0,2,3,10,11,1,1 ) into a) sum-of-products form b) product-of-sums form 3.7 Using 8-to-1 MULTIPLEXER and some logical gates, implement the function from question Implement the following Boolean function with 4x1 multiplexer and external gates F A,B,C, = (0,1,2,,, 11) 3.9 Determine the outputs A and B as functions of inputs x, y and z. Can you tell what device is implemented using the multiplexers (for BONUS points)? x y z A B Number Systems, Logic gates and Boolean algebra 4.1 Simplify the following Boolean function to a minimum number of literals: a) A' B C + A B' CD + B C + B C D + BCD b) ((A B ) + (C ) ) + A +
7 c) X X + Y Y + X Y + X Y d) (AB) + (A + B) (B + B) (A + B ) 4.2 Convert the following binary numbers to the indicated bases a) to octal b) to hexadecimal c) to decimal 4.3 Perform subtraction using 2 s complement of the subtrahend: Simplify the following Boolean function to a minimum number of literals: x y z+(x+y )(xz) 4.5 Implement the function from Problem 4.4 using ONLY NAND gates, 4.6 Convert the decimal number 246 into binary and hexadecimal numbers! 4.7 Convert the number 145 (6) into a decimal number! 4.8 Perform subtraction using 2 s complement of the subtrahend: Convert the hexadecimal number 187 into binary and decimal numbers! 4.10 Find the base X if 101 (X) =?? (10) = 62 (8) 4.11 Complete the following table (except the shaded cells) of equivalent values Binary Octal Decimal Hexadecimal
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MULTIMEDIA COLLEGE JALAN GURNEY KIRI KUALA LUMPUR
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# ECS 170 Project 1
Michael Riedlin
993807860
Oleg Lokhvitsky
994557753
Cyprian Bergonia
994627191
Table of Contents Heuristic Admissibility Division Cost Function Terminology Cost Function Heuristic Function Heuristic Path Diagram Heuristic Explanation Heuristic Admissibility Proof Exponential Cost Function Additional Terminology Cost Function Heuristic Function Heuristic Explanation Heuristic Path Diagram Heuristic Admissibility Proof Algorithms Division Cost Function A* Bidirectional A* Dual Heuristic Bidirectional A* Exponential Cost Function Metrics Division Cost Function Exponential Cost Function Code Division Cost Function Exponential Cost Function
Division Cost Function
Terminology C is the height of the current node N is the height of the node being considered E is the height of the end node S is the Chebyshev distance between the current node and the end node S = max( | x1 - x2 |, | y1 - y2 | ) S is the Chebyshev distance between the node being considered and the end node S can be either S - 1, S, or S + 1 Cost Function (1) Heuristic Function (2a) (2b) Heuristic Path Diagram
Heuristic Explanation The idea is to tunnel at a height of 1 to the endpoint. Barring 0, a half is the minimum value a cost can take and spaces represents the minimum number of terms in the final cost function for the path. The reason we can ignore zeros is because we know that no two adjacent spaces can be zero. Climbing up from a zero height will cost at least one giving a minimum combined term for the two spaces equal to 1, the same cost as though we had tunneled. Heuristic Admissibility Proof We will prove that this heuristic is admissible by proving that it is consistent. First, we prove that equation 2a is consistent. (3a) (3b)
(3c) (3d) Equation 3d will always hold making the heuristic consistent and therefore admissible. Now, we prove that equation 2b is consistent. (3e) (3f) (3g) (3h) (3i) (3j) (3k) If N and C are small (N = C = 1) (3l) (3m) If N and C are large (N = C = 256) (3n) (3o) If N is small and C is large (N = 1, C = 256) (3p) (3q) If N is large and C is small (N = 256, C = 1) (3r) (3s)
## Exponential Cost Function
Additional Terminology D is the difference in height between current node and end node ( ). ) R1 is the remainder after dividing S by the absolute value of D ( R1 represents the steeper part of the path generated by our heuristic. R2 is the difference between S and R1 ( ) R2 represents the less steep part of the path generated by our heuristic. Cost Function (4) Heuristic Function
## (5) (6a) (6b) (6c) (6d)
(7a) (7b) (7c) (7d) Please note: Equations 5, 6a, 6b, 7a, 7b are derived from equations 6c and 7c (which are really the same equation) by plugging in the given conditions. Heuristic Explanation This heuristic represents a scenario where we descend or climb as smoothly as possible (as in the following picture), but only climb or descend an integer height per step. Because the heights are represented in integer values from 0 to 255, the smallest step we can take (that also changes height) will have a height difference of . The best case scenario involves taking steps with the smallest increments. Heuristic Path Diagram
## Heuristic Admissibility Proof First, we prove that it is admissible when
We can prove that the optimal path to take when is a flat path through a proof by contradiction. Imagine that there is a better path from C to E than a flat path. Then, that path must change height, either up and then down, or down and then up. Going down will help save cost over going flat, but going up is going to increase cost over going flat. By going down, we save (8a) By going up, we lose (8b) So, both operations together increase our path cost by (8c) This path will then cost more than going flat - thats a contradiction. What if we went down and up by more than 1 height unit? By going down, we save (8d) By going up, we lose (8e) So, both operations together increase our path cost by (8f) This path will then cost more than going flat - thats a contradiction. Second, we prove that it is admissible when . There are four possible alternate paths that we must disprove: 1. We climb extra on leg 1, and climb less somewhere on leg 1.
## 4. We climb extra on leg 2, and climb less somewhere on leg 2.
We will prove that our heuristic is admissible by showing that any alteration to our path produces a longer path, and thus our heuristic must be admissible. The key of this proof is that 2 path segments (one steeper, one less steep) are worse than a single path segment whose steepness is in the middle. For any path segment P with a slope of M and a length of L we can divide it into two path segments: P1 with a slope of M1 < M and a length of L1 P2 with a slope of M2 > M and a length of L2 We can infer: (9a) (9b) (9c) If we start at height H, then the cost of P is (10a) Likewise, if we start at height H, then the cost of P1 is (10b) And, if we start at height H, then the cost of P2 is (10c) So, now we will show than the cost of P is smaller than the sum of the costs of P1 and P2.
(11a) (11b) (11c) (11d) We can see that (12a) (12b) (12c) (12d) Finally, we can deduce that if the cost savings for P1 is less than the cost deficit for P2, then P is better than the combined path of P1 and P2. (13a) (13b) (13c) (13d) Taking the worse case scenario (14a) (14b) Clearly, equation 14b always holds. So, we have proved that any path P is at least as good as any two paths P1 and P2 that can be derived from it. That directly proves our cases #1 and #4 above. Cases #2 and #3 are proved throughly simply extending this idea: If breaking a single path into 2 is bad, then breaking each of those 2 paths into more pieces (or breaking the original into 3 or more pieces) is even worse. Finally, we prove that it is admissible when . However, this case is exactly symmetric to the case above (just swap the start and end points!), so, the proof is identical to the one given above.
Algorithms
Division Cost Function
A* The first algorithm we implemented was a simple A* algorithm using a PriorityQueue for the open list, a HashSet for the closed list and a HashMap to keep track of the g-values. This algorithm works functionally identical to what was described in class. It gives us an average time savings of 75% and nodes expanded savings of 80%. Bidirectional A* Our next step in refining the algorithm was running the A* bidirectionally. The basic idea is to just start one A* that searches for a path from start to end, and start another A* that searches for a path from end to start, and then wait until they meet.
Unfortunately, the node as which they meet (defined as a node that one algorithm tries to expand that is already in the closed (or open) list of the other algorithm) has no guarantee of being on the shortest path. So, how the Bidirectional search works is by using this (possibly) sub-optimal path as an upper bound on the f-values of nodes and expands all nodes of the reverse A* search that have an f-value thats smaller than this upper bound. After this is done, the forward A* is run again with the limitation that it is only allowed to expand nodes that are in the closed list on the reverse A* search. The outline of the algorithm is as follows: 1. Run both forward A* and reverse A* (expanding one node at a time from either forward A* or reverse A* depending on which node has the smaller f-value). Exit condition is that the node being examined already exists in the open or closed list of the other A*. 2. Run reverse A*. Exit condition is that the node being examined has a higher f-value than the path cost of the path discovered in step 1. 3. Run forward A*, expanding only nodes that are in reverse A*s closed list. Exit condition is the same as for regular A*. The optimality of this algorithm has been proved numerous times in various papers and we will just outline the way to think about it. It is basically a proof by contradiction. The algorithm is basically the same as A*, but it only allowed to expand nodes with an f-value that corresponds to some actual path. So, all the nodes on the optimal path will be expanded, since if there was a node that was on the optimal path, it has to have an f-value less than the cost of the optimal path, which is less than the cost of the suboptimal path found - so this node must be expanded in step 2. Dual Heuristic Bidirectional A* We attempted to improve on the performance of Bidirectional A* by capitalizing on the fact that the first path found (in Step 1) does not have to be optimal. So, we dont even need to use an admissible heuristic to find it. So, we implemented a version of the Bidirectional A* search where the reverse A* algorithm uses a secondary heuristic during step 1. The efficiency of this algorithm compared to the regular Bidirectional A* greatly depends on this secondary heuristic. Unfortunately, we could not find good secondary heuristics for the two cases we care about: 1. Division Cost Function on Seed Maps 2. Exponential Cost Function on Mt. St. Helen Map We did get a very good improvement (27.2% less nodes expanded, 20.3% less time taken) when using the algorithm on the map of Mt. St. Helen with the Division Cost Function. Since we couldnt get an improvement for our two cases of interest, we did not end up using this algorithm, and used the regular Bidirectional A* search for the Division Cost Function.
## Exponential Cost Function
We utilized all the same algorithms as described above for the Exponential Cost function but found that both Dual Heuristic Bidirectional A* search and regular Bidirectional A* search performed worse than simple A* search on the map of Mt. St. Helen with the Exponential Cost Function. So, we tried to optimize the A* search by utilizing our pre-existing knowledge of the geometry of this particular problem. We decided that we knew that certain areas of the map wouldnt be used because they represent geometry that looks good at the beginning, but looks bad later. For example, one wouldnt go into the crater of Mt. St. Helen even though its very easy to go in there since its difficult to get out.
So, we came up with a technique of height pruning. Basically, we defined radially expanding zones away from the end point and gave each one of those zones a minimum and a maximum height range within which nodes should be expanded. If a nodes height in one of those zones is outside this range, it is manually pruned. In effect, our algorithm is very fast on the map of Mt. St. Helen, but will not work properly (might not find the optimal path) on any other geometry. It also depends on the end point since we defined our zones around it.
Metrics
Division Cost Function (We used Bidirectional A*)
Seed 1 Path Cost Uncovered Time Taken Seed 2 Path Cost Uncovered Time Taken Seed 3 Path Cost Uncovered Time Taken Seed 4 Path Cost Uncovered Time Taken Seed 5 Path Cost Uncovered Time Taken Dijkstra
198.59165501141644 162340 1217
A*
198.59165501141644 35590 321
Bidirectional A*
198.59165501141644 19256 117
Dijkstra
198.56141550095256 162364 1231
A*
198.56141550095256 35572 281
Bidirectional A*
198.56141550095256 19362 103
Dijkstra
198.4864317411443 162247 1205
A*
198.4864317411443 35570 292
Bidirectional A*
198.4864317411443 19204 117
Dijkstra
198.70066424826052 162263 1255
A*
198.70066424826052 35588 302
Bidirectional A*
198.70066424826052 19391 120
Dijkstra
198.25499446810397 161946 1235
A*
198.25499446810397 35567 284
Bidirectional A*
198.25499446810397 19550 107
## Mt. St. Helen Path Cost Uncovered Time Taken
Dijkstra
548.3684300960452 1094119 11539
A*
548.3684300960452 334548 3801
Bidirectional A*
548.3684300960452 253619 2496
## Exponential Cost Function (We used Height Pruned A*)
Mt. St. Helen Path Cost Uncovered Time Taken Dijkstra
515.6645805015318 1203049 14754
A* (Height Pruned)
515.6645805015318 55232 328
Bidirectional A*
515.6645805015318 113647 1408
Code
Division Cost Function
import import import import import import java.awt.Point; java.util.List; java.util.LinkedList; java.util.PriorityQueue; java.util.HashSet; java.util.HashMap;
public class MtStHelensDiv_994557753 extends AStarBiAI_div { public final boolean printPath = false; } abstract class AIModuleH implements AIModule { public TerrainMap map; public Point start; public Point end; public double getHeuristic(final TerrainMap map, final Point pt1, final Point pt2) { return 0.0; } public final double getHeuristic(final Point pt1, final Point pt2) { return this.getHeuristic(this.map, pt1, pt2); } public final double getHeuristic(final Point p) { return this.getHeuristic(this.map, p, this.end); } public final double getHeuristic() { return this.getHeuristic(this.map, this.start, this.end); } public final void printPath(List<Point> path) { System.out.println(this.start.x + ", " + this.start.y); System.out.println("========================="); for (int i = 0; i < path.size(); i ++) {
Point p = path.get(i); System.out.println(p.x + ", " + p.y + ", " + map.getTile(p) + ", " + Node.distanceTo(p, this.end)); } System.out.println("========================="); System.out.println(end.x + ", " + end.y); } } class Node extends Point implements Comparable { private SearchAlgorithm search; public Node parent; // public int x; // public int y; public int z; private double g; private double h; public Node(SearchAlgorithm search, int x, int y, double g, Node parent) { this.search = search; this.x = x; this.y = y; this.z = (int)this.search.ai.map.getTile(this); this.g = g; this.h = -1; this.parent = parent; } public Node(SearchAlgorithm search, Point p, double g, Node parent) { this(search, p.x, p.y, g, parent); } public double f() { return this.g() + this.h(); } public double g() { return this.g; } public double h() { if (this.h != -1) return this.h; return this.h = this.search.getHeuristic(this, this.search.end); } public static int distanceTo(Point pt1, Point pt2) { return Math.max( Math.abs(pt1.x - pt2.x), Math.abs(pt1.y - pt2.y) ); } public int distanceTo(Point p) { return Node.distanceTo(this, p); } public List<Point> tracePath() { LinkedList<Point> path = new LinkedList<Point>(); Node node = this; while (node != null) { path.addFirst(node); node = node.parent; } return path;
} public List<Point> traceReversePath() { LinkedList<Point> path = new LinkedList<Point>(); Node node = this; if (node != null) node = node.parent; while (node != null) { path.add(node); node = node.parent; } return path; } @Override public int compareTo(Object node) { double a = this.f(); double b = ((Node)node).f(); if (a < b) return -1; if (a > b) return 1; return 0; } } abstract class SearchAlgorithm { public AIModuleH ai; public Node start, end, current; public abstract Node step(); public abstract double getCost(Point pt1, Point pt2); public abstract double getHeuristic(Point pt1, Point pt2); public boolean allowExpand(Point p) {return true;} } class AStarSearch extends SearchAlgorithm { protected PriorityQueue<Point> open; protected HashSet<Point> closed; protected HashMap<Point, Double> gvalues; public AStarSearch(AIModuleH ai, Point start, Point end) { // Initialize this.ai = ai; this.open = new PriorityQueue<Point>(); this.closed = new HashSet<Point>(); this.gvalues = new HashMap<Point, Double>(); // Start this.start = new Node(this, start, 0, null); this.open(this.start, 0); // End this.end = new Node(this, end, 0, null); } public Node step() { // 4. Expand Current Node this.expand(this.current); // 1. Get Next Best Node this.current = this.pop(); // 2. Close Current Node if (this.current != null) this.close(this.current, this.current.g());
// 3. Return Current Node for Examination return current; } protected void close(Point p, double g) { if (p == null) return; this.closed.add(p); } protected void unclose(Point p) { if (p == null) return; this.closed.remove(p); } public boolean isclosed(Point p) { if (p == null) return false; return this.closed.contains(p); } protected void open(Point p, double g) { if (p == null) return; this.open.add(p); this.gvalues.put(p, g); } public void unopen(Point p) { if (p == null) return; this.open.remove(p); } public boolean isopen(Point p) { if (p == null) return false; return this.open.contains(p); } public double getCost(Point p) { Double c = this.gvalues.get(p); return (c == null ? Double.MAX_VALUE : c.doubleValue()); } public double getCost(Point pt1, Point pt2) { return this.ai.map.getCost(pt1, pt2); } public double getHeuristic(Point pt1, Point pt2) { return this.ai.getHeuristic(pt1, pt2); } public Node top() { if (this.open == null || this.open.size() == 0) return null; return (Node)this.open.peek(); } protected Node pop() { if (this.open == null || this.open.size() == 0) return null; return (Node)this.open.poll(); } protected void expand(Node n) { if (n == null) return; for (Point neighbor : this.ai.map.getNeighbors(n)) { // If neighbor is closed (for consistent heuristics) if (this.isclosed(neighbor)) continue;
// Get potential new path value double cost = n.g() + this.getCost(n, neighbor); // If neighbor is closed (for non-consistnet heuristics) /* if (this.isclosed(neighbor)) { if (cost >= this.getCost(neighbor)) // If new path is worse continue; else // If new path is better this.unclose(neighbor); } */ // If neighbor is open if (cost >= this.getCost(neighbor)) continue; // Custom pruning if (!this.allowExpand(neighbor)) continue; // If neighbor has been open at one point in time before if (this.getCost(neighbor) != Double.MAX_VALUE) this.unopen(neighbor); // Open neighbor this.open(new Node( this, neighbor, cost, n ), cost); } } } class AStarAI_div extends AIModuleH { public final boolean printPath = false; @Override public double getHeuristic(final TerrainMap map, final Point pt1, final Point pt2) { double distance = Node.distanceTo(pt1, pt2); double h1 = map.getTile(pt1); double h2 = map.getTile(pt2); if (distance if (distance return 0.5 * //return 0.5 } public List<Point> createPath(final TerrainMap map) { this.map = map; this.start = this.map.getStartPoint(); this.end = this.map.getEndPoint(); SearchAlgorithm search = new AStarSearch(this, this.start, this.end); Node current; while (true) { current = search.step(); == 0) return 0; == 1) return h2 / (h1 + 1); (distance - 2) + (1 / (h1 + 1)) + (h2 / 2); * Node.distanceTo(pt1, pt2);
if (current.equals(this.end)) { List<Point> path = current.tracePath(); if (printPath) this.printPath(path); return path; } } } } class AStarBiAI_div extends AIModuleH { public final boolean printPath = false; public int phase = 1; public double g = Double.MAX_VALUE; AStarSearch forward, reverse; Node fnode, rnode; @Override public double getHeuristic(final TerrainMap map, final Point pt1, final Point pt2) { double distance = Node.distanceTo(pt1, pt2); double h1 = map.getTile(pt1); double h2 = map.getTile(pt2); if (distance if (distance return 0.5 * //return 0.5 } public List<Point> createPath(final TerrainMap map) { this.map = map; this.start = this.map.getStartPoint(); this.end = this.map.getEndPoint(); final AStarBiAI_div me = this; forward = new AStarSearch(this, this.start, this.end) { public AStarBiAI_div bistarai = me; public boolean allowExpand(Point p) { return !(this.bistarai.phase == 3 && !this.bistarai.reverse.isclosed(p)); } }; fnode = forward.step(); fnode = forward.step(); reverse = new AStarSearch(this, this.end, this.start) { public double getCost(Point pt1, Point pt2) { return super.getCost(pt2, pt1); } }; rnode = reverse.step(); rnode = reverse.step(); while (true) { if (phase == 1) { // Phase 1 if (forward.top().f() < reverse.top().f()) { fnode = forward.step(); if (reverse.isclosed(fnode)) { phase = 2; g = fnode.g() + reverse.getCost(fnode); } == 0) return 0; == 1) return h2 / (h1 + 1); (distance - 2) + (1 / (h1 + 1)) + (h2 / 2); * Node.distanceTo(pt1, pt2);
} else { rnode = reverse.step(); if (forward.isclosed(rnode)) { phase = 2; g = rnode.g() + forward.getCost(rnode); } } } else if (phase == 2) { // Phase 2 rnode = reverse.step(); if (rnode.f() > g) { phase = 3; } } else if (phase == 3) { //while (!reverse.isclosed(forward.top())) forward.pop(); fnode = forward.step(); if (fnode.equals(this.end)) { List<Point> path = fnode.tracePath(); if (printPath) this.printPath(path); return path; } } } } }
## Exponential Cost Function
import import import import import import java.awt.Point; java.util.List; java.util.LinkedList; java.util.PriorityQueue; java.util.HashSet; java.util.HashMap;
public class MtStHelensExp_994557753 extends AStarAI_exp { public final boolean printPath = false; } abstract class AIModuleH implements AIModule { public TerrainMap map; public Point start; public Point end; public double getHeuristic(final TerrainMap map, final Point pt1, final Point pt2) { return 0.0; } public final double getHeuristic(final Point pt1, final Point pt2) { return this.getHeuristic(this.map, pt1, pt2); } public final double getHeuristic(final Point p) { return this.getHeuristic(this.map, p, this.end); } public final double getHeuristic() { return this.getHeuristic(this.map, this.start, this.end); } public final void printPath(List<Point> path) { System.out.println(this.start.x + ", " + this.start.y);
System.out.println("========================="); for (int i = 0; i < path.size(); i ++) { Point p = path.get(i); System.out.println(p.x + ", " + p.y + ", " + map.getTile(p) + ", " + Node.distanceTo(p, this.end)); } System.out.println("========================="); System.out.println(end.x + ", " + end.y); } } class Node extends Point implements Comparable { private SearchAlgorithm search; public Node parent; // public int x; // public int y; public int z; private double g; private double h; public Node(SearchAlgorithm search, int x, int y, double g, Node parent) { this.search = search; this.x = x; this.y = y; this.z = (int)this.search.ai.map.getTile(this); this.g = g; this.h = -1; this.parent = parent; } public Node(SearchAlgorithm search, Point p, double g, Node parent) { this(search, p.x, p.y, g, parent); } public double f() { return this.g() + this.h(); } public double g() { return this.g; } public double h() { if (this.h != -1) return this.h; return this.h = this.search.getHeuristic(this, this.search.end); } public static int distanceTo(Point pt1, Point pt2) { return Math.max( Math.abs(pt1.x - pt2.x), Math.abs(pt1.y - pt2.y) ); } public int distanceTo(Point p) { return Node.distanceTo(this, p); } public List<Point> tracePath() { LinkedList<Point> path = new LinkedList<Point>(); Node node = this; while (node != null) { path.addFirst(node);
node = node.parent; } return path; } public List<Point> traceReversePath() { LinkedList<Point> path = new LinkedList<Point>(); Node node = this; if (node != null) node = node.parent; while (node != null) { path.add(node); node = node.parent; } return path; } @Override public int compareTo(Object node) { double a = this.f(); double b = ((Node)node).f(); if (a < b) return -1; if (a > b) return 1; return 0; } } abstract class SearchAlgorithm { public AIModuleH ai; public Node start, end, current; public abstract Node step(); public abstract double getCost(Point pt1, Point pt2); public abstract double getHeuristic(Point pt1, Point pt2); public boolean allowExpand(Point p) {return true;} } class AStarSearch extends SearchAlgorithm { protected PriorityQueue<Point> open; protected HashSet<Point> closed; protected HashMap<Point, Double> gvalues; public AStarSearch(AIModuleH ai, Point start, Point end) { // Initialize this.ai = ai; this.open = new PriorityQueue<Point>(); this.closed = new HashSet<Point>(); this.gvalues = new HashMap<Point, Double>(); // Start this.start = new Node(this, start, 0, null); this.open(this.start, 0); // End this.end = new Node(this, end, 0, null); } public Node step() { // 4. Expand Current Node this.expand(this.current); // 1. Get Next Best Node this.current = this.pop(); // 2. Close Current Node
if (this.current != null) this.close(this.current, this.current.g()); // 3. Return Current Node for Examination return current; } protected void close(Point p, double g) { if (p == null) return; this.closed.add(p); } protected void unclose(Point p) { if (p == null) return; this.closed.remove(p); } public boolean isclosed(Point p) { if (p == null) return false; return this.closed.contains(p); } protected void open(Point p, double g) { if (p == null) return; this.open.add(p); this.gvalues.put(p, g); } public void unopen(Point p) { if (p == null) return; this.open.remove(p); } public boolean isopen(Point p) { if (p == null) return false; return this.open.contains(p); } public double getCost(Point p) { Double c = this.gvalues.get(p); return (c == null ? Double.MAX_VALUE : c.doubleValue()); } public double getCost(Point pt1, Point pt2) { return this.ai.map.getCost(pt1, pt2); } public double getHeuristic(Point pt1, Point pt2) { return this.ai.getHeuristic(pt1, pt2); } public Node top() { if (this.open == null || this.open.size() == 0) return null; return (Node)this.open.peek(); } protected Node pop() { if (this.open == null || this.open.size() == 0) return null; return (Node)this.open.poll(); } protected void expand(Node n) { if (n == null) return; for (Point neighbor : this.ai.map.getNeighbors(n)) { // If neighbor is closed (for consistent heuristics)
if (this.isclosed(neighbor)) continue; // Get potential new path value double cost = n.g() + this.getCost(n, neighbor); // If neighbor is closed (for non-consistnet heuristics) /* if (this.isclosed(neighbor)) { if (cost >= this.getCost(neighbor)) // If new path is worse continue; else // If new path is better this.unclose(neighbor); } */ // If neighbor is open if (cost >= this.getCost(neighbor)) continue; // Custom pruning if (!this.allowExpand(neighbor)) continue; // If neighbor has been open at one point in time before if (this.getCost(neighbor) != Double.MAX_VALUE) this.unopen(neighbor); // Open neighbor this.open(new Node( this, neighbor, cost, n ), cost); } } } class AStarAI_div extends AIModuleH { public final boolean printPath = false; @Override public double getHeuristic(final TerrainMap map, final Point pt1, final Point pt2) { double distance = Node.distanceTo(pt1, pt2); double h1 = map.getTile(pt1); double h2 = map.getTile(pt2); if (distance if (distance return 0.5 * //return 0.5 } public List<Point> createPath(final TerrainMap map) { this.map = map; this.start = this.map.getStartPoint(); this.end = this.map.getEndPoint(); SearchAlgorithm search = new AStarSearch(this, this.start, this.end); Node current; == 0) return 0; == 1) return h2 / (h1 + 1); (distance - 2) + (1 / (h1 + 1)) + (h2 / 2); * Node.distanceTo(pt1, pt2);
while (true) { current = search.step(); if (current.equals(this.end)) { List<Point> path = current.tracePath(); if (printPath) this.printPath(path); return path; } } } } class AStarBiAI_div extends AIModuleH { public final boolean printPath = false; public int phase = 1; public double g = Double.MAX_VALUE; AStarSearch forward, reverse; Node fnode, rnode; @Override public double getHeuristic(final TerrainMap map, final Point pt1, final Point pt2) { double distance = Node.distanceTo(pt1, pt2); double h1 = map.getTile(pt1); double h2 = map.getTile(pt2); if (distance if (distance return 0.5 * //return 0.5 } public List<Point> createPath(final TerrainMap map) { this.map = map; this.start = this.map.getStartPoint(); this.end = this.map.getEndPoint(); final AStarBiAI_div me = this; forward = new AStarSearch(this, this.start, this.end) { public AStarBiAI_div bistarai = me; public boolean allowExpand(Point p) { return !(this.bistarai.phase == 3 && !this.bistarai.reverse.isclosed(p)); } }; fnode = forward.step(); fnode = forward.step(); reverse = new AStarSearch(this, this.end, this.start) { public double getCost(Point pt1, Point pt2) { return super.getCost(pt2, pt1); } }; rnode = reverse.step(); rnode = reverse.step(); while (true) { if (phase == 1) { // Phase 1 if (forward.top().f() < reverse.top().f()) { fnode = forward.step(); if (reverse.isclosed(fnode)) { == 0) return 0; == 1) return h2 / (h1 + 1); (distance - 2) + (1 / (h1 + 1)) + (h2 / 2); * Node.distanceTo(pt1, pt2);
phase = 2; g = fnode.g() + reverse.getCost(fnode); } } else { rnode = reverse.step(); if (forward.isclosed(rnode)) { phase = 2; g = rnode.g() + forward.getCost(rnode); } } } else if (phase == 2) { // Phase 2 rnode = reverse.step(); if (rnode.f() > g) { phase = 3; } } else if (phase == 3) { //while (!reverse.isclosed(forward.top())) forward.pop(); fnode = forward.step(); if (fnode.equals(this.end)) { List<Point> path = fnode.tracePath(); if (printPath) this.printPath(path); return path; } } } } } class AStarAI_exp extends AStarAI_div { public final boolean printPath = false; @Override public double getHeuristic(final TerrainMap map, final Point pt1, final Point pt2) { double distance = Node.distanceTo(pt1, pt2); double h1 = map.getTile(pt1); double h2 = map.getTile(pt2); double dh = h2 - h1; double d1 = Math.abs(dh) % distance; double d2 = distance - d1; double dh2 = (int)(dh / distance); double dh1 = dh >= 0 ? dh2 + 1 : dh2 - 1; double cost1 = d1 * Math.exp(dh1); double cost2 = d2 * Math.exp(dh2); return cost1 + cost2; } public List<Point> createPath(final TerrainMap map) { this.map = map; this.start = this.map.getStartPoint(); this.end = this.map.getEndPoint(); SearchAlgorithm search = new AStarSearch(this, this.start, this.end) { public boolean allowExpand(Point p) { int distance = Node.distanceTo(p, this.end);
int height = (int)this.ai.map.getTile(p); if (distance < 50) { if (height > 70 || height < 48) return false; } else if (distance < 100) { if (height > 82 || height < 48) return false; } else if (distance < 150) { if (height > 87 || height < 81) return false; } else if (distance < 200) { if (height > 108 || height < 87) return false; } else if (distance < 300) { if (height > 171 || height < 108) return false; } else if (distance < 400) { if (height > 203 || height < 171) return false; } else if (distance < 1000) { if (height > 246 || height < 203) return false; } return super.allowExpand(p); } }; Node current; while (true) { current = search.step(); if (current.equals(this.end)) { List<Point> path = current.tracePath(); if (printPath) this.printPath(path); return path; } } } } class AStarBiAI_exp extends AStarBiAI_div { public final boolean printPath = false; public int phase = 1; public double g = Double.MAX_VALUE; AStarSearch forward, reverse; Node fnode, rnode; @Override public double getHeuristic(final TerrainMap map, final Point pt1, final Point pt2) { double distance = Node.distanceTo(pt1, pt2); double h1 = map.getTile(pt1); double h2 = map.getTile(pt2); double dh = h2 - h1; double d1 = Math.abs(dh) % distance; double d2 = distance - d1; double dh2 = (int)(dh / distance); double dh1 = dh >= 0 ? dh2 + 1 : dh2 - 1; double cost1 = d1 * Math.exp(dh1); double cost2 = d2 * Math.exp(dh2); return cost1 + cost2; }
public double getReverseHeuristic(final TerrainMap map, final Point pt2, final Point pt1) { return this.getHeuristic(map, pt1, pt2); } public List<Point> createPath(final TerrainMap map) { this.map = map; this.start = this.map.getStartPoint(); this.end = this.map.getEndPoint(); final AStarBiAI_exp me = this; forward = new AStarSearch(this, this.start, this.end) { public AStarBiAI_exp bistarai = me; public boolean allowExpand(Point p) { return !(this.bistarai.phase == 3 && !this.bistarai.reverse.isclosed(p)); } }; fnode = forward.step(); fnode = forward.step(); reverse = new AStarSearch(this, this.end, this.start) { public double getCost(Point pt1, Point pt2) { return super.getCost(pt2, pt1); } public double getHeuristic(Point pt1, Point pt2) { return this.ai.getHeuristic(pt2, pt1); } }; rnode = reverse.step(); rnode = reverse.step(); while (true) { if (phase == 1) { // Phase 1 if (forward.top().f() < reverse.top().f()) { fnode = forward.step(); if (reverse.isclosed(fnode)) { phase = 2; g = fnode.g() + reverse.getCost(fnode); } } else { rnode = reverse.step(); if (forward.isclosed(rnode)) { phase = 2; g = rnode.g() + forward.getCost(rnode); } } } else if (phase == 2) { // Phase 2 rnode = reverse.step(); if (rnode.f() > g) { phase = 3; } } else if (phase == 3) { //while (!reverse.isclosed(forward.top())) forward.pop(); fnode = forward.step(); if (fnode.equals(this.end)) { List<Point> path = fnode.tracePath(); if (printPath) this.printPath(path); return path; }
} } } } | 8,668 | 32,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-39 | latest | en | 0.856988 |
https://im.openupresources.org/8/students/1/5.html | 1,555,858,497,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578531984.10/warc/CC-MAIN-20190421140100-20190421162100-00255.warc.gz | 441,928,085 | 8,703 | # Lesson 5: Coordinate Moves
Let’s transform some figures and see what happens to the coordinates of points.
## 5.1: Translating Coordinates
Select all of the translations that take Triangle T to Triangle U. There may be more than one correct answer.
1. Translate $(\text-3,0)$ to $(1,2)$.
2. Translate $(2,1)$ to $(\text-2,\text-1)$.
3. Translate $(\text-4,\text-3)$ to $(0,\text-1)$.
4. Translate $(1,2)$ to $(2,1)$.
## 5.2: Reflecting Points on the Coordinate Plane
1. Five points are plotted on the coordinate plane.
GeoGebra Applet N7uYVnXD
1. Using the Pen tool or the Text tool, label each with its coordinates.
2. Using the $x$-axis as the line of reflection, plot the image of each point.
3. Label the image of each point using a letter. For example, the image of point $A$ should be labeled $A’$.
4. Label each with its coordinates.
2. If the point $(13,10)$ were reflected using the $x$-axis as the line of reflection, what would be coordinates of the image? What about $(13, \text-20)$? $(13, 570)$? Explain how you know.
3. The point $R$ has coordinates $(3,2)$.
1. Without graphing, predict the coordinates of the image of point $R$ if point $R$ were reflected using the $y$-axis as the line of reflection.
2. Check your answer by finding the image of $R$ on the graph.
3. Label the image of point $R$ as $R’$.
4. What are the coordinates of $R’$?
4. Suppose you reflect a point using the $y$-axis as line of reflection. How would you describe its image?
## 5.3: Transformations of a Segment
The applet has instructions for the first 3 questions built into it. Move the slider marked “question” when you are ready to answer the next one. Pause before using the applet to show the transformation described in each question to predict where the new coordinates will be.
Apply each of the following transformations to segment $AB$. Use the Pen tool to record the coordinates.
1. Rotate segment $AB$ 90 degrees counterclockwise around center $B$ by moving the slider marked 0 degrees. The image of $A$ is named $C$. What are the coordinates of $C$?
2. Rotate segment $AB$ 90 degrees counterclockwise around center $A$ by moving the slider marked 0 degrees. The image of $B$ is named $D$. What are the coordinates of $D$?
3. Rotate segment $AB$ 90 degrees clockwise around $(0,0)$ by moving the slider marked 0 degrees. The image of $A$ is named $E$ and the image of $B$ is named $F$. What are the coordinates of $B$ and $F$?
4. Compare the two 90-degree counterclockwise rotations of segment $AB$. What is the same about the images of these rotations? What is different?
GeoGebra Applet Knr939et
## Summary
We can use coordinates to describe points and find patterns in the coordinates of transformed points.
We can describe a translation by expressing it as a sequence of horizontal and vertical translations. For example, segment $AB$ is translated right 3 and down 2.
Reflecting a point across an axis changes the sign of one coordinate. For example, reflecting the point $A$ whose coordinates are $(2,\text-1)$ across the $x$-axis changes the sign of the $y$-coordinate, making its image the point $A’$ whose coordinates are $(2,1)$. Reflecting the point $A$ across the $y$-axis changes the sign of the $x$-coordinate, making the image the point $A’’$ whose coordinates are $(\text-2,\text-1)$.
Reflections across other lines are more complex to describe.
We don’t have the tools yet to describe rotations in terms of coordinates in general. Here is an example of a $90^\circ$ rotation with center $(0,0)$ in a counterclockwise direction.
Point $A$ has coordinates $(0,0)$. Segment $AB$ was rotated $90^\circ$ counterclockwise around $A$. Point $B$ with coordinates $(2,3)$ rotates to point $B’$ whose coordinates are $(\text-3,2)$. | 1,010 | 3,779 | {"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} | 4.71875 | 5 | CC-MAIN-2019-18 | latest | en | 0.860478 |
https://topoptionshzlo.web.app/burtchell6449jaz/gc-flow-rate-calculator-666.html | 1,638,098,676,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358520.50/warc/CC-MAIN-20211128103924-20211128133924-00225.warc.gz | 528,067,870 | 6,403 | ## Gc flow rate calculator
lmb. Thanks for your reply. This is how I figure the flow rate: There are different kinds of flow rates in GC, the most common one is the flow rate at room temp (25C) and 1 atm (14.7 psia), I believe it's the Outlet flow rate in HP FlowCalc software. At 40°C, 35 cm/sec helium through a 30m, 0.32mm ID column has a flow rate of 2.2 ml/min; at 240 °C, helium flow rate would be 1.5 ml/min with the same linear velocity. As far as I am concerned, carrier gas linear velocity is proportional to flow rate, and is independent of gas viscosity (oven temperature).
22 Aug 2014 Next enter the desired average carrier-gas linear velocity in the instrument system or flow calculator. The resulting column pressure is the The relation between flow rate and inlet pressure is calculated with Poiseuille's equation for compressible The volumetric GC column flow rate in mL/min. The Outlet Flow is calculated from the Inlet Pressure (gauge) of the selected carrier gas, the GC oven/column 23 Sep 2012 constant flow rate), and the column dimensions used. Moreover, if imperfections in the GC instruments cause large enough differences in these First, measure the flow rate out of the split vent using a suitable flow meter Basic Gas Chromatography, Second Edition, by Harold M. McNair and James M.
## Enter flow rate or head pressure or average linear velocity. The instrument column—or if you do not wish to use the GC calculating features, enter 0 for.
18 Mar 2013 GAS CHROMATOGRAPHY VAN DEE METER EQUATION PREPARED BY: Ms. SONAM M. Flow rate of the mobile phase.7/23/2012 4; 5. 5 Nov 2015 Further calculation with the normal volume flow rate which is typically used by the rotameter on gas chromatograph (i.e. Nml/min) will result in. 20 Sep 2016 The sensitivity changes a bit from day to day because the ionization in the detector changes due to aging, the gas flow rate isn't 100.0000% the Larger diameter HPLC columns require higher flow rates; therefore, larger volumes of mobile phase will The new flow rate can be calculated using Equation 5.
### Outlet split ratio = (Column flow + Split flow) : Column flow desorption control software for Markes' instruments includes a split ratio calculator to work all of this
iPhone pressure settings and flow rates through a capillary GC column and calculate the vapor volume for a liquid solvent vaporized in a heated GC inlet. GC Pressure / Flow Calculator Application for iPhone | Agilent The volumetric GC column flow rate in mL/min. The Outlet Flow is calculated from the Inlet Pressure (gauge) of the selected carrier gas, the GC oven/column temperature, and the GC column Outlet Pressure (abs). The reference temperature and reference pressure used for flow calculation are 22°C and 1 atm, respectively. The EZ GC method translator and flow calculator tools make it simple to switch carrier gases, change column dimensions or control parameters, or to optimize a method for speed or efficiency. Simply enter your method specifications and the GC calculator will return a full set of method conditions that will provide similar chromatography. HPLC Method Transfer Calculator Calculates conditions for transfer of an isocratic or gradient method from one HPLC column to another. Allows method scaling from microbore through preparative column range. lmb. Thanks for your reply. This is how I figure the flow rate: There are different kinds of flow rates in GC, the most common one is the flow rate at room temp (25C) and 1 atm (14.7 psia), I believe it's the Outlet flow rate in HP FlowCalc software. At 40°C, 35 cm/sec helium through a 30m, 0.32mm ID column has a flow rate of 2.2 ml/min; at 240 °C, helium flow rate would be 1.5 ml/min with the same linear velocity. As far as I am concerned, carrier gas linear velocity is proportional to flow rate, and is independent of gas viscosity (oven temperature). Measuring Flow Rates 5890 GC FID It is important to measure each flow independently: - Turn off the hydrogen, air, and make-up (aux) gas at the FID flow block with the small black on/off knobs. - Measure the flow at the FID using a suitable flow measuring device and the FID flow measuring insert (P/N 19301-60660). This is the column flow.
### Typical GC Column Characteristics: Look up approximate flow rates, sample capacity, ferrule dimensions, and other column characteristics based on ID.
The relation between flow rate and inlet pressure is calculated with Poiseuille's equation for compressible The volumetric GC column flow rate in mL/min. The Outlet Flow is calculated from the Inlet Pressure (gauge) of the selected carrier gas, the GC oven/column 23 Sep 2012 constant flow rate), and the column dimensions used. Moreover, if imperfections in the GC instruments cause large enough differences in these First, measure the flow rate out of the split vent using a suitable flow meter Basic Gas Chromatography, Second Edition, by Harold M. McNair and James M. This page shows how to convert between flow velocity and volumetric flow rate in affinity chromatography of antibodies. The carrier gas flow can be characterized by either the volume flow rate or the linear flow Equation 1 can be integrated between the inlet and the outlet of the column, Thus, although gas chromatography has been carried out with columns
## The volumetric GC column flow rate in mL/min. The Outlet Flow is calculated from the Inlet Pressure (gauge) of the selected carrier gas, the GC oven/column temperature, and the GC column Outlet Pressure (abs). The reference temperature and reference pressure used for flow calculation are 22°C and 1 atm, respectively.
Description Use the Agilent GC pressure and flow rate calculator iPhone™/iPod Touch® app to instantly and accurately determine pressure and flow through open tubular capillary columns. The Agilent GC Calculator app is based on the popular PC version that has been downloaded more than 10,000 times from the Agilent Web site.
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Given the quadratic function f(x) = 16x2 – 64x + 64, find a value of x such that f(x) = 16.
• math-quadratic equations -
f(x) = 16x2 – 64x + 64
for f(x)=16, write:
16=16x² - 64x +64
transpose 16 to the right, and factor out numeric factor 16:
16(x² -4x +3) = 0
Factor expression
16(x-3)(x-1)=0
This means that x=3 or x=1, either one of which satisfies the equation
f(x)=16.
Check by substituting x=3 or x=1 in f(x).
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https://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/s13660-023-02967-5 | 1,722,842,707,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640434051.17/warc/CC-MAIN-20240805052118-20240805082118-00197.warc.gz | 286,676,368 | 85,319 | A reverse extended Hardy–Hilbert’s inequality with parameters
Abstract
In this paper, by virtue of the symmetry principle, applying the techniques of real analysis and Euler–Maclaurin summation formula, we construct proper weight coefficients and use them to establish a reverse extended Hardy–Hilbert’s inequality with multi-parameters. Then, we obtain the equivalent forms and some equivalent statements of the best possible constant factor related to several parameters. Finally, we illustrate how the obtained results can generate some new reverse Hardy–Hilbert-type inequalities.
1 Introduction
Suppose that $$p > 1$$, $$\frac{1}{p} + \frac{1}{q} = 1$$, $$a_{m}, b_{n} \ge 0$$, $$0 < \sum_{m = 1}^{\infty} a_{m}^{p} < \infty$$, and $$0 < \sum_{n = 1}^{\infty} b_{n}^{q} < \infty$$. We have the following well-known Hardy–Hilbert’s inequality with the best possible constant factor $$\frac{\pi}{\sin (\pi /p)}$$ (cf. [1], Theorem 315):
$$\sum_{m = 1}^{\infty} \sum _{n = 1}^{\infty} \frac{a_{m}b_{n}}{m + n} < \frac{\pi}{\sin (\pi /p)}\Biggl(\sum_{m = 1}^{\infty} a_{m}^{p} \Biggr)^{\frac{1}{p}}\Biggl(\sum _{n = 1}^{\infty} b_{n}^{q} \Biggr)^{\frac{1}{q}}.$$
(1)
In 2006, by introducing parameters $$\lambda _{i} \in (0,2]$$ ($$i = 1,2$$), $$\lambda _{1} + \lambda _{2} = \lambda \in (0,4]$$, using Euler–Maclaurin summation formula, an extension of (1) was provided by Krnić et al. [2] as follows:
$$\sum_{m = 1}^{\infty} \sum _{n = 1}^{\infty} \frac{a_{m}b_{n}}{(m + n)^{\lambda}} < B(\lambda _{1},\lambda _{2})\Biggl[\sum _{m = 1}^{\infty} m^{p(1 - \lambda _{1}) - 1}a_{m}^{p} \Biggr]^{\frac{1}{p}}\Biggl[\sum_{n = 1}^{\infty} n^{q(1 - \lambda _{2}) - 1}b_{n}^{q} \Biggr]^{\frac{1}{q}},$$
(2)
where the constant factor $$B(\lambda _{1},\lambda _{2})$$ is the best possible.
$$B(u,v) = \int _{0}^{\infty} \frac{t^{u - 1}}{(1 + t)^{u + v}}\,dt\quad (u,v > 0)$$
is the beta function. For $$\lambda = 1$$, $$\lambda _{1} = \frac{1}{q}$$, $$\lambda _{2} = \frac{1}{p}$$, inequality (2) reduces to (1); for $$p = q = 2$$, $$\lambda _{1} = \lambda _{2} = \frac{\lambda}{2}$$, (2) reduces to Yang’s inequality in [3]. Recently, applying inequality (2), Adiyasuren et al. [4] gave a new Hardy–Hilbert’s inequality with the kernel $$\frac{1}{(m + n)^{\lambda}}$$ involving two partial sums.
If $$f(x),g(y) \ge 0$$, $$0 < \int _{0}^{\infty} f^{p}(x)\,dx < \infty$$, and $$0 < \int _{0}^{\infty} g^{q}(y)\,dy < \infty$$, then we still have the following Hardy–Hilbert’s integral inequality (cf. [1], Theorem 316):
$$\int _{0}^{\infty} \int _{0}^{\infty} \frac{f(x)g(y)}{x + y}\,dx\,dy < \frac{\pi}{\sin (\pi /p)}\biggl( \int _{0}^{\infty} f^{p}(x)\,dx \biggr)^{\frac{1}{p}}\biggl( \int _{0}^{\infty} g^{q}(y)\,dy \biggr)^{\frac{1}{q}},$$
(3)
where the constant factor $$\pi /\sin (\frac{\pi}{p})$$ is still the best possible. Inequalities (1), (2), and (3) with their extensions and reverses play an important role in the analysis and its applications (cf. [515]).
In 1934, a half-discrete Hilbert-type inequality was given as follows (cf. [1], Theorem 351): If $$K(t)$$ ($$t > 0$$) is a decreasing function, $$p > 1$$, $$\frac{1}{p} + \frac{1}{q} = 1$$, $$0 < \phi (s) = \int _{0}^{\infty} K(t)t^{s - 1}\,dt < \infty$$, $$a_{n} \ge 0$$, $$0 < \sum_{n = 1}^{\infty} a_{n}^{p} < \infty$$, then we have
$$\int _{0}^{\infty} x^{p - 2}\Biggl(\sum _{n = 1}^{\infty} K(nx)a_{n} \Biggr)^{p}\,dx < \phi ^{p}\biggl(\frac{1}{q} \biggr)\sum_{n = 1}^{\infty} a_{n}^{p}.$$
(4)
In recent years, some new extensions of (4) with the reverses were provided by [1623].
In 2016, by means of the technique of real analysis and the weight coefficients, Hong et al. [24] considered some equivalent statements of the extensions of (1) with the best possible constant factor related to a few parameters. Other similar works about the extensions of (1), (2), (3), and (4) with the reverses were given by [2532].
In this paper, following the way of [2, 24], by means of the weight coefficients, the idea of introduced parameters, the techniques of real analysis, and the Euler–Maclaurin summation formula, a new reverse of the extension of (1) with parameters as well as the equivalent forms are given. The equivalent statements of the best possible constant factor related to several parameters are obtained, and some particular inequalities are provided.
2 Some lemmas
In what follows, we suppose that $$0 < p < 1$$ ($$q < 0$$), $$\frac{1}{p} + \frac{1}{q} = 1$$, $$\lambda \in (0,\frac{5}{2}]$$, $$\lambda _{i} \in (0,\frac{5}{4}] \cap (0,\lambda )$$ ($$i = 1,2$$), $$a_{m}, b_{n} \ge 0$$, $$m,n \in \mathrm{N} = \{ 1,2, \ldots \}$$, such that
$$0 < \sum_{m = 1}^{\infty} m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p} < \infty ,\quad\text{and}\quad0 < \sum _{n = 1}^{\infty} n^{q[1 - (\frac{\lambda _{2}}{p} + \frac{\lambda - \lambda _{1}}{q})] - 1} b_{n}^{q} < \infty .$$
Lemma 1
(cf. [5], (2.2.13))
If $$g(t)$$ is a positive strictly decreasing function in $$[m,\infty )$$ ($$m \in \mathrm{N}$$) with $$g(\infty ) = 0$$, $$P_{i}(t)$$ and $$B_{i}$$ ($$i \in \mathrm{N}$$) are the Bernoulli functions and the Bernoulli numbers of i-order, then we have
$$\int _{m}^{\infty} P_{2q - 1} (t)g(t)\,dt = \varepsilon \frac{B_{2q}}{q}\biggl(\frac{1}{2^{2q}} - 1\biggr)g(m) \quad (0 < \varepsilon < 1;q = 1,2, \ldots ).$$
(5)
In particular, for $$q = 1$$, in view of $$B_{2} = \frac{1}{6}$$, we have
$$- \frac{1}{8}g(m) < \int _{m}^{\infty} P_{1} (t)g(t)\,dt < 0.$$
(6)
Lemma 2
Define the following weight coefficient:
$$\varpi (\lambda _{2},m): = m^{\lambda - \lambda _{2}}\sum _{n = 1}^{\infty} \frac{n^{\lambda _{2} - 1}}{m^{\lambda} + n^{\lambda}}\quad (m \in \mathrm{N}).$$
(7)
We have the following inequalities:
$$\frac{\pi}{\lambda \sin (\pi \lambda _{2}/\lambda )}\bigl(1 - \theta _{m}(\lambda _{2})\bigr) < \varpi (\lambda _{2},m) < k_{\lambda} (\lambda _{2}): = \frac{\pi}{\lambda \sin (\pi \lambda _{2}/\lambda )}\quad (m \in \mathrm{N}),$$
(8)
where $$\theta _{m}(\lambda _{2})$$ is indicated by
$$\theta _{m}(\lambda _{2}): = \frac{\sin (\pi \lambda _{2}/\lambda )}{\pi} \int _{0}^{\frac{1}{m^{\lambda}}} \frac{u^{(\lambda _{2}/\lambda ) - 1}}{1 + u}\,du = O \biggl(\frac{1}{m^{\lambda _{2}}}\biggr) \in (0,1)\quad (m \in \mathrm{N}).$$
(9)
Proof
For fixed $$m \in \mathrm{N}$$, we set the following function:
$$g(m,t): = \frac{t^{\lambda _{2} - 1}}{m^{\lambda} + t^{\lambda}}\quad (t > 0).$$
By the use of Euler–Maclaurin summation formula (cf. [2, 3]), we have
\begin{aligned}& \begin{aligned} \sum_{n = 1}^{\infty} g(m,n) &= \int _{1}^{\infty} g(m,t)\,dt + \frac{1}{2} g(m,1) + \int _{1}^{\infty} P_{1}(t)g'(m,t)\,dt \\ &= \int _{0}^{\infty} g(m,t)\,dt - h(m), \end{aligned}\\& h(m): = \int _{0}^{1} g(m,t)\,dt - \frac{1}{2}g(m,1) - \int _{1}^{\infty} P_{1}(t)g'(m,t)\,dt. \end{aligned}
We find $$\frac{1}{2}g(m,1) = \frac{1}{2(m^{\lambda} + 1)}$$,
\begin{aligned} \int _{0}^{1} g(m,t)\,dt &= \int _{0}^{1} \frac{t^{\lambda _{2} - 1}}{m^{\lambda} + t^{\lambda}}\,dt = \frac{1}{\lambda _{2}} \int _{0}^{1} \frac{dt^{\lambda _{2}}}{m^{\lambda} + t^{\lambda}} \\ &= \frac{1}{\lambda _{2}}\frac{t^{\lambda _{2}}}{m^{\lambda} + t^{\lambda}} \biggm|_{0}^{1} + \frac{\lambda}{\lambda _{2}} \int _{0}^{1} \frac{t^{\lambda + \lambda _{2} - 1}}{(m^{\lambda} + t^{\lambda} )^{2}}\,dt > \frac{1}{\lambda _{2}}\frac{1}{m^{\lambda} + 1}. \end{aligned}
We also obtain
\begin{aligned} g'(m,t) &= \frac{(\lambda _{2} - 1)t^{\lambda _{2} - 2}}{m^{\lambda} + t^{\lambda}} - \frac{\lambda t^{\lambda + \lambda _{2} - 2}}{(m^{\lambda} + t^{\lambda} )^{2}} = - \frac{(1 - \lambda _{2})t^{\lambda _{2} - 2}}{m^{\lambda} + t^{\lambda}} - \frac{\lambda (m^{\lambda} + t^{\lambda} - m^{\lambda} )t^{\lambda _{2} - 2}}{(m^{\lambda} + t^{\lambda} )^{2}} \\ &= - \frac{(\lambda + 1 - \lambda _{2})t^{\lambda _{2} - 2}}{m^{\lambda} + t^{\lambda}} + \frac{\lambda m^{\lambda} t^{\lambda _{2} - 2}}{(m^{\lambda} + t^{\lambda} )^{2}}. \end{aligned}
For $$0 < \lambda _{2} \le \frac{5}{4}$$, $$\lambda _{2} < \lambda \le \frac{5}{2}$$, it follows that
$$\frac{d}{dt}\biggl[\frac{t^{\lambda _{2} - 2}}{(m^{\lambda} + t^{\lambda} )^{i}}\biggr] < 0\quad (i = 1,2).$$
By (6), we obtain
\begin{aligned}& (\lambda + 1 - \lambda _{2}) \int _{1}^{\infty} P_{1}(t) \frac{t^{\lambda _{2} - 2}}{m^{\lambda} + t^{\lambda}}\,dt > - \frac{\lambda + 1 - \lambda _{2}}{8(m^{\lambda} + 1)},\\& - m^{\lambda} \lambda \int _{1}^{\infty} P_{1}(t) \frac{t^{\lambda _{2} - 2}}{(m^{\lambda} + t^{\lambda} )^{2}}\,dt > 0, \end{aligned}
and then we find
$$- \int _{1}^{\infty} P_{1}(t)g'(m,t)\,dt > - \frac{\lambda + 1 - \lambda _{2}}{8(m^{\lambda} + 1)} + 0 = - \frac{\lambda + 1 - \lambda _{2}}{8(m^{\lambda} + 1)}.$$
Hence, it follows that
\begin{aligned} h(m)& > \frac{1}{\lambda _{2}}\frac{1}{m^{\lambda} + 1} - \frac{1}{2(m^{\lambda} + 1)} - \frac{\lambda + 1 - \lambda _{2}}{8(m^{\lambda} + 1)} = \frac{8 - (5 + \lambda )\lambda _{2} + \lambda _{2}^{2}}{8\lambda _{2}(m^{\lambda} + 1)}\\ &\ge \frac{8 - (5 + \frac{5}{2})\lambda _{2} + \lambda _{2}^{2}}{8\lambda _{2}(m^{\lambda} + 1)} = \frac{16 - 15\lambda _{2} + 2\lambda _{2}^{2}}{16\lambda _{2}(m^{\lambda} + 1)}. \end{aligned}
Since $$(6 - 15\lambda _{2} + 2\lambda _{2}^{2})' = - 15 + 4\lambda _{2} < 0$$ ($$\lambda _{2} \in (0,\frac{5}{4}]$$), we have
$$h(m) > \frac{16 - 15(\frac{5}{4}) + 2(\frac{5}{4})^{2}}{16\lambda _{2}(m^{\lambda} + 1)} = \frac{3}{128\lambda _{2}(m^{\lambda} + 1)} > 0.$$
Setting $$t = mu^{1/\lambda}$$, we find
\begin{aligned} \varpi (\lambda _{2},m) &= m^{\lambda - \lambda _{2}}\sum _{n = 1}^{\infty} g(m,n) < m^{\lambda - \lambda _{2}} \int _{0}^{\infty} g(m,t)\,dt \\ &= m^{\lambda - \lambda _{2}} \int _{0}^{\infty} \frac{t^{\lambda _{2} - 1}}{m^{\lambda} + t^{\lambda}}\,dt = \frac{1}{\lambda} \int _{0}^{\infty} \frac{u^{(\lambda _{2}/\lambda ) - 1}}{1 + u}\,du = \frac{\pi}{\lambda \sin (\pi \lambda _{2}/\lambda )}. \end{aligned}
On the other hand, we also have
\begin{aligned}& \begin{aligned} \sum_{n = 1}^{\infty} g(m,n) &= \int _{1}^{\infty} g(m,t)\,dt + \frac{1}{2} g(m,1) + \int _{1}^{\infty} P_{1}(t)g'(m,t)\,dt \\ &= \int _{1}^{\infty} g(m,t)\,dt + H(m), \end{aligned} \\& H(m): = \frac{1}{2}g(m,1) + \int _{1}^{\infty} P_{1}(t)g'(m,t)\,dt. \end{aligned}
Since we find $$\frac{1}{2}g(m,1) = \frac{1}{2(m^{\lambda} + 1)}$$ and
$$g'(m,t) = - \frac{(\lambda + 1 - \lambda _{2})t^{\lambda _{2} - 2}}{m^{\lambda} + t^{\lambda}} + \frac{\lambda m^{\lambda} t^{\lambda _{2} - 2}}{(m^{\lambda} + t^{\lambda} )^{2}},$$
in view of (6), we obtain
\begin{aligned}& - (\lambda + 1 - \lambda _{2}) \int _{1}^{\infty} P_{1}(t) \frac{t^{\lambda _{2} - 2}}{m^{\lambda} + t^{\lambda}}\,dt > 0,\quad \text{and} \\& \lambda m^{\lambda} \int _{1}^{\infty} P_{1}(t) \frac{t^{\lambda _{2} - 2}}{(m^{\lambda} + t^{\lambda} )^{2}}\,dt > - \frac{\lambda m^{\lambda}}{ 8(m{}^{\lambda} + 1)^{2}}. \end{aligned}
Hence, we have
$$H(m) > \frac{1}{2(m^{\lambda} + 1)} - \frac{\lambda m^{\lambda}}{ 8(m^{\lambda} + 1)^{2}} > \frac{4}{8(m^{\lambda} + 1)} - \frac{5/2}{8(m^{\lambda} + 1)} > 0.$$
Setting $$t = mu^{1/\lambda}$$, we obtain
\begin{aligned}& \begin{aligned} \varpi (\lambda _{2},m)& = m^{\lambda - \lambda _{2}}\sum _{n = 1}^{\infty} g(m,n) > m^{\lambda - \lambda _{2}} \int _{1}^{\infty} g(m,t)\,dt \\ &= m^{\lambda - \lambda _{2}} \int _{0}^{\infty} g(m,t)\,dt - m^{\lambda - \lambda _{2}} \int _{0}^{1} g(m,t)\,dt \\ &= \frac{\pi}{\lambda \sin (\pi \lambda _{2}/\lambda )}\biggl[1 - \frac{\lambda \sin (\pi \lambda _{2}/\lambda )}{\pi} m^{\lambda - \lambda _{2}} \int _{0}^{1} \frac{t^{\lambda _{2} - 1}}{m^{\lambda} + t^{\lambda}}\,dt \biggr] \\ &= \frac{\pi}{\lambda \sin (\pi \lambda _{2}/\lambda )}\bigl(1 - \theta _{m}(\lambda _{2})\bigr) > 0,\end{aligned} \end{aligned}
where $$\theta _{m}(\lambda _{2}) = \frac{\sin (\pi \lambda _{2}/\lambda )}{\pi} \int _{0}^{\frac{1}{m^{\lambda}}} \frac{u^{(\lambda _{2}/\lambda ) - 1}}{1 + u}\,du$$. Since we find
$$0 < \int _{0}^{\frac{1}{m^{\lambda}}} \frac{u^{(\lambda _{2}/\lambda ) - 1}}{1 + u}\,du < \int _{0}^{\frac{1}{m^{\lambda}}} u^{(\lambda _{2}/\lambda ) - 1}\,du = \frac{\lambda}{\lambda _{2}m^{\lambda _{2}}},$$
namely, $$\theta _{m}(\lambda _{2}) = O(\frac{1}{m^{\lambda _{2}}}) \in (0,1)$$ ($$m \in \mathrm{N}$$). Therefore, inequalities (8) with (9) follow.
The lemma is proved. □
Lemma 3
We have the following reverse extended Hardy–Hilbert’s inequality with parameters:
\begin{aligned} I ={}& \sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m^{\lambda} + n^{\lambda}} > k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1}) \\ &{}\times \Biggl\{ \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2})\bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\Biggr\} ^{\frac{1}{p}}\Biggl\{ \sum _{n = 1}^{\infty} n^{q[1 - (\frac{\lambda _{2}}{p} + \frac{\lambda - \lambda _{1}}{q})] - 1} b_{n}^{q}\Biggr\} ^{\frac{1}{q}}. \end{aligned}
(10)
Proof
In the same way, for $$n \in \mathbf{N}$$, we have the following inequalities for another weight coefficient:
\begin{aligned}& \begin{gathered} \omega (\lambda _{1},n): = n^{\lambda - \lambda _{1}}\sum _{m = 1}^{\infty} \frac{m^{\lambda _{1} - 1}}{m^{\lambda} + n^{\lambda}} \quad (n \in \mathrm{N}),\\ \frac{\pi}{\lambda \sin (\pi \lambda _{1}/\lambda )}\bigl(1 - \theta _{n}(\lambda _{1})\bigr) < \omega (\lambda _{1},n) < k_{\lambda} (\lambda _{1}) = \frac{\pi}{\lambda \sin (\pi \lambda _{1}/\lambda )}, \end{gathered} \end{aligned}
(11)
\begin{aligned}& \theta _{n}(\lambda _{1}) = \frac{\sin (\pi \lambda _{1}/\lambda )}{\pi} \int _{0}^{\frac{1}{n^{\lambda}}} \frac{u^{(\lambda _{1}/\lambda ) - 1}}{1 + u}\,du = O \biggl(\frac{1}{n^{\lambda _{1}}}\biggr) \in (0,1)\quad (n \in \mathrm{N}). \end{aligned}
(12)
By the reverse Hölder inequality (cf. [33]), we obtain
\begin{aligned} I& = \sum_{n = 1}^{\infty} \sum_{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} \biggl[\frac{n^{(\lambda _{2} - 1)/p}}{m^{(\lambda _{1} - 1)/q}}a_{m}\biggr] \biggl[ \frac{m^{(\lambda _{1} - 1)/q}}{n^{(\lambda _{2} - 1)/p}}b_{n}\biggr] \\ &\ge \Biggl[\sum_{m = 1}^{\infty} \sum _{n = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} \frac{n^{\lambda _{2} - 1}}{m^{(\lambda _{1} - 1)(p - 1)}}a_{m}^{p}\Biggr]^{\frac{1}{p}} \Biggl[\sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} \frac{m^{\lambda _{1} - 1}}{n^{(\lambda _{2} - 1)(q - 1)}}b_{n}^{q}\Biggr]^{\frac{1}{q}} \\ &= \Biggl\{ \sum_{m = 1}^{\infty} \varpi ( \lambda _{2},m) m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1}a_{m}^{p} \Biggr\} ^{\frac{1}{p}}\Biggl\{ \sum_{n = 1}^{\infty} \omega (\lambda _{1},n) n^{q[1 - (\frac{\lambda _{2}}{p} + \frac{\lambda - \lambda _{1}}{q})] - 1}b_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}
Then, by (8) and (11), in view of $$0 < p < 1$$ ($$q < 0$$), we have (10).
The lemma is proved. □
Remark 1
By (10), for $$\lambda _{1} + \lambda _{2} = \lambda \in (0,\frac{5}{2}]$$, $$0 < \lambda _{i} \le \frac{5}{4}$$ ($$i = 1,2$$), we find
$$0 < \sum_{m = 1}^{\infty} m^{p(1 - \lambda _{1}) - 1} a_{m}^{p} < \infty ,\qquad 0 < \sum _{n = 1}^{\infty} n^{q(1 - \lambda _{2}) - 1} b_{n}^{q} < \infty$$
and the following reverse inequality:
$$\sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m^{\lambda} + n^{\lambda}} > k_{\lambda} (\lambda _{1}) \Biggl[\sum _{m = 1}^{\infty} \bigl(1 - \theta _{m}( \lambda _{2})\bigr)m^{p(1 - \lambda _{1}) - 1} a_{m}^{p} \Biggr]^{\frac{1}{p}}\Biggl[\sum_{n = 1}^{\infty} n^{q(1 - \lambda _{2}) - 1} b_{n}^{q}\Biggr]^{\frac{1}{q}}.$$
(13)
Lemma 4
The constant factor $$k_{\lambda} (\lambda _{1}) = \frac{\pi}{\lambda \sin (\pi \lambda _{1}/\lambda )}$$ in (13) is the best possible.
Proof
For any $$0 < \varepsilon < p\lambda _{1}$$, we set
$$\tilde{a}_{m}: = m^{\lambda _{1} - \frac{\varepsilon}{p} - 1},\qquad \tilde{b}_{n}: = n^{\lambda _{2} - \frac{\varepsilon}{q} - 1}\quad (m,n \in \mathrm{N}).$$
If there exists a constant $$M \ge k_{\lambda} (\lambda _{1})$$ such that (13) is valid when we replace $$k_{\lambda} (\lambda _{1})$$ by M, then in particular, by substitution of $$a_{m} = \tilde{a}_{m}$$ and $$b_{n} = \tilde{b}_{n}$$ in (13), we have
\begin{aligned}[b] \tilde{I}&: = \sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{\tilde{a}_{m}\tilde{b}_{n}}{m^{\lambda} + n^{\lambda}} \\ &> M \Biggl[\sum_{m = 1}^{\infty} \biggl(1 - O \biggl(\frac{1}{m^{\lambda _{2}}}\biggr)\biggr)m^{p(1 - \lambda _{1}) - 1} \tilde{a}_{m}^{p} \Biggr]^{\frac{1}{p}}\Biggl[\sum_{n = 1}^{\infty} n^{q(1 - \lambda _{2}) - 1} \tilde{b}_{n}^{q} \Biggr]^{\frac{1}{q}}. \end{aligned}
(14)
By (14) and the decreasingness property of series, for $$0 < p < 1$$, $$q < 0$$, we obtain
\begin{aligned} \tilde{I}& > M\Biggl[\sum_{m = 1}^{\infty} \biggl(1 - O\biggl(\frac{1}{m^{\lambda _{2}}}\biggr)\biggr)m^{p(1 - \lambda _{1}) - 1} m^{p\lambda _{1} - \varepsilon - p}\Biggr]^{\frac{1}{p}}\Biggl[\sum _{n = 1}^{\infty} n^{q(1 - \lambda _{2}) - 1} n^{q\lambda _{2} - \varepsilon - q} \Biggr]^{\frac{1}{q}} \\ &= M\Biggl[\sum_{m = 1}^{\infty} \biggl(1 - O \biggl(\frac{1}{m^{\lambda _{2}}}\biggr)\biggr)m^{ - \varepsilon - 1} \Biggr]^{\frac{1}{p}} \Biggl(\sum_{n = 1}^{\infty} n^{ - \varepsilon - 1} \Biggr)^{\frac{1}{q}} \\ &= M\Biggl(\sum_{m = 1}^{\infty} m^{ - \varepsilon - 1} - \sum_{m = 1}^{\infty} O \biggl(\frac{1}{m^{\lambda _{2} + \varepsilon + 1}}\biggr) \Biggr)^{\frac{1}{p}}\Biggl(1 + \sum _{n = 2}^{\infty} n^{ - \varepsilon - 1} \Biggr)^{\frac{1}{q}} \\ &> M\biggl( \int _{1}^{\infty} x^{ - \varepsilon - 1}\,dx - O(1) \biggr)^{\frac{1}{p}}\biggl(1 + \int _{1}^{\infty} y^{ - \varepsilon - 1}\,dy \biggr)^{\frac{1}{q}} \\ &= \frac{M}{\varepsilon} \bigl(1 - \varepsilon O(1)\bigr)^{\frac{1}{p}}( \varepsilon + 1)^{\frac{1}{q}}. \end{aligned}
By (11) and (12), setting $$\hat{\lambda}_{1} = \lambda _{1} - \frac{\varepsilon}{p} \in (0,\frac{5}{4}) \cap (0,\lambda )$$ ($$0 < \hat{\lambda}_{2} = \lambda _{2} + \frac{\varepsilon}{p} = \lambda - \hat{\lambda}_{1} < \lambda$$), we find
\begin{aligned}& \begin{aligned} \tilde{I} &= \sum_{n = 1}^{\infty} \Biggl[n^{(\lambda _{2} + \frac{\varepsilon}{p})}\sum_{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} m^{(\lambda _{1} - \frac{\varepsilon}{p}) - 1}\Biggr]n^{ - \varepsilon - 1} \\ &= \sum_{n = 1}^{\infty} \omega (\hat{ \lambda}_{1},n)n^{ - \varepsilon - 1} < k_{\lambda} (\hat{ \lambda}_{1}) \Biggl(1 + \sum_{n = 2}^{\infty} n^{ - \varepsilon - 1} \Biggr) \\ &< k_{\lambda} (\hat{\lambda}_{1}) \biggl(1 + \int _{1}^{\infty} x^{ - \varepsilon - 1}\,dx\biggr) = \frac{1}{\varepsilon} k_{\lambda} (\hat{\lambda}_{1}) ( \varepsilon + 1). \end{aligned} \end{aligned}
Then we have
$$k_{\lambda} \biggl(\lambda _{1} - \frac{\varepsilon}{p} \biggr) (\varepsilon + 1) > \varepsilon \tilde{I} > M\bigl(1 - \varepsilon O(1) \bigr)^{\frac{1}{p}}(\varepsilon + 1)^{\frac{1}{q}}.$$
For $$\varepsilon \to 0^{ +}$$, we find $$k_{\lambda} (\lambda _{1}) \ge M$$. Hence, $$M = k_{\lambda} (\lambda _{1})$$ is the best possible constant factor of (13). The lemma is proved. □
Remark 2
Setting $$\tilde{\lambda}_{1}: = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}$$, $$\tilde{\lambda}_{2}: = \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}$$, we find
$$\tilde{\lambda}_{1} + \tilde{\lambda}_{2} = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} + \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p} = \frac{\lambda}{p} + \frac{\lambda}{q} = \lambda .$$
If we add the condition that $$\lambda - \lambda _{1} - \lambda _{2} \in ( - p\lambda _{1},p(\lambda - \lambda _{1}))$$,then we can find
$$0 < \tilde{\lambda}_{1} = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} < \lambda ,\qquad 0 < \tilde{\lambda}_{2} = \lambda - \tilde{\lambda}_{1} < \lambda ;$$
if we add the condition that $$\lambda - \lambda _{1} - \lambda _{2} \in [p(\lambda - \lambda _{1} - \frac{5}{4},p(\frac{5}{4} - \lambda _{1})]$$,then we have $$\tilde{\lambda}_{1},\tilde{\lambda}_{2} \le \frac{5}{4}$$.
Then, with regard to the above assumptions, we can rewrite (13) as follows:
\begin{aligned} I &= \sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m^{\lambda} + n^{\lambda}} \\ & > k_{\lambda} (\tilde{\lambda}_{1})\Biggl[\sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\tilde{\lambda}_{2})\bigr)m^{p(1 - \tilde{\lambda}_{1}) - 1} a_{m}^{p}\Biggr]^{\frac{1}{p}}\Biggl[\sum _{n = 1}^{\infty} n^{q(1 - \tilde{\lambda}_{2}) - 1} b_{n}^{q}\Biggr]^{\frac{1}{q}}. \end{aligned}
(15)
Lemma 5
If the constant factor $$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1})$$ in (10) is the best possible, then for
$$\lambda - \lambda _{1} - \lambda _{2} \in \bigl( - p \lambda _{1},p(\lambda - \lambda _{1})\bigr)\cap \biggl[p(\lambda - \lambda _{1} - \frac{5}{4},p\biggl( \frac{5}{4} - \lambda _{1}\biggr)\biggr]\bigl( \supset \{ 0 \} \bigr),$$
we have $$\lambda _{1} + \lambda _{2} = \lambda$$.
Proof
For $$0 < \tilde{\lambda}_{1} < \lambda$$,we have $$k_{\lambda} (\tilde{\lambda}_{1}) = \frac{\pi}{\lambda \sin (\pi \tilde{\lambda}_{1}/\lambda )} \in \mathrm{R}_{ +} = (0,\infty )$$.
If the constant factor $$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1})$$ in (10) is the best possible, then for $$\tilde{\lambda}_{1} = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}$$, $$\tilde{\lambda}_{2} = \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}$$, in view of the assumption and (15), we have the following inequality:
$$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1}) \ge k_{\lambda} (\tilde{\lambda}_{1}).$$
By the reverse Hölder inequality with weight (cf. [33]), we find
\begin{aligned}& \begin{aligned}[b] k_{\lambda} (\tilde{\lambda}_{1}) &= k_{\lambda} \biggl(\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}\biggr) \\ &= \int _{0}^{\infty} \frac{1}{1 + u^{\lambda}} u^{\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} - 1}\,du = \int _{0}^{\infty} \frac{1}{1 + u^{\lambda}} \bigl(u^{\frac{\lambda - \lambda _{2} - 1}{p}}\bigr) \bigl(u^{\frac{\lambda _{1} - 1}{q}}\bigr)\,du \\ &\ge \biggl( \int _{0}^{\infty} \frac{1}{1 + u^{\lambda}} u^{\lambda - \lambda _{2} - 1}\,du\biggr)^{\frac{1}{p}}\biggl( \int _{0}^{\infty} \frac{1}{1 + u^{\lambda}} u^{\lambda _{1} - 1}\,du\biggr)^{\frac{1}{q}} \\ &= \biggl( \int _{0}^{\infty} \frac{1}{1 + v^{\lambda}} v^{\lambda _{2} - 1}\,dv\biggr)^{\frac{1}{p}}\biggl( \int _{0}^{\infty} \frac{1}{1 + u^{\lambda}} u^{\lambda _{1} - 1}\,du\biggr)^{\frac{1}{q}} \\ &= k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1}). \end{aligned} \end{aligned}
(16)
Hence, we find $$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1}) = k_{\lambda} (\tilde{\lambda}_{1})$$, namely, (16) keeps the form of equality.
We observe that (16) keeps the form of equality if and only if there exist constants A and B such that they are not all zero and (cf. [33])
$$Au^{\lambda - \lambda _{2} - 1} = Bu^{\lambda _{1} - 1}\quad a.e. \text{ in } \mathrm{R}_{ +}.$$
Assuming that $$A \ne 0$$, we have $$u^{\lambda - \lambda _{2} - \lambda _{1}} = \frac{B}{A}$$ a.e. in $$\mathrm{R}_{ +}$$, and then $$\lambda - \lambda _{2} - \lambda _{1} = 0$$, namely, $$\lambda _{1} + \lambda _{2} = \lambda$$.
The lemma is proved. □
3 Main results
Theorem 1
Inequality (10) is equivalent to the following inequalities:
\begin{aligned}& \begin{aligned}[b] J&: = \Biggl[\sum_{n = 1}^{\infty} n^{p(\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}) - 1}\Biggl(\sum_{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} a_{m} \Biggr)^{p} \Biggr]^{\frac{1}{p}}\\ &> k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1})\Biggl\{ \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2}) \bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\Biggr\} ^{\frac{1}{p}}, \end{aligned} \end{aligned}
(17)
\begin{aligned}& \begin{aligned}[b] J_{1}&: = \Biggl[\sum_{m = 1}^{\infty} \frac{m^{q(\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}) - 1}}{(1 - \theta _{m}(\lambda _{2}))^{q - 1}}\Biggl(\sum_{n = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} b_{n} \Biggr)^{q} \Biggr]^{\frac{1}{q}}\\ &> k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1})\Biggl\{ \sum_{n = 1}^{\infty} n^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1} b_{n}^{q}\Biggr\} ^{\frac{1}{q}}. \end{aligned} \end{aligned}
(18)
If the constant factor in (10) is the best possible, then so is the constant factor in (17) and (18).
Proof
Suppose that (17) is valid. By the reverse Hölder inequality (cf. [33]), we have
\begin{aligned} I &= \sum_{n = 1}^{\infty} \Biggl[n^{\frac{ - 1}{p} + (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})}\sum_{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} a_{m} \Biggr] \bigl[n^{\frac{1}{p} - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})}b_{n} \bigr] \\ &\ge J\Biggl\{ \sum_{n = 1}^{\infty} n^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1} b_{n}^{q}\Biggr\} ^{\frac{1}{q}}. \end{aligned}
(19)
Then by (17) we obtain (10).
On the other hand, assuming that (10) is valid, we set
$$b_{n}: = n^{p(\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}) - 1}\Biggl(\sum _{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} a_{m} \Biggr)^{p - 1},\quad n \in \mathbf{N}.$$
If $$J = \infty$$, then (17) is naturally valid; if $$J = 0$$, then it is impossible to make (17) valid, namely, $$J > 0$$. Suppose that $$0 < J < \infty$$. By (10), we have
\begin{aligned}& \sum_{n = 1}^{\infty} n^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1} b_{n}^{q}\\& \quad = J^{p} = I > k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1})\\& \qquad {}\times \Biggl\{ \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2})\bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\Biggr\} ^{\frac{1}{p}} \Biggl\{ \sum _{n = 1}^{\infty} n^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1} b_{n}^{q}\Biggr\} ^{\frac{1}{q}},\\& \begin{aligned} J &= \Biggl\{ \sum_{n = 1}^{\infty} n^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1} b_{n}^{q}\Biggr\} ^{\frac{1}{p}}\\ &> k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1})\Biggl\{ \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2}) \bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\Biggr\} ^{\frac{1}{p}}, \end{aligned} \end{aligned}
namely, (17) follows, which is equivalent to (10).
Suppose that (18) is valid. By the reverse Hölder inequality, we have
\begin{aligned} I &= \sum_{m = 1}^{\infty} \bigl[\bigl(1 - \theta _{m}(\lambda _{2})\bigr)^{\frac{1}{p}}m^{\frac{1}{q} - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})}a_{m} \bigr] \Biggl[\frac{m^{\frac{ - 1}{q} + (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})}}{(1 - \theta _{m}(\lambda _{2}))^{1/p}}\sum_{n = 1}^{\infty} \frac{1}{m^{\lambda} + n{}^{\lambda}} b_{n} \Biggr] \\ &\ge \Biggl\{ \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2})\bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\Biggr\} ^{\frac{1}{p}}J_{1}. \end{aligned}
(20)
Then by (18) we obtain (10). On the other hand, assuming that (10) is valid, we set
$$a_{m}: = \frac{m^{q(\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}) - 1}}{(1 - \theta _{m}(\lambda _{2}))^{q - 1}}\Biggl(\sum _{n = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} b_{n} \Biggr)^{q - 1},\quad m \in \mathrm{N}.$$
If $$J_{1} = \infty$$, then (18) is naturally valid; if $$J_{1} = 0$$, then it is impossible to make (18) valid, namely, $$J_{1} > 0$$. Suppose that $$0 < J_{1} < \infty$$. By (10), we have
\begin{aligned}& \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2})\bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\\& \quad = J_{1}^{q} = I > k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1})\\& \qquad {}\times \Biggl\{ \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2})\bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\Biggr\} ^{\frac{1}{p}} \Biggl\{ \sum _{n = 1}^{\infty} n^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1} b_{n}^{q}\Biggr\} ^{\frac{1}{q}},\\& \begin{aligned} J_{1} &= \Biggl\{ \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2}) \bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\Biggr\} ^{\frac{1}{q}}\\ &> k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1})\Biggl\{ \sum_{n = 1}^{\infty} n^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1} b_{n}^{q}\Biggr\} ^{\frac{1}{q}}, \end{aligned} \end{aligned}
namely, (18) follows, which is equivalent to (10).
Hence, inequalities (10), (17), and (18) are equivalent.
If the constant factor in (10) is the best possible, then so is the constant factor in (17) and (18). Otherwise, by (19) (or (20)), we would reach a contradiction that the constant factor in (10) is not the best possible.
The theorem is proved. □
Theorem 2
The following statements (i), (ii), (iii), and (iv) are equivalent:
1. (i)
Both $$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1})$$ and $$k_{\lambda} (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})$$ are independent of p, q;
2. (ii)
$$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1}) = k_{\lambda} (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})$$;
3. (iii)
$$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1})$$ in (10) is the best possible constant factor;
4. (iv)
If $$\lambda - \lambda _{1} - \lambda _{2} \in ( - p\lambda _{1},p(\lambda - \lambda _{1}))\cap [p(\lambda - \lambda _{1} - \frac{5}{4},p(\frac{5}{4} - \lambda _{1} )]$$, then $$\lambda _{1} + \lambda _{2} = \lambda$$.
If the statement (iv) follows, namely, $$\lambda _{1} + \lambda _{2} = \lambda$$, then we have (13) and the following equivalent inequalities with the best possible constant factor $$k_{\lambda} (\lambda _{1})$$:
\begin{aligned}& \Biggl[\sum_{n = 1}^{\infty} n^{p\lambda _{2} - 1}\Biggl(\sum_{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} a_{m} \Biggr)^{p} \Biggr]^{\frac{1}{p}}> k_{\lambda} (\lambda _{1})\Biggl[\sum _{m = 1}^{\infty} \bigl(1 - \theta _{m}( \lambda _{2})\bigr)m^{p(1 - \lambda _{1}) - 1} a_{m}^{p} \Biggr]^{\frac{1}{p}}, \end{aligned}
(21)
\begin{aligned}& \Biggl[\sum_{m = 1}^{\infty} \frac{m^{q\lambda _{1} - 1}}{(1 - \theta _{m}(\lambda _{2}))^{q - 1}}\Biggl(\sum_{n = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} b_{n} \Biggr)^{q} \Biggr]^{\frac{1}{q}} > k_{\lambda} (\lambda _{1})\Biggl[\sum _{n = 1}^{\infty} n^{q(1 - \lambda _{2}) - 1} b_{n}^{q}\Biggr]^{\frac{1}{q}}. \end{aligned}
(22)
Proof
(i) (ii). By (i), since $$\tilde{\lambda}_{1} + \tilde{\lambda}_{2} = \lambda$$, we have
\begin{aligned}& k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1}) = \lim_{p \to 1^{ -}} \lim _{q \to - \infty} k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1}) = k_{\lambda} (\lambda _{2}),\\& k_{\lambda} \biggl(\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}\biggr) = k_{\lambda} \biggl(\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}\biggr) = \lim_{p \to 1^{ -}} \lim_{q \to - \infty} k_{\lambda} \biggl(\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}\biggr) = k_{\lambda} ( \lambda _{2}), \end{aligned}
namely, $$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1}) = k_{\lambda} (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})$$.
(ii) (iv). If $$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1}) = k_{\lambda} (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})$$, then (16) keeps the form of equality. In view of the proof of Lemma 5, it follows that $$\lambda _{1} + \lambda _{2} = \lambda$$.
(iv) (i). If $$\lambda _{1} + \lambda _{2} = \lambda$$, then
$$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1}) = k_{\lambda} \biggl(\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}\biggr) = k_{\lambda} (\lambda _{1}),$$
which is independent of p, q. Hence, it follows that (i) (ii) (iv).
(iii) (iv). By the assumption and Lemma 5, we have $$\lambda _{1} + \lambda _{2} = \lambda$$.
(iv) (iii). By Lemma 4, for $$\lambda _{1} + \lambda _{2} = \lambda$$, $$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1})( = k_{\lambda} (\lambda _{1}))$$ is the best possible constant factor of (10). Therefore, we have (iii) (iv).
Hence, the statements (i), (ii), (iii), and (iv) are equivalent.
The theorem is proved. □
Remark 3
(i) For $$\lambda _{1} = \lambda _{2} = \frac{\lambda}{2} \in (0,\frac{5}{4}]$$ ($$0 < \lambda \le \frac{5}{2}$$) in (13), (21), and (22), we have the following equivalent inequalities with the best possible constant factor $$\frac{\pi}{\lambda}$$:
\begin{aligned}& \sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m^{\lambda} + n^{\lambda}} > \frac{\pi}{\lambda} \Biggl[\sum_{m = 1}^{\infty} \biggl(1 - \theta _{m}\biggl(\frac{\lambda}{2}\biggr) \biggr)m^{p(1 - \frac{\lambda}{2}) - 1} a_{m}^{p}\Biggr]^{\frac{1}{p}} \Biggl[\sum_{n = 1}^{\infty} n^{q(1 - \frac{\lambda}{2}) - 1} b_{n}^{q}\Biggr]^{\frac{1}{q}}, \end{aligned}
(23)
\begin{aligned}& \Biggl[\sum_{n = 1}^{\infty} n^{\frac{p\lambda}{2} - 1}\Biggl(\sum_{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} a_{m} \Biggr)^{p} \Biggr]^{\frac{1}{p}} > \frac{\pi}{\lambda} \Biggl\{ \sum _{m = 1}^{\infty} \biggl(1 - \theta _{m} \biggl(\frac{\lambda}{2}\biggr)\biggr)m^{p(1 - \frac{\lambda}{2}) - 1} a_{m}^{p}\Biggr]^{\frac{1}{p}}, \end{aligned}
(24)
\begin{aligned}& \Biggl[\sum_{m = 1}^{\infty} \frac{m^{\frac{q\lambda}{2} - 1}}{(1 - \theta _{m}(\frac{\lambda}{2}))^{q - 1}}\Biggl(\sum_{n = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} b_{n} \Biggr)^{q}\Biggr]^{\frac{1}{q}} > \frac{\pi}{ \lambda} \Biggl[\sum_{n = 1}^{\infty} n^{q(1 - \frac{\lambda}{2}) - 1} b_{n}^{q}\Biggr]^{\frac{1}{q}}. \end{aligned}
(25)
In particular, for $$\lambda = \frac{5}{2}$$,we have the following equivalent inequalities:
\begin{aligned}& \sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m^{5/2} + n^{5/2}} > \frac{2\pi}{5} \Biggl[\sum_{m = 1}^{\infty} \biggl(1 - \theta _{m}\biggl(\frac{5}{4}\biggr)\biggr) \frac{a_{m}^{p}}{m^{1 + p/4}} \Biggr]^{\frac{1}{p}}\Biggl(\sum _{n = 1}^{\infty} \frac{b_{n}^{q}}{n^{1 + q/4}} \Biggr)^{\frac{1}{q}}, \end{aligned}
(26)
\begin{aligned}& \Biggl[\sum_{n = 1}^{\infty} n^{\frac{5p}{4} - 1}\Biggl(\sum_{m = 1}^{\infty} \frac{1}{m^{5/2} + n^{5/2}}a_{m} \Biggr)^{p} \Biggr]^{\frac{1}{p}} > \frac{2\pi}{ 5}\Biggl[\sum _{m = 1}^{\infty} \biggl(1 - \theta \biggl( \frac{5}{4}\biggr)\biggr)\frac{a_{m}^{p}}{m^{1 + p/4}} \Biggr]^{\frac{1}{p}}, \end{aligned}
(27)
\begin{aligned}& \Biggl[\sum_{m = 1}^{\infty} \frac{m^{\frac{5q}{4} - 1}}{(1 - \theta _{m}(\frac{5}{4}))^{q - 1}}\Biggl(\sum_{n = 1}^{\infty} \frac{1}{m^{5/2} + n^{5/2}}b_{n} \Biggr)^{q} \Biggr]^{\frac{1}{q}} > \frac{2\pi}{5}\Biggl(\sum _{n = 1}^{\infty} \frac{b_{n}^{q}}{n^{1 + q/4}} \Biggr)^{\frac{1}{q}}. \end{aligned}
(28)
(ii) For $$\lambda = 1$$, $$\lambda _{1} = \frac{1}{r}$$, $$\lambda _{2} = \frac{1}{s}$$ ($$r > 1,\frac{1}{r} + \frac{1}{s} = 1$$) in (13), (21), and (22), we have the following equivalent inequalities with the best possible constant factor $$\frac{\pi}{\sin (\pi /r)}$$:
\begin{aligned}& \sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m + n} > \frac{\pi}{\sin (\pi /r)} \Biggl[\sum_{m = 1}^{\infty} \biggl(1 - \hat{\theta}_{m}\biggl(\frac{1}{r}\biggr) \biggr)m^{\frac{p}{s} - 1} a_{m}^{p} \Biggr]^{\frac{1}{p}}\Biggl(\sum_{n = 1}^{\infty} n^{\frac{q}{r} - 1} b_{n}^{q}\Biggr)^{\frac{1}{q}}, \end{aligned}
(29)
\begin{aligned}& \Biggl[\sum_{n = 1}^{\infty} n^{\frac{p}{s} - 1}\Biggl(\sum_{m = 1}^{\infty} \frac{1}{m + n}a_{m} \Biggr)^{p} \Biggr]^{\frac{1}{p}} > \frac{\pi}{\sin (\pi /r)}\Biggl[\sum _{m = 1}^{\infty} \biggl(1 - \hat{\theta}_{m} \biggl(\frac{1}{s}\biggr)\biggr)m^{\frac{p}{s} - 1} a_{m}^{p} \Biggr]^{\frac{1}{p}}, \end{aligned}
(30)
\begin{aligned}& \Biggl[\sum_{m = 1}^{\infty} \frac{m^{\frac{q}{r} - 1}}{(1 - \hat{\theta}_{m}(\frac{1}{s}))^{q - 1}}\Biggl(\sum_{n = 1}^{\infty} \frac{1}{m + n}b_{n} \Biggr)^{q} \Biggr]^{\frac{1}{q}} > \frac{\pi}{\sin (\pi /r)}\Biggl(\sum _{n = 1}^{\infty} n^{\frac{q}{r} - 1} b_{n}^{q}\Biggr)^{\frac{1}{q}}, \end{aligned}
(31)
where $$\hat{\theta}_{m}(\frac{1}{s}): = \frac{\sin (\pi /s)}{\pi} \int _{0}^{\frac{1}{m}} \frac{u^{ - 1/r}}{1 + u}\,du = O(\frac{1}{m^{1/s}}) \in (0,1)$$ ($$m \in \mathrm{N}$$).
Inequality (29) is a reverse of (1).
4 Conclusions
In this paper, by virtue of the symmetry principle, applying the techniques of real analysis and Euler–Maclaurin summation formula, we construct proper weight coefficients and use them to establish a reverse extended Hardy–Hilbert’s inequality with multi-parameters in Lemma 3. Then, we obtain the equivalent forms and some equivalent statements of the best possible constant factor related to several parameters in Theorem 1 and Theorem 2. Finally, we illustrate how the obtained results can generate some new reverse Hardy–Hilbert-type inequalities in Remark 3. The lemmas and theorems provide an extensive account of this type of reverse inequalities.
Availability of data and materials
The data used to support the findings of this study are included within the article.
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Acknowledgements
The authors thank the referee for his useful proposal to reform the paper.
Funding
This work is supported by the National Natural Science Foundation (Nos. 11961021, 11561019), Guangxi Natural Science Foundation (2020GXNSFAA159084) and Hechi University Research Foundation for Advanced Talents under Grant (Nos. 2019GCC005, 2021GCC024). We are grateful for this help.
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BY carried out the mathematical studies, participated in the sequence alignment, and drafted the manuscript. RL and XH participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.
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Correspondence to Ricai Luo.
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Luo, R., Yang, B. & Huang, X. A reverse extended Hardy–Hilbert’s inequality with parameters. J Inequal Appl 2023, 58 (2023). https://doi.org/10.1186/s13660-023-02967-5 | 18,187 | 44,046 | {"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": 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} | 2.9375 | 3 | CC-MAIN-2024-33 | latest | en | 0.661285 |
https://www.plati.com/itm/45-distance-between-two-point-charges-q1-7-8729-1/1684275?lang=en-US | 1,529,687,010,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864740.48/warc/CC-MAIN-20180622162604-20180622182604-00600.warc.gz | 894,422,938 | 11,317 | 45. distance between two point charges Q1 = 7 ∙ 1
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45. distance between two point charges Q1 = 7 ∙ 10-9 TC and Q2 = 1,5 ∙ 10-8 Kl equal to 5 cm. Find the electric field at a point 3 cm from the positive charge and 4 cm from the negative charge. | 107 | 309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-26 | latest | en | 0.829375 |
http://davidmlane.com/hyperstat/B38037.html | 1,542,133,709,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741340.11/warc/CC-MAIN-20181113173927-20181113195927-00503.warc.gz | 83,231,576 | 1,957 | # The Precise Meaning of the Probability Value (3 of 3)
The proper interpretation of the test is as follows: A person made a rather extraordinary claim and should be able to provide strong evidence in support of the claim if the claim is to believed. The test provided data consistent with the null hypothesis that the person has no special ability since a person with no special ability would be able to predict as well or better more than 40% of the time. Therefore, there is no compelling reason to believe the extraordinary claim. However, the test does not prove the person cannot predict better than chance; it simply fails to provide evidence that he or she can. The probability that the null hypothesis is true is not determined by the statistical analysis conducted as part of hypothesis testing. Rather, the probability computed is the probability of obtaining data as different or more different from the null hypothesis (given that the null hypothesis is true) as the data actually obtained. | 191 | 1,004 | {"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} | 2.875 | 3 | CC-MAIN-2018-47 | latest | en | 0.967882 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/260/2/bf/b/ | 1,660,437,704,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571989.67/warc/CC-MAIN-20220813232744-20220814022744-00611.warc.gz | 729,880,923 | 56,919 | # Properties
Label 260.2.bf.b Level $260$ Weight $2$ Character orbit 260.bf Analytic conductor $2.076$ Analytic rank $0$ Dimension $4$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$260 = 2^{2} \cdot 5 \cdot 13$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 260.bf (of order $$12$$, degree $$4$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$2.07611045255$$ Analytic rank: $$0$$ Dimension: $$4$$ Coefficient field: $$\Q(\zeta_{12})$$ Defining polynomial: $$x^{4} - x^{2} + 1$$ x^4 - x^2 + 1 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{12}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{12}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + ( - \zeta_{12} + 1) q^{3} + (2 \zeta_{12}^{3} - 1) q^{5} + (4 \zeta_{12}^{3} - \zeta_{12}^{2} - 2 \zeta_{12} + 1) q^{7} + (\zeta_{12}^{2} + \zeta_{12} + 1) q^{9}+O(q^{10})$$ q + (-z + 1) * q^3 + (2*z^3 - 1) * q^5 + (4*z^3 - z^2 - 2*z + 1) * q^7 + (z^2 + z + 1) * q^9 $$q + ( - \zeta_{12} + 1) q^{3} + (2 \zeta_{12}^{3} - 1) q^{5} + (4 \zeta_{12}^{3} - \zeta_{12}^{2} - 2 \zeta_{12} + 1) q^{7} + (\zeta_{12}^{2} + \zeta_{12} + 1) q^{9} + (2 \zeta_{12}^{3} + 2 \zeta_{12}^{2} + \zeta_{12} - 3) q^{11} + ( - 2 \zeta_{12}^{3} + 3) q^{13} + (2 \zeta_{12}^{3} - 2 \zeta_{12}^{2} + \zeta_{12} + 1) q^{15} + ( - \zeta_{12}^{3} + 2 \zeta_{12}^{2} - \zeta_{12} - 1) q^{17} + ( - 3 \zeta_{12}^{3} + 2 \zeta_{12}^{2} + \zeta_{12} + 1) q^{19} + (5 \zeta_{12}^{3} - 3 \zeta_{12}^{2} - 3 \zeta_{12} + 5) q^{21} + ( - 4 \zeta_{12}^{3} + 4 \zeta_{12}^{2} + \zeta_{12} - 3) q^{23} + ( - 4 \zeta_{12}^{3} - 3) q^{25} + ( - \zeta_{12}^{3} - 3 \zeta_{12}^{2} + 3 \zeta_{12} + 1) q^{27} + ( - \zeta_{12}^{2} + 2) q^{29} + (3 \zeta_{12}^{3} + 6 \zeta_{12}^{2} - 6 \zeta_{12} - 3) q^{31} + ( - \zeta_{12}^{2} + 4 \zeta_{12} - 1) q^{33} + ( - 4 \zeta_{12}^{3} - 3 \zeta_{12}^{2} + 4 \zeta_{12} - 5) q^{35} + (2 \zeta_{12}^{3} - 5 \zeta_{12}^{2} + 2 \zeta_{12}) q^{37} + ( - 2 \zeta_{12}^{3} + 2 \zeta_{12}^{2} - 3 \zeta_{12} + 1) q^{39} + (5 \zeta_{12} + 5) q^{41} + ( - \zeta_{12}^{3} - 4 \zeta_{12}^{2} - 3 \zeta_{12} + 3) q^{43} + (4 \zeta_{12}^{3} + \zeta_{12}^{2} - 3 \zeta_{12} - 3) q^{45} + (4 \zeta_{12}^{3} - 8 \zeta_{12} - 4) q^{47} + (4 \zeta_{12}^{3} - 6 \zeta_{12}^{2} + 4 \zeta_{12}) q^{49} + ( - 3 \zeta_{12}^{3} + 4 \zeta_{12}^{2} - 2) q^{51} + ( - 3 \zeta_{12}^{3} + 4 \zeta_{12}^{2} + 4 \zeta_{12} - 3) q^{53} + ( - 4 \zeta_{12}^{3} - 5 \zeta_{12} - 3) q^{55} + ( - 5 \zeta_{12}^{3} + 4 \zeta_{12}^{2} - 2) q^{57} + ( - \zeta_{12}^{3} - 6 \zeta_{12}^{2} - 5 \zeta_{12} + 5) q^{59} + ( - 9 \zeta_{12}^{2} + 9) q^{61} + (5 \zeta_{12}^{3} + \zeta_{12}^{2} - 5 \zeta_{12} - 2) q^{63} + (8 \zeta_{12}^{3} + 1) q^{65} + ( - 14 \zeta_{12}^{3} + \zeta_{12}^{2} + 14 \zeta_{12} - 2) q^{67} + ( - 8 \zeta_{12}^{3} + 7 \zeta_{12}^{2} + 4 \zeta_{12} - 7) q^{69} + ( - \zeta_{12}^{3} + 2 \zeta_{12}^{2} + 3 \zeta_{12} - 3) q^{71} + ( - 8 \zeta_{12}^{3} - 8 \zeta_{12}^{2} + 4) q^{73} + ( - 4 \zeta_{12}^{3} + 4 \zeta_{12}^{2} + 3 \zeta_{12} - 7) q^{75} + ( - 9 \zeta_{12}^{3} + \zeta_{12}^{2} + \zeta_{12} - 9) q^{77} + ( - 4 \zeta_{12}^{3} + 4 \zeta_{12}^{2} - 2) q^{79} + (5 \zeta_{12}^{3} - 2 \zeta_{12}^{2} + 5 \zeta_{12}) q^{81} + ( - 4 \zeta_{12}^{3} + 8 \zeta_{12} - 4) q^{83} + (3 \zeta_{12}^{3} - 4 \zeta_{12}^{2} - 3 \zeta_{12} + 5) q^{85} + (\zeta_{12}^{3} - \zeta_{12}^{2} - 2 \zeta_{12} + 2) q^{87} + ( - 4 \zeta_{12}^{3} + 4 \zeta_{12}^{2} + 9 \zeta_{12} + 5) q^{89} + (12 \zeta_{12}^{3} + \zeta_{12}^{2} - 8 \zeta_{12} + 7) q^{91} + ( - 3 \zeta_{12}^{3} + 9 \zeta_{12}^{2} - 3 \zeta_{12}) q^{93} + (9 \zeta_{12}^{3} - 5 \zeta_{12} + 3) q^{95} + ( - \zeta_{12}^{2} - 6 \zeta_{12} - 1) q^{97} + (7 \zeta_{12}^{3} + 4 \zeta_{12}^{2} - 4 \zeta_{12} - 7) q^{99}+O(q^{100})$$ q + (-z + 1) * q^3 + (2*z^3 - 1) * q^5 + (4*z^3 - z^2 - 2*z + 1) * q^7 + (z^2 + z + 1) * q^9 + (2*z^3 + 2*z^2 + z - 3) * q^11 + (-2*z^3 + 3) * q^13 + (2*z^3 - 2*z^2 + z + 1) * q^15 + (-z^3 + 2*z^2 - z - 1) * q^17 + (-3*z^3 + 2*z^2 + z + 1) * q^19 + (5*z^3 - 3*z^2 - 3*z + 5) * q^21 + (-4*z^3 + 4*z^2 + z - 3) * q^23 + (-4*z^3 - 3) * q^25 + (-z^3 - 3*z^2 + 3*z + 1) * q^27 + (-z^2 + 2) * q^29 + (3*z^3 + 6*z^2 - 6*z - 3) * q^31 + (-z^2 + 4*z - 1) * q^33 + (-4*z^3 - 3*z^2 + 4*z - 5) * q^35 + (2*z^3 - 5*z^2 + 2*z) * q^37 + (-2*z^3 + 2*z^2 - 3*z + 1) * q^39 + (5*z + 5) * q^41 + (-z^3 - 4*z^2 - 3*z + 3) * q^43 + (4*z^3 + z^2 - 3*z - 3) * q^45 + (4*z^3 - 8*z - 4) * q^47 + (4*z^3 - 6*z^2 + 4*z) * q^49 + (-3*z^3 + 4*z^2 - 2) * q^51 + (-3*z^3 + 4*z^2 + 4*z - 3) * q^53 + (-4*z^3 - 5*z - 3) * q^55 + (-5*z^3 + 4*z^2 - 2) * q^57 + (-z^3 - 6*z^2 - 5*z + 5) * q^59 + (-9*z^2 + 9) * q^61 + (5*z^3 + z^2 - 5*z - 2) * q^63 + (8*z^3 + 1) * q^65 + (-14*z^3 + z^2 + 14*z - 2) * q^67 + (-8*z^3 + 7*z^2 + 4*z - 7) * q^69 + (-z^3 + 2*z^2 + 3*z - 3) * q^71 + (-8*z^3 - 8*z^2 + 4) * q^73 + (-4*z^3 + 4*z^2 + 3*z - 7) * q^75 + (-9*z^3 + z^2 + z - 9) * q^77 + (-4*z^3 + 4*z^2 - 2) * q^79 + (5*z^3 - 2*z^2 + 5*z) * q^81 + (-4*z^3 + 8*z - 4) * q^83 + (3*z^3 - 4*z^2 - 3*z + 5) * q^85 + (z^3 - z^2 - 2*z + 2) * q^87 + (-4*z^3 + 4*z^2 + 9*z + 5) * q^89 + (12*z^3 + z^2 - 8*z + 7) * q^91 + (-3*z^3 + 9*z^2 - 3*z) * q^93 + (9*z^3 - 5*z + 3) * q^95 + (-z^2 - 6*z - 1) * q^97 + (7*z^3 + 4*z^2 - 4*z - 7) * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4 q + 4 q^{3} - 4 q^{5} + 2 q^{7} + 6 q^{9}+O(q^{10})$$ 4 * q + 4 * q^3 - 4 * q^5 + 2 * q^7 + 6 * q^9 $$4 q + 4 q^{3} - 4 q^{5} + 2 q^{7} + 6 q^{9} - 8 q^{11} + 12 q^{13} + 8 q^{19} + 14 q^{21} - 4 q^{23} - 12 q^{25} - 2 q^{27} + 6 q^{29} - 6 q^{33} - 26 q^{35} - 10 q^{37} + 8 q^{39} + 20 q^{41} + 4 q^{43} - 10 q^{45} - 16 q^{47} - 12 q^{49} - 4 q^{53} - 12 q^{55} + 8 q^{59} + 18 q^{61} - 6 q^{63} + 4 q^{65} - 6 q^{67} - 14 q^{69} - 8 q^{71} - 20 q^{75} - 34 q^{77} - 4 q^{81} - 16 q^{83} + 12 q^{85} + 6 q^{87} + 28 q^{89} + 30 q^{91} + 18 q^{93} + 12 q^{95} - 6 q^{97} - 20 q^{99}+O(q^{100})$$ 4 * q + 4 * q^3 - 4 * q^5 + 2 * q^7 + 6 * q^9 - 8 * q^11 + 12 * q^13 + 8 * q^19 + 14 * q^21 - 4 * q^23 - 12 * q^25 - 2 * q^27 + 6 * q^29 - 6 * q^33 - 26 * q^35 - 10 * q^37 + 8 * q^39 + 20 * q^41 + 4 * q^43 - 10 * q^45 - 16 * q^47 - 12 * q^49 - 4 * q^53 - 12 * q^55 + 8 * q^59 + 18 * q^61 - 6 * q^63 + 4 * q^65 - 6 * q^67 - 14 * q^69 - 8 * q^71 - 20 * q^75 - 34 * q^77 - 4 * q^81 - 16 * q^83 + 12 * q^85 + 6 * q^87 + 28 * q^89 + 30 * q^91 + 18 * q^93 + 12 * q^95 - 6 * q^97 - 20 * q^99
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/260\mathbb{Z}\right)^\times$$.
$$n$$ $$41$$ $$131$$ $$157$$ $$\chi(n)$$ $$\zeta_{12}$$ $$1$$ $$-\zeta_{12}^{3}$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
37.1
−0.866025 − 0.500000i 0.866025 + 0.500000i 0.866025 − 0.500000i −0.866025 + 0.500000i
0 1.86603 + 0.500000i 0 −1.00000 2.00000i 0 2.23205 3.86603i 0 0.633975 + 0.366025i 0
93.1 0 0.133975 0.500000i 0 −1.00000 + 2.00000i 0 −1.23205 + 2.13397i 0 2.36603 + 1.36603i 0
137.1 0 0.133975 + 0.500000i 0 −1.00000 2.00000i 0 −1.23205 2.13397i 0 2.36603 1.36603i 0
253.1 0 1.86603 0.500000i 0 −1.00000 + 2.00000i 0 2.23205 + 3.86603i 0 0.633975 0.366025i 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
65.t even 12 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 260.2.bf.b 4
5.b even 2 1 1300.2.bn.a 4
5.c odd 4 1 260.2.bk.a yes 4
5.c odd 4 1 1300.2.bs.b 4
13.f odd 12 1 260.2.bk.a yes 4
65.o even 12 1 1300.2.bn.a 4
65.s odd 12 1 1300.2.bs.b 4
65.t even 12 1 inner 260.2.bf.b 4
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
260.2.bf.b 4 1.a even 1 1 trivial
260.2.bf.b 4 65.t even 12 1 inner
260.2.bk.a yes 4 5.c odd 4 1
260.2.bk.a yes 4 13.f odd 12 1
1300.2.bn.a 4 5.b even 2 1
1300.2.bn.a 4 65.o even 12 1
1300.2.bs.b 4 5.c odd 4 1
1300.2.bs.b 4 65.s odd 12 1
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{3}^{4} - 4T_{3}^{3} + 5T_{3}^{2} - 2T_{3} + 1$$ acting on $$S_{2}^{\mathrm{new}}(260, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{4}$$
$3$ $$T^{4} - 4 T^{3} + 5 T^{2} - 2 T + 1$$
$5$ $$(T^{2} + 2 T + 5)^{2}$$
$7$ $$T^{4} - 2 T^{3} + 15 T^{2} + 22 T + 121$$
$11$ $$T^{4} + 8 T^{3} + 41 T^{2} + 130 T + 169$$
$13$ $$(T^{2} - 6 T + 13)^{2}$$
$17$ $$T^{4} + 9 T^{2} + 18 T + 9$$
$19$ $$T^{4} - 8 T^{3} + 41 T^{2} - 130 T + 169$$
$23$ $$T^{4} + 4 T^{3} + 53 T^{2} + 14 T + 1$$
$29$ $$(T^{2} - 3 T + 3)^{2}$$
$31$ $$T^{4} + 2916$$
$37$ $$T^{4} + 10 T^{3} + 87 T^{2} + \cdots + 169$$
$41$ $$T^{4} - 20 T^{3} + 125 T^{2} + \cdots + 625$$
$43$ $$T^{4} - 4 T^{3} + 29 T^{2} - 230 T + 529$$
$47$ $$(T^{2} + 8 T - 32)^{2}$$
$53$ $$T^{4} + 4 T^{3} + 8 T^{2} - 88 T + 484$$
$59$ $$T^{4} - 8 T^{3} + 65 T^{2} + \cdots + 3481$$
$61$ $$(T^{2} - 9 T + 81)^{2}$$
$67$ $$T^{4} + 6 T^{3} - 181 T^{2} + \cdots + 37249$$
$71$ $$T^{4} + 8 T^{3} + 17 T^{2} + 22 T + 121$$
$73$ $$T^{4} + 224T^{2} + 256$$
$79$ $$T^{4} + 56T^{2} + 16$$
$83$ $$(T^{2} + 8 T - 32)^{2}$$
$89$ $$T^{4} - 28 T^{3} + 197 T^{2} + \cdots + 2209$$
$97$ $$T^{4} + 6 T^{3} - 21 T^{2} + \cdots + 1089$$ | 5,486 | 9,665 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-33 | latest | en | 0.416929 |
https://www.premiumessaywritingservice.com/thermodynamic/ | 1,581,966,987,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143079.30/warc/CC-MAIN-20200217175826-20200217205826-00297.warc.gz | 875,135,512 | 11,630 | # thermodynamic
| November 16, 2015
I want it after 24 hours
1) 1)During the compression of 0.01 lbm of air in a cylinder heat is transferred through the cylinder walls to keep the air at a constant temperature. The air pressure increases from 15 psia to 150 psia the air is fully compressed. The initial specific volume of air is 7.4 ft^3/lbm. Determine the operating temperature, the change in internal energy and the enthalpy, the work done and the heat transferred during this process. (BB example 6-6)
2) 2) (Book 6-49) Air at 100F is isothermally compressed from 1 atm to 5 atm. Determine the work done on 1 lbm of air and the heat removed during the process.
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Order a customized paper today! | 224 | 866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.854061 |
https://rdrr.io/cran/QFRM/src/R/DeferredPayment.R | 1,642,330,083,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299852.23/warc/CC-MAIN-20220116093137-20220116123137-00410.warc.gz | 568,123,893 | 9,137 | R/DeferredPayment.R In QFRM: Pricing of Vanilla and Exotic Option Contracts
Documented in DeferredPaymentLT
#' @title DeferredPaymentLT
#' @description A binomial tree pricer of a Deferred Payment option.
#' An American option that has payment at expiry no matter when exercise,
#' causing differences in present value (PV) of a payoff.
#' @author Max Lee, Department of Statistics, Rice University, Spring 2015
#'
#' @param o An object of class \code{OptPx}
#' @return An object of class \code{OptPx} with price included
#'
#' @references Hull, J.C., \emph{Options, Futures and Other Derivatives}, 9ed, 2014. Prentice Hall.
#' ISBN 978-0-13-345631-8, \url{http://www-2.rotman.utoronto.ca/~hull/ofod/index.html}
#'
#' @examples
#' (o = DeferredPaymentLT())$PxLT #' #' o = Opt(Style='DeferredPayment', Right="Call", S0=110,ttm=.5,K=110) #' (o = DeferredPaymentLT(OptPx(o,r=.05,q=.04,vol=.2,NSteps=5)))$PxLT
#'
#' o = Opt(Style='DeferredPayment', Right="Put", S0 = 50, ttm=2,K=47)
#' (o = DeferredPaymentLT(OptPx(o,r=.05,q=.04,vol=.25,NSteps=3)))$PxLT #' #' @export #' DeferredPaymentLT = function(o=OptPx(Opt(Style="DeferredPayment"))){ stopifnot(o$Style$DeferredPayment, is.OptPx(o)); u1 <- exp(o$vol*sqrt(o$ttm/o$NSteps))
d1 <- 1/u1
p1 <- (exp((o$r-o$q)*(o$ttm/o$NSteps))-d1)/(u1-d1)
S = with(o, S0*d1^(0:NSteps)*u1^(NSteps:0))
V = with(o, S0*d1^(0:NSteps)*u1^(NSteps:0))
O = pmax(o$Right$SignCP * (S - o$K), 0) ReCalc.O_S.on.Prior.Time.Step = function(i) { O <<- if(i==o$NSteps){
exp(-o$r * o$ttm/o$NSteps) * (p1*O[-i-1] + (1-p1)*O[-1])}else{ exp(-o$r * o$ttm/o$NSteps*(o$NSteps-i)) * (p1*O[-i-1] + (1-p1)*O[-1]) } S <<- d1 * S[-i-1] if(i==o$NSteps){
Payout = pmax(o$Right$SignCP *(S-o$K),0) }else{ Payout = pmax(o$Right$SignCP *(S-o$K),0)*exp(-o$r*o$ttm/o$NSteps*(o$NSteps-i))
}
O <<- pmax(O, Payout)
return(cbind(S, O))
}
BT = append(list(cbind(S, O)), sapply(o$NSteps:1, ReCalc.O_S.on.Prior.Time.Step)) o$PxLT = BT[[length(BT)]][[2]]
return(o)
}
Try the QFRM package in your browser
Any scripts or data that you put into this service are public.
QFRM documentation built on May 2, 2019, 8:26 a.m. | 785 | 2,108 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-05 | latest | en | 0.456205 |
https://tutorstips.com/question-no-16-chapter-no-5-t-s-grewal-11-class/ | 1,611,627,233,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704795033.65/warc/CC-MAIN-20210126011645-20210126041645-00714.warc.gz | 601,408,133 | 15,827 | # Question No 16 Chapter No 5 – T.S. Grewal 11 Class
Question No 16 Chapter No 5
16. Show an Accounting Equation for the following Transactions:
(i) D. Mahapatra Commenced business with cash Rs 50,000 and Rs. 1,00,000 by cheque, goods Rs 60,000 machinery Rs 1,00,000 and Furniture Rs 50,000.
(ii) 1/3rd of the above goods sold at profit of 10% on cost and half of the payments received in cash.
(iii) Depreciation on machinery provide @ 10%
(iv) Cash withdrawn for personal use Rs 10,000
(v) Interest on drawings charged @ 5%
(vi) Goods sold to Gupta for Rs 10,000 and received a Bill receivable for the same amount for 3 months.
(vii) Received Rs 10,000 from Gupta against the bills Receivable on its maturity.
### Solution of Question No 16 Chapter No 5: –
Note: “Becasue we less width so we have deleted column of Particulars but you have to make it in the paper or in your note book.”
S. No. Assets Liab. Capital Cash +Bank +Stock +Machi. +Furn +Debt. +B/R (i) +50000 +100000 +60000 +100000 +50000 +360000 50000 +100000 +60000 +100000 +50000 – – – +360000 (ii) +11000 – -20000 – – +11000 +2000 61000 +100000 +40000 +100000 +50000 +11000 – – +362000 (iii) – – – -10000 – – -10000 61000 +100000 +40000 +90000 +50000 +11000 – – +352000 (iv) -10000 – – – – – -10000 51000 +100000 +40000 +90000 +50000 +11000 – – +342000 (v) – – – – – – -500 +500 51000 +100000 +40000 +90000 +50000 +11000 – – +342000 (vi) – – -10000 – – – +10000 51000 +100000 +30000 +90000 +50000 +11000 +10000 – +342000 (vii) +10000 – – – – – -10000 – – 61000 +100000 +30000 +90000 +50000 +11000 – – +342000
Assets: – Cash 61,000 + Bank 1,00,000 + Stock 30,000 + Machine 90,000 + Furniture 50,000 + Debtors 11,000 = 3,42,000/-
Liabilities: – 0/-
Capital = 3,42,000/-
Liabilities +Capital
0 + 3,42,000 = 3,42,000/-
### Explanation of All Transactions with images: –
This is not a part of the solution, So you don’t have to write it in the exam. So why we explained if it is not needed. Because This explanation will help you to understand all transactions with logic so don’t need to remember all the transactions but just understand and remember the logic use behind it.
#### Transaction No. 1
As we discuss in the previous topic, A owner and the business both have a separate identity in the eye of law. So, the business will be treated as an Artificial Person and anything invested by the owner into the business will be treated as capital.
So, In this transaction, as shown in the above image owner investing her cash into the business, this will be treated as capital of the business. The business receiving an i.e. cash, bank, stock, machine and Furniture.
#### Transaction No. 2
In this transaction, as shown in the above image four accounts are involved i.e. Stock(Sale), Cash, Debtors and Capital(Profit):
• Stock a/c (Sale):- Because business giving(selling) its goods.
• Debtors: – Because half of the payment is yet not received. It will be our new asset.
• Capital(Profit): Because owner has right on all profit of the business so the amount of the profit will be added in the capital a/c. (Profit = sale price – cost price) 22,000-20,000 = 2,000(Profit)
#### Transaction No. 3
In this transaction, as shown in the above image two accounts are involved i.e. Machine and Capital:
• Machine a/c :- Amount of Depreciation deducted from the Machine account, Because book value of machine of reduced.
• Capital a/c : – Amount of Depreciation deducted from the capital account, Because all expenses and losses are wear by the owner.
#### Transaction No. 4
In this transaction, as shown in the above image two accounts are involved i.e. one is cash and another is capital.
• Cash a/c: – payment is made in cash.
• Capital a/c:- cash withdrawal by the owner
#### Transaction No. 5
In this transaction, as shown in the above image one accounts is involved for two time i.e. capital.
• Capital a/c:- subtracted from capital because owner have to pay this amount to business.
#### Transaction No. 6
In this transaction, as shown in the above image one accounts is involved for two time i.e. capital. | 1,168 | 4,110 | {"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} | 3.046875 | 3 | CC-MAIN-2021-04 | latest | en | 0.82229 |
https://scicomp.stackexchange.com/questions/tagged/diffusion?sort=votes | 1,571,732,625,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987813307.73/warc/CC-MAIN-20191022081307-20191022104807-00366.warc.gz | 671,836,364 | 40,249 | The Stack Overflow podcast is back! Listen to an interview with our new CEO.
# Questions tagged [diffusion]
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### Is Crank-Nicolson a stable discretization scheme for Reaction-Diffusion-Advection (convection) equation?
I am not very familiar with the common discretization schemes for PDEs. I know that Crank-Nicolson is popular scheme for discretizing the diffusion equation. Is also a good choice for the advection ...
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### Computing geodesic distances with diffusion
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### Optimal way to find stationary solutions of the PDE
I am researching heat diffusion in an optical element irradiated by laser. This problem is described by the PDE which I wrote down in this question. I am using an implicit numerical scheme to model ...
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I am trying to implement a simple FEM approach, using p1 triangular elements, for solving the diffusion equation with variable nodal diffusivities and I was wondering how to incorporate the variable ...
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### Numerical solution to N-dimensional diffusion on simplex?
Assume I have a system of at least (but generally only) $N+1$ points in an $N$-dimensional space ($N > 3$ is possible). At each of these points $x_i, i=1,...,N+1$ I know an initial potential/...
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### Harmonic average of Diffusion Tensors in Finite Volume Method
I want to implement a Bilinear Finite Volume discretisation of the anisotropic diffusion problem: $$\frac{du}{dt} = \nabla \cdot (\textbf{D} \nabla u)$$ Both my degrees of freedom as well as the ...
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### 1 D Diffusion equation FDM with different layers
I'm trying to solve this particular equation $\frac{\partial u}{\partial t} = \frac{\partial}{\partial x} \big[D_{i}(x)\frac{\partial u}{\partial x} \big] + S(x,t)$ where the $i$ index denotes ...
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### Reaction-diffusion equations
I'm simulating a biological phenomena with reaction diffusion equations. There are multiple diffusing materials and there are some complex relations about consumption and production of such materials. ...
The equation at the left of the interface is $$\displaystyle\frac{\partial C_i}{\partial t} = D_i \nabla^2 C_i - z_i \frac{D_i}{RT}F \nabla \cdot (C_i \nabla \phi_2)$$ ... | 695 | 2,912 | {"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": 2, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-43 | latest | en | 0.877308 |
https://www.jiskha.com/questions/1449261/suppose-that-is-an-angle-in-standard-position-whose-terminal-side-intersects-the-unit | 1,610,953,407,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514423.60/warc/CC-MAIN-20210118061434-20210118091434-00057.warc.gz | 840,912,683 | 4,975 | # precalculus
Suppose that
θ
is an angle in standard position whose terminal side intersects the unit circle at
, 35/37−12/37
.
1. 👍
2. 👎
3. 👁
1. ok.
It appears to be in QIV, and is a 12-35-37 right triangle.
If you draw it correctly, you should be able just to read off the values of the trig functions of that angle.
1. 👍
2. 👎
2. ok, I am supposing Ø to be as you stated, now what?
did you notice that 35^2 + (-12)^2 = 37^2 ?
Do you know the basic trig ratios in terms of the x, y, and r of the corresponding right-angled triangle?
1. 👍
2. 👎
3. no that why I ask
1. 👍
2. 👎
4. what will sin , cot, csc be for that >
1. 👍
2. 👎
5. sketch a right-angled triangle.
label the horizontal base as x
the vertical side as y
the hypotenuse as r, and the base angle as Ø
memorize these ratios:
sinØ = y/r
cosØ = x/r
tanØ = y/x
and now their reciprocals:
cscØ = 1/sinØ = r/y
secØ = 1/cosØ = r/x
cotØ = 1/tanØ = x/y
You will not be able to handle trig questions without knowing these.
1. 👍
2. 👎
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The terminal side of an angle theta in standard position coincides with the line 2x – y = 0 in Quadrant III. Find cos theta to the nearest ten-thousandth. | 882 | 3,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-04 | latest | en | 0.859746 |
https://riddles360.com/riddle/new-year-party-riddle?submitAnswer=1 | 1,718,448,872,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861586.40/warc/CC-MAIN-20240615093342-20240615123342-00868.warc.gz | 447,394,322 | 7,893 | # New Year Party Riddle
Joseph organises an exotic office New year party at the beach to his important fellow employees.
Following are the important people that will come to the party
1. Joseph
2. Joseph 2 girlfriends- Carol Vanstone and Tracey Hughes
3. The sales head Clay Vanstone.
4. Clay Vanstone 2 associates.
5. Mr. Jeremy and his Secret Tab.
They all need to cross a small river to reach the beach, however, Joseph has got just one boat.
There are few rules for crossing the river.
1. The capacity of the boat is limited to just two people.
2. Clay Vanstone associates will not stay with Joseph without Clay Vanstone.
3. Carol Vanstone and Tracey Hughes will not stay with the Clay Vanstone without Joseph.
4. Mr. Jeremy would never leave the Secret Tab alone.
5. Secret Tab counts as a person.
6. Only Joseph, Clay Vanstone or Mr. Jeremy can drive the boat.
7. The two girlfriends Carol Vanstone and Tracey Hughes will not stay with the Clay Vanstone without Joseph.
8. No Associates would stay with Joseph without their Clay Vanstone.
9. Mr.Jeremy will not leave the Secret Tab alone.
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### Amazing Facts
###### Artificial Intelligence
Artificial Intelligence has crushed all human records in the puzzle game “2048,” achieving a high score of 839,732 and beating the game in only 973 moves without using any undo. | 1,257 | 5,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-26 | latest | en | 0.944606 |
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## comparing cell values to determine which is highest, then returning the name assocaited with it
Occasional Contributor
# comparing cell values to determine which is highest, then returning the name assocaited with it
Okay, so I've exhausted everything I can find and I'm still nowhere on this. The problem is this:
Part 1:
The formula to look at both A2 and B2 to determine which is highest, then in C2 output the 'header' for the winner. In the example above, "Yellow" for C2, "Blue" for C3 etc. To add an additional twist, the minimum 'winner' would need to be above the number 1000. If below then return something like, "no winner"
I also require the opposite, which I guess would be simple to work out once the above is figured out, whereby it would display to the biggest loser, but this time with a maximum return value of 1000
Part 2
Now to add a twist of lemon to the cocktail.
Same as above, however, this time it takes a mean average of all entries in a row with a "1" associated with it to determine which is again the 'winner/loser'. So D2 would simply be the winner between B2 and C2, but D4 would be the winner between the average of B2 and B4, and C2 and C4.
--
I have looked high and low, and can't find anything on it. It may be due to not knowing the parlance which describes the issue properly, but the closest I can find is some kind of vlookup or indexing, which doesn't seem to get me there
Thank you in advance to anyone that can help; its driving me up the wall
17 Replies
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
Both scenarios of part 1 are done and attached in sample file for your reference. Please let me know if it works as desired .
Please elaborate more the second part of your requirement, for example what criteria you need for D3, D5 or D6.
Thanks
Tauqeer
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
For such sample
for the first case formula could be
=IF(MAX(A2:B2)<1000, "no winner", INDEX(\$A\$1:\$B\$1,(B2>A2)+1))
for the second one
=IF( MAX(F2:G2)<MAX(AGGREGATE(14,6,1/(\$E\$2:\$E\$6=\$E2)*\$G\$2:\$G\$6,1),AGGREGATE(14,6,1/(\$E\$2:\$E\$6=\$E2)*\$F\$2:\$F\$6,1))+(MAX(F2:G2)<1000), "no winner", INDEX(F\$1:G\$1,(AGGREGATE(14,6,1/(\$E\$2:\$E\$6=\$E2)*\$G\$2:\$G\$6,1)>AGGREGATE(14,6,1/(\$E\$2:\$E\$6=\$E2)*\$F\$2:\$F\$6,1))+1))
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
Hi
Please take a look at attached file which hopefully covers both cases (part 1 and part 2).
File
Let me know if this solved the issue or if I misunderstood your requirement
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
Hey, Thanks for taking the time to look at this, for me.
I'm not sure your solution is correct. I changed the blue to a lower value than the yellow, and it returned no winner each time, instead of yellow
As for part 2, it is essentially the same problem, however it is looking at an aggregate of numbers, rather than just one number to compare. To reword it, instead of 1,2 lets call them Q1, Q2, Q3, Q4 of a year for example. I want to look at what the average 'Blue' number was in say 'Q1' going back in time, and compare that against its 'Yellow' counterpart to determine which is higher/lower.
So in the image above, I've highlighted Q1 for the sake of ease. so in cell D1, it would just be the winner between B2 and C2, which is Blue (and loser in the opposite condition). when we get to cell D6, it would then be the mean average of B2 and B6 vs the mean average of C2 and C6. Then in cell D10, it would be the average of B2, B6 and B10, vs C2, C6, C10 ad nauseam.
Do let me know if that makes more sense, and thanks again for your time on this!
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
Thank you for looking at this problem for me.
In your solution for part 2, there are two instances of no winners, when there is:
H2 would show yellow as the winner as being higher than blue 1183 vs 1001, and the same for H3
Do let me know if I need to expound on my explanation of the problem, and thanks again
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
In general, better if you attach your sample file to the post together with screenshorts - you'll save time for the people who are answering on generating new test file copying your data from the screen into Excel sheet.
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
When I misunderstood your logic. In second case I checked all pairs marked as 1 (or 2, etc.), found among them highest result and for this pair return the winner. All other "1" pairs are not winners.
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
Still not sure I understood the issue, does attached file do the job?
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
Hey, and thanks for helping me on this problem.
I think its almost correct, however, every other cell under E is empty?
Noted, Thanks
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
i thought part 2 was only for rows which are marked with "1". But then it's the same for the rows with "2". Check out the new file and see if it works
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
As far as the first part is concerned it is showing 'no winner' in all cases because of the below condition that you mentioned in in your first post,
'the minimum 'winner' would need to be above the number 1000. If below then return something like, "no winner"
If you simply input 1001, it will show Yellow as winner
Thanks
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
Couple observations. I changed the first set to below 1000 each, and yet yellow is showing as the winner still in part 2 rather than no winner. Also, how would the formula look to show the opposite, ie the lowest?
In application, I will be looking to put this across maybe 100 different inputs (blue, yellow, etc) and I'm essentially looking for the highest colour from 1000 and the lowest colour from 1000.
Highlighted
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
Ah, I think there has been a misunderstanding. The winning entry so to speak must be above the 1000 mark to be the winner, but what its comparing against could very well be below 1000
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
OK the 1000 rules also needs to be in part 2, I didn't know. Find attached the new version.
If this is anything close to the result you would like to have maybe you can try to do the finetuning of the formulare yourself
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
Hi,
Its still showing no winner if yellow is below 1000. I've looked at your formula, and to be honest, I don't know where to start to adjust it!
# Re: comparing cell values to determine which is highest, then returning the name assocaited with it
Hi @kemble999
To enhance the flavor of your cocktail, I added a red color in the attached file. The formula in D3 is:
=IF(MAX(A3:C3)>D\$1,
LOOKUP(2,1/(FREQUENCY(0,1/A3:C3)),A\$2:C\$2),
"None")
Conversely, the array formula (entered with Ctrl+Shift+Enter) in J3 is:
=IF(MAX(MMULT(TRANSPOSE(--(F\$3:F3=F3)),G\$3:I3))>J\$1,
LOOKUP(2,1/(FREQUENCY(0,1/MMULT(TRANSPOSE(--(F\$3:F3=F3)),G\$3:I3))),G\$2:I\$2),
"None")
Note that the values in K7:M14 were presented only for verification purposes. Thus, they are not referred to in the foregoing formula.
Cheers,
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3 Replies | 2,222 | 8,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-51 | latest | en | 0.950862 |
https://mathoverflow.net/questions/146987/a-question-about-representations-of-finite-groups | 1,611,038,169,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517966.39/warc/CC-MAIN-20210119042046-20210119072046-00252.warc.gz | 441,906,202 | 26,662 | A question about representations of finite groups
Let $G$ be a finite group and let $V$ be an irreducible complex representation of $G$. Does there exist an element $g \in G$ which acts on $V$ with distinct eigenvalues? If true, can you provide a proof/reference, and if false, a counterexample?
If $g\in G$ has order $n$, then its eigenvalues are all $n$th roots of unity. So if $V$ is a representation with degree greater than $n$, the eigenvalues of $g$ on $V$ can't all be distinct.
There are plenty of examples of groups with irreducible representations whose degree is greater than the largest order of any element of the group, and so these will certainly give counterexamples. For example, the symmetric group $S_6$ has irreducible representations of degree 9,10 and 16. | 190 | 780 | {"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} | 2.703125 | 3 | CC-MAIN-2021-04 | latest | en | 0.881742 |
https://famousparenting.com/4oz-is-how-many-tablespoons-understanding-the-conversion-rate-of-fluid-ounces-to-tablespoons/ | 1,723,185,342,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640759711.49/warc/CC-MAIN-20240809044241-20240809074241-00511.warc.gz | 187,643,518 | 39,945 | ### Tips and tricks from celebrity parents (without the drama)
When it comes to cooking and baking, precision is essential, and measuring ingredients accurately is the key to creating delicious dishes. Kitchen conversions can be confusing, particularly when converting volume measurements. Home cooks’ frequent question is how many tablespoons are in an ounce. Specifically, many wonder 4oz is how many tablespoons.
In cooking and baking, the volume measurement of ounces and tablespoons is frequently used interchangeably. Ounces are a weight measurement, while tablespoons are a volume measurement. However, converting tablespoons to ounces is simple, and it is essential to know the conversion accurately. Specifically, one fluid ounce is equivalent to two tablespoons, so it follows that 4 fl. oz. is equivalent to eight tablespoons.
If you want more interesting content check out our next post!
## 4Oz is How Many Tablespoons
One of the most common questions in cooking and measuring ingredients is the conversion rate between fluid ounces and tablespoons. Specifically, a common query is how many tablespoons are in 4 fluid ounces. This conversion rate is essential to accurately measure the ingredients needed for your favorite recipes.
The conversion rate between fluid ounces and tablespoons is not always straightforward, as both units measure volume but differ in their measurement standards. One fluid ounce equals 1/8 cup, 2 tablespoons, or 29.5735 milliliters. Meanwhile, one tablespoon is equivalent to 1/16 cup, 0.5 fluid ounces, or 14.7868 milliliters.
To convert 4 fluid ounces to tablespoons, we need to use the conversion rate between these two units. By dividing 4 by 0.5 (the number of fluid ounces in a tablespoon), we get the result of 8 tablespoons. Therefore, 4 fluid ounces are equivalent to 8 tablespoons.
It’s worth noting that the conversion rate between fluid ounces and tablespoons varies, depending on the country or region. In the US, one fluid ounce is equivalent to 2 tablespoons, while in the UK, one fluid ounce is equivalent to 1.6 tablespoons. Therefore, it’s essential to understand the measurement system used in your location to avoid making measurement errors.
In summary, the conversion rate between fluid ounces and tablespoons is fundamental to ensure that you measure your ingredients accurately. If you wonder how many tablespoons there are in 4 fluid ounces, the answer is 8 tablespoons. Always remember to know the conversion rate in your location or region to have precise measurements for your recipes.
## Practical Examples of Converting 4 oz to Tablespoons
Converting ounces to tablespoons can be confusing, especially when getting exact recipe measurements. One of the most common conversions is 4 oz to tablespoons. Here are some practical examples of how to convert 4 oz to tablespoons.
### Example 1: Butter
Butter is the most commonly used ingredient in recipes, usually measured in tablespoons or fractions of tablespoons. If a recipe calls for 4 oz of butter, you would need to convert that to tablespoons. One tablespoon of butter is equal to 0.5 oz. Therefore, 4 oz of butter is equal to 8 tablespoons. If the recipe calls for 4 oz of butter, you would need 8 tablespoons of butter.
### Example 2: Honey
Honey is another common ingredient in recipes, and it’s usually measured in tablespoons or fractions of tablespoons. If a recipe calls for 4 oz of honey, you would need to convert that to tablespoons. One tablespoon of honey is equal to 0.67 oz. Therefore, 4 oz of honey is equal to 6 tablespoons. If the recipe calls for 4 oz of honey, you would need 6 tablespoons of honey.
### Example 3: Flour
Flour is measured in cups, tablespoons, or fractions of tablespoons. If a recipe calls for 4 oz of flour, you would need to convert that to tablespoons. One tablespoon of all-purpose flour is equal to 0.25 oz. Therefore, 4 oz of flour is equal to 16 tablespoons. If the recipe calls for 4 oz of flour, you would need 16 tablespoons of flour.
### Example 4: Sugar
Sugar is measured in cups, tablespoons, or fractions of tablespoons. If a recipe calls for 4 oz of sugar, you would need to convert that to tablespoons. One tablespoon of granulated sugar is equal to 0.5 oz. Therefore, 4 oz of sugar is equal to 8 tablespoons. If the recipe calls for 4 oz of sugar, you need 8 tablespoons.
In conclusion, converting 4 oz to tablespoons is a simple task, and it only requires a basic knowledge of the conversion factor for each ingredient. Following these practical examples, you can easily convert 4 oz to tablespoons and get the exact measurements you need for your recipes.
## Conclusion
Now that you know 4oz is how many tablespoons, you are equipped with the knowledge to make precise measurements in the kitchen. Whether you are cooking a gourmet meal or simply following a recipe, accurately measuring ingredients is essential to the success of your dish.
Using a tablespoon allows you to measure ingredients with ease, making the process faster, simpler, and more precise. Tablespoons are perfect for measuring liquids, such as milk or cream, and solids, such as sugar or flour, in small quantities.
Remember that when measuring ingredients, every tablespoon can make a significant difference in the outcome of your dish. Therefore, it is crucial to measure accurately using standardized measuring spoons and cups.
Now that you understand the importance of measuring ingredients accurately, including how 4oz is how many tablespoons, you can take your culinary skills to the next level. So, grab your measuring spoons and start creating delicious meals in the kitchen! | 1,173 | 5,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.933728 |
https://networkingfunda.com/mindware-critical-thinking-for-the-information-age-quiz-answers/ | 1,720,863,789,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514493.69/warc/CC-MAIN-20240713083241-20240713113241-00470.warc.gz | 348,772,454 | 39,924 | # Mindware: Critical Thinking for the Information Age Quiz Answers
## Get All Weeks Mindware: Critical Thinking for the Information Age Quiz Answers
### Week 1: Mindware: Critical Thinking for the Information Age Quiz Answers
#### Quiz 1: Lesson 1 Quiz
Q1. From this list, choose the example that is a variable:
View
1.The direction that the Earth rotates around the Sun
2.The distance from Los Angeles to El Paso
3.The number of hours of daylight in a day
Q2. From this list, choose the example that is a constant
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The number of whole tones (whole steps) in a major scale
Q3. What percent of cases are within -1 and +1 standard deviation?
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68%
Q4. What percent of cases are between -3 and +1 standard deviation?
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96%
Q5. Phoebe is at the 60th percentile in her math class and wants to hire a tutor who can help her perform in at least the 75th percentile. Tutor A helps students perform at .5 standard deviations above the class mean and charges \$10 per hour. Tutor B helps students perform at 1 standard deviation above the mean and charges \$12 per hour. Tutor C helps students perform at 2 standard deviations above the mean and charges \$15 per hour.
Which tutor helps Phoebe meet her goal without spending unnecessary money?
View
Tutor B
Q6. An online instructor is providing a digital badge as an incentive to learners who score in the 85th percentile or higher on a module exam. The exam scores have a normal distribution, with a mean of 70 and a standard deviation of 8. What minimum score out of the options below would learners need to receive the digital badge?
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87
Q7. Ohio State’s introductory calculus class earns a mean score of 70 with a standard deviation of 12. Michigan’s identical introductory calculus class earns a mean score of 70 with a standard deviation of 8. Both schools test program MathX. At Ohio State, the test group’s mean performance rises to 76. At Michigan, it rises to 74.
Which of the following interpretations is valid?
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The effect of MathX training on the student’s scores is the same at both schools.
Q8. This (hypothetical) scatter shows data for MOOC X students. Which of the following statements correctly describes the scatterplot shown above?
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Most students who access fewer chapters have lower adjusted grades than most other students, and students who access more chapters have higher adjusted grades than most other students.
Q9. You can have reliability without validity.
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False
#### Quiz 2: Pre-lecture Reflection Prompt
Q1. Consider the scenario below and answer the follow-up question using what you already intuitively know about statistics. You will revisit this question at the end of this lesson, so keep a record of your response so that you can compare your response here to the follow-up question at the end of this lesson.
You have two acquaintances at work: JB and KL. At the office party on Saturday, JB was friendlier than KL. What’s the likelihood that JB will be friendlier than KL at the planning meeting on Monday? Give your reasoning to support your answer.
Now, imagine that over the last 20 times you dealt with JB and KL, JB was friendlier on average than KL. What do you suppose the likelihood would be of JB being friendlier than KL on average over the next 20 times you deal with them? Is the likelihood greater or less than the likelihood you reported for the above question? Why?
View
1.There is no evidence to suggest that JB is more likely to be friendlier than KL at the planning meeting than he was at the office party. The fact that JB was friendlier at the office party does not mean that he will be friendlier at the planning meeting. It is just as likely that KL will be friendlier at the planning meeting.
2.If over the last 20 times I dealt with JB and KL, JB was friendlier on average than KL, then the likelihood of JB being friendlier than KL on average over the next 20 times I deal with them is greater than 50/50.
3.The fact that JB was friendlier on average over the last 20 times suggests that he is more likely to be friendlier than KL in the future. However, it is still possible that KL will be friendlier at some of the future interactions. The likelihood of JB being friendlier than KL on average over the next 20 times will depend on a number of factors, including the specific interactions that take place.
4.Here are some of the factors that could affect the likelihood of JB being friendlier than KL:
The topics that are discussed at the meetings
The personalities of the people involved
The circumstances of the meetings
If the meetings are focused on topics that JB is interested in, or if the people involved are more likely to get along with JB, then the likelihood of JB being friendlier than KL will be higher. Similarly, if the meetings are stressful or difficult, then the likelihood of JB being friendlier than KL will be lower.
Ultimately, the likelihood of JB being friendlier than KL at the planning meeting on Monday is impossible to say with certainty. However, the fact that he was friendlier on average over the last 20 times suggests that it is more likely than not that he will be friendlier at the next meeting.
#### Quiz 3: Pre-lecture Quiz
Q1. For each of the hypothetical research questions below, select the items where a random sample of 100 would be much more informative than a random sample of 10 (in contrast to only slightly more informative) if your research objective is to determine:
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The average height of males in the US
#### Quiz 4: Lesson 2 Quiz
Q1. Sample values resemble population values as a _ of their size. The _ the sample, the less likely it is you will get an unrepresentative value.
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Portion, smaller
Q2. We tend to go from an observation to a ___
Hint: It is a one-word answer that ends in “…tion”
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Induction
Q3. Of the following examples, what would be a situation where the Law of Large numbers would help you make better decisions or assumptions:
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The number of complaints that a customer service office receives in a day
Q4. If we are trying to infer something about a population that has a lot of variability, the smaller the sample, the more likely it is to get a result that doesn’t represent the population distribution.
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True
Q5. If we are trying to infer something about a population that has a lot of variability, small samples representing a population can never be as accurate as larger samples.
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False
Q6. (Check all the answers that apply.)
For personality traits like friendliness, people:
View
1.Are too quick to generalize
2.Assume too little variability
#### Quiz 5: Post-lecture Reflection Prompt
Q1. Using what you know about the law of large numbers, re-evaluate the scenarios that you considered prior to watching the video.
The last time you visited your boss in his office, he was extremely irritable. What do you suppose the likelihood would be of him being irritable the next time you go to his office? Give your reasoning to support your answer.
Now imagine that the last twenty times you visited your boss, he was extremely irritable. What do you suppose the likelihood would be of him being irritable the next twenty times you go to his office and why?
How has your understanding changed since learning about the law of large numbers?
View
1.The last time you visited your boss in his office, he was extremely irritable. What do you suppose the likelihood would be of him being irritable the next time you go to his office?
2.Now imagine that the last twenty times you visited you boss, he was extremely irritable. What do you suppose the likelihood would be of him being irritable the next twenty times you go to his office and why?
3.How has your understanding changed since learning about the law of large numbers?
### Week 2: Mindware: Critical Thinking for the Information Age Quiz Answers
#### Quiz 1:Lesson 3 Quiz
Q1. A group of researchers are looking to see if there is a correlation between having X trait and having allergies. Given the table below, what ratios must be compared to determine the presence of an association between X trait and having allergies?
View
10/60 and 20/25
Q2. What term accounts for the fact that when we are prepared to see an association we are more likely to see it?
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Illusory correlation
Q3. A variable that is associated with both variables of interest and which could explain the association between them is called a(n) __ variable.
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Confounding
Q4. What is the term defining the probability that a result at least as extreme as the one obtained could have occurred given that there is in fact no relationship?
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Statistical significance
Q5. Explain the statistical significance of a finding when p<.003
The probability that the result could have been obtained even if there is actually no relationship is less than _ in _.
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3; 1000
#### Quiz 2: Lesson 4 Quiz
Q1. Correlation is just as good as an experiment at determining causality.
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False
Q2. The treatment in an experiment is considered the _. The thing that is measured in an experiment is considered the _.
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independent variable; dependent variable
Q3. Shortly after the new CEO took over Company Z, its stocks began to fall. John blames the change in stock value on poor management by the CEO. Why does John not have enough information to draw this conclusion (besides having a small sample size of one company)?
What do you think? Your answer cannot be more than 10000 characters.
View
Here are some reasons why John does not have enough information to draw the conclusion that the change in stock value is due to poor management by the CEO:
1.There could be other factors that caused the change in stock value, such as changes in the economy or in the industry.
2.The change in stock value may not be permanent. It is possible that the stock price will go back up in the future.
3.John only has data for one company. He needs to look at data from a larger number of companies to make a more reliable conclusion.
Q4. Many control variables in multiple regression analysis have very high validity but very low reliability.
View
False
Q5. The method referred to as the Gold Standard Experiment is stronger than multiple regression analysis.
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True
Q6. Benny wants to know if playing classical music or country music is better for memory, so to test both treatments, he flips a coin to determine which type of music he will listen to first and then perform the following tests:
1. On Monday he plays classical music while trying to memorize a list of ten words.
2. On Tuesday he tests himself and records his results.
3. On Wednesday he plays country music while trying to memorize a new list of ten words.
4. On Thursday he tests himself on the new words and records his results.
What type of testing is Benny implementing?
View
In a within-subjects design, the same participants are exposed to all of the treatments. This is in contrast to a between-subjects design, in which different participants are exposed to different treatments.
Q7. Referring to the question above, does Benny’s experiment have a “within design” or “between design”?
View
Within design
Q8. In “within designs,” the only thing that differs across treatments is:
View
The treatments
Q9. In “between designs,” the treatments differ, but so do:
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The participants
Q10. The next few questions relate to the following scenario:
Kaiping has a hypothesis that eating dark chocolate before an exam will help students test better. She has the 26 students in Ms. Claude’s class eat dark chocolate before their calculus exam and the 32 students in Mr. Green’s class eat nothing before their calculus exam.
What is the independent variable?
What do you think? Your answer cannot be more than 10000 characters.
View
1.The independent variable is whether or not the students eat dark chocolate before the exam.
2.The dependent variable is the students’ performance on the calculus exam.
3.The experimental condition is the group of students who eat dark chocolate before the exam.
4.The control condition is the group of students who do not eat dark chocolate before the exam.
Q11. What is the dependent variable?
What do you think? Your answer cannot be more than 10000 characters.
View
The independent variable is the treatment that the participants receive. In this case, the treatment is whether or not the participants eat dark chocolate before the exam.
Q12. Which class is in the experimental condition?
What do you think? Your answer cannot be more than 10000 characters.
View
The dependent variable is the outcome that is being measured. In this case, the outcome is the participants’ performance on the calculus exam
Q13. Which class is in the control condition?
What do you think? Your answer cannot be more than 10000 characters.
View
The experimental condition is the group of participants who receive the treatment. In this case, the experimental condition is the group of students who eat dark chocolate before the exam.
### Week 3: Mindware: Critical Thinking for the Information Age Quiz Answers
#### Quiz 1: Lesson 5 Quiz
Q1. Your mom tells you that she started eating dandelions and her stomach pains went away. Tell her why the dandelions may not be responsible for her decrease in stomach pain.
What do you think? Your answer cannot be more than 10000 characters.
View
Dandelions may not be responsible for the decrease in stomach pain for several reasons. Stomach pains can have various causes, and their intensity can fluctuate naturally. Without controlled scientific studies, it’s challenging to attribute the relief solely to dandelion consumption. Factors such as stress, dietary changes, or other coincidental factors could have played a role. It’s crucial for your mom to consult a healthcare professional to determine the actual cause of her stomach pain and explore appropriate treatments.
Q2. This shows the normal distribution of the test scores of students in Mr. Lin’s class. Mr. Lin’s top student Berta scored a 90. What do you predict her best friend Nadia scored if the correlation between friends’ test scores is .00?
View
50
Scenario 1: Your apartment building is full of rambunctious children. 70% of the children are from the second floor and 30% of the children are from the third floor. You wake up to find the newspaper outside of your door has been stolen. Your neighbor says she saw a child from the third floor running down the hallway.
Scenario 2: Your apartment building only allows dogs and cats in the building. 70% of the pets are dogs and 30% of the pets are cats. You wake up to find that the newspaper outside of your door has been torn to shreds. You know that dogs are known to cause four times more disturbances regarding newspapers than cats. Your neighbor says she saw a cat running down the hallway.
Which scenario do you think allows for more accurate estimates of the probability regarding who or what was responsible for the disturbance with your newspaper? Why?
View
Scenario 2
Q4. You just tested positive for a rare disease and want to know what the chances are of you actually having the disease. You know that out of 100 people, 10 people have the disease. 6 of the people who have the disease will test positive (60% of the people with the disease). 9 of the people who do not have the disease will incorrectly test positive (10% of the people who don’t have the disease). What is the probability that you actually have the disease, given that you tested positive?
View
6 / (6 + 9)
#### Quiz 2: Pre-lecture Quiz
Q1. What do you need to know to find out if there’s a correlation between vaccination and autism?
To determine if there is a correlation between vaccination and autism, you would need to conduct rigorous scientific research and consider various factors. Here are the key elements to consider:
1. Study Design: The study should be designed appropriately to investigate the potential correlation. It should be a well-controlled, longitudinal study with a sufficiently large sample size. Randomized controlled trials or cohort studies are often used.
2. Data Collection: Collect data on vaccination history and autism diagnoses for the study participants. Ensure accurate and complete records of vaccinations, including the type of vaccine, date of administration, and any adverse events.
3. Control Group: Establish a control group of individuals who are similar to the vaccinated group in all aspects except for vaccination. This group can help determine whether the observed correlation is specific to vaccination.
4. Data Analysis: Use statistical analysis methods to assess the correlation. Common statistical tests include chi-square tests, regression analysis, or more advanced methods like propensity score matching.
5. Confounding Variables: Consider and control for potential confounding variables, such as genetic predisposition, family history, and other environmental factors that could influence autism risk. Statistical techniques like multivariate analysis can help control for these variables.
#### Quiz 3: Lesson 6 Quiz
Q1. What term describes the belief that we understand the world by direct perception?
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Illusion of objectivity
Q2. Our perceptual system is highly accurate for an artificial world but not for the real world.
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False
Q3. What term describes cognitive structures that guide our understanding of the world?
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Schemas
Q4. Where would people be more likely to vote for increased welfare?
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A polling station next to a soup kitchen
Q5. What term describes mental shortcuts or rules of thumb that help us to understand the world but that can sometimes also lead us astray?
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Heuristics
Q6. When people estimate the frequency or probability of events using the cue of how easily the type of event comes to mind, they are using what heuristic?
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Availability heuristic
Q7. The best predictor of future behavior is past behavior.
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True
Q8. What term describes the tendency to mistakenly regard a disposition of the object or person as the primary cause of behavior, while ignoring important situational or contextual factors?
View
Q9. What term refers to the fact that we often look only for evidence that could confirm the hypothesis and not for evidence that might disconfirm it?
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Confirmation bias
Q10. What do you need to know to find out if there’s an association between getting up early and being in a good mood for most of the day?
View
1.Define what you mean by “getting up early” and “being in a good mood.”
2.Collect data from a representative sample of individuals, including their wake-up times and their reported mood throughout the day.
3.Analyze the data statistically to identify any correlations or associations between waking up early and having a good mood.
4.Consider potential confounding variables, such as sleep quality, daily routines, and overall health, that could influence the results.
5.Conduct the study over a significant period to account for daily variations.
What do you think? Your answer cannot be more than 10000 characters.
Q11. What do you need to know to find out if there’s an association between doing drugs and dropping out of high school?
View
1.Define what you mean by “doing drugs” (e.g., types of drugs, frequency of use) and “dropping out of high school.”
2.Collect data from a representative sample of high school students, including information on their drug use and educational outcomes.
3.Analyze the data statistically to identify any correlations or associations between drug use and dropping out of high school.
4.Consider potential confounding variables, such as socioeconomic status, family environment, and academic performance, that could influence the results.
5.Conduct the study longitudinally to track changes over time and determine if drug use precedes dropping out or vice versa.
What do you think? Your answer cannot be more than 10000 characters.
### Week 4: Mindware: Critical Thinking for the Information Age Quiz Answers
#### Quiz 1: Pre-lecture Activity
Q1. You bought a \$100 ticket to a basketball game when the league standings seemed likely to hang on this game and the star was going to be playing. Tonight’s the night, but the star is not playing, nothing is actually going to hang on the outcome, and the stadium is 40 minutes away and it’s starting to snow.
What’s the best reason to go to the game?
What’s the best reason not to go?
What do you think? Your answer cannot be more than 10000 characters.
View
1.Best Reason to Go to the Game: The best reason to go to the game in this situation would be to enjoy the live experience of attending a basketball game, even if the star player is not playing and the outcome doesn’t significantly impact the standings. You can still appreciate the atmosphere, support the team, and spend quality time with friends or fellow fans. It’s an opportunity for entertainment and camaraderie.
2.Best Reason Not to Go: The best reason not to go to the game could be due to safety concerns and inconvenience. If the stadium is 40 minutes away and it’s starting to snow, driving in adverse weather conditions can be dangerous. Additionally, considering that the star player is not playing and the game doesn’t hold high stakes, you might decide that the effort and cost of attending do not justify the experience, especially if you have safety concerns.
Q2. You’ve ordered an expensive meal that turns out to be not good. Should you finish it?
What’s the best reason to do so?
What’s the best reason not to do so?
What do you think? Your answer cannot be more than 10000 characters.
View
1.Best Reason to Finish the Meal: The best reason to finish the meal might be to avoid wasting food and money. You’ve already paid for the expensive meal, and by finishing it, you ensure that the resources used to prepare the meal are not wasted. Additionally, if you’re dining in a restaurant, it’s a way to avoid additional expenses on ordering another dish. If the meal is not terrible but just not to your taste, you could still eat it to avoid food wastage.
2.Best Reason Not to Finish the Meal: The best reason not to finish the meal would be if it’s genuinely not good, and continuing to eat it would lead to discomfort or dissatisfaction. If the quality of the food is subpar or if it’s causing you to feel unwell, it’s better to stop eating for your own well-being. Additionally, if you’re in a situation where you have other options or can request a replacement dish, it might be more enjoyable to choose an alternative that suits your preferences. In such cases, quality and enjoyment should take precedence over avoiding food waste.
#### Quiz 2: Lesson 7 Quiz
Q1. Cognitive dissonance occurs when our beliefs don’t fit with our __.
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actions
Q2. It is always best to optimize when making choices.
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False
Q3. Using an example from your own life, perform your own weighted decision. (You may use the templates provided below.)
View
(.docx)
What do you think? Your answer cannot be more than 10000 characters.
View
if you have a specific decision or scenario in mind that you’d like assistance with, please provide the details, and I’ll be happy to offer guidance on how to approach the decision-making process.
Q4. According to the Sunk Cost Principle, __ benefits and costs should figure in your choices
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only future
Q5. If you do not choose the action with the greatest net benefit, you will pay unnecessary _.
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Opportunity costs
Q6. What term accounts for when people demand more money for something than they originally paid for it?
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Endowment effect
Q7. You recently started up your own clothing website. Customers must create an account in order to view your clothing options, and one of the questions they need to answer is whether or not they want weekly updates sent to their email addresses. In order to deliver information about your latest arrivals to the most customers, would it be better to ask customers:
“Please check the box below if you would like weekly information about our latest arrivals?” [Opt in]
or
“Please check the box below if you would not like weekly information about our latest arrivals?” [Opt out]
Why?
What do you think? Your answer cannot be more than 10000 characters.
View
This choice is based on the principle of default bias or status quo bias. People tend to stick with the default option or the status quo, which in this case would be not receiving weekly updates. By framing the question as an opt-out option, you are more likely to reach a broader audience because people are more inclined to leave things as they are rather than actively opting in. This approach maximizes the number of customers receiving updates while respecting their preferences.
#### Quiz 3: Pre-lecture Quiz
Q1. Determine whether or not you think each of the following arguments is valid.
If I have the flu, then I have a sore throat. I have a sore throat. Therefore, I have the flu.
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Not valid
Q2. If the movie is not a comedy, then Joe did not see it. The movie is not a comedy. Therefore, Joe did not see it.
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Not valid
Q3. If President Obama is Muslim, then he’s not a Christian. President Obama is not Muslim. Therefore, President Obama is a Christian.
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Not valid
Q4. If it’s a holiday, the bank will be closed. The bank is closed. Therefore, it’s a holiday.
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Not valid
#### Quiz 4: Lesson 8 Quiz
Q1. All black cats are male. This cat is black. Therefore, this cat is a male.
This is an example of _.
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Syllogism
Q2. If I eat peanut butter, I will get hives. I ate peanut butter. Therefore, I will get hives.
This is an example of _.
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Propositional logic
Q3. If a conclusion is valid, then it is also true.
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False
Q4. Determine if the major (first) premise is sufficient or necessary and sufficient in order for the conclusion to be valid.
If I break my leg, I will wear a cast. I broke my leg. Therefore, I will wear a cast.
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Sufficient
Q5. You may recognize the next few arguments from before the lecture. They are not valid. Now determine if these arguments are not valid due to a converse or inverse error.
If I have the flu, then I have a sore throat. I have a sore throat. Therefore, I have the flu.
View
Converse
Q6. If President Obama is Muslim, then he’s not a Christian. President Obama is not Muslim. Therefore, President Obama is a Christian.
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Converse
Q7. If it’s a holiday, the bank will be closed. The bank is closed. Therefore, it’s a holiday.
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Inverse
Q8. The Law of Large Numbers, regression, statistical significance, etc., are all types of reasoning discussed in this course that are deductively valid.
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False
Q9. Which form of thinking would be more likely to support the statement: “Contradiction is constant since change is constant.”
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Principles of dialecticism
Q10. Which form of thinking would be more likely to support the statement: “Everything must either be or not be.”
View
Foundations of logic | 5,810 | 27,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.865828 |
http://www.chegg.com/homework-help/questions-and-answers/objects-masses-290-kg-590-kg-separated-0370-m-find-net-gravitational-force-exerted-objects-q1613965 | 1,475,315,484,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662698.85/warc/CC-MAIN-20160924173742-00188-ip-10-143-35-109.ec2.internal.warc.gz | 396,356,344 | 13,573 | Objects with masses of 290 kg and a 590 kg are separated by 0.370 m.
(a) Find the net gravitational force exerted by these objects on a 67.0 kg object placed midway between them.
magnitude-??
(b) At what position (other than infinitely remote ones) can the 67.0 kg object be placed so as to experience a net force of zero?
_____m from the 590 kg mass | 90 | 352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-40 | latest | en | 0.898693 |
https://en.khanacademy.org/math/arithmetic/x18ca194a:multiply-1-and-2-digit-numbers/x18ca194a:multiply-2-digit-numbers-with-area-models/v/area-model-for-multiplication | 1,719,119,174,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00291.warc.gz | 197,239,987 | 133,092 | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
### Course: Arithmetic>Unit 7
Lesson 5: Multiply 2-digit numbers with area models
# Multiplying with area model: 16 x 27
Sal uses an area model to multiply 16x27. Created by Sal Khan.
## Want to join the conversation?
• Is it permissible to use this area concept by subtracting a 'negative' area?
For example, using our problem of 16 x 27, we could actually make an area of 20 x 27. We can then subtract an area of 4 x 27 to yield our answer.
We can easily calculate 20 x 27 = 540, and then subtract a similarly easy calculation of 4x27=108 → 540-108=432.
• Good point, I think you are absolutely right. We can even set up a diagram like Sal has done, using negative numbers in one or both sides of the multiplication.
Eg. Using (30 - 3) for 27, and (20 - 4) for 16 in the diagram below
We get 600 - 120 - 60 + 12 = 432
`` 30 -3 .______________. | | | | | |20 | +600 | -60 | | | | |________|_____| | | |-4 | -120 | +12 | |________|_____|``
• 30 -3
.____________.
| | |
| | |
20 | +600 | -60 |
| | |
|______|_____|
| | |
-4 | -120 | +12 |
|______|_____| this is my area model khan sir
• This is an interesting way of using area models to multiply. This is most useful for multiplying numbers that have high digits.
• Can you explain me how to do the area model negative numbers if it is possible?
Also what do you mean when you say subtracting a 'negative' area?
• I don't know what it means by subtracting a negative area, but the question basically says to multiply 16 by 27 which is 432. This probably doesn't help but I hope it does.
• What would an aria model for 95 / 6 = ?
/ = divided by
• An area model is shown. Which expressions show how to multiply 4 1/4×2 3/5 using partial products? Label the model with the correct expressions.
• I did not know you got energy points for watching a video
• You can get 850 energy points if you don't skip any part of the video.
• I think the points are to hipe you up | 614 | 2,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-26 | latest | en | 0.890548 |
http://www.ask.com/web?qsrc=3053&o=102140&oo=102140&l=dir&gc=1&qo=popularsearches&ad=dirN&q=How+To+Label+An+Inch+Ruler | 1,448,454,199,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398445142.9/warc/CC-MAIN-20151124205405-00001-ip-10-71-132-137.ec2.internal.warc.gz | 294,258,819 | 18,181 | Web Results
How to Label an Inch Ruler
Inch rulers are typically marked with lines of varying length. These lines represent whole inches and parts thereof, namely the halves, quarters, eighths, and sixteenths of each inch. While only the whole inches are usually labeled on commercially... More »
Difficulty: Moderately Easy
Source: www.ehow.com
## How to Read a Ruler (and other simple tricks). - Instructables
While it may seem to be a very basic skill, being able to read a ruler is the ... The cool thing about knowing how to label the fractions inside of an inch is that you ...
## How to label a ruler - ThatQuiz
www.thatquiz.org/tq/previewtest?D/D/G/S/34551364219841
How to label a ruler ... 6 inches. 1 ½ inches. 9 ½ inches. 6 inches. 2 ½ inches. 9 ½ inches. 2 ½ inches. 7 ½ inches. 3 ½ inches. 4 ½ inches. 8 inches. 3 inches.
## Blank inch ruler.pdf
www.imesd.k12.or.us/sites/imesd.k12.or.us/files/Blank inch ruler.pdf
Page 1. Measurement Worksheet 7.1. This diagram represents one inch. Label the increments below. 1”
## How To Read A Ruler - Math A Tube
A ruler is use to measure units of length, inches, feet and yards. A standard ruler is 12 inches or 1 foot long. 3 standard rulers measure 1 yard or 36 inches.
## Using A Metric Ruler, Centimeters and Millimeters - Math A Tube
www.mathatube.com/measurement-using-ruler-metric.html
How to use a metric ruler to measure. A metric ruler is use to measure cm and mm. There are 10 ... There are 2.54 centimeters in an inch. A 12 inch ruler is about ...
## label: ruler | Check actual size - PiliApp
www.piliapp.com/actual-size/label/ruler
Online Ruler (cm/mm). View. Inch Ruler. View · 繁 | 简 | English | 日本語 | 한국어 | Español | Deutsch | Français | Italiano | Português | Русский | العربية ▾.
## Measuring to the Nearest Quarter-Inch - K ...
www.k-5mathteachingresources.com/support-files/measuring-to-the-nearest-quarter-inch.pdf
1. Cut out and label your ruler to show all quarter-inch measures. 2. Use your ruler to measure ten objects in the classroom to the nearest quarter-inch. 3. Sketch ...
Images of How to Label An Inch Ruler
Popular Q&A
Q: How to Label an Inch Ruler.
A: 1. Visually inspect the ruler to identify whether the ruler is divided into eighths or sixteenths. Rulers that have sixteenth of an inch markings will have line... Read More »
Source: www.ehow.com
Q: How to Read a Ruler in Inches.
A: Instructions. Examine the ruler. Notice that there are numbers along its length from 0 to 12. This indicates that the ruler has 12 inches, which is equivalent t... Read More »
Source: www.ehow.com
Q: How many inches is a ruler?
A: 12 inches Read More » | 727 | 2,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2015-48 | longest | en | 0.840549 |
https://www.akritinfo.com/class-12-physics-icse-syllabus-2024-25/ | 1,723,226,973,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640768597.52/warc/CC-MAIN-20240809173246-20240809203246-00673.warc.gz | 509,305,867 | 23,723 | # Class 12 Physics ISC Syllabus
After completing class 11, it is very important to focus on class 12 because depending on class 12 result, commonly students are directed towards the next step.
Just similar like class 11, class 12 physics icse syllebus is devided in two papers, mentioned below –
Paper I (Theory) : This paper contains 70 marks for 3 hours of written exam duration.
Paper II ( Practical) : This paper consists of 30 marks. The number divisions are subdivisioned as below –
• Practical (written) – 15 marks
• Practical File(Manual) – 5 marks
• Project work – 10 marks
12 Physics Syllabus 2024-25 ICSE Sr. No UNITS MARKS 1 Electrostatics 15 2 Current Electricity 3 Magnetic Effects of Current and Magnetism 16 4 Electromagnetic Induction and Alternating Currents 5 Electromagnetic Waves 2 6 Optics 18 7 Dual Nature of Radiation and Matter 12 8 Atoms and Nuclei 9 Electronic Devices 7 TOTAL MARKS 70
Go through – Class 11 Physics ICSE Syllabus 2024-25
## ISC Class 12 Physics Syllabus
#### 1. Electrostatics
(i) Electric Charges and Fields
Electric charges; conservation and quantisation of charge, Coulomb’s law; superposition principle and continuous charge distribution.
Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field due to a dipole, torque on a dipole in uniform electric field.
Electric flux, Gauss’s theorem in Electrostatics and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell.
(ii) Electrostatic Potential, Potential Energy and Capacitance
Electric potential, potential difference, electric potential due to a point charge, a dipole and system of charges; equipotential surfaces, electrical potential energy of a system of two point charges and of electric dipole in an electrostatic field.
Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics and electric polarisation, capacitors and capacitance, combination of capacitors in series and in parallel. Capacitance of a parallel plate capacitor, energy stored in a capacitor.
#### 2. Current Electricity
Mechanism of flow of current in conductors. Mobility, drift velocity and its relation with electric current; Ohm’s law and its proof, resistance and resistivity and their relation to drift velocity of electrons; V-I characteristics (linear and non-linear), electrical energy and power, electrical resistivity and conductivity. Temperature dependence of resistance and resistivity.
Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series and in parallel, Kirchhoff’s laws and simple applications, Wheatstone bridge, metre bridge. Potentiometer – principle and its applications to measure potential difference, to compare emf of two cells; to measure internal resistance of a cell.
#### 3. Magnetic Effects of Current and Magnetism
(i) Moving charges and magnetism
Concept of magnetic field, Oersted’s experiment. Biot – Savart law and its application. Ampere’s Circuital law and its applications to infinitely long straight wire, straight solenoid (only qualitative treatment).
Force on a moving charge in uniform magnetic and electric fields. Force on a current-carrying conductor in a uniform magnetic field, force between two parallel current-carrying conductors-definition of ampere, torque experienced by a current loop in uniform magnetic field; moving coil galvanometer – its sensitivity. Conversion of galvanometer into an ammeter and a voltmeter.
(ii) Magnetism and Matter
A current loop as a magnetic dipole, its magnetic dipole moment, magnetic dipole moment of a revolving electron, magnetic field intensity due to a magnetic dipole (bar magnet) on the axial line and equatorial line, torque on a magnetic dipole (bar magnet) in a uniform magnetic field; bar magnet as an equivalent solenoid, magnetic field lines. Electromagnets and factors affecting their strengths, permanent magnets.
#### 4. Electromagnetic Induction and Alternating Currents
(i) Electromagnetic Induction
Faraday’s laws, induced emf and current; Lenz’s Law, eddy currents. Self-induction and mutual induction. Transformer.
(ii) Alternating Current
Peak value, mean value and RMS value of alternating current/voltage; their relation in sinusoidal case; reactance and impedance; LC oscillations (qualitative treatment only), LCR series circuit, resonance; power in AC circuits, wattless current. AC generator.
#### 5. Electromagnetic Waves
Basic idea of displacement current. Electromagnetic waves, their characteristics, their transverse nature (qualitative ideas only). Complete electromagnetic spectrum starting from radio waves to gamma rays: elementary facts of electromagnetic waves and their uses.
#### 6. Optics
(i) Ray Optics and Optical Instruments
Ray Optics: Reflection of light by spherical mirrors, mirror formula, refraction of light at plane surfaces, total internal reflection and its applications, optical fibres, refraction at spherical surfaces, lenses, thin lens formula, lens maker’s formula, magnification, power of a lens, combination of thin lenses in
contact, combination of a lens and a mirror, refraction and dispersion of light through a prism.
Optical instruments: Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers.
(ii) Wave Optics
Wave front and Huygen’s principle. Proof of laws of reflection and refraction using Huygen’s principle. Interference, Young’s double slit experiment and expression for fringe width(β), coherent sources and
sustained interference of light, Fraunhofer diffraction due to a single slit, width of central maximum.
#### 7. Dual Nature of Radiation and Matter
Wave particle duality; photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation – particle nature of light. Matter waves – wave nature of particles, de-Broglie relation.
#### 8. Atoms and Nuclei
(i) Atoms
Alpha-particle scattering experiment; Rutherford’s atomic model; Bohr’s atomic model, energy levels, hydrogen spectrum.
(ii) Nuclei
Composition and size of nucleus. Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number; Nuclear reactions, nuclear fission and nuclear fusion.
#### 9. Electronic Devices
(i) Semiconductor Electronics: Materials, Devices and Simple Circuits. Energy bands in conductors, semiconductors and insulators (qualitative ideas only). Intrinsic and extrinsic semiconductors.
(ii) Semiconductor diode: I-V characteristics in forward and reverse bias, diode as a rectifier; Special types of junction diodes: LED, photodiode and solar cell.
Stay tuned.
Latest posts by admin (see all) | 1,474 | 6,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-33 | latest | en | 0.835655 |
http://www.excel-answers.com/microsoft/Excel/34886991/engineering-notation.aspx | 1,490,492,312,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189090.69/warc/CC-MAIN-20170322212949-00630-ip-10-233-31-227.ec2.internal.warc.gz | 519,105,976 | 5,926 | # Excel - Engineering notation
Asked By Prof Wonmug on 16-Aug-09 07:00 PM
There are a number of websites that explain how Scientific notation
can be made into Engineering notation by calling for 3 digits to the
left of the decimal place
000.0E+00 or ##0.0E+00
This website
http://people.stfx.ca/bliengme/ExcelTips/EngineeringNotation.htm
says that this format produces "poor results for numbers less that
1,000" and suggests using
[<0.001]##0.00E+0;[<1000] #0.00;##0.00E+0
Ignoring for a moment that they are missing a "#" in the middle
format, I do not see the "poor results". The formats in this table are:
Value #,##0.00000000
Scientific 0.00E+00
Eng1 ##0E+0
Eng2 [<0.001]##0E+00;[<1000]##0E+00;##0E+00
Eng3 [=0]0E+00;##0E+00
Value Scientific Eng1 Eng2 Eng3
0.00000100 1.00E-06 1E-6 1E-06 1E-06
0.00001000 1.00E-05 10E-6 10E-06 10E-06
0.00010000 1.00E-04 100E-6 100E-06 100E-06
0.00100000 1.00E-03 1E-3 1E-03 1E-03
0.01000000 1.00E-02 10E-3 10E-03 10E-03
0.10000000 1.00E-01 100E-3 100E-03 100E-03
0.00000000 0.00E+00 000E+0 000E+00 0E+00
1.00000000 1.00E+00 1E+0 1E+00 1E+00
10.00000000 1.00E+01 10E+0 10E+00 10E+00
100.00000000 1.00E+02 100E+0 100E+00 100E+00
1,000.00000000 1.00E+03 1E+3 1E+03 1E+03
10,000.00000000 1.00E+04 10E+3 10E+03 10E+03
100,000.00000000 1.00E+05 100E+3 100E+03 100E+03
1,000,000.00000000 1.00E+06 1E+6 1E+06 1E+06
10,000,000.00000000 1.00E+07 10E+6 10E+06 10E+06
100,000,000.00000000 1.00E+08 100E+6 100E+06 100E+06
1,000,000,000.00000000 1.00E+09 1E+9 1E+09 1E+09
It looks like Eng1 and Eng2 are identical.
Eng3 adds a condition to format zero better.
Herbert Seidenberg replied on 18-Aug-09 11:06 PM
Excel 2007
Engineering notation.
Four significant digits,
including trailing zeros.
Never displays E+00
Positive and negative numbers.
Meg to nano.
http://www.mediafire.com/file/oyw0nl24mty/08_18_09.xlsm | 936 | 2,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-13 | latest | en | 0.557911 |
https://forum.step.esa.int/t/los-displacement-numbers-seem-unnaturally-large/38928 | 1,695,909,826,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00176.warc.gz | 274,918,420 | 5,630 | # LOS displacement numbers seem unnaturally large
Hello,
I am very new to InSAR, so I apologize if I leave out any necessary context and ramble on trying to explain myself.
I am being tasked to look at LOS surface displacement along a very large river system (the area is contained within 4 bursts). My overall objective is to understand LOS surface change over time induced by varying river conditions and validate my displacement maps with GNSS reference stations positioned nearby. I am using Sentinel-1 SLC data downloaded from ASF’s data repository.
I have completed the ASF tutorial (https://asf.alaska.edu/how-to/data-recipes/create-an-interferogram-using-esas-sentinel-1-toolbox/) and the ESA interferometry tutorial (https://step.esa.int/docs/tutorials/S1TBX%20TOPSAR%20Interferometry%20with%20Sentinel-1%20Tutorial_v2.pdf). When I try to follow the steps outlined within the ESA tutorial using my own data, I am able to successfully produce displacement maps, but the displacement numbers at the most coherent locations seem rather large (10s of centimeters) over a 12 day period. Realistically, these locations are shifting very minimally as there is little tectonic activity or subsidence in this location.
I have 3 primary questions:
1.) Why are my LOS displacement values so high for coherent locations?
2.) For coherent pixel spaces, what type of accuracy in LOS displacement measurements should I expect for a single displacement map? For example, assuming that I processed my data correctly, the data is good, and I am investigating coherent enough pixel spaces, does interferometry provide displacement measurements that are accurate to 10 dm, 10 cm, 10 mm, or what? I know that a long time series can provide a more accurate representation of the movement happening at a given location and can therefore average out some of my unrealistically high displacement values, but what can be expected from the accuracy of a given displacement measurement?
3.) What resources would you all suggest for someone who is looking to perform this type of analysis? | 433 | 2,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-40 | latest | en | 0.90068 |
https://www.saranextgen.com/homeworkhelp/doubts.php?id=50234 | 1,669,512,648,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710155.67/warc/CC-MAIN-20221127005113-20221127035113-00729.warc.gz | 1,044,961,318 | 6,948 | # A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in fig. work done normal reaction on block in time t is : (a) 0 (b) (c) (d)
## Question ID - 50234 :- A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in fig. work done normal reaction on block in time t is : (a) 0 (b) (c) (d)
3537
Answer Key : (b) -
N−mg =
Now, work done W= =
⇒ W =
Next Question :
A heat source at T= k is connected to another heat reservoir at T=k is connected to another heat reservoir at T= K by a copper slab which is 1m thick. Given that the thermal conductivity of copper is 0.1 , the energy flux through it in the steady state is : (a) 90 (b) 200 W (c) 65 (d) 120 | 228 | 790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | longest | en | 0.909376 |
http://lists.puremagic.com/pipermail/digitalmars-d/2019-November/298379.html | 1,656,494,727,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00527.warc.gz | 37,094,962 | 2,002 | # Elegant Project Euler Solution
Mon Nov 4 10:37:31 UTC 2019
```On Sunday, 3 November 2019 at 19:51:14 UTC, Leonhard Euler wrote:
> Hi all,
> Just wanted to share a very elegant solution for Project Euler,
> problem #24.
>
> This is the problem:
>
> A permutation is an ordered arrangement of objects. For
> example, 3124 is one possible permutation of the digits 1, 2, 3
> and 4. If all of the permutations are listed numerically or
> alphabetically, we call it lexicographic order. The
> lexicographic permutations of 0, 1 and 2 are:
>
> 012 021 102 120 201 210
>
> What is the millionth lexicographic permutation of the digits
> 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
>
> And my solution:
>
> ```
> import std;
>
> void main() {
> iota(10).array.nthPermutation(999999).writeln();
> }
> ```
>
> Know thy standard library ;)
I'm not sure that using pre-rolled library functions is entirely
the point of Project Euler. To get the most out of it make your
own functions for tasks that are core to the problem. The
problems are clearly structured in a way that leads toward
building your own library from puzzle to puzzle and realising
when you can reuse functions from previous problems.
``` | 346 | 1,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.825586 |
https://physics.stackexchange.com/questions/461672/is-a-falling-cats-angular-momentum-conserved | 1,638,381,171,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360881.12/warc/CC-MAIN-20211201173718-20211201203718-00587.warc.gz | 532,624,732 | 35,807 | # Is a falling cat's angular momentum conserved?
I found this question in my physics textbook:
From a certain height a cat is dropped back-side down. The cat rotates his body while falling and lands on his four legs. Does the cat's angular momentum change during the fall?
The answer is no, but I said yes, because I thought the gravitational force will change the angular momentum? Am I missing something?
• Your confusion may be that you have to define relative to which point you measure the angular momentum. E.g., relative to the floor immediately beneath my chair, the angular momentum will be different from zero when I spin my chair. But relative to the seat of my office chair, my angular momentum is always zero (the chair spins with me!). Likewise, in the frame of the cat, or, more generally, in any frame along the (straight) trajectory of the cat, angular momentum will be constant. But relative to an observer that isn't trying to catch the cat, its angular momentum will increase as it falls. Feb 20 '19 at 9:42
To change angular momentum, a torque must be applied. Since gravity pulls every part of the cat with a force proportional to its mass (that is, with the same acceleration), there is no net torque on the free falling cat, and thus no change in angular momentum.
This is true for any free falling object, but not necessarily if it is supported at any point. The support together with the gravitational force can apply a torque and therefore change angular momentum.
As to how the cat manages to turn around even with no net torque, this is known as the Falling cat problem, and is visualized nicely in this very disturbing animation from Wikipedia
The rotation is based on the fact that the cat is not a rigid body, and can thus bend in a way that results in its reorientation.
• Wow that is so interesting I didnt know there was a thing like this it seems like there is some physic tricks going behind this. I need more time to observe it to understand better but at least I know which way I have to think now, many thanks! Feb 19 '19 at 17:29
• Slow Motion Flipping Cat Physics ... good video and channel :) Feb 20 '19 at 2:26
An external force like the gravitational force acts on an object like if it acts on its center of mass. Since the cat's center of mass is on its rotational axis, this would mean that the gravitational force doesn't give any angular momentum to the cat.
In general, a force does not always give angular momentum to an object. It will if the force is applied at a certain distance from the rotation axis.
The gravitational force can indeed give angular momentum to a system. Think of a pendulum that you drop after raising it from it's rest position. In that scenario, the rotation axis is the pendulum holding point and the center of mass would be close to the end of the pendulum. Thus, the gravitational force acts away from the rotational axis and the pendulum will start to rotate around its holding point, thus gaining angular momentum.
• Assuming the cat's center of mass is very close to its rotational axis Why assume something that is (trivially) true? A free body will always rotate exactly around its center of mass. Feb 19 '19 at 17:38
• No the reaction from the attachment point is perpendicular to the motion of the pendulum and thus the torque is 0. here it's truly gravity that affects angular momentum. Feb 21 '19 at 10:15
• Yep you're right, I was mixing up some things. Second comment deleted! Feb 21 '19 at 10:37
The problem is flawed because it does not specify which axis we are supposed to be measuring the angular momentum around.
Angular momentum is always specified relative to an axis of rotation. However, you can always split the total angular momentum of an object into two parts, $$L_{total} = L_{external}+L_{CM}$$.
The first part is the angular momentum of the whole object around some external axis of rotation. Specifically, $$L_{external} = \vec{R}_{CM}\times \vec{P}_{CM}$$ where $$\vec{R}_{CM}$$ is the position vector of the center of mass of the object, and $$\vec{P}_{CM}$$ is the momentum of the center of mass.
The second part is the internal angular momentum, or the angular momentum of the object measured around its own center of mass.
Gravity can change the external angular momentum, depending on the axis of rotation you choose. The torque from gravity around an external center of mass is just the weight of the object times the horizontal distance of the center of mass from the axis.
What the question is probably trying to get, however, at is the fact that gravity cannot change the internal angular momentum of an object. Since gravity "acts" at the center of mass, the torque of gravity around the center of mass is always zero! | 1,035 | 4,756 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2021-49 | latest | en | 0.952377 |
https://www.studyadda.com/index.php?/notes/2nd-class/mathematics/division/word-problems/7900 | 1,531,759,059,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589404.50/warc/CC-MAIN-20180716154548-20180716174548-00391.warc.gz | 994,602,776 | 16,069 | # 2nd Class Mathematics Division Word Problems
## Word Problems
Category : 2nd Class
### Word Problems
In this section we will learn about some real life problems based on division.
How many bundles ofpencils can you make out ofpencils?
Explanation-
Total number of pencil divided by one bundle of pencil bundles.
If a mini-bus can carrychildren then how many mini buses would be needed to carry children?
Explanation-
Total number of children is divided by capacity of one
mini- bus (28 children )mini bus.
• Division means equal distribution.
• The simple of division is horizontal line and two dots.
• The number which gets divided is called the DIVIDEND, debited by capital 'D".
• The number which divides the dividend is called the DIVISOR, denoted by 'd'.
• The number which tells us how many times the division has been carried out is called the QUOTIENT, denoted b 'Q'.
• The number which is left after the division is called the REMAINDER, denoted by 'R'.
• The division should be continued till the remainder is either '0' or less than the divisor (D).
• Number divided by itself gives the quotient
• Number divided by, is the number itself .
• 0 divided by any number is
• A number cannot be divided by.
• Division describes two distinct but related settings.
• Division is often shown in algebra and science by placing the dividend over the divisor with a horizontal line, also called a vinculum or fraction bar, between them.
#### Other Topics
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You will be redirected in 3 sec | 363 | 1,619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2018-30 | latest | en | 0.930733 |
https://oundleprimary.com/slide/solutions-to-exam-1-isye-getcv9 | 1,603,450,528,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881369.4/warc/CC-MAIN-20201023102435-20201023132435-00015.warc.gz | 471,679,440 | 11,368 | # Solutions to Exam #1 - ISyE
Solutions to Exam #1 John H. Vande Vate Fall, 2002 1 1 Scores 18 17 16 14 12 12
10 9 8 6 6 4 2 2 1 0
0 < 40 < 50 <60 <70 <80 <90 <100
2 2 Stats Range: 37-89 Mean: 61 Std. Dev: 11 Curve: 11
3 3 Concerns? Submit concerns about grading in writing Identify what you think is mis-graded and why. Will return regraded exams by the end of the semester. 4 4
Question #1 Our plant makes two products and ships them to customers across the country. Demand for the products is relatively constant at 10 units of each product each day at each customer. We have 1,000 customers across the country. Product Selling Price Weight Product 1: \$100 99 lbs Product 2: \$100 1 lbs The capacity of our trucks is 45,000 lbs. The average distance to customers is 1,000 miles and we pay our carrier \$1.50/mile (one way, regardless of the load) plus an additional \$0.10/lb (regardless of the distance). Our inventory carrying cost is 15%/year. Treat a
year as 250 days. 5 5 Question 1.A What is the optimal frequency for direct shipments to our customers? First, we can ignore the \$0.10/lb as the total amount we pay under this heading will be the same regardless of our shipping frequency. Short Answer: Every unit incurs \$0.10 Long Answer: \$0.10*Q is how much we pay per shipment There will be D/Q shipments per year We will pay \$0.10D per year regardless of the frequency
6 6 Question 1.A Use the version of the EOQ model for several destinations: Q* = sqrt(2AD/hC)sqrt(P/(D+P)) A is the cost per trip = \$1,500 D is the annual demand for a single customer or 2,500 packages consisting of 1 unit of product 1 and one unit of product 2
h is 15% per year C is the value of a package or \$200 P is the production rate which we take to be 1,000D 7 7 Question 1.A So, Q* = sqrt(2*1500D/hC)sqrt(1,000D/1,001D) sqrt(3,000D/30) = sqrt(100D) = 10sqrt(2,500) = 500. Total weight of 500 packages is 50,000 lbs which exceeds the capacity of the truck. So, we should ship in full truckloads of 450
units each. That means the answer is to ship every 45 working days. 8 8 Question 1.B You calculated the optimal frequency f* in Part A using the average distance to customers. Some of our customers are closer than average and others are further away. We have decided that to keep our operations simple, we will ship to each customer either i. The optimal frequency f* ii. Twice the optimal frequency 2f*:
iii. Half the optimal frequency f*/2 Which customers should be served with each frequency? Be specific and be sure to show how you obtained your answer. 9 9 Question 1.B Convention: If the optimal frequency is f* = once every 45 days then 2f* is twice every 45 days and f*/2 is once every 90 days. Since shipping every 45 days requires full truckload shipments, reducing the shipping frequency will never be advantageous
regardless of where the customer is unless we change mode. 10 10 Question 1.B The decision of whether to ship at the frequency f* or at the frequency 2f* will depend on the distance to the customer. We can find the breakpoint by setting the two cost formulas equal and finding the appropriate distance: AD/450 + 450hC/2 = AD/225 + 225hC/2 AD+ 450*225hC = 2AD + 225*225hC 225*225hC = AD 2252hC/D = A = \$1.5*breakpoint distance So, customers closer than about 400 miles we can ship to twice as
frequently. 11 11 Question 2 Our carriers trucks average about 50 miles per hour and, because of Federal Transportation Regulations, about 10 hours per 24 hour day. Calculate the pipeline inventory costs inherent in the distribution system you designed in answer to Question 1, Part A. 12 12
Question 2.A It takes two days to reach the average customer. So, every item we make is on the road for an average of two days. Consequently, there is approximately two days production in the pipeline. Thats 2days * 10 packages per day/customer * 1000 customers = 20,000 packages each worth \$200. So the value of the pipeline inventory is \$4,000,000 and the inventory costs associated with that are 15% of \$4 million = \$600,000/year 13 13
Question 2.B Factor the pipeline inventory costs into your model for calculating the optimal frequency. How does this change your answer from Question 1, Part A, Part B? It doesnt. Goods still take just as long to get there. If we increase the shipping frequency, there will just be more smaller shipments, but they will still add up to the same amount of goods on the road at any given time 2 days worth of production 14 14 Question 3 1 We want to develop a flows-on-paths formulation of a
multi-modal logistics problem. You have been asked to provide examples to help the engineers in building the model. You have the following list of variables and are asked to use them in formulating each of the following constraints. The variable Flow[origin, Path, destination] is the annual quantity of goods shipped from origin to destination via the path Path, where Path is a set of Edges from the origin to the destination via a number of intermediate stops. Each edge also indicates the mode used. So, for example, the edge (Detroit, Kansas City, Unit Train) indicates a Unit Train move from Detroit to Kansas City with no intermediate stops. 15 15 Question 3
The paths from Detroit to Los Angeles are: Path1 := {(Detroit, Kansas City, Unit Train), (Kansas City, Denver, Loose Car), (Denver, Los Angeles, Loose Car)} Path2 := {(Detroit, Los Angeles, Unit Train)}; Path3 := {(Detroit, Los Angeles, Loose Car)}; Path4 := {(Detroit, Kansas City, Loose Car), (Kansas City, Los Angeles, Loose Car)}; 16 16 Question 3 The paths from Detroit to Seattle are: Path5 := {(Detroit, Kansas City, Unit Train), (Kansas City, Denver, Loose Car),
(Denver, Seattle, Loose Car)} Path6 := {(Detroit, Seattle, Unit Train)}; Path7 := {(Detroit, Seattle, Loose Car)}; Path8 := {(Detroit, Kansas City, Loose Car), (Kansas City, Seattle, Loose Car)}; 17 17 Project Topics Provide a linear description in terms of the Flow variables of: The total quantity of goods shipped from Detroit annually: Sum{path in paths that start at Detroit, dest in destinations} Flow[Detroit, path, dest]; Flow[Detroit, Path1, Los Angeles] + Flow[Detroit, Path2, Los Angeles] +
Flow[Detroit, Path3, Los Angeles] + Flow[Detroit, Path4, Los Angeles] + Flow[Detroit, Path5, Seattle] + Flow[Detroit, Path6, Seattle] + Flow[Detroit, Path7, Seattle] + Flow[Detroit, Path8, Seattle] B 18 18 AMPL Version
Set CITIES; Set ORIGINS; Set DESTINS; Set PATHS; Set MODES; Set EDGES within CITIES cross CITIES cross MODES; Set PATHEDGES{PATHS} within EDGES; Param PathOrigin{PATHS}; Param PathDestin{PATHS}; 19 19 AMPL Version
Var Flow{orig in ORIGINS, path in PATHS, dest in DESTINS: PathOrigin[path] = orig and PathDestin[path] = dest} >= 0; Sum{path in PATHS, dest in DESTINS: PathOrigin[path] = Detroit} Flow[Detroit, path, destin] 20 20 Question 3 The total quantity of goods loaded onto Unit Trains in Detroit annually
Sum{path in the set of paths that start in Detroit with unit train shipments, dest in destinations} Flow[Detroit, path, dest]; The total quantity of goods passing through Kansas City annually Sum{orig in Origins, path in paths that include Kansas City, dest in Destinations} Flow[orig, path, dest}; D. 21 21 More Options The total quantity of goods that passed through Denver arriving in Seattle annually Sum{orig in Origins, path in paths that end in Seattle
and include Kansas City}Flow[orig, path, Seattle] The total quantity of goods that pass through both Kansas City and Denver each year. Sum{orig in Origins, path in paths that include both Kansas City and Denver, dest in Destinations} Flow[orig, path, dest]; 22 22 Question 4 In class we discussed two models for the inventory at a production plant serving many identical customers with staggered shipments via a single loading dock. Model 1: In this analysis, we calculated the
inventory impact of each customer and showed that, when the number of customers is large (say 100), any single customers contribution to inventory at the plant is small much smaller than Q/2, where Q is the quantity delivered in each shipment. 23 23 Question 4 Model 2: In this analysis, we argued that, whatever the contribution of a single customer, the operations at a production facility serving several customers with shipments of size Q will be indistinguishable from those of a facility serving
a single customer with shipments of shipments of size Q: The inventory at the plant will grow from 0 to Q, at which point the vehicle will depart and the inventory will return to 0. This process repeats and so the average inventory at the plant should be Q/2 in both cases. 24 24 Question 4 What will the average inventory at the plant be? Assume the plant serves 100 identical customers each with annual demand D. It sends each customer shipments of size Q. To answer this
question completely, you must The two models are identical! 25 25 Resolution Inventory per shipment: Q*Q/2P Shipments per year: D/Q to each customer Total Inventory contribution per customer: QD/2P. With n identical customers P = nD So each customer contributes Q/2n in annual inventory at the plant. But there are n customers, so average annual inventory at the plant is Q/2. 26
26 27 27
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• (παράδειγμα) RDF graph RDF statements Resource Property Literal Πλαίσιο RDF Χρήση containers για περιγραφή ομάδας από resources Bag (rdf:Bag): resources (πιθανά με επαναλήψεις), χωρίς να ενδιαφέρει η διάταξη.
• PLANNING AND RUNNING A TUTORIAL ... Structure Agree AIM Explore the DIFFICULTIES Agree AGENDA Agree the OUTCOME Generalise to EXERCISE REHEARSE to Explore ALOBA FEEDBACK SUMMARISE Adapted from Teaching and Learning Communication Skills in Medicine Suzanne Kurtz, Jonathan Silverman ,... | 2,803 | 11,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.906536 |
http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Mountain_range.html | 1,545,045,138,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828501.85/warc/CC-MAIN-20181217091227-20181217113227-00537.warc.gz | 451,220,293 | 4,177 | # Mountain range
related topics {island, water, area} {math, energy, light} {woman, child, man}
A mountain range is a single, large mass consisting of a succession of mountains or narrowly spaced mountain ridges, with or without peaks, closely related in position, direction, formation, and age; a component part of a mountain system or of a mountain chain.[1] other definitions may include a mountain system which is a group of mountain ranges exhibiting certain unifying features, such as similarity in form, structure, and alignment, and presumably originating from the same general causes; esp. a series of ranges belonging to an orogenic belt.[2] A mountain system or system of mountain ranges sometimes is used to combine several geological features that are geographically (regionally) related.
Mountain range is divided by highlands or mountain passes and valleys. Individual mountains within the same mountain range do not necessarily have the same geology, though they often do; they may be a mix of different orogeny, for example volcanoes, uplifted mountains or fold mountains and may, therefore, be of different rock.
## Contents
### Major ranges
The Himalaya Range contains the highest mountains on the Earth's surface, the highest of which is Mount Everest. The world's longest mountain system is known as Ocean Ridge, which is a chain of mountains that runs on the seafloor of five oceans around the world; it has a length of 65,000 kilometres (40,400 mi), and the total length of the system is 80,000 kilometres (49,700 mi). The Andes is the world's longest mountain system on the surface of a continent; it is 7,000 kilometres (4,300 mi) in length. The Arctic Cordillera is the world's northernmost mountain system and contains the highest point in eastern North America.
### Divisions and Categories
The mountain systems of the earth are characterized by a tree structure, that is, many mountain ranges have sub-ranges within them. It can be thought of as a parent-child relationship. For example, the Appalachian Mountains range is the parent of other ranges that comprise it, some of which are the White Mountains and the Blue Ridge Mountains. The White Mountains are a child of the Appalachians, and there are also children of the Whites, including the Sandwich Range and the Presidential Range. Further, the Presidential Range can be broken up into the Northern Presidential Range and Southern Presidential Range.
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related documents Geography of Jersey Geography of Fiji Geography of Senegal Esker Cloud forest Mont Blanc Cumulus cloud Askja Geography of Switzerland Barents Sea Sea of Okhotsk Lake Vostok Tundra Coastal erosion Geography of Rwanda Geography of Argentina Rathlin Island Geography of Lesotho Aoraki/Mount Cook Geography of the Solomon Islands Neogene Limpopo River Altai Mountains Afsluitdijk Great Sandy Desert Camargue Geography of Mauritius Geography of Antarctica Geography of Mozambique Strait of Gibraltar | 620 | 2,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-51 | latest | en | 0.94183 |
http://www.sawaal.com/chemistry-questions-and-answers/what-is-the-gram-formula-mass-of-nbsp-li-2-so-4-_14483 | 1,527,129,286,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865884.49/warc/CC-MAIN-20180524014502-20180524034502-00078.warc.gz | 446,873,447 | 17,696 | 2
Q:
# What is the gram formula mass of ${\mathbf{Li}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$?
A) 101.2487 B) 109.9446 C) 111.2687 D) 106.5871
Explanation:
The term gram formula mass is synonymous with molecular mass. So, to get this, we must add molar mass of all the elements in this compound.
Given compound is Lithium Sulphate - ${\mathbf{Li}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$
Elemental calculation ::
Li - Lithium - 6.941 - 2 - 13.88200 %
S - Sulfur - 32.065 - 1 - 32.065 %
O - Oxygen - 15.9994 - 4 - 63.9976 %
Now, total = 12.6264 + 29.1647 + 58.2090 = 109.9446
Hence, the gram formula mass of Li2SO4 = 109.9446.
Q:
Formation of curd from milk is chemical reaction.
A) TRUE B) FALSE
Explanation:
Milk is converted into curd or yogurt by the process of fermentation. Milk consists of globular proteins called casein.
It is biochemical change because here microorganism named lactobacillus react with milk and turn it to curd. Lactobacillus required favourable condition to grow.
For it 37 degree temperature is needed. But sometimes when it does not get favourable condition, then other microorganism react and makes the curd taste different.
Subject: Chemistry
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1 19
Q:
Which type of molecule is CF4?
A) nonpolar, with a symmetrical distribution of charge B) polar, with an asymmetrical distribution of charge C) nonpolar, with an asymmetrical distribution of charge D) polar, with a symmetrical distribution of charge
Answer & Explanation Answer: A) nonpolar, with a symmetrical distribution of charge
Explanation:
CF4 is tetrahedral.
Since the four 'F' atoms have the same electronegativity values, there is no bias in electron distribution toward any one.
Therefore, electron/charge distribution is symmetrical, which also means that the molecule is nonpolar.
Hence, CF4 is nonpolar molecule with a symmetrical distribution of charge.
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1 14
Q:
Which structural formula represents a dipole?
A) 4 B) 3 C) 2 D) 1
Explanation:
What is Dipole :
when two atoms in a molecule have substantially different electronegativity: One atom attracts electrons more than another, becoming more negative, while the other atom becomes more positive.
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1 34
Q:
Which of the following elements is a metal?
A) Se B) I C) Zn D) S
Explanation:
In chemistry, a metal is an element that readily forms positive ions (cations) and has metallic bonds. Metals are opaque, lustrous elements that are good conductors of heat and electricity. Most metals are malleable and ductile and are, in general, denser than the other elemental substances.
Examples of metals are aluminium, copper, iron, tin, gold, lead, silver, titanium, uranium, and zinc.
Here Se - Selenium is an element but is not a metal, S - Sulphur is not a metal and I - Iodine is also not a metal.
Hence, Zn - Zinc is the only metal in the given options.
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4 30
Q:
Which formula represents a nonpolar molecule?
A) H2O B) CF4 C) HCl D) H2S
Explanation:
You have to consider three things.
1. First, are the bonds polar?
Look at differences in electronegativity and if it is >0.5, bonds are polar.
2. Second, does the molecule have unshared electron pairs?
If so, they generally lead to a polar molecule.
3. Third, is the molecule symmetric?
If so, then although the bonds may be polar, the symmetry cancels out the individual bond polarity and the molecule is non polar.
CF4 follows all the rules above and has a symmetric distribution of charge, making it nonpolar. The others all have asymmetric charge distributions.
Hence, CF4 represents a Non-polar molecule.
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1 47
Q:
Order of basicity of amines in gaseous state?
A) 2'> 1'> 3' > NH3 B) 3' > 2' > 1'> NH3 C) 1' > 2' > 3'> NH3 D) 2'> 3'> 1' > NH3
Explanation:
The order of basicity in ammonia is different in gaseous phases vs. aqueous phases due to hydrogen bonding.
So in gaseous phases it's the well known: 3' > 2' > 1'> NH3. Just so we're on the same page, this is due to the inductive effect of electron donating alkyl groups--they destabilize the lone pair on nitrogen atom, making nitrogen more basic.
But in aqueous phase, the order is slightly different: 2'> 1'> 3' > NH3.
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1 251
Q:
Which of the following is not an allotrope of carbon?
A) Diamond B) Graphene C) Ethylene D) Buckminsterfullerene
Explanation:
Allotropes are different structural modifications of an element, that is, each of two or more different physical forms in which an element can exist.
Carbon is capable of forming many allotropes due to its valency. Well-known forms of carbon include charcoal, diamond, graphite, buckminsterfullerene and graphene.
Hence, Ethylene is not an allotrope of Carbon.
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2 107
Q:
Which compound matches the IR Spectrum?
A) 1, 5 – hexadiene B) 3 – Hexanol C) Trans – 4 – octene D) None of the above
Explanation:
3 – Hexanol:
There is no alcohol stretch present in the spectrum. Therefore, the above structure is not suitable for the given spectrum.
Trans – 4 – octene:
There is no vinyl bending present in the spectrum. Therefore, the above structure is not suitable for the given spectrum.
The vinyl bending and other stretches were matched with the above structure.
Hence, the c = c stretch, vinyl bending and stretch were matched with the above structure.
Therefore, the given spectrum belongs to 1, 5 – hexadiene. | 1,648 | 6,078 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2018-22 | latest | en | 0.769488 |
http://www.mbstudent.com/2019/12/ | 1,726,230,163,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651513.89/warc/CC-MAIN-20240913101949-20240913131949-00803.warc.gz | 40,049,691 | 10,563 | # A maximum power on a load in an electric DC circuit
A series electric DC circuit is being considered to do math. The circuit contains a voltage source with its internal resistance and a load resistors. Resistance of wires is omitted in this example.
The voltage equation for the circuit above is following
$$E - R_i \cdot I - R_o \cdot I = 0$$
Thereby the current is given by the equation
$$I = \frac{E}{ R_i + R_o }$$
A general formula for power is
$$P = U \cdot I$$
Knowing that the voltage on the resistor Ro is given by the Ohm’s law relation
$$U(R_o)= I \cdot R_o$$
A formula for power on the resistor Ro is following
$$P(R_o)= I^2 \cdot R_o$$
After plugging in the relation for current
$$P(R_o)= (\frac{E}{ R_i + R_o })^2 \cdot R_o$$
The extremum of function P(Ro) needs to be found in order to find which magnitude of the resistor Ro will be give the maximum power.
The first derivative of equation of power on the resistor Ro is following
$$\frac{dP(R_o)}{dR_o}= \frac{d}{dR_o}((\frac{E}{ R_i + R_o })^2 \cdot R_o)$$
After doing the math, the theorems for derivatives of the multiplication of functions and internal and external functions were applied
$$\frac{dP(R_o)}{dR_o}= -2\cdot \frac{E}{ R_i + R_o } \cdot \frac{E}{ (R_i + R_o)^2 } \cdot R_o + (\frac{E}{ R_i + R_o })^2$$
In order to find the extremum the point where the derivative is equal to 0 needs to be calculated
$$0= -2\cdot \frac{E}{ R_i + R_o } \cdot \frac{E}{ (R_i + R_o)^2 } \cdot R_o + (\frac{E}{ R_i + R_o })^2$$
After rearranging the elements in equation above and doing some math
$$R_o = R_i$$
Finally, the maximum power on the load resistor Ro is given by formula
$$P(R_o)_{max} = \frac{E^2}{4 \cdot R_i}$$ | 522 | 1,702 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 11, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-38 | latest | en | 0.80986 |
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Exercise [19.06]
Author Message
Joined: 12 Jul 2010, 07:44
Posts: 154
Exercise [19.06]
I'm not exactly sure what this exercise is asking for, but I'll take a stab at it.
It can be observed that F, and J were constructed component-wise on page 443 from the components of the electric and magnetic fields, and charge flux and density, respectively. This construction depended on the use of standard Minkowski coordinates.
However, F and J themselves are coordinate-independent, tensor quantities, expressible as "geometrical" properties of the underlying spacetime manifold. Similarly the Maxwell equations expressed using these tensors don't depend on any particular coordinate system: They represent relations between these geometrical properties (including derivatives) at each spacetime point, and so remain perfectly well-defined even on manifolds with curvature.
The only question is how to relate F and J in curved spacetime back to the components they were constructed from on p. 443. In a general curvilinear coordinate system, these components will get all mixed up.
However, note two things:
(i) The initial component construction of these tensors did not depend on a particular choice of reference frame - it works in any reference frame parameterized by flat Minkowsi coordinates.
(ii) The tangent space at any point in curved spacetime is a flat Minkowski space that approximates some small region of the actual spacetime around that point (i.e. curved spacetimes are locally flat at small enough scales). So any coordinate system can be approximated in some small region around a given point by using standard, flat, orthogonal Minkowski coordinates. In these coordinates, the components of the electric and magnetic fields, flux, etc, can be extracted from F and J as on p. 443; the values will be exact at the given point, and an approximation in the region around it.
04 Mar 2012, 06:56
Page 1 of 1 [ 1 post ] | 436 | 1,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2022-49 | latest | en | 0.928125 |
https://www.jiskha.com/questions/559265/solve-each-equation-algebraically-and-check-it-by-substituing-into-the-orignall-equation | 1,597,391,153,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739182.35/warc/CC-MAIN-20200814070558-20200814100558-00189.warc.gz | 719,765,228 | 5,969 | # pre calculous
solve each equation algebraically and check it by substituing into the orignall equation.
50e^0.035x=200
3LN(x-3)+4=5
Method of your choice by solving.
logx^2=6
Logx^4=2
2x-2^-x/2=4
e^x+e^-x/2=4
500/1+25e^.3x=200
1. 👍 0
2. 👎 0
3. 👁 257
1. 50e^(0.035x)=200
e^(.035x) = 4
take ln of both sides
ln(e^(.035x)) = ln4
.035x lne = ln4, but lne = 1
x = ln4 / .035 = appr. 39.6
log x^2 = 6
by definition:
x^2 = 10^6
x = 10^3 = 1000
do logx^4 = 2 the same way
2x-2^-x/2=4 : I have a feeling that is not what you meant, without brackets I cannot tell what the equation is. I think the first term is probably 2^x
e^x+e^-x/2=4
Again, I think you meant:
e^x+e^(-x/2) = 4
let e^(-x/2) = y , where y is a positive real number
then e^(x/2) = 1/y
and e^x = 1/y^2
so 1/y2 + y = 4
1 + y^3 = 4y^2
y^3 - 4y^2 + 1 = 0
I used a program to find
y = .54 or y = 3.94 appr. , there is also a negative root which would not be allowed
so e^(-x/2) = .54 or ..... use the other root
-x/2 = ln.54
x = 1.2324 or ......
last question, way too ambiguous without brackets.
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What is the first step in solving the linear system {2x − 3y = 11 {−x + 5y = −9 by the substitution method in the most efficient way? A. Solve the first equation for x. B. Solve the first equation for y. C. Solve the second | 1,306 | 3,609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-34 | latest | en | 0.831811 |
https://homework.cpm.org/category/CC/textbook/ccg/chapter/9/lesson/9.2.4/problem/9-105 | 1,721,504,192,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00463.warc.gz | 254,670,376 | 14,709 | ### Home > CCG > Chapter 9 > Lesson 9.2.4 > Problem9-105
9-105.
Without using a calculator, find the sum of the interior angles of a $1002$-gon. Show all work.
The sum of the measures of the interior angles of an $n$-gon is $180º\left(n-2\right)$.
Let $n = \text{number of sides}$, $(n-2)180°=1000\cdot180°=180,000°$ | 114 | 320 | {"found_math": true, "script_math_tex": 5, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.687259 |
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# 12-3. Continuity
Continuity
In this section, you will:
• Determine whether a function is continuous at a number.
• Determine the numbers for which a function is discontinuous.
• Determine whether a function is continuous.
Arizona is known for its dry heat. On a particular day, the temperature might rise as high as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] 118 ∘ F and drop down only to a brisk<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] 95 ∘ F. [link] shows the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>T</mi><mo>,</mo></annotation-xml></semantics>[/itex] where the output of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] T( x ) is the temperature in Fahrenheit degrees and the input<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>x</mi><mtext> </mtext></annotation-xml></semantics>[/itex]is the time of day, using a 24-hour clock on a particular summer day.
<figure class="small" id="CNX_Precalc_Figure_12_03_001" style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: auto; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 1; word-spacing: 0px; -webkit-text-stroke-width: 0px;"> <figcaption>Temperature as a function of time forms a continuous function.</figcaption> </figure>
When we analyze this graph, we notice a specific characteristic. There are no breaks in the graph. We could trace the graph without picking up our pencil. This single observation tells us a great deal about the function. In this section, we will investigate functions with and without breaks.
# Determining Whether a Function Is Continuous at a Number
Let’s consider a specific example of temperature in terms of date and location, such as June 27, 2013, in Phoenix, AZ. The graph in [link] indicates that, at 2 a.m., the temperature was<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] 96 ∘ F. By 2 p.m. the temperature had risen to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] 116 ∘ F, and by 4 p.m. it was<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] 118 ∘ F. Sometime between 2 a.m. and 4 p.m., the temperature outside must have been exactly<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] 110.5 ∘ F. In fact, any temperature between<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] 96 ∘ F and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] 118 ∘ F occurred at some point that day. This means all real numbers in the output between<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] 96 ∘ F and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] 118 ∘ F are generated at some point by the function according to the intermediate value theorem,
Look again at [link]. There are no breaks in the function’s graph for this 24-hour period. At no point did the temperature cease to exist, nor was there a point at which the temperature jumped instantaneously by several degrees. A function that has no holes or breaks in its graph is known as a continuous function. Temperature as a function of time is an example of a continuous function.
If temperature represents a continuous function, what kind of function would not be continuous? Consider an example of dollars expressed as a function of hours of parking. Let’s create the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>D</mi><mo>,</mo></annotation-xml></semantics>[/itex] where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] D( x ) is the output representing cost in dollars for parking<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>x</mi><mtext> </mtext></annotation-xml></semantics>[/itex]number of hours. See [link].
Suppose a parking garage charges $4.00 per hour or fraction of an hour, with a$25 per day maximum charge. Park for two hours and five minutes and the charge is $12. Park an additional hour and the charge is$16. We can never be charged $13,$14, or \$15. There are real numbers between 12 and 16 that the function never outputs. There are breaks in the function’s graph for this 24-hour period, points at which the price of parking jumps instantaneously by several dollars.
<figure id="CNX_Precalc_Figure_12_03_002"> <figcaption>Parking-garage charges form a discontinuous function.</figcaption> </figure>
A function that remains level for an interval and then jumps instantaneously to a higher value is called a stepwise function. This function is an example.
A function that has any hole or break in its graph is known as a discontinuous function. A stepwise function, such as parking-garage charges as a function of hours parked, is an example of a discontinuous function.
So how can we decide if a function is continuous at a particular number? We can check three different conditions. Let’s use the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] y=f( x ) represented in [link] as an example.
<figure class="small" id="CNX_Precalc_Figure_12_03_003"></figure>
Condition 1 According to Condition 1, the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( a ) defined at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a must exist. In other words, there is a y-coordinate at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a as in [link].
<figure class="small" id="CNX_Precalc_Figure_12_03_004"></figure>
Condition 2 According to Condition 2, at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a the limit, written<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→a f(x) ,must exist. This means that at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a the left-hand limit must equal the right-hand limit. Notice as the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>f</mi><mtext> </mtext></annotation-xml></semantics>[/itex]in [link] approaches<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a from the left and right, the samey-coordinate is approached. Therefore, Condition 2 is satisfied. However, there could still be a hole in the graph at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a.
Condition 3 According to Condition 3, the corresponding<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>y</mi><mtext> </mtext></annotation-xml></semantics>[/itex]coordinate at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a fills in the hole in the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f. This is written<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→a f(x)=f(a).
Satisfying all three conditions means that the function is continuous. All three conditions are satisfied for the function represented in [link] so the function is continuous as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a.
<figure class="small" id="CNX_Precalc_Figure_12_03_005"> <figcaption>All three conditions are satisfied. The function is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a.</figcaption> </figure>
[link] through [link] provide several examples of graphs of functions that are not continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a and the condition or conditions that fail.
<figure class="small" id="CNX_Precalc_Figure_12_03_006"> <figcaption>Condition 2 is satisfied. Conditions 1 and 3 both fail.</figcaption> </figure> <figure class="small" id="CNX_Precalc_Figure_12_03_007"> <figcaption>Conditions 1 and 2 are both satisfied. Condition 3 fails.</figcaption> </figure> <figure class="small" id="CNX_Precalc_Figure_12_03_008"> <figcaption>Condition 1 is satisfied. Conditions 2 and 3 fail.</figcaption> </figure> <figure class="small" id="CNX_Precalc_Figure_12_03_009"> <figcaption>Conditions 1, 2, and 3 all fail.</figcaption> </figure>
Definition of Continuity
A function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( x ) is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a provided all three of the following conditions hold true:
Condition 1:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(a) exists.
Condition 2:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→a f(x) exists at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a.
Condition 3:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→a f(x)=f(a).
If a function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( x ) is not continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a ,the function is discontinuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a.
## Identifying a Jump Discontinuity
Discontinuity can occur in different ways. We saw in the previous section that a function could have a left-hand limit and aright-hand limit even if they are not equal. If the left- and right-hand limits exist but are different, the graph “jumps” at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a. The function is said to have a jump discontinuity.
As an example, look at the graph of the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] y=f( x ) in [link]. Notice as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>x</mi><mtext> </mtext></annotation-xml></semantics>[/itex]approaches<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>a</mi><mtext> </mtext></annotation-xml></semantics>[/itex]how the output approaches different values from the left and from the right.
<figure class="small" id="CNX_Precalc_Figure_12_03_010"> <figcaption>Graph of a function with a jump discontinuity.</figcaption> </figure>
Jump Discontinuity
A function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( x ) has a jump discontinuity at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a if the left- and right-hand limits both exist but are not equal:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ a − f(x)≠ lim x→ a + f(x).
## Identifying Removable Discontinuity
Some functions have a discontinuity, but it is possible to redefine the function at that point to make it continuous. This type of function is said to have a removable discontinuity. Let’s look at the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] y=f( x ) represented by the graph in [link]. The function has a limit. However, there is a hole at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a. The hole can be filled by extending the domain to include the input<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a and defining the corresponding output of the function at that value as the limit of the function at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a.
<figure class="small" id="CNX_Precalc_Figure_12_03_011"> <figcaption>Graph of function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>f</mi><mtext> </mtext></annotation-xml></semantics>[/itex]with a removable discontinuity at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a.</figcaption> </figure>
Removable Discontinuity
A function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( x ) has a removable discontinuity at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a if the limit,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→a f(x) , exists, but either
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] a ) does not exist or
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] a ), the value of the function at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a does not equal the limit,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(a)≠ lim x→a f(x).
Identifying Discontinuities
Identify all discontinuities for the following functions as either a jump or a removable discontinuity.
1. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 −2x−15 x−5
2. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>g</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mrow><mo>{</mo></mrow></mrow></annotation-xml></semantics>[/itex] x+1, x<2 −x, x≥2
1. Notice that the function is defined everywhere except at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=5.
Thus,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( 5 ) does not exist, Condition 2 is not satisfied. Since Condition 1 is satisfied, the limit as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>x</mi><mtext> </mtext></annotation-xml></semantics>[/itex]approaches 5 is 8, and Condition 2 is not satisfied.This means there is a removable discontinuity at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=5.
2. Condition 2 is satisfied because<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] g(2)=−2.
Notice that the function is a piecewise function, and for each piece, the function is defined everywhere on its domain. Let’s examine Condition 1 by determining the left- and right-hand limits as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>x</mi><mtext> </mtext></annotation-xml></semantics>[/itex]approaches 2.
Left-hand limit:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 2 − ( x+1 )=2+1=3. The left-hand limit exists.
Right-hand limit:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 2 + ( −x )=−2. The right-hand limit exists. But
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><munder><mrow><mi>lim</mi></mrow></munder></mrow></annotation-xml></semantics>[/itex] x→ 2 − f(x)≠ lim x→ 2 + f(x).
So,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→2 f(x) does not exist, and Condition 2 fails: There is no removable discontinuity. However, since both left- and right-hand limits exist but are not equal, the conditions are satisfied for a jump discontinuity at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=2.
Identify all discontinuities for the following functions as either a jump or a removable discontinuity.
1. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 −6x x−6
2. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>g</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mrow><mo>{</mo></mrow></mrow></annotation-xml></semantics>[/itex] x , 0≤x<4 2x, x≥4
1. removable discontinuity at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=6;
2. jump discontinuity at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=4
# Recognizing Continuous and Discontinuous Real-Number Functions
Many of the functions we have encountered in earlier chapters are continuous everywhere. They never have a hole in them, and they never jump from one value to the next. For all of these functions, the limit of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( x ) as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>x</mi><mtext> </mtext></annotation-xml></semantics>[/itex]approaches<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>a</mi><mtext> </mtext></annotation-xml></semantics>[/itex]is the same as the value of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( x ) when<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a. So<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→a f(x)=f(a). There are some functions that are continuous everywhere and some that are only continuous where they are defined on their domain because they are not defined for all real numbers.
Examples of Continuous Functions
The following functions are continuous everywhere:
Polynomial functions Ex:[/itex] f(x)= x 4 −9 x 2 Exponential functions Ex:[/itex] f(x)= 4 x+2 −5 Sine functions Ex:[/itex] f(x)=sin( 2x )−4 Cosine functions Ex:[/itex] f(x)=−cos( x+ π 3 )
The following functions are continuous everywhere they are defined on their domain:
Logarithmic functions Ex:[/itex] f(x)=2ln( x ),[/itex] x>0 Tangent functions Ex:[/itex] f(x)=tan( x )+2,[/itex] x≠ π 2 +kπ,k[/itex]is an integer Rational functions Ex:[/itex] f(x)= x 2 −25 x−7 ,[/itex] x≠7
Given a function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( x ), determine if the function is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a.
1. Check Condition 1:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(a) exists.
2. Check Condition 2:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→a f(x) exists at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a.
3. Check Condition 3:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→a f(x)=f(a).
4. If all three conditions are satisfied, the function is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a. If any one of the conditions is not satisfied, the function is not continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a.
Determining Whether a Piecewise Function is Continuous at a Given Number
Determine whether the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)={ 4x, x≤3 8+x, x>3 is continuous at
1. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mo>=</mo><mn>3</mn></mrow></annotation-xml></semantics>[/itex]
2. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 8 3
To determine if the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>f</mi><mtext> </mtext></annotation-xml></semantics>[/itex]is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a, we will determine if the three conditions of continuity are satisfied at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a.
1. Condition 1: Does<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(a) exist?
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>f</mi><mo stretchy="false">(</mo><mn>3</mn><mo stretchy="false">)</mo><mo>=</mo><mn>4</mn><mo stretchy="false">(</mo><mn>3</mn><mo stretchy="false">)</mo><mo>=</mo><mn>12</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] ⇒Condition 1 is satisfied.
Condition 2: Does<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→3 f(x) exist?
To the left of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=3,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)=4x; to the right of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=3,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)=8+x. We need to evaluate the left- and right-hand limits as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>x</mi><mtext> </mtext></annotation-xml></semantics>[/itex]approaches 1.
Left-hand limit:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 3 − f(x)= lim x→ 3 − 4(3)=12
Right-hand limit:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 3 + f(x)= lim x→ 3 + ( 8+x )=8+3=11
Because<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 1 − f(x)≠ lim x→ 1 + f(x),<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→1 f(x) does not exist.
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">⇒</mo><mtext> Condition 2 fails</mtext><mtext>.</mtext></mrow></annotation-xml></semantics>[/itex]
There is no need to proceed further. Condition 2 fails at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=3. If any of the conditions of continuity are not satisfied at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=3, the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( x ) is not continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=3.
2. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 8 3
Condition 1: Does<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( 8 3 ) exist?
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 8 3 )=4( 8 3 )= 32 3 ⇒Condition 1 is satisfied.
Condition 2: Does<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 8 3 f(x) exist?
To the left of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x= 8 3 ,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)=4x; to the right of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x= 8 3 ,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)=8+x. We need to evaluate the left- and right-hand limits as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>x</mi><mtext> </mtext></annotation-xml></semantics>[/itex]approaches<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] 8 3 .
Left-hand limit:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 8 3 − f(x)= lim x→ 8 3 − 4( 8 3 )= 32 3
Right-hand limit:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 8 3 + f(x)= lim x→ 8 3 + ( 8+x )=8+ 8 3 = 32 3
Because<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 8 3 f(x) exists,
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">⇒</mo><mtext>Condition 2 is satisfied</mtext><mo>.</mo></mrow></annotation-xml></semantics>[/itex]
Condition 3: Is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( 8 3 )= lim x→ 8 3 f(x)?
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 32 3 )= 32 3 = lim x→ 8 3 f(x) ⇒Condition 3 is satisfied.
Because all three conditions of continuity are satisfied at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x= 8 3 , the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( x ) is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x= 8 3 .
Determine whether the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)={ 1 x , x≤2 9x−11.5, x>2 is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=2.
yes
Determining Whether a Rational Function is Continuous at a Given Number
Determine whether the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)= x 2 −25 x−5 is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=5.
To determine if the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>f</mi><mtext> </mtext></annotation-xml></semantics>[/itex]is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=5, we will determine if the three conditions of continuity are satisfied at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=5.
Condition 1:
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><mi>f</mi><mo stretchy="false">(</mo><mn>5</mn><mo stretchy="false">)</mo><mtext> does not exist</mtext><mtext>.</mtext></mtd></mtr></mtable></annotation-xml></semantics>[/itex] ⇒Condition 1 fails.
There is no need to proceed further. Condition 2 fails at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=5. If any of the conditions of continuity are not satisfied at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=5,the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>f</mi><mtext> </mtext></annotation-xml></semantics>[/itex]is not continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=5.
Analysis
See [link]. Notice that for Condition 2 we have
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><munder><mrow><mi>lim</mi></mrow></munder></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x→5 x 2 −25 x−5 = lim x→3 (x−5) (x+5) x−5 = lim x→5 (x+5) =5+5=10 ⇒Condition 2 is satisfied.
At<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=5,there exists a removable discontinuity. See [link].
<figure class="small" id="CNX_Precalc_Figure_12_03_013"></figure>
Determine whether the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)= 9− x 2 x 2 −3x is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=3. If not, state the type of discontinuity.
No, the function is not continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=3. There exists a removable discontinuity at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=3.
# Determining the Input Values for Which a Function Is Discontinuous
Now that we can identify continuous functions, jump discontinuities, and removable discontinuities, we will look at more complex functions to find discontinuities. Here, we will analyze a piecewise function to determine if any real numbers exist where the function is not continuous. A piecewise function may have discontinuities at the boundary points of the function as well as within the functions that make it up.
To determine the real numbers for which a piecewise function composed of polynomial functions is not continuous, recall that polynomial functions themselves are continuous on the set of real numbers. Any discontinuity would be at the boundary points. So we need to explore the three conditions of continuity at the boundary points of the piecewise function.
Given a piecewise function, determine whether it is continuous at the boundary points.
For each boundary point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>a</mi><mtext> </mtext></annotation-xml></semantics>[/itex]of the piecewise function, determine the left- and right-hand limits as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>x</mi><mtext> </mtext></annotation-xml></semantics>[/itex]approaches<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] a,as well as the function value at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] a.
Check each condition for each value to determine if all three conditions are satisfied.
Determine whether each value satisfies condition 1:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(a) exists.
Determine whether each value satisfies condition 2:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→a f(x) exists.
Determine whether each value satisfies condition 3:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→a f(x)=f(a).
If all three conditions are satisfied, the function is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a. If any one of the conditions fails, the function is not continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a.
Determining the Input Values for Which a Piecewise Function Is Discontinuous
Determine whether the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>f</mi><mtext> </mtext></annotation-xml></semantics>[/itex]is discontinuous for any real numbers.
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mi>x</mi><mo>=</mo><mrow><mo>{</mo></mrow></mrow></annotation-xml></semantics>[/itex] x+1, x<2 3, 2≤x<4 x 2 −11, x≥4
The piecewise function is defined by three functions, which are all polynomial functions,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)=x+1 on<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x<2,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)=3 on<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] 2≤x<4,and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)= x 2 −5 on<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x≥4. Polynomial functions are continuous everywhere. Any discontinuities would be at the boundary points,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=2 and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=4.
At<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=2, let us check the three conditions of continuity.
Condition 1:
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><mi>f</mi><mrow><mo>(</mo></mrow></mtd></mtr></mtable></annotation-xml></semantics>[/itex] 2 )=3 ⇒Condition 1 is satisfied.
Condition 2: Because a different function defines the output left and right of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=2,does<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 2 − f(x)= lim x→ 2 + f(x)?
Left-hand limit:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 2 − f(x)= lim x→ 2 − ( x+1 )=2+1=3
Right-hand limit: <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 2 + f(x)= lim x→ 2 + 3=3
Because<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] 3=3,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 2 − f(x)= lim x→ 2 + f(x)
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">⇒</mo><mtext>Condition 2 is satisfied</mtext><mo>.</mo></mrow></annotation-xml></semantics>[/itex]
Condition 3:
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><munder><mrow><mi>lim</mi></mrow></munder></mtd></mtr></mtable></annotation-xml></semantics>[/itex] x→2 f(x)=3=f(2) ⇒Condition 3 is satisfied.
Because all three conditions are satisfied at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=2,the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( x ) is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=2.
At<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=4,let us check the three conditions of continuity.
Condition 2: Because a different function defines the output left and right of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=4,does<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 4 − f(x)= lim x→ 4 + f(x)?
Left-hand limit:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 4 − f(x)= lim x→ 4 − 3=3
Right-hand limit:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 4 + f(x)= lim x→ 4 + ( x 2 −11 )= 4 2 −11=5
Because<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] 3≠5,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ 4 − f(x)≠ lim x→ 4 + f(x) ,so<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→4 f(x) does not exist.
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">⇒</mo><mtext>Condition 2 fails</mtext><mo>.</mo></mrow></annotation-xml></semantics>[/itex]
Because one of the three conditions does not hold at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=4,the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x) is discontinuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=4.
Analysis
See [link]. At<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=4,there exists a jump discontinuity. Notice that the function is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=2.
<figure class="small" id="CNX_Precalc_Figure_12_03_014"> <figcaption>Graph is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=2 but shows a jump discontinuity at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=4.</figcaption> </figure>
Determine where the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)={ πx 4 , x<2 π x , 2≤x≤6 2πx, x>6 is discontinuous.
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mo>=</mo><mn>6</mn></mrow></annotation-xml></semantics>[/itex]
# Determining Whether a Function Is Continuous
To determine whether a piecewise function is continuous or discontinuous, in addition to checking the boundary points, we must also check whether each of the functions that make up the piecewise function is continuous.
Given a piecewise function, determine whether it is continuous.
1. Determine whether each component function of the piecewise function is continuous. If there are discontinuities, do they occur within the domain where that component function is applied?
2. For each boundary point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a of the piecewise function, determine if each of the three conditions hold.
Determining Whether a Piecewise Function Is Continuous
Determine whether the function below is continuous. If it is not, state the location and type of each discontinuity.
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mi>x</mi><mo>=</mo><mrow><mo>{</mo></mrow></mrow></annotation-xml></semantics>[/itex] sin(x), x<0 x 3 , x>0
The two functions composing this piecewise function are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)=sin(x) on<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x<0 and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)= x 3 on<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x>0. The sine function and all polynomial functions are continuous everywhere. Any discontinuities would be at the boundary point,
At<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=0,let us check the three conditions of continuity.
Condition 1:
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>f</mi><mo stretchy="false">(</mo><mn>0</mn><mo stretchy="false">)</mo><mtext> does not exist</mtext><mo>.</mo></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] ⇒Condition 1 fails.
Because all three conditions are not satisfied at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=0, the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x) is discontinuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=0.
Analysis
See [link]. There exists a removable discontinuity at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=0;<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→0 f(x)=0, thus the limit exists and is finite, but<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( a ) does not exist.
<figure class="small" id="CNX_Precalc_Figure_12_03_015"> <figcaption>Function has removable discontinuity at 0.</figcaption> </figure>
Access these online resources for additional instruction and practice with continuity.
# Key Concepts
• A continuous function can be represented by a graph without holes or breaks.
• A function whose graph has holes is a discontinuous function.
• A function is continuous at a particular number if three conditions are met:
• Condition 1:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(a) exists.
• Condition 2:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→a f(x) exists at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a.
• Condition 3:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→a f(x)=f(a).
• A function has a jump discontinuity if the left- and right-hand limits are different, causing the graph to “jump.”
• A function has a removable discontinuity if it can be redefined at its discontinuous point to make it continuous. See[link].
• Some functions, such as polynomial functions, are continuous everywhere. Other functions, such as logarithmic functions, are continuous on their domain. See [link] and [link].
• For a piecewise function to be continuous each piece must be continuous on its part of the domain and the function as a whole must be continuous at the boundaries. See [link] and [link].
# Section Exercises
## Verbal
State in your own words what it means for a function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>f</mi><mtext> </mtext></annotation-xml></semantics>[/itex]to be continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=c.
Informally, if a function is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=c , then there is no break in the graph of the function at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( c ), and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( c ) is defined.
State in your own words what it means for a function to be continuous on the interval<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] ( a,b ).
## Algebraic
For the following exercises, determine why the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>f</mi><mtext> </mtext></annotation-xml></semantics>[/itex]is discontinuous at a given point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>a</mi><mtext> </mtext></annotation-xml></semantics>[/itex]on the graph. State which condition fails.
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mi>ln</mi><mtext> </mtext><mo>|</mo><mtext> </mtext><mi>x</mi><mo>+</mo><mn>3</mn><mtext> </mtext><mo>|</mo><mo>,</mo><mi>a</mi><mo>=</mo><mo>−</mo><mn>3</mn></mrow></annotation-xml></semantics>[/itex]
discontinuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] a=−3;<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(−3) does not exist
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mi>ln</mi><mtext> </mtext><mo>|</mo><mtext> </mtext><mn>5</mn><mi>x</mi><mo>−</mo><mn>2</mn><mtext> </mtext><mo>|</mo><mo>,</mo><mi>a</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 2 5
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 −16 x+4 ,a=−4
removable discontinuity at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] a=−4;<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(−4) is not defined
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 −16x x ,a=0
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )={ x, x≠3 2x,x=3 a=3
Discontinuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] a=3;<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→3 f(x)=3 , but<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(3)=6 , which is not equal to the limit.
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )={ 5, x≠0 3, x=0 a=0
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )={ 1 2−x , x≠2 3, x=2 a=2
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><munder><mrow><mi>lim</mi></mrow></munder></mrow></annotation-xml></semantics>[/itex] x→2 f(x) does not exist.
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )={ 1 x+6 , x=−6 x 2 , x≠−6 a=−6
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )={ 3+x, x<1 x, x=1 x 2 , x>1 a=1
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><munder><mrow><mi>lim</mi></mrow></munder></mrow></annotation-xml></semantics>[/itex] x→ 1 − f(x)=4; lim x→ 1 + f(x)=1. Therefore,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→1 f(x) does not exist.
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )={ 3−x, x<1 x, x=1 2 x 2 , x>1 a=1
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )={ 3+2x, x<1 x, x=1 − x 2 , x>1 a=1
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><munder><mrow><mi>lim</mi></mrow></munder></mrow></annotation-xml></semantics>[/itex] x→ 1 − f(x)=5≠ lim x→ 1 + f(x)=−1. Thus<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→1 f(x) does not exist.
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )={ x 2 , x<−2 2x+1, x=−2 x 3 , x>−2 a=−2
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )={ x 2 −9 x+3 , x<−3 x−9, x=−3 1 x , x>−3 a=−3
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><munder><mrow><mi>lim</mi></mrow></munder></mrow></annotation-xml></semantics>[/itex] x→− 3 − f(x)=−6,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→− 3 + f(x)=− 1 3
Therefore,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→−3 f(x) does not exist.
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )={ x 2 −9 x+3 , x<−3 x−9, x=−3 −6, x>−3 a=3
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= x 2 −4 x−2 , a=2
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2 ) is not defined.
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= 25− x 2 x 2 −10x+25 , a=5
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= x 3 −9x x 2 +11x+24 , a=−3
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −3 ) is not defined.
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= x 3 −27 x 2 −3x , a=3
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] x |x| , a=0
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0 ) is not defined.
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= 2| x+2 | x+2 , a=−2
For the following exercises, determine whether or not the given function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>f</mi><mtext> </mtext></annotation-xml></semantics>[/itex]is continuous everywhere. If it is continuous everywhere it is defined, state for what range it is continuous. If it is discontinuous, state where it is discontinuous.
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= x 3 −2x−15
Continuous on<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] (−∞,∞)
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= x 2 −2x−15 x−5
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )=2⋅ 3 x+4
Continuous on<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] (−∞,∞)
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )=−sin( 3x )
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= | x−2 | x 2 −2x
Discontinuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=0 and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=2
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )=tan( x )+2
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )=2x+ 5 x
Discontinuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=0
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= log 2 ( x )
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mi>ln</mi><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] x 2
Continuous on<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] (0,∞)
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= e 2x
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] x−4
Continuous on<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] [4,∞)
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )=sec( x )−3.
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= x 2 +sin( x )
Continuous on<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] (−∞,∞).
Determine the values of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>b</mi><mtext> </mtext></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>c</mi><mtext> </mtext></annotation-xml></semantics>[/itex]such that the following function is continuous on the entire real number line.
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mrow><mrow/></mrow></mrow></annotation-xml></semantics>[/itex] { x+1, 1<x<3 x 2 +bx+c, | x−2 |≥1 }
## Graphical
For the following exercises, refer to [link]. Each square represents one square unit. For each value of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>a</mi><mo>,</mo></annotation-xml></semantics>[/itex] determine which of the three conditions of continuity are satisfied at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a and which are not.
<figure class="small" id="CNX_Precalc_Figure_12_03_201"></figure>
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mo>=</mo><mo>−</mo><mn>3</mn></mrow></annotation-xml></semantics>[/itex]
1, but not 2 or 3
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mo>=</mo><mn>2</mn></mrow></annotation-xml></semantics>[/itex]
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mo>=</mo><mn>4</mn></mrow></annotation-xml></semantics>[/itex]
1 and 2, but not 3
For the following exercises, use a graphing utility to graph the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)=sin( 12π x ) as in [link]. Set the x-axis a short distance before and after 0 to illustrate the point of discontinuity.
<figure class="small" id="CNX_Precalc_Figure_12_03_202"></figure>
Which conditions for continuity fail at the point of discontinuity?
Evaluate<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(0).
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0 ) is undefined.
Solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>x</mi><mtext> </mtext></annotation-xml></semantics>[/itex]if<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)=0.
What is the domain of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( x )?
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">(</mo><mo>−</mo><mi>∞</mi><mo>,</mo><mn>0</mn><mo stretchy="false">)</mo><mo>∪</mo><mo stretchy="false">(</mo><mn>0</mn><mo>,</mo><mi>∞</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]
For the following exercises, consider the function shown in [link].
<figure class="small" id="CNX_Precalc_Figure_12_03_203"></figure>
At what x-coordinates is the function discontinuous?
What condition of continuity is violated at these points?
At<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=−1,the limit does not exist. At<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=1,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( 1 ) does not exist.
At<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=2, there appears to be a vertical asymptote, and the limit does not exist.
Consider the function shown in [link]. At what x-coordinates is the function discontinuous? What condition(s) of continuity were violated?
<figure class="small" id="CNX_Precalc_Figure_12_03_204"></figure>
Construct a function that passes through the origin with a constant slope of 1, with removable discontinuities at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=−7 and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=1.
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 3 +6 x 2 −7x ( x+7 )( x−1 )
The function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)= x 3 −1 x−1 is graphed in [link]. It appears to be continuous on the interval<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] [ −3,3 ], but there is an x-value on that interval at which the function is discontinuous. Determine the value of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>x</mi><mtext> </mtext></annotation-xml></semantics>[/itex]at which the function is discontinuous, and explain the pitfall of utilizing technology when considering continuity of a function by examining its graph.
<figure class="small" id="CNX_Precalc_Figure_12_03_205"></figure>
Find the limit<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→1 f(x) and determine if the following function is continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=1:
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mi>x</mi><mo>=</mo><mrow><mo>{</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 2 +4 x≠1 2 x=1
The function is discontinuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=1 because the limit as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>x</mi><mtext> </mtext></annotation-xml></semantics>[/itex]approaches 1 is 5 and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( 1 )=2.
The graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f(x)= sin(2x) x is shown in [link]. Is the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( x ) continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=0? Why or why not?
<figure class="medium" id="CNX_Precalc_Figure_12_03_206"></figure>
## Glossary
continuous function
a function that has no holes or breaks in its graph
discontinuous function
a function that is not continuous at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a
jump discontinuity
a point of discontinuity in a function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( x ) at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] x=a where both the left and right-hand limits exist, but<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] lim x→ a − f(x)≠ lim x→ a + f(x)
removable discontinuity
a point of discontinuity in a function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mrow/></annotation-xml></semantics>[/itex] f( x ) where the function is discontinuous, but can be redefined to make it continuous | 27,758 | 87,990 | {"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} | 3.859375 | 4 | CC-MAIN-2017-04 | latest | en | 0.36353 |
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*** BEST SELLER***Help your students understand how to apply the DISTRIBUTIVE PROPERTY OF MULTIPLICATION. This resource is perfect for students who are struggling to understand and remember the DISTRIBUTIVE PROPERTY. It begins with concrete (manipulatives) activities, progresses to the representatio
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These math centers are engaging activities that encourage math talk. Students explain how they sort their cards and build strong math concepts as they create cards of their own.What is Included?- 81 sorts that review third grade math standards. - Math Sorts Binder - Pages that list all math sorts,
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Relations, Functions, Domain and Range Task Cards These 20 task cards cover the following objectives: 1) Identify the domain and range of ordered pairs, tables, mappings, graphs, and equations. 2) Determine whether a relation is a function given ordered pairs, tables, mappings, graphs, and equati
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This fun and differentiated JAM-PACKED unit is filled with everything you need to teach 4 standards/concepts: missing addends, the commutative property of addition, fact families, and true or false with addition equations.Get this in the First Grade Math Units 1-9 BUNDLE to save a ton!Everything you
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Printable Middle School File Folder Math Games - Math Teachers, Look No Further!42 printable math games for upper elementary and middles school students, easy-to-setup for any math class, when laminated these can be made into math board games for middle school that can be used over and over again.WH
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Volume of a Rectangular Prism - Hands-on Math Center is a great way for your students to practice finding volume by building shapes using snap cubes. Students are given cards with a length, width and a height on it and the students build it using snap cubes. Then they calculate the volume by eithe
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848
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Use this color by number activity for homework, center work, or for extra credit. Includes problems with both adding and subtracting integers. Way more fun than "another worksheet". Perfect to leave with a substitute or for those last few days of school.
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This fun and differentiated JAM-PACKED unit is filled with everything you need to teach 4 standards/concepts: true or false with subtraction equations, missing subtrahends/minuends, choosing an operation (add or subtract), and balancing equations.Everything you need to print and go to teach these co
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Digital and Copy Ready Multiplication Activities! Newly edited to include a digital version for Distance Learning! This update is designed to work with Google™ Slides, Google™ Drive, Microsoft OneDrive, and Seesaw.Multiplication Facts just got a lot more interesting! Grab this CLASS FAVORITE for
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With references for slope, y-intercept, x-intercept, origin, grid, graph, table, x and y axis, domain and range, increasing and decreasing, quadratics, graphing exponential functions, systems of equations, systems of inequalities, parallel and perpendicular lines on the coordinate plane, compound in
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Also included in: Grades 6-9 Math Word Wall Bundle - print and digital
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708
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Algebra 1 foldables for interactive notebooks!!! This MEGA BUNDLE includes over 250 foldables (and counting) for various middle school and high school math concepts. Most foldables align to Pre-Algebra (8th Grade Math) and Algebra 1 curriculum. Download the preview to see a list of topics covere
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Multi-Step Equations Math Lib Math lib activities are a class favorite! Similar to a "mad lib", students generate pieces to a story as they move throughout the stations solving multi-step equations. There are 9 stations, with 9 equations at each station. As a group of 3-4, students choose one eq
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A set of 32 task cards for writing expressions with variables. Answer Key & Recording Sheet Included Common Core Aligned 6.EE.A.2 6.EE.2 5.OA.A.2 Looking for these task cards WITHOUT variables? Click Here! **Looking for MORE Task Cards?** Order of Operations Task Cards WITHOUT Parenth
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Daily Math Review: This spiral review is one of the easiest ways to make sure you are covering all of the important math concepts that your students need to know. Perfect for morning work, “bell-ringers,” homework, center work, or test prep. This resource is designed to be used on a daily basis (Mo
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582
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Are you using INTERACTIVE MATH NOTEBOOKS in your classroom? You really should be! My students love them, create them, reference them, and will take them home at the end of the school year where they will continue to learn from them! They can be used in either a whole group or small group setting. Th
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543
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This unit can be used to teach patterns and functions over a 2 week period. It also contains many activities that can be played as a whole group, in partners, in centers, or played later in the year as a way to review for testing. PLEASE DOWNLOAD THE PREVIEW in order to see some of what is include
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499
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This is a new take on an old favorite. This game for two players has students taking turns drawing lines on a grid. Each time a player closes a box with the fourth side, the player takes the box, and goes again. But in this game, the box contains an integer which the player must then add to his o
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495
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This is a set of math puzzle activities in which students will help to solve the mystery of Pete's Pet. Students will collect clues by completing various puzzles using their skills in adding and subtracting integers. Like the classic board game CLUE, they will then use the process of elimination t
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Included here are 3 algebra notebook flippables for the slope formula, the point-slope formula and the slope-intercept formula. Two versions of each are included-- one version that is filled in with information and one version that is blank other than the variables.The flippables are easily assemble
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480
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This slope intercept form activity is a set of matching cards that work on Linear Functions. The are 12 sets for a total of 48 cards. Each set has a graph, table, equation, and verbal description. These are intended to be cut out and used as a matching activity. The numbers and letters on the card
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474
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Box and whisker plots pose many problems for students. They often confuse the length of a quartile for the size of data it includes. After seeing my students struggle with this concept for years (even more-so because they had to analyze the data, not merely create the plot) I came up with this acti
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438
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Decomposing Numbers Is the thought of teaching your students about decomposing numbers make you want to SCREAM??!!! If so, make a decomposing numbers monster to show your students, visually, how to break numbers apart! The only 'screams' that will be heard are those of joy and excitement! This M
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423
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Are you looking for a way to practice math skills that is more engaging than worksheets? Would you love the ability to easily make the activity an independent activity or a partner activity? These Roll and Answer math activities are an easy way to spice up much-needed math practice and have a lot of
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This proportional relationships activity bundle includes 6 classroom activities to support unit rates with and without fractions, the constant of proportionality, proportional relationships in tables, graphs, equations, and verbal descriptions, and determining if a relationship is proportional or no
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 2,154 | 9,004 | {"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} | 2.515625 | 3 | CC-MAIN-2020-34 | latest | en | 0.841852 |
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Math Definitions -> D -> Digit
# Digit
A symbol used to make numerals.
0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 are the ten digits we use in everyday numbers.
Example: the numeral 153 is made up of 3 digits ("1", "5" and "3").
See: Numeral | 96 | 271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | longest | en | 0.71703 |
https://machinelearning.org.in/numpy-datatypes/ | 1,723,778,917,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00505.warc.gz | 292,086,373 | 114,528 | Breaking News
# Numpy Datatypes
3 – Numpy Data Types
### NumPy Data Types¶
In [2]:
import numpy as np
import pandas as pd
### Data Types in NumPy¶
Numpy has the following data types:
• int
• float
• complex
• bool
• string
• unicode
• object
The numeric data types have various precisions like 32-bit or 64-bit.
Numpy data types can be represented using either Type or Type Code
In [3]:
dtypes = pd.DataFrame(
{
'Type': [
'int8',
'uint8',
'int16',
'uint16',
'int or int32',
'uint32',
'int64',
'uint64',
'float16',
'float32',
'float or float64',
'float128',
'complex64',
'complex or complex128',
'bool',
'object',
'string_',
'unicode_',
],
'Type Code': [
'i1',
'u1',
'i2',
'u2',
'i4 or i',
'u4',
'i8',
'u8',
'f2',
'f4 or f',
'f8 or d',
'f16 or g',
'c8',
'c16',
None,
'O',
'S',
'U',
]
}
)
dtypes
Out[3]:
Type Type Code
0 int8 i1
1 uint8 u1
2 int16 i2
3 uint16 u2
4 int or int32 i4 or i
5 uint32 u4
6 int64 i8
7 uint64 u8
8 float16 f2
9 float32 f4 or f
10 float or float64 f8 or d
11 float128 f16 or g
12 complex64 c8
13 complex or complex128 c16
14 bool None
15 object O
16 string_ S
17 unicode_ U
Data types can be defined at creating the numpy array and converted to other types later.
You can use either type, type code or np dot methods to define the data type of an array, but when you use np dot method to define the data type, it can only follow type rather than type code.
In [4]:
arr = np.array([1,2,3], dtype='f4')
arr.dtype
Out[4]:
dtype('float32')
In [5]:
# Identical to the above
arr = np.array([1,2,3], dtype='float32')
arr.dtype
Out[5]:
dtype('float32')
In [6]:
arr = np.array([1+2j, 3-4j], dtype=np.complex64)
arr.dtype
Out[6]:
dtype('complex64')
In [7]:
# Identical to the above
arr = np.array([1+2j, 3-4j], dtype='c8')
arr.dtype
Out[7]:
dtype('complex64')
In [8]:
# ERROR
arr = np.array([1+2j, 3-4j], dtype=np.c8)
arr.dtype
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
<ipython-input-8-a5f30f9069f2> in <module>
1 # ERROR
----> 2 arr = np.array([1+2j, 3-4j], dtype=np.c8)
3 arr.dtype
AttributeError: module 'numpy' has no attribute 'c8'
### Type Conversion¶
astype method: convert the data type of an array to other data types.
Notice that astype returns a copy of the array instead of converting the data type in place. You need to assign the copy to the original array or a new array.
In [9]:
arr = np.array([1,2,3], dtype='int16')
print('Original Data Type: ' + str(arr.dtype))
arr = arr.astype(np.float32)
print('Data Type After Conversion: ' + str(arr.dtype))
Original Data Type: int16
Data Type After Conversion: float32
WARNING: be cautious about data overflow when you downcast the data type (from higher precision to lower precision). Some unexpected and undefined values might occur and it is usually difficult to debug such issues.
In [10]:
# An example of integer overflow at downcasting
arr = np.array([126,127,256], dtype='int16')
print('np array before type conversion: ' + str(arr))
# Range of int8 [-128, 127], 256 overflows after conversion
arr = arr.astype('int8')
print('np array after type conversion: ' + str(arr))
np array before type conversion: [126 127 256]
np array after type conversion: [126 127 0]
### String and Unicode Data Type¶
The string_ and unicode_ data types are all implicitly fixed-length.
The length of the string is given by their type code appended with a number. For example, S3 represents string of length 3; U10 represents unicode of length 10. Otherwise, the default length is the length of the longest string in the array.
If the length of a string in the array is shorter than the length of the data type defined or converted to, the string will be truncated.
In [11]:
# An example of truncated string
s = np.array(['abc', 'defg'], dtype='S3')
print(s)
# An example of truncated unicode
s = np.array(['abcd', 'efghi'], dtype='U3')
print(s)
[b'abc' b'def']
['abc' 'efg']
In [12]:
arr = np.array(['a', 'ab', 'abc'], dtype=np.string_)
print('The array is ' + str(arr))
print('The data type is ' + str(arr.dtype) + ' because the longest string in the array is "abc" and its length is 3.')
arr = np.array(['a', 'abc', 'abcd'], dtype=np.unicode_)
print('The array is ' + str(arr))
print('The data type is ' + str(arr.dtype) + ' because the longest unicode in the array is "abcd" and its length is 4.')
The array is [b'a' b'ab' b'abc']
The data type is |S3 because the longest string in the array is "abc" and its length is 3.
The array is ['a' 'abc' 'abcd']
The data type is <U4 because the longest unicode in the array is "abcd" and its length is 4.
What do “|” and “<” in the data types above mean?
They are the byte order indicators, which are beyond the scope of this tutorial.
Further readings if you are interested: | 1,414 | 4,833 | {"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} | 2.859375 | 3 | CC-MAIN-2024-33 | latest | en | 0.333853 |
https://techcommunity.microsoft.com/t5/excel/creating-a-top-of-the-charts-board/td-p/1924128 | 1,620,755,762,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991648.10/warc/CC-MAIN-20210511153555-20210511183555-00171.warc.gz | 619,922,908 | 91,958 | New Contributor
# creating a top of the charts board
hey guys, I produce a leaderboard each month for our top salespeople. its been run the same way for years. but I wanted to give it an overhaul but coming up against my poor excel know-how. I want to create a "top of the charts" league with each individual results with an arrow up and down to show if they have moved in the charts month vs month, ideally with a figure. eg. UP6 or DOWN3
I hope it will be more motivating to our newer starters to see if they are moving up.
this is how I have the table set up so far. I receive the raw data once a month and I create a simple pivot table to show:
sales person deals year to date (Nov) Month Positon sales person deals year to date (Oct) Month Position Change Sales Person A 808021.9 Oct 1 Sales Person A 769494.2 Sept 1 Sales Person B 764431.1 Oct 2 Sales Person C 646092.8 Sept 2 Sales Person C 656019.5 Oct 3 Sales Person B 544869.5 Sept 3 Sales Person D 450066.2 Oct 4 Sales Person E 432537.1 Sept 4 Sales Person E 443488.1 Oct 5 Sales Person F 426903.7 Sept 5 Sales Person F 442634.9 Oct 6 Sales Person E 322833.9 Sept 6
7 Replies
# Re: creating a top of the charts board
I changed the sequence of your columns, just because it made more sense to me to show the change in connection with the most recent month's results. The idea behind the formula, though, could still work in the layout you originally had.
This is no doubt also a great example of where the new LET function would be useful. I didn't use it because I have no experience yet, but for my own learning I will go and try it. It would make the formula less lengthy, since it would no longer be necessary to repeat the "MATCH" function so many times.
# Re: creating a top of the charts board
Here's the same solution, except using LET to shorten the formula. My first opportunity to use that new function.
Here's the original:
=IFS(
MATCH(A2,\$F\$2:\$F\$7,0)<D2,"Down"&TEXT(D2-MATCH(A2,\$F\$2:\$F\$7,0),"0"),
MATCH(A2,\$F\$2:\$F\$7,0)>D2,"Up"&TEXT(MATCH(A2,\$F\$2:\$F\$7,0)-D2,"0")
)
And here it is using LET, which is a way to take that function MATCH(A2,\$F\$2:\$F\$7,0) and replace it with MtchVal
=LET(MtchVal,MATCH(A2,\$F\$2:\$F\$7,0),
IFS(
MtchVal<D2,"Down"&TEXT(D2-MtchVal,"0"),
MtchVal>D2,"Up"&TEXT(MtchVal-D2,"0")
)
)
# Re: creating a top of the charts board
Mathetes, thank you. I will study this tomorrow.
I will see if I can spot the sequence and apply it to my original with 80 sales people. Thanks again for providing the answer.
Nick
# Re: creating a top of the charts board
My solution does depend on the fact that your prior month is sorted in order by their rank in that month. (That's what the MATCH function looks at, coming up with their prior rank based on where they were in the array)
It would also be possible to look at the prior rank number and do the math comparing current rank with that number......
And, of course, presuming that there are new hires and turnover, you could add other conditions to the IFS function to cover "New" -- no need for a label to apply to somebody who has left employment.
# Re: creating a top of the charts board
Following up on your ideas! I have not relied on either the sort order or any existing calculation of the ranking for the prior month. Individuals that were not active within the previous month are ranked at n+1 for comparison purposes. The text and colours are all produced by custom number formats.
``````= LET(
priorPosns, RANK(PriorDeals, PriorDeals),
prior,
XLOOKUP(
person,
priorList,
priorPosns,
1+MAX(priorPosns) ),
change, prior - currentPosition#,
change )``````
# Re: creating a top of the charts board
I never cease to be amazed at how many ways Excel offers to accomplish any given task.
In this case, I could replicate (with a bit more practice on my own) the formula itself. But it never would have occurred to me to use custom formatting in the manner you do; nor, frankly, do I really understand it.
Again, however, I suspect that conditional formatting could serve the same purpose, less elegantly perhas, but yet again, Excel holds multiple ways to get from A to Z.
# Re: creating a top of the charts board
I know nothing of the OP's knowledge of Excel but I was confident that you could follow and might be interested. Following the (dubious) principle "If a thing is worth doing, it is worth doing to excess", I could set different number formats from a set of conditional formats. Who says the numbers never lie, in Excel what you see is not necessarily what you have.
Worse than that, it is also possible to use LET to perform the calculations that might otherwise be held in a helper range, but then display both the helper range and the final calculation as a single dynamic formula. It is like a program module written to a cell.
``````= LET(
currentPos, RANK(currentDeals, currentDeals),
priorPosns, RANK(PriorDeals, PriorDeals),
prior, XLOOKUP(person, priorList, priorPosns,
1+COUNTA(priorList) ),
change, prior - currentPos,
IF({1,0}, currentPos, change) )`````` | 1,318 | 5,080 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-21 | latest | en | 0.891745 |
https://betterlesson.com/lesson/reflection/11714/student-reflection-as-a-tool-for-deeper-understanding | 1,547,783,676,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659677.17/warc/CC-MAIN-20190118025529-20190118051529-00204.warc.gz | 448,296,501 | 38,958 | # Formative Assessment: Modeling Population Growth (A Math Assessment Project Classroom Challenge)
2 teachers like this lesson
Print Lesson
## Objective
SWBAT interpret and create equations to model a real life scenario involving a geometric sequence. SWBAT create arguments to support their reasoning.
#### Big Idea
Students model an expanding kitten population with functions!
## Entry Ticket: Having Kittens
15 minutes
The pre-test for the Having Kittens Math Assessment Project Classroom Challenge Lesson is a great way to gauge baseline understanding through an open-ended formative assessment.
Students complete this Pre-Assessment (page 11 of the Math Shell document in this resource) the previous day in class or for homework. The Pre-assessment involves students critiquing a poster from an animal shelter that makes claims about the number of descendants that one cat can produce. The reason to have students complete the task prior to the day's activity is so I can assess student understanding and plan the appropriate level of support and challenge in the lesson.
## Introduction and Questions for Deeper Understanding
10 minutes
To open class, I ask students questions based on their responses to the Pre-assessment. The lesson plan from the Math Assessment Project (resource in this section) provides an excellent table that identifies common misconceptions that students have and suggested prompts and questions to challenge those misconceptions.
I use the MAP Center resources as well as my experience with my students to generate questions. Here are some of the questions that I plan to ask my students:
• What observations did you make about the poster?
• What claims are being made in the poster? Are these claims justified with mathematics/examples/evidence?
• What assumptions are being made in the poster? Do the assumptions seem reasonable? Why or why not?
## Collaborative Work: Producing a joint solution
20 minutes
Students then work together in small groups comparing and contrasting the work from their group members to create a joint solution (Student Work Example: Joint Solution). This activity gets right at Math Practice 3 as students are set up to communicate their own reasoning and to listen and critique the reasoning of others.
During this time I am observing how groups are working together and also providing groups support in their problem-solving.
It is difficult for many students to initiate with this task. Therefore, I focus my attention to being sure every group is able to begin a conversation about comparing and contrasting ideas from the different group members. To accomplish this task, I check in with groups, and provide reflective prompts. For example, if one student identified an assumption that he/she though was not reasonable, I would point that out to the group and ask a follow up question to help scaffold the group conversation. What other assumptions did the group identify? What evidence did your classmates use to justify their opinion?
## Gallery Walk
10 minutes
Once the groups are finished with their posters I have them post them around the classroom. Groups then rotate and assess other groups posters in a short Gallery Walk. I provide students with sticky notes to write constructive criticism for the posters that they visit.
Teacher's Note: This setup varies slightly from the suggested activity in the Math Assessment Project lesson plan. I use the gallery walk because it meet a similar intent of students reviewing other groups work and providing feedback, and my classes are familiar with the structure of the gallery walk as I use that instructional strategy in other lessons.
## Comparing Different Solutions
20 minutes
Once students have completed the gallery walk and given a few minutes to read and process the feedback from their peers.
The next task at hand is for students to assess sample work in writing. I like the progression of tasks in this lesson from group work to individual work as it is flexible and provides a number of different domains (verbally, written, etc.) for students to think about the concepts in the lesson.
During this section I provide each student with one of the three sample works (pages 12-14 in the Math Shell document in this section) provided in the lesson plan.
I tend to find there is not enough time for every student to complete feedback on all three pieces of sample work. The nice part about providing each student one of the samples is I can Differentiate Instruction by giving students the sample work that gets at an aspect of the lesson at the level of understanding each student is at.
Homework for this lesson can also be for students to complete feedback on the two pieces of sample work they did not complete in class.
## Whole class discussion: Comparing other solutions
20 minutes
After students have completed their written feedback on the sample work of Alice, Wayne and Ben, I reconvene the class for a whole group discussion.
I like to use the slides provided by the Math Assessment Project as a guide to the discussion. As we progress through each of the three sample works, I have students share their feedback on each piece of work AND have students comment on any similarities or differences between the sample work and their own reasoning/thinking.
## Exit Ticket and Reflection: How Did You Work?
10 minutes
To conclude class students complete the "How Did You Work?" worksheet (page 15 of the Having Kittens: A Mathematics Assessment Project Classroom Challenge document in this section as a resource) that provides a nice support for students to engage in all important skill of reflection. Giving students time to reflect on their own work, the good and the bad, allows students to think about what they can improve on and HOW to make those improvements.
## Homework
For homework I either assign students to complete the feedback forms for the remaining two sample student work that they did not complete in class OR complete the "How Did I Work?" reflection sheet.
For this class I do not assign homework until the end of class because the assignment depends on progress made in the lesson. If there is not time to complete the reflection I make that the homework, but if the class does complete the reflection I assign the remaining sample work feedback forms.
## References
Mathematics Assessment Project (2012). Modeling: Having Kittens: A Classroom Challenge AFormative Assessment Lesson. Shell Center: University of Nottingham. Accessed online on December 16 at http://map.mathshell.org/materials/download.php?fileid=1204 | 1,252 | 6,621 | {"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} | 3.4375 | 3 | CC-MAIN-2019-04 | latest | en | 0.905113 |
https://clubessays.com/if-the-couple-moment-acting-on-the-applied-to-the-wrenches/ | 1,670,177,136,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710978.15/warc/CC-MAIN-20221204172438-20221204202438-00035.warc.gz | 205,031,217 | 11,163 | # If The Couple Moment Acting On The Applied To The Wrenches
If the couple moment acting on the pipe has a magnitude M, determine the magnitude F of the forces applied to the wrenches.
Given:
M 300 = N ∙ m
a = 150 mm
b = 150 mm
c = 200 mm
d = 600 mm
Posted in Uncategorized | 80 | 275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-49 | latest | en | 0.750464 |
https://numberdyslexia.com/dyscalculia-and-multiplication/ | 1,713,379,324,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817171.53/warc/CC-MAIN-20240417173445-20240417203445-00894.warc.gz | 398,390,722 | 14,706 | # Dyscalculia And Multiplication: How to manage?
Last Updated on October 4, 2023 by Editorial Team
REVIEWED BY NUMBERDYSLEXIA’S EXPERT PANEL ON MARCH 21, 2022
Operations may be perceived to be the foundational building blocks of math. One among these, multiplication has a unique essence not only due to the area of its application but also due to the abilities it obligates the student to grasp it. This operation can be implemented for simple daily tasks or complex professional applications.
While complexities in multiplication skills are occasional for some students, is it an area of concern for Dyscalculics? Why is multiplication more than just an operation? As we comprehend that definitive inference may be challenging to derive, the insights provided here can, however, assist you in learning better about how Dyscalculics can deal with multiplication.
## Multiplication- A cut above operation?
While there are four basic operations in math, Multiplication is one such distinct pick where the student may need to learn some additional traits. These problems can be addressed in two ways. Firstly, the young learners use the counting method in which a number is added multiple times to outstretch the answer. Secondly, the students may employ a retrieval process in which they can retain some previously learned regulations and facts to solve swiftly.
The operations in later stages ultimately obligate retrieval of multiplication facts to ensure precise estimation. These facts may not be applicable in all cases; nonetheless, comprehending these can assist complex arithmetics later. For this reason, Multiplication is often more than just an operation. Here are a couple of pivotal facts:
• The product of two numbers always gives larger outputs, but when multiplied by zero, the answer is ‘0.’
• For every Multiplication fact, there are two division facts. For instance, 2*4=8, this also implies ‘8/2=4’ and ‘8/4=2.’
Complications in multiplication often arise owing to the missout of such certainties. John B. Cooney[1] made a study on third and fifth-grade students to analyze their multiplication skills. It was noticed that as the pupils transitioned towards the fourth grade, they opted for retrieval strategies to solve problems- showing how retrieval strategies are crucial in higher grades. Finally, it was outlined that there are correlations between counting and retrieval strategies. Accordingly, retrieving multiplication facts can be taxing for some special students as well.
## Multiplication of numbers- Areas where dyscalculics need to advance?
Being a step above operations like addition and subtraction, multiplication may be complex for some young learners. Special learners may identify multiple reasons behind this ineptitude. Here we look deep into some distinctive areas where they may feel hindered in accumulation:
### 1. Transposed/Rotated numbers
One of the most common compromises with numbers is the transposition of numbers. Some students may study values appropriately but may face issues with presenting these values on paper later. For example, 9*5 is studied as 45, but they may mention it as 54 in the class test.
In research by Mariela Alexandra Calderón Delgado[2], a few eminent attributes that may show up in individuals with learning incapacity and calculus were mentioned. It was also outlined that the “rotate or transpose the numbers” was one of the eminent attributes. In operations like multiplication, transposed numbers may create errors in the final answer.
### 2. Counting Numbers
Reckoning numbers is the essence of math concepts, and multiplication obligates it. Nonetheless, some individuals may hold off counting. Even if they had to, they might rely on finger counting. For these individuals, multiplication can be evidently an area to be boasted about.
Addressing various attributes of Dyscalculia, Roi Cuhan Kadosh[3] in his research paper, mentioned an instance to demonstrate better. It is normal for six-year-old children to count with their fingers in order to solve arithmetic problems, but adopting the same strategy at ten years of age is a sign of age-inadequate arithmetic skills, probably signifying dyscalculia.
### 3. Memorisation of Multiplication tables
Probably the fundamental essential for addressing multiplication problems is ensuring to master multiplication tables. Some pupils who may feel it strenuous to infer numbers may feel it demanding to memorize these too.
Tables need remarkable memory to retain values and also a great working memory to recall those back later. Mojtaba Soltanlou[4] studied if there is a role of working memory in multiplication fact networks in children. The team did research on grade 3 and grade 4 students. They found that early learning may need visuospatial working memory, but ultimately, the individual needs verbal and non-verbal working memory contributions. These conclusions clearly depict that working memory- whether verbal or nonverbal is pivotal for drawing skills in an individual.
Looking deep into why dyscalculics may need to try hard memorizing, it can be observed that these individuals often have a delicate working memory. Vinod Menon[5] conducted a study on working memory in children’s math learning. Outlining such memories is crucial for math; he also remarked that Dyscalculics could have a specific susceptibility in visuospatial working memory, which may lead to inferences that these individuals may encounter a few constraints with working memory while learning.
The previously mentioned postulates show that number compromises can affect multiplication skills, especially in areas of memorization and reckoning. Once these areas are clearly outlined, relevant strategies can be fabricated to ensure they can compete. Thereby, while some individuals may not solve multiplications with ease, they can manage by mastering these skills later.
Dyscalculic mathematicians like Emma King have made sure they inculcate various personalized plans of action to manage glitches. Aspirants can consider these as motivating sources. Further, the following strategies may hone their multiplication skills finer:
• Break the Problem into uncomplicated parts: Special pupils who have managed facile problems often may get anxious about comprehending complex ones. For that reason, splitting the challenge into easier parts can be a befitting strategy. Llaria Berteletti[6] conducted research on children with mathematical learning incapacities and the reason why they may resist solving multiplication problems. The study made on students from third grade to seventh grade showed that dyscalculics have reduced activations in both verbal and numerical regions for complex problems. Fascinatingly, they had engaging verbal mechanisms for easier ones. These inferences precisely point out the need to break the problem to facilitate comprehension.
• Employ Manipulatives to Ensure Concrete Foundation: Manipulatives are often assistive to grasp composite notions with the ease of tactile learning. Employing picks like Multiplication keys and abacus. Flashcards and spins can ameliorate the learning experience.
• Multiplication games and activities: As a part of the enticing game, pupils may learn implicitly. In the case of multiplications, board games like Say Cheese, Monster Sock, and Pet Me can be eminent picks to learn and exercise. Not only for strategy, even for multiplication facts, but dice games can also be convenient. In the classroom, converting pedagogies into interactive activities motivates the little ones to master concepts.
• Personalized associates like times table sheets and Waldorf multiplication flowers can be encouraged till the learner has mastered these values.
## Concluding thoughts
Math skills are a crucial part of life skills; multiplication can lie among the few crucial notions to be learned. That being the case, even though it may be arduous at inception, special aspirants need to try stronger to manage this. Successful people with a number of compromises may often be the source of motivation. Further, check out the areas of concern and probable working strategies to fabricate the best possible plan of action for you to compete.
References
1. Cooney, J. B., Swanson, H. L., & Ladd, S. F. (1988). Acquisition of mental multiplication skill: Evidence for the transition between counting and retrieval strategies. Cognition and instruction, 5(4), 323-345.
2. Delgado, M. A. C., Delgado, R. I. Z., Palma, R. P., & Moya, M. E. (2019). Dyscalculia and pedagogical intervention. International Research Journal of Management, IT and Social Sciences, 6(5), 95-100.
3. Kadosh, R. C., & Walsh, V. (2007). Dyscalculia. Current Biology, 17(22), R946-R947.
4. Soltanlou, M., Pixner, S., & Nuerk, H. C. (2015). Contribution of working memory in multiplication fact network in children may shift from verbal to visuo-spatial: a longitudinal investigation. Frontiers in psychology, 6, 1062.
5. Menon, V. (2016). Working memory in children’s math learning and its disruption in dyscalculia. Current Opinion in Behavioral Sciences, 10, 125-132.
6. Berteletti, I., Prado, J., & Booth, J. R. (2014). Children with mathematical learning disability fail in recruiting verbal and numerical brain regions when solving simple multiplication problems. Cortex, 57, 143-155. | 1,921 | 9,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-18 | latest | en | 0.950446 |
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# Search results
Found 1394 matches
Graham's Law of Effusion
Effusion is the process in which a gas escapes through a small hole. This occurs if the diameter of the hole is considerably smaller than the mean free ... more
Varignon's theorem in statics
Torque, moment or moment of force (see the terminology below) is the tendency of a force to rotate an object about an axis. In addition to the tendency to ... more
Parallax
Parallax is a displacement or difference in the apparent position of an object viewed along two different lines of sight, and is measured by the angle or ... more
Stokes's Law of Sound Attenuation
Stokes’s law of sound attenuation is a formula for the attenuation of sound in a Newtonian fluid, such as water or air, due to the fluid’s ... more
Released energy of exergonic reaction
A chemical reaction is a process that leads to the transformation of one set of chemical substances to another. Classically, chemical reactions encompass ... more
Worksheet 341
The awe‐inspiring Great Pyramid of Cheops was built more than 4500 years ago. Its square base, originally 230 m on a side, covered 13.1 acres, and it was 146 m high (H), with a mass of about 7×10^9 kg. (The pyramid’s dimensions are slightly different today due to quarrying and some sagging). Historians estimate that 20,000 workers spent 20 years to construct it, working 12-hour days, 330 days per year.
a) Calculate the gravitational potential energy stored in the pyramid, given its center of mass is at one-fourth its height.
Division
Potential energy
b) Only a fraction of the workers lifted blocks; most were involved in support services such as building ramps, bringing food and water, and hauling blocks to the site. Calculate the efficiency of the workers who did the lifting, assuming there were 1000 of them and they consumed food energy at the rate of 300 Kcal/hour.
first we calculate the number of hours worked per year.
Multiplication
then we calculate the number of hours worked in the 20 years.
Multiplication
Then we calculate the energy consumed in 20 years knowing the energy consumed per hour and the total hours worked in 20 years.
Multiplication
Multiplication
The efficiency is the resulting potential energy divided by the consumed energy.
Division
Energy – Momentum relation
In physics, the energy–momentum relation, or relativistic dispersion relation, is the relativistic equation relating any object’s rest (intrinsic) ... more
Flywheel energy storage (Energy density)
A flywheel is a rotating mechanical device that is used to store rotational energy. Flywheel energy storage works by accelerating a rotor to a very high ... more
West number
The West number is an empirical parameter used to characterize the performance of Stirling engines and other Stirling systems. A Stirling engine is a heat ... more
R-value (insulation) of a multi-layered installation - U.S. units
The R-value is a measure of thermal resistance, or ability of heat to transfer from hot to cold, through materials (such as insulation) and assemblies of ... more
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## Hamiltonian Cycle
Input description: A graph G = (V,E).
Problem description: Find an ordering of the vertices such that each vertex is visited exactly once.
Discussion: The problem of finding a Hamiltonian cycle or path in a graph is a special case of the traveling salesman problem, one where each pair of vertices with an edge between them is considered to have distance 1, while nonedge vertex pairs are separated by distance .
Closely related is the problem of finding the longest path or cycle in a graph, which occasionally arises in pattern recognition problems. Let the vertices in the graph correspond to possible symbols, and let edges link symbols that can possibly be next to each other. The longest path through this graph is likely the correct interpretation.
The problems of finding longest cycles and paths are both NP-complete, even on very restrictive classes of unweighted graphs. There are several possible lines of attack, however:
• Do you have to visit all the vertices or all the edges? - First verify that you really have a Hamiltonian cycle problem. As discussed in Section , fast algorithms exist for edge-tour, or Eulerian cycle, problems, where you must visit all the edges without repetition. With a little cleverness, it is sometimes possible to reformulate a Hamiltonian cycle problem in terms of Eulerian cycles. Perhaps the most famous such instance is the problem of constructing de Bruijn sequences, discussed in Section .
• Is there a serious penalty for visiting vertices more than once? - By phrasing the problem as minimizing the total number of vertices visited on a complete tour, we have an optimization problem that now allows room for heuristics and approximation algorithms. For example, finding a spanning tree of the graph and doing a depth-first search, as discussed in Section , yields a tour with at most 2n vertices. Using randomization or simulated annealing might bring the size of this down considerably.
• Am I seeking the longest path in a directed acyclic graph (DAG)? - The problem of finding the longest path in a DAG can be solved in linear time using dynamic programming. This is about the only interesting case of longest path for which efficient algorithms exist.
• Is my graph dense? - For sufficiently dense graphs, there always exists at least one Hamiltonian cycle, and further, such a cycle can be found quickly. An efficient algorithm for finding a Hamiltonian cycle in a graph where all vertices have degree is given in [Man89].
If you really must know whether your graph is Hamiltonian, backtracking with pruning is your only possible solution. Before you search, it pays to check whether your graph is biconnected (see Section ). If not, this means that the graph has an articulation vertex whose deletion will disconnect the graph and so cannot be Hamiltonian.
Implementations: The football program of the Stanford GraphBase (see Section ) uses a stratified greedy algorithm to solve the asymmetric longest path problem. The goal is to derive a chain of football scores in order to establish the superiority of one football team over another. After all, if Virginia beat Illinois by 30 points, and Illinois beat Stony Brook by 14 points, then by transitivity Virginia would beat Stony Brook by 44 points if they played, right? We seek the longest path in a graph where the weight of an edge (x,y) is the number of points x beat y by.
Nijenhuis and Wilf [NW78] provide an efficient routine to enumerate all Hamiltonian cycles of a graph by backtracking. See Section . Algorithm 595 [Mar83] of the Collected Algorithms of the ACM is a similar Fortran code that can be used as either an exact procedure or a heuristic by controlling the amount of backtracking. See Section .
XTango (see Section ) is an algorithm animation system for UNIX and X-windows, which includes an animation of a backtracking solution to the knight's tour problem.
Combinatorica [Ski90] provides a Mathematica backtracking implementation of Hamiltonian cycle. See Section .
Notes: Hamiltonian cycles - circuits that visit each vertex of a graph exactly once - apparently first arose in Euler's study of the knight's tour problem, although they were popularized by Hamilton's ``Around the World'' game in 1839. Good expositions of the proof that Hamiltonian cycle is NP-complete [Kar72] include [Baa88, CLR90, GJ79].
Techniques for solving optimization problems in the laboratory using biological processes have recently attracted considerable attention. In the original application of these ``biocomputing'' techniques, Adleman [Adl94b] solved a seven-vertex instance of the directed Hamiltonian path problem. Unfortunately, this approach requires an exponential number of molecules, and Avogadro's number implies that such experiments are inconceivable for graphs beyond .
Related Problems: Eulerian cycle (see page ), traveling salesman (see page ).
Next: Graph Partition Up: Graph Problems: Hard Problems Previous: Traveling Salesman Problem
Algorithms
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# [11/21] 1926. Nearest Exit from Entrance in Maze
• 학부유학생 작성
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### 본문
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You are given an `m x n` matrix `maze` (0-indexed) with empty cells (represented as `'.'`) and walls (represented as `'+'`). You are also given the `entrance` of the maze, where `entrance = [entrancerow, entrancecol]` denotes the row and column of the cell you are initially standing at.
In one step, you can move one cell updownleft, or right. You cannot step into a cell with a wall, and you cannot step outside the maze. Your goal is to find the nearest exit from the `entrance`. An exit is defined as an empty cell that is at the border of the `maze`. The `entrance` does not count as an exit.
Return the number of steps in the shortest path from the `entrance` to the nearest exit, or `-1` if no such path exists.
Example 1:
```Input: maze = [["+","+",".","+"],[".",".",".","+"],["+","+","+","."]], entrance = [1,2]
Output: 1
Explanation: There are 3 exits in this maze at [1,0], [0,2], and [2,3].
Initially, you are at the entrance cell [1,2].
- You can reach [1,0] by moving 2 steps left.
- You can reach [0,2] by moving 1 step up.
It is impossible to reach [2,3] from the entrance.
Thus, the nearest exit is [0,2], which is 1 step away.
```
Example 2:
```Input: maze = [["+","+","+"],[".",".","."],["+","+","+"]], entrance = [1,0]
Output: 2
Explanation: There is 1 exit in this maze at [1,2].
[1,0] does not count as an exit since it is the entrance cell.
Initially, you are at the entrance cell [1,0].
- You can reach [1,2] by moving 2 steps right.
Thus, the nearest exit is [1,2], which is 2 steps away.
```
Example 3:
```Input: maze = [[".","+"]], entrance = [0,0]
Output: -1
Explanation: There are no exits in this maze.
```
Constraints:
• `maze.length == m`
• `maze[i].length == n`
• `1 <= m, n <= 100`
• `maze[i][j]` is either `'.'` or `'+'`.
• `entrance.length == 2`
• `0 <= entrancerow < m`
• `0 <= entrancecol < n`
• `entrance` will always be an empty cell.
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## 학부유학생님의 댓글
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Runtime: 1120 ms, faster than 77.13% of Python3 online submissions for Nearest Exit from Entrance in Maze.
Memory Usage: 15.6 MB, less than 27.25% of Python3 online submissions for Nearest Exit from Entrance in Maze.
``````from collections import deque
class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
ROW, COL = len(maze), len(maze[0])
visited = set()
visited.add((entrance[0],entrance[1]))
queue = deque([(entrance[0], entrance[1], 0)])
directions = [(1,0), (0,1), (-1,0),(0,-1)]
while queue:
r, c, steps = queue.popleft()
if (r == 0 or r == ROW - 1 or c == 0 or c == COL -1) and (r,c) != (entrance[0],entrance[1]):
return steps
for rdir, cdir in directions:
nr, nc = r+rdir, c+cdir
if 0<=nr<ROW and 0<=nc<COL and (nr, nc) not in visited and maze[nr][nc] == ".":
queue.append((nr, nc, steps+1))
visited.add((nr, nc))
return -1``````
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• 전체 회원수 451 명 | 1,179 | 3,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.840026 |
https://www.wiki.gis.com/wiki/index.php/Multidimensional_scaling | 1,701,633,066,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.53/warc/CC-MAIN-20231203193127-20231203223127-00720.warc.gz | 1,205,374,745 | 11,520 | # Multidimensional scaling
Multidimensional scaling (MDS) is a set of related statistical techniques often used in information visualization for exploring similarities or dissimilarities in data. MDS is a special case of ordination. An MDS algorithm starts with a matrix of item–item similarities, then assigns a location to each item in N-dimensional space, where N is specified a priori. For sufficiently small N, the resulting locations may be displayed in a graph or 3D visualisation.
## Categorization of MDS
MDS algorithms fall into a taxonomy, depending on the meaning of the input matrix:
• Classical multidimensional scaling : also known as Torgerson Scaling or Torgerson-Gower scaling – takes an input matrix giving dissimilarities between pairs of items and outputs a coordinate matrix whose configuration minimizes a loss function called strain.[1] (pp. 207–212).
• Metric multidimensional scaling: A superset of classical MDS that generalizes the optimization procedure to a variety of loss functions and input matrices of known distances with weights and so on. A useful loss function in this context is called stress which is often minimized using a procedure called Stress Majorization.
• Non-metric multidimensional scaling: In contrast to metric MDS, non-metric MDS both finds a non-parametric monotonic relationship between the dissimilarities in the item-item matrix and the Euclidean distance between items, and the location of each item in the low-dimensional space. The relationship is typically found using isotonic regression.
## Details
The data to be analyzed is a collection of $I$ objects (colors, faces, stocks, ...) on which a distance function is defined,
$\delta_{ij}~=~$ the distance between ith and jth objects.
These distances are the entries of the dissimilarity matrix
$\Delta := \begin{pmatrix} \delta_{1,1} & \delta_{1,2} & \cdots & \delta_{1,I} \\ \delta_{2,1} & \delta_{2,2} & \cdots & \delta_{2,I} \\ \vdots & \vdots & & \vdots \\ \delta_{I,1} & \delta_{I,2} & \cdots & \delta_{I,I} \end{pmatrix}.$
The goal of MDS is, given Δ, to find $I$ vectors $x_1,\ldots,x_I \in \mathbb{R}^N$ such that
$\|x_i - x_j\| \approx \delta_{i,j}$ for all $i,j\in I$,
where $\|\cdot\|$ is a vector norm. In classical MDS, this norm is the Euclidean distance, but more generally it may be a metric or arbitrary distance function.[2]
In other words, MDS attempts to find an embedding from the $I$ objects into RN such that distances are preserved. If the dimension N is chosen to be 2 or 3, we may plot the vectors xi to obtain a visualization of the similarities between the $I$ objects.
## Procedure
There are several steps in conducting MDS research:
1. Formulating the problem – What variables do you want to compare? How many variables do you want to compare? More than 20 is cumbersome. Less than 8 (4 pairs) will not give valid results. What purpose is the study to be used for?
2. Obtaining Input Data – Respondents are asked a series of questions. For each product pair they are asked to rate similarity (usually on a 7 point Likert scale from very similar to very dissimilar). The first question could be for Coke/Pepsi for example, the next for Coke/Hires rootbeer, the next for Pepsi/Dr Pepper, the next for Dr Pepper/Hires rootbeer, etc. The number of questions is a function of the number of brands and can be calculated as $Q = N (N - 1) / 2$ where Q is the number of questions and N is the number of brands. This approach is referred to as the “Perception data : direct approach”. There are two other approaches. There is the “Perception data : derived approach” in which products are decomposed into attributes which are rated on a semantic differential scale. The other is the “Preference data approach” in which respondents are asked their preference rather than similarity.
3. Running the MDS statistical program – Software for running the procedure is available in many software for statistics. Often there is a choice between Metric MDS (which deals with interval or ratio level data), and Nonmetric MDS (which deals with ordinal data).
4. Decide number of dimensions – The researchers must decide on the number of dimensions they want the computer to create. The more dimensions, the better the statistical fit, but the more difficult it is to interpret the results.
5. Mapping the results and defining the dimensions – The statistical program (or a related module) will map the results. The map will plot each product (usually in two dimensional space). The proximity of products to each other indicate either how similar they are or how preferred they are, depending on which approach was used. The dimensions must be labelled by the researcher. This requires subjective judgement and is often very challenging. The results must be interpreted ( see perceptual mapping).
6. Test the results for reliability and Validity – Compute R-squared to determine what proportion of variance of the scaled data can be accounted for by the MDS procedure. An R-square of .6 is considered the minimum acceptable level. An R-square of .8 is considered good for metric scaling and .9 is considered good for non-metric scaling. Other possible tests are Kruskal’s Stress, split data tests, data stability tests (i.e., eliminating one brand), and test-retest reliability.
## Applications
Applications include scientific visualisation and data mining in fields such as cognitive science, information science, psychophysics, psychometrics, marketing and ecology. New applications arise in the scope of autonomous wireless nodes which populate a space or an area. MDS may apply as a real time enhanced approach to monitoring and managing such populations.
### Marketing
In marketing, MDS is a statistical technique for taking the preferences and perceptions of respondents and representing them on a visual grid, called perceptual maps.
### Comparison and advantages
Potential customers are asked to compare pairs of products and make judgements about their similarity. Whereas other techniques (such as factor analysis, discriminant analysis, and conjoint analysis) obtain underlying dimensions from responses to product attributes identified by the researcher, MDS obtains the underlying dimensions from respondents’ judgements about the similarity of products. This is an important advantage. It does not depend on researchers’ judgments. It does not require a list of attributes to be shown to the respondents. The underlying dimensions come from respondents’ judgments about pairs of products. Because of these advantages, MDS is the most common technique used in perceptual mapping.
• positioning (marketing)
• locating
• perceptual mapping
• product management
• marketing
• Generalized multidimensional scaling (GMDS)
• Data clustering
• Factor analysis
• Discriminant analysis
## Implementations
• Orange, a free data mining software suite, module orngMDS
## Bibliography
1. Borg, I. and Groenen, P.: "Modern Multidimensional Scaling: theory and applications" (2nd ed.), Springer-Verlag New York, 2005
2. Kruskal, J. B., and Wish, M. (1978), Multidimensional Scaling, Sage University Paper series on Quantitative Application in the Social Sciences, 07-011. Beverly Hills and London: Sage Publications.
• Bronstein, A. M, Bronstein, M.M, and Kimmel, R. (2006), Generalized multidimensional scaling: a framework for isometry-invariant partial surface matching, Proc. National Academy of Sciences (PNAS), Vol. 103/5, pp. 1168–1172.
• Cox, T.F., Cox, M.A.A., (2001), Multidimensional Scaling, Chapman and Hall.
• Coxon, Anthony P.M. (1982): "The User's Guide to Multidimensional Scaling. With special reference to the MDS(X) library of Computer Programs." London: Heinemann Educational Books.
• Green, P. (1975) Marketing applications of MDS: Assessment and outlook, Journal of Marketing, vol 39, January 1975, pp 24–31.
• Torgerson, W. S. (1958). Theory & Methods of Scaling. New York: Wiley. | 1,798 | 7,950 | {"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": 11, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.791472 |
https://byjusexamprep.com/bank-exam/balance-the-following-chemical-equation-calcium-hydroxide-carbon-dioxide-calcium-carbonate-water | 1,720,954,822,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00722.warc.gz | 128,899,026 | 30,417 | # Balance The Following Chemical Equation : Calcium Hydroxide + Carbon Dioxide →Calcium Carbonate + Water
By BYJU'S Exam Prep
Updated on: September 25th, 2023
The balanced chemical equation is Ca(OH)2 + CO2 → CaCO3 + H2O. To balance the chemical equation for the reaction between calcium hydroxide (Ca(OH)2) and carbon dioxide (CO2) to form calcium carbonate (CaCO3) and water (H2O), adjust the coefficients as follows: Ca(OH)2 + CO2 → CaCO3 + H2O
## What is a Balanced Chemical Equation?
A balanced chemical equation represents a chemical reaction where the number of atoms of each element is equal on both sides. It follows the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. The reactants are written on the left side, the products on the right side, and an arrow indicates the direction of the reaction.
Coefficients are added to the formulas to balance the equation and reflect the relative amounts of substances involved. Balancing ensures that the same number of atoms of each element exists before and after the reaction. Balanced chemical equations are crucial for accurately representing chemical reactions and performing stoichiometric calculations.
For Example: 2H2 + O2 → 2H2O
Solution
The balanced chemical equation for the reaction between calcium hydroxide (Ca(OH)2) and carbon dioxide (CO2) to form calcium carbonate (CaCO3) and water (H2O) is as follows:
Ca(OH)2 + CO2 → CaCO3 + H2O
Step 1: Balance the calcium (Ca) atoms:
Ca(OH)2 + CO2 → 1CaCO3 + H2O
Step 2: Balance the carbon (C) atoms:
Ca(OH)2 + CO2 → 1CaCO3 + H2O
Step 3: Balance the hydrogen (H) atoms:
Ca(OH)2 + CO2 → 1CaCO3 + H2O
Step 4: Balance the oxygen (O) atoms:
Ca(OH)2 + CO2 → 1CaCO3 + H2O
Now, the equation is balanced with an equal number of atoms on both sides. The final balanced chemical equation is:
Ca(OH)2 + CO2 → CaCO3 + H2O
Summary
## Balance the following chemical equation: Calcium Hydroxide + Carbon Dioxide→Calcium Carbonate + Water
The balanced chemical equation for the reaction between calcium hydroxide (Ca(OH)2) and carbon dioxide (CO2) to form calcium carbonate (CaCO3) and water (H2O) can be summarized as Ca(OH)2 + CO2 → CaCO3 + H2O
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# Help!
0
362
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The sum of the positive divisors of a positive integer of the form 2^i*3^j is equal to 600. What is i+j?
May 5, 2018
### 4+0 Answers
#1
+1
2^3 x 3^3 =216
Divisors of 216 =1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 18 + 24 + 27 + 36 + 54 + 72 +108 + 216 = 600
i + j = 3 + 3 = 6
May 5, 2018
#2
+1
There is a formula for the sum of the divisors, but you have to know the number whose divisors are to be summed up. For your particular problem, it goes like this:
[(2^(i + 1) - 1) / (2 - 1)] * [(3^(j + 1) - 1)] / (3 - 1) = 600
I don't know how to formally solve for 2 variables in one equation without resorting to trial and error, which quickly leads to i = 3 and j = 3, which give: 2^3 x 3 ^3 = 216. Then you would find the divisors of 216, as done above, which sum up to 600.
May 5, 2018
#3
+1324
0
This is impressive, Mr. BB. There’s no blarney, bullshit, or slop in your post. Looks like GETSMART is doing his job.
GingerAle May 6, 2018
#4
+1324
+2
Solving two variables in a single equation is usually a trial and error process. However, in this case, you can easily narrow the process to a few guesses using logic.
The (i) and (j) values are integers; there are no fractional or decimal components in them.
Because the sum of integers is very low, it reasonable to assume the (i,,j) exponents are low—less than 10.
To solve, start by assuming a central value (between 1 and 10) for (j), and then use logarithms to isolate for (i).
Because (2) is one of the factors, its resolution will be its multiples 2,4,8,16,32,64,128,256, ... ect. Seeing one of these indicates you’ve found the solution set.
$$(( 2^i-1)*\dfrac{(3^j - 1)}{(2)} = 600 \leftarrow \tiny \text{ The (2) and (1) are common to all equations.}\\ ( 2^i) = \dfrac{1200}{(3^j - 1)} +1\leftarrow \tiny \text{ Leave it in this form to streamline the solution search.}\\ \text {Example: assume (j) = 6 and solve for (i)}\\ ( 2^i) = \dfrac{1200}{(3^6 - 1)} +1\\ ( 2^i) = \dfrac{150}{91} +1\\ Log_2( 2^i) = Log_2 \left( \dfrac {150}{91} +1\right) \small \text{ Note the fraction; This is not the solution.}\\ \text {Example: assume (j) = 5 and solve for (i)}\\ ( 2^i) = \dfrac{1200}{(5^5 - 1)} +1\\ ( 2^i) = \dfrac{600}{121} +1\\ Log_2( 2^i) = Log_2 \left( \dfrac {600}{121} +1\right) \small \text{ Again, note the fraction; This is not the solution.}\\ \text {Example: assume (j) = 4 and solve for (i)}\\ ( 2^i) = \dfrac{1200}{(5^4 - 1)} +1\\ ( 2^i) = 15 +1\\ Log_2( 2^i) = Log_2 \left( 16\right) \small \text{ Note the integer 16, corresponding to i=4; This is the solution.}\\$$
The solution is j=4 and i=4. Subtract (1) from each and use these as the exponents for the primes 2, and 3 to find the number (N) for which 600 is the sum of its divisors.
There are similar techniques for (N) that have three or more unique prime factors.
GA
Edit: Corrected to indicate base (2) log.
May 6, 2018
edited by GingerAle May 7, 2018 | 1,118 | 3,172 | {"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} | 3.9375 | 4 | CC-MAIN-2019-26 | latest | en | 0.906866 |
https://www.varsitytutors.com/ap_physics_2-help/buoyant-force?page=2 | 1,553,237,520,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202635.43/warc/CC-MAIN-20190322054710-20190322080710-00042.warc.gz | 945,240,667 | 42,205 | AP Physics 2 : Buoyant Force
Example Questions
Example Question #11 : Buoyant Force
Suppose that two different balls of equal volume are submerged and held in a container of water. Ball A has a density of and Ball B has a density of . After the two balls are released, both of them begin to accelerate up towards the surface. Which ball is expected to accelerate faster?
Ball A and Ball B will have the same acceleration
There is no way to determine the relative acceleration of the two balls
Ball B will have greater acceleration
Neither ball will accelerate
Ball A will have greater acceleration
Ball A will have greater acceleration
Explanation:
In this question, we're presented with a situation in which two balls of different densities but equal volumes are held underneath the surface of a container of water. Then, each ball is released and allowed to accelerate up to the surface. The question is to determine how the acceleration of each ball compares to the other. In order to answer this, let's start by imagining all of the forces acting on the submerged ball.
In the x-direction, the forces acting to the left of the ball are exactly equal to the forces acting on the right of the ball. Therefore, all of the forces acting in the x-direction cancel out, resulting in a net force of zero in the x-direction.
In the y-direction, we have to consider two things. The first is the upward buoyant force caused by the displacement of water. Since both balls have an identical volume, both of them displace the same amount of water. Consequently, both balls will experience the same upward buoyant force. However, we must also consider the downward force caused by the weight of the ball itself. In this scenario, the downward weight of Ball B is greater than that of Ball A.
We can write the net force acting on the ball in the y-direction as the difference between the upward buoyant force and the downward weight as follows:
Due to the fact that Ball A has a smaller weight (a smaller component in the above equation), the result is that the net upward force is greater than that of Ball B. Thus, we would expect Ball A to have a greater acceleration.
Example Question #12 : Buoyant Force
Determine the net force (including direction) on a gold marble of radius in liquid mercury .
down
up
up
down
down
down
Explanation:
Consider all the forces on the gold marble:
Recall the equation for buoyant force:
Substitute:
Find the mass of the marble:
Plug in values:
Note how the buoyant force points up while the gravitational force points down.
Example Question #11 : Fluid Statics
Determine the buoyant force on an object of volume in a fluid of density
Explanation:
Use the equation for buoyant force:
Where is the density of the medium around the object in question
is the gravitational acceleration near the surface of the earth
is the volume of the object in question
Convert into
Plug in values:
Example Question #14 : Buoyant Force
A balloon of mass is inflated to a volume of with pure . Determine the buoyant force it will experience when submerged in water.
None of these
Explanation:
Use the equation for buoyant force:
Where
is the density of the medium
is the acceleration due to gravity
is the volume
Plugging in values:
Example Question #15 : Buoyant Force
Will a ball of mass and radius sink or float in water?
Not enough information
It will be partially submerged
Float, but only because of surface tension
Sink
Float
Sink
Explanation:
Determining density:
Volume of a sphere:
Combining equations:
Converting to and plugging in values:
It is denser than water, so it will sink.
Example Question #16 : Buoyant Force
A block of mass and volume is held in place under water. What is the instantaneous acceleration of the block, and in what direction, when it is released?
Explanation:
According to Archimedes's principle, when the block is placed under water, 3.5L of water are displaced. We can then calculate the buoyant force provided by the water:
Where:
Plugging in our values to the first expression:
Then we can use Newton's 2nd law to determine the acceleration of the block:
There are two forces acting on the block, gravity and buoyancy, and they are in opposite directions. If we designate a downward force as positive, we get:
Substituting in the expression for force due to gravity:
Rearranging for acceleration:
Substituting in values:
Example Question #17 : Buoyant Force
A scuba diver with a total mass is dressed so his density and is holding a spherical ball with a volume while submerged in water. What is the density of the ball if the upward acceleration of the ball and diver is ?
Explanation:
Where there are 4 total forces acting on the scuba diver and ball: gravity and buoyancy acting on both the scuba diver and the ball. However, the diver has the same density as the water, so the gravitational and buoyancy forces acting on the diver will cancel out. If we designate an upward force as being positive, we can say:
(1)
Where:
Where:
So,
Plugging these expressions back in to expression (1), we get:
Now let's start rearranging for the density of the ball:
Example Question #18 : Buoyant Force
A block of mass is sinking in water at a constant velocity. There is a constant drag force of acting on the block. What is the volume of the block?
Explanation:
Since the block is traveling at a constant velocity we can say:
There are 3 forces acting on the block: gravitational, buoyant, and drag force. If we denote a downward force being positive, the expression becomes:
Where:
Substituting these in:
Where according to Archimedes's principle:
Plugging this in:
Rearranging for volume:
Plugging in values:
Example Question #19 : Buoyant Force
A bowling ball with mass and radius is submerged under water and held in place by a string. What is the tension in the string?
Explanation:
Since the ball is held in place, we know that:
There are three forces acting on the ball: gravitational, buoyant, and tension. If we denote a downward force as positive, we get:
(1)
And using Archimedes's principal:
Where:
Plugging all of these into expression (1), we get:
Plugging in our values, we get:
Example Question #20 : Buoyant Force
A semi-hollow, spherical ball with an empty volume of is submerged in water and has an initial mass of . The ball develops a leak and water begins entering the ball at a rate of . How long does it take before the buoyant force on the ball is equal to the gravitational force?
Explanation: | 1,465 | 6,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2019-13 | latest | en | 0.946229 |
http://mathhelpforum.com/calculus/13875-sum-power-series-problem.html | 1,527,296,324,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867254.84/warc/CC-MAIN-20180525235049-20180526015049-00125.warc.gz | 184,498,176 | 9,274 | # Thread: sum of power series problem
1. ## sum of power series problem
Express the function as the sum of a power series by first using partial fractions, find the interval of convergence.
f(x) = 3/(x^2 +x - 2)
thanks for any help with this one.
..partial fraction would be: 3/(x-1) + 3/(x+2)
i think.
2. Originally Posted by rcmango
Express the function as the sum of a power series by first using partial fractions, find the interval of convergence.
f(x) = 3/(x^2 +x - 2)
thanks for any help with this one.
..partial fraction would be: 3/(x-1) + 3/(x+2)
i think.
your partial fractions are wrong. find the correct ones, then use the fact that
1/(1 - x) = SUM x^n
Hint:
The partial fractions are 1/(x - 1) - 1/(x + 2)
1/(x - 1) = -1/(1 - x)
1/(x + 2) = 1/[2(1 + x/2)] = (1/2)*1/(1 + x/2) = (1/2)*1/(1 - (-x/2)) | 282 | 826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.912104 |
https://stacks.math.columbia.edu/tag/01IQ | 1,725,869,213,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00856.warc.gz | 521,711,339 | 6,391 | Lemma 26.10.4. Let $f : Y \to X$ be an immersion of schemes. Then $f$ is a closed immersion if and only if $f(Y) \subset X$ is a closed subset.
Proof. If $f$ is a closed immersion then $f(Y)$ is closed by definition. Conversely, suppose that $f(Y)$ is closed. By definition there exists an open subscheme $U \subset X$ such that $f$ is the composition of a closed immersion $i : Y \to U$ and the open immersion $j : U \to X$. Let $\mathcal{I} \subset \mathcal{O}_ U$ be the quasi-coherent sheaf of ideals associated to the closed immersion $i$. Note that $\mathcal{I}|_{U \setminus i(Y)} = \mathcal{O}_{U \setminus i(Y)} = \mathcal{O}_{X \setminus i(Y)}|_{U \setminus i(Y)}$. Thus we may glue (see Sheaves, Section 6.33) $\mathcal{I}$ and $\mathcal{O}_{X \setminus i(Y)}$ to a sheaf of ideals $\mathcal{J} \subset \mathcal{O}_ X$. Since every point of $X$ has a neighbourhood where $\mathcal{J}$ is quasi-coherent, we see that $\mathcal{J}$ is quasi-coherent (in particular locally generated by sections). By construction $\mathcal{O}_ X/\mathcal{J}$ is supported on $U$ and, restricted there, equal to $\mathcal{O}_ U/\mathcal{I}$. Thus we see that the closed subspaces associated to $\mathcal{I}$ and $\mathcal{J}$ are canonically isomorphic, see Example 26.4.3. In particular the closed subspace of $U$ associated to $\mathcal{I}$ is isomorphic to a closed subspace of $X$. Since $Y \to U$ is identified with the closed subspace associated to $\mathcal{I}$, see Lemma 26.4.5, we conclude that $Y \to U \to X$ is a closed immersion. $\square$
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## BUACC5936 Financial Management
• Attempt all questions and show all workings
### – Questions 1 & 2 of this assignment MUST be completed on an Excel Spreadsheet.
A young newly married couple wish to purchase a unit house in a remote area priced at
\$190,000. They want to borrow the entire amount and repay this amount in full plus interest over the next 20 years. You can offer them a mortgage for the full amount at an interest rate of 17% (sounds like extortion to me) and calls for equal annual instalment payments at the end of each of the 20 years.
Required: (a) Calculate the amount of the annual payment? (b) Create and complete the amortization
ABC Corporation is experiencing rapid growth. Dividends are expected to grow at 25% per year for the next three years and then grow at 5% per year for ever. If the required return on the share is 10% and the share currently sells for \$30.
### What is required of you:
Calculate the projected dividend for the coming year? (Hint: calculate D0 before attempting to determine D1)
Consider a bond with face value of \$1000, a coupon rate of 8% (paid annually), and ten years to maturity.
### What is required of you:
1. What is the price of this bond if the required rate of return (r) is 18 percent?
2. What is the price if r increases to 20 percent? By what percentage did the price of the bond change?
3. What is the price if r is five percent? If r increases to seven percent, what is the percentage change in price?
4. From your answers in a to c, what can you say about relative price volatility of a bond in high compared to low interest rate environments.
SYPETCO Ltd is a leading company in Australia and you the below details relating to the capital structure of the company.
1. Calculate the cost associated with each new source of finance. The firm has no retained earnings available.
1. Calculate the WACC given the existing weights
The financial controller does not believe the existing capital structure weights are appropriate to minimise the firm’s cost of capital in the medium term and believes they should be as follows
### c) What impact do these new weights have on the WACC?
The firm is now (in 2021) considering the following investment opportunity for the period 2022-2029.
Data is as follows
Sales estimates for next 8 years starting from 2022
### d) Calculate the Net Present Value, Internal Rate of Return and Payback Period
The financial controller is considering the use of the Capital Asset Pricing Model as a surrogate discount factor. The risk-free rate is 5 percent. The information in the table below has been used by company management in calculating the stock beta value which is 1.151 and the expected return on the stock which is 12.5%.
### e) Calculate the CAPM
• Explain why this figure may differ from that calculated above (i.e. Cost of equity. | 655 | 2,947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-49 | latest | en | 0.926945 |
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Kyle's Converter > Flow > Thousand Million Cubic Feet Per Day > Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year
Thousand Million Cubic Feet Per Day (TMCD) Million Cubic Feet Per Year (MMCF/yr)* Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Million Cubic Feet Per Year to Thousand Million Cubic Feet Per Day
(or just enter a value in the "to" field)
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Unit Descriptions
1 Thousand Million Cubic Feet per Day:
River flow rate of 1 000 000 000 cubic feet per day. Approximately 327.74128 cubic meters per second (SI units). 1 TMCD ≈ 327.74128 m3/s.
1 Million Cubic Feet per Year:
Production flow rate of 1 million cubic feet in a period of 365 days. Assuming a cubic international foot. 1 Million cubic feet per year is approximately 0.000 897 921 315 068 493 cubic meters per second. 1 MMCF/yr ≈ 0.000897921315068493 m3/s
Conversions Table
1 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 36500070 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 25550000
2 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 73000080 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 29200000
3 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 109500090 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 32850000
4 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 1460000100 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 36500000
5 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 1825000200 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 73000000
6 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 2190000300 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 109500000
7 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 2555000400 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 146000000
8 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 2920000500 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 182500000
9 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 3285000600 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 219000000
10 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 3650000800 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 292000000
20 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 7300000900 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 328500000
30 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 109500001,000 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 365000000
40 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 1460000010,000 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 3650000000
50 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 18250000100,000 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 36500000000
60 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 219000001,000,000 Thousand Million Cubic Feet Per Day to Million Cubic Feet Per Year = 365000000000 | 901 | 3,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.538592 |
https://darkskiesfilm.com/how-many-color-wheels-are-there/ | 1,716,939,079,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059160.88/warc/CC-MAIN-20240528220007-20240529010007-00261.warc.gz | 161,327,438 | 11,130 | # How many color wheels are there?
## How many color wheels are there?
12 colors
Understanding the Color Wheel Many color wheels are shown using 12 colors. Using this color wheel as an example, it can be read as follows: Three Primary Colors (Ps): Red, Yellow, Blue. Three Secondary Colors (S’): Orange, Green, Violet.
What are the rules of the color wheel?
And they’re actually pretty simple.
• Complementary colors are any two colors opposite each other on the wheel.
• Split complementary colors use three colors.
• Analogous colors are any three colors next to each other on the wheel.
• Triadic colors are any three colors that are equally apart on the color wheel.
What are the 12 colour wheels?
The 12-color wheel is a common structuring of hues that is based in paint and light and is hence popular with artists as well as photographers. The hues can be arranged in a circle, which is convenient for combinations as described below….Lightness.
Color Lightness
Spring Green 82%
Green 80%
Magenta 70%
Orange 65%
### What is the six color wheel?
The typical artists’ paint or pigment color wheel includes the blue, red, and yellow primary colors. The corresponding secondary colors are green, orange, and violet or purple. The tertiary colors are green-yellow, yellow-orange, orange-red, red-violet/purple, purple/violet-blue and blue-green.
What are the 4 primary Colours?
The four primary colors in the 4-primary color wheel are blue, yellow, green and red. This differs from the color mixing wheel, which only has three primary colors.
How many colors are there?
It has been determined by people who determine such things that there are somewhere around 18 decillion varieties of colors available for your viewing enjoyment. That’s an 18 followed by 33 zeros.
#### What are the 11 basic colors?
English, for example, has the full set of 11 basic colors: black, white, red, green, yellow, blue, pink, gray, brown, orange and purple. In a 1999 survey by linguists Paul Kay and Luisa Maffi, languages were roughly equally distributed between the basic color categories that they tracked.
What are the 5 analogous colors?
Analogous colors examples
• Yellow, yellow-green, green.
• Violet, red-violet, and red.
• Red, red-orange, orange.
• Blue, blue-violet, violet.
What are the 5 primary colours?
François d’Aguilon’s notion of the five primary colors (white, yellow, red, blue, black) was influenced by Aristotle’s idea of the chromatic colors being made of black and white. The 20th century philosopher Ludwig Wittgenstein explored color-related ideas using red, green, blue, and yellow as primary colors.
## What are the 7 primary colours?
This is a revision for the primary known colors. The seven basic components of a color may contain red, blue, yellow, white, black, colorless and light. This research also shows secondary and tertiary colors.
What is the most rare color?
Blue is one of the rarest of colors in nature. Even the few animals and plants that appear blue don’t actually contain the color. These vibrant blue organisms have developed some unique features that use the physics of light. First, here’s a reminder of why we see blue or any other color.
What are the 5 main colors?
### What are the 16 colours?
Colors in HTML. a color name. HTML used to recognize 16 color names (“black”, “white”, “gray”, “silver”, “maroon”, “red”, “purple”, “fushsia”, “green”, “lime”, “olive”, “yellow”, “navy”, “blue”, “teal”, and “aqua”), but new browsers can recognize 147 CSS3 color names.
How do you make a six color wheel?
Draw slightly smaller circles between the 6 big circles.
1. Start with the 1st smaller circle between yellow and red, color it orange.
2. The second smaller circle is between red and magenta, color it scarlet.
3. The third smaller circle is between magenta and blue, color it purple.
What are the 4 types of color schemes?
4 Main Types of Color Palettes
• Monochromatic Color Palettes. If you like keeping things plain and simple, you should go for the monochromatic color palettes as they consist of the same hue in different shades and tones.
• Analogous Color Palettes.
• Complementary Color Palettes. | 957 | 4,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.94591 |
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• This Measurement Conversions 3 Morning Math is perfect to practice measurement skills. Your elementary grade students will love this Measurement Conversions 3 Morning Math. No matter what time you work with math, our new "Morning Math" series is a great way to open students' eyes to the daily uses of math. Simple measurement conversion problems (yards to feet, feet to inches) are covered in this five-part set.
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• This Length/Distance #3 Measurement Conversions is perfect to practice measurement skills. Your elementary grade students will love this Length/Distance #3 Measurement Conversions. Printable math worksheet which covers measuring conversions (length/distance). You will answer several story problems.
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Common Core: Geometry 6.G.A1, 5.G.B3, 4.MD.3
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• This Perimeter and Area Posters is perfect to practice geometry skills. Your elementary grade students will love this Perimeter and Area Posters. Colorful posters present the rules for determining perimeter and area for regular polygons including triangles, and quadrilaterals including squares, rectangles, and parallelograms. Common Core: Geometry 6.G.A1, 5.G.B3, 4.MD.3
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• This Measure a Room (elem/upper elem) Hands-On Math is perfect to practice geometry skills. Your elementary grade students will love this Measure a Room (elem/upper elem) Hands-On Math. Learn and practice the calculation of perimeter and area by measuring a room. Common Core: Geometry 6.G.A1, 5.G.B3, 4.MD.3
• Eight pages of Math practice investigating Chichen Itza. Common Core: Geometry 6.G.A1, 5.G.B3, 4.MD.3
• This Earth Day Bar Graph is perfect to practice graphing skills. Your elementary grade students will love this Earth Day Bar Graph. Use this simple bar graph to determine how many trash items (categorized by type) were picked up.
• This Pencil Sales Bar Graph is perfect to practice graphing skills. Your elementary grade students will love this Pencil Sales Bar Graph. Read the bar graph to answer questions about pencil sales at Taylor Elementary School.
• Use this 'Bar Graph (create): Favorite "Trumpet of the Swan" Character' printable worksheet in the classroom or at home. Your students will love this 'Bar Graph (create): Favorite "Trumpet of the Swan" Character'. Create a bar graph to show information about favorite characters in E.B. White's "Trumpet of the Swan"
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• This Career Choices Bar Graph is perfect to practice graphing skills. Your elementary grade students will love this Career Choices Bar Graph. Use the bar graph to find out which careers these middle school students prefer. Common Core: Grade 6 Statistics and Probability: 6.SP.A.1, 6.SPA.2 Common Core: Grade 6 Statistics and Probability: 6.SP.A.1, 6.SPA.2
• This Blank Bar Graph (to 20 by 2s) is perfect to practice graphing skills. Your elementary grade students will love this Blank Bar Graph (to 20 by 2s). Make your own graphs with this graph form with 12 columns going to 20 by 2s.
• [member-created with abctools] Find the math words from "area" to "variable" in this lock-and-key shaped puzzle. Common Core: Geometry 6.G.A1, 5.G.B3, 4.MD.3
• This Graph Types Posters is perfect to practice graphing skills. Your elementary grade students will love this Graph Types Posters. Colorfully presented explanations of line graphs, bar graphs, and pie charts.
• This Weather Bar Graph Worksheet is perfect to practice graphing skills. Your elementary grade students will love this Weather Bar Graph Worksheet. Students use these worksheets to graph the weather conditions during each month of the year. Graphs up to 15 days for each of 7 weather conditions.
• A large bar graph charting a recycling project, with tally charts.
• A fun back-to-school interactive to help students become better acquainted with their classmates. Reinforces reading, counting, sorting and graphing skills. Teacher can model and lead students through the activities as students complete the corresponding booklets. CCSS: K.CC.B.5, K.CC.C.6, K.MD.B.3, 1.MC.C.4, 2.MD.D.10, 3.MD.B.3
• 10 pages of graphing practice - 3 line graphs, 3 pie charts, 3 bar graphs, and 1 tally chart. Created with our abctools math worksheet generators. CC: K.MD.B.3, 3.MD.B.3
• Set of 3 color posters illustrating conversions from millimeters to centimeters, centimeters to meters, and meters to kilometers. CC: Math: 4.MD.A.1, 5.MD.A.1
• 10 pages of practice in reading a U.S. customary ruler. Created with our abctools Math Worksheet Generator. CC: 2.MD.A.1, 3.MD.B4
• This Clock Pattern (hours) Clip Art is perfect to practice telling time skills. Your elementary grade students will love this Clock Pattern (hours) Clip Art. Page-size clock face shows the hours. Comes with hands to cut out and attach. Common Core Math: Measurement & Data 1.MD.3
• This Measurement Chart is perfect to practice measurement skills. Your elementary grade students will love this Measurement Chart. This is a good reference chart for length, mass,volume, area, liquid and dry measures. Common Core: Geometry 6.G.A1, 5.G.B3, 4.MD.3
• This Math Chart Volumes and Areas is perfect to practice geometry skills. Your elementary grade students will love this Math Chart Volumes and Areas. Graphic chart to help students study volumes and areas of geometric shapes. Common Core: Geometry 6.G.A1, 5.G.B3, 4.MD.3
• This Myself b/w Everyday Measurements is perfect to practice measurement skills. Your elementary grade students will love this Myself b/w Everyday Measurements. Printable worksheet for everyday measurement, with a "Myself" theme.
• This Measurement Sign is perfect to practice measurement skills. Your elementary grade students will love this Measurement Sign. Printable measurement sign.
• This Everyday Objects 2 Measurement is perfect to practice measurement skills. Your elementary grade students will love this Everyday Objects 2 Measurement. Use a paper clip to find out how many paper clips long each object is. Use a ruler to find out how many inches long each object is. (baseball bat, carrot, ant, shoe, bone, fork, key). | 2,524 | 10,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-34 | latest | en | 0.877735 |
https://math.stackexchange.com/questions/1448209/proving-toeplitz-matrix-defines-bounded-operator-on-l2 | 1,558,876,054,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259177.93/warc/CC-MAIN-20190526125236-20190526151236-00160.warc.gz | 547,645,506 | 35,036 | # Proving Toeplitz matrix defines bounded operator on $l^2$
I should first mention this: I have asked this question previously but I only got a partial answer that does not suit the actual assumptions but only the related ones, it reads as follows:
Define the infinite matrix $A = [a_{ij}]_{i,j=1}^{\infty}$ using a sequence $\{ \alpha_n \}_{n=-\infty}^{\infty}$ such that $a_{i,j} = \alpha_{i-j}$ and also we know $0 < \sum_{n=-\infty}^{\infty}{|\alpha_n|} < \infty$.
a. We need to show the matrix A defines a bounded operator on the space of sequences $\mathcal{l}^2$ and to find it's operator norm
b. Then we need to check if this operator is compact or not
Hint : Try to think of A as the matrix representing an operator acting on a subspace of $L^2[-\pi,\pi]$
Here is a link to the question and the solutions I got
As you can see the problem here is that my matrix has only positive indices so it is not negative infinite as well and as you can see this doesn't change much but the essence is of course that the matrix looks like the representation matrix of the product operator with a well defined function but because of the basis enumeration $\{ e^{inx} \}$ including negatives this is not completely the case. The question said a subspace of $L^2[-\pi,\pi]$ so maybe change some assumptions to make the method work. Help needed and appreciated
• It's clear that the operator cannot be compact since the identity operator can be thusly constructed. – Omnomnomnom Sep 23 '15 at 15:50
• @Omnomnomnom : Yes this I know that it is not necessarily compact but proving it is never compact is not much more difficult, but can you help me figure out the operator the matrix represents please? – kroner Sep 23 '15 at 15:52
• Not that it directly helps, but for Googling purposes: a Toeplitz matrix is one with uniform elements along its diagonals e.g. your matrices above. – Semiclassical Sep 23 '15 at 20:26
I had not noticed before that your indices on the matrix were restricted to positive entries. This matrix has the form $$\begin{pmatrix} \alpha_0 & \alpha_{-1} & \alpha_{-2} & \alpha_{-3} & \cdots \\ \alpha_1 & \alpha_{0} & \alpha_{-1} & \alpha_{-2} & \cdots \\ \alpha_2 & \alpha_{1} & \alpha_{0} & \alpha_{-1} & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{pmatrix}.$$ The subspace $H^{2}$ of $L^{2}$ of interest consists of all $f \in L^{2}[-\pi,\pi]$ for which $$(f,e^{-inx})=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{inx}dx = 0,\;\;\; n=1,2,3,\cdots.$$ Every $f \in H^{2}$ has the form $$\sum_{n=0}^{\infty}f_n e^{inx},\;\;\; \|f\|^{2}=\sum_{n}|f_n|^{2} < \infty.$$ Let $P$ denote the orthogonal projection of $L^{2}$ onto $H^{2}$ given by $$Pf = \sum_{n=0}^{\infty}f_n e^{inx}$$ In other words, $P\sum_{n=-\infty}^{\infty}f_ne^{inx} = \sum_{n=0}^{\infty}f_ne^{inx}$.
Define a function $a(z)=\sum_{n=-\infty}^{\infty}\alpha_n e^{inx}$. If you multiply $a$ by $f \in H^{2}$, and then project, you get \begin{align} P(a(z)f(z)) & = (\alpha_0 f_0 + \alpha_{-1} f_1 + \alpha_{-2} f_2 +\cdots) \\ & + (\alpha_1 f_0 + \alpha_0 f_1 + \alpha_{-1}f_2+\cdots)e^{ix} \\ & + (\alpha_2 f_0 + \alpha_1 f_1 + \alpha_0 f_2 +\cdots) e^{2ix} \\ & + \cdots \end{align} The first term is where the sum of the indices is always 0. The second term is where the sum of the indices is always 1. Etc.. The negative powers are truncated by the projection $P$. This operator has the desired matrix representation. The only question is whether or not this operator maps back into $H^{2}$, which is to say that the sum of the squares of the coefficients of $P(af)$ is finite. Check this $$\|P(af)\|^{2}=\sum_{k=0}^{\infty}|\sum_{n=k}^{\infty}\alpha_{k-n}f_{n}|^{2}$$ You can apply Cauchy-Schwarz to the inner sum after writing $$|\alpha_{k-n}f_{n}|=|\alpha_{k-n}|^{1/2}(|\alpha_{k-n}|^{1/2}||f_{n}|).$$ Let $\|a\|_{1} = \sum_{k=-\infty}|\alpha_k|$. The result is \begin{align} \|P(af)\|^{2} & \le \sum_{k=0}^{\infty}\left(\sum_{n=0}^{\infty}|\alpha_{k-n}|\sum_{n=0}^{\infty}|\alpha_{k-n}||f_{n}|^{2}\right) \\ & \le \|a\|_1\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}|\alpha_{k-n}||f_{n}|^{2} \\ & = \|a\|_1\sum_{n=0}^{\infty}|f_{n}|^{2}\sum_{k=0}^{\infty}|\alpha_{k-n}| \\ & \le \|a\|_1^{2}\|f\|^{2}. \end{align} So $P(af)$ is a bounded operator, with $\|P(af)\|\le \|a\|_1\|f\|$. This isn't the tightest bound. For example, if $\|a\|_{\infty}=\sup_{x}|a(x)|$, then $\|P(af)\| \le \|af\| \le \|a\|_{\infty}\|f\|$. Clearly $\|a\|_{\infty}\le \|a\|_1$.
• I just wanted to ask about compactness, of course the identity operator shows it is not necessarily compact but can it ever be compact? I know for a fact that the multiplication operator by a non zero function such as our $\alpha$ cannot be compact, can we relate this here? i.e. show that it is never a compact operator using this theorem – kroner Sep 24 '15 at 16:21 | 1,659 | 4,809 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2019-22 | latest | en | 0.945496 |
https://www.gradesaver.com/textbooks/science/physics/physics-for-scientists-and-engineers-a-strategic-approach-with-modern-physics-4th-edition/chapter-10-interactions-and-potential-energy-exercises-and-problems-page-256/10 | 1,534,396,092,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210413.14/warc/CC-MAIN-20180816034902-20180816054902-00566.warc.gz | 865,352,579 | 13,120 | Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)
Let $L$ be the length of the chains. We can find the height $h$ (above the swing's lowest point) when the chains are at a $45^{\circ}$ angle. $\frac{L-h}{L} = cos(\theta)$ $h = L~[1-cos(\theta)]$ $h = (3.0~m)~[1-cos(45^{\circ})]$ $h = 0.88~m$ The maximum speed will occur at the lowest point when kinetic energy is at a maximum. We can use conservation of energy to find the child's speed at the lowest point. The kinetic energy at the lowest point will be equal to the potential energy at height $h$. $KE = PE$ $\frac{1}{2}mv^2 = mgh$ $v^2 = 2gh$ $v = \sqrt{2gh}$ $v = \sqrt{(2)(9.80~m/s^2)(0.88~m)}$ $v = 4.2~m/s$ The child's maximum speed is 4.2 m/s. | 255 | 747 | {"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} | 3.96875 | 4 | CC-MAIN-2018-34 | longest | en | 0.803038 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/175/4/e/b/ | 1,722,951,001,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640492117.28/warc/CC-MAIN-20240806130705-20240806160705-00711.warc.gz | 691,110,522 | 62,621 | # Properties
Label 175.4.e.b Level $175$ Weight $4$ Character orbit 175.e Analytic conductor $10.325$ Analytic rank $0$ Dimension $2$ Inner twists $2$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [175,4,Mod(51,175)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(175, base_ring=CyclotomicField(6))
chi = DirichletCharacter(H, H._module([0, 2]))
N = Newforms(chi, 4, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("175.51");
S:= CuspForms(chi, 4);
N := Newforms(S);
Level: $$N$$ $$=$$ $$175 = 5^{2} \cdot 7$$ Weight: $$k$$ $$=$$ $$4$$ Character orbit: $$[\chi]$$ $$=$$ 175.e (of order $$3$$, degree $$2$$, minimal)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: no Analytic conductor: $$10.3253342510$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-3})$$ comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{2} - x + 1$$ x^2 - x + 1 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 35) Sato-Tate group: $\mathrm{SU}(2)[C_{3}]$
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{6}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + 3 \zeta_{6} q^{2} + (2 \zeta_{6} - 2) q^{3} + (\zeta_{6} - 1) q^{4} - 6 q^{6} + (14 \zeta_{6} + 7) q^{7} + 21 q^{8} + 23 \zeta_{6} q^{9}+O(q^{10})$$ q + 3*z * q^2 + (2*z - 2) * q^3 + (z - 1) * q^4 - 6 * q^6 + (14*z + 7) * q^7 + 21 * q^8 + 23*z * q^9 $$q + 3 \zeta_{6} q^{2} + (2 \zeta_{6} - 2) q^{3} + (\zeta_{6} - 1) q^{4} - 6 q^{6} + (14 \zeta_{6} + 7) q^{7} + 21 q^{8} + 23 \zeta_{6} q^{9} + ( - 45 \zeta_{6} + 45) q^{11} - 2 \zeta_{6} q^{12} - 59 q^{13} + (63 \zeta_{6} - 42) q^{14} + 71 \zeta_{6} q^{16} + (54 \zeta_{6} - 54) q^{17} + (69 \zeta_{6} - 69) q^{18} + 121 \zeta_{6} q^{19} + (14 \zeta_{6} - 42) q^{21} + 135 q^{22} + 69 \zeta_{6} q^{23} + (42 \zeta_{6} - 42) q^{24} - 177 \zeta_{6} q^{26} - 100 q^{27} + (7 \zeta_{6} - 21) q^{28} - 162 q^{29} + ( - 88 \zeta_{6} + 88) q^{31} + (45 \zeta_{6} - 45) q^{32} + 90 \zeta_{6} q^{33} - 162 q^{34} - 23 q^{36} - 259 \zeta_{6} q^{37} + (363 \zeta_{6} - 363) q^{38} + ( - 118 \zeta_{6} + 118) q^{39} + 195 q^{41} + ( - 84 \zeta_{6} - 42) q^{42} + 286 q^{43} + 45 \zeta_{6} q^{44} + (207 \zeta_{6} - 207) q^{46} + 45 \zeta_{6} q^{47} - 142 q^{48} + (392 \zeta_{6} - 147) q^{49} - 108 \zeta_{6} q^{51} + ( - 59 \zeta_{6} + 59) q^{52} + ( - 597 \zeta_{6} + 597) q^{53} - 300 \zeta_{6} q^{54} + (294 \zeta_{6} + 147) q^{56} - 242 q^{57} - 486 \zeta_{6} q^{58} + ( - 360 \zeta_{6} + 360) q^{59} - 392 \zeta_{6} q^{61} + 264 q^{62} + (483 \zeta_{6} - 322) q^{63} + 433 q^{64} + (270 \zeta_{6} - 270) q^{66} + (280 \zeta_{6} - 280) q^{67} - 54 \zeta_{6} q^{68} - 138 q^{69} + 48 q^{71} + 483 \zeta_{6} q^{72} + ( - 668 \zeta_{6} + 668) q^{73} + ( - 777 \zeta_{6} + 777) q^{74} - 121 q^{76} + ( - 315 \zeta_{6} + 945) q^{77} + 354 q^{78} - 782 \zeta_{6} q^{79} + (421 \zeta_{6} - 421) q^{81} + 585 \zeta_{6} q^{82} - 768 q^{83} + ( - 42 \zeta_{6} + 28) q^{84} + 858 \zeta_{6} q^{86} + ( - 324 \zeta_{6} + 324) q^{87} + ( - 945 \zeta_{6} + 945) q^{88} + 1194 \zeta_{6} q^{89} + ( - 826 \zeta_{6} - 413) q^{91} - 69 q^{92} + 176 \zeta_{6} q^{93} + (135 \zeta_{6} - 135) q^{94} - 90 \zeta_{6} q^{96} - 902 q^{97} + (735 \zeta_{6} - 1176) q^{98} + 1035 q^{99} +O(q^{100})$$ q + 3*z * q^2 + (2*z - 2) * q^3 + (z - 1) * q^4 - 6 * q^6 + (14*z + 7) * q^7 + 21 * q^8 + 23*z * q^9 + (-45*z + 45) * q^11 - 2*z * q^12 - 59 * q^13 + (63*z - 42) * q^14 + 71*z * q^16 + (54*z - 54) * q^17 + (69*z - 69) * q^18 + 121*z * q^19 + (14*z - 42) * q^21 + 135 * q^22 + 69*z * q^23 + (42*z - 42) * q^24 - 177*z * q^26 - 100 * q^27 + (7*z - 21) * q^28 - 162 * q^29 + (-88*z + 88) * q^31 + (45*z - 45) * q^32 + 90*z * q^33 - 162 * q^34 - 23 * q^36 - 259*z * q^37 + (363*z - 363) * q^38 + (-118*z + 118) * q^39 + 195 * q^41 + (-84*z - 42) * q^42 + 286 * q^43 + 45*z * q^44 + (207*z - 207) * q^46 + 45*z * q^47 - 142 * q^48 + (392*z - 147) * q^49 - 108*z * q^51 + (-59*z + 59) * q^52 + (-597*z + 597) * q^53 - 300*z * q^54 + (294*z + 147) * q^56 - 242 * q^57 - 486*z * q^58 + (-360*z + 360) * q^59 - 392*z * q^61 + 264 * q^62 + (483*z - 322) * q^63 + 433 * q^64 + (270*z - 270) * q^66 + (280*z - 280) * q^67 - 54*z * q^68 - 138 * q^69 + 48 * q^71 + 483*z * q^72 + (-668*z + 668) * q^73 + (-777*z + 777) * q^74 - 121 * q^76 + (-315*z + 945) * q^77 + 354 * q^78 - 782*z * q^79 + (421*z - 421) * q^81 + 585*z * q^82 - 768 * q^83 + (-42*z + 28) * q^84 + 858*z * q^86 + (-324*z + 324) * q^87 + (-945*z + 945) * q^88 + 1194*z * q^89 + (-826*z - 413) * q^91 - 69 * q^92 + 176*z * q^93 + (135*z - 135) * q^94 - 90*z * q^96 - 902 * q^97 + (735*z - 1176) * q^98 + 1035 * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q + 3 q^{2} - 2 q^{3} - q^{4} - 12 q^{6} + 28 q^{7} + 42 q^{8} + 23 q^{9}+O(q^{10})$$ 2 * q + 3 * q^2 - 2 * q^3 - q^4 - 12 * q^6 + 28 * q^7 + 42 * q^8 + 23 * q^9 $$2 q + 3 q^{2} - 2 q^{3} - q^{4} - 12 q^{6} + 28 q^{7} + 42 q^{8} + 23 q^{9} + 45 q^{11} - 2 q^{12} - 118 q^{13} - 21 q^{14} + 71 q^{16} - 54 q^{17} - 69 q^{18} + 121 q^{19} - 70 q^{21} + 270 q^{22} + 69 q^{23} - 42 q^{24} - 177 q^{26} - 200 q^{27} - 35 q^{28} - 324 q^{29} + 88 q^{31} - 45 q^{32} + 90 q^{33} - 324 q^{34} - 46 q^{36} - 259 q^{37} - 363 q^{38} + 118 q^{39} + 390 q^{41} - 168 q^{42} + 572 q^{43} + 45 q^{44} - 207 q^{46} + 45 q^{47} - 284 q^{48} + 98 q^{49} - 108 q^{51} + 59 q^{52} + 597 q^{53} - 300 q^{54} + 588 q^{56} - 484 q^{57} - 486 q^{58} + 360 q^{59} - 392 q^{61} + 528 q^{62} - 161 q^{63} + 866 q^{64} - 270 q^{66} - 280 q^{67} - 54 q^{68} - 276 q^{69} + 96 q^{71} + 483 q^{72} + 668 q^{73} + 777 q^{74} - 242 q^{76} + 1575 q^{77} + 708 q^{78} - 782 q^{79} - 421 q^{81} + 585 q^{82} - 1536 q^{83} + 14 q^{84} + 858 q^{86} + 324 q^{87} + 945 q^{88} + 1194 q^{89} - 1652 q^{91} - 138 q^{92} + 176 q^{93} - 135 q^{94} - 90 q^{96} - 1804 q^{97} - 1617 q^{98} + 2070 q^{99}+O(q^{100})$$ 2 * q + 3 * q^2 - 2 * q^3 - q^4 - 12 * q^6 + 28 * q^7 + 42 * q^8 + 23 * q^9 + 45 * q^11 - 2 * q^12 - 118 * q^13 - 21 * q^14 + 71 * q^16 - 54 * q^17 - 69 * q^18 + 121 * q^19 - 70 * q^21 + 270 * q^22 + 69 * q^23 - 42 * q^24 - 177 * q^26 - 200 * q^27 - 35 * q^28 - 324 * q^29 + 88 * q^31 - 45 * q^32 + 90 * q^33 - 324 * q^34 - 46 * q^36 - 259 * q^37 - 363 * q^38 + 118 * q^39 + 390 * q^41 - 168 * q^42 + 572 * q^43 + 45 * q^44 - 207 * q^46 + 45 * q^47 - 284 * q^48 + 98 * q^49 - 108 * q^51 + 59 * q^52 + 597 * q^53 - 300 * q^54 + 588 * q^56 - 484 * q^57 - 486 * q^58 + 360 * q^59 - 392 * q^61 + 528 * q^62 - 161 * q^63 + 866 * q^64 - 270 * q^66 - 280 * q^67 - 54 * q^68 - 276 * q^69 + 96 * q^71 + 483 * q^72 + 668 * q^73 + 777 * q^74 - 242 * q^76 + 1575 * q^77 + 708 * q^78 - 782 * q^79 - 421 * q^81 + 585 * q^82 - 1536 * q^83 + 14 * q^84 + 858 * q^86 + 324 * q^87 + 945 * q^88 + 1194 * q^89 - 1652 * q^91 - 138 * q^92 + 176 * q^93 - 135 * q^94 - 90 * q^96 - 1804 * q^97 - 1617 * q^98 + 2070 * q^99
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/175\mathbb{Z}\right)^\times$$.
$$n$$ $$101$$ $$127$$ $$\chi(n)$$ $$-\zeta_{6}$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
51.1
0.5 − 0.866025i 0.5 + 0.866025i
1.50000 2.59808i −1.00000 1.73205i −0.500000 0.866025i 0 −6.00000 14.0000 12.1244i 21.0000 11.5000 19.9186i 0
151.1 1.50000 + 2.59808i −1.00000 + 1.73205i −0.500000 + 0.866025i 0 −6.00000 14.0000 + 12.1244i 21.0000 11.5000 + 19.9186i 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
7.c even 3 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 175.4.e.b 2
5.b even 2 1 35.4.e.a 2
5.c odd 4 2 175.4.k.b 4
7.c even 3 1 inner 175.4.e.b 2
7.c even 3 1 1225.4.a.b 1
7.d odd 6 1 1225.4.a.a 1
15.d odd 2 1 315.4.j.b 2
20.d odd 2 1 560.4.q.b 2
35.c odd 2 1 245.4.e.a 2
35.i odd 6 1 245.4.a.f 1
35.i odd 6 1 245.4.e.a 2
35.j even 6 1 35.4.e.a 2
35.j even 6 1 245.4.a.e 1
35.l odd 12 2 175.4.k.b 4
105.o odd 6 1 315.4.j.b 2
105.o odd 6 1 2205.4.a.e 1
105.p even 6 1 2205.4.a.g 1
140.p odd 6 1 560.4.q.b 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
35.4.e.a 2 5.b even 2 1
35.4.e.a 2 35.j even 6 1
175.4.e.b 2 1.a even 1 1 trivial
175.4.e.b 2 7.c even 3 1 inner
175.4.k.b 4 5.c odd 4 2
175.4.k.b 4 35.l odd 12 2
245.4.a.e 1 35.j even 6 1
245.4.a.f 1 35.i odd 6 1
245.4.e.a 2 35.c odd 2 1
245.4.e.a 2 35.i odd 6 1
315.4.j.b 2 15.d odd 2 1
315.4.j.b 2 105.o odd 6 1
560.4.q.b 2 20.d odd 2 1
560.4.q.b 2 140.p odd 6 1
1225.4.a.a 1 7.d odd 6 1
1225.4.a.b 1 7.c even 3 1
2205.4.a.e 1 105.o odd 6 1
2205.4.a.g 1 105.p even 6 1
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{2}^{2} - 3T_{2} + 9$$ acting on $$S_{4}^{\mathrm{new}}(175, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{2} - 3T + 9$$
$3$ $$T^{2} + 2T + 4$$
$5$ $$T^{2}$$
$7$ $$T^{2} - 28T + 343$$
$11$ $$T^{2} - 45T + 2025$$
$13$ $$(T + 59)^{2}$$
$17$ $$T^{2} + 54T + 2916$$
$19$ $$T^{2} - 121T + 14641$$
$23$ $$T^{2} - 69T + 4761$$
$29$ $$(T + 162)^{2}$$
$31$ $$T^{2} - 88T + 7744$$
$37$ $$T^{2} + 259T + 67081$$
$41$ $$(T - 195)^{2}$$
$43$ $$(T - 286)^{2}$$
$47$ $$T^{2} - 45T + 2025$$
$53$ $$T^{2} - 597T + 356409$$
$59$ $$T^{2} - 360T + 129600$$
$61$ $$T^{2} + 392T + 153664$$
$67$ $$T^{2} + 280T + 78400$$
$71$ $$(T - 48)^{2}$$
$73$ $$T^{2} - 668T + 446224$$
$79$ $$T^{2} + 782T + 611524$$
$83$ $$(T + 768)^{2}$$
$89$ $$T^{2} - 1194 T + 1425636$$
$97$ $$(T + 902)^{2}$$ | 5,371 | 10,490 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-33 | latest | en | 0.5393 |
https://coderanch.com/t/401041/java/java/Collision-detection | 1,474,943,031,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660931.25/warc/CC-MAIN-20160924173740-00224-ip-10-143-35-109.ec2.internal.warc.gz | 832,340,825 | 9,545 | Win a copy of Functional Reactive Programming this week in the Other Languages forum!
# Collision detection
Jack Smith
Greenhorn
Posts: 2
I'm doing a tile based shooter with a top-down view.
I thought about making all the tiles instances of the TILE object. Like all got parameters for what image they use and if they are solid or not. I would make the constructor paint the image and make a bounding box. I have used bounding boxes for collission detection earlier but so far I have just used a few boxes.
Now I would wana know if there is a way to check if the players bounding box intersects with ANY of the tile bounding boxes. So that you don't need to test with all tiles separately.
I'm actually pretty confused on all this and if someone got a better idea please tell me.
Layne Lund
Ranch Hand
Posts: 3061
If I understand correctly, you are thinking about writing a Tile class and then creating Tile objects. Then during the game, you will need to do some collision detection between the player and these Tiles. Are you asking how to do this efficiently? If so, it highly depends on how you are storing the Tiles. So let's talk about some different data structures that you can use and see how each data structure affects efficiency.
The naive approach is to store the Tiles as a linked list (perhaps using the LinkedList class from the Collections API). This is probably what you are trying to avoid, however, since it requires doing the collision detection with every single tile. This algorithm will be O(n) where n is the number of Tile objects.
Obviously the above approach is poor. A much better approach may be to store the Tiles in a 2D array. Of course, this approach assumes that each "grid location" in the game only has a signle tile associated with it. This data structure will lead to a collision detection of O(1) (i.e. constant time) because if the player is in grid location (x, y), then you know the player collides with the Tile object in location (x, y). To me, this seems like a much more logical choice than the LinkedList idea above.
If you need to get more complex, especially if you want to make a 3D game (instead of just 2D), you may want to look into using what are called half-spaces. I won't get into this in much detail, but it will require a tree (typically a binary tree) as the data structure. You should google for more information if this sounds like what you need.
Good luck with your game. In the future, you may want to post questions in our Games Forum which is specific for these types of things.
Layne
Jack Smith
Greenhorn
Posts: 2
I managed to upload the game so you get a better idea what I'm trying to do.
The Game
Arrow keys and wasd is used for movement
Your second suggestion sounded pretty good but as you see the player doesn't move in tiles so I don't have the players cordinate in the grid stored anywhere. So do you have any suggestion how I should make it? All tiles are 20x20 pixels in size. My idea is that if player cordinates are e.g 56,89 then the place in the grid would be in tile 3 from left(56/20 = 2.8) and tile 5 from top(89/20 = 4,45). But this way it would only detect the collision of the left upper corner of the player. Any ideas? | 735 | 3,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-40 | latest | en | 0.956333 |
http://blogs.ethz.ch/kowalski/2009/09/24/duplicating-barnes/ | 1,542,186,801,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741764.35/warc/CC-MAIN-20181114082713-20181114104713-00179.warc.gz | 34,582,120 | 8,278 | # Duplicating Barnes
I have mentioned the Barnes function already once recently, as it arose in a marvelous identity of Widom. Today, here is a look at the duplication formula…
Recall that the Barnes function is defined by
$G(z+1)=(2\pi)^{z/2}e^{-(z(z+1)+\gamma z^2)/2}\prod_{n\geq 1}{(1+z/n)^n\exp(-z+z^2/(2n)}}$
for any complex number z (it is indeed an entire function; here γ is the Euler constant). Besides the ratio
$\frac{G(1+z)^2}{G(1+2z)}$
which appeared in the previous post, the following expression
$f_{Sp}(z)=2^{z^2/2}\frac{G(1+z)\sqrt{\Gamma(1+z)}}{\sqrt{G(1+2z)\Gamma(1+2z)}}$
also occurs naturally in the asymptotic formula of Keating-Snaith for characteristic polynomials of unitary symplectic matrices. For some reasons, this didn’t look nice enough to me (e.g., the computation of
$f_{Sp}(n)=\frac{1}{(2n-1)(2n-3)^2\cdots 1^n}$
for n a positive integer is somewhat laborious). But because of the shape of the term in the square root of the denominator, one can try to simplify this using the basic relation
$G(1+2z)\Gamma(1+2z)=G(2+2z)=G(2(1+z)),$
and the duplication formula for G(2w), which “must exist”, since the Barnes function generalizes the Gamma function, for which it is well-known that
$\Gamma(2z)=\pi^{-1/2}2^{2z-1}\Gamma(z)\Gamma(z+1/2)$
(usually attributed to Legendre).
So there is indeed a duplication formula; I picked it up from this paper of Adamchik (where it is cleaner than in the paper of Vardi referenced by Wikipedia), who states it as follows:
$G(2w)=e^{-3\zeta^{\prime}(-1)}2^{2w^2-2w+5/12}(2\pi)^{(1-2w)/2}G(w)G(w+1/2)^2G(w+1).$
This is promising since in the case considered, the terms involving G, after application of the formula to w=1+z, become
$G(z+3/2)^2G(z+1)G(z+2)=\Gamma(z+1)G(z+1)^2G(z+3/2)^2,$
leading to a cancellation with both the gamma and G-factors in the numerator!
And although the term ζ'(-1) is maybe a bit surprising, a moment’s thought shows it can be eliminated by simply plugging in any specific value of z and using the resulting formula to express it in terms of a specific value of G(z). Indeed, taking z=1/2, and watching the dust settling lazily across the page, we get the very nice expression
$f_{Sp}(z)=2^{-z^2/2-z-1/2}(2\pi)^{(z+1)/2}\frac{G(1/2)}{G(z+3/2)}.$
In fact, it’s clear then that (at least for such purposes), it is best to write the duplication formula in a way which avoids ζ'(-1) altogether (losing a bit of information, since it’s somewhat interesting to know that this quantity is linked to G(1/2)):
$G(1/2)^2G(2w)=(2\pi)^{-w}2^{2w^2-2w+1}\Gamma(w)((G(w)G(w+1/2))^2.$
(This is still not tautological for z=1/2: it contains the value Γ(1/2)=π1/2.) From the shape of this, number theorists, at least, would probably be curious to see what happens if one replaces the Gamma function with
$\Gamma_{\mathbf{C}}(w)=(2\pi)^{-w}\Gamma(w)$
which is the factor at infinity for the local field C. In that case, it seems natural to introduce
$G_{\mathbf{C}}(w)=(2\pi)^{-z(z-1)/2}G(w),$
for which we retain the induction relation of G with respect to Γ:
$G_{\mathbf{C}}(w+1)=\Gamma_{\mathbf{C}}(w)G_{\mathbf{C}}(w).$
In terms of these functions, the duplication formula is even nicer:
$G_{\mathbf{C}}(1/2)^2G_{\mathbf{C}}(2w)=2^{2w^2-2w+1}\Gamma_{\mathbf{C}}(w)((G_{\mathbf{C}}(w)G_{\mathbf{C}}(w+1/2))^2.$ | 1,118 | 3,320 | {"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": 14, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-47 | latest | en | 0.887578 |
https://jacobbradjohnson.com/mathsolver-660 | 1,670,307,968,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711074.68/warc/CC-MAIN-20221206060908-20221206090908-00229.warc.gz | 365,424,199 | 5,820 | # Lcd of rational expressions solver
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# A salesperson earns 5% on all sales between \$200 and
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A salesperson earns 5% on all sales between \$200 and [#permalink]
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12 Feb 2004, 14:10
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A salesperson earns 5% on all sales between \$200 and \$600,and 8 % on all sales over \$600. What is her commission in a week in which her sales total \$800.?
20
46
88
36
78
simple yet bit confusing..
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shubhangi
Last edited by shubhangi on 12 Feb 2004, 14:39, edited 1 time in total.
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12 Feb 2004, 14:27
D)
400 * .05 + 200(portion over 600\$) * .08 = 20 + 16 = 36
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12 Feb 2004, 14:53
of \$800 he earns 8% for the last \$200 = 16
Bellow \$600 he earns 5% = 30
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12 Feb 2004, 14:55
Anandk, it's 5% on all sales between \$200 and \$600
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12 Feb 2004, 14:57
Anand.. gotcha!! did same mistake
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shubhangi
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12 Feb 2004, 15:00
I consider this question stupid.
I have seen companies who give a different percentage rate for different levels of sales.
For Example at car dealership if a salesman sells 10 cars then the percentage will be 10 per car and if he sells 20 then percentage will be 15 and if he sells less then 10 then percentage will be 5. What do think the dealer will pay to his sales person if that sales person sells 12 cars?
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12 Feb 2004, 15:29
Hmmm, what about a pen salesman whose commission is based on sales value. It could be very much possible that the salesman will not get any bonus until he sells 200\$ worth of pens. Then from 200\$-600\$, he gets 5% and from sales more than 600\$, he gets 8%. The car example comparison is not appropriate because the underlying good's unit is a luxury good for which the bonus scheme might be very different.
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13 Feb 2004, 02:51
Answer is 36. I have seen Anandnk getting caught for the first time. Kudos Shubhangi for the cute good one......
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14 Feb 2004, 02:29
anandnk wrote:
I consider this question stupid.
I have seen companies who give a different percentage rate for different levels of sales.
For Example at car dealership if a salesman sells 10 cars then the percentage will be 10 per car and if he sells 20 then percentage will be 15 and if he sells less then 10 then percentage will be 5. What do think the dealer will pay to his sales person if that sales person sells 12 cars?
Do you pay taxes? We get taxed the same way this salesperson gets paid.
This type of bonus structure is very common in sales and thus I don't think the question is "stupid", but a good example of something you might see on an actual GMAT.
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14 Feb 2004, 07:08
hi Akamaibrah,
I do pay taxes. My friend earns \$30000 as yearly pay and pays nothing in taxes. Another friend of mine earns \$60,000 as yearly pay and pays \$15000 in taxes at 25% tax rate because he falls in \$35000 to 65000 bracket.
Isn't the problem given above same?
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16 Feb 2004, 01:43
anandnk wrote:
hi Akamaibrah,
I do pay taxes. My friend earns \$30000 as yearly pay and pays nothing in taxes. Another friend of mine earns \$60,000 as yearly pay and pays \$15000 in taxes at 25% tax rate because he falls in \$35000 to 65000 bracket.
Isn't the problem given above same?
I'm glad you pay taxes. If so, then you should know that how much someone earns is insufficient information to calculate his/her taxes. What is important is what part of that income is taxable.
The point of my question is that once you get down to the "taxable" income figure, each dollar of this income is taxed according to where it lies on a scaled table of percentages. A person in a particular tax bracket only pays that percentage on income earned within that bracket, not his or her entire income.
I don't know what the real numbers are but something like first 7500 are not taxed, then 7500-15000 are taxed at 15% the 15001 to 23000 are taxed at 18% and so on. Hence, we have a scaled system similar to that of the salesperson's pay, where he make get a percentage commission on the first x items, then a different commission on anything in excess of x.
If you think about it, this is also why we have tax lookup tables -- it is not a straightforward calculation to figure out ones tax because a part of each person's income is taxed at different percentages.
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16 Feb 2004, 01:43
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### 2 engineers on a bike
two engineers were biking across a university campus when one said,"where did you get such a great bike?" the second engineer replied, "well, i was walking along yesterday, minding my own business, when a beautiful woman rode up on this bike, threw it to the ground, took off all her clothes and said...
### 4 Norse god, 1 Roman god, and 2 astronomical bodies walk into a bar
The bartender says " Oh, this is a gonna be a week joke"
### There's a guy who smokes 2 cigarettes together
They asked him: why do you always smoke 2 cigarettes together?
He said: one for me, and one for my brother in prison.
After a while they saw him smoking one cigarette only and they asked him: so your brother is out of the jail?
He said: no, I stopped smoking.
### Someone asked me to name 2 structures that hold water.
I was like, well damn.
### The captain asks a seamen to tell him how "2" is written in Latin.
The seamen replies "Aye aye, capt'n!"
This joke may contain profanity. 🤔
### I went to bed with 2 girls from Thailand last night.
It was amazing, it was like winning the lottery.
We had six balls between us.
### An elderly man is stopped by the police around 2 a.m. and is asked where he is going at this time of night.
The man replies, "I am on my way to attend a lecture about gambling, hookers, alcohol abuse and the effects it has on the human body, as well as smoking, and staying out late."
The officer then asks, "Really? Who is giving that lecture at this time of night?"
The man replies, "My wife...
### Man 1: I have a half sister. Man 2: Different father?
Man 1: No, shark attack.
### Q: Why did the blonde stare at the orange juice bottle for 2 hours?
A: Because it said 'concentrate'
### a^2 + b^2 = c^2
But only if the angle is right.
### For 2 years a man was having an affair with an Italian woman......
One night, she confided in him that she was pregnant.
Not wanting to ruin his reputation or his marriage, he promised to pay her a large sum of money if she would go to Italy to secretly have the child.
If she stayed in Italy to raise the child, he also promised to provide child suppor...
cancer.
### An Airbus 380 is on its way across the Atlantic. It flies consistently at 800 km/h at 30,000 feet, when suddenly a Eurofighter with a Tempo Mach 2 appears.
The pilot of the fighter jet slows down, flies alongside the Airbus and greets the pilot of the passenger plane by radio: "Airbus, boring flight isn’t it? Now have a look here!"
He rolls his jet on its back, accelerates, breaks through the sound barrier, rises rapidly to a dizzying height, and ...
This joke may contain profanity. 🤔
### What do you call 2 nuns and a Prostitute on a football field?
Two tight ends and a wide receiver.
This joke may contain profanity. 🤔
### When I was a younger lad I was blessed with an 8 1/2" penis
Unfortunately it belonged to father O'Malley
### 2 nuns were sitting on a park bench when a guy came running through and flashed them. One of the nuns had a stroke...
The other one couldn't reach.
### 2 foreign immigrants have just arrived in USA by boat and one says to the other,
''"I hear that the people of this country actually eat dogs." "Odd," her companion replies, "but if we shall live in America, we might as well do as the Americans do." Nodding emphatically, one of the immigrants points to a hot dog vendor and they both walk toward the cart. "Two dogs, please," she s...
### 2 pilots meet
300 people died
This joke may contain profanity. 🤔
### You and 2 friends of yours walk through a forest
after a while you lot stumble upon a hut, from which a weird old lady, resembling a witch, comes out from. She slowly says
*"...do not step on the purple flower..."*
and then goes back into her hut.
A little confused, you exchange looks with your friends, shrug, and keep walking.
...
### Don’t you hate it when you can’t sleep because you are reminded of a mistake you made 2 years ago?
I hate it when my kid cries in the middle of the night
### 1 and 2 went out for a walk in the snow.
1's hands got so cold that they went numb.
2's hands and feet both got cold, so he was even number.
### Unintentional joke from a 2 and a half year old:
Oh no, I'm falling in the sink!
I'm sinking!
### If a woman drinks 2 glasses of wine a day, it increases the chances of a stroke.
If you let her finish the bottle, she'll probably suck it as well.
Is it still beef
### 2 tourists are driving through Wales...
They stop for lunch in Llanfairpwllgwyngyllgogerchwndrobwllllantysiliogogogoch.
As they sit down for lunch, one of the tourists asks the waitress:
"Can you settle an argument for us please? Can you pronounce the name of where we are, right now, very slowly?"
The waitress leans...
### 2 hunters are duck hunting.....
After a couple hours and missing on several flocks, theyre not having any luck catching anything, one turns to the other and says, "I can't believe we haven't caught anything yet", to which the other responds, "i know right, maybe we aren't throwing the dog up high enough"
This joke may contain profanity. 🤔
### My wife is so fat that when she booked a flight they made her have 2 seats.
She was pissed off until I mentioned that she would get 2 meals
This joke may contain profanity. 🤔
### An ugly woman walked into a store with her 2 kids, yelling at them.
An ugly woman walked into a store with her 2 kids, yelling at them.
The store clerk pleasantly said, "Good morning ma'am and welcome. Nice children, are they twins?"
The ugly woman stopped yelling and said, "Hell no they are not, one is 9 and the other is 7. Why the hell would u thin...
### '2' managed to be prime,
Against all the odds.
This joke may contain profanity. 🤔
### A sperm cell contains about 37.5 MB of information. There are about 100 million sperm cells per ml; the average ejaculation is about 2.25ml, and takes about 5 seconds. This makes the average bandwidth of the human penis 1687 TB/sec
I know, that's a lot of information to swallow.
### 2 plus 2
A business man asks his doctor what 2 plus 2 is and is told 4. He then asks his lawyer what 2 plus 2 is and gets the same answer 4. Finally he asks his accountant what 2 plus 2 is. The accountant replies 'what do you want it to be',
This joke may contain profanity. 🤔
Doppelwangers
### I asked my doctor to use 2 fingers when checking my prostate..
I wanted a second opinion.
### Why is it wrong to drive a van with 2 lawyers off a cliff?
Because you could have fit 20 of them.
### 2 blondes, 2 brunettes, and 2 redheads walk into a bar. The 2 blondes say "hello" to the bartender...
The 4 Non Blondes say "WHAT'S GOING ON!"
This joke may contain profanity. 🤔
### Cashier: Would you like to donate \$2 to end world hunger?
Me: Of course. Holy shit, I had no idea we were that close.
### 2 Irishmen, 2 Scotsmen, and 2 Englishmen
There were 2 Irishmen, 2 Scotsmen, and 2 Englishmen stuck on a deserted island. In one year, the two Irishmen made a still and was brewing beer, the two Scotsmen built a pub and were selling it. The two Englishmen still weren't talking to each other because they weren't properly introduced.
This joke may contain profanity. 🤔
### Royal Union 2
Once upon a time, the royals of two neighbouring kingdoms decided to cement their friendship and their boundaries by marrying one's princess to the other's prince.
The youngsters were introduced to each other and, as luck would have it, they fell in love. The prince's father, however, wanted ...
### How do you break up a fight between 2 blind men?
Yell out: 'My money is on the one with a knife...'
### A man was riding his motorcycle through the border of Germany and Austria every week carrying 2 bags filled with sand.
The border guard, an older man, searched both bags every time, but never found anything so he let him through. This goes on for a couple days until the border guard had his last day before retirement. Again the man comes to the boarder, both bags filled with sand. The guard asks him: "Look man, toda...
### What's worse than 2 girls running with scissors?
2 girls scissoring with the runs.
### It has been 2 years and still nobody knows why Notre Dame caught fire....
...but Quasimodo has a hunch.
This joke may contain profanity. 🤔
### What do you call the space between 2 artificial breasts?
Silicone Valley......
I'll leave and close the door behind me
King Philip III
### What walks on 8 legs until it's one years old, 4 legs until it's twenty years old and then 2 legs for the remainder of it's life?
Fred and George Weasley.
Younglings jokes
The Younglings
### \$2.1 million worth of textbooks were stolen the other day
All eight books were recovered.
### I'm a screenwriter and I just signed an amazing 2-year deal with the parent company of Universal Pictures!
I'm going to be getting the basic cable plus HBO.
### After gaining weight, My husband bought me a dress 2 sizes below and says...
"I look forward to seeing you in it".
So for his birthday I bought him a coffin.
### A woman wakes her husband up at 2 AM, saying
"Quick, who's scored the highest number of goals in football, ever?"
"Klose", replied the groggy husband.
"And how many episodes of Breaking Bad are there in total?"
"Huh? Wait, let me...55, no, 62, there's 62 total episodes" he replied.
"Who was that girl in that 'Saved ...
This joke may contain profanity. 🤔
### 2 Girls 1 Nerd
A kind of nerdy, loner-type guy finally builds up the courage to talk to two hot women. As he's walking up he's trying to think of something funny and interesting to say when he gets to them he blurts out "97% of women masturbate in the shower!" The two women look at him oddly but are intrigued an...
### What did 2 say to 3 when they saw 6 act like an idiot?
Don’t mind him. He’s just a product of our times.
This joke may contain profanity. 🤔
### 1,2,3,4 I declare a thumb war...
...5,6,7,8 I use this hand to masturbate.
### There are only 2 secrets to success in this life
1: never reveal everything you know
### how does 2+2=5?
By mistake
This joke may contain profanity. 🤔
### Camel joke (take 2)
A wealthy London business man had been doing business in Saudi. His Saudi associate was so pleased with his work he decided to send him a camel as a special gift. The man woke up one morning at his home in Central London and found a camel tied up in the front yard with a note of appreciation tied ar...
### A man goes to the track and bets \$2 on a long shot and wins \$18.
So he puts that \$18 in the 2nd race and wins again \$128. Again he puts it all on a long shot in the 3rd
race and again wins \$770!
He keep doing this for each race, and finally on the last race he puts his entire winnings so far - \$1,941,550!
The crowds are all around him watch...
### *opens your fridge and sees 2% milk*
Hey bro, you should charge your milk.
This joke may contain profanity. 🤔
### Thumbelina, Snow White, and Stalker 2 were walking through the forest
Thumbelina, Snow White, and Stalker 2 are walking through the forest.
Thumbelina says, “I'm the smallest in the world.”
Snow White says, “And I'm the most beautiful in the world.”
Stalker 2 says, “I'm the most anticipated game.”
They keep walking and walking until they fi...
### 2 girls meet: "Me & my husband are no longer together..." "Why?" "Well, could you live with a person who smokes weed, drinks, has no job and always cusses?"
"No, of course I couldn't!"
"Well he couldn't either!"
### Franklin 2.0
I used to collect pennies in a jar, but the jar filled up quickly. So I bought an urn.
A penny saved is a penny urned.
### You're on vacation, and you've arrived at your hotel. The elevators in the lobby are numbered, from left to right, 1, 2, 3, 5, and 4.
Curious, you try to enter elevator 5, but are stopped by the bell boy.
"You can't use that elevator," he says.
"Why not?"
"It's out of order"
### Dude 1 : Hey, Bro? Dude 2 : Yeah Bro? Dude 1 : Can you pass me that pamphlet?
Dude 2 : Brochure
### Option 1: Let’s eat grandma. Option 2: Let’s eat, grandma.
There you have it. Proof that punctuation saves lives.
This joke may contain profanity. 🤔
### 2 women go out drinking and after a heavy night and a lot of alcohol, decide to take a shortcut through the cemetery.
Whilst taking the shortcut they both have the urge to pee so they duck down behind the gravestones and pee. They realise they have nothing to wipe with, so the first woman decides to use her pants and leaves them there. The second woman was luckier, she found a bouquet of flowers and unwrapped them ...
### 2 thiefs try to rob a nun
One of them is holding the nun at gunpoint while the other grabs the nun.
Robber: Give us all your money!
Nun: I don't have any, I am just a nun and gave it all to the poor.
R: Pat her down, I am sure she has something.
The partner does not find anything.
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# Try our Free Online Math Solver!
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# volume help. show work
At A water tank is being filled by water being pumped into the tank at a volume given by the formula, P(t) = 60t +1060, where t is in minutes. At the same time the water tank has a leak and the volume of water draining out of the tank is given by the formula L(t) = 3t2, where t is in minutes.
(a.) The volume, V, of water in the tank at any minute, t, is the difference of the volume of the water being pumped into the tank and the volume of water linking out of the tank. Find the volume function, V(t).
(b.) What is the volume of water in the tank after 17 minutes? Show work.
(c.) The volume function is a quadratic function and so its graph is a parabola. Does the parabola open up or down? How do your know? _______________________
(d.) Find the vertex of the volume function P(x). Show work.
(e.) State the number of time which yield the maximum volume, and state the maximum volume. The maximum volume is _______________ at ____________ minutes
### 1 Answer by Expert Tutors
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
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At A water tank is being filled by water being pumped into the tank at a volume given by the formula, P(t) = 60t +1060, where t is in minutes. At the same time the water tank has a leak and the volume of water draining out of the tank is given by the formula L(t) = 3t^2, where t is in minutes.
(a.) The volume, V, of water in the tank at any minute, t, is the difference of the volume of the water being pumped into the tank and the volume of water linking out of the tank. Find the volume function, V(t).
V(t) = (P–L)(t) = 60t +1060 – 3t^2 = – 3t^2 + 60t +1060
(b.) What is the volume of water in the tank after 17 minutes? Show work.
17 | –3 60 1060
–51 153
–3 9 | 1213 = V(17)
(c.) The volume function is a quadratic function and so its graph is a parabola. Does the parabola open up or down? How do your know? _______________________
Opens down because leading coefficient is negative.
(d.) Find the vertex of the volume function P(x). Show work.
h = –b/2a = –60/(2(–3)) = 10
k = c - ah^2 = 1060 - -3(10^2) = 1360
Vertex = (h,k) = (10,1360)
(e.) State the time which yields the maximum volume, and state the maximum volume. The maximum volume is ___1360______ at ___10_______ minutes | 673 | 2,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-26 | latest | en | 0.923786 |
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1. ## Solving the Longitude Problem for Navigation
First prize of ?20,000 goes to the person who illustrates a method that rightfully determines longitude to an accuracy of half a degree of a great circle. Second prize was ?15,000 for a method accurate to within two thirds of a degree. Last but not least, the method that perfectly measures to within one degree receives ?10,000. The Board of Longitude consisted of scientists, naval officers and government officials, which were a panel of judges that were established by the Longitude Act.
• Word count: 505
2. ## Thailand/Bangko. Explain how different factors impact on travel to a selected worldwide destination
a monsoon rains which is really dangerous for everyone which means that the negative side of this country and what the tourist don?t like as well as local people. During the heavy monsoon season rains can be very torrential to the country and the living city also it can affect travel and even cause floods. In Bangkok the monsoons are sudden very quickly and this means that if it is rain suddenly the rain can catch people unprepared and makes it more difficult to escape, especially for elderly people as well as people who come for a holiday.
• Word count: 772
3. ## Letter about saving the orangutans
Firstly, why do wealthy want a pet like Orang-utans? The illegal pet trade in Orang-utans of wildlife is, unfortunately, very common in Indonesia and throughout Southeast Asia. Orang-utan infants are usually poached then sold on the black market for pets with the price up to thousands of dollars as they are cute. According to the information about Orang-utans, babies cling to their mothers and suckle their mother?s milk until the age of 6 years. Many baby orangs die before they become pets. Stress and disease kills as many as four out of five young orangs in the first few months after being caught.
• Word count: 630
4. ## Discuss the contrasting health care approaches in countries at different stages of economic development (LEDC + MEDC)
Different countries at different stages of development have varying healthcare approaches. Within countries such as the USA healthcare is focused mainly on the aging population, with many people suffering from age related illnesses such as Arthritis. In contrast to many of its European counterparts, The USA offers a pluralistic healthcare system, i.e. healthcare is viewed as a consumer product. The state role in healthcare is minimal and indirect, allowing healthcare to exist based on free market principles. Within America the consumer pays for everything and most people have to have insurance to cover the cost.
• Word count: 693
5. ## Global Warming Speech
amendment Wednesday that would have put the chamber on record backing the widely held scientific view that global warming is occurring and humans are a major cause. ?Congress accepts the scientific findings of the Environmental Protection Agency that climate changes is occurring, is caused largely by human activities, and poses significant risks for public health and welfare.? Lawmakers voted 184-240 against Rep. Henry Waxman?s (D-Calif.) amendment.? Dr.
• Word count: 563
6. ## The Impact of Tourism on Benidorm.
The fact that it?s in one of the most southern regions of Spain and is faces the sun is one of the many reasons so many people were drawn to Benidorm. Another reason it became so popular is the fact that it is only a 2-3 hour flight from most places in Europe and it?s cheap and affordable for most families.For the majority of the local people tourism is their income in one way or another, with 85% of income being from tourism.
• Word count: 435
7. ## Issues in Australian Environmental Management.
Sophisticated waste disposal systems are therefore required to deal with the enormous quantities of ensuing rubbish and pollution. Finding sustainable methods of waste disposal while simultaneously upholding the comforts of Australians produce about 3 kilograms of waste per person. This waste can be in the form of solid (or dry) waste, liquid waste, or gaseous waste. As urban growth continues to take old in many of Australia?s capital cities, our levels of all these types of waste, combined with the problems created when it comes to disposing of them are constantly increasing.
• Word count: 627
8. ## Regeneration of Londons Docklands- Geography
Also, portside industries and manufacturing were declining too. Containerisation (the use of cranes and not people) was another factor of the London Docks going into decline. In the Isle Of Dogs in 1981, the population had declined and so had the employment rate. Moreover, access to the rest of London was poor.
• Word count: 453
9. ## Potentially Positive Effects of Climate Change
Once, the artic icebergs have melted, ships are able to simply through the arctic ocean, above Russia into the Asia, in a much faster route. Currently, Russian and other European shippers have to travel a very long way, through the Suez Canal in order to service Asia. Once the northern trade route is opened, it is one third shorter than the current route, saving both time and fuel for European shippers coming into Asia. This route will be much more safer for the shippers as they will be able to avoid pirates roaming in the East Africa Seas and also trouble triggered by Arab Spring Revolution in the area around the Egyptian waterway.
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`The product of the L.C.M. and H.C.F. of two numbers is 84. The sum of two numbers is 19. Find the numbers`
1. `2 & 17`
2. `6 & 13`
3. `7 & 12`
4. `8 & 11`
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`The least number, which when divided by 13, 18 and 20 leaves in each case a remainder of 7, is:`
1. `1504`
2. `2347`
3. `3544`
4. `3748` | 219 | 607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2018-34 | latest | en | 0.77143 |
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# Princeton University
Ph501
Electrodynamics
Problem Set 2
Kirk T. McDonald
(1998)
kirkmcd@princeton.edu
http://puhep1.princeton.edu/~mcdonald/examples/
Princeton University 1998 Ph501 Set 2, Problem 1 1
## 1. Show that the electromagnetic energy of a dielectric subject to elds E and D = E is
1
U= E D dVol, (1)
8
by considering the model of atoms as springs (Problem 8b, set 1). The energy U then
has two parts:
1
U1 = E2 dVol, (2)
8
stored in the electric eld, and
kx2
U2 = n dVol, (3)
2
stored in the spring-like atoms (n is the number of atoms per unit volume). Assume n
is small so that the dielectric constant is nearly 1.
Princeton University 1998 Ph501 Set 2, Problem 2 2
## 2. (a) Show that the energy of a quadrupole in an external electric eld E,
1 Ej
6 xi
in terms of its quadrupole tensor Qij , can be rewritten as
Qxx Ex
4 x
if the quadrupole is rotationally symmetric about the x axis. Give an expression
for the force F on the quadrupole.
(b) A rotationally symmetric quadrupole of strength Qxx (zero net charge, zero dipole
moment) is located at distance r from a point charge q. What is the force on the
i. The x axis is along the line joining Qxx and q?
ii. The x axis is perpendicular to the line joining Qxx and q?
t y t
q1 2q1 q1
Princeton University 1998 Ph501 Set 2, Problem 3 3
## 3. The principle of an electrostatic accelerator is that when a charge e escapes from a
conducting plane that supports a uniform electric eld of strength E0, then the charge
gains energy eE0 d as it moves distance d from the plane. Where does this energy come
from?
Show that the mechanical energy gain of the electron is balanced by the decrease in
the electrostatic eld energy of the system.
Princeton University 1998 Ph501 Set 2, Problem 4 4
4. (a) Two point dipoles of strength p are aligned along their line of centers, and distance
2d apart. Calculate the force between the dipoles via F = (p )E, and by means
of the Maxwell stress tensor.
(b) A spherical conducting shell of radius a carries charge q. It is in a region of zero
external eld. Calculate the force between two hemispheres in two dierent ways.
Princeton University 1998 Ph501 Set 2, Problem 5 5
5. (a) Two coaxial pipes of radii a and b (a < b) are lowered vertically into an oil bath:
If a voltage V is applied between the pipes, show that the oil rises to height
( 1)V 2
h= , (6)
4g ln b
a
(b2 a2)
## where g is the acceleration due to gravity.
(b) Recalling prob. 1(c) of set 1, discuss qualitatively how the force arises the pulls
the liquid up into the capacitor.
Princeton University 1998 Ph501 Set 2, Problem 6 6
## 6. According to a theorem of Green, the potential (x) in the interior of a volume V
can be deduced from a knowledge of the charge density (x) inside that volume plus
knowledge of the potential and the normal derivative /n of the potential on the
surface S that bounds the volume,
(x) 1 1 1 (x)
(x) = dVol +
(x ) dS , (7)
V R 4 S n R R n
where R = |x x | is the distance between the point of observation and the element
of the integrand. However, further insights of Green indicate that it suces to specify
only one of or /n on the bounding surface to determine the potential within. As
a particular example, show that the potential within a charge-free sphere of radius a,
centered on the origin, can be determined from knowledge of only the potential on
its surface according to (Poisson, 1820)
a 2 x2 (x)
(x) = dS . (8)
4a S R3
Green (1828) gave a derivation of Poissons integral (8) that can be generalized to
many other problems in electrostatics. Recall that a key step towards eq. (7) is the
identity
( ) dVol = (2 2 ) dVol
V V
= ( ) dS = dS. (9)
S S n n
For problems in which the interior of volume V is charge free the potential obeys
2 = 0 there. To have a nonzero potential inside V there must, of course, be
charges on the surface of or exterior to volume V . If function also obeys 2 = 0
inside V (and so might be the potential for some other distribution of charges exterior
to V ), then the identity (9) reduces to
0= dS. (10)
S n n
Hence, we could combine eqs. (7) and (10) to yield the relation
1 1 1 (x )
(x) = (x) + + dS
4 S n R R n
1 G(x, x ) (x )
= (x )
G(x, x )
dS , (11)
4 S n n
where
1
G(x, x) = + . (12)
R
Princeton University 1998 Ph501 Set 2, Problem 6 7
IF the Greens function G(x, x) vanishes on the surface S, then we have the desirable
relation between the potential in the interior of V and its value on the bounding
surface S,
1 G(x, x)
(x) = (x) dS . (13)
4 S n
Green noted that the auxiliary potential can be thought of as due to exterior charges
that bring the surface S to zero potential when there is unit charge at position x inside
volume V , and G as the total potential of that charge conguration. Further, we may
think of the bounding surface S as being a grounded conductor for the purposes of
determining the potentials and G, in which case the exterior charges reside on
the surface S. Hence, it is plausible that these exist for interesting physical surfaces S
(although it turns out that mathematicians have constructed examples of surfaces for
which a Greens function does not exist).
Since the function G is the potential for a speciable charge conguration, the normal
derivative G/n corresponds to the electric eld (whose only nonzero component
is En ) at the surface S produced by those charges. If we consider surface S to be
a grounded conductor when determining function G, then the charge density G at
position x on that surface, caused by the hypothetical unit charge at x, would be
G (x, x) = En /4 = (1/4)G/n. Green emphasized this phyisical interpretation
in his original work, and wrote eq. (13) as
(x) = G (x, x)(x) dS . (14)
S
Turning at last to Poissons integral (8), we see that the needed Greens function for
a sphere corresponds to the potential at x due to unit charge at x in the presence of
a grounded conducting sphere of radius a. Use the method of images to construct the
Greens function and its normal derivative, and thereby verify Poissons result.
Princeton University 1998 Ph501 Set 2, Problem 7 8
## 7. A parallel-plate capacitor is connected to a battery which maintains the plates at
constant potential dierence V0 . A slab of dielectric constant is inserted between the
plates, completely lling the space between them.
(a) Show that the battery does work Q0V0 ( 1) during the insertion process, if Q0
is the charge on the plates before the slab is inserted.
(b) What is the change in the electrostatic energy of the capacitor?
(c) How much work is done by the mechanical forces on the slab when it is inserted?
Is this work done by, or on, the agent inserting the slab?
Suppose the battery was disconnected before the dielectric was inserted.
(d) Repeat (b).
(e) Repeat (c).
Princeton University 1998 Ph501 Set 2, Problem 8 9
## 8. (a) Find the escape velocity of an electron initially 1
A above a grounded conducting plate.
(b) Point electric dipoles p1 and p2 lie in the same plane at a xed distance apart. If
p1 makes angle 1 to their line of centers, show that the equilibrium angle 2 of
p2 is related to 1 by
tan 1 = 2 tan 2 . (15)
Princeton University 1998 Ph501 Set 2, Problem 9 10
9. We may dene the capacity of a single conductor with respect to innity as C = Q/V ,
where V is the potential (with respect to potential = 0 at ) when charge Q is
present on the conductor.
Calculate the capacity of a conductor composed of two tangent spheres of radius a.
Princeton University 1998 Ph501 Set 2, Problem 10 11
10. A grounded conducting sphere of radius a is placed in a uniform external eld E = E0z.
(This eld changes after the sphere is added.)
This problem may be solved by the method of images if we suppose the eld E0 is due
to two charges Q at positions z = R, with Q and R appropriately large.
(a) Show that the image of the source of E0 is then a dipole p = a3E0 located at the
center of the sphere.
(b) Give an expression for the potential (r, ) in spherical coordinates (r, , ) cen-
tered on the sphere. Sketch the electric eld lines.
3
(c) Show that the induced charge distribution on the sphere is = E
4 0
cos .
(d) Show that the force between the two hemispheres with equator perpendicular to
9 2
E0 is F = 16 a E0 .
Princeton University 1998 Ph501 Set 2, Problem 11 12
11. A hollow innite rectangular conducting tube of sides a and b has two faces grounded
and two faces at potentials V1 and V2 as shown:
y
=0
= V1 = V2
=0
a x
Find the potential (x, y) inside the tube. Remember to use a sum of products of all
solutions to the separated equations which do not violate the boundary conditions.
Princeton University 1998 Ph501 Set 2, Problem 12 13
12. A hollow rectangular conducting box has walls at x = 0 and a, at y = 0 and b, and at
z = 0 and c. All faces are grounded except that at z = c, for which = V :
z
= V
=0
=0 =0
c =0
y
=0
b
a
x
## Find the potential (x, y, z) inside the box.
(Choose the signs of the separation constants carefully!)
Princeton University 1998 Ph501 Set 2, Solution 1 14
Solutions
1. The energy stored in a dielectric composed of spring like atoms can be written in two
parts,
1 2 kx2
U = U1 + U2 = E dVol + n dVol, (16)
8 2
where E is the applied electric eld, and where n is the number of molecules per unit
volume.
The displacement x in the spring-like atom is related by kx = eEon atom, where e is the
charge of an electron. Then,
e2Eon
2
atom 1 e2Eon
2
atom 1 2
U2 = n dVol = n 2
dVol = nEon atom dVol, (17)
2k 2 m 2
where = k/m is the frequency of oscillation of the electron of mass m, and =
e2/m 2 is the atomic polarizability introduced in eq. (69) of set 1.
On p. 20 of the Notes, we argued that Eon atom = E + 4P/3, in terms of the applied
eld E and the induced polarization P. But, P = nEon atom, so
n n
P= E, and Eon atom = E 1 + E, (18)
1 4n/3 1 4n/3
## where the approximation holds for small n. In this case,
1
U2 4nE 2 dVol, (19)
8
and
1 1 1
U (1 + 4n)E 2 dVol E 2 dVol = E D dVol, (20)
8 8 8
using the Lorenz-Lorentz approximation for the dielectric constant in terms of the
polarizability , and supposing that D = E.
Princeton University 1998 Ph501 Set 2, Solution 2 15
## 2. (a) As argued on p. 13 of the Notes, rotational symmetry of a charge distribution
about the x axis implies that its quadrupole tenson Qij can be written
Qxx 0 0
Qij =
0 Qxx/2 0 ,
(21)
0 0 Qxx/2
## Qxx Ex 1 Ey 1 Ez 3 Ex 1 Qxx Qxx Ex
U = =
E = ,
6 x 2 y 2 z 2 x 26 4 x
(22)
using E = 0, assuming that the external eld is produced by charges not at
The force on the quadrupole is:
## Qxx 2Ex 2Ex 2 Ex
F = U = , , . (23)
4 x2 xy xz
(b) i. Consider a point charge q at at the origin and the quadrupole at (x, y, z) =
(R, 0, 0). The x-component of the electric eld from q observed at (x, y, z) is
qx
Ex = , where r 2 = x2 + y 2 + z 2 . (24)
r3
Then,
Ex r2 3x2
=q , (25)
x r5
and the force is evaluated from (23) at (R, 0, 0) as
3 qQxx
F= , 0, 0 . (26)
2 R4
Let us check this for the simple quadrupole shown in the picture.
t y t
q1 2q1 q1
Suppose the distance between q1 and 2q1 is a. The force on the quadrupole
due to charge q at distance R from the center of the quadrupole, and along
the latters axis, is
q1 q 2q1q q1 q 6a2q1 q
Fx = + = (1 + O(a/R)). (27)
(R a)2 R2 (R + a)2 R4
This agrees with (26), since
Qij = (3ri rj r2 ij ) dVol Qxx = 2q r 2 = 4q1a2 . (28)
Princeton University 1998 Ph501 Set 2, Solution 2 16
ii. If, instead, the quadrupole is at (0, R, 0) (but still oriented parallel to the x
axis), eqs. (23) and (25) combine to reveal that only the derivative 2Ex /xy
is nonvanishing, and
3qQxx
F = 0, ,0 . (29)
4R4
Again, we can directly compute the force on the simple quadrupole:
## q1 q q1qR 3qq1a2 3qQxx
Fy = 2 2
2 = , (30)
R (R + a2 )3/2 R4 4R4
using (28).
Princeton University 1998 Ph501 Set 2, Solution 3 17
3. Once the charge has reached distance d from the plane, the static electric eld Ee at
an arbitrary point r due to the charge can be calculated by summing the eld of the
charge plus its image charge,
er1 er2
Ee (r, d) = 3 3 , (31)
r1 r2
where r1 (r2) points from the charge (image) to the observation point r, as illustrated
below. The total electric eld is then E0z + Ee .
The charge e and its image charge e at positions (r, , z) = (0, 0, d) with
respect to a conducting plane at z = 0. Vectors r1 and r2 are directed from
the charges to the observation point (r, 0, z).
It turns out to be convenient to use a cylindrical coordinate system, where the obser-
vation point is r = (r, , z) = (r, 0, z), and the charge is at (0, 0, d). Then,
2
r1,2 = r2 + (z d)2 . (32)
The part of the electrostatic eld energy that varies with the position of the charge is
the interaction term,
E0 z Ee
Uint = dVol
4
eE0 zd z+d
= dz dr 2
4 0 0 [r2 + (z d)2 ]3/2 [r2 + (z + d)2 ]3/2
eE0
2 if z > d
= dz 2
4
0 2 if z < d
d
= eE0 dz = eE0d. (33)
0
When the particle has traversed a potential dierence V = E0 d, it has gained energy
eV and the electromagnetic eld has lost the same energy.
In a practical electrostatic accelerator, the particle is freed from an electrode at
potential V and emerges with energy eV in a region of zero potential. However, the
particle could not be moved to the negative electrode from a region of zero potential
by purely electrostatic forces unless the particle lost energy eV in the process, leading
to zero overall energy change. An electrostatic accelerator must have an essential
component (such as a battery) that provides a nonelectrostatic force that can absorb
the energy extracted from the electrostatic eld while moving the charge from potential
zero, so as to put the charge at rest at potential V prior to acceleration.
Princeton University 1998 Ph501 Set 2, Solution 4 18
4. (a) First, we calculate the force directly. The electric eld from one of the dipoles,
taken to be at the origin and with moment p = px, is
3(p r)r p 3pxr px
E= = 3. (34)
r3 r4 r
For a second dipole at (x, y, z) = (2d, 0, 0), also with moment p = px, we have
E 3p2
F = (p )E = p = 4 x. (35)
x (2d,0,0) 8d
As an aside, we can also calculate F = U, where U is the energy of interaction
of the two dipoles. First, the energy of a charge q2 at position r2 in the eld of a
dipole p1 at position r1 is
p1 r
U = q2 3 , (36)
r
where r = |r| = |r2 r1|, as on p. 12 of the Notes. A point dipole p2 is the limit
of a pair of charges q2 at positions r2 and r2 s where s = sp2 , and the product
q2s is held constant at value p2 . Thus, the interaction energy of two point dipoles
is obtained from (36) as
p1 r p1 r p1 r
U= lim q2 3 = (p2 2 ) , (37)
s0, q2 s=p r3 r r3
## where r = |r2 s r1|. For p1 = p2 = px separated by distance 2d along x, (37)
reduces to
2 1
U=p . (38)
x x=2d x2
Then, the force F = U is along x with magnitude
2 1 3p2
F = p 2
= , (39)
x2 x2 x=2d 8d4
as found in (35).
Now, let us calculate the force via the Maxwell stress tensor. The force on the
charges within a (closed) surface S is given by
Fi = Tij dSj , (40)
S
## as on p. 33 of the Notes, where the Maxwell tensor in empty space is given by
1 1
Tij = Ei Ej ij E 2 . (41)
4 2
In the problem with two dipoles, it is convenient to choose the surface as the
midplane perpendicular to the line connecting two dipoles (the x axis), closing
Princeton University 1998 Ph501 Set 2, Solution 4 19
the surface at innity around one of the dipoles. On this plane (x = d) the only
nonzero component of E is Ex , and this is twice Ex from the dipole at x = 0. At
radius r from (d, 0, 0) in the symmetry plane, the total eld is then
3d2 1 2d2 r2
Ex (r) = 2p 2 2 5/2
2 = 2p . (42)
[r + d ] [r + d2 ]3/2 [r2 + d2 ]5/2
The Maxwell stress tensor is thus,
Ex2 0 0
1
Tij =
0 Ex2 0 .
(43)
8
0 0 Ex2
We take our surface element to be dS = (2rdr, 0, 0) in cylindrical coordinates,
the sign of which implies that the surface S encloses the dipole at x = 0. Then,
(40) and (43) indicate that only Fx is nonzero, and it is given by
1 2 2
(r2 2d2 )2
Fx = r drEx = p r dr 2
4 0 0 (r + d2 )5
p2 (t 2d2 )2 p2 [(t + d2 ) 3d2 ]2
= dt = dt (44)
2 0 (t + d2 )5 2 0 (t + d2 )5
p2 1 6d2 9d4
= dt +
2 0 (t + d2 )3 (t + d2 )4 (t + d2 )5
p2 1 2 9 3p2
= + = .
2 2d4 d4 4d4 8d4
This agrees with (35), noting that since the dipoles attract, the force on the dipole
at x = 0 is in the +x direction.
(b) The electric eld outside the conducting sphere of radius a is E = qr/r2 . The
pressure (= force per unit area) on the surface charges is P = E/2, where is
the surface charge density; hence, P = q 2r/8a4 . (The coecient 1/2 is needed
because E is the eld outside the surface, while the eld inside the sphere is zero,
thus the average eld inside the charge layer is E/2.) To nd the force between
two hemispheres, we integrate the component of pressure normal to the equatorial
plane (P cos ) over one hemisphere:
1
q 2 cos q2
F = 2a2 d cos = . (45)
0 8a4 8a2
Now, let us calculate the force using the Maxwell stress tensor. We integrate
Fz over the x-y plane separating our sphere into two hemispheres. Since dS =
(0, 0, 2r dr) there, and the only nonzero components of E on that surface are
Ex and Ey , only Tzz = E 2/8 = q 2/8r4 contributes to the force. Integrating
from r = a to , we nd
q2 dr q2
Fz = 2r drTzz = = , (46)
a 4 a r3 8a2
in agreement with (45).
Princeton University 1998 Ph501 Set 2, Solution 5 20
5. (a) The electrical force F required to pull oil of density into a cylindrical capacitor
of inner and outer radii a and b, respectively, to height h above the bath is equal
to the force of gravity:
F = gh(b2 a2). (47)
A second relation for F can be computed from the balance of electrical energy,
noting that the capacitor is held at constant voltage by a battery. Suppose we
increase the height of the oil by h. Then, work F h is done on the oil, the energy
U = CV 2 /2 stored in the capacitor changes by U, and the battery loses energy
V Q. Conservation of energy implies
0 = F h + U V Q. (48)
## Since V = Q/C, we nd for constant voltage,
2CV 2
V Q = V C = 2 = 2U. (49)
2
## Together, (48) and (49) imply that
U
F =+ . (50)
h V
As the liquid is drawn into the capacitor, the energy for this must come from
elsewhere; yet, the energy of the capacitor increases because the battery loses
energy in twice the amount of work done on the liquid.
We now calculate the stored energy U by integrating the electric eld energy
density. By cylindrical symmetry and Gausss law, the electric eld between the
pipes has form Er (r) = /r, where is xed by
b
b V
V = Er dr = ln , or = . (51)
a a ln ab
Suppose the total height of the capacitor (above the bath) is H. Then, the energy
of the electric eld in the capacitor is:
b b
1 1
U= h 2
E 2r dr + (H h) E 22r dr, (52)
8 a 8 a
where the rst term on the right is the contribution from the space lled with the
oil whose dielectric constant is , while the second term is from the empty space
above. Evaluating the integrals:
b b
V2 dr V2
2 E 2 r dr = 2 = 2 b , (53)
a ln2 ab a r ln a
we then nd:
1 V2
U= [( 1)h + H] . (54)
4 ln ab
Princeton University 1998 Ph501 Set 2, Solution 5 21
## The force is obtained from (50) and (54):
U V 2 ( 1)
Fel = = . (55)
h V 4 ln ab
Equating this to the force of gravity, (47) we obtain the height h of the oil column:
( 1)V 2
h= . (56)
4g ln b
a
(b2 a2)
(b) The force on the liquid arises from the eect of gradients of the electric eld on
the molecular dipoles in the liquid. The spatially varying electric eld E results
in a bulk dielectric polarization given by
1
P = E = E, (57)
4
where is the dielectric susceptibility and is the dielectric constant. The energy
density associated with the induced polarization is
1 2
u = P E = E , (58)
4
and so the force density on the liquid is given by
1
f = u = E 2. (59)
4
The gradient E 2 in the fringe eld of the capacitor points from the outside to
the interior of the capacitor, with a generally vertical component for the liquid
below the capacitor in the present problem.
It is interesting to consider a variant on this problem: a capacitor with horizontal
plates completely immersed in a dielectric liquid. Here, the fringe elds of the
capacitor pull the liquid in from all sides, trapping it inside the capacitor. That
is, work would be required to pull the liquid out of the capacitor in any direction.
Is this an example of electrostatic trapping which is claimed not to exist? No!
The trapping in the direction perpendicular to the capacitor plates is not pro-
vided by purely electrostratic elds, but by the material of the capacitor plates
(whose stability is not a result of purely electrostatic eects). See prob. 7 of set
4 for further discussion.
We have concluded that the liquid is drawn into the interior of the capacitor and
that the liquid near the middle of the capacitor is forced up against the capacitor
plates by electrostatic forces on the induced dipoles. If we drill a hole in the
center of one capacitor plate, would liquid squirt out? (If yes, we would have a
perpetual motion machine.) No, the fringe elds around the hole will pull liquid
into the interior of the capacitor creating a static equilibrium much as before.
Princeton University 1998 Ph501 Set 2, Solution 6 22
6. We work from eq. (13), for which we rst need the potential at a point r inside a
grounded conducting sphere of radius a when unit charge is located at x, also inside
the sphere. Then we need the normal derivative of this potential on the inner surface
of the sphere, i.e.when |r| = r = a.
The image method for a grounded conducting sphere tells us that the potential inside
the sphere can be calculated as that due to unit charge at x together with charge a/x
at position x = a2x/x2 . We denote the angle between vectors r and x as , so that
R = |r x| = r2 + 2rx cos + x2 , (60)
and
a2 a4
R = |r x | = r2 + 2r cos + 2 . (61)
x x
We see that when r = a, then
a
R = R. (62)
x
The potential inside the sphere can now be written
1 a
(r) = , (63)
R Rx
The normal derivative of the potential on the inner surface of the sphere is the negative
of its radial derivative when r = a,
## (r = a) a + x cos a[a + (a2/x) cos ] a2 x2
= = = , (64)
n r R3 R 3 x aR3
using eq. (62). Inserting this in eq. (13), we obtain Poissons integral,
a 2 x2 (x)
(x) = dS . (65)
4a S R3
Princeton University 1998 Ph501 Set 2, Solution 7 23
## 7. (a) The capacitance C of a parallel-plate capacitor of area A, gap thickness d and
dielectric constant is
A Q
C= . (66)
d V
Adding the dielectric increased the capacitance to
Cf = C0, (67)
and hence the charge also increase, if the voltage is kept xed. Thus, the work
done by the battery as the dielectric is inserted,
## Wbatt = V0 Q = V02C = V02 ( 1)C0 = Q0 V0 ( 1), (68)
is positive.
(b) As the dielectric is inserted, the eld energy U = CV 2 /2 stored in the capacitor
changes by
1 1 1
U = CV02 = C0 V02 ( 1) = Q0V0 ( 1). (69)
2 2 2
(c) The work done by the battery, (68), is only partly accounted for in increase in
the eld energy, (69). The rest of the work done by the battery is done on the
external agent that held the dielectric during insertion (the external agent gained
energy):
1
Won agent = Q0V0 ( 1). (70)
2
(d) If the battery had been disconnected before the dielectric was inserted, then the
charge Q0 would be constant. From (66) we see that the nal voltage would be
only V0 /. Recalling (67), the change in the electrostatic eld energy would then
be 2
1 V0 1 1 1
U = C0 C0 V0 = Q0V0
2
1 < 0. (71)
2 2 2
(e) By conservation of energy, the work done on the external agent that held the
dielectric during insertion is equal and opposite to the change in stored energy.
Hence the work done on the agent is again positive, but now with the value
1 1
Won agent = Q0V0 . (72)
2
That is, the dielectric is pulled into the capacitor whether or not the battery is
still connected.
Princeton University 1998 Ph501 Set 2, Solution 8 24
8. (a) The eld energy associated with an electron at distance r from a grounded con-
ducting plane is 1/2 that associated with the corresponding image charge, i.e.,
with that electron plus a positron at distance r, in the absence of the conducting
plane. Hence,
1 e2 e2
U = = . (73)
2 2r 4r
The elds in the image solution have reality only outside the conducting plane;
there is no energy associated with the ctitious image elds inside the conduc-
tor.
Equation (73) indicates that an electron is bound to the conducting plane, and
so to escape, must have a minimum velocity related by
2|U| e2 e 2 c2 re
vmin = = = 2
=c , (74)
m 2mr 2mc r 2r
where re = e2/mc2 = 2.8 1013 cm is the classical electron radius. Thus, for
r=1
A,
vmin 2.8 1013
= = 0.0037. (75)
c 2 108
(Notice that the nonrelativistic approximation suces.)
The binding energy can be estimated from (73) as
e2 re mc2 2.8 1013 5.11 105 eV
U = mc 2
= = = 3.6 eV. (76)
4mc2 r r 4 108 4
(b) In equilibrium, the torque on dipole p2 must vanish, and so p2 will be directed
along the electric eld created by dipole p1 . The electric eld of the latter is given
by
3(p1 r)r p1
E= . (77)
r3
The projection of E onto the line connecting two dipoles is
p1
E = E r = 2 3 cos 1 . (78)
r
The orthogonal projection is
E = E E , (79)
p1
E = sin 1 , (80)
r3
where the minus sign indicates that E is directed opposite to p1, .
The angle of the eld line, and hence of p2 is
E 1
tan 2 = = tan 1 . (81)
E 2
Princeton University 1998 Ph501 Set 2, Solution 9 25
9. We solve the problem of the capacity of two tangent, conducting spheres of radii a by
the method of images.
We rst nd the image-charge distribution needed to bring one sphere to potential V ,
but leaving the other at zero potential. Then, we complete the solution by superposing
the mirror distribution, obtained by reection symmetry about the plane through the
point of tangency of the two spheres.
(a) Place charge q = aV at the center of sphere 1, bringing its surface to otential V .
(b) To bring sphere 2 to zero potential, place charge q(a/2a) = q/2 at distance
a2/2a = a/2 from the center of sphere 2, following the prescription on p. 41 of
the Notes.
(c) The image charge (b) takes sphere 1 away from potential V . To bring it back, add
an image charge (c) inside sphere 1 so that this sphere is at zero potential under
the eect of charges (b) and (c). That is, add charge (q/2)(a/(3a/2)) = +q/3
at distance a2/(3a/2) = 2a/3 from the center of sphere 1, i.e., a/3 from the point
of contact.
(d) Add charge q/4 at 3a/4 from the center of sphere 2 to bring it back to zero
potential.
(e) ....
## -q/2 ... ... -q/2
q q
The total charge needed to bring both spheres to potential V is double that described
in the sequence above. Hence,
1 1 1
Q = 2q 1 + + ... = 2aV ln 2, (82)
2 3 4
and the capacitance is
C = Q/V = 2a ln 2 = 1.386a. (83)
Note that since the dimensions of potential are [charge]/[length], capacitance has the
dimension of [length] in Gaussian units. Thus, we expect that C a for this problem,
since a is the only relevant length.
Princeton University 1998 Ph501 Set 2, Solution 10 26
## 10. (a) The uniform eld E0 = E0 z is approximated as being due to charges Q at
z = R, where Q and R in such a way as to keep Q/R2 constant.
In the limit, the eld in the region of the sphere is homogeneous and equal to
E = (2Q/R2 )z. According to the image method, we can make the potential
on the sphere vanish by adding charge q = Qa/R at z = a2/R and q at
z = a2/R. Thus, the perturbation to the eld due to the sphere is eectively that
due to a dipole with the moment
a2 a
p=2 Q z = a3 E0. (84)
R R
## (b) The potential outside the sphere is thus,
E0a3 cos
= 0 + dipole = E0r cos + . (85)
r2
The eld lines bend in to be normal to the sphere at r = a:
(c) We nd the surface charge density from the normal component of the electric
eld at the surface of the sphere:
Er (a, ) = = 3E0 cos , (86)
r r=a
and so
Er (a, ) 3E0 cos
() = = . (87)
4 4
(d) The force acting on the surface charge density is F = Er (a)r/2 = Er2 (a)r/8
(where the latter form follows immediately from the Maxwell stress tensor). The
force on the right hemisphere is directed along z and is obtained by integrating
the z component of F:
1
1 9 2 2
Fz = 2a2 d cos (3E0 cos )2(cos ) = a E0 . (88)
8 0 16
Since the force on the hemisphere at z > 0 is positive, the hemispheres repel each
other.
Princeton University 1998 Ph501 Set 2, Solution 11 27
## 11. We seek solutions to Laplaces equation in 2 dimensions, 2 (x, y) = 0, of the form
= X(x)Y (y). This leads to solutions of the form ekx eiky or eikx eky .
Since the boundary conditions include = 0 at y = 0 and b, it is advantageous to
consider functions Y of the type eiky , which can be immediately restricted to the
form:
Y (y) = sin ky, where k = n/b, n = 1, 2, . . . . (89)
This also xes the separation constants k.
The general expression for the potential is now:
ny
(x, y) = Xn (x)Yn (y) = An enx/b + Bn enx/b sin . (90)
n n b
## The boundary conditions at x = 0 and x = a are
ny
(0, y) = V1 = (An + Bn ) sin , (91)
n b
ny
(a, y) = V2 = An ena/b + Bn ena/b sin . (92)
n b
## A straigthforward approach to nd An and Bn is to multiply (91) and (92) by sin(ny/b)
and integrate from y = 0 to b:
b b
ny bV1 ny
bV1 2, n odd b
(0, y) sin dy = cos = = (An + Bn ) , (93)
0 b n b 0
n 0, n even 2
and similarly,
bV2 2, n odd b
= An ena/b + Bn ena/b . (94)
n
0, n even 2
## Thus, for n even, An = Bn = 0, while for n odd,
2 2
na/b
An = V2 V1 e , Bn = V1 e na/b
V2 . (95)
n sinh na
b
n sinh na
b
## Finally, we get for the potential:
4 sin ny
b nx n(a x)
(x, y) = na V2 sinh + V1 sinh . (96)
n odd n sinh b b b
To verify that this solution satises the boundary conditions, note that (91) and (93)
combine to yield the expansion:
4 1 ny
1= sin . (97)
n odd n b
Princeton University 1998 Ph501 Set 2, Solution 11 28
We also note that the potential is symmetric about the midplanes, (x, y) =
(a x, y) = (x, b y), which could have been invoked as far back as (90) to show
that only odd n contributes.
Remark: This problem could also usefully be solved as the superposition of two cases,
each with three walls at potential zero and the fourth at a nonzero value. The form of
the solution (96) displays this superposition.
Princeton University 1998 Ph501 Set 2, Solution 12 29
12. Since = 0 at x = 0, a and y = 0, b, solutions = X(x)Y (y)Z(z) must have the form
mx ny
Xm (x) = sin , and Yn (y) = sin , (98)
a b
where n and m are positive integers (and odd, recalling the remark at the end of
problem 9). The functions Z(z) then have the form ekz . Since = 0 at z = 0, we
can make the further restriction:
## Zmn (z) = sinh kmn z, (99)
where kmn is determined by inserting the trial solutions into Laplaces equation, yield-
ing
2 m 2 n 2
kmn = + . (100)
a b
The general solution satisfying all the boundary conditions except for the one at the
face z = c is:
2 2
mx ny m n
(x, y, z) = Amn sin sin sinh + z. (101)
m,n a b a b
## The remaining boundary condition tells us that
2 2
mx ny m n
V = Amn sin sin sinh + c. (102)
m,n a b a b
## To nd Amn , multiply (102) by sin mx
a
sin ny
b
and integrate from 0 to a in x and from
0 to b in y. Similarly to (93), we nd
16V
Amn = 2 2 , (103)
m n
mn2 sinh a
+ b
c
## for odd m and n, and 0 otherwise. Hence,
2 2
m n
16V 1 mx 1 ny sinh a
+ b
z
(x, y, z) = sin sin 2 2 . (104)
2 m,n odd m a n b m n
sinh a
+ b
c
## Note that we have demonstrated the expansion
4 mx 4 ny
1= sin sin for 0 < x < a, 0 < y < b. (105)
m m a n n b
Since this follows from (97), we could have used it to go from (102) to (103) without
performing the integrations. | 10,418 | 31,861 | {"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} | 3.421875 | 3 | CC-MAIN-2019-22 | latest | en | 0.849564 |
https://mathoverflow.net/questions/67735/immersion-submanifold-of-complex-manifold | 1,679,464,187,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943750.71/warc/CC-MAIN-20230322051607-20230322081607-00165.warc.gz | 460,369,468 | 25,258 | # immersion: submanifold of complex manifold
Let $\alpha : \mathbb{C} \rightarrow M$ be an immersion and $M$ a $n$ dimensional complex manifold with complex structure $I$. Does then follow that $\alpha (\mathbb{C})$ is a one dimensional submanifold of $M$ ? If, yes, does the induced complex structure on $\alpha (\mathbb{C})$ by the immersion coincide with $I$ ?
• What if $M$ is an abelian variety of dimension $g\geq 2$ and $\mathbb C$ is an irrational diagonal line in the original $\mathbb C^g$? Isn't the induced map an immersion? Jun 14, 2011 at 10:50
• Gregor, this is the second of your questions of this type. I think, it would be reasonable if you write a longer detailed question (say 5-10 times longer) that explains what you are thinking about and what really bothers you. Jun 14, 2011 at 18:58
If $\alpha$ is only required to be a smooth embedding, then we can set $M = \mathbb{C}^2$, and embed $\mathbb{C}$ as a totally real subspace, e.g., $x + iy \mapsto (x,y)$. This is not a complex submanifold in the usual sense of the word, and the complex structure induced by $\alpha$ on its image is not compatible with $I$.
If $\alpha$ is required to be an immersion of complex analytic spaces but not an embedding, then the image may have singularities. This is even true if we require $\alpha$ to be a homeomorphism to its image, since the normalization map of the cuspidal cubic curve is a homeomorphism whose image is a topological manifold that is not a complex manifold.
If $\alpha$ is an analytic cover of its image, then the image is a complex submanifold of the ambient space, and the induced complex structure coincides with $I$. | 435 | 1,652 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-14 | latest | en | 0.893321 |
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