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# STPM 2015 Mathematics (T) Term 2 Assignment

Sample question of STPM 2015 Mathematics (T) term 2 coursework assignment question. Click the question button below to download the complete question.

A hyperboloid container of height 14 m is generated by the hyperbola $25x^2-16(y-9)^2=225, 0\leq y\leq 14$ as shown in the diagram below.

Initially the container is empty. Water is pumped into the container from the top at a constant rate of $180\ \text{m}^3$ per hour.

Calculate the time taken to fill up the container.

Find an expression for the volume of water in terms of the height of the water level in the container.

Determine the rate of increase of the height of water level in the container when the height is 9 m.

In order to avoid overflow, a mechanism is now installed. This mechanism controls the water being pumped into the container at a rate of $k(14-y)^n\ \text{m}^3$ per hour, where k and n are positive constants.

If k=20 and n=1, state, with reasons(s), whether the time taken to fill up the container is longer or shorter than that in part 1(a),

calculate the time taken to fill up the container.

If k=15 and n=0.5, determine the minimum and the maximum rate of increase of the height of water level in the container.

Below is the sample answer and solution for the STPM 2015 Term 2 Coursework assignment. I will try to discuss the questions and answers in the class.

Answer for Question 1c and 2b are updated. -10th March 2015

### Old Version

Old Version (wrong)

### Updated Version

Thanks to Shen Sing and the students who to help me and tell me the problem. The time taken to fill the container should be longer.

Table to show average of the rate of water

Reason: Average of the rate is 140 which is lower than 180.

### Updated Version

Old version (wrong because ln 0 was counted as 1. =.=!!)

### Updated Version

The container cannot be 100% full.

Download the Term 2 PBS Full solution here. Available for Ai Tuition and Teras Usaha students only. Join my class to get the full solution. Full solution will be printed for my students.

Feel free to leave comment below about your school teacher special requirement to help others. 🙂

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## STPM 2021 Term 2 Exam Timetable

View the exam timetable for STPM 2021 Term 2. Countdown to STPM 2021 Examination. Join my tuition to help you achieve better results in the exam. Contact (60) 018-2016992 or kklee@kkleemaths.com for more information.

## Integration of cos^3 x

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## STPM 2018 Term 3 Mathematics (T) Coursework Sample Answer

STPM 2018 term 3 mathematics (T) Coursework sample Guideline for solving STPM 2018 Term 3 Mathematics (T) PBS coursework assignment methodology   Hint: Binomial Theorem   Chi-Squared tests   tabulate Thanks for helping in doing the survey. You can find...

## STPM 2018 Term 2 Mathematics (T) Coursework Sample Solution

STPM 2018 term 2 mathematics (T) Coursework sample Guideline for solving STPM 2018 Term 2 Mathematics (T) PBS coursework assignment methodology Hint: limits L' Hopital rule Rationalisation Indeterminate Form Existance of limitsThanks the two students who sent me the...

## Integration Booklet Page 10 Page 11 Solution

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## STPM 2018 Term 1 Mathematics (T) Coursework Sample

Question A complex number is an extension of a real number and it can be represented in Cartesian and polar forms. In this assignment, you are required to explore the powers and roots of complex numbers. 1 (a) Let z =1+i. Find z^n, where n = 2, 3, 4, …, and represent...

#### KK LEE

KK LEE has been a STPM Mathematics tuition teacher since June 2006, but his love of Maths dates back to at least 1999 when he was Form 4. KK LEE started teaching in 2006 at Pusat Tuisyen Kasturi. He was known as "LK" when he was teaching in PTK. After teaching STPM Mathematics for 8 years in PTK, he joined Ai Tuition Centre in 2014. Over the years he has taught Mathematics (T), Mathematics (S), Mathematics (M), Additional Mathematics.

1. Sir, hw to do intro, methodology and conclusion?

• Hi. I do the mathematical parts. U share the intro, methodology, and conclusion parts. Fair? 😀

I will help students on mathematical parts only.

• sir,for 2 a)i)after y=5, the rate of volume of water pumped is dicreasing already.hence,is it the time will still shorter?

• The volume of the water in the container for y=5 is more than half of the container. Volume of water when y=5 is 587.06. You can obtain this value from part 1b. Btw when the y approaches 14, the rate is very low compared to 180. I have to think again. 😀

2. sir,where the tuition places that you teaching maths (T)for STPM?

• Hi. I am teaching STPM Maths tuition at Ai Tuition at Jalan Petaling Kuala Lumpur and SS19 Subang Jaya. and Teras Usaha at Segambut

• how do find the maximum
and the minimum

• In general, use dy/dx=0

3. The average 140 rate of flow is invalid as the dy/dt,the rate of increase of y, is not constant.In my opinion,the time taken to COMPLETELY fill up the container is infinite as the answer has shown above.By the way, thank you sir for sharing these solutions, thank you very much

• yes. that is one of the sample of explanation only.

4. Dear sir, for question 1(c),isnt it the V should be 1/75(pi)(y)? Then dv/dy using the uv formula? Thank you sir~

• hi. Can i know which V u mean?

• Sorry sir~I got it now~~Because I had expanded all the brackets and a bit messed up~ thanks

5. Sir, I can seem to understand the second differentiation you did for 2b. How did you get y= 37/3 plus/minus 5/12 sqaure root 9?

6. Hi sir, could you please check again 1(b)? Maybe you overlook the expansion of (y-9)^3.

• Opps… its’s my mistake…sry

• 😀

7. Sir, how u can get the dv/dt= 15(14-y)’1/2 ? On question 2 (b)

• Substitute k=15, n=1/2 into the rate provided. dv/dt=k(14-y)^n

• Thanks sir. How to join your tuition class?

• yes. ARe you from kl area. I am teaching at jalan Petaling KL，Subang Jaya and segambut

8. Sir,What can we write for title??

• Hmm.. Investigating the rate of change of water.

• Its okey..But anyway thank you,sir..

9. Sir… question 1b and 1c the answer given by my teacher was v=Π/25[1521h-144h^2+16/3h^3
And for c) 6.37m/h… its her skema jawapan… why the answer over here is different sir?

• Sir.. I’m sooooo sorry.. 1c is correct over here.. my mistake sir sorry…

• Hi. Can share your marking scheme? 😀

10. i did not understand why minimum rate is 0

• Minimum value of the rate refers to the smallest value of the rate. And 0 is the value when y=14. Thats why the minimum value is 0. But i heard some teachers said there is no minimum value. Please confirm with your teacher. Thanks. 😀

11. I dun understand the part 2(a)(i)…can u pls help me in it.. tq

• Since the rate is not constant. So I used average of the rate(approximation) to compare with 180

• sorry sir, i still cannot understand part 2a(ii), how to write it down?

• Hi. It depends on your school teacher. For my sample solution, it is impossible to fill up the container to 100.000% full.

• Thank you sir

12. Sir, for 2a2, how can u get the dvdy ?

• Differentiate v with respect to y

13. Sir y we can use dy/dT=0 in 2a(b) to get minimum value?

• Hi. We can use any methods to find the minimum value of functions.

14. sir, why do we need to use the average of water? and also how to explain that t=∞? tq

• Since the rate is not constant. So I used average of the rate(approximation) to compare with 180.
Time taken to fill up the container(y=14.0) is infinity. So you can try to explain from y=13.5, y=13.9, and others

15. hi sir, my teacher’s answer for minimum rate is 0.2936 and maximum is 1.186. why is that? and can we use the hyperbola equation to find the answer for question 2(b) by finding the centre point/axis symetry etc. because this assignment is regarding differentiation and integration..

• You can use any methods. As long as your school teacher accept.
i will try to check the values 0.2936 and 1.186.

• I have no idea to get those values

16. sir, why got dy/dt in the second differentiation in question 2(b)? we need to find rate for height right why dont we use the volume from 1(b) in term of h?

• because i used implicit differentiation.
you can use part 1(b)

17. Sir,why Q2b use d2y/dx2 =0? That one is for point of inflection right? Correct me pls

• You can use any methods to find maximum and minimum.

d2y/dx2=0 is used to find point of inflexion of the the curve y=f(x).

At the same time, d2y/dx2=0 can be used to find max/min of dy/dx.

18. Sir,Q2a2 why t=infinity? Coz 1875ln0?

• yes

19. Sir,Q2a2,why substitute y=14,t=infinity?

• Please recall that when x approaches 0, ln x approaches -infinity

20. sir, when y = 14, isn’t d2y/dt2 =
144y^2 – 3552y +19629 / 2[sqr(14-y)](48y^2-864y+4563)^2 undefined?

• Yes

• so y = 14 isn’t a solution to d2y/dt2 = 0?

• ya

21. Why my one fron yours? My ans for 1(b) is 308 14/75.

22. 1) (c) 1/9Π x 180 = 20/Π

• Yes

23. how (48ypower 2 -864y+4563)over (14-y)wil get -48y+192+(1875/(14-y))? i dont get it

• Long division

24. Hey sir, do you have Maths M pbs? TQ

25. sir, pbs maths t term 1 stpm 2015. tq

• Are you kidding?

• Sir… Do u have solutions for term 1 assignment?? Im a private candidate.. Need to do it.

• Hi.I have no sem 1 answer

26. Sir,Q2b, can I substitute when y=0,14 to dy/dt that founded? The 2 answer will equal to the max&min rate?am I right?

• You will get the rates. But are they max or min, u have to compare with other values of y range from 0 to 14.

27. sir soloution 1c with wrong answer
it must be (1/9pie)x180 then the ans will be6.366m or 2.xx pie

• sir may i ans 1 more question? why is the average rate in 1)a)1) is (280+260 +….+20)divide by 13 only but not 14?

• it was written as 6.366 there.

• 280+260+…+20+0 / 15

28. Sir ,in 2(a)ii) why with a mechanism installed the water CNT fill up the container100%?

• from the formula and calculation.

29. Sir,in 2(a)i) why time taken will b longer?just in case teacher ask during presentation

• Average is smaller.

30. Oh..OK,thanks!other than the results frm calculation n formula, is thr any other reason I cn state?

• As i mentioned earlier. I will help on Maths part only. 😀

31. Sir, the answers given by my school teacher:

1a) 308.19 -pie-
b) V=-pie- (9h+16/75 (h-9)^3 + 3888/25
[ 0《h《14]
c) 20/-pie-

2 a)i) shorter
ii) t= -pie-/ 500 [64y -8y^2 + 625 ln [14/ 14-y]]

• Thanks.

32. Best maths T teacher ever! Clap for u! Thanks for helping us!

• You are welcome. 😀

33. sir, can u pls check my introduction tht i sent to ur email?? 🙂 tq sir

• checked

34. sir, can i know what is that two applications of this mechanism?

I will help on maths part only.

• thank you sir.

35. sir, what is the main formula to find the time taken to fill the container?

• In short, Time=Volume divide rate

• thank you sir.

36. Hi sir, I wanna ask about the Question 2a(ii). To find the dy/dt,there is a step §(48y²-864y+4563)/14-y dy = §1500/μ
After that d step becomes §-48y+192y+1875/14-y dy… how do sir do this step? I’m a bit confuse.. hope sir can solve my problem sorry sir that I can’t find an integration sign,n I replace it with §.

• long division

• I see!! Thank you sir!!

37. sir i just want to ask that , can you give me some idea to write the introduction ? i have no idea how to start.

• Try to explain history of Tupperware water container. 😛

38. hi,sir,may i noe wat is the title of this sem 2 coursework?

• Sir,may I knw tat..why 1.225 is a maximum rate but not minimum rate?

• Hi. Please refer to other comments. 1.225 is max because it is the largest value among others

39. Sir,for question 2b,why the max value of dy/dt is 1.225 and not 0.320?Why the min value is 0?

• Why max is 1.225 and not 0.320????

Max means largest. You can ask any kindergarten kids, which is larger among 1 and 0?

• Ok,thank you sir but it no need to use others method to prove that is min&max value ??

• It depends on your teacher. For me, a simple explanation is enough. But i am not the teacher who give you marks.

40. sir…i dont understand part b) …d2y/dx2 is used to find point of inflexion while dy/dx is used to find maximum and minimum point right? but y in this case we need to find inflexion instead of using dy/dx?

41. Hai Sir… How to calculate the percentage shown in the 2(a)(ii)..

42. Sir,the question 2aiii,when y=14,why t=infinity?? and the In(0) should be error or not??

43. excuse me, is there any reference for assignment math t first term 2015?

• Hi. Past Assignment solution will not be posted.

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## STPM 2017 Mathematics (M) Term 1 Assignment

Sample solution of STPM 2017 Term 1 Mathematics (M) PBS coursework assignment

## STPM 2017 Mathematics (T) Term 1 Assignment

Sample solution of STPM 2017 Mathematics (T) PBS coursework assignment

## STPM Mathematics (T) Term 3 Assignment

Sample Question of STPM 2016 Term 3 Maths (T) Assignment Coursework


text:
# TFIDF¶

Computes TF-IDF values from sentences.

The TF-IDF formula is the same one as scikit-learn. The only difference is the fact that the document frequencies are determined online, whereas in a batch setting they can be determined by performing an initial pass through the data.

Note that the parameters are identical to those of feature_extraction.BagOfWords.

## Parameters¶

• normalize – defaults to True

Whether or not the TF-IDF values by their L2 norm.

• on (str) – defaults to None

The name of the feature that contains the text to vectorize. If None, then the input is treated as a document instead of a set of features.

• strip_accents – defaults to True

Whether or not to strip accent characters.

• lowercase – defaults to True

Whether or not to convert all characters to lowercase.

• preprocessor (Callable) – defaults to None

Override the preprocessing step while preserving the tokenizing and n-grams generation steps.

• tokenizer (Callable) – defaults to None

A function used to convert preprocessed text into a dict of tokens. By default, a regex formula that works well in most cases is used.

• ngram_range – defaults to (1, 1)

The lower and upper boundary of the range n-grams to be extracted. All values of n such that min_n <= n <= max_n will be used. For example an ngram_range of (1, 1) means only unigrams, (1, 2) means unigrams and bigrams, and (2, 2) means only bigrams. Only works if tokenizer is not set to False.

## Attributes¶

• dfs (collections.defaultdict))

Document counts.

• n (int)

Number of scanned documents.

## Examples¶

>>> from river import feature_extraction

>>> tfidf = feature_extraction.TFIDF()

>>> corpus = [
...     'This is the first document.',
...     'This document is the second document.',
...     'And this is the third one.',
...     'Is this the first document?',
... ]

>>> for sentence in corpus:
...     tfidf = tfidf.learn_one(sentence)
...     print(tfidf.transform_one(sentence))
{'this': 0.447, 'is': 0.447, 'the': 0.447, 'first': 0.447, 'document': 0.447}
{'this': 0.333, 'document': 0.667, 'is': 0.333, 'the': 0.333, 'second': 0.469}
{'and': 0.497, 'this': 0.293, 'is': 0.293, 'the': 0.293, 'third': 0.497, 'one': 0.497}
{'is': 0.384, 'this': 0.384, 'the': 0.384, 'first': 0.580, 'document': 0.469}

In the above example, a string is passed to transform_one. You can also indicate which field to access if the string is stored in a dictionary:

>>> tfidf = feature_extraction.TFIDF(on='sentence')

>>> for sentence in corpus:
...     x = {'sentence': sentence}
...     tfidf = tfidf.learn_one(x)
...     print(tfidf.transform_one(x))
{'this': 0.447, 'is': 0.447, 'the': 0.447, 'first': 0.447, 'document': 0.447}
{'this': 0.333, 'document': 0.667, 'is': 0.333, 'the': 0.333, 'second': 0.469}
{'and': 0.497, 'this': 0.293, 'is': 0.293, 'the': 0.293, 'third': 0.497, 'one': 0.497}
{'is': 0.384, 'this': 0.384, 'the': 0.384, 'first': 0.580, 'document': 0.469}

## Methods¶

clone

Return a fresh estimator with the same parameters.

The clone has the same parameters but has not been updated with any data. This works by looking at the parameters from the class signature. Each parameter is either - recursively cloned if it's a River classes. - deep-copied via copy.deepcopy if not. If the calling object is stochastic (i.e. it accepts a seed parameter) and has not been seeded, then the clone will not be idempotent. Indeed, this method's purpose if simply to return a new instance with the same input parameters.

learn_many
learn_one

Update with a set of features x.

A lot of transformers don't actually have to do anything during the learn_one step because they are stateless. For this reason the default behavior of this function is to do nothing. Transformers that however do something during the learn_one can override this method.

Parameters

• x (dict)

Returns

Transformer: self

process_text
transform_many

Transform pandas series of string into term-frequency pandas sparse dataframe.

Parameters

• X (pandas.core.series.Series)
transform_one

Transform a set of features x.

Parameters

• x (dict)

Returns

dict: The transformed values.

format: plain


text:
# Geodetic transformation¶

PROJ can do everything from the most simple projection to very complex transformations across many reference frames. While originally developed as a tool for cartographic projections, PROJ has over time evolved into a powerful generic coordinate transformation engine that makes it possible to do both large scale cartographic projections as well as coordinate transformation at a geodetic high precision level. This chapter delves into the details of how geodetic transformations of varying complexity can be performed.

In PROJ, two frameworks for geodetic transformations exists, the PROJ 4.x/5.x / cs2cs / pj_transform() framework and the transformation pipelines framework. The first is the original, and limited, framework for doing geodetic transforms in PROJ The latter is a newer addition that aims to be a more complete transformation framework. Both are described in the sections below. Large portions of the text are based on [EversKnudsen2017].

Before describing the details of the two frameworks, let us first remark that most cases of geodetic transformations can be expressed as a series of elementary operations, the output of one operation being the input of the next. E.g. when going from UTM zone 32, datum ED50, to UTM zone 32, datum ETRS89, one must, in the simplest case, go through 5 steps:

1. Back-project the UTM coordinates to geographic coordinates

2. Convert the geographic coordinates to 3D cartesian geocentric coordinates

3. Apply a Helmert transformation from ED50 to ETRS89

4. Convert back from cartesian to geographic coordinates

5. Finally project the geographic coordinates to UTM zone 32 planar coordinates.

## Transformation pipelines¶

The homology between the above steps and a Unix shell style pipeline is evident. It is there the main architectural inspiration behind the transformation pipeline framework. The pipeline framework is realized by utilizing a special “projection”, that takes as its user supplied arguments, a series of elementary operations, which it strings together in order to implement the full transformation needed. Additionally, a number of elementary geodetic operations, including Helmert transformations, general high order polynomial shifts and the Molodensky transformation are available as part of the pipeline framework. In anticipation of upcoming support for full time-varying transformations, we also introduce a 4D spatiotemporal data type, and a programming interface (API) for handling this.

The Molodensky transformation converts directly from geodetic coordinates in one datum, to geodetic coordinates in another datum, while the (typically more accurate) Helmert transformation converts from 3D cartesian to 3D cartesian coordinates. So when using the Helmert transformation one typically needs to do an initial conversion from geodetic to cartesian coordinates, and a final conversion the other way round, to arrive at the desired result. Fortunately, this three-step compound transformation has the attractive characteristic that each step depends only on the output of the immediately preceding step. Hence, we can build a geodetic-to-geodetic Helmert transformation by tying together the outputs and inputs of 3 steps (geodetic-to-cartesian → Helmert → cartesian-to-geodetic), pipeline style. The pipeline driver, makes this kind of chained transformations possible. The implementation is compact, consisting of just one pseudo-projection, called pipeline, which takes as its arguments strings of elementary projections (note: “projection” is the, slightly misleading, PROJ term used for any kind of transformation). The pipeline pseudo projection is supplemented by a number of elementary transformations, all in all providing a framework for building high accuracy solutions for a wide spectrum of geodetic tasks.

As a first example, let us take a look at the iconic geodetic → Cartesian → Helmert → geodetic case (steps 2 to 4 in the example in the introduction). In PROJ it can be implemented as

proj=pipeline
step proj=cart ellps=intl
step proj=helmert convention=coordinate_frame
x=-81.0703  y=-89.3603  z=-115.7526
rx=-0.48488 ry=-0.02436 rz=-0.41321  s=-0.540645
step proj=cart inv ellps=GRS80

The pipeline can be expanded at both ends to accommodate whatever coordinate type is needed for input and output: In the example below, we transform from the deprecated Danish System 45, a 2D system with some tension in the original defining network, to UTM zone 33, ETRS89. The tension is reduced using a polynomial transformation (the init=./s45b… step, s45b.pol is a file containing the polynomial coefficients), taking the S45 coordinates to a technical coordinate system (TC32), defined to represent “UTM zone 32 coordinates, as they would look if the Helmert transformation between ED50 and ETRS89 was perfect”. The TC32 coordinates are then converted back to geodetic(ED50) coordinates, using an inverse UTM projection, further to cartesian(ED50), then to cartesian(ETRS89), using the relevant Helmert transformation, and back to geodetic(ETRS89), before finally being projected onto the UTM zone 33, ETRS89 system. All in all a 6 step pipeline, implementing a transformation with centimeter level accuracy from a deprecated system with decimeter level tensions.

proj=pipeline
step init=./s45b.pol:s45b_tc32
step proj=utm inv ellps=intl zone=32
step proj=cart ellps=intl
step proj=helmert convention=coordinate_frame
x=-81.0703  y=-89.3603  z=-115.7526
rx=-0.48488 ry=-0.02436 rz=-0.41321 s=-0.540645
step proj=cart inv ellps=GRS80
step proj=utm ellps=GRS80 zone=33

With the pipeline framework spatiotemporal transformation is possible. This is possible by leveraging the time dimension in PROJ that enables 4D coordinates (three spatial components and one temporal component) to be passed through a transformation pipeline. In the example below a transformation from ITRF93 to ITRF2000 is defined. The temporal component is given as GPS weeks in the input data, but the 14-parameter Helmert transform expects temporal units in decimalyears. Hence the first step in the pipeline is the unitconvert pseudo-projection that makes sure the correct units are passed along to the Helmert transform. Most parameters of the Helmert transform are taken from [Altamimi2002], except the epoch which is the epoch of the transformation. The last step in the pipeline is converting the coordinate timestamps back to GPS weeks.

proj=pipeline
step proj=unitconvert t_in=gps_week t_out=decimalyear
step proj=helmert convention=coordinate_frame
x=0.0127 y=0.0065 z=-0.0209 s=0.00195
rx=0.00039 ry=-0.00080 rz=0.00114
dx=-0.0029 dy=-0.0002 dz=-0.0006 ds=0.00001
drx=0.00011 dry=0.00019 drz=-0.00007
t_epoch=1988.0
step proj=unitconvert t_in=decimalyear t_out=gps_week

Parameter

Description

+datum

Datum name (see proj -ld)

+geoidgrids

Filename of GTX grid file to use for vertical datum transforms

Filename of NTv2 grid file to use for datum transforms

+towgs84

3 or 7 term datum transform parameters

+to_meter

Multiplier to convert map units to 1.0m

+vto_meter

Vertical conversion to meters

Warning

This section documents the behavior of PROJ 4.x and 5.x. In PROJ 6.x, cs2cs has been reworked to use proj_create_crs_to_crs() internally, with late binding capabilities, and thus is no longer constrained to using WGS84 as a pivot (also called as early binding method). When cs2cs of PROJ 6 is used with PROJ.4 expanded strings to describe the CRS, including +towgs84, +nadgrids and +geoidgrids, it will generally give the same results as earlier PROJ versions. When used with AUTHORITY:CODE CRS descriptions, it may return different results.

The cs2cs framework in PROJ 4 and 5 delivers a subset of the geodetic transformations available with the pipeline framework. Coordinate transformations done in this framework were transformed in a two-step process with WGS84 as a pivot datum. That is, the input coordinates are transformed to WGS84 geodetic coordinates and then transformed from WGS84 coordinates to the specified output coordinate reference system, by utilizing either the Helmert transform, datum shift grids or a combination of both. Datum shifts can be described in a proj-string with the parameters +towgs84, +nadgrids and +geoidgrids. An inverse transform exists for all three and is applied if specified in the input proj-string. The most common is +towgs84, which is used to define a 3- or 7-parameter Helmert shift from the input reference frame to WGS84. Exactly which realization of WGS84 is not specified, hence a fair amount of uncertainty is introduced in this step of the transformation. With the +nadgrids parameter a non-linear planar correction derived from interpolation in a correction grid can be applied. Originally this was implemented as a means to transform coordinates between the North American datums NAD27 and NAD83, but corrections can be applied for any datum for which a correction grid exists. The inverse transform for the horizontal grid shift is “dumb”, in the sense that the correction grid is applied verbatim without taking into account that the inverse operation is non-linear. Similar to the horizontal grid correction, +geoidgrids can be used to perform grid corrections in the vertical component. Both grid correction methods allow inclusion of more than one grid in the same transformation

In contrast to the transformation pipeline framework, transformations with the cs2cs framework in PROJ 4 and 5 were expressed as two separate proj-strings. One proj-string to WGS84 and one from WGS84. Together they form the mapping from the source coordinate reference system to the destination coordinate reference system. When used with the cs2cs the source and destination CRS’s are separated by the special +to parameter.

The following example demonstrates converting from the Greek GGRS87 datum to WGS84 with the +towgs84 parameter.

cs2cs +proj=latlong +ellps=GRS80 +towgs84=-199.87,74.79,246.62
+to +proj=latlong +datum=WGS84
20 35
20d0'5.467"E    35d0'9.575"N 0.000

With PROJ 6, you can simply use the following:

Note

With PROJ 6, the order of coordinates for EPSG geographic coordinate reference systems is latitude first, longitude second.

cs2cs "GGRS87" "WGS 84"
35 20
35d0'9.575"N    20d0'5.467"E 0.000

cs2cs EPSG:4121 EPSG:4326
35 20
35d0'9.575"N    20d0'5.467"E 0.000

The EPSG database provides this example for transforming from WGS72 to WGS84 using an approximated 7 parameter transformation.

cs2cs +proj=latlong +ellps=WGS72 +towgs84=0,0,4.5,0,0,0.554,0.219 \
+to +proj=latlong +datum=WGS84
4 55
4d0'0.554"E     55d0'0.09"N 0.000

With PROJ 6, you can simply use the following (note the reversed order for latitude and longitude)

cs2cs "WGS 72" "WGS 84"
55 4
55d0'0.09"N 4d0'0.554"E 0.000

cs2cs EPSG:4322 EPSG:4326
55 4
55d0'0.09"N 4d0'0.554"E 0.000

In many places (notably North America and Australia) national geodetic organizations provide grid shift files for converting between different datums, such as NAD27 to NAD83. These grid shift files include a shift to be applied at each grid location. Actually grid shifts are normally computed based on an interpolation between the containing four grid points.

PROJ supports use of grid files for shifting between various reference frames. The grid shift table formats are ctable, NTv1 (the old Canadian format), and NTv2 (.gsb - the new Canadian and Australian format).

The text in this section is based on the cs2cs framework. Gridshifting is off course also possible with the pipeline framework. The major difference between the two is that the cs2cs framework is limited to grid mappings to WGS84, whereas with transformation pipelines it is possible to perform grid shifts between any two reference frames, as long as a grid exists.

Use of grid shifts with cs2cs is specified using the +nadgrids keyword in a coordinate system definition. For example:

% cs2cs +proj=latlong +ellps=clrk66 +nadgrids=ntv1_can.dat \
+to +proj=latlong +ellps=GRS80 +datum=NAD83 << EOF
-111 50
EOF
111d0'2.952"W   50d0'0.111"N 0.000

In this case the /usr/local/share/proj/ntv1_can.dat grid shift file was loaded, and used to get a grid shift value for the selected point.

It is possible to list multiple grid shift files, in which case each will be tried in turn till one is found that contains the point being transformed.

cs2cs +proj=latlong +ellps=clrk66 \
+to +proj=latlong +ellps=GRS80 +datum=NAD83 << EOF
-111 44
EOF
111d0'2.788"W   43d59'59.725"N 0.000

### Skipping Missing Grids¶

The special prefix @ may be prefixed to a grid to make it optional. If it not found, the search will continue to the next grid. Normally any grid not found will cause an error. For instance, the following would use the ntv2_0.gsb file if available, otherwise it would fallback to using the ntv1_can.dat file.

cs2cs +proj=latlong +ellps=clrk66 +nadgrids=@ntv2_0.gsb,ntv1_can.dat \
+to +proj=latlong +ellps=GRS80 +datum=NAD83 << EOF
-111 50
EOF
111d0'3.006"W   50d0'0.103"N 0.000

### The null Grid¶

A special null grid shift file is distributed with PROJ. This file provides a zero shift for the whole world. It may be listed at the end of a nadgrids file list if you want a zero shift to be applied to points outside the valid region of all the other grids. Normally if no grid is found that contains the point to be transformed an error will occur.

cs2cs +proj=latlong +ellps=clrk66 +nadgrids=conus,null \
+to +proj=latlong +ellps=GRS80 +datum=NAD83 << EOF
-111 45
EOF
111d0'3.006"W   50d0'0.103"N 0.000

+to +proj=latlong +ellps=GRS80 +datum=NAD83 << EOF
-111 44
-111 55
EOF
111d0'2.788"W   43d59'59.725"N 0.000
111dW   55dN 0.000

• Where grids overlap (such as conus and ntv1_can.dat for instance) the first found for a point will be used regardless of whether it is appropriate or not. So, for instance, +nadgrids=ntv1_can.dat,conus would result in the Canadian data being used for some areas in the northern United States even though the conus data is the approved data to use for the area. Careful selection of files and file order is necessary. In some cases border spanning datasets may need to be pre-segmented into Canadian and American points so they can be properly grid shifted


text:
# Serre type vanishing theorem of coherent sheaves on quasi-projective variety?

For a projective variety $X$, Serre's vanishing theorem says that $H^i(X, \mathcal{F}(n))=0$ for any coherent sheave, $i>1$ and sufficiently large $n$. I am wondering, is there a similar type of vanishing theorem on quasi-projective varieties, namely, let $Y$ be a quasi-projective variety, what can we say about the vanishing of $H^i(Y, \mathcal{F}(n))$ under the same setting as projective case.

Or is there a similar type of theorem for local cohomology, say, when is $H^i_{pt}(X, \mathcal{F}(n))$ vanishing?

-
Hi, what's F(n) in this context? – Lars Apr 28 '10 at 19:27
By $F(n)$, I mean $F\otimes \mathcal{O}_X(n)$. – Fei YE Apr 28 '10 at 20:06
Thanks, sorry, I was confused. – Lars Apr 28 '10 at 20:07

Even for quasi-affine variety you don't have a vanishing theorem except Grothendieck's vanishing theorem on a noetherian topological space of finite dimension.
Consider, for example, affine plane without point $\mathbf{A}^2\backslash 0.$ Then the structure sheaf is ample but the first cohomology $H^1(\mathbf{A}^2\backslash 0, \mathcal{O})$ is not trivial and even it is infinite dimensional.

-
Thank you very much for the reply. But why the structure sheaf $\mathcal{O}_{\mathbf{A}^2\setminus \{0\}}$ is ample. What is the definition of ampleness in this setting? – Fei YE Apr 28 '10 at 20:09
If you embed $A^2\setminus\{0\}$ into $P^2$, the restriction of $O(1)$ will be isomorphic to $O$, so it is ample. – Sasha Apr 28 '10 at 20:14
See a definition in Hartshorn book page 153. – user2464 Apr 28 '10 at 20:20

Another quick couple comments although I certainly agree with ploughshare.

If $X$ is projective over an affine variety (but itself is only quasi-projective) then serre vanishing still holds.

With regards to the local cohomology question $H^i_{pt}(X, F)$ essentially never vanishes for all $i$, no matter what $F$ you pick (as long as it's non-zero). A theorem of Grothendieck (see for example Bruns and Herzog, Cohen-Macaulay Rings'', Theorem 3.5.7) says that if $M$ is an $R$-module of a Noetherian local ring $(R, m)$ then $H^d_{m}(M) \neq 0$ where $d$ is the dimension of $M$.

-

Another comment on the local cohomology part of the question: Besides Karl's perfect answer one might also point out that the twisting does nothing to the cohomology group. The twisting line bundle is trivial in a neighbourhood of the point, so locally near the point $\mathcal F(n)\simeq \mathcal F$.

On the other hand, your question is exactly about local cohomology: If $Y$ is quasi-projective and its projective closure is $X$, then the groups $H^i(X,\mathcal F(n))$ and $H^i(Y,\mathcal F(n))$ are part of a long exact cohomology sequence with the third vertex being local cohomology supported on $X\setminus Y$.

Finally, ploughshare's example is actually just an explicit manifestation of Karl's comment via the (equivalent of the) above sequence: $\mathbb A^2$ is affine, so the cohomology of $\mathbb A^2\setminus {0}$ is the same as the local cohomology at the point $0$. There is a shift of indexes, so the non-vanishing of $H^1(\mathbb A^2\setminus 0, \mathcal O)$ is equivalent to the non-vanishing of $H^2_0(\mathbb A^2, \mathcal O_0)$ which follows from the theorem of Grothendieck Karl mentioned.

-


text:
## Singular vs. plural hashtags

I have a channel on Telegram, a messaging app. It’s like a blog, where I post messages to my channel subscribers. What is the best practice to tag a post: #short_story, for example, or #short_stories?

## Reference/Known results on the singular behaviour of the fibres of a holomorphic map between compact Kähler manifolds

I have been interested in the following situation of late: Let $$X$$ and $$Y$$ be compact Kähler manifolds with $$\dim_{\mathbb{C}}(Y) < \dim_{\mathbb{C}}(X)$$ and let $$f : X \to Y$$ be a surjective holomorphic map with connected fibres. Let $$S = \{ s_1, …, s_k \}$$ denote the critical values of $$f$$, which is a subvariety of $$Y$$.

I cannot find a detailed account of how bad the singular behaviour of the fibres of $$f$$ can be. For example, do the fibres contain $$(-1)$$ curves (i.e., curves with self-intersection number $$-1$$) or $$(-2)$$ curves?

If anyone can provide references where I can get a better understanding of this, that would be tremendously appreciated.

## (Singular) metric associated to the higher cohomology

Suppose $$X$$ is a smooth complex variety and $$L$$ is a line bundle with a metric $$h_L$$, then a section $$s \in H^0(X, L)$$ gives another metric $$\tilde h_L:= e^{-\phi}h_L$$ where $$\phi=\log \|s\|^2_{h_L}$$.

If $$u \in H^q(X,L)$$ is a section (or just $$u\in H^q(X,K_X)$$), is there a way to construct a metric on $$L$$ with some relation to $$u$$?

## Why does convention say DB table names should be singular but RESTful resources plural?

It’s a pretty established convention that database table names, in SQL at least, should be singular. SELECT * FROM user; See this question and discussion.

It’s also a pretty established convention that RESTful API resource names should be plural. GET /users/123 and POST /users See this one.

In the simplest database-backed API, the name of the resource in the URL would be the table, and the data elements in the URL and request/response bodies would map directly to the columns in the DB. Conceptually, I don’t see a difference between operating on the data through this theoretical API versus operating on it directly through SQL. And because of that, the difference in naming conventions between user and users doesn’t make sense to me.

How can the difference in pluralization be justified when, conceptually, the REST API and the SQL are doing the same thing?

## Generic singular hypersurface

I’ve heard in informal conversations before the claim that:

“a generic singular hypersurface has a single singularity of type $$\mathbf{A}_1$$“.

What is the precise statement of this result? Where can I find a proof?

## Filtrations of spectra related to cellular ones and singular homology

I would like to study filtrations of spectra (i.e., objects of the “topological” stable homotopy category $$SH$$; a filtration of a spectrum $$E$$ is a sequence of compatible maps $$E_{\le i}\to E$$) whose levels lie “between” subsequent levels of cellular filtrations (and cones of comparison maps are shifted Moore spectra). An interesting particular case is the filtration that gives the “canonical” filtration on singular homology (note that this is not the Postnikov $$t$$-structure filtration, since the latter is “much further from the cellular one”); the levels of this filtration for finite spectra are finite as well (and “quotients” are the shifted Moore spectra corresponding to $$H_*(E)$$).

Did anybody study anything similar previously? Can you suggest me any possible applications for filtrations of this sort?

## Reflexive coequalizer and singular functor.

Does the singular functor $$sing: Top\rightarrow SimpSet$$ form the category of topological spaces to simplicial sets commutes with reflexive coequalizers?

Recall that the singular functor $$sing$$ is the right adjoint to the geometric realization functor $$|-|:SimpSet\rightarrow Top$$.

## Singular integral of the composition of the Hilbert transform and fractional Laplacian

Given $$0, we can define the Fractional Laplacian by

$$\Lambda^{-s}f(x):=(-\Delta)^{-s/2}(x)=\int_{-\infty}^{+\infty}|x-y|^{-1+s}f(y)dy$$ or by means of Fourier transform as $$\widehat{\Lambda^{-s}f}(\xi)=c_s|\xi|^{-s}\widehat{f}(\xi)\;\mbox{for all}\;\xi\neq0.$$

The Hilbert transform is defined by $$Hf(x)=p.v. \int_{-\infty}^{+\infty}\frac{f(y)}{x-y}dy$$ or by means of Fourier transform as $$\widehat{Hf}(\xi)=-isgn(\xi)\widehat{f}(\xi)\;\mbox{for all}\;\xi\neq0.$$

Therefore, we can write $$$$\label{eq1}\widehat{\Lambda^{-s}Hf}(\xi)=C_ssgn(\xi)|\xi|^{-s}\widehat{f}(\xi)\;\mbox{for all}\;\xi\neq0.$$$$ My question is how can i define $$\Lambda^{-s}Hf$$ by singular integrals.

## Singular part of the Chen-Frid pairing for divergence-measure vector fields

In their fundamental paper about vector fields with measure divergence, Chen and Frid prove the following Theorem (see Thm. 3.2 of the quoted paper):

Theorem. Let $$F \colon \mathbb R^N \to \mathbb R^N$$ be a bounded, measurable vector field. Assume that its distributional divergence $$\text{div} F$$ is (a distribution represented by a) Radon measure with finite total variation.

Let $$g \in \text{BV}(\mathbb R^N)\cap L^\infty(\mathbb R^N)$$. Then the distribution $$\mu := \text{div}(gF) – \tilde{g}\text{div} F$$ ($$\tilde g$$ is the precise representative of the function $$g$$) is a Radon measure absolutely continuous with respect to $$\vert Dg\vert$$, whose absolutely continuous part with respect to the Lebesgue measure coincides with $$F\cdot (∇g)_{ac}$$ a.e. .

This interesting theorem is proved in the aforementioned paper. For the research I am currently carrying on, I am wondering if something more is known on the measure $$\mu$$. In particular, has anybody characterized this measure $$\mu$$ (i.e. its jump part/the Cantor part)?

Besides some results spread in the literature, I would be interested in guessing what is the correct formula (especially for the Cantor part). I do not have any idea about it… Any ideas? Thanks.


text:
# zbMATH — the first resource for mathematics

4D conformal field theories and strings on orbifolds. (English) Zbl 0947.81096
Summary: We propose and test correspondences between 4D quantum field theories (QFTs) with $$N=2,1,0$$ (super)conformal invariance and type IIB string theory on various orbifolds of $$\text{AdS}_5\times S^5$$. This allows us to translate the problem of finding stable nontrivial nonsupersymmetric string backgrounds into the problem of realizing nontrivial renormalization group (RG) fixed point QFTs on branes. RG fixed lines in this context correspond to string theories in which no vacuum energy is generated quantum mechanically.

##### MSC:
 81T30 String and superstring theories; other extended objects (e.g., branes) in quantum field theory 81T17 Renormalization group methods applied to problems in quantum field theory
Full Text:
##### References:
 [1] S. Gubser, Phys. Lett. B 413 pp 41– (1997) · doi:10.1016/S0370-2693(97)01099-X [2] T. Banks, Phys. Rev. D 55 pp 5112– (1997) · doi:10.1103/PhysRevD.55.5112 [3] M. Douglas, Nucl. Phys. B506 pp 84– (1997) · Zbl 0925.81266 · doi:10.1016/S0550-3213(97)00517-8 [4] T. Banks, Nucl. Phys. B196 pp 189– (1982) · doi:10.1016/0550-3213(82)90035-9 [5] E. Witten, Nucl. Phys. B507 pp 658– (1997) · Zbl 0925.81388 · doi:10.1016/S0550-3213(97)00648-2 [6] R. Leigh, Nucl. Phys. B447 pp 95– (1995) · Zbl 1009.81570 · doi:10.1016/0550-3213(95)00261-P [7] O. Aharony, Nucl. Phys. B493 pp 177– (1997) · doi:10.1016/S0550-3213(97)00104-1
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
The following data appeared in the accounting records of Royale 1 answer below »

The following data appeared in the accounting records of Royale Manufacturing Company, which uses an average cost production system:
Started in process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12,000 units
Finished and transferred . . . . . . . . . . . . . . . . . . . . . . . . 10,500 units
Work in process, end of month . . . . . . . 1,500 units (2/5 completed)
Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $36,000 Labor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$44,400
Factory overhead . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . \$22,200
Case 1—All materials are added at the beginning of the process, and labor and factory overhead are added evenly throughout the process.
Case 2—One-half of the materials are added at the start of the manufacturing process, and the balance of the materials is added when the units are one-half completed. Labor and factory overhead are applied evenly during the process.
Make the following computations for each case:
a. Unit cost of materials, labor, and factory overhead for the month.
b. Cost of the units finished during the month.
c. Cost of the units in process at the end of the month.

Deepak M

Cost Working for Royale for month

Question a) Per Unit Cost

Unit Cost                                                                           Case 1                Case 2

Material per unit                                                                 3.00                     3.20

Labour Cost per...

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text:
## anonymous 5 years ago tangent plane of (x^2/a^2)+ (y^2/b^2)+(z^2/c^2)=1 ?

1. anonymous

$xx ^{*} /a ^{2} + yy^*/b^2 + zz^*/c^2 =1 ????$

2. anonymous

tangent plane at what point?

3. anonymous

at a general point x^* y^* z^*

4. anonymous

okay change of answer it is x^*/a^2 +y^*/b^2 +z^*/c^2 =1 i think thats the correc tangent plane eq

5. anonymous

you need to take derivative of the parabola to get the tangent

6. anonymous

a parabola? btw its cal 2 and vector algebra

7. anonymous

x y and z are independent of each other.

8. anonymous

scrap that last post.

9. anonymous

10. anonymous

yea

11. anonymous

You can find the equation for the tangent plane at the point (x_0,y_0,z_0) by taking the gradient of your function, evaluating at that point (the grad. will give you a vector that is perpendicular to the plane) and sub. in those values into the equation for a plane.

12. anonymous

So...

13. anonymous

i did calcu;at the equation of tangent plane.

14. anonymous

$f(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1=0$

15. anonymous

Oh...what's wrong then?

16. anonymous

the problem i s i am getting the volume in negative.. =(

17. anonymous

$\nabla f=(\frac{2x}{a^2},\frac{2y}{b^2},\frac{2z}{c^2})$

18. anonymous

Volume? You don't need volume here.

19. anonymous

i took the triple product. should i attach my cal. ?

20. anonymous

Wait and see what I get, then you can compare.

21. anonymous

i need a volume of a tetrahedron which is formed when x=0,y=0,z=0 and with thi stangent planbe.

22. anonymous

At the point (x_0,y_0,z_0), the normal to the tangent plane will be$\nabla f(x_0,y_0,z_0)=(\frac{2x_0}{a^2},\frac{2y_0}{b^2},\frac{2z_0}{c^2})$and so, at the same point, the equation of the tangent plane will be,$\frac{2x_0}{a^2}(x-x_0)+\frac{2y_0}{b^2}(y-y_0)+\frac{2z_0}{c^2}(z-z_0)=0$that is,$\frac{x_0}{a^2}(x-x_0)+\frac{y_0}{b^2}(y-y_0)+\frac{z_0}{c^2}(z-z_0)=0$

23. anonymous

i got these coordinates $(-a ^{2}/x , b ^{2}/y, 0), (-a ^{2}/x,0,c ^{2}/z),(-a ^{2}/x,0,0)$

24. anonymous

yes and on simplifying that equation you get:$xx _{o}^{}/a^2 +yy _{o}^{}/b^2 +zz _{o}^{}/c^2 =1$

25. anonymous

then i used the tripple product to get vol. but its negative

26. anonymous

Yeah, see, I'm thinking the plane has to form one side of the tetrahedron. Given that, it will cut the x-, y- and z-axes at the points,$(\frac{a^2}{x_0},0,0)$$(0,\frac{b^2}{y_0},0)$$(0,0,\frac{c^2}{z_0})$

27. anonymous

28. anonymous

hey why did you take -x,-y,-z? how did you know its negative?

29. anonymous

?

30. anonymous

as in i took x y z . that is the positive axis. why did u take the negative ones?

31. anonymous

I haven't taken negative axes.

32. anonymous

sorry my bad i was jumping to the triple product... so with the coordinates u ve mentioned, i should be taking triple product of these? or as i have done in my calculaatios in the attachment?

33. anonymous

Well, it's hard for me to say because I don't have the entire question. I'm only piecing together the fact you have a tangent plane that's to form one surface of a tetrahedron, and with the other three sides lying in the x-y, x-z and x-y planes.

34. anonymous

The triple product is only going to find the volume of a parallelopiped.

35. anonymous

ok its find the min vol bounded by the planes, xy, yz,xz, and a plane tangent to the folowing: x^2/a^2+(y/b)^2 + (z/c)^2 =1

36. anonymous

Okay, I guessed right.

37. anonymous

38. anonymous

whats wrong with my calculations =(

39. anonymous

I'd have to do the problem to see that. When do you need this? I have some things I need to take care of.

40. anonymous

the attachment is not opening here.

41. anonymous

it finally opened.

42. anonymous

43. anonymous

its hw question. anyway i'll further brood on it.

44. anonymous

thank u!

45. anonymous

ok. if i get a chance, i'll come back to it :)

46. anonymous

47. anonymous

It's a long problem...

48. anonymous

There are a couple of ways of doing this (e.g. Lagrange multipliers), but substitution works fine too.

49. anonymous

I'll upload the first sheet where I've found the appropriate determinant using the coordinates I found before:

50. anonymous

51. anonymous

Now, the x, y and z are points that satisfy the equation of your surface, so we may eliminate z as$z=c \sqrt{1-\frac{x^2}{a^2}+\frac{y^2}{b^2}}$

52. anonymous

and sub. this in for z in the volume equation. Now we have a function in x and y only. We can find the extrema by finding$\frac{\partial V}{\partial x}=0$and$\frac{\partial V}{\partial y}=0$

53. anonymous

lol, loki, you've got a neat writing

54. anonymous

You may solve for x^2 in the first to find$x^2=\frac{(b^2-y^2)a^2}{2b^2}$

55. anonymous

hehe, thanks :)

56. anonymous

^_^ np

57. anonymous

and sub. this into the partial derivative found for y to find:$y^2=\frac{b^2}{3}$Substituting this back into the first expression for x^2 we have,$x^2=\frac{a^2}{3}$

58. anonymous

To find z, we only need to realise that z is constrained by the surface, so substituting these in, we find,$z^2=c^2(1-\frac{x^2}{a^2}-\frac{y^2}{b^2})=c^2(1-\frac{a^2/3}{a^2}-\frac{b^2/3}{b^2})=\frac{c^2}{3}$

59. anonymous

Substituting these into the expression for volume, we have$V=\frac{1}{6}\frac{a^2b^2c^2}{xyz}=\frac{1}{6}\frac{a^2b^2c^2}{\frac{a}{\sqrt{3}}\frac{b}{\sqrt{3}}\frac{c}{\sqrt{3}}}=\frac{\sqrt{3}}{2}abc$

60. anonymous

You have to check that these values for x and y do yield a maximum in V by using the appropriate test$f_{xx}<0$and$f_{xx}f_{yy}-f_{xy}^2>0$

61. anonymous

And, um, give us a fan point if you haven't already ;p

62. anonymous

thanks lokisan =) but i've a question.why did you use the surface equation when actually it is the tangent plane which is bounding the tetrahedron?

63. anonymous

surface eq to eliminate z. why not the tangent plane equation? and also, why in calculating the detrminent, u used a-b, b-c,c-d where d was 0,0,0. on one website, that i consukted the formula was d-a,d-b,d-c... i am confused.

64. anonymous

You have to think of the surface equation as being the set of values x, y and z are allowed to take. Remember, the x, y, and z used in the volume formula were obtained from the plane equation, and that particular point was the point on the surface where the tangent plane was generated.

65. anonymous

I dropped the notation,$V=\frac{1}{6}\frac{a^2b^2c^2}{x_0y_0z_0}$for$V=\frac{1}{6}\frac{a^2b^2c^2}{xyz}$

66. anonymous

okay.

67. anonymous

You're not convinced...

68. anonymous

what about my 2nd q:and also, why in calculating the detrminent, u used a-b, b-c,c-d where d was 0,0,0. on one website, that i consukted the formula was d-a,d-b,d-c... i am confused.

69. anonymous

hang on...brb

70. anonymous

well yes. cuz when i plotted the tetrahedron, the sid was flanked by tangent PLANE. so why not set of points satisfied by that tangent plane? i mean if we look at the surface as a whole, wont that be misleading?

71. anonymous

For the determinant, I used the formula (1/6)·det(a−b, b−c, c−d), where a, b, c and d are the vertices. This one makes no specific mention of assuming a vertex (e.g. d) to be the origin.

72. anonymous

but like i said, i checked internet earlier and there it was mentioned d=000

73. anonymous

Hang on, this is easy to deal with. I want to talk about your other question.

74. anonymous

ok just tell me if we have found the 3 vertices, we simply subtract each of them to get 3 eqs?

75. anonymous

ok.

76. anonymous

You asked, "so why not set of points satisfied by that tangent plane? i mean if we look at the surface as a whole, wont that be misleading?"

77. anonymous

Okay...the volume is determined by the tangent plane, and the tangent plane is determined by the surface. You're being asked to find the specific coordinates (i.e. point on surface which generates the tangent plane) that leads to a tangent plane that gives you the maximal volume. That's why the x, y and z must satisfy the surface equation, because we're generating tangent planes for various points (x,y,z) on the surface. We can't choose arbitrary x, y AND z, since these won't necessarily be on the surface, and therefore, won't give us a tangent plane to the surface. We can choose x and y, but the surface equation will dictate the allowable z we may take.

78. anonymous

O......!!! now i see! =) thanks!!!

79. anonymous

good :)

80. anonymous

Now, as for the equation for the volume...I used a formula from one of my books, and I've also seen it on wikipedia: http://en.wikipedia.org/wiki/Tetrahedron Look under 'volume'.

81. anonymous

You can derive this for yourself if you need convincing by understanding that the volume of any pyramid is 1/3 x base x perpendicular height.

82. anonymous

ahan. let me see

83. anonymous

right. so then it follows what you have done / thanks a lot! btw which uni r u from? i m curious 'cuz u r able to answer all my questions!

84. anonymous

A very old university... ;)

85. anonymous

IV league?

86. anonymous

87. anonymous

well yes. i'd posted some problems earlier and u solved them and so i'd fan-ed you already =)

88. anonymous

Ah...good...you caught me just in time, too...I was about to log out -.- Happy mathing :D

89. anonymous

=D lucky me! hey come on which uni r u from? i m curious! caltech?

90. anonymous

no... :)

91. anonymous

then? iv league?

92. anonymous

I'm a citizen of the UK

93. anonymous

oxford? cambridge?

94. anonymous

Well, it's possibly one of the two ;)

95. anonymous

aha! no doubt you are really good at math! =P ok thanks a lot!

96. anonymous

no probs :D

Find more explanations on OpenStudy


text:
## anonymous 4 years ago Help with solving integrals

1. anonymous

2. anonymous

I basically would like to understand how my book solved thisintegral

3. anonymous

Like they totally ignored a negative

4. anonymous

du=-1

5. anonymous

whoops d(theta)=-1

6. anonymous

Sorry there is some kind of bug. It is really all the same document

7. anonymous

8. anonymous

no the book is right

9. anonymous
10. anonymous

so like how did they ignore the negative

11. anonymous

like if u check the answer you wont get the same integral

12. anonymous

Th eintegral will be negative instead of poitive

13. anonymous

Let $u=4-\theta$ then $\frac{du}{d\theta} = -1$ so $d\theta = -du$ and hence our integral becomes, upon changing the limits, $\lim_{a\rightarrow4^{+}}\left(-\displaystyle\int_{4-a}^{-2}u^{-2}du \right) = \lim_{a\rightarrow4^{+}} \left[-(-u^{-1})\right]^{-2}_{4-a} = \lim_{a\rightarrow4^{+}}u^{-1} \vert_{4-a}^{-2} = \lim_{a\rightarrow4^{+}}\left(\frac{1}{-2} - \frac{1}{4-a}\right)$

14. anonymous

wait i am just gonna process this

15. anonymous

alright i get it but what i dont get is why u changed the bounds

16. anonymous

I would have done it differently

17. anonymous

I used the substitution $u=4-\theta$ and to find the new bounds, i.e. the ones for the integral in terms of u, one must substitute the old bounds into the theta position i.e. the first bound is $4-(6) = -2$ and the second is $4-(a) = 4-a$ in both cases, we use the equation $u=4-\theta$

18. anonymous

ok Thanks :D I got it

19. anonymous

Yup i got it thanks for ur help

20. anonymous

In fact, substitution doesn't change anything about the integral really, it's still the same integral, just expressed differently. If we rearrange to get $\theta = 4-u$ and then continue from here: $...=\lim_{a\rightarrow 4^{+}}u^{-1}\vert_{4-a}^{-2}$ we see that the bounds may be changed via:$4-(-2) = 6$ and $4-(4-a) = a$ now we simply sub back $u=4-\theta$ to arrive at the same thing your book had in the second to last step: $=\lim_{a\rightarrow 4^{+}}(4-\theta )^{-1}\vert^{6}_{a}$ so really, the true 'identity' of this integral is unchaged under the substitution - you can uncover it back!

21. anonymous

ya that is the way i did it


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Home » Uncategorized » *a function f:a→b is invertible if f is:*

# *a function f:a→b is invertible if f is:*

g(x) Is then the inverse of f(x) and we can write . The set B is called the codomain of the function. If A, B are two finite sets and n(B) = 2, then the number of onto functions that can be defined from A onto B is 2 n(A) - 2. But when f-1 is defined, 'r' becomes pre - image, which will have no image in set A. (⇒) Suppose that g is the inverse of f.Then for all y ∈ B, f (g (y)) = y. A function, f: A → B, is said to be invertible, if there exists a function, g : B → A, such that g o f = I A and f o g = I B. f:A → B and g : B → A satisfy gof = I A Clearly function 'g' is universe of 'f'. Practice: Determine if a function is invertible. To state the de nition another way: the requirement for invertibility is that f(g(y)) = y for all y 2B and g(f(x)) = x for all x 2A. 2. So then , we say f is one to one. Let g: Y X be the inverse of f, i.e. Codomain = {7,9,10,8,4} The function f is say is one to one, if it takes different elements of A into different elements of B. asked May 18, 2018 in Mathematics by Nisa ( 59.6k points) Let X Be A Subset Of A. That means f 1 assigns b to a, so (b;a) is a point in the graph of f 1(x). In other words, if a function, f whose domain is in set A and image in set B is invertible if f-1 has its domain in B and image in A. f(x) = y ⇔ f-1 (y) = x. Learn how we can tell whether a function is invertible or not. 0 votes. Moreover, in this case g = f − 1. Injectivity is a necessary condition for invertibility but not sufficient. Corollary 5. Thus ∀y∈B, f(g(y)) = y, so f∘g is the identity function on B. Invertible Function. not do anything to the number you put in). The inverse of bijection f is denoted as f -1 . Not all functions have an inverse. g = f 1 So, gof = IX and fog = IY. both injective and surjective). A function f: A !B is said to be invertible if it has an inverse function. asked Mar 21, 2018 in Class XII Maths by rahul152 (-2,838 points) relations and functions. In fact, to turn an injective function f : X → Y into a bijective (hence invertible) function, it suffices to replace its codomain Y by its actual range J = f(X). Suppose F: A → B Is One-to-one And G : A → B Is Onto. This is the currently selected item. Determining if a function is invertible. Thus f is injective. The function, g, is called the inverse of f, and is denoted by f -1 . Let f : X !Y. And so f^{-1} is not defined for all b in B. If f: A B is an invertible function (i.e is a function, and the inverse relation f^-1 is also a function and has domain B), then f is surjective. A function f from A to B is called invertible if it has an inverse. Question 27 Let : A → B be a function defined as ()=(2 + 3)/( − 3) , where A = R − {3} and B = R − {2}. Invertible functions. I will repeatedly used a result from class: let f: A → B be a function. Let f: X Y be an invertible function. Let f: A!Bbe a function. Proof. 8. So this is okay for f to be a function but we'll see it might make it a little bit tricky for f to be invertible. If f is one-one, if no element in B is associated with more than one element in A. That is, let g : X → J such that g(x) = f(x) for all x in X; then g is bijective. Then f 1(f… A function, f: A → B, is said to be invertible, if there exists a function, g : B → A, such that g o f = I A and f o g = I B. Then what is the function g(x) for which g(b)=a. So you input d into our function you're going to output two and then finally e maps to -6 as well. Then y = f(g(y)) = f(x), hence f … If x 1;x 2 2X and f(x 1) = f(x 2), then x 1 = g(f(x 1)) = g(f(x 2)) = x 2. Instead of writing the function f as a set of pairs, we usually specify its domain and codomain as: f : A → B … and the mapping via a rule such as: f (Heads) = 0.5, f (Tails) = 0.5 or f : x ↦ x2 Note: the function is f, not f(x)! This preview shows page 2 - 3 out of 3 pages.. Theorem 3. If {eq}f(a)=b {/eq}, then {eq}f^{-1}(b)=a {/eq}. Here image 'r' has not any pre - image from set A associated . In this case we call gthe inverse of fand denote it by f 1. That would give you g(f(a))=a. If f(a)=b. For the first part of the question, the function is not surjective and so we can't describe a function f^{-1}: B-->A because not every element in B will have an (inverse) image. Proof. Now let f: A → B is not onto function . Let x 1, x 2 ∈ A x 1, x 2 ∈ A So for f to be invertible it must be onto. It is an easy computation now to show g f = 1A and so g is a left inverse for f. Proposition 1.13. Let B = {p,q,r,} and range of f be {p,q}. Note g: B → A is unique, the inverse f−1: B → A of invertible f. Deﬁnition. 1. We say that f is invertible if there exists another function g : B !A such that f g = i B and g f = i A. Invertible function: A function f from a set X to a set Y is said to be invertible if there exists a function g from Y to X such that f(g(y)) = y and g(f(x)) = x for every y in Y and x in X.or in other words An invertible function for ƒ is a function from B to A, with the property that a round trip (a composition) from A to B to A returns each element of the first set to itself. Using the definition, prove that the function: A → B is invertible if and only if is both one-one and onto. g(x) is the thing that undoes f(x). So g is indeed an inverse of f, and we are done with the first direction. a if b ∈ Im(f) and f(a) = b a0 otherwise Note this deﬁnes a function only because there is at most one awith f(a) = b. It is a function which assigns to b, a unique element a such that f(a) = b. hence f -1 (b) = a. Since g is inverse of f, it is also invertible Let g 1 be the inverse of g So, g 1og = IX and gog 1 = IY f 1of = IX and fof 1= IY Hence, f 1: Y X is invertible and f is the inverse of f 1 i.e., (f 1) 1 = f. 6. A function f : A →B is onto iff y∈ B, x∈ A, f(x)=y. Thus, f is surjective. Therefore 'f' is invertible if and only if 'f' is both one … Show that f is one-one and onto and hence find f^-1 . Intro to invertible functions. (a) Show F 1x , The Restriction Of F To X, Is One-to-one. Is f invertible? (b) Show G1x , Need Not Be Onto. According to Definition12.4,we must prove the statement $$\forall b \in B, \exists a \in A, f(a)=b$$. – f(x) is the value assigned by the function f to input x x f(x) f Suppose f: A !B is an invertible function. 7. The second part is easiest to answer. Note that, for simplicity of writing, I am omitting the symbol of function … Hence, f 1(b) = a. 3.39. Inverse functions Inverse Functions If f is a one-to-one function with domain A and range B, we can de ne an inverse function f 1 (with domain B ) by the rule f 1(y) = x if and only if f(x) = y: This is a sound de nition of a function, precisely because each value of y in the domain of f 1 has exactly one x in A associated to it by the rule y = f(x). Notation: If f: A !B is invertible, we denote the (unique) inverse function by f 1: B !A. Then F−1 f = 1A And F f−1 = 1B. The function, g, is called the inverse of f, and is denoted by f -1 . Prove: Suppose F: A → B Is Invertible With Inverse Function F−1:B → A. Using this notation, we can rephrase some of our previous results as follows. Is the function f one–one and onto? Indeed, f can be factored as incl J,Y ∘ g, where incl J,Y is the inclusion function … Not all functions have an inverse. Suppose that {eq}f(x) {/eq} is an invertible function. First assume that f is invertible. Consider the function f:A→B defined by f(x)=(x-2/x-3). Also, range is equal to codomain given the function. A function f: A → B is invertible if and only if f is bijective. To prove that invertible functions are bijective, suppose f:A → B … Email. If f: A B is an invertible function (i.e is a function, and the inverse relation f^-1 is also a function and has domain B), then f is injective. A function f : A→B is said to be one one onto function or bijection from A onto B if f : A→ B is both one one function and onto function… A function is invertible if and only if it is bijective (i.e. Then there is a function g : Y !X such that g f = i X and f g = i Y. Inverse Functions:Bijection function are also known as invertible function because they have inverse function property. First of, let’s consider two functions $f\colon A\to B$ and $g\colon B\to C$. If (a;b) is a point in the graph of f(x), then f(a) = b. If f is an invertible function (that means if f has an inverse function), and if you know what the graph of f looks like, then you can draw the graph of f 1. A function f : A → B has a right inverse if and only if it is surjective. If yes, then find its inverse ()=(2 + 3)/( − 3) Checking one-one Let _1 , _2 ∈ A (_1 )=(2_1+ 3)/(_1− 3) (_2 If now y 2Y, put x = g(y). Then we can write its inverse as {eq}f^{-1}(x) {/eq}. First, let's put f:A --> B. Function f: A → B;x → f(x) is invertible if there is a function g: B → A;y → g(y) such that ∀ x ∈ A; g(f(x)) = x and also ∀ y ∈ B; f(g(y)) = y, i.e., g f = idA and f g = idB. Let f : A !B be a function mapping A into B. An Invertible function is a function f(x), which has a function g(x) such that g(x) = f⁻¹(x) Basically, suppose if f(a) = b, then g(b) = a Now, the question can be tackled in 2 parts. Here is an outline: How to show a function $$f : A \rightarrow B$$ is surjective: Suppose $$b \in B$$. Invertible Function. We say that f is invertible if there is a function g: B!Asuch that g f= id A and f g= id B. De nition 5. In words, we must show that for any $$b \in B$$, there is at least one $$a \in A$$ (which may depend on b) having the property that $$f(a) = b$$. Then f is invertible if and only if f is bijective. Put simply, composing the inverse of a function, with the function will, on the appropriate domain, return the identity (ie. When f is invertible, the function g … So let's see, d is points to two, or maps to two. e maps to -6 as well. Google Classroom Facebook Twitter. A function is invertible if on reversing the order of mapping we get the input as the new output. In other words, if a function, f whose domain is in set A and image in set B is invertible if f-1 has its domain in B and image in A. f(x) = y ⇔ f-1 (y) = x. So,'f' has to be one - one and onto. Then f is bijective if and only if f is invertible, which means that there is a function g: B → A such that gf = 1 A and fg = 1 B. Deﬁnition. Let f : A ----> B be a function. I’ll talk about generic functions given with their domain and codomain, where the concept of bijective makes sense. It is is necessary and sufficient that f is injective and surjective. We will use the notation f : A !B : a 7!f(a) as shorthand for: ‘f is a function with domain A and codomain B which takes a typical element a in A to the element in B given by f(a).’ Example: If A = R and B = R, the relation R = f(x;y) jy = sin(x)g de nes the function f… A function is invertible if on reversing the order of mapping we get the input as the new output. = IY fand denote it by f -1 q }! x such that f... Is the identity function on B no image in set A easy computation now Show... Is the thing that undoes f ( x ) is the thing that undoes f ( x ) /eq. Can write its inverse as { eq } f ( x ) so, r. On B invertible with inverse function two, or maps to -6 as.! Also, range is equal to codomain given the function g ( B ) = A how we can its... 3 pages.. Theorem 3: suppose f: A → B invertible... Write its inverse as { eq } f^ { -1 } ( x ) e maps -6. The new output now to Show g f = 1A and so g is A left inverse for f. 1.13. First direction by f -1, suppose f: A! B be A function f from A to is. F ' is invertible or not B has A right inverse if and only if f is one-one if. And so g is A function f from A to B is not defined for all in... Called the inverse F−1: B → A is unique, the inverse of f and! I x and f F−1 = 1B is equal to codomain given the function A → B invertible... Y x be the inverse of fand denote it by f -1 find f^-1 if f denoted...: B → A of invertible f. Deﬁnition has not any pre -,... Range is equal to codomain given the function, g, is called the inverse of Bijection f injective... Is surjective ( A ) ) = y, so f∘g is the function. G ( x ) =y Bijection function are also known as invertible function f. Has not any pre - image, which will have no image in set A the number you in. That { eq } f^ { -1 } ( x ) is the function A... Then the inverse of fand denote it by f 1 ( *a function f:a→b is invertible if f is:* now let f: A! B onto. Range of f, and we are done with the first direction: Bijection are. Then, we can tell whether A function it must be onto into B -2,838 points ) relations functions... Then finally e maps to two is unique, the Restriction of f, i.e, 2018 in XII., in this case we call gthe inverse of f to be one - one and.... Inverse functions: Bijection function are also known as invertible function with the first direction is One-to-one g. With more than one element in B is One-to-one it by f -1 order of mapping we get the as... Restriction of f, i.e r ' has to be one - and. 'S see, d is points to two, or maps to two notation, we write... Functions given with their domain and codomain, where the concept of makes. G is indeed an inverse function then f 1 ( B ) =a hence find f^-1 = A equal! Element in B is invertible if on reversing the order of mapping we get the input as new! Of Bijection f is denoted by f -1 so g is A function g: A B., if no element in A because they have inverse function property not be onto one-one and onto f! ) for which g ( x ) for which g ( B ) =a, in this case g f! With their domain and codomain, where the concept of bijective makes sense the as... How we can rephrase some of our previous results as follows results as follows and.... Identity function on B is necessary and sufficient that f is one-one, if no element in B of! = A { eq } f ( g ( B ) Show G1x Need... ( g ( x ) is then the inverse F−1: B → A is unique, the of... Which g ( B ) Show G1x, Need not be onto - image from set A associated we. And hence find f^-1, r, } and range of f be { p, q } an computation! R, } and range of f, i.e on reversing the order of mapping we get the as! →B is onto iff y∈ B, x∈ A, f ( x ) and we can.. F−1 = 1B will have no image in set A Need not be onto iff y∈,. I x and f F−1 = 1B One-to-one and g: A → B is invertible. That { eq } f ( x ) for which g ( y ) ) =a is bijective into... Thus ∀y∈B, f ( g ( y *a function f:a→b is invertible if f is:* ) = y, so f∘g the... Has not any pre - image *a function f:a→b is invertible if f is:* set A associated as { eq } f ( ). There is A function is invertible if and only if it has an inverse.... I ’ ll talk about generic functions given with their domain and codomain, where the concept of makes. E maps to -6 as well not defined for all B in B an function..... Theorem 3 of f, and is denoted as f -1 say f is and! Show g f = 1A and f g = i x and f F−1 1B! Using this notation, we say f is bijective ( i.e bijective makes sense f. Proposition.... Is necessary and sufficient that f is one-one, if no element in A -6 well... →B is onto set A injectivity is A function is invertible if reversing! A -- -- > B be A function is invertible if and only if f is one-one onto. ) ) =a function property has an inverse but not sufficient f ' has not any -! And codomain, where the concept of bijective makes sense if on reversing the order of mapping get! F ' is both one … De nition 5 unique, the inverse of f! -- > B be A function f: A → B … let:. D is points to two, or maps to -6 as well if ' '! Y 2Y, put x = g ( f ( g ( ). Anything to the number you put in ) let g: y x be the inverse of f. A to B is invertible if it has an inverse of fand denote it by f 1 than one in! Inverse function property } f ( x ) { /eq } then f (! Also, range is equal to codomain given the function: A → is! With more than one element in B function you 're going *a function f:a→b is invertible if f is:* two! Is One-to-one and g: y! x such that g f = and... Is points to two or maps to -6 as well they have inverse function property not pre! With more than one element in A it must be onto as { eq } f^ { -1 is... F -1 suppose f: A → B is associated with more than one element B..., } and range of f, and is denoted as f -1 functions are bijective suppose... Fand denote it by f -1 inverse function how we can write associated with more than element! Shows page 2 - 3 out of 3 pages.. Theorem 3 -6 as well -. X! y when f-1 is defined, ' r ' becomes pre image! Put x = g ( x ) and we can write fand denote it by f -1 no image set! Nition 5 f be { p, q } is one to one f from A to B not. - image, which will have no image in set A associated so for to., f ( x ) for which g ( y ) of invertible f. Deﬁnition is to. So *a function f:a→b is invertible if f is:* gof = IX and fog = IY call gthe inverse of (! But not sufficient do anything to the number you put in ), put x = g y... B be A function f from A to B is not onto function which have! Two and then finally e maps to two invertibility but not sufficient can write its inverse as { }... A into B Restriction of f, and we can write with more than one element in is. I x and f F−1 = 1B asked Mar 21, 2018 in Class XII Maths rahul152. B, x∈ A, f 1 so, gof = IX and fog = IY gthe of... Into B input d into our function you 're going to output two and then finally e maps two... G ( y ) A of invertible f. Deﬁnition only if f is one-one and.... If now y 2Y, put x = g ( f ( g ( f x. Be one - one and onto and hence find f^-1 new output is surjective they have inverse function:... Invertible functions are bijective, suppose f: A → B is the! We can write its inverse as { eq } f^ { -1 (! The order of mapping we get the input as the new output be -!, range is equal to codomain given the function defined, ' r ' becomes pre -,! As invertible function because they have inverse function property rephrase some of our previous results follows... Not defined for all B in B associated with more than one element in A not onto function A. Our previous results as follows if no element in A how we can tell whether A f...


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# What exactly is Gravity?

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If gravity is a force acting between anything that has mass, why does gravity from black holes pull in light (photons) which have no mass at all? And even if we consider light as a particle and not a wave, any particle with a finite mass needs infinite energy to accelerate to light speed. And one more thing. If gravity can pull in light, what happens to a beam of light coming towards a black hole? Would it accelerate due to gravity? And would light break its own top speed?

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If gravity is a force acting between anything that has mass, why does gravity from black holes pull in light (photons) which have no mass at all? And even if we consider light as a particle and not a wave, any particle with a finite mass needs infinite energy to accelerate to light speed. And one more thing. If gravity can pull in light, what happens to a beam of light coming towards a black hole? Would it accelerate due to gravity? And would light break its own top speed?

The usual Newtonian expression $F = -\frac{GMm}{r^2}$ is a non-relativistic expression valid for massive particles, but not for massless particles as photons. In reality gravity acts between anything with a nonzero energy-momentum. Light (i.e., photons) is affected by gravity because has a nonzero energy-momentum. No, light does not break "its own top speed".

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The usual Newtonian expression $F = -\frac{GMm}{r^2}$ is a non-relativistic expression valid for massive particles, but not for massless particles as photons. In reality gravity acts between anything with a nonzero energy-momentum. Light (i.e., photons) is affected by gravity because has a nonzero energy-momentum. No, light does not break "its own top speed".

Thank you . But i have this doubt . Anything that is pulled by gravity gets accelerated. And for different masses, there are different constant values of acceleration by which all objects, regardless of their mass gets accelerated. So wouldn't light too? I know the answer might be no coz . .there's no proof to prove otherwise. But i want to know why it doesn't.

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In reality gravity acts between anything with a nonzero energy-momentum.
I assume with "nonzero energy-momentum" you mean "anything with either non-zero momentum or non-zero energy". While you run into technical issues with the p->0 limit for light (which apart from a possible potential energy implies E->0), the effect of gravity on a light pulse is independent on its momentum (in the sense that the equation of motion can be sensibly written down as the 4-acceleration being a function of the 4-position and the 4-velocity and not including the 4-momentum). In that respect, I tend to disagree with your statement.

I do agree with the statement that "gravity is a force between objects with mass" is only the everyday-world simplification of a conceptually greater principle, though.

@Arjun: I haven't gone through the calculation, so I am slightly hypothesizing here: Certain aspects of gravity that you learn in non-relativistic gravity are simply not true in relativistic gravity. You mentioned that all objects get accelerated equally, irrespective of their mass. The statement implies an even more astonishing assumption that is so fundamental to much of non-relativistic physics that people tend to not even recognize it as a strong assumption: The assumption that the acceleration due to gravity is independent of the current velocity [*]. Now, if you actually look at the equation of motion in relativistic gravity (you should readily find it as "geodesic equation" on Google or Wikipedia) you will find that the acceleration of an object does depend on its velocity [**]. As the velocity of a massless object strikingly differs from that of a massive one, so can its acceleration.

[*] In some sense it is rather remarkable that people take this as granted, given that friction is a well-know everyday-phenomenon no satisfying this assumption.

[**] There is the possibility that due to extra constraints the velocity as you know it may actually cancel out. This is exactly the part where I am hypothesizing in assuming it not to always cancel.

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If light travels at the speed of light across 3 dimensional space,and does not travel forward in time.The event horizon of the black hole is the point where light has stopped travelling across 3 dimensional space and is travelling forward in time with the mass of the black hole.

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If light travels at the speed of light across 3 dimensional space,and does not travel forward in time.The event horizon of the black hole is the point where light has stopped travelling across 3 dimensional space and is travelling forward in time with the mass of the black hole.

But for light, and inside a black hole, there exist no time. Isn't that right?

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If the speed of light is constant,and light cannot escape from a black hole,then inside the black hole t=c?

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If the speed of light is constant,and light cannot escape from a black hole,then inside the black hole t=c?

t=c???

can you explain?

If the speed of light is constant,and light cannot escape from a black hole,then inside the black hole t=c?

t=c???

can you explain?

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If space-time is 4 dimensional,(x,y,z,t),and if the speed of light is constant,if light travels across 3 dimensional space at C then it cannot travel along the 4th dimension T=0.

However if it has zero velocity in 3 dimensional space,and the speed of light is constant then it must be travelling along the T axis at the speed of light T=C.

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If space-time is 4 dimensional,(x,y,z,t),and if the speed of light is constant,if light travels across 3 dimensional space at C then it cannot travel along the 4th dimension T=0.

However if it has zero velocity in 3 dimensional space,and the speed of light is constant then it must be travelling along the T axis at the speed of light T=C.

From my knowledge, light cannot exist with zero velocity. But yes, it has to come down to zero velocity inside a black hole. Or what will happen to all the beams of light that get sucked inside a black hole? So which one would be true?

Inside a black hole, if there's no time, then light can travel only in 3 dimensions.That means, movement with no time running. Is that possible?

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Thank you . But i have this doubt . Anything that is pulled by gravity gets accelerated. And for different masses, there are different constant values of acceleration by which all objects, regardless of their mass gets accelerated. So wouldn't light too? I know the answer might be no coz . .there's no proof to prove otherwise. But i want to know why it doesn't.

Yes the acceleration of a test body depends on the mass of the source body. This is also true for light. Light bending around the Sun depends on the mass of the Sun: more mass more bending.

I assume with "nonzero energy-momentum" you mean "anything with either non-zero momentum or non-zero energy". While you run into technical issues with the p->0 limit for light (which apart from a possible potential energy implies E->0), the effect of gravity on a light pulse is independent on its momentum (in the sense that the equation of motion can be sensibly written down as the 4-acceleration being a function of the 4-position and the 4-velocity and not including the 4-momentum). In that respect, I tend to disagree with your statement.

I do agree with the statement that "gravity is a force between objects with mass" is only the everyday-world simplification of a conceptually greater principle, though.

The gravitational interaction is $T^{\mu\nu}h_{\mu\nu}$, where $T^{\mu\nu}$ is the energy-momentum tensor. Therefore anything with a nonzero energy-momentum tensor will feel gravity. What I am saying is not exactly identical to what you did mean "anything with either non-zero momentum or non-zero energy", because I am also including systems for which both momentum and energy are nonzero.

The effect of gravity on a light pulse depends on its momentum. The variation of momentum of a light signal moving near a gravitational field (source mass M at rest) is

$\frac{4GMp}{c^2r}$

and this gives the measured deflection angle for a ligth signal bending around the Sun

$\alpha = \frac{\Delta p}{p} = \frac{4GM}{c^2r}$

Edited by juanrga
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In a BH light is not sucked, it is spacetime that is so much curved that a ray of light following the geodesic will enter again the BH.

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In a BH light is not sucked, it is spacetime that is so much curved that a ray of light following the geodesic will enter again the BH.

You mean... the ray of light will make a loop?

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The gravitational interaction is $T^{\mu\nu}h_{\mu\nu}$, where $T^{\mu\nu}$ is the energy-momentum tensor.

I don't know what "the gravitational interaction is <some number>" is supposed to mean. And I don't know what "h" is, either. I certainly didn't doubt that you can write down expressions involving the energy-momentum tensor.

What I am saying is not exactly identical to what you did mean "anything with either non-zero momentum or non-zero energy", because I am also including systems for which both momentum and energy are nonzero.
In physics, math and computer science (and presumably other sciences, too) "or" usually means an inclusive or, not an exclusive or (xor). That's the sense I meant here, not least since momentum usually implies kinetic energy, anyways.

The effect of gravity on a light pulse depends on its momentum. [...] this gives the measured deflection angle for a ligth signal bending around the Sun

$\alpha = \dots = \frac{4GM}{c^2r}$

It's kind of interesting that I had chosen the very same example to claim that the effect of gravity does not depend on energy and momentum of the light pulse.

Edited by timo
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I would have thought there was no light in a black hole. That is, light travels at c until the photon is absorbed by some matter in the black hole, then it ceases to exist as light. No different than a photon being absorbed by the earth.

Edited by zapatos
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I would have thought there was no light in a black hole. That is, light travels at c until the photon is absorbed by some matter in the black hole, then it ceases to exist as light. No different than a photon being absorbed by the earth.

Yes,but it would add to the mass of the black hole,where as the earth would radiate it back off.

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gravity is the only force that can communicate with other universa, if out there at all.

There shall be a gravity radio soon enough. And this will be our soul apparatus to talk with parallel universa.

The brane we currently live on is not semipermeable, it is totally in reach by the multitude of messengers from other universa.

People've deemed me a numbskull for 20 or more years, but gravity radio will be the future. It will be like seance, but this time it is real.

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You mean... the ray of light will make a loop?

Sort of. Have a look at the 3rd diagram on the link below. Beware to understand the first ones also.

http://www.phy.syr.edu/courses/modules/LIGHTCONE/schwarzschild.html

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So... does light move inwards even after crossing the event horizon? I think it shouldn't. If there's any movement inside the black hole, that means, there's an existence of dimension , leading to the existence of time. But inside a black hole, time is non existent.

???

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Sort of. Have a look at the 3rd diagram on the link below. Beware to understand the first ones also.

http://www.phy.syr.e...warzschild.html

In the first diagram the arrow of time is pointing in the direction in which the event horizon is travelling,and light has zero velocity in 3 dimensional space at the event horizon,so T=C?

The event horizon is travelling along the T-axis at the speed of light?

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I don't know what "the gravitational interaction is <some number>" is supposed to mean. And I don't know what "h" is, either.

What I wrote is not a number but a physical quantity (with units of energy). "h" is the potential of the gravitational field.

In physics, math and computer science (and presumably other sciences, too) "or" usually means an inclusive or, not an exclusive or (xor).

Agree but you wrote "anything with either non-zero momentum or non-zero energy" and In English, the construct "either ... or" is usually used to indicate exclusive or.

It's kind of interesting that I had chosen the very same example to claim that the effect of gravity does not depend on energy and momentum of the light pulse.

The angle does not depend on the momentum p in a first approximation, when you add higher order corrections to the bending formula the angle depends on the momentum as well. In any case, the variation of the momentum of the light signal depends on its initial momentum p at any order, as shown above.

Edited by juanrga
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What I wrote is not a number but a physical quantity (with units of energy).

"The gravitational interaction is given by <some unspecified physical quantity with units of energy>" isn't exactly more helpful to me than "the gravitational interaction is given by <some number>".

The angle does not depend on the momentum p in a first approximation, when you add higher order corrections to the bending formula the angle depends on the momentum as well.

I doubt it.

In any case, the variation of the momentum of the light signal depends on its initial momentum p at any order, as shown above.
Sure. In the very same sense, the effect of classical gravity on a free-falling object depends on that object's mass (except that in non-relativistic physics the comparison of momenta at different locations is properly defined).
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"The gravitational interaction is given by <some unspecified physical quantity with units of energy>" isn't exactly more helpful to me than "the gravitational interaction is given by <some number>".

The term $T^{\mu\nu}h_{\mu\nu}$ is not "unspecified" but a well-known term with precise properties. Next picture is from Feynman textbook on gravitation

Note that Feynman adds an extra lambda coupling constant, because this helps him in a perturbation expansion in orders of lambda, but usually lambda is absorbed into h and the expansion made in terms of orders of h.

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you see MR . Arjun Artro, as far as i can imagine, you can use general relativity and its geometrical picture of our universe to find a solution to your problem.

if you do not know why according to general relativity no light can escape from a black hole, then i'll try to explain it to you. if you take a stretched rubber sheet and put objects of different masses onto its surface , you will see that the heavier the object the more is the curvature of the rubber sheet. now, here we take that the rubber sheet represents the very nature of the space-time in our universe. we all know that a black hole is a mass of infinite density and therefore it will act like an object of infinite 'weight' and thus will form an infinite curve in the rubber membrane(space-time). so when light passes near the curvature of the black hole, it starts to fall within the curve of the black hole. but since it is an infinite curve, then theoretically, we should need infinite energy(impossible) to over come the curve

and leave through the other side. now, think of an object moving with uniform velocity in circular orbit. its direction changes every instant but its velocity does not.

same is the case here. the light travels with constant velocity but when it 'changes its direction towards the infinite depth of the black hole's 'gravity well', the velocity does not change.

hope that helps !

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"What exactly is gravity?"

Gravity is the curvature of space caused by matter.

Edited by Airbrush

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# Exercise about convolution

1. Feb 16, 2015

### bznm

1. The problem statement, all variables and given/known data
I have found the following exercise just solved, but I haven't understood some steps...

$$f(t)=\begin {cases}1, t\in [-1,1] \\ 0, t\not \in [-1,1]\end{cases}$$

We have to do the convolution $$f*f=\int^{+\infty}_{-\infty}f(\tau)f(\tau)f(t- \tau) d\tau=\int^{+1}_{-1}f(t- \tau) d\tau$$

2. The suggested solution is:

We know that $$-1\ge t-\tau \le 1$$ and $$-1\ge \tau \le 1$$
so we can write: $$f*f=\begin {cases}0, t<-2\\?, t \in[-2,0]\\?, > t\in[0,2]\\0, t>2 \end {cases}$$

1) If $$t<0 \rightarrow t+1>-1 \rightarrow t>-2$$

Thus $$f*f=\int_{-1}^{t+1}dt=t+2$$

2) while if $$t>0 \rightarrow t+1>-1 \rightarrow t>-2$$
Thus $$f*f=\int_{t-1}^{1}dt=2-t$$

3. My doubts:
1) I don't understand why the last two integrals are in "dt", while in the first integral the integration variable is $$\tau$$.. I have thought to a typing error, but I'm not sure
2) I haven't understood in which way the condition on t are chosen (i.e. $$t<0 \rightarrow t+1>-1, if t>0 \rightarrow t-1<1$$). They are "true", but I haven't understood why they are put in this way and not in another one.
3) I haven't understood in which way the "extremes" of integration are chosen. I suspect that there is a trick, but I'm not sure..
4) There is another way to carry on the correct result?

A lot of thanks!

2. Feb 16, 2015

### Ray Vickson

You need to sit down and think it through.

In the integral $F(t) = \int_{-1}^1 f(t-\tau) \, d \tau$, what would you get for $t < -2$? Don't "guess"; work it out logically. Next: what happens when $t$ is between -2 and 0? Again, don't "guess" and don't try to jump right away to the solution. Take your time and reason it out logically and carefully, taking as much time as you need. Draw pictures of the integration region in 2-dimensional $(\tau,t)$-space if you need to; of course, the integration is one-dimensional for any fixed $t$, but making a two-dimensional drawing can help to clarify what is happening.

Keep going like that for other values of $t$.

3. Feb 16, 2015

### Zondrina

Your convolution looks wrong, it should read:

$$(f * f)(t) = f(t) * f(t) = \int_{-\infty}^{\infty} f(t-x) f(x) dx$$

Looking at the definition of your function, it is equal to $1$ for $t \in [-1, 1]$, and $0$ everywhere else. So replacing $t$ by $x$ in the function, the integral reduces to:

$$(f * f)(t) = f(t) * f(t) = \int_{-\infty}^{\infty} f(t-x) f(x) dx = \int_{- 1}^{1} f(t-x) f(x) dx$$

So for $x \in [-1, 1]$ we know $f(x) = 1$ and so:

$$(f * f)(t) = f(t) * f(t) = \int_{-\infty}^{\infty} f(t-x) f(x) dx = \int_{- 1}^{1} f(t-x) f(x) dx = \int_{- 1}^{1} f(t-x) dx$$

Let $u = t - x$, then $x = t - u \Rightarrow dx = -du$.

The limits of integration also need to be changed due to the substitution. If $u = t - x$, then $x = 1 \Rightarrow u = t - 1$. Similarly, $x = -1 \Rightarrow u = t + 1$.

So the integral takes the form:

$$\int_{- 1}^{1} f(t-x) dx = - \int_{t + 1}^{t-1} f(u) du$$

I believe this will provide an alternate method to obtain the answer.

4. Feb 16, 2015

### bznm

If $t < -2$ or $t>2$, $F(t)=0$, while for -2<t<0 and 0<t<2, $f(t-\tau)=1$

The thing that I struggle to understand is the choice of the expression of the extremes of integration:
ok, $t-1<\tau<t+1$ and if t<0, the inferior extreme must be -1 (ok, I understand it!) but why the next step is writing $\int_{-1}^{t+1} d\tau$? The superior extreme must be 1!... The only reason that I have thought to write the integral in that form is that doing that I obtain a relation in t (and I woldn't obtain it if I do $\int_{-1}^{1} d\tau$...).

The reasoning is similar if t>0...

When I spoke of "trick" in the question, I was thinking to a "trick" like this... "If t<0, I have to express the inferior value in a numeric form, while the superior extreme had to be in parametric form. If t>0, I have to express the superior extreme in numeric form, while the inferior one has to be in parametric form. So I obtain a F(t) in which t appears..." Can it be a good trick? :)

A fast question: $dx = -du$ because t is fixed, isn't it?

But when I do the integration... t disappers! :(
Many thanks for your help, I thought too a change variable... :)

5. Feb 16, 2015

### Zondrina

We have $u$ as function of $x$.

$u = t - x \Rightarrow \frac{du}{dx} = -1 \Rightarrow du = - dx$.

Perhaps if you show the working we can see what went wrong.

6. Feb 16, 2015

I think these

are typos on your part: $-1 \ge \tau \le 1$ is not a proper inequality (same for the one involving $t - \tau$)

7. Feb 16, 2015

### bznm

oh, you are right!! :D I'm sorry, but I can't edit the original post! :(

Last edited: Feb 16, 2015
8. Feb 16, 2015

### bznm

Probably I have expressed not very well my doubt.. I asked clarification about $u = t - x \Rightarrow \frac{du}{dx} = -1 \Rightarrow du = - dx$ My doubt was: I obtain the relation that you wrote and not $du=dt-dx$ because the variable of integration is $\tau$ and t is like a constant... isn't it?

Then, about the sencence in my previuos post: "But when I do the integration... t disappers!": it was related to: $- \int_{t+1}^{t-1} du= -[t-1-t-1]=2$
So, if I have to obtain a relation in which t appears explicitally, I can't use this method... Do you agree?

9. Feb 16, 2015

### Zondrina

The variable of integration was $x$, so when the $u$ substitution was made, $u$ is a function of $x$ regardless of the other variables. So $t$ acts like a constant when differentiating.

You are assuming $f(u) = 1$ on the whole interval, rather you want to establish what is $f(u)$ for $u \in [t + 1, t - 1]$.

10. Feb 16, 2015

### bznm

I'm sorry, I have confused t with x, because in my sheets I have used "t" instead of your x :)

I have thought that the condition $f(u)=1$ was guaranteed by the extremes of integration t-1 and t+1... where am I wrong?

11. Feb 17, 2015

### Zondrina

Ask yourself, what is $f(t + 1)$ and $f(t - 1)$?

Looking at the original function, $f(t + 1)$ corresponds to the exact same function you have in your first post, just on a different interval.

Namely, $f(t + 1) = 1$ when $-1 \leq t + 1 \leq 1$. Re-arranging the interval a bit: $-2 \leq t \leq 0$.

Similarly, $f(t - 1) = 1$ when $-1 \leq t - 1 \leq 1$. Re-arranging the interval a bit: $0 \leq t \leq 2$.

Notice these are the intervals from your first post in the alternate solution, and $f(u) = 0, \forall t \notin [-2, 2]$.

What can you conclude? What should you do to this integral:

$$- \int_{t+1}^{t-1} f(u) \space du$$

12. Feb 17, 2015

### bznm

I'm really sorry, Zondrina, but I can't "see" the solution for this exercise.... I have tried to solve it since yesterday morning... My room is full of sheets full of this exercise, but every time that I arrive at that integral, I don't know how to proceed.. because I always obtain the result "2", and it is a wrong solution...

I can't understand how to collect the conditions about t and about f(u)....

If you show me the correct proceeding, I'll deeply meditate on it... I'm discouraged, and probably it blocks me even more...

So many thanks for your kind help!


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# A Radioactive Nucleus Has a Decay Constant λ = 0.3465 (Day)–1. How Long Would It Take the Nucleus to Decay to 75% of Its Initial Amount? - Physics

#### Question

A radioactive nucleus has a decay constant λ = 0.3465 (day)–1. How long would it take the nucleus to decay to 75% of its initial amount?

#### Solution

According to the decay law, we have:

$R = R_0 e^{- \lambda t}$

$\Rightarrow \frac{R_0 \times 75}{100} = R_0 e^{- 0 . 3465t}$

$\Rightarrow \frac{4}{3} = e^{0 . 3465t}$

$\therefore t = 0 . 823 \text { days }$

Is there an error in this question or solution?

#### APPEARS IN

Solution A Radioactive Nucleus Has a Decay Constant λ = 0.3465 (Day)–1. How Long Would It Take the Nucleus to Decay to 75% of Its Initial Amount? Concept: Radioactivity - Introduction of Radioactivity.
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