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# Concavity and Curve Sketching (Graphing Functions Using Derivatives)
Video Series: Calculus 1 (Explore, Discover, Learn) Series
(D4M) — Here is the video transcript for this video.
00:00
[Music] what’s the difference between the functions
taking the square root of a number or squaring a number
both of these functions are increasing but one of them is
increasing much faster than the other comparing the growth
of functions is the idea behind concavity
and is foundational in the subjects like computer science let’s get started
[Music] alright everyone welcome back to the calculus one
00:01
explore discover learn series uh this episode is inflection points and concavity
curve sketching and we are going to uh talk about
uh concavity test in this uh episode and we’re going to talk about inflection
points um and then we’ll talk about uh vertical tangents and cusps
and once we get those two items out of the way then we’ll be able to talk about
curve sketching where we apply and combine everything that we’ve talked about
first derivative test concavity test that we’re learning today
vertical tangents because and also of course horizontal and vertical tangent uh
horizontal and vertical isotopes and so the previous uh episodes are very
important to to watch first so i have a playlist in the description below
where i uh have video an episode over um isotopes and
the first derivative test and the extreme value theorem
so those videos or are of course uh useful
00:02
to watch before this one right here but the main topic of this one is the
concavity test and that’s we’re going to cover first let’s go
okay so up first is the concavity test of course to understand the concavity
test what is concavity and why is it important so first thing is
we’re going to need a number in our domain
so we’re going to say c is in the domain of a function
we have a function given and we’re interested in a number
and we’re going to call that number c a first order critical number
if the derivative is zero or the derivative does not exist
now that should look familiar to you because that’s just what we’ve been
calling a critical number we talked about critical numbers um i
think in the episode extreme value theorem first
and so you know make sure and understand what critical number is
00:03
but the reason why we’re calling them now instead of just critical numbers
now we’re also calling them second order critical numbers
um because we need to have different uh derivative tests one will be a
concavity test and one will be the first derivative test
so for the first derivative test we use the
first order critical numbers and if you look back at that episode we just call
them critical numbers but now we’re going to have a concavity
test in the concavity test is going to use the second order
critical numbers i want to distinguish between
the two types of critical numbers the first order in the second order
the second order is the same as the first order critical numbers
it’s just the only difference is we’re looking at the second derivative now
if the second derivative is zero or if the second derivative doesn’t exist
then they’re called critical number second order critical numbers
of course you always start off with a number in the domain of the function
00:04
all right so um what does concavity mean well suppose we have a function that’s
differentiable and we’re looking at the um first
derivative here if the first derivative is increasing then we call the function
concave upward on that interval if the first derivative is decreasing
then we call the graph of f concave downward on that interval
and so let’s see what that means let’s see what that means
so if the graph of f lies above all of its tangents
on an interval then it is concave upward on i
on the enter on that interval if the graph of f
lies below of its tangent is concave downward on that interval and
if you have a change in concavity if if if at a number c to the left of it
00:05
to the left of c it’s and then to the right of c it’s a different concavity
then we say we have a change in concavity so for example to the left of c
might be concave upward but then to the right of c
the graph might be concave downward and then we would have an inflection point
because it would be a change in concavity
so let’s look at here this graph here so you can see what i mean
you know when we have this part right here from a to b
here so i’m looking on the interval right here a to b
and i’m looking at the shape of the graph right here
so this is an example of a concave downward graph
and the reason why is because if you pick a point any point
and you were to draw the tangent line the tangent line would be above the graph
for example if i pick the point right here pick a point right here along the
x-axis pick a number and they come up here to the graph and
draw the tangent line the tangent line would be above the graph
00:06
now from b to c on this interval the graph is now concave upward looks
like the shape of a bowl so if i pick a point between b and c say one right here
and i go up and i draw the tangent line the tangent line is now below the graph
so what about c to d is it concave downward or is it concave upward
so if i come over here and i can see that’s concave downward
and here it’s concave upward [Music] and here it’s concave upward and here’s
concave downward notice there’s a inflection point at p
because to the left of p we’re concave upward
and to the right of p we’re concave downward so there’s a change in
concavity at p notice that b is also a inflection point because to the left
00:07
where concave down and to the right were concave up and same thing
for point c here we’re concave up and then to the right of point c here
now we’re concave down so there’s a change in concavity at c
so c point c is a an inflection point so you know notice the definition of an
inflection point says the we have an inflection point
provided f has a tangent line at c so notice that at d here there is a
sharp corner and so there’s no derivative at d um so right here also
at this there’s no inflection point at e where concave down we’re concave down
there’s no inflection point at this number here e
okay so if the graph lies above all this tangent on an interval is
00:08
concave upward if the graph lie lies below all this tangent it’s concave
downward and we have the definition of here of an inflection point
so here’s what the concavity test is because the concavity test is going to
allow us to break the number line the horizontal axis
into pieces into a finite number of pieces
and we’ll be able to test each of these pieces each of these intervals
and determine whether it’s concave up or concave down
so to use a concavity test though you need to know that your function is twice
differentiable in other words the derivative has a derivative
now if you have that then what you can say is that
if the second derivative is positive then the graph of f is concave upward
so if you know the second derivative exists on the interval and it’s positive
00:09
then the graph has to be concave upward if the second derivative is negative
on an interval then the graph is concave downward
and so let’s get started with an example so you can see
all this put together so we want to find all the inflection points
all the local extrema and we want to sketch the curve
so here we go with our first example here
so when i’m looking at this example here we need to take the first and second
derivative test now it says to find local extrema right
there right so that’s the first derivative test
we can use the first derivative test on that and to find the inflection points
that’s the concavity test which we just talked about and we’re going to now
understand better so here we go so the first derivative three x square
00:10
and then minus three and we can factor out three and we can factor all this
x minus one x plus one and i’m gonna set it equal to zero and so
x equals plus or minus one um before we would say these are critical
numbers now let’s say these are first order critical numbers
just to make sure i distinguish them between the second order critical
numbers i’m about to find so first order critical numbers
and so we could go apply the first derivative test now
um let’s go ahead and take our second derivative now when i’m taking the
second derivative i’m not looking here i’m going to look right here
so this will be 6x and that’s it and so i’m going to set this equal to 0
and solve and so x equals 0 is the second order critical number [Music]
00:11
second order critical number now make sure that you always keep in mind
the domain of the original function we have a polynomial
it is defined everywhere is continuous everywhere
is differentiable everywhere so we have our first order critical numbers
and he says and this is the second order critical number here
so in order to find the local extrema and inflection points
i’m going to find i’m going to work out the first derivative test and the
concavity test so first i’ll do the first derivative test
so i’m looking at the uh first order critical numbers here
so i’m going to look less than minus one between minus one and one
and then and then let’s go ahead and put equal to minus one
and then let’s put between and then equal to one and then greater than one
and so these are the first order critical numbers minus one and one
00:12
right so i broke up the number line i’m less than i’m equal i’m between
i’m equal or greater than and so here i’m gonna have a column for f
and the derivative and i’m gonna make some conclusions
right so we covered this in detail in the uh episode that derivative test
monotonic and derivative test all right so i’m going to be testing
this this right up here and if i choose a number less than minus
one for example minus two then what’s happening over here we’re
gonna get a negative and a negative what’s a negative times a negative we’re
gonna get a positive [Music] and let’s test the number in here for example zero
we test zero into our first derivative so we’re going to get a negative times a
positive so we’re going to get this whole thing as negative
now i’m going to test the number greater than one for example
two just test the two here positive positive so the whole thing’s positive
00:13
all right so now we know that this is increasing we know this is increasing
we know we’re decreasing right here [Music]
all right so what’s happening at these points here do we have any relative
extrema so we’re changing from increasing to decreasing
so at -1 we have a change in monotonicity we’re changing from
increasing to decreasing so think of it like this we’re
increasing and then we start to decrease right so that’s going to be a relative
max so at minus one we have a relative max
now we’re we have a change also a change in ma
monotonicity so we’re decreasing and then we start to increase that’s going
to be a relative min here so at at one we have a relative min
now because we have to go and sketch the graph here
00:14
we probably should just go ahead and plug in these points here what’s happening
at minus one what’s happening at one so at minus one
we’re having minus one to the third minus three times minus one and then
plus one and so that’s what minus one uh plus three plus one so that is three so
at minus one we have a three and then what’s happening at a one
we have one to the third minus three times one plus one and so that’s one
and one that’s two so minus three plus two so that’s a minus one
all right so we have a relative max the value is three
and occurs at minus one we have a relative min
of minus one and it occurs at x equals one
and so there’s our first derivative test it gives us our real
00:15
it gives us our local extrema so you can put local max here or local men
it says local extrema here so maybe i should use the word local but
to me they’re interchangeable okay so now let’s go do
the concavity test so um we only have one second order critical number so it’s
going to be a lot easier easier to do the concavity test now when
you have multiple tests maybe you might want to put here in this
box here first derivative test so now when i go make the concavity test
i’ll have a ct here for concavity test and we’re looking less than zero equal
to zero and greater than zero because those are the only choices x
equals zeros are only second order critical number
of the concavity test here i’m gonna look at the function
in the second derivative and i’m gonna make a conclusion here
all right so that says conclusion there all right so
00:16
what’s happening to the second derivative i’m going to test right here
at minus six so i’m going to choose something less
than zero for example minus one and i get a minus six that’s negative so
the second derivative is negative that means concave down so concave down
and some people draw you know this right here showing it’s concave down
like i draw the arrows here some people accept only this some people accept only
that just depends upon who you’re communicating with let’s look greater than one
let’s say a two we tried to the second derivative is positive without concave up
and so do we have a change in concavity and the answer is yes on the left side
of zero or concave down on the right side we’re concave up so
00:17
this right here is an inflection point so just like here we had to change in
monotonicity so we had a relative extrema
here we have a change in concavity so we have an inflection point
now what is that point the inflection point happens at x equals zero but what
is the actual output value so if we go but look back at the
original function it’s at one okay so now if we look at all this
information here we should be able to sketch the graph
so let’s see here i’m going to try to sketch it right here
and see if you guys can see all of this here we need to take into account the
first derivative test so i need to break it up between minus
one and one something’s happening at those points right there
so i’m gonna stay on minus one is here and a one is here
and i’m gonna be increasing and then i’m gonna reach that height of three and
00:18
then i’m gonna be decreasing so i’m not gonna pay attention to this
yet i’m going to just be increasing to minus one
i’m gonna hit a height of three at minus one
so let’s say minus one is here i have a height of three here
so i’m going to be increasing to there and then so here’s minus one
here’s minus one three that’s the point i have right here
so i’m going to be increasing to minus 1 3 that’s where i have my relative max
right here so now i’m going to be decreasing between
-1 and 1 and where do i cross the y-axis
that’s this point right here zero one so let’s put a one here
so i’m going to be increasing make sure you don’t have a sharp corner there
because it’s not because the derivative is
exists so it’s nice and smooth and then i’m going to be
00:19
at one here we have a relative min so that’s at minus one so
let’s say a minus one is here so i need it to be nice and smooth through there
and then after one so nice and smooth through here
we don’t want to make a sharp corner there now at one
greater than one then we start to increase we know this doesn’t come back
down because it’s going to be increasing and we know this only gets here it’s
increasing here and then it’s decreasing here
and it hits somewhere in here we don’t know exactly where yet
alright so this is increasing and then decreasing and then increasing
and i have my relative max and my relative min and that all comes from the
first derivative test now let’s see if i sketched it right
according to the concavity test because this is increasing here but
there’s different ways to increase wait wait a minute what do you mean
00:20
there’s a different way of increasing well just as a quick aside example here
think about these two functions here i was like to have these two functions in
my mind when i’m thinking about concavity and
and the expression there are different ways of increasing right
when you have x squared it’s just a parabola going through here
when you have y equals square root of x it’s just going like this
so both of these functions are forever increasing when you’re greater than zero
it’s just increasing and this right here when you’re greater
than zero you’re just increasing but this one right here increases much
slower this one increases much faster so how do you how do you
like quantify that well concavity remember takes into account
the second derivative this is concave up here concave up is
going to increase much faster this is concave down right here and this
is going to increase much slower in any case back to our problem [Music]
00:21
when i’m less than zero i’m concave down so here’s zero right here did i draw a
concave down yes when i’m greater than zero
oh that says one when i’m greater than zero sorry if that confused anybody
when i’m greater than zero it’s going to be concave up
and so do i have it concave up yes and so right here at 0 1 is an inflection
point and that tells me exactly where i changed concavity
so this is a relative max relative min and that’s a change in concavity right
there so it looks like i’m concave up looks like i’m concave down
and if you don’t see it quite if you’re not happy with it too much
you know make it make it more obvious ah
make it more obvious you know don’t make it so straight it’s nice and smooth
and here this is you almost want to like draw it first
00:22
and then label everything this is one one minus one yeah so there there’s a
nice smooth curve and you can see the relative max and the relative extrema
but what’s new is you can see the inflection point right there at 0 1.
all right so there’s our first example there and we can see it now
so that’s what we came up with right there this is a nice smooth curve going
through those points and we have a minus one three is our relative max
we have one minus one is our relative min and we have our inflection point right
there as zero one all right let’s look at the next example
okay so determine where this curve is concave upward
where it is concave downward and what are the inflection points
00:23
so this one here we just need to do the concavity test let’s do the concavity
test here so to do the concavity test we need the second order critical numbers
so let’s first start off with the first derivative
um so let’s see here we’re going to get 4x to the third
and then minus 12x squared and we don’t really need to set this
equal to zero and solve it because we don’t need the first order critical
numbers if you’re looking for relative extrema
and doing the first derivative test of course but here i’m going to go on to
the second derivative so this will be 12x squared minus 24x
and to find the second order critical numbers i’m gonna look to see where this
is zero and where this is undefined now look at our original function
it’s a polynomial it’s continuous and differentiable everywhere
and the second order uh the second derivative is also continuous
and differentiable everywhere so i just need to find out where this is zero
00:24
so i’m gonna set this equal to zero and let’s solve so we can factor out an
x and get an x minus two and so we get x equals zero and x equals two
and there’s our [Music] second order critical numbers here so
let’s go and do the concavity test so i’m gonna look less than zero [Music]
equal to zero between zero and two equal to two and greater than two
so this will be my concavity test and i’m going to be testing the function f
and to do that i’m going to be using the second derivative and i’m going to make
some conclusions [Music] all right so let’s look less than zero
so when i’m testing my derivative i want to look right here
this is the easy one to test the second derivative is right here [Music]
00:25
so i usually like to underline it just so it’s easier to find but
there’s my second derivative right there
i’m going to test it’s nice and factored for us
i’m testing less than zero for example minus one that’ll be a negative and a
negative so multiply we get a positive so it’s concave up
uh between zero and two i test the number for example one
i get a positive times a negative so that’ll be negative so it’ll be concave
down now is there a change in concavity yes punk gave up concave down
so at zero we have an inflection point now you you can’t just say um
if it was concave up and concave up there would be no inflection point
or if it was concave down and then concave down there would be no
inflection point there has to be a change in concavity
all right so now let’s test greater than two for example a three
so i go to my second derivative in test three so here is positive and positive
00:26
multiplied together we get positive so it’s concave up we have another
change in concavity so we have another inflection point here
so here’s an example where we have two inflection points
we have an inflection point zero what happens when we substitute zero into our
graph we have zero what happens when we substitute in two
so two to the fourth minus four times eight all right so that’s what uh
16 minus 48 and what is that minus 32 okay so we have the inflection points here
in here and um we’ve determined where it’s concave up
it’s concave up less than zero and greater than two
and we determine where it’s concave downward that is between zero and two
00:27
and we determine what the inflection points are so we got everything labeled
here in this table and we found these these outputs here make sure you find
the outputs because you need the inflection point so you need an x and a y and
everything is here and so now we can go look at the graph
all right so there we go so we can see the concavity that is concave
up less than zero and that is concave down between zero and two
and then you can see that it’s concave up when you’re greater than two
so all that matches the the concavity test that we just saw all right very good
let’s go on to the next example uh now let’s look at vertical tangent
some customs all right so now i’m going to cover vertical tangents and cusps
00:28
but not not all calculus courses do that in fact if you look in some calculus
books some of them cover vertical tangents some of them don’t
i’m going to go ahead and put it in here so we have a complete um
understanding of of how to make really good sketches um at least
in terms of concavity and tangents and cusps and i actually find them quite fun
now when we’re looking at the derivative um there’s four possibilities i’m going
to show you so we’re going to assume we have a continuous function
we’re going to assume we have a continuous function and
we are going to talk about the four possibilities so
two of the possibilities we’re going to have a vertical tangent and the other
two we’re going to have a cusps so to find the vertical tangent
00:29
i’m going to look where the derivatives um on the left and on the right
are both matching and they’re both positive infinity
if that happens you have a vertical tangent at c
if the both the derivatives the limits as approaching left and right
if they both match again but this time they’re both minus infinity
then we also have a vertical tangent and so that’s two
of the possibilities the other possibility is
if they don’t match so we’re taking the limit of the derivative
from the right and if that’s positive infinity and then from the left is
minus infinity then we have something called a cusp
at c and then the last final possibility is
if the derivative the limit from the uh right of the derivative
is minus infinity and then from the other side
the derivative approaches positive infinity then we also have a cusp so notice
in all of those cases we are looking at the
00:30
limit of the derivative so in order to find out if you have a vertical tangent
or vertical or cusp you have to look at the limit of the derivative
and so let’s look at some examples so we can get a
good feeling for what this means here so here’s our first example here now
when i’m looking here at the uh curve here we have to ask ourselves
where do we start what do we what do we need so let’s go back for a second here
how do we find like like like let’s look at number one for a second
how do we know what the seas are how do we know where to look
how do we know when the derivative is going to be infinite
so we covered infinite limits um i think last episode we talked about
horizontal and vertical isotopes but we also talked about limits
involving infinity but in a way a way to get an infinite
00:31
limit whether it’s positive or negative infinity
is when you have a fixed number in the numerator over something that’s going to
zero and so we’ll look for that case we’ll look for those cases
but first of course we need to find our derivative alright so here we go
so our derivative is going to be all right so we’re going to use the
product rule here so i’m going to take the derivative of
the first piece so it’ll be nine fifths x to the three-fifths
minus one and then times the second one which is five minus x minus four x
squared plus now the first one so three x to the three fifths
times the derivative of the second one which is minus one minus eight x
00:32
and then times the derivative of the uh yeah that’s good all right there we go
yeah so um the first piece times the derivative of the second
and that’s it all right so now um in order to see
where what the seas are where i’m going to be taking my limits of the derivative
i need to get a common denominator so let’s put all this together
so this has a negative exponent here so and we have a 9 up here so let’s write
this as 9 times five minus x minus four x squared
all over five and then we have x to the two fifths power plus
and now all of this is in the numerator um so let’s just write it down again
and right so i just i didn’t change anything just write it down again
think of this as being over one but we need a common denominator
00:33
so i need to multiply by five x to the two fifths over five x to the two fifths
and so now we have a common denominator which is five x to the two fifths
and let’s see what’s happening in the numerator so here we’re getting 45
minus nine x minus 36 x squared 36 x squared
all right and what’s happening over here here we’re getting a
three times a five so let’s call this um 15 x um
so i’m just taking this times this part here
and i’m going to say 15 and then an x right because that’s three fifths
plus the two fifths that’s five fifths which is one
and i’m just going to leave this out here for right now [Music]
00:34
all right so i’ll distribute that next and let’s see if we get anything
interesting [Music] so let’s distribute that we’re going to get minus 15x
and then we’re going to get a minus a 15 times a which is obviously 120
x squared all over five x to the two fifths [Music] all right and that’s a 15.
[Music] okay so so far so good and the that can be simplified more but
if the x is zero here remember this is ours um first derivative here [Music]
00:35
and right so that’s the function is f so there’s our first derivative and
um we can simplify that but what c are we going to be looking for so remember
when we are looking at the um when we’re looking for vertical tangents
and cuts we need those the limit of the derivative to go to blow up to
positive or negative infinity so i need to find those c’s to look at
it has to be some c where the function is continuous so
when i’m looking here i’m looking at the c
equals zero here because the derivative here at at x equals zero is not defined
so it and when i plug in zero i’m gonna get all those
zeros there i’m gonna get 45 over zero so i know that if i take the limit as we
approach zero we’re going to get um some infinity so now i need to look at
00:36
the limit as we approach zero from the right
of the derivative and i need to look at the limit
as we approach zero from the left of the derivative
so i’m looking for vertical tangents or vertical cuts
and i look where the derivative is going to blow up we approach zero of this
derivative right here we’re approaching from the right so
what is this what is this limit going to be if i try to plug in 0
i’m going to get 45 over 0 right so i know it’s going to be some infinity
same same for both of these so here it’s very crucial that if you
don’t understand this you go back and look at the episode over
uh limits involving infinity the horizontal and vertical isotope
uh episode but as we approach zero from the right we’re going to get 45 over
zero the question is is it positive or is it negative infinity
00:37
so as you can see all these other terms over here don’t really matter
what matters is as you’re approaching zero from the right
this is getting extremely small these are getting really small
think of x for example is like point zero zero zero one
so that’s really close to zero these are all really close to zero
the 45 is dominating so it’s going to be positive
and if i choose a uh any any number right here whether it’s pot
to the right of zero to the left to zero that x and that’s going to be a squared
and then i take the fifth root this is always going to be positive out
here so both of these limits are positive infinity right here
and so what that means is that we’re gonna have a vertical tangent [Music]
at x equals zero we have a vertical tangent there [Music]
so that’s what these two limits give us right here and now if i try to sketch
00:38
the graph it’s gonna there’s gonna be a vertical tangent here
so when i’m approaching zero the function
the derivative the slope of the tangent line is increasing without bound so
something like this right here [Music] so if i draw a tangent line for example
right here the slope is positive but if i draw
another tangent line say right here the slope is still positive but it’s
it’s greater than the slope that it was right here the closer we get
to zero the tangent lines increase the tangent lines keep
increasing and so we have a vertical tangent there
the slopes of the tangents are just growing
unbounded as you get closer and closer zero in fact let me try to make it even
00:39
a little bit more obvious let’s see if i can do that
i mean it’s almost like a vertical line there i’m going to come in
and look really close [Music] so it’s still a function
but there’s a vertical tangent right there so if i start taking um slopes of
tangent lines that you see the slopes keep increasing
when you’re really close to zero the tangent line is almost vertical
here the tangent line is positive when i get a little closer the tangent line is
still positive alright so here’s what graph would look like
right there uh and you can see the vertical tangent at x equals zero
so there’s an example of a vertical tangent and we found it by looking where the
first derivative was going to blow up [Music]
00:40
so let’s look at another example now so an easy way to generate vertical
tangents and cusp is to be looking at fractional powers
and so here’s another one let’s look to see if we have a vertical tangent or
cusp on this one here so here we go let’s find the derivative and see where
the derivative blows up so here we go the derivative is all right so
derivative of the first one is two thirds x to the two thirds minus one
and times the second one [Music] plus now leave the first one alone
times the derivative of the second one which is two x
plus five all right so now because i have a negative exponent here
00:41
i know i need to get a common denominator here
so i’m gonna write this as a two x squared and then it
plus 10x and then a minus 40 and then in the denominator i have three
x to the one third and over here i just going to write down the same thing
one and i want to multiply and get a common denominator
so i’m gonna do three x to the one third and then three x to the one third
all right so now we have a common denominator
our denominator is 3x to the one-third and now let’s see what we get on top now
i see a 2x and a 10x let me go double check that
i think that should be 2x squared i forgot my squared here
so this will be 2x squared plus 10x minus 40.
00:42
and what’s happening over here we have x to the two thirds
times an x to the one-third so that’ll be x to the three-thirds which is just
an x so i’m going to get a 3x multiplying that part in that part
and then i’ll just leave this alone for for now
in fact there’s really no point in expanding this out because
you know this is our first derivative here
right that’s the first derivative and we need to look where this blows up
right we don’t need to go simplify this the the goal of this exercise isn’t to
find the first derivative it’s just to find the vertical tangents and cusps
oh in fact they actually tell us uh sorry just looked at the problem
the i’m asking why there is a cusp at zero all right so
pretend you didn’t know that where is the derivative blowing up
00:43
so if i look at zero makes the bottom zero the denominator zero
what about the numerator that’s going to zero that’s going to zero
minus forty and then because of that zero right there that’s all going to zero
so this looks like it’s going to minus 40 over 0
so that’s going to blow up so i’m going to look again at zero so i’m going to
look to the left of zero of the derivative make sure don’t put
the function there we’re looking at the behavior of the derivative [Music]
and then as we approach zero from the right [Music]
so because when we’re approaching zero from either side these are going to zero
and that’s going to zero so the top is going to
a fixed number and the denominator is going to zero
so from those two things i know that this has to be some infinity here it has
has to be some kind of blow up the question is is it
positive or negative infinity so i look at where i’m approaching i’m approaching
00:44
from the left that means the x’s are negative so
the top is going to be dominated by -40 that’s
really close to zero that’s really close to zero and that’s all really close to
zero and then minus forty right so the top is going to be negative
and then now what about the bottom approaching zero from the left means
we’re like negative point one so when i do that fractional exponent a negative
will still be a negative so i’m looking at negative over negative
looking at negative but now i’ll be looking at positive down
here so negative over positive will be negative
now because there’s a change here this satisfies the condition of a cusp
at x equals zero if there’s if there’s if one of those positive in one of those
negative infinity then there’s a cusp if they both match then that’s a
00:45
vertical tangent so now we see why we have a cusp at zero
and if we look at the graph we can see that so there’s a cusp
at x equals zero there um as you approach zero from the left the
slope of the tangent lines are going to infinity and if we approach
zero from the left the slope of the tangent lines are approaching minus infinity
so i’ll try to illustrate what i mean by that looking back here
so if we sketch the graph here it’s coming in like this and it’s going
back out like this and so what happens what is this right here mean
right so again the derivative is the slope of the tangent line
we’re approaching zero so we’re close to zero so let’s say we’re right here
00:46
right what’s the x value right there not too close to zero but it’s a little
bit the tangent line is right here so what’s that slope
now let’s look a little bit closer here now the slope of the tangent line is
is greater so as we move closer and closer the slopes of the tangent line
are getting greater and greater and greater so the slope of the tangent
lines are going to positive infinity now as we approach zero from the right
like take this number here it’s a little bit close to zero now what
does the slope of the tangent line look like it’s going down
it’s negative and if i move a little bit closer it’s still negative it’s a
greater negative if i keep moving closer and closer the negative
becomes more and more negative right there so you can see that one right there
is also holding so those are the tangent lines there
and that’s the behavior there and you can see that we have a cusp right here
[Music] all right very good so let’s look at one more
00:47
and this time i’m not telling you which one it is
we’ll have to we’ll have to work it out and find out together
so here we go we’re going to look at the first derivative
and see where the first derivative blows up if if any places at all all right so
our first derivative is so derivative of the first one so two thirds
minus one and then x minus one to the one third plus
now x to the two thirds and times the derivative of the second one
so derivative of the second one is one-third x to the minus one and then now my
one-third minus one and then times the derivative of x minus one
00:48
which is just one all right so on this one right here [Music]
we’re going to write this as two and then x minus one to the one third
over three and then we have x to the one third [Music]
plus now here we have x to the two thirds on top [Music]
three’s in the denominator and then x minus one to the two thirds
okay so here we need to get a common denominator we cannot look at these
individually and determine where the derivative blows up we have to
look at one numerator one denominator to get that kind of behavior
so what’s going to be our common denominator
i’ll show this step out i’ve seen this um not be the friendliest thing for
students they both have three so i’ll see this is x minus one
00:49
um and i’m gonna multiply x minus one over x minus one to the two thirds here
[Music] all right so multiply numerator and
denominator by x minus one to the two thirds because that’s what’s missing
here this one has it but this one doesn’t plus x to the two thirds
over three and then leave this one alone and then what are
we missing over here this one doesn’t have
the x to the one third so to multiply by x to the one third
over x to the one third all right so that could be your next step
if that’s not your next step that you’re
written then you’re probably visualizing that
whatever case may be what do we get um we get the common denominator right
three and then what’s our common denominator x to the one third
00:50
and then x minus one to the two thirds there’s our common denominator sorry
there’s our common denominator there and so what we end up with here x
x minus one to the three thirds so x minus one so we’re gonna get two times
x minus one plus and here we’re just gonna get an x and so
our derivative simplifies quite nicely it’s a two x plus x so that’s a three x
minus three x minus two over three and then x to the one third and
then x minus one to the two thirds all right so there’s our derivative
so now where are we going to look now we have two places to look
where x is zero because why x equals zero well it makes the
denominator zero and makes the numerator non-zero
00:51
so the derivative is going to blow up at zero same thing at one
when i try to use x equals one in here i get zero down here and i get a
non-zero up here so we’re going to have unbounded behavior
so let’s go look at these four limits we’re going to approach versus 0 right
here we’re going to look at the left and we’re gonna look at the right [Music]
make sure we have derivative here so what are these limits
so again we’re approaching zero so the numerator is going to go to minus 2
and the denominator is going to go to 0 so we know that these are some type of
infinity here so when i’m looking less than 0 say a negative 0.1
well i’m going to get a negative and then a negative
and then this will be positive because of that square right here this term
00:52
right here will always be a positive and so i’m going to use a negative 0.1
so i’ll get a negative over a negative so that’ll be positive [Music]
and when i look at 0 from the right of this derivative right here so say
like 0.1 so that negative 2 will still dominate so i’ll get negative
and then i’ll get a positive and this will be positive
so that’ll be negative over positive so this will be negative infinity right
there so we have a cusp at x equals zero now what about at the minus one also
so i need to look at these two limits also limit as we approach uh one from the
left of the derivative and the limit as we approach
one from the right of the derivative so what are these limits
derivative all right so one from the left so again this is always gonna be
00:53
positive because that’s square so when i’m looking at one from the left
like .9 well that’s positive and if i’m looking at one from the right like 1.1
that’s still positive so for either one of these
the denominator is positive and for either one of these
well let’s look at it let’s look at this one we’re approaching one from the left
like point nine that is going to be three that that’s really close to three
minus two that’s positive so i’m getting positive over positive
so i’m getting positive infinities for these
for both of these for example one point one
that’s a little bit greater than three but that’ll be
minus two that’ll still be positive so we’re gonna have a vertical tangent here
00:54
vertical tangent at x equals one [Music] so this is what we found for the
derivative and by looking at the derivative right here
we’re able to find these limits right here where there’s blow ups
and we’re able to find these uh limits here and we can determine if they’re cuts
because these limits don’t agree and these two limits agrees
is a vertical tangent here are very good let’s look at the graph
okay so then you can see we have two different
uh types of unbounded growth for the derivative
at zero you can see that we have a cusp and at one you can see that we have a
vertical tangent all right so let’s go on now next part
00:55
okay so now we’re going to talk about curve sketching
curve sketching so now we’re going to put everything together
that we’ve talked about and i’m going to put it all like in a checklist format
for us so we can see what we have accomplished so far [Music]
and putting it all together putting all together might seem boring
but at the same time if you’re new to all this
when you work it all out in one problem then you can kind of see how
they’re different from each other and how they
how it all works together to give you a really good
sketch so i think it’s really fun to do at least a couple of them all together
but after a while it gets you know kind of a long problem to do
but anyways if you’re given a function you know one of the first things you
want to look at is domain and range and of course you want to simplify the
00:56
function if possible um now i say algebraically simplify the
function if possible but you know also trigonometrically
so for example if you if you were given um hey go graph the function for us
sine x squared plus cosine squared x right graph that function for us
well hopefully you would know the trig identity that no matter what x is
that this is always one right so you’re just going to be graphing the horizontal
line y equals one and there’s really nothing to do there
so not only algebraically simplify but also
trigonometrically simplify or just flat out simplify the function um
but make sure that you don’t alter the function so for example
you don’t want to go overboard in your simplification
00:57
in the sense that you’re actually changing your function for example
look at these two functions here so these are two different functions this
is a function right here that looks like the line y equals one
and this is a function right here that looks like y equals one but in fact
they’re not both the same graph this one has a hole at one
uh sorry this one has a hole right here at zero
right so if someone says here’s your function right here
well i would not simplify it to this one because these functions have different
domains this function the domain is all real numbers this function is all real
numbers except 0. so you want to simplify your functions
but not alter your function all right in any case
step three determine where the function is continuous
so we have a large collection of continuous functions that we know
about for example polynomial and rational functions are
00:58
continuous on their domain and trigonometric functions are
continuous on their domain and exponential and logarithmic so and
then we know how to combine them together using addition subtraction
multiplication and composition so we know a lot about
continuous functions and and we have building blocks we know how to create
more continuous functions so having a good idea where your function
is continuous and not continuous is always the important part when you’re
sketching of course you know as you did in
precalculus spend some time looking at your intercepts
many cases you’ll be able to do that and make sure
that you’ve watched the episode over asymptotes make sure you know how to
make sure you understand how to find those using limits
because when you’re sketching the graph it’s the limits that are actually
00:59
uh very very important all right and so then the heavier uh techniques are
finding the critical numbers and using the first derivative test
finding the second order critical numbers and using the concavity test
and then what we just talked about a little bit ago vertical tangents any cusps
a lot import plot important points for example
any points that you might find that are extrema those are obviously important
any points that are tangent vertical tangents or cuz
intercepts you know plot those important points and then
finally sketch in between all that information
and make sure your sketch is reasonable in the sense that don’t put any sharp
corners where there shouldn’t be if you put a sharp corner that means
you’re telling the reader that the function is not differentiable
there so if you know the function is differentiable
01:00
make sure you make it nice and smooth so
if i was to sketch the graph of y equals x squared versus the graph of y equals
absolute value of x and i was just lazy this graph right here has a sharp corner
it’s not differentiable at zero and if i was to get lazy and just sketch that
graph and you know can you tell which one that is right so
that’s all i’m saying is make sure that you don’t put any unnecessary sharp
corners where there are not any all right so let’s begin that procedure
now and let’s look at this function right here
and so we’re going to sketch the graph here and i’m going to um
look at we’re going to look at this graph together here
and try to figure out what’s going on so i’m going to start
01:01
as is my habit of finding the first derivative so here we go so we have low
derivative high which is 4x minus low all over 4x wait
so this will be 2x squared and then this will be 2x
all over x squared minus 1 squared so there’s our first derivative let’s
see if we can simplify this so this will be 4x to the third
minus 4x and this will be minus 4x to the third also okay so
01:02
in other words our derivative is just the minus 4x the 4x to the thirds
add up to zero okay so i’m going to erase this part right here
because we’re going to need all the space we can get
and just write the first derivative here is minus 4x
over x squared minus 1 squared okay so there’s our first derivative
now we can go find the local extrema if there are any we can look where it’s
increasing we can look where it’s decreasing
inflection points and stuff like that so maybe we’ll take a look at the second
derivative so here we go um low times derivative high
01:03
minus high times derivative low which will be two x squared minus one
so 2x squared minus 1 and then times another 2x
and then all over x squared minus 1 to the fourth
all right and let’s see if this is going to be something we can work with here
[Music] i see i have x squared minus 1 squared
and over here we have an x squared minus 1 but only to the first power so
what can we factor out from see this minus right here is separating
things so out of this we can factor the minus four and one of the x squared
minus ones i’m gonna write that here minus four and then x squared minus one
and then what do we have left we still have one of these x squared
01:04
minus ones left we factored out the minus four now we have a minus
now i factored out that minus four so we still have an x
left and it’s going to be times and i factored out that x squared minus
one so it’ll be times four x and then all over in fact let’s just put
it over here all over x squared minus one to the fourth
okay so let’s see here what we’re going to get
we’re going to get minus 4 x squared minus 1.
and then what are we getting here we already have an x squared and we’re
going to have a minus 4x squared so i’m going to say minus 3x squared
01:05
and then minus 1 and then all over x squared minus 1 to the fourth okay so
now i’m going to look at the minus 4 those are minuses and i’m going to
cancel here and so what i’m going to end up with here is a 4
and then the 3x squared plus 1 and then that’s going to cancel with one
of those so x squared minus 1 to the third okay so
what are the second order critical numbers now when i’m looking at my second
derivative here i’m looking at hey it’s undefined at plus or minus one
but the original was undefined at plus or minus one also
right so we’re gonna need to be careful at plus or minus one because
um you know what’s happening there is we have vertical isotopes there
01:06
so let’s write our second derivative over here um yeah so let’s do that
right here our second derivative is four times three x squared plus one all over
the x squared minus one to the third all right so this numerator is always
positive um and so we’re going to be looking here
at this third here there may be a change in concavity there
all right so now that we’ve done that grunt work we found the first and second
derivative now we can start trying to find the uh relative extrema
and the concavity the points of inflection and all that stuff here so
01:07
let’s do the first derivative test first um so
yeah okay let’s just do that all right so i’m looking at the
critical numbers what are the critical numbers
um actually before we do the first derivative test let’s just go and take
care of the isotopes right so we need some limits remember to
do the vertical isotopes we need some limits
so i’m looking at the plus or minus one here
so what’s happening is we take the limit as we approach the
um one of the function here and the limit as we approach one
from the left and from the right and what about these limits here when we
approach minus one from the left and minus one from the right
so we need to know these limits here because we’re getting
01:08
unbounded growth there we need to be able to have that on our sketch
so when we look at one from the left of this function up here so notice when
we look at that function up there the top is always positive
but when we’re approaching one we’re going to get two over zero so we’re gonna
so all of these are going to be under bounded growth
because all of these limits the numerator is approaching a non-zero number
and the denominator is approaching zero so we’re getting unbounded growth
so the question is is it positive or negative so when i’m
approaching one from the left like for example like 0.9 now if i do
0.9 squared minus 1 then that’s going to be
negative so when i’m looking at 1 from the left this will be negative infinity
01:09
when i look at 1 from the right for example like 1.1 now 1.1 squared
minus 1 that makes that positive and a positive to the third power will still
be positive the numerator is always positive i’m
sorry yeah the numerator is always positive 2x squared right
all right so we got those now what happens if we approach
minus one from the left you know something like minus one from the left
would be like minus 1.1 and when i square that
i’m going to get something greater than 1 and then minus 1. so that will be
positive so when i approach -1 from the left this is positive
and when i approach minus one from the right like
um you know something like negative point nine
and then but if i square that and then subtract one
that’ll be negative so when i’m approaching minus one
from the right i get minus infinity so clearly we have a vertical isotope at one
01:10
and vertical isotope at minus one but it’s not just that we have isotopes
we actually need to understand all this behavior
because we need to put this on the graph [Music]
all right so that’s the behavior we needed to understand that and now when i
break up and look at the uh where it’s increasing and decreasing and all that
i’m going to take into account um the zero right here
and you know when when x is zero that’s in the domain of our original and
it’s a critical number here because it makes the derivative zero when x is zero
we’re going to get zero over one so that’s so we’re going to look at
so this will be less than minus one um between minus one and zero equal to zero
01:11
and then between zero and one [Music] and then greater than one
and so i’m going to be looking at the function the derivative
and i’m going to be making a conclusion all right so let’s do our first
derivative test first derivative test so when i’m looking
less than -1 what’s happening to our derivative
the derivative is right here now notice on our derivative the denominator is
always squared so it’s always positive so when i’m looking less than minus one
i am looking at say like minus two what’s happening at minus two it’s
because it’s positive so it’s increasing and when i look between
minus one and zero for example negative one half
well that’s that’s positive so negative one half so that’s positive still
01:12
that’s positive so still increasing and so there’s you know no
at -1 there’s no relative extrema which we knew already because there’s a
vertical isotope there but what about between 0 and 1 say a half
so now let’s test a half so now we’re going to get negative over a positive
so that’ll be negative and then now what about greater than one
so greater than one for example say two and that’ll still be negative
and so this will be decreasing so we know it’s decreasing on these
intervals here increasing here and increasing here
all right what’s happening at zero it was increasing and then it was decreasing
right so it was increasing and then decreasing that’s a relative max
01:13
we have a relative max at zero now what is the function
at zero when we plug in zero into the original function we get out zero
zero over minus one which is just zero so that’s our relative max
and now we can go start trying to look for concavity um
before we do that though where is the derivative blowing up
is the derivative ever blowing up remember we need to find those
values that make the denominator 0 in the numerator
non-zero so the numbers that make the denominator zero which are plus or minus
one um right so so you know we’re not going to have any vertical
isotopes or cuts at 1 or -1 because we already know we
have vertical isotopes there so we don’t need to worry about vertical
01:14
tangents and cuts because we already have vertical isotope there
now what about horizontal isotopes let’s look at horizontal isotopes before
we go do the concavity test so let’s look at horizontal isotopes are
there any so i need to find the limit as we approach positive infinity
of the function two x squared over x squared minus one
and in order to find this limit here i’m going to divide by the highest power
so i’m going to divide by x squared on top and x squared on the bottom [Music]
so what is this limit here so we have 2x squared over x squared x
squared over x squared minus 1 over x squared i divided everything by x squared
01:15
and that limit is after simplifying it 2 over 1 minus 1 over x squared
so this limit is 2 right because that’s going to go to 0
so we have 2 over 1 that’s 2. so we know that y equals 2 is a horizontal isotope
is it the only horizontal isotope what about if we put a minus here
we go from here to here we’re just dividing everything by x squared
so that’ll still work even if you have a minus there
and now i’m simplifying x squared over x squared is 1
and then x squared over x squared is another one and then we have 1 over x
squared now even if this is a minus infinity
this still goes to zero we still have two so y equals two is the only
horizontal asymptote all right so good we have a we have the
two vertical asymptotes we have the horizontal isotope we know
01:16
where it’s increasing decreasing in the relative extrema and so
we need now the concavity test so when we’re looking here at this second
derivative here um all we have to look at are the plus or minus ones
and we know that there’s no inflection point
at those but we need to figure out the concavity so let’s look at [Music]
less than -1 between -1 and one and greater than one
and so we’re looking at the function the second derivative
we’re going to make our conclusion so this will be the concavity test
so when we’re looking less than -1 for example -2
what’s happening to the derivative the derivative is all positive up here
01:17
but when we’re looking at -2 here that’ll be a positive to the third
so that’ll be positive so it’s all positive so this is concave up here on
this interval here so to the left at -1 remember there’s a
vertical isotope at -1 but to the left of it we know it’s concave up
and in between minus one and one what’s happening there so
between minus one and one i can choose zero so when i choose zero
everywhere now i get minus one to the third so that’s negative
so i’m gonna get concave down here so between minus one and one is concave
down now greater than one it’s going to be like a two
so that’ll be positive here and it’s always positive here
so that’ll be positive it’s concave up again so you know you have a change in
01:18
concavity so you you might be tempted to say oh
there’s an inflection point there but remember that x equals negative one
is not a second order critical number um because at x equals minus one
the function is not defined that’s technically why it’s not a
critical second order critical number what’s happening at minus one we already
know we have a vertical isotope there so of course there cannot be an
inflection point because there is simply no point
when x is -1 it’s not on the function okay but now we know where the function
is concave up and now we know where it’s concave down
okay so putting all that together the first derivative test the concavity
test the horizontal and vertical isotopes and knowing that this function doesn’t
have any vertical tangents or vertical cusps
we can put all that together and look at a sketch
01:19
and so that’s what it basically looks like um
except for the fact that those vertical those vertical asymptotes should be
um dashed in okay but less than minus one we can see that we’re concave up
between minus one and one we can see where concave down
and greater than one we can see that we’re concave up
and less than minus one we can see that we’re increasing
and between minus one and zero we can see that we’re still increasing
and between zero and one we can see we’re decreasing
and we and greater than one we can see that we’re decreasing
and since there’s a change in monotonicity
at zero in other words to the left we’re increasing into the right of zero we’re
decreasing at zero zero there has to be a relative max and we can see that
and then also i don’t have on their dash the horizontal isotope
01:20
of y equals two um so we could see that at y equals two
there’s going to be a horizontal isotope
and we should dash that in to finish off the sketch
okay so let’s go on to the next one [Music] and let’s get started on this one
so on this one right here we’re going to um are there going to be any vertical
isotopes on this one right here that might be what we’re
looking at first here is this denominator every 0
the answer is no the x squared plus 1 would never be 0. on the previous one we
had x squared minus one but if you have x squared plus one that’s never going to
be zero so there are no vertical isotopes um
what about horizontal isotopes we’ll just put here no
01:21
vertical isotopes so what about horizontal isotopes
so let’s take the limit as we go to infinity and i’m looking at the
x square plus one so that’s sometimes known as x squared plus two x plus one
all over x squared plus one and i’m going to divide by the highest power
so here i’m going to divide by x squared and 2x by x squared and the 1 by an x
squared the next square by an x squared and the 1 by an x squared
and i’m going to find this limit now all these are going to zero
as we saw in the episode uh horizontal vertical isotopes
um but this is x squared and that’s an x squared
01:22
so this is going to one over one which is just one
and it doesn’t matter if we have plus or minus infinity here
so from this step to this step all i did was divide everything by x squared
now when i find the limit here this is still going to zero and zero and zero
and this is still a one and a one so it doesn’t really matter if you’re going to
plus or minus infinity all of these squares on here they’re the
same this is a one and that’s a one so this is all going to one so y equals
one is the horizontal isotope and there are no vertical isotopes
all right so now let’s look at the first derivative so here we go we have low
derivative high and i’m going to look at this and i’m taking the derivative of
that numerator you could just use the chain rule if you
want but actually x plus one squared is just sitting right here
01:23
and so what is the derivative of it it’s just two x plus two
and then all right so low times derivative high minus high so x plus 1 squared
times derivative of the denominator which is 2x
and then all 1 plus x squared over squared okay so what can we factor out here
i’m going to think of this as 2 times x plus one
so i have an x plus one squared here [Music] um
so probably it’s just easier to just multiply it all out one times the two x
plus two x squared so that would be two x to the third
01:24
and then plus two x squared and then minus there’s a squared there
so that’ll be negative and then this right here is x squared plus 2x plus 1
all times 2x right so i’m working on this part right here
so this will be minus and then 2x to the third and then minus the 4x
and then minus 2x now this will be third and this will be a squared here
and then minus the 2x so by just by multiplying that out i get those four
and then we’re going to get these three right here minus 2x to the third
minus 4x squared minus 2x all over 1 plus x squared squared
and so then simplifying that okay so i just multiplied all that out and all
01:25
that out and what do we get here these add up to zero
and here we’re going to get 2x squared and a minus 4x squared so i’m going to
get a minus 2x squared and then these add up to zero and then i get a plus two
[Music] and then all over one plus x squared squared so that’ll be minus 2
and then x squared minus 1 1 plus x squared squared
so minus 2 and then x minus one x plus one [Music]
all right so that’s what i like my first derivative to look like right there
so we just had to expand all that out and then combine it together and
01:26
simplify it factor it and now this one is easy to test it looks like
something’s happening at minus one and one where we have the
first derivative here um on the other hand this one’s kind of handy here
because we could take the second derivative here or actually even here so
let’s go up here and erase this and put our first derivative here so
it is minus two now let’s write this one minus 2x squared plus 2
over 1 plus x squared squared and then that’s also equal to this one x minus one
x plus one all over one plus x squared the reason why i like two is one’s easy
01:27
to test but still we need our second derivative so
there we go now we can erase all that let me do it here all right so we got
our first derivative accomplished we can look at this right here for
testing we can look at this right here for the second derivative now
so here we go low [Music] times derivative high and that’ll be minus 4x
and then minus and then minus 2x squared times the derivative here which will be
2 1 plus x squared to the first power times another 2x
and then now this is 1 plus x squared to the squared to the squared
01:28
so to the fourth okay so what can we factor out all this
how can we simplify this we could just go expand it out again
but i want to can’t i want to keep one plus x square and see if i can
cancel with one of these this minus right here is separating all
that that minus and then we have all of that
so think in terms of this one and this one
and what do they have in common one plus x squares
and then they have some twos in common where we are fours we have a four and a
four and this one has a minus and this one has a minus so let’s factor that out
let’s factor out the um minus 4x and then the 1 plus x squared
only to the first power though all right so what are we going to have left here
we factored out that we factored out one of these
so we have a 1 plus x squared left we’re factoring out that minus
01:29
and that 2 and that 2x and we get that and so what do we have left here we have
a plus and then we have and we factored that out too
so we really just have left a minus 2x squared plus 2.
and we’re not touching the bottom yet so let’s double check this
so i’m thinking about we have that and that two different things and i
factored this out and one of these out so i still have one of these left
now i factored out the 2x the 4 and the minus sign so i have all that out
and i factored that out so i have this left right here with a positive in front
of it so i have a positive and then it’s a minus two x squared plus two
01:30
so i think that looks good there and then so obviously we still have the
same denominator all right so now we look like we have a minus x squared here
and then a three and so let’s go with that minus four x one plus x square
we’re gonna cancel these one plus x squares
we’re gonna get minus four 4x here and then we’re going to get a 3
minus x squared all over 1 plus x squared and then now to the third power
all right so there’s our second derivative there all right so
after we do all that work now let’s erase all that and move our second
01:31
derivative up here so here we go and let’s move this up so minus 4x
over uh 3 minus x squared and so we can see what the second order
and first order critical numbers are now so we can start making our derivative
test here in our concavity test so let’s do the first derivative test
first so we’re looking at 1 and -1 as first order critical numbers if we
substitute them into our function they’re both defined there and they make
the derivative zero is there anywhere where the derivative is going to blow up
and the answer is no this is never going to be zero down here
so there’s not going to be any vertical tangents there’s not going to be any
cusps on this problem and so we can just go do the first
derivative test so i’m looking less than minus 1 equal to minus 1
01:32
between -1 and 1 equal to 1 and greater than one and i’m going to
look at the function the derivative and i’m gonna make a conclusion [Music]
and so here we go when we’re less than minus one
for example a minus two now keep in mind the denominator is always positive
and this is always negative so if i look at minus two
that’ll be a negative a negative so i have three negatives and so that’ll be a
negative it’s gonna be decreasing now between
minus one and one for example is zero then this is going to be negative
positive so what’s a negative times a negative
it’ll be positive so now i’m increasing so decreasing increasing
and greater than 1 say for example 10 right positive positive that’s that
01:33
negative there is still there so it’s negative so it’s decreasing so decreasing
all right so is there a change between decreasing and increasing
yeah so this is going to be a relative min at -1 and at one we’re going from
increasing to decreasing so this is going to be a relative max [Music]
now what happens when we plug in minus one
um when we plug in minus one we’re going to get zero we plug in one
substitute in one into our original function and we get out two oops
two and here we put a zero here in the wrong place
all right so at the point -1 0 we have a relative min
and at the point 1 2 we have a relative max
01:34
and so we know where it’s increasing and where it’s decreasing
and so this was the first derivative test and so now let’s look at our second
derivative here and let’s make our concavity test so when we say it’s
inc when we say it’s increasing we’ll know exactly how
it’s increasing and when it’s decreasing we’ll know exactly how it’s decreasing
so um i’m looking here at zero and i’m looking here
at plus or minus um well what makes that zero right
so plus or minus square root of three so we need to um
look at those places there at zero so i’ll just put it on a number line
mine like that so we need to look at one two three four intervals there
and we need to start doing some testing so we’ll look at less than square root
01:35
of three equal to negative square root of three between the
negative square root of three and zero equal to zero and between zero
and square root of three equal to square root of three
and greater than square root of 3. and so those are the places that we need
to look and we’re going to look at the function the second derivative
and we’re going to make our conclusion all right so here’s our derivative we’re
testing right here now keep in mind square root of three it’s
about 1.7 ish so think of that as negative 1.7 but when i’m less than it
like say like negative 10 when i square negative 10 that’ll be
like like that’ll be like a really big number but it’ll be 3 minus that number
01:36
so that negative will dominate and when i use the negative right here
that’ll be negative times the negative so that’ll be positive so this will all
be positive here so um actually if this is a negative
here and that would be a negative times a negative times a negative
that’ll be three negatives sorry i missed that two negatives there
right so think of that as negative 10 so negative 4 times negative 10
and then this is negative because it’s 3 minus negative 10 would be like 100. so
what’s 3 minus a hundred right so that’s negative so negative times
negative times negative you can’t look at that x and assume it’s
positive the x is less than a negative square root of three so you know that x
is negative all right so we know it’s concave down here so concave down
01:37
all right and then let’s see what’s happening between negative
square root of three and zero so that’s like negative
one 1.7 so think of like negative one so what’s happening when we look at the
derivative second derivative at negative one
so negative four times negative one so that’s positive
now when i use a negative one squared that’s a one so what’s three minus one
that’s positive so i’m going to get a negative times a negative times a positive
so i end up with a positive right here so this is concave up [Music]
and then between zero and three so square root of three is about one
point seven so think of the x as being a one
so when x is a one here so that would be negative four times a one so that’ll be
negative and when i use x is one here that’ll be three minus one
01:38
so that’s a positive so we’re going to get a negative times a positive times a
positive so now it’s negative so now it’s concave down again and then um
greater than square root of 3 for example a 10.
so if i look at 10 up here now if i look at 3 minus
10 square right that’s 3 minus 100 right so that’s a negative
and a positive and a negative so that’ll be negative times a positive times a
negative so that’s that’s positive so let’s concave up again
so is there a change in concavity square root of three
and is square root of three in the domain of the function
yes so this right here is an inflection point inflection point
01:39
and that minus square root of three there’s a change in concavity
this is also an inflection point all right so um and what’s happening at zero
um concave up concave down is x equals zero on the graph yes so we
still have another inflection point so we have three inflection points
when x is zero we’re going to get out of one when x is square root of three and
negative square root of three we get out some numbers
um for example we’re going to get out 4 in the denominator so
we can write those numbers over here f of square root of three is
01:40
square three plus one over one plus square root of three square so w four
so let’s square that and then f of minus square root of three is minus
square root of three plus one over four again and this will be uh
squared so i’ll just call this one m1 and this one m2
and so we have m2 right here and m1 right here and that way we can say
yeah we found those numbers they’re over there they’re not exactly
we could go find those as decimals but all right so we
found the horizontal isotopes we found the no vertical isotopes
we did the first derivative test we did the concavity test
and we noticed that this first derivative is never going to blow up
so there’s no vertical tangent so there’s no there’s no cusp
01:41
all right so let’s see the graph all right so there we go and let’s make
sure everything makes sense let’s look at the first derivative test
that we found so if we’re less than -1 we should be decreasing
at -1 we should have a relative minimum between -1 and 1
it should be increasing and then at 1 2 we should have a relative max and then
greater than 1 we should be decreasing now what about the inflection and the
concavity when we’re at square root of three and zero
and minus square root of three we should
have some inflection points there and we can see those on the graph
when we’re less than minus square root of three we’re concave
down um between minus square root of three and zero
we’re concave up and you can see that on the graph
01:42
between zero and square root of three we’re concave down
and you can see that and then greater than square root of three
we are concave up all right so yeah everything checks out
oh yeah and then the last thing is the horizontal isotope is y equals one that
we found and you can see that on the graph right there and then
of course there is no vertical isotopes and we plotted the x-intercept and we
plotted the y-intercept and so that looks like a pretty good
shape shape of the graph there okay [Music] so let’s look at one more example
01:43
here we go so this time it has fractional exponents so i’m anticipating some
vertical tangents and or cusps happening but let’s see what we get so here’s the
first derivative and so here we go we have um two thirds x to the two thirds
minus one times the second one and then plus x to the two thirds
times um one third six minus x to the two thirds minus one
and then with the minus one out here okay so there’s the product rule there
right and now we have some negative exponents so
let’s get a common denominator we have a negative exponent here
01:44
meaning we have a denominator and we have a denominator here
and we need to put them together so first i’ll write this as
two and then six minus x to the one third and then this will be over three x to
the one third and then i’m gonna save some space here and then
what do we have going on over here we have a minus
one so we have a minus x to the two thirds
and we have a three and then we have six minus x to the two thirds
what’s going to be our common denominator they both have threes
so i’m going to use six minus x to the two thirds here
over six minus x to the two thirds now we have the and then over here i’m
going to use a x to the one-third and then x to the one-third
01:45
okay so now we have a common denominator everywhere
so our common denominator is x to the one third
three times x to the one third and then six times six minus x to the two thirds
okay so far so good right there’s our common denominator
now what are we going to get in the numerator this is 6 minus x and 6 minus
x we’re going to add the exponents one-third plus two-thirds three-thirds
so we’re gonna get two times six minus x
to the first power and then we have plus and then here we have x and x so we add
the exponents we get three thirds we get an x
so we’re just going to get minus x out of that
okay and let’s just simplify this a little bit this is going to be
12 and then minus 2x minus another x so minus 3x and this will be over
01:46
three we have a three here and we’re the arrows already have three all right so
three and then x to the one third and then six minus x to the two thirds
so simplifying some threes here we have four minus x over x to the one third
six minus x to the two thirds so that looks like the best first derivative there
so right we can simplify some freeze there so i like that so
let’s see if we can rewrite this first derivative up here and save us some space
so let’s see here it’s 4 minus x over x to the one third
01:47
times six minus x to the two thirds all right so there’s our first
derivative there and let’s get the reddish through the rest of this work here
all right we’re doing good so far now um you know when we look at this
uh original function up here the domain is
all real numbers um because we’re going to square an x we can always square an x
we can always cube root so um you know it’s continuous everywhere it’s
not differentiable everywhere it’s not differentiable at six
it’s not differentiable zero so we got some things happening there
um in fact we have some blow ups don’t we when x is zero
right we’re going to get four over zero right that’s going to be a blow up
when x is six we’re going to get another blow up
when x is six we’re gonna get four minus six so we’re gonna get
a minus two we’re gonna get a non-zero over zero
01:48
so we’re gonna have two potential places where we have vertical tangent or
cusp do we have any vertical isotopes or horizontal isotopes
the answer is no if you take the limit as x goes to infinity
you’re just going to get minus infinity um when you take the limit as x goes to
positive infinity and negative infinity they’re just going to
blow up to infinity so we’re not going to get any
vertical or horizontal isotopes so but we are going to get a vertical
tangent or cusp one of the two maybe one of each so let’s look at the limit
as we approach zero from the left of the first derivative
and the limit as we approach zero from the right of the first derivative
right because that’s going to make that zero four over zero so it’s going to
01:49
blow up there and let’s also look at the limit as we approach 6 from the left
and the limit as we approach 6 from the right
of the first derivative those are also blow ups because at six we’re going to
get -2 over zero so non-zero over zero that’s going to be a blow up
all right so now we need to find the sine of these infinities
now if i approach zero from the left that means i’m negative i’m a little bit
negative so this will dominate the four the four will dominate four minus
something really really close to zero that’s going to be really close to
positive 4 so that’s going to be positive there but
if i’m approaching 0 from the left this will be negative
and this will be positive so this whole thing right here will be negative
so we’re approaching 0 from the right now
01:50
something like 0.1 again the 4 is going to dominate so positive
0.1 this will be positive this will be positive so
positive over positive times positive all that’s positive there
okay so now what about as we approach six from the left now we approach six from
the left this is going to be getting really close to six
so this will be four minus something really close to six so this is
all really close to negative two so for both of these the numerator is
going to be negative now if we’re approaching six either from
the left or from the right if you’re 5.9 or 6.1
if you cube root those you’re still positive
so this will be negative and this will be positive
so if you square something it’s always positive so i’m getting negative
01:51
over a positive times a positive so both of these are negatives
so here’s where we’re going to have a vertical tangent
vertical tangent at x equals six and we’re going to have a cusp
at x equals zero so whatever the sketch that we come up with
it needs to have that cusp and it needs to have this vertical tangent
all right so let’s remember those now let’s go and do our first derivative test
um four is a first order critical number right because four’s in the domain it
makes the derivative zero um what about x equals zero it’s in the domain
it makes the derivative undefined but we already know what’s happening at x
equals zero right but we still need to check to the left and to the right of it
01:52
six is also a first order critical number the function is defined at six
the output is zero at six the derivative is undefined so we have um zero
four and six is our critical numbers so here we go less than zero
equal to zero between zero and four equal to four between four and six
equal to six and greater than six and so we’re going to have the function
that we’re testing and the first derivative and a conclusion
so first derivative test there we go [Music] and i’m just gonna
01:53
write out the derivative over here so that’s the first derivative that
we’re testing there all right so let’s test it less than
zero so something like -1 so we’re going to get a positive and
we’re going to get a negative and this expression is always positive
because we’re squaring that right there so that’s always positive
so when we’re looking at say -1 i’m going to get a positive over a
negative so that’s all negative so it’s going to be decreasing right there
now between 0 and 4 for example a 3 what’s happening we’re gonna get a
positive positive a positive so it’s all positive so we’re gonna be increasing
so between zero and four we’re increasing and then what’s happening at
between four and six so between four and six
say say a five five so now i’m getting negative
01:54
five so we’re getting positive positive so negative
over a positive is negative so i’m gonna be decreasing
and what about greater than six or like for example like a ten so i’m looking at
a negative positive positive so it’s negative
so at six here we’re not going to have a relative max or min
but in fact we already know what’s happening at six we have a vertical
tangent at six and at zero we already know we have a cusp at
here so you know we can see it’s not changing monotonicity there’s
no relative extrema here in fact we have a cusp there now what
about it four though at four we’re increasing to the left and then
increasing to the right so this is a relative max
and we plug in 4 to get out that number into the function
01:55
so we can look at the function right here we plug in a four we get um four
to the two thirds and then six minus four so that’ll be minus two
to the one third and that may simplify i’m just going to call it an m
so at 4 we get this number m here and i would definitely get a decimal
approximation of that so that you can kind of see where where it is
this simplifies to two to the five thirds so maybe you didn’t know that
um so we can say this is two to the four thirds
01:56
and we can write this as minus one and then two to the one third
so actually where am i getting that minus from if i use four
it’s six minus four that’s a positive two
so this will be two to the one third all right so it’s just two to the five
thirds that’s our m i guess we’ll just write that two to the five thirds
all right so we plug in four and we got our crazy
two to the five thirds out that’s just the cube root of 32. [Music]
in any case that’s our relative max and now we can go find our um
second derivative of this and look for the concavity um
01:57
however the video is running long and i don’t think i’m going to take the
second derivative of that here’s the graph we can see what we’re going to get
if we do take the derivative of that and get our second derivative it’ll be gone
it’ll be concave down less than zero it’ll be concave down between zero and six
and concave up when we’re greater than six so
we see that the first derivative that we got it’s in
decreasing left left left of zero between 0 and 4 it’s increasing
and then between 4 and 6 is decreasing and then greater than 6 is still
decreasing and you can see the vertical tangent vertical tangent right here
and you can see the cusp right here at zero that we found
01:58
um and so definitely want to check that concavity out where it’s
where it’s concave down here and it’s concave down here
and then it’s concave up right here so at six here we’re going to have a point
of inflection and if we get the second derivative of that
we’re still going to get these expressions right here
where to go take the derivative of that we’d still get these expressions right
here in the denominator with some powers on them
so it’s not too hard to see that six would be an inflection point
all right so um that was our last example so let’s look at some exercises
01:59
[Music] so as you can see graphing using that curved sketching procedure
can be quite long um because there’s just so many
things that we’ve learned and we’ve done and you know when your functions get
more complicated to show a variety of of uh
features you know can you come up with a function that has
a vertical tangent has some isotopes right so then your derivatives start
getting more complex and the problems keep getting longer and
longer but in any case here i have a good selection of problems for you right
here some of them are straightforward for example one’s a polynomial
number two only has one fractional exponent
but anyways i definitely want to check out these problems here
and see if you can solve them and see how that goes
and then i got some more from you here and number two all the information’s
been figured out all you gotta do is sketch the graph
02:00
and then on number three here um give you a function right here
if you looked at that episode on horizontal isotopes
that function right there you might watch out for how many horizontal
isotopes it has on number four here we have another fractional exponent
so watch out for vertical tangents and tests
uh number five here five years kind of interesting because
the the functions not fully given to you you got some parameters
you want to find the parameters so that you have a vertical isotope
and you have a horizontal isotope so those type of functions
are kind of fun those type of problems are kind of fun
and then we got six seven and eight where we’re concentrating on
vertical tangents and vertical isotopes and then um nine and ten here the last
02:01
two um more curse curve sketching and finding all the information out that
they want and so yeah if you uh get stuck on any of those ten problems
or you have any questions for me i look forward to hearing
from you in the comments and hope you enjoyed this video
so i just want to say thank you for watching and if you like this video please
subscribe and like below and on the next video we’re going to talk about applied
optimization problems so we’ll be solving a lot of word problems
and we’ll be using all the tools that we’ve developed so far like the first
derivative test the rel the extreme value theorem and the second derivative test
will be big so i look forward to uh seeing you there
and i’ll see you next time if you like this video
please press this button and subscribe to my channel
now i want to turn it over to you math can be difficult
02:02
because it requires time and energy to become skills
i want you to tell everyone what you do to succeed in your studies
either way let us know what you think in the comments
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Discover a lot of information on the number 4973: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 4973
Is 4973 a prime number? Yes
Is 4973 a perfect number? No
Number of divisors 2
List of dividers 1, 4973
Sum of divisors 4974
Prime factorization 4973
Prime factors 4973
## How to write / spell 4973 in letters?
In letters, the number 4973 is written as: Four thousand nine hundred and seventy-three. And in other languages? how does it spell?
4973 in other languages
Write 4973 in english Four thousand nine hundred and seventy-three
Write 4973 in french Quatre mille neuf cent soixante-treize
Write 4973 in spanish Cuatro mil novecientos setenta y tres
Write 4973 in portuguese Quatro mil novecentos setenta e três
## Decomposition of the number 4973
The number 4973 is composed of:
1 iteration of the number 4 : The number 4 (four) is the symbol of the square. It represents structuring, organization, work and construction.... Find out more about the number 4
1 iteration of the number 9 : The number 9 (nine) represents humanity, altruism. It symbolizes generosity, idealism and humanitarian vocations.... Find out more about the number 9
1 iteration of the number 7 : The number 7 (seven) represents faith, teaching. It symbolizes reflection, the spiritual life.... Find out more about the number 7
1 iteration of the number 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3
Other ways to write 4973
In letter Four thousand nine hundred and seventy-three
In roman numeral MMMMCMLXXIII
In binary 1001101101101
In octal 11555
In US dollars USD 4,973.00 (\$)
In euros 4 973,00 EUR (€)
Some related numbers
Previous number 4972
Next number 4974
Next prime number 4987
## Mathematical operations
Operations and solutions
4973*2 = 9946 The double of 4973 is 9946
4973*3 = 14919 The triple of 4973 is 14919
4973/2 = 2486.5 The half of 4973 is 2486.500000
4973/3 = 1657.6666666667 The third of 4973 is 1657.666667
49732 = 24730729 The square of 4973 is 24730729.000000
49733 = 122985915317 The cube of 4973 is 122985915317.000000
√4973 = 70.519500849056 The square root of 4973 is 70.519501
log(4973) = 8.5117785587147 The natural (Neperian) logarithm of 4973 is 8.511779
log10(4973) = 3.6966184592322 The decimal logarithm (base 10) of 4973 is 3.696618
sin(4973) = 0.14070219793438 The sine of 4973 is 0.140702
cos(4973) = -0.99005196403847 The cosine of 4973 is -0.990052
tan(4973) = -0.14211597274192 The tangent of 4973 is -0.142116
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http://www.pinkmonkey.com/studyguides/subjects/algebra/chap9/a0909103.asp
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Support the Monkey! Tell All your Friends and Teachers
Home MonkeyNotes Printable Notes Digital Library Study Guides Study Smart Parents Tips College Planning Test Prep Fun Zone Help / FAQ How to Cite New Title Request
Example Three consecutive even natural numbers are such that five times of the smallest exceeds three times the largest by 28. Find the middle number. Solution : You are looking for the middle number. Let ’ x ’ be the required middle number Hence the three consecutive even natural numbers are x - 2, x, x+2 Setting up equation as - Five times of the smallest exceeds three times the largest by28 5 ( x - 2) = 3 ( x + 2 ) + 28 \ 5 x - 10 = 3 x + 6 +28 \ 5 x - 10 = 3 x + 34 \ 5 x - 3 x = 34 + 10 \ 2 x = 44 \ x = 22 Therefore, the middle number is 22. Index 9.1 - Definition and Solving Techniques9.2 - Use of Simultaneous Linear Equations Chapter 1
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https://edubuddie.com/c-programing-general-preprocessing-items-use-mnemonic-variables-of-3-characters-or-more-use-void/
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# C Programing – General Preprocessing items: Use mnemonic variables of 3 characters or more. Use void
C Programing
– General Preprocessing items: Use mnemonic variables of 3 characters or more. Use void main (void) or void main ( ) in your program and leave off the return 0. Use double as the choice for all floating point values. Add heading items using as printf statement. Watch for instructions that specify a symbolic constant and assign that symbolic constant in all capital letters and using the #define.
Problem:
Write a program that calculates the distance from the roof top to the ground for a free falling ball that is only influenced by gravity as it falls. The program should ask the user to input the height of the building. It is great weather on the roof, so the initial velocity is 0 meter / second. This gives a distance equation shown below and a Velocity equation shown below. The program should assign the value of gravity as a symbolic constant. Gravity = 9.81 meters / seconds-squared. The program must use the built in function(s),such as A????1powA????1 and or A????1sqrtA????1, from the Math header file (library) to calculate values. The program should calculate the time and print the answer to show 5 decimal place(s). The program should calculate the velocity and print the answer to show 4 decimal place(s). The program lets the user know how many valid tries and how many error tries were completed. The program continues until it is told to stop when -999 is entered. Use the given data and reproduce the sample output as a comment in your program.
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Question 48
# If the roots $$x_1$$ and $$x_2$$ are the roots of the quadratic equation $$x^2 -2x+c=0$$ also satisfy the equation $$7x_2 - 4x_1 = 47$$, then which of the following is true?
Solution
$$x_1 + x_2 = 2$$
and $$7x_2 - 4x_1 = 47$$
So $$x_1 = -3$$ and $$x_2 = 5$$
And $$c = x_1 \times x_2 = -15$$
### Video Solution
Create a FREE account and get:
• All Quant CAT complete Formulas and shortcuts PDF
• 35+ CAT previous year papers with video solutions PDF
• 5000+ Topic-wise Previous year CAT Solved Questions for Free
Boost your Prep!
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http://www.cultracing.com/2019/05/span-of-two-vectors/write-v-sum-two-vectors-u-e-span-u-u-u-assume-ul-u-orthogonal-basis-r-q/
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# Write V Sum Two Vectors U E Span U U U Assume Ul U Orthogonal Basis R Q
This post categorized under Vector and posted on May 16th, 2019.
This Write V Sum Two Vectors U E Span U U U graphigraphice Ul U Orthogonal Basis R Q has 1024 x 816 pixel resolution with jpeg format. Span Of Vectors In R3, Span And Basis Of A Vector graphice, Span Of Vectors Calculator, Linear Combination Of Vectors, What Does It Mean To Be In The Span Of Two Vectors, Can 3 Vectors Span R2, How To Determine If A Set Of Vectors Span A graphice, Span Of A Single Vector, Span Of Vectors Calculator, What Does It Mean To Be In The Span Of Two Vectors, How To Determine If A Set Of Vectors Span A graphice was related topic with this Write V Sum Two Vectors U E Span U U U graphigraphice Ul U Orthogonal Basis R Q. You can download the Write V Sum Two Vectors U E Span U U U graphigraphice Ul U Orthogonal Basis R Q picture by right click your mouse and save from your browser.
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# The Central Limit Theorem
## Presentation on theme: "The Central Limit Theorem"β Presentation transcript:
The Central Limit Theorem
The Central Limit Theorem tells us that for a population with any distribution, the distribution of the sample mean approaches a normal distribution as the sample size increases. Furthermore, if the original distribution has mean π and standard deviation π, the mean of the sample means will be π and the standard deviation of the sample means will be π π , where π is the sample size.
The Central Limit Theorem
Principles to use the Central Limit Theorem For a population with any distribution, if π>30, then the sample means will have a distribution that can be approximated by a normal distribution with mean π and standard deviation π π . If πβ€30 and the original population has a normal distribution, then the sample means have a normal distribution with mean π and standard deviation π π . If πβ€30 and the original population does not have a normal distribution, then we cannot apply the central limit theorem! There is a cool Chart on page 288 summarizing how to use the Central Limit Theorem.
The Central Limit Theorem
Notation for the Sampling Distribution of π If all possible random samples of size n are selected from a population with mean π and standard deviation π, the sample means is denoted by π π₯ , so π π =π
The Central Limit Theorem
Notation for the Sampling Distribution of π If all possible random samples of size n are selected from a population with mean π and standard deviation π, the sample means is denoted by π π , so π π =π Also the standard deviation of the sample means is denoted by π π , so π π = π π π π₯ is called the standard error of the mean.
The Central Limit Theorem
Lets Look at example 1.
The Central Limit Theorem
Lets Look at example 1. Note: Individual value: When working with individual values from a normally distributed population, use the methods from last class. Use π= πβπ π Sample of values: When working with a mean for some sample (or group), be sure to use the value of π/ π for the standard deviation of the sample means. Use π= π βπ π π .
The Central Limit Theorem
Lets Look at example 1. Note: Individual value: When working with individual values from a normally distributed population, use the methods from last class. Use π= πβπ π or normalcdf(lower, upper, mean, stdev) Sample of values: When working with a mean for some sample (or group), be sure to use the value of π/ π for the standard deviation of the sample means. Use π= π βπ π π or normalcdf(lower, upper, mean, π π ). Now Lets do example 2 on page 290.
The Central Limit Theorem
A water taxi sank in Baltimoreβs Inner Harbor. Assume the weights of men is are normally distributed with a mean of 172 lb. and a standard deviation of 29 lb. Find the probability that if an individual man is randomly selected, his weight will be greater than 175 lb.
The Central Limit Theorem
A water taxi sank in Baltimoreβs Inner Harbor. Assume the weights of men is are normally distributed with a mean of 172 lb. and a standard deviation of 29 lb. Find the probability that if an individual man is randomly selected, his weight will be greater than 175 lb. Find the probability that 20 randomly selected men will have a mean weight that is greater than 175 lb.
The Central Limit Theorem
Recall the Rare Event rule for inferential Statistics If under a given assumption, the probability of a particular observed event is exceptionally small (such as less than 0.05), we conclude that the assumption is probably not correct.
The Central Limit Theorem
The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. If 1 pregnant woman is randomly selected, find the probability that her length of pregnancy is less than 260 days.
The Central Limit Theorem
The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. If 1 pregnant woman is randomly selected, find the probability that her length of pregnancy is less than 260 days. If 25 randomly selected women are put on a special diet just before they become pregnant, find the probability that their lengths of pregnancy have a mean that is less than 260 days.
The Central Limit Theorem
The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. If 1 pregnant woman is randomly selected, find the probability that her length of pregnancy is less than 260 days. If 25 randomly selected women are put on a special diet just before they become pregnant, find the probability that their lengths of pregnancy have a mean that is less than 260 days. If the 25 women do have a mean of less that 260 days, does it appear that the, does it appear that the diet has an effect on the length of pregnancy?
The Central Limit Theorem
Membership in Mensa requires and IQ score of above Nine candidates take IQ tests, and their summary results indicated that their mean IQ score is (IQ scores are normally distributed with a mean of 100 and a standard deviation of 15). If 1 person is randomly selected from the general population, find the probability of getting someone with an IQ score of at least 133.
The Central Limit Theorem
Membership in Mensa requires and IQ score of above Nine candidates take IQ tests, and their summary results indicated that their mean IQ score is (IQ scores are normally distributed with a mean of 100 and a standard deviation of 15). If 1 person is randomly selected from the general population, find the probability of getting someone with an IQ score of at least 133. If 9 people are randomly selected, find the probability that their mean IQ is at least 133.
The Central Limit Theorem
Membership in Mensa requires and IQ score of above Nine candidates take IQ tests, and their summary results indicated that their mean IQ score is (IQ scores are normally distributed with a mean of 100 and a standard deviation of 15). If 1 person is randomly selected from the general population, find the probability of getting someone with an IQ score of at least 133. If 9 people are randomly selected, find the probability that their mean IQ is at least 133. Although the summary results are available, the individual scores have been lost. Can is be concluded that all 9 candidates have IQ scores above so that they are all eligible for Mensa membership?
Homework!! 6-5:1-9, 11 β 19 odd.
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IsSzasz - Maple Help
# Online Help
###### All Products Maple MapleSim
Magma
IsSzasz
test whether a given magma is a Szasz magma
Calling Sequence IsSzasz( m )
Parameters
m - Array representing the Cayley table of a finite magma
Description
• A Szasz magma M is one in which there is exactly one nonassociative ordered triple of members of the magma.
• The IsSzasz command returns true if the given magma is a Szasz magma. It returns false otherwise.
Examples
> $\mathrm{with}\left(\mathrm{Magma}\right):$
> $m≔⟨⟨⟨1|2|3⟩,⟨2|3|1⟩,⟨3|1|2⟩⟩⟩$
${m}{≔}\left[\begin{array}{ccc}{1}& {2}& {3}\\ {2}& {3}& {1}\\ {3}& {1}& {2}\end{array}\right]$ (1)
> $\mathrm{IsSzasz}\left(m\right)$
${\mathrm{false}}$ (2)
> $m≔⟨⟨⟨1|1|3⟩,⟨1|2|3⟩,⟨1|2|3⟩⟩⟩$
${m}{≔}\left[\begin{array}{ccc}{1}& {1}& {3}\\ {1}& {2}& {3}\\ {1}& {2}& {3}\end{array}\right]$ (3)
> $\mathrm{IsSzasz}\left(m\right)$
${\mathrm{true}}$ (4)
Compatibility
• The Magma[IsSzasz] command was introduced in Maple 15.
• For more information on Maple 15 changes, see Updates in Maple 15.
See Also
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http://www.teacherled.com/2011/09/01/mars-rover/
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This is the second in the series of Mission to Mars resources for use in a space themed set of lessons. A more detailed description of how I use this resource and the thinking behind its creation can be found at the supporting site teacherledplus.
This is another IWB resource to allow you to easily build in a thinking skills aspect to your lessons. This resource encourages children to display persistence and develop their thinking strategies as well as record progress. The mission is detailed on the opening screen of the resource. In essence the challenge is to visit each of the squares in the grid with the mars rover in 12 moves (the absolute minimum). Restrictions are enforced in that the rover can only turn in multiples of 45 degrees.
Tap a square that you want the mars rover to move to. If it is a valid square it will move if not it won’t. You can click on already visited squares. Every square the mars rover explores by moving through will be illuminated. If you illuminate all squares and return to the start using no more than 12 waypoints the challenge is completed with 100% efficiency. This is possible but will usually require a lot of effort to achieve. Doing it in 13 is not too difficult and their are a number of ways of doing that but 12 is a great challenge. Ideas for making this more achievable for all children are given on the supporting site mentioned above.
Restart will clear the current progress. Briefing will show the mission constraints.
Go to the the IWB resource.
If you need the solution please ask in the comments. It will not be sent for a couple of days to avoid homework cheating. If you use what is clearly a staff email address I will try to send it quicker but cannot guarantee this.
## 8 thoughts on “Flash Mars Rover”
1. Hi
I am a year 6/7 teacher in Australia and my kids have done it in 13, but I would love the answer for 12 moves.
Thanks
2. Dryburgh says:
Hi
My class have eventually worked out 13 but we are struggling for 12. Can we have the solution please?
3. Bill Reece says:
The best we can do with our 5th graders is 13 moves. Could I get the solution please? Thanks so much
4. Liam Kenney says:
Could I please get the solution?
5. Jo says:
My class have loved this please could you send the solution. Thanks
6. Dave Walsh says:
Hi I’m a Year 6/7 teacher in Australia and my kids can’t get past 13. Can you send me the solution please?
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https://homebrew.stackexchange.com/questions/7891/can-i-derive-final-gravity-from-original-gravity-and-abv/7894
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# Can I derive Final Gravity from Original Gravity and ABV?
I think the answer is yes. Since we use OG and FG to determine ABV ... I would assume we could do the same with ABV and OG to determine FG, right?
• Out of curiosity, in what case would you know the ABV but not the FG? Is this for a commercial brew that you are trying to clone? Oct 30 '12 at 17:22
You can calculate ABV using the below equation:
``````ABV = (OG - FG) * 131
``````
So, with the use of some Algebra, you can calculate FG using the below equation:
``````FG = -1*(ABV/131 - OG)
``````
Yes, it's a simple calculation:
ABV = (OG - FG) * 131
For example:
ABV = (1.055 - 1.012) * 131
ABV = (0.043) * 131 = 5.633%
So to reverse it,
FG = (131 * OG - ABV) / 131
For example:
FG = (131*1.055 - 5.633) / 131
FG = (138.205 - 5.633) / 131 = 1.012
I have seen 129 used as the scaling factor as well, which would drop the ABV to 5.547%, but for homebrewing purposes either seem ok, or split the difference and use 130...
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Class 11
MATHS
Functions
# If 2f(x)-3f(1/x)=x^2(x!=0), then f(2) is equal to (a)-7/4("b")""5/2 (c) -1 (d) none of these
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 24-5-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
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## Very Important Questions
Class 11th
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2f(x) - 3f(1/x) = x^2<br> When x = 2,<br> 2f(2)-3f(1/2) = 4 ->(1)<br> When x = 1/2,<br> 2f(1/2) - 3f(2) = 1/4->(2)<br> Nw, multiplying (1) with 2 and (2) with 3 and then adding (1) and (2),<br> 4f(2)-6f(1/2)+6f(1/2)-9f(2) = 8+3/4<br> =>-5f(2) = 35/4<br> =>f(2) = -7/4<br> So, option (a) is the correct ooption.<br>
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# Problems with Boundary Profile
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February 22, 2005, 09:12 Problems with Boundary Profile #1 hermann Guest Posts: n/a Dear fellow Fluent users, I am simulating the air flow and temperature in a room. Due to memory restrictions I simulated a small zone (box) around the air ducts in a seperate modell with the boundaries as pressure outlets. From this modell I wrote the x,y and z velocities, the temperature, turb. kin. energy and dissipation rate in a Profile file. However after importing the profile into the model of the whole room there are significant variations in the velocity vectors and minor variations in the mass fluxes. To be more specific, the velo magnitude an the edges of the box are about 1.3 m/s, the corresponding value of the boundary profile in the room model is about 0.4 m/s!! (the mesh of the small box is 4 times finer then the mesh of the room, so there will be some interpolation error). I searched the forum for similar problems but couldn't find a solution. Any advice would be greatly appreciated.. Regards K.D. Hermann
March 2, 2005, 12:00 Re: Problems with Boundary Profile #2 reda gad Guest Posts: n/a i would ask you a question What is the boundary condition at the inlet of the room? i think that you have to use veloity inlet bc. and use profiles of flow direction cosines (ni , nj, and nk) in addition to your profiles to specify flow direction at inlet of room.
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Heat transfer by convection causes the spiral over the light bulb to
## Heat transfer by convection causes the spiral over the light bulb to
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# Generalized Null Bertrand Curves In Minkowski Space-Time
Generalized Null Bertrand Curves In Minkowski Space-Time Coken and Ciftci proved that a null Cartan curve in Minkowski space¸¨ ¸ time E4 is a null Bertrand curve if and only if k2 is nonzero constant and k3 is zero. 1 That is, the null curve with non-zero curvature k3 is not a Bertrand curve in Minkowski space-time E4 . 1 So, in this paper we defined a new type of Bertrand curve in Minkowski space-time E4 for a null curve with non-zero curvature k3 by using the similar idea of generalized 1 Bertrand curve given by Matsuda and Yorozu and we called it a null (1, 3)-Bertrand curve. Also, we proved that if a null curve with non-zero curvatures in Minkowski spacetime E4 is a null (1, 3)-Bertrand curve then it is a null helix. We give an example of such 1 curves. Mathematics Subject Classification 2010: 53C50, 53B30. Key words: Minkowski space-time, null curve, Frenet vectors, Bertrand curves. 1. Introduction In the classical differential geometry of curves in Euclidean space, Saint Venant (see [15]) proposed the question whether upon the surface generated by the principal normal of a curve, a second curve can exist which has for its principal normal the principal normal of the given curve. This question was answered by Bertrand in 1850 in a paper (see [3]) in which he showed that a necessary and sufficient condition for the existence of such a second curve is that a linear relationship with constant coefficients exists between the first and second curvatures of the given original curve. In other ¨ 490 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN words, if we denote first and second curvatures of a given curve by k1 and k2 , respectively then we have k1 + µk2 = 1 where , µ R. Since Bertrand published his paper, curve pairs of this type have been called Conjugate Bertrand Curves or more commonly Bertrand Curves (see [11]). In 1888, Bioche [4] gave a new theorem to obtain Bertrand curves by using the given two curves C1 and C2 in Euclidean 3-space. Later, Burke [6] gave a theorem related with Bioche's theorem on Bertrand curves. The following properties of Bertrand curves are well known: If two curves have the same principal normals: (i) corresponding points are a fixed distance apart; (ii) the tangents at corresponding points are at a fixed angle. These well known properties of Bertrand curves in Euclidean 3space were extended by Pears in [14] to Riemannian n-space and found general results for Bertrand curves. When we apply these general results to Euclidean n-space, it is easily found that either k2 or k3 is zero. In other words, Bertrand curves in En (n > 3) are degenerate curves. This result was restated in [12] by Matsuda and Yorozu. They proved that there is no special Bertrand curve in En (n > 3) and they defined new type which is called (1, 3)-type Bertrand curve in 4-dimensional Euclidean space. In differential geometry of curves in Minkowski space, there are three different kinds of curves called spacelike, timelike and null ( lightlike) depending on their causal characters. Many of the classical results from Riemannian geometry have Lorentz counterparts. In fact, spacelike curves or timelike curves can be studied by a similar approach to that in positive definite Riemannian geometry. However, since the induced metric of a null curve is degenerate, this case is much more complicated and also different from a non-degenerate case. The presence of null curves often causes important and interesting differences as will be the case in the present study. In Minkowski 3-space (also in a Lorentzian manifold), spacelike and timelike Bertrand curves and their characterizations were studied in [2, 7, 9, 10]. Null Bertrand curves in Minkowski 3- space were studied by Balgetir, Bektas and Inoguchi in [1] and they proved the following theorem for a ¸ null Cartan curve to be a Bertrand curve: Theorem A. Let be a Cartan framed null curve. Then it is a Bertrand curve if and only if is a null geodesic or a Cartan framed null curve with constant second curvature k2 . Null Bertrand curves were studied in a Lorentzian manifold by Jin [10]. Coken and Ciftci [5] proved the following theorem for Bertrand curves ¸ ¸ in Minkowski space-time Theorem B. A Cartan curve in Minkowski space-time E4 is a null 1 Bertrand curve if and only if the curvature k2 is a non-zero constant and k3 is zero. That is, the null curve with non-zero curvature k3 is not a Bertrand ¨¸ curve in Minkowski space-time E4 . Also, Gocmen and Keles [8] obtained ¸ 1 some new results for Cartan framed null Bertrand curves in R4 . 1 In this paper, we define a new type of Bertrand curve in Minkowski space-time E4 for a null curve with non-zero curvature k3 by using the 1 similar idea of generalized Bertrand curve given by Matsuda and Yorozu [12] and we called it a null (1, 3)-Bertrand curve. Also, we prove that if a null curve with non-zero curvatures in Minkowski space-time E4 is a null 1 (1, 3)-Bertrand curve then it is a null helix. Also, some properties of null (1, 3)-Bertrand curves in Minkowski space-time are given. We complete the paper with an example of such curves. 2. Preliminaries The Lorentzian 4 -space E4 is the Euclidean 4-space E4 equipped with 1 indefinite flat metric given by g= -dx2 1 i=2 dx2 , i where (x1 , x2 , x3 , x4 ) is a rectangular coordinate system of E4 . Recall that 1 a vector v E4 \{0} is spacelike if g(v, v) > 0, timelike if g(v, v) < 0 and 1 null (lightlike) if g(v, v) = 0 and v = 0. In particular, the vector v = 0 is a spacelike vector. The norm of a vector v is given by ||v||L = |g(v, v)| and two vectors v and w are said to be orthogonal if g(v, w) = 0. An arbitrary curve in E4 can locally be spacelike, timelike or null (lightlike) 1 if all of its velocity vectors (s) are spacelike, timelike or null, respectively. Recall that a spacelike curve in E4 is called pseudo-null curve or partially1 null curve if its principal normal vector is null and its first binormal vector is null, respectively. A spacelike or a timelike curve has unit speed, if g( (s), (s)) = ±1 ([13]). A null curve is parametrized by arclength function s if g( (s), (s)) = 1. In particular, a pseudo-null or a partiallynull curve has unit speed if g( (s), (s)) = 1. ¨ 492 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN Let {T, N1 , N2 , N3 } be the moving Frenet frame along a null curve in E4 . Then the Frenet frame field of the curve satisfies the following Frenet 1 equations: T N1 N2 N3 = = = = k1 N1 , k2 T - k1 N2 , -k2 N1 + k3 N3 , -k3 T (2.1) where the first curvature k1 (s) = 0 if is a straight line or k1 (s) = 1 in all other cases [16]. Such curve has two non zero curvatures k2 (s) and k3 (s) . Moreover, this moving Frenet frame {T, N1 , N2 , N3 } satisfies the following conditions: g(T, T ) = g(N2 , N2 ) = 0, g(N1 , N1 ) = g(N3 , N3 ) = 1, g(T, N1 ) = g(T, N3 ) = g(N1 , N3 ) = g(N1 , N2 ) = g(N2 , N3 )=0, g(T, N2 )=1. In this study we consider the curve is not a straight line, that is, the first curvature of the curve is equal to one. 3. Null (1, 3)-bertrand curves in minkowski space-time From [5], we know that a null curve with non-zero curvature k3 is not a Bertrand curve in Minkowski space-time E4 . In this section, we give the 1 definition of null (1, 3)- Bertrand curve for a null curve with non-zero curvature k3 in E4 . Also we give some characterizations of such curves. 1 Definition 3.1. Let : I R E4 and : I R E4 be null curves 1 1 with curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and k1 ((s)), k 2 ((s)), k3 ((s)), respectively, where : I I, s = (s) is a regular C -function such that each point (s) of the curve corresponds to the point (s) = ((s)) of the curve for all s I. If the Frenet (1, 3)-normal plane at each point (s) of the curve coincides with the Frenet (1, 3)-normal plane at corresponding point (s) = ((s)) of the curve for all s I then is called a null (1, 3)-Bertrand curve in E4 and is called a null (1, 3)-Bertrand mate of 1 the curve . Theorem 3.1. Let : I R E4 be a null curve with curvature 1 functions k1 (s) = 1, k2 (s) and k3 (s) = 0. Then is a null (1, 3)-Bertrand curve if and only if there exist constant real numbers , , and µ = 0 satisfying (3.1-a) (3.1-b) (3.1-c) (3.1-d) for all s I. Proof. We assume that is a null (1, 3)-Bertrand curve parametrized by arc-length s and is the null (1, 3)-Bertrand mate of the curve with arc-length s. Then we can write the curve as (3.2) (s) = ( (s)) = (s) + (s)N1 (s) + µ(s)N3 (s) = 0, 1 + k2 (s) - µk3 (s) = 0, (k2 (s))2 + (k3 (s))2 = - k2 (s) = , k3 (s) 2 , 4 for all s I where (s) and µ(s) are C -functions on I. Differentiating (3.2) with respect to s and by using the Frenet equations, we have (3.3) T ( (s)) (s) = [1 + (s) k2 (s) - µ (s) k3 (s)] T (s) + (s)N1 (s) -(s)k1 (s)N2 (s) + µ (s)N3 (s) for all s I. Since the plane spanned by N1 (s) and N3 (s) coincides with the plane spanned by N 1 ( (s)) and N 3 ( (s)), we can write (3.4) (3.5) N 1 ( (s)) = cos (s) N1 (s) + sin (s) N3 (s), N 3 ( (s)) = - sin (s) N1 (s) + cos (s) N3 (s). And then by using (3.4) and (3.5), we have g(N 1 ( (s)), T ( (s)) (s)) = (s) cos (s) + µ (s) sin (s) = 0, g(N 3 ( (s)), T ( (s)) (s)) = - (s) sin (s) + µ (s) cos (s) = 0. Thus we get (s) = 0, µ (s) = 0. That is, and µ are constant functions on I. So, we can rewrite (3.2) and (3.3) for all s I as follows: (3.6) (s) = ( (s)) = (s) + N1 (s) + µN3 (s) ¨ 494 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN and (3.7) T ( (s)) (s) = [1 + k2 (s) - µk3 (s)] T (s) - k1 (s)N2 (s). (s) = 1 + k2 (s) - µk3 (s) k1 (s) , (s) = - (s) (s) If we denote (3.8) for all s I. We can easily obtain that (3.9) T ( (s)) = (s)T (s) + (s)N2 (s), where (s) and (s) are C -functions on I. Since T ( (s)) , T (s) and N2 (s) are lightlike vectors, we get (3.10) (s)(s) = 0. That is, 1 + k2 (s) - µk3 (s) = 0 or = 0. We assume that = 0 and 1 + k2 (s) - µk3 (s) = 0. In that case, we can write T ( (s)) = (s)T (s) and if we differentiate the last equation with respect to s, we get dT ((s)) (s) = ds (s)T (s) + (s)T (s). By using the Frenet equations of and null curves, we have k1 ((s)))N 1 ( (s)) (s) = (s)T (s)+ (s)k1 (s)N1 (s). From (3.4), it holds (s) = 0. That is, (s) is non-zero constant function on I. So we get k1 ((s))N 1 ( (s)) (s) = k1 (s)N1 (s), where (s) = || . Since the null curves and are not straight lines, the principal curvature functions of the null curves and are equal to one. That is, k1 ((s)) = 1 and k1 (s) = 1. So we have N 1 ( (s)) = ±N1 (s), for all s I. This implies that is a null Bertrand curve. But by Theorem B, this fact is a contradiction. Thus we must consider only the case of = 0 and 1 + k2 (s) - µk3 (s) = 0. Then we obtain the relations (3.1-a) and (3.1-b). Hence we can write (3.11) T ( (s)) = (s)N2 (s). Differentiating (3.11) and by using the Frenet equations, we obtain (3.12) k1 ((s)))N 1 ( (s)) (s) = -(s)k2 (s)N1 (s) + (s)N2 (s) + (s)k3 (s)N3 (s). Since N 1 ( (s)) is expressed by linear combination of N1 (s) and N3 (s), it holds that (s) = 0. That is, (s) is a non-zero constant function. Also from (3.8), we can write (3.13) (s) = - . Since = 0, it follows (s) = 0. Hence there exists a regular map : I I defined by s = (s) = - s + , where is a real constant. We can rewrite (3.12) as k1 ((s)))N 1 ( (s)) (s) = -k2 (s)N1 (s) + k3 (s)N3 (s). and we can easily see that (3.14) (s) k1 ((s)) = 2 (k2 (s))2 + (k3 (s))2 . By substituting (3.13) into (3.14) and using k1 ((s)) = 1, we obtain the relation (3.1-c). From (3.12), we have N 1 ( (s)) = - where (3.15) cos (s) = - k2 (s) , (s) sin (s) = k3 (s) . (s) k2 (s) k3 (s) N1 (s) + N3 (s), (s) (s) Differentiating (3.4) with respect to s and using the Frenet equations, we obtain (3.16) (s) k2 ((s))T ((s)) - (s) k1 ((s))N 2 ( (s)) = (cos (s)) N1 (s) + (sin (s)) N3 (s) + (cos (s) k2 (s) - sin (s) k3 (s)) T (s) - cos (s) k1 (s)N2 (s) for all s I. From the above fact, it holds (3.17) (cos (s)) = 0, (sin (s)) = 0. That is, is a constant function on L with value 0 . Let = (cos 0 ) (sin 0 )-1 k2 (s) be a constant number. So from (3.15) , we get k3 (s) = -. Thus we obtain the relation (3.1-d). Conversely, we assume that : I R E4 is a null curve with cur1 vature functions k1 (s) = 1, k2 (s) and k3 (s) = 0 satisfying the relation (3.1 - a) , (3.1 - b) , (3.1 - c) and (3.1 - d) for constant numbers , , and µ. Then we define a null curve : I R E4 such as 1 (3.18) (s) = (s) + N1 (s) + µN3 (s) for all s I. Differentiating (3.18) with respect to s and by using the Frenet equations, we have d (s) = (1 + k2 (s) - µk3 (s)) T (s) + (-k1 (s)) N2 (s). ds ¨ 496 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN By using (3.1-b), we obtain d (s) = -k1 (s)N2 (s) ds for all s I. From (3.1-a), it follows d(s) = 0. We consider that the curve ds is not a straight line, that is, the first curvature of the curve is equal to one. So we can write (3.19) (s) = -N2 (s). (s) = (k2 (s)N1 (s) - k3 (s)N3 (s)) . There exists a regular map : I I defined by Differentiating (3.19) with respect to s, we get s = (s) = g (t), (t) dt, (s I) where s denotes the pseudo-arc length parameter of the curve . From (3.1-c), we obtain (3.20) (s) = for all s I, where : I I is a regular C -function and = 1, -1, >0 . <0 Thus the curve is rewritten as follows: (3.21) (s) = ((s)) = (s) + N1 (s) + µN3 (s). Differentiating (3.21) with respect to s, we obtain (s) From (3.1-b) (3.22) T ((s)) = - N2 (s), (s) d (s) ds = (1 + k2 (s) - µk3 (s)) T (s) - N2 (s). s=(s) 9 where T ((s)) = (3.23) d(s) ds . By substituting (3.20) into (3.22), we get T ((s)) = -N2 (s) for all s I. Differentiating (3.23) with respect to s and by using the Frenet equations, we have dT ((s)) k2 (s) k3 (s) = N1 (s) - N3 (s) ds (s) (s) and (3.24) 2 (k2 (s))2 + (k3 (s))2 dT ((s)) = . ds ( (s))2 Since the curve is a null (1, 3)-Bertrand curve, the curvatures of satisfy (3.1-c). If we substitute (3.1-c) into (3.24), we get (3.25) k1 ((s)) = dT ((s)) = 1. ds Then we can define a unit vector field N 1 ((s)) along the curve by N 1 ((s)) = = 1 dT ((s)) ds k1 ((s)) k2 (s) k3 (s) N1 (s) - N3 (s) . (s) (s) Since N 1 ((s)) is expressed by linear combination of N1 (s) and N3 (s), we can put N 1 ((s)) = cos (s) N1 (s) + sin (s) N3 (s) , where (3.26) cos (s) = k2 (s) (s) sin (s) = - k3 (s) (s) for all s I and (s) is a C -function. (3.1-c) and (3.1-d) imply that the curvatures of the curve k2 (s) and k3 (s) are constants. On the other hand, (s) = =constant. Thus cos (s) and sin (s) are constants, that is, (s) = 0 =constant. We can rewrite as (3.27) N 1 ((s)) = cos 0 N1 (s) + sin 0 N3 (s) . ¨ 498 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN Differentiating (3.27) with respect to s and by using the Frenet equations, we have dN 1 ((s)) (s) = (k2 cos 0 - k3 sin 0 ) T (s) - cos 0 N2 (s) . ds By using (3.26) in the above equality, we get k2 dN 1 ((s)) = T (s) - N2 (s) ds ( (s))2 for all s I. From the Frenet equations (3.28) 1 k2 ((s)) = - g 2 dN 1 ((s)) dN 1 ((s)) , ds ds = k2 . ( (s))2 Thus we can define a unit vector field N 2 ((s)) along the curve by N 2 ((s)) = k2 ((s))T ((s)) - that is, N 2 ((s)) = - T (s) dN 1 ((s)) , ds Next we can define a unit vector N 3 ( (s)) along the curve by N 3 ((s)) = - sin 0 N1 (s) + cos 0 N3 (s) , that is, N 3 ((s)) = Thus we obtain (3.29) k3 ((s)) = -g dN 3 (s) , N 2 (s) ds =- k3 (s) . ( (s))2 k2 (s) k3 (s) N1 (s) + N3 (s) . (s) (s) Notice that g T , T = g N 2 , N 2 = 0, g N 1 , N 1 = g N 3 , N 3 = 1 and g T , N 1 = g T , N 3 = g N 1 , N 3 = g N 1 , N 2 = g N 2 , N 3 = 0, g T , N 2 = 1 for all s I, where T , N 1 , N 2 , N 3 is moving Frenet 4 frame along null curve in E1 . And it is trivial that the Frenet (1,3)normal plane at each point (s) of the curve coincides with the Frenet (1,3)-normal plane at corresponding point (s) of the curve . Hence is a null (1,3)- Bertrand curve in E4 . This completes the proof. 1 Corollary 3.1. Let : I R E4 be a null (1, 3)-Bertrand curve 1 with curvature functions k1 (s) = 1, k2 (s), k3 (s) = 0 and : I R E4 be 1 a null (1, 3)-Bertrand mate of the curve with curvature functions k1 (s), k2 (s), k3 (s) where s and s denote the arc-length parameter of the curves and , respectively. Then the relations between these curvature functions are k2 (s) k3 (s) k1 ((s)) = 1, k 2 ((s)) = , 2 , k 3 ((s)) = - ( (s)) ( (s))2 where : I I, s = (s) is a regular C -function for all s I. Proof. It is obvious from the Theorem (3.1). Corollary 3.2. Let : I R E4 be a null (1, 3)-Bertrand curve 1 with curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and be a null (1, 3)-Bertrand mate of the curve and : I I, s = (s) is a regular C -function such that each point (s) of the curve corresponds to the point (s) = ((s)) of the curve for all s I. Then the distance between the points (s) and (s) is constant for all s I. Proof. Let : I R E4 be a null (1, 3)-Bertrand curve with 1 curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and be a null (1, 3)-Bertrand mate of the curve . We assume that is distinct from . Let the pairs of (s) and (s) = ((s)) (where : I I, s = (s) is a regular C -function) be corresponding points of the curves and . Then we can write (s) = ( (s)) = (s) + N1 (s) + µN3 (s), where and µ are non-zero constants. Thus we can write (s) - (s) = N1 (s) + µN3 (s) and (s) - (s) = 2 + µ2 . So, d ( (s) , (s)) = constant. This completes the proof. Corollary 3.3. Let : I R E4 be a null (1, 3)-Bertrand curve 1 with curvature functions k1 (s) = 1, k2 (s), k3 (s) = 0 and be a null (1, 3)Bertrand mate of the curve with curvature functions k1 ((s)), k2 ((s)), k3 ((s)). Then the curvatures of the curves and are constants. Proof. We assume that : I R E4 is a null (1, 3)-Bertrand curve 1 with curvatures k1 (s) = 1, k2 (s) = 0, k3 (s) = 0 and is a null (1, 3)Bertrand mate of the curve . In that case, the relations (3.1-c) and (3.1-d) imply that k2 (s), k3 (s) are constant. From the Corollary 3.1, it can be easily seen that the curvatures of the curve are constant curvatures. ¨ 500 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN Corollary 3.4. If : I R E4 is a null (1, 3)-Bertrand curve with 1 non-zero curvatures and is a null (1, 3)-Bertrand mate of then and are null helices. Proof. We assume that : I R E4 is a null (1, 3)-Bertrand curve 1 with curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and is a null (1, 3)-Bertrand mate of the curve . Then both the curvatures k2 (s), k3 (s) belong to the curve and the curvatures k2 ((s)), k3 ((s)) belong to the curve are nonzero constants. Hence they are null helices. Corollary 3.5. Let : I R E4 be a null (1, 3)-Bertrand curve 1 and be a null (1, 3)-Bertrand mate of . Then the curvatures k2 (s), k3 (s) belong to the curve and the curvatures k 2 ((s)), k3 ((s)) belong to the curve satisfy k 2 k3 + k2 k3 = 0. Proof. It is obvious from (3.28) and (3.29). 4 Example 3.1. Let be a null curve in E1 given by 1 (s) = (sinh (s) , cosh (s) , sin (s) , cos (s)) . 2 The Frenet frame of the curve is given by T (s) = N1 (s) = N2 (s) = N3 (s) = 1 (cosh (s) , sinh (s) , cos (s) , - sin (s)) , 2 1 (sinh (s) , cosh (s) , - sin (s) , - cos (s)) , 2 1 (- cosh s, - sinh s, cos (s) , - sin (s)) , 2 1 (sinh (s) , cosh (s) , sin (s) , cos (s)) . 2 Then we get the curvatures of as follows k1 (s) = 1, k2 (s) = 0, k3 (s) = -1. By using the Theorem B, the curve is not a null Bertrand curve. But if we take constants , , and µ as = 1, = 1, = 0, µ = -1, then it is trivial that the relations (3.1-a), (3.1-b), (3.1-c) and (3.1-d) hold. Therefore, the curve is a null (1, 3)-Bertrand curve in E4 . In this case, the 1 null (1, 3)-Bertrand mate of the curve is given by 1 (s) = (sinh (s) , cosh (s) , - sin (s) , - cos (s)) . 2 By using (3.25), (3.28), (3.29), we obtain the curvatures of the curve as k1 = 1, k2 = 0 and k 3 = 1. Acknowledgement. The authors would like to express their sincere gratitude to the referee for valuable suggestions to improve the paper. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Annals of the Alexandru Ioan Cuza University - Mathematics de Gruyter
# Generalized Null Bertrand Curves In Minkowski Space-Time
, Volume 60 (2) – Nov 24, 2014
14 pages
/lp/de-gruyter/generalized-null-bertrand-curves-in-minkowski-space-time-Ny0CGQoi0a
# References (16)
Publisher
de Gruyter
ISSN
1221-8421
eISSN
1221-8421
DOI
10.2478/aicu-2013-0031
Publisher site
See Article on Publisher Site
### Abstract
Coken and Ciftci proved that a null Cartan curve in Minkowski space¸¨ ¸ time E4 is a null Bertrand curve if and only if k2 is nonzero constant and k3 is zero. 1 That is, the null curve with non-zero curvature k3 is not a Bertrand curve in Minkowski space-time E4 . 1 So, in this paper we defined a new type of Bertrand curve in Minkowski space-time E4 for a null curve with non-zero curvature k3 by using the similar idea of generalized 1 Bertrand curve given by Matsuda and Yorozu and we called it a null (1, 3)-Bertrand curve. Also, we proved that if a null curve with non-zero curvatures in Minkowski spacetime E4 is a null (1, 3)-Bertrand curve then it is a null helix. We give an example of such 1 curves. Mathematics Subject Classification 2010: 53C50, 53B30. Key words: Minkowski space-time, null curve, Frenet vectors, Bertrand curves. 1. Introduction In the classical differential geometry of curves in Euclidean space, Saint Venant (see [15]) proposed the question whether upon the surface generated by the principal normal of a curve, a second curve can exist which has for its principal normal the principal normal of the given curve. This question was answered by Bertrand in 1850 in a paper (see [3]) in which he showed that a necessary and sufficient condition for the existence of such a second curve is that a linear relationship with constant coefficients exists between the first and second curvatures of the given original curve. In other ¨ 490 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN words, if we denote first and second curvatures of a given curve by k1 and k2 , respectively then we have k1 + µk2 = 1 where , µ R. Since Bertrand published his paper, curve pairs of this type have been called Conjugate Bertrand Curves or more commonly Bertrand Curves (see [11]). In 1888, Bioche [4] gave a new theorem to obtain Bertrand curves by using the given two curves C1 and C2 in Euclidean 3-space. Later, Burke [6] gave a theorem related with Bioche's theorem on Bertrand curves. The following properties of Bertrand curves are well known: If two curves have the same principal normals: (i) corresponding points are a fixed distance apart; (ii) the tangents at corresponding points are at a fixed angle. These well known properties of Bertrand curves in Euclidean 3space were extended by Pears in [14] to Riemannian n-space and found general results for Bertrand curves. When we apply these general results to Euclidean n-space, it is easily found that either k2 or k3 is zero. In other words, Bertrand curves in En (n > 3) are degenerate curves. This result was restated in [12] by Matsuda and Yorozu. They proved that there is no special Bertrand curve in En (n > 3) and they defined new type which is called (1, 3)-type Bertrand curve in 4-dimensional Euclidean space. In differential geometry of curves in Minkowski space, there are three different kinds of curves called spacelike, timelike and null ( lightlike) depending on their causal characters. Many of the classical results from Riemannian geometry have Lorentz counterparts. In fact, spacelike curves or timelike curves can be studied by a similar approach to that in positive definite Riemannian geometry. However, since the induced metric of a null curve is degenerate, this case is much more complicated and also different from a non-degenerate case. The presence of null curves often causes important and interesting differences as will be the case in the present study. In Minkowski 3-space (also in a Lorentzian manifold), spacelike and timelike Bertrand curves and their characterizations were studied in [2, 7, 9, 10]. Null Bertrand curves in Minkowski 3- space were studied by Balgetir, Bektas and Inoguchi in [1] and they proved the following theorem for a ¸ null Cartan curve to be a Bertrand curve: Theorem A. Let be a Cartan framed null curve. Then it is a Bertrand curve if and only if is a null geodesic or a Cartan framed null curve with constant second curvature k2 . Null Bertrand curves were studied in a Lorentzian manifold by Jin [10]. Coken and Ciftci [5] proved the following theorem for Bertrand curves ¸ ¸ in Minkowski space-time Theorem B. A Cartan curve in Minkowski space-time E4 is a null 1 Bertrand curve if and only if the curvature k2 is a non-zero constant and k3 is zero. That is, the null curve with non-zero curvature k3 is not a Bertrand ¨¸ curve in Minkowski space-time E4 . Also, Gocmen and Keles [8] obtained ¸ 1 some new results for Cartan framed null Bertrand curves in R4 . 1 In this paper, we define a new type of Bertrand curve in Minkowski space-time E4 for a null curve with non-zero curvature k3 by using the 1 similar idea of generalized Bertrand curve given by Matsuda and Yorozu [12] and we called it a null (1, 3)-Bertrand curve. Also, we prove that if a null curve with non-zero curvatures in Minkowski space-time E4 is a null 1 (1, 3)-Bertrand curve then it is a null helix. Also, some properties of null (1, 3)-Bertrand curves in Minkowski space-time are given. We complete the paper with an example of such curves. 2. Preliminaries The Lorentzian 4 -space E4 is the Euclidean 4-space E4 equipped with 1 indefinite flat metric given by g= -dx2 1 i=2 dx2 , i where (x1 , x2 , x3 , x4 ) is a rectangular coordinate system of E4 . Recall that 1 a vector v E4 \{0} is spacelike if g(v, v) > 0, timelike if g(v, v) < 0 and 1 null (lightlike) if g(v, v) = 0 and v = 0. In particular, the vector v = 0 is a spacelike vector. The norm of a vector v is given by ||v||L = |g(v, v)| and two vectors v and w are said to be orthogonal if g(v, w) = 0. An arbitrary curve in E4 can locally be spacelike, timelike or null (lightlike) 1 if all of its velocity vectors (s) are spacelike, timelike or null, respectively. Recall that a spacelike curve in E4 is called pseudo-null curve or partially1 null curve if its principal normal vector is null and its first binormal vector is null, respectively. A spacelike or a timelike curve has unit speed, if g( (s), (s)) = ±1 ([13]). A null curve is parametrized by arclength function s if g( (s), (s)) = 1. In particular, a pseudo-null or a partiallynull curve has unit speed if g( (s), (s)) = 1. ¨ 492 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN Let {T, N1 , N2 , N3 } be the moving Frenet frame along a null curve in E4 . Then the Frenet frame field of the curve satisfies the following Frenet 1 equations: T N1 N2 N3 = = = = k1 N1 , k2 T - k1 N2 , -k2 N1 + k3 N3 , -k3 T (2.1) where the first curvature k1 (s) = 0 if is a straight line or k1 (s) = 1 in all other cases [16]. Such curve has two non zero curvatures k2 (s) and k3 (s) . Moreover, this moving Frenet frame {T, N1 , N2 , N3 } satisfies the following conditions: g(T, T ) = g(N2 , N2 ) = 0, g(N1 , N1 ) = g(N3 , N3 ) = 1, g(T, N1 ) = g(T, N3 ) = g(N1 , N3 ) = g(N1 , N2 ) = g(N2 , N3 )=0, g(T, N2 )=1. In this study we consider the curve is not a straight line, that is, the first curvature of the curve is equal to one. 3. Null (1, 3)-bertrand curves in minkowski space-time From [5], we know that a null curve with non-zero curvature k3 is not a Bertrand curve in Minkowski space-time E4 . In this section, we give the 1 definition of null (1, 3)- Bertrand curve for a null curve with non-zero curvature k3 in E4 . Also we give some characterizations of such curves. 1 Definition 3.1. Let : I R E4 and : I R E4 be null curves 1 1 with curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and k1 ((s)), k 2 ((s)), k3 ((s)), respectively, where : I I, s = (s) is a regular C -function such that each point (s) of the curve corresponds to the point (s) = ((s)) of the curve for all s I. If the Frenet (1, 3)-normal plane at each point (s) of the curve coincides with the Frenet (1, 3)-normal plane at corresponding point (s) = ((s)) of the curve for all s I then is called a null (1, 3)-Bertrand curve in E4 and is called a null (1, 3)-Bertrand mate of 1 the curve . Theorem 3.1. Let : I R E4 be a null curve with curvature 1 functions k1 (s) = 1, k2 (s) and k3 (s) = 0. Then is a null (1, 3)-Bertrand curve if and only if there exist constant real numbers , , and µ = 0 satisfying (3.1-a) (3.1-b) (3.1-c) (3.1-d) for all s I. Proof. We assume that is a null (1, 3)-Bertrand curve parametrized by arc-length s and is the null (1, 3)-Bertrand mate of the curve with arc-length s. Then we can write the curve as (3.2) (s) = ( (s)) = (s) + (s)N1 (s) + µ(s)N3 (s) = 0, 1 + k2 (s) - µk3 (s) = 0, (k2 (s))2 + (k3 (s))2 = - k2 (s) = , k3 (s) 2 , 4 for all s I where (s) and µ(s) are C -functions on I. Differentiating (3.2) with respect to s and by using the Frenet equations, we have (3.3) T ( (s)) (s) = [1 + (s) k2 (s) - µ (s) k3 (s)] T (s) + (s)N1 (s) -(s)k1 (s)N2 (s) + µ (s)N3 (s) for all s I. Since the plane spanned by N1 (s) and N3 (s) coincides with the plane spanned by N 1 ( (s)) and N 3 ( (s)), we can write (3.4) (3.5) N 1 ( (s)) = cos (s) N1 (s) + sin (s) N3 (s), N 3 ( (s)) = - sin (s) N1 (s) + cos (s) N3 (s). And then by using (3.4) and (3.5), we have g(N 1 ( (s)), T ( (s)) (s)) = (s) cos (s) + µ (s) sin (s) = 0, g(N 3 ( (s)), T ( (s)) (s)) = - (s) sin (s) + µ (s) cos (s) = 0. Thus we get (s) = 0, µ (s) = 0. That is, and µ are constant functions on I. So, we can rewrite (3.2) and (3.3) for all s I as follows: (3.6) (s) = ( (s)) = (s) + N1 (s) + µN3 (s) ¨ 494 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN and (3.7) T ( (s)) (s) = [1 + k2 (s) - µk3 (s)] T (s) - k1 (s)N2 (s). (s) = 1 + k2 (s) - µk3 (s) k1 (s) , (s) = - (s) (s) If we denote (3.8) for all s I. We can easily obtain that (3.9) T ( (s)) = (s)T (s) + (s)N2 (s), where (s) and (s) are C -functions on I. Since T ( (s)) , T (s) and N2 (s) are lightlike vectors, we get (3.10) (s)(s) = 0. That is, 1 + k2 (s) - µk3 (s) = 0 or = 0. We assume that = 0 and 1 + k2 (s) - µk3 (s) = 0. In that case, we can write T ( (s)) = (s)T (s) and if we differentiate the last equation with respect to s, we get dT ((s)) (s) = ds (s)T (s) + (s)T (s). By using the Frenet equations of and null curves, we have k1 ((s)))N 1 ( (s)) (s) = (s)T (s)+ (s)k1 (s)N1 (s). From (3.4), it holds (s) = 0. That is, (s) is non-zero constant function on I. So we get k1 ((s))N 1 ( (s)) (s) = k1 (s)N1 (s), where (s) = || . Since the null curves and are not straight lines, the principal curvature functions of the null curves and are equal to one. That is, k1 ((s)) = 1 and k1 (s) = 1. So we have N 1 ( (s)) = ±N1 (s), for all s I. This implies that is a null Bertrand curve. But by Theorem B, this fact is a contradiction. Thus we must consider only the case of = 0 and 1 + k2 (s) - µk3 (s) = 0. Then we obtain the relations (3.1-a) and (3.1-b). Hence we can write (3.11) T ( (s)) = (s)N2 (s). Differentiating (3.11) and by using the Frenet equations, we obtain (3.12) k1 ((s)))N 1 ( (s)) (s) = -(s)k2 (s)N1 (s) + (s)N2 (s) + (s)k3 (s)N3 (s). Since N 1 ( (s)) is expressed by linear combination of N1 (s) and N3 (s), it holds that (s) = 0. That is, (s) is a non-zero constant function. Also from (3.8), we can write (3.13) (s) = - . Since = 0, it follows (s) = 0. Hence there exists a regular map : I I defined by s = (s) = - s + , where is a real constant. We can rewrite (3.12) as k1 ((s)))N 1 ( (s)) (s) = -k2 (s)N1 (s) + k3 (s)N3 (s). and we can easily see that (3.14) (s) k1 ((s)) = 2 (k2 (s))2 + (k3 (s))2 . By substituting (3.13) into (3.14) and using k1 ((s)) = 1, we obtain the relation (3.1-c). From (3.12), we have N 1 ( (s)) = - where (3.15) cos (s) = - k2 (s) , (s) sin (s) = k3 (s) . (s) k2 (s) k3 (s) N1 (s) + N3 (s), (s) (s) Differentiating (3.4) with respect to s and using the Frenet equations, we obtain (3.16) (s) k2 ((s))T ((s)) - (s) k1 ((s))N 2 ( (s)) = (cos (s)) N1 (s) + (sin (s)) N3 (s) + (cos (s) k2 (s) - sin (s) k3 (s)) T (s) - cos (s) k1 (s)N2 (s) for all s I. From the above fact, it holds (3.17) (cos (s)) = 0, (sin (s)) = 0. That is, is a constant function on L with value 0 . Let = (cos 0 ) (sin 0 )-1 k2 (s) be a constant number. So from (3.15) , we get k3 (s) = -. Thus we obtain the relation (3.1-d). Conversely, we assume that : I R E4 is a null curve with cur1 vature functions k1 (s) = 1, k2 (s) and k3 (s) = 0 satisfying the relation (3.1 - a) , (3.1 - b) , (3.1 - c) and (3.1 - d) for constant numbers , , and µ. Then we define a null curve : I R E4 such as 1 (3.18) (s) = (s) + N1 (s) + µN3 (s) for all s I. Differentiating (3.18) with respect to s and by using the Frenet equations, we have d (s) = (1 + k2 (s) - µk3 (s)) T (s) + (-k1 (s)) N2 (s). ds ¨ 496 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN By using (3.1-b), we obtain d (s) = -k1 (s)N2 (s) ds for all s I. From (3.1-a), it follows d(s) = 0. We consider that the curve ds is not a straight line, that is, the first curvature of the curve is equal to one. So we can write (3.19) (s) = -N2 (s). (s) = (k2 (s)N1 (s) - k3 (s)N3 (s)) . There exists a regular map : I I defined by Differentiating (3.19) with respect to s, we get s = (s) = g (t), (t) dt, (s I) where s denotes the pseudo-arc length parameter of the curve . From (3.1-c), we obtain (3.20) (s) = for all s I, where : I I is a regular C -function and = 1, -1, >0 . <0 Thus the curve is rewritten as follows: (3.21) (s) = ((s)) = (s) + N1 (s) + µN3 (s). Differentiating (3.21) with respect to s, we obtain (s) From (3.1-b) (3.22) T ((s)) = - N2 (s), (s) d (s) ds = (1 + k2 (s) - µk3 (s)) T (s) - N2 (s). s=(s) 9 where T ((s)) = (3.23) d(s) ds . By substituting (3.20) into (3.22), we get T ((s)) = -N2 (s) for all s I. Differentiating (3.23) with respect to s and by using the Frenet equations, we have dT ((s)) k2 (s) k3 (s) = N1 (s) - N3 (s) ds (s) (s) and (3.24) 2 (k2 (s))2 + (k3 (s))2 dT ((s)) = . ds ( (s))2 Since the curve is a null (1, 3)-Bertrand curve, the curvatures of satisfy (3.1-c). If we substitute (3.1-c) into (3.24), we get (3.25) k1 ((s)) = dT ((s)) = 1. ds Then we can define a unit vector field N 1 ((s)) along the curve by N 1 ((s)) = = 1 dT ((s)) ds k1 ((s)) k2 (s) k3 (s) N1 (s) - N3 (s) . (s) (s) Since N 1 ((s)) is expressed by linear combination of N1 (s) and N3 (s), we can put N 1 ((s)) = cos (s) N1 (s) + sin (s) N3 (s) , where (3.26) cos (s) = k2 (s) (s) sin (s) = - k3 (s) (s) for all s I and (s) is a C -function. (3.1-c) and (3.1-d) imply that the curvatures of the curve k2 (s) and k3 (s) are constants. On the other hand, (s) = =constant. Thus cos (s) and sin (s) are constants, that is, (s) = 0 =constant. We can rewrite as (3.27) N 1 ((s)) = cos 0 N1 (s) + sin 0 N3 (s) . ¨ 498 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN Differentiating (3.27) with respect to s and by using the Frenet equations, we have dN 1 ((s)) (s) = (k2 cos 0 - k3 sin 0 ) T (s) - cos 0 N2 (s) . ds By using (3.26) in the above equality, we get k2 dN 1 ((s)) = T (s) - N2 (s) ds ( (s))2 for all s I. From the Frenet equations (3.28) 1 k2 ((s)) = - g 2 dN 1 ((s)) dN 1 ((s)) , ds ds = k2 . ( (s))2 Thus we can define a unit vector field N 2 ((s)) along the curve by N 2 ((s)) = k2 ((s))T ((s)) - that is, N 2 ((s)) = - T (s) dN 1 ((s)) , ds Next we can define a unit vector N 3 ( (s)) along the curve by N 3 ((s)) = - sin 0 N1 (s) + cos 0 N3 (s) , that is, N 3 ((s)) = Thus we obtain (3.29) k3 ((s)) = -g dN 3 (s) , N 2 (s) ds =- k3 (s) . ( (s))2 k2 (s) k3 (s) N1 (s) + N3 (s) . (s) (s) Notice that g T , T = g N 2 , N 2 = 0, g N 1 , N 1 = g N 3 , N 3 = 1 and g T , N 1 = g T , N 3 = g N 1 , N 3 = g N 1 , N 2 = g N 2 , N 3 = 0, g T , N 2 = 1 for all s I, where T , N 1 , N 2 , N 3 is moving Frenet 4 frame along null curve in E1 . And it is trivial that the Frenet (1,3)normal plane at each point (s) of the curve coincides with the Frenet (1,3)-normal plane at corresponding point (s) of the curve . Hence is a null (1,3)- Bertrand curve in E4 . This completes the proof. 1 Corollary 3.1. Let : I R E4 be a null (1, 3)-Bertrand curve 1 with curvature functions k1 (s) = 1, k2 (s), k3 (s) = 0 and : I R E4 be 1 a null (1, 3)-Bertrand mate of the curve with curvature functions k1 (s), k2 (s), k3 (s) where s and s denote the arc-length parameter of the curves and , respectively. Then the relations between these curvature functions are k2 (s) k3 (s) k1 ((s)) = 1, k 2 ((s)) = , 2 , k 3 ((s)) = - ( (s)) ( (s))2 where : I I, s = (s) is a regular C -function for all s I. Proof. It is obvious from the Theorem (3.1). Corollary 3.2. Let : I R E4 be a null (1, 3)-Bertrand curve 1 with curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and be a null (1, 3)-Bertrand mate of the curve and : I I, s = (s) is a regular C -function such that each point (s) of the curve corresponds to the point (s) = ((s)) of the curve for all s I. Then the distance between the points (s) and (s) is constant for all s I. Proof. Let : I R E4 be a null (1, 3)-Bertrand curve with 1 curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and be a null (1, 3)-Bertrand mate of the curve . We assume that is distinct from . Let the pairs of (s) and (s) = ((s)) (where : I I, s = (s) is a regular C -function) be corresponding points of the curves and . Then we can write (s) = ( (s)) = (s) + N1 (s) + µN3 (s), where and µ are non-zero constants. Thus we can write (s) - (s) = N1 (s) + µN3 (s) and (s) - (s) = 2 + µ2 . So, d ( (s) , (s)) = constant. This completes the proof. Corollary 3.3. Let : I R E4 be a null (1, 3)-Bertrand curve 1 with curvature functions k1 (s) = 1, k2 (s), k3 (s) = 0 and be a null (1, 3)Bertrand mate of the curve with curvature functions k1 ((s)), k2 ((s)), k3 ((s)). Then the curvatures of the curves and are constants. Proof. We assume that : I R E4 is a null (1, 3)-Bertrand curve 1 with curvatures k1 (s) = 1, k2 (s) = 0, k3 (s) = 0 and is a null (1, 3)Bertrand mate of the curve . In that case, the relations (3.1-c) and (3.1-d) imply that k2 (s), k3 (s) are constant. From the Corollary 3.1, it can be easily seen that the curvatures of the curve are constant curvatures. ¨ 500 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN Corollary 3.4. If : I R E4 is a null (1, 3)-Bertrand curve with 1 non-zero curvatures and is a null (1, 3)-Bertrand mate of then and are null helices. Proof. We assume that : I R E4 is a null (1, 3)-Bertrand curve 1 with curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and is a null (1, 3)-Bertrand mate of the curve . Then both the curvatures k2 (s), k3 (s) belong to the curve and the curvatures k2 ((s)), k3 ((s)) belong to the curve are nonzero constants. Hence they are null helices. Corollary 3.5. Let : I R E4 be a null (1, 3)-Bertrand curve 1 and be a null (1, 3)-Bertrand mate of . Then the curvatures k2 (s), k3 (s) belong to the curve and the curvatures k 2 ((s)), k3 ((s)) belong to the curve satisfy k 2 k3 + k2 k3 = 0. Proof. It is obvious from (3.28) and (3.29). 4 Example 3.1. Let be a null curve in E1 given by 1 (s) = (sinh (s) , cosh (s) , sin (s) , cos (s)) . 2 The Frenet frame of the curve is given by T (s) = N1 (s) = N2 (s) = N3 (s) = 1 (cosh (s) , sinh (s) , cos (s) , - sin (s)) , 2 1 (sinh (s) , cosh (s) , - sin (s) , - cos (s)) , 2 1 (- cosh s, - sinh s, cos (s) , - sin (s)) , 2 1 (sinh (s) , cosh (s) , sin (s) , cos (s)) . 2 Then we get the curvatures of as follows k1 (s) = 1, k2 (s) = 0, k3 (s) = -1. By using the Theorem B, the curve is not a null Bertrand curve. But if we take constants , , and µ as = 1, = 1, = 0, µ = -1, then it is trivial that the relations (3.1-a), (3.1-b), (3.1-c) and (3.1-d) hold. Therefore, the curve is a null (1, 3)-Bertrand curve in E4 . In this case, the 1 null (1, 3)-Bertrand mate of the curve is given by 1 (s) = (sinh (s) , cosh (s) , - sin (s) , - cos (s)) . 2 By using (3.25), (3.28), (3.29), we obtain the curvatures of the curve as k1 = 1, k2 = 0 and k 3 = 1. Acknowledgement. The authors would like to express their sincere gratitude to the referee for valuable suggestions to improve the paper.
### Journal
Annals of the Alexandru Ioan Cuza University - Mathematicsde Gruyter
Published: Nov 24, 2014
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# GMAT Arithmetic
GMAT arithmetic forms the basis for lots of quantitative problems. Work through the resources on this page for a complete GMAT arithmetic review that will give you the foundations that you need to ace the exam!
## Most Popular Posts About GMAT Arithmetic
GMAT arithmetic problems are rather straightforward. Read on to find out the specific arithmetic topics covered, tips for success, and practice problems!
There's no easy method for solving questions that involve adding and subtracting exponents but learn some strategies here to help you on test day!
What do you need to know about unit digit problems for GMAT Quant? Get the expert answer here, along with practice problems and explanations.
Consecutive integer problems on GMAT Quant can be tricky. Try out some practice problems and review explanations to brush up on your knowledge before test day!
What exponent problems will you see on test day? What's an exponent, a base, a power, and exponentiation? Try your hand at some practice problems here.
Need to refresh prime factorization on GMAT? “How many odd factors does 210 have?” If questions like these make you cringe, this post is for you!
## Most Recent Posts About GMAT Arithmetic
What is number sense and how can you recognize number sense problems on the GMAT? Try practice problems & play a number sense prep game.
GMAT number properties will comprise a large portion of the Quantitative section. Here are the basic number properties on the GMAT needed to succeed!
Is 1 a prime number? Nope, 1 is neither prime nor composite. Learn more about 1 and the Fundamental theorem of arithmetic here.
Learn how to simplify these seemingly devilishly complicated GMAT Quant problems! First, consider these problems 3) Consider these three quantities Rank these three quantities from least to greatest. (A) I, II, III (B) I, III, II (C) II, I, III (D) II, III, I (E) III, I, II These are challenging […]
Consider these two practice GMAT Quantitative problems: 1) Given f(x) = 3x – 5, for what value of x does 2*[f(x)] – 1 = f(3x – 6) (A) 0 (B) 4 (C) 6 (D) 7 (E) 13 (A) –2 (B) 5/3 (C) 1 (D) 2 (E) 8 If you find these questions completely incomprehensible, then […]
Fact: one of the most tested categories of concepts on the GMAT Quantitative section is Integer Properties What are the “properties of integers”? Probably none of these are brand new to you —- in fact, you probably learned about all these in grade school. Here’s a list. a) factors and multiples; GCF and LCM b) […]
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# Search by Topic
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### Go Moku
##### Age 7 to 11 Challenge Level:
A game for two players on a large squared space.
### Counter Roundup
##### Age 7 to 11 Challenge Level:
A game for 1 or 2 people. Use the interactive version, or play with friends. Try to round up as many counters as possible.
##### Age 7 to 11 Challenge Level:
How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!
### World of Tan 17 - Weather
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of the watering can and man in a boat?
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##### Age 7 to 11 Challenge Level:
Can you work out what kind of rotation produced this pattern of pegs in our pegboard?
### Endless Noughts and Crosses
##### Age 7 to 11 Challenge Level:
An extension of noughts and crosses in which the grid is enlarged and the length of the winning line can to altered to 3, 4 or 5.
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##### Age 7 to 11 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
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##### Age 7 to 11 Challenge Level:
Which of these dice are right-handed and which are left-handed?
### Square to L
##### Age 7 to 11 Challenge Level:
Find a way to cut a 4 by 4 square into only two pieces, then rejoin the two pieces to make an L shape 6 units high.
### World of Tan 19 - Working Men
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of this shape. How would you describe it?
### World of Tan 18 - Soup
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of Mai Ling and Chi Wing?
### World of Tan 16 - Time Flies
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of the candle and sundial?
### Put Yourself in a Box
##### Age 7 to 11 Challenge Level:
A game for 2 players. Given a board of dots in a grid pattern, players take turns drawing a line by connecting 2 adjacent dots. Your goal is to complete more squares than your opponent.
### World of Tan 15 - Millennia
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of the workmen?
### World of Tan 14 - Celebrations
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming and Little Fung dancing?
### Turning Cogs
##### Age 7 to 11 Challenge Level:
What happens when you turn these cogs? Investigate the differences between turning two cogs of different sizes and two cogs which are the same.
### Makeover
##### Age 5 to 11 Challenge Level:
Exchange the positions of the two sets of counters in the least possible number of moves
### Painted Faces
##### Age 7 to 11 Challenge Level:
Imagine a 3 by 3 by 3 cube made of 9 small cubes. Each face of the large cube is painted a different colour. How many small cubes will have two painted faces? Where are they?
### World of Tan 20 - Fractions
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of the chairs?
### World of Tan 8 - Sports Car
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of this sports car?
### Midpoint Triangle
##### Age 7 to 11 Challenge Level:
Can you cut up a square in the way shown and make the pieces into a triangle?
### Fractional Triangles
##### Age 7 to 11 Challenge Level:
Use the lines on this figure to show how the square can be divided into 2 halves, 3 thirds, 6 sixths and 9 ninths.
### World of Tan 30 - Logical Thinking
##### Age 7 to 11 Challenge Level:
Can you logically construct these silhouettes using the tangram pieces?
### World of Tan 29 - the Telephone
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of this telephone?
### World of Tan 28 - Concentrating on Coordinates
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming playing the board game?
### World of Tan 27 - Sharing
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of Little Fung at the table?
### World of Tan 26 - Old Chestnut
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of this brazier for roasting chestnuts?
### Dicey
##### Age 7 to 11 Challenge Level:
A game has a special dice with a colour spot on each face. These three pictures show different views of the same dice. What colour is opposite blue?
### World of Tan 21 - Almost There Now
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of the lobster, yacht and cyclist?
### Green Cube, Yellow Cube
##### Age 7 to 11 Challenge Level:
How can you paint the faces of these eight cubes so they can be put together to make a 2 x 2 cube that is green all over AND a 2 x 2 cube that is yellow all over?
### World of Tan 25 - Pentominoes
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of these people?
### World of Tan 24 - Clocks
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of these clocks?
### Three Squares
##### Age 5 to 11 Challenge Level:
What is the greatest number of squares you can make by overlapping three squares?
### World of Tan 22 - an Appealing Stroll
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of the child walking home from school?
### More Building with Cubes
##### Age 7 to 11 Challenge Level:
Here are more buildings to picture in your mind's eye. Watch out - they become quite complicated!
### World of Tan 3 - Mai Ling
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of Mai Ling?
### World of Tan 9 - Animals
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of this goat and giraffe?
### The Development of Spatial and Geometric Thinking: the Importance of Instruction.
##### Age 5 to 11
This article looks at levels of geometric thinking and the types of activities required to develop this thinking.
### Taking Steps
##### Age 7 to 11 Challenge Level:
In each of the pictures the invitation is for you to: Count what you see. Identify how you think the pattern would continue.
### World of Tan 13 - A Storm in a Tea Cup
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of these convex shapes?
### Reef and Granny
##### Age 7 to 11 Challenge Level:
Have a look at what happens when you pull a reef knot and a granny knot tight. Which do you think is best for securing things together? Why?
### Square Surprise
##### Age 5 to 11 Challenge Level:
Why do you think that the red player chose that particular dot in this game of Seeing Squares?
### Circles, Circles
##### Age 5 to 11 Challenge Level:
Here are some arrangements of circles. How many circles would I need to make the next size up for each? Can you create your own arrangement and investigate the number of circles it needs?
### World of Tan 10 - Butterflies
##### Age 7 to 11 Challenge Level:
Can you use the interactive to complete the tangrams in the shape of butterflies?
### How Would We Count?
##### Age 5 to 11 Challenge Level:
An activity centred around observations of dots and how we visualise number arrangement patterns.
### World of Tan 6 - Junk
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the silhouette of the junk?
### Flip
##### Age 7 to 11 Challenge Level:
Can you picture where this letter "F" will be on the grid if you flip it in these different ways?
### Making Maths: Rolypoly
##### Age 5 to 11 Challenge Level:
Paint a stripe on a cardboard roll. Can you predict what will happen when it is rolled across a sheet of paper?
### Cube Drilling
##### Age 7 to 11 Challenge Level:
Imagine a 4 by 4 by 4 cube. If you and a friend drill holes in some of the small cubes in the ways described, how many will not have holes drilled through them?
### Move Those Halves
##### Age 7 to 11 Challenge Level:
For this task, you'll need an A4 sheet and two A5 transparent sheets. Decide on a way of arranging the A5 sheets on top of the A4 sheet and explore ...
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SystemsDifferentialEquationsForPrint
# SystemsDifferentialEquationsForPrint - Systems of...
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Systems of Differential Equations S. Bonnot S. Bonnot Systems of Differential Equations
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Examples coming from physics Goal : Physical systems where separate parts interact on each other create automatically systems of differential equations. Example : a system with 2 consecutive springs. The motion of the first mass is described by m 1 x 00 1 = - k 1 x 1 + k 2 ( x 2 - x 1 ) . and the motion of the second mass at the bottom is given by m 2 x 00 2 = - k 2 ( x 2 - x 1 ) . System of two equations combined together : m 1 x 00 1 = - k 1 x 1 + k 2 ( x 2 - x 1 ) m 2 x 00 2 = - k 2 ( x 2 - x 1 ) S. Bonnot Systems of Differential Equations
Examples, continued Electric Circuits: L di 1 dt + Ri 2 = E R di 2 dt + 1 C ( i 2 - i 1 = 0 We get a system of two first order linear equations with constant coefficients . How to solve it? Eliminate one variable (say i 2 ) to obtain a single differential equation in i 1 . We know Ri 2 = - Li 0 1 + E , so we deduce Ri 0 2 = - Li 00 1 + E 0 . Replace i 2 and i 0 2 in the second equation, and obtain ( - Li 00 1 + E 0 ) + 1 RC ( - Li 0 1 + E ) - 1 C = 0. Solve this equation to get i 1 and then use Ri 2 = - Li 0 1 + E to obtain i 2 . S. Bonnot Systems of Differential Equations
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General case: system of first order equ. with constant coefficients Given the system a 1 x 0 + b 1 y 0 + c 1 x + d 1 y = f 1 ( t ) a 2 x 0 + b 2 y 0 + c 2 x + d 2 y = f 2 ( t ) let’s try to eliminate y 0 by taking b 2 . ( Line 1 ) - b 1 ( Line 2 ) . we obtain ( a 1 b 2 - a 2 b 1 ) x 0 + ( c 1 b 2 - c 2 b 1 ) x + ( d 1 b 2 - d 2 b 1 ) y = b 2 f 1 ( t ) - b 1 f 2 ( t ) this equation is of type Ax 0 + Bx + Cy = F ( t ) . Case C = 0. Solve Ax 0 + Bx = F ( t ) then substitute in either equation to get y . Case C 6 = 0. We know y = ( - Ax 0 - Bx + F ( t )) / C , so we know y 0 and we can substitute y , y 0 in Equ.1 or Equ.2. S. Bonnot Systems of Differential Equations
Solving systems of first order equ. with constant coefficients Example : 3 x 0 + 2 y 0 + x + y = e t 6 x 0 + 4 y 0 + x = t We eliminate y 0 by taking Line 2 - 2 Line 1, to obtain - x - 2 y = t - 2 e t . Observe that by accident we also eliminated x 0 . So far we know y = 1 2 ( - x - t + 2 e t ) , so we also deduce y 0 = 1 2 ( - x 0 - 1 + 2 e t ) . Substitute y , y 0 in Equ.1 to get a first order equ. in x alone: 3 x 0 + ( - x 0 - 1 + 2 e t ) + x + ( 1 2 ( - x - t + 2 e t )) = e t . See textbook for an example where x 0 doesn’t disappear accidentally. S. Bonnot Systems of Differential Equations
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From higher order equ. to systems of first order equ.
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https://stridetutoring.com/articles/calculating-the-surface-area-of-a-cone/
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## Introduction
Cones are not only mathematical shapes but also objects that we encounter in our daily lives. From the delicious ice cream cone to party hats and even traffic cones, understanding the surface area of a cone can have practical applications. In this blog post, we will explore the step-by-step process of calculating the surface area of a cone, providing you with a solid foundation for tackling this mathematical concept.
## Understanding the Surface Area Formula
To calculate the surface area of a cone, we use the following formula:
Surface Area = π * r * (r + l)
In this formula, ‘r’ represents the radius of the base of the cone, and ‘l’ represents the slant height.
Now, let’s dive into the step-by-step guide for calculating the surface area of a cone.
Step 1: Measure the Radius and Height
To begin, measure the radius (r) of the base of the cone. This measurement should be taken from the center of the base to its outer edge. Next, measure the height (h) of the cone, which is the distance from the apex (top point) to the base.
Step 2: Calculate the Slant Height
The slant height (l) of a cone can be found using the Pythagorean theorem. It is the distance from the apex to a point on the circumference of the base. To calculate the slant height, use the formula:
l = √(r^2 + h^2)
Step 3: Calculate the Base Area
The base area (A) of a cone is given by the formula:
A = π * r^2
Substitute the value of the radius (r) into the formula to calculate the base area.
Step 4: Calculate the Lateral Surface Area
The lateral surface area (LSA) of a cone refers to the curved surface area excluding the base. It can be calculated using the formula:
LSA = π * r * l
Substitute the values of the radius (r) and the slant height (l) into the formula to calculate the lateral surface area.
Step 5: Calculate the Total Surface Area
The total surface area (TSA) of a cone is the sum of the base area and the lateral surface area. Use the following formula to calculate the total surface area:
TSA = A + LSA
Substitute the values of the base area (A) and the lateral surface area (LSA) into the formula to calculate the total surface area.
## Example Problem
Imagine you have a cone with a radius of 3 cm and a height of 4 cm. Our goal is to calculate the surface area of this cone using the steps outlined above.
Given Values:
• Radius (r) = 3 cm
• Height (h) = 4 cm
Step 1: Confirming the Given Values
We start with the radius of the base, 3 cm, and the height of the cone, 4 cm, as given.
Step 2: Calculating the Slant Height (l)
Using the Pythagorean theorem:
\(l = √(r^2 + h^2) = √(3^2 + 4^2) = √(9 + 16) = √25 = 5\) cm
Step 3: Calculating the Base Area (A)
Next, we calculate the base area:
\(A = π * r^2 = π * 3^2 = 9π\) cm²
Step 4: Calculating the Lateral Surface Area (LSA)
The lateral surface area is found using the formula:
\(LSA = π * r * l = π * 3 * 5 = 15π\) cm²
Step 5: Calculating the Total Surface Area (TSA)
Finally, we calculate the total surface area by adding the base area to the lateral surface area:
\(TSA = A + LSA = 9π + 15π = 24π\) cm²
Therefore, the total surface area of the cone is \(24π\) cm², which is approximately \(75.36\) cm².
## Real-Life Applications
Understanding the surface area of a cone has practical applications in various everyday objects. Let’s explore a few examples:
• Ice Cream Cone: The surface area calculation helps determine the amount of edible material needed to cover the ice cream cone with chocolate or sprinkles.
• Party Hat: Calculating the surface area of a cone can help estimate the amount of paper or fabric required to create a party hat of a specific size.
• Traffic Cone: Surface area calculations are used to determine the amount of reflective material needed to ensure visibility and safety on traffic cones.
## Conclusion
Calculating the surface area of a cone may seem challenging at first, but by following the step-by-step guide outlined in this blog post, you can master this mathematical concept. Understanding the surface area of a cone has real-life applications in various objects we encounter regularly. From ice cream cones to party hats and traffic cones, the ability to calculate surface area opens doors to practical problem-solving. So, embrace the process, practice your calculations, and enjoy exploring the world of cones from a mathematical perspective.
Looking for more math help? Check out our available math tutors for additional support!
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# A certain jar contains only w white marbles b black marbles
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Manager
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A certain jar contains only w white marbles b black marbles [#permalink]
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30 Jul 2007, 00:21
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
A certain jar contains only w white marbles b black marbles and r red marbles. If one marble is chosen at random what is the probability that the marble chosen red greater than the probability that the marble chosen white ?
(1) (r/(b+w)) > (b/(r+w))
(2) b-w > r
Q2.
In a certain group of 10 members , 4 members teach only french and the rest teach only spanish and german. If the group is to choose a 3 member comitte, which must have at least 1 member who teaches French, How many different committee member can be chosen.
(a) 40
b) 50
c 64
d)80
e) 100
Plse help asap
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30 Jul 2007, 00:27
Figured out Q2 can someone help on Q1 thx....
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Re: EMERGENCY HELP two questions asap test soon [#permalink]
### Show Tags
30 Jul 2007, 00:27
tlntd42 wrote:
In a certain group of 10 members , 4 members teach only french and the rest teach only spanish and german. If the group is to choose a 3 member comitte, which must have at least 1 member who teaches French, How many different committee member can be chosen.
(a) 40
b) 50
c 64
d)80
e) 100
Getting E.
Total no. of possible committees = 10C3 = 120
No. of possible committees without any French teacher = 6C3 = 20
No. of committees with at least one French teacher = 120-20 = 100.
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Re: EMERGENCY HELP two questions asap test soon [#permalink]
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30 Jul 2007, 00:31
tlntd42 wrote:
A certain jar contains only w white marbles b black marbles and r red marbles. If one marble is chosen at random what is the probability that the marble chosen red greater than the probability that the marble chosen white ?
(1) (r/(b+w)) > (b/(r+w))
(2) b-w > r
Discussed before.
Check here:
http://www.gmatclub.com/phpbb/viewtopic.php?t=44112&highlight=black++white++red
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30 Jul 2007, 00:32
And all the best for your test !!
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30 Jul 2007, 00:32
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Substitution of variables in differential equations
• To: mathgroup at smc.vnet.net
• Subject: [mg6294] Substitution of variables in differential equations
• From: Michael Hartl <michael.hartl at uni-konstanz.de>
• Date: Sat, 8 Mar 1997 00:26:16 -0500 (EST)
• Organization: Univ. Konstanz
• Sender: owner-wri-mathgroup at wolfram.com
```How can I solve ordinary differential equations by the method of
substitution of variables? (I am using Mathematica 2.2.)
In the following example I want to simplify an ODE (function y, variable
x, differentiation with respect to x) substituting x by t (with
x=Cot[t]):
In[1]:= dgl[y_] = Dt[y,{x,2}]+1/(1+x^2)^2 y == 0
y
Out[1]= --------- + Dt[y, {x, 2}] == 0
2 2
(1 + x )
In[2]:= dgl[f[x]] /. x->Cot[t]
f[Cot[t]]
Out[2]= -------------- + f''[Cot[t]] == 0
2 2
(1 + Cot[t] )
In[3]:= % //Simplify
4
Out[3]= f[Cot[t]] Sin[t] + f''[Cot[t]] == 0
The correct result should look similar to:
Sin[t]^4 (y''[t] + 2 Cot[t] y'[t] + y[t])
Mike
PS: In some textbooks on Mathematica (even ones dedicated to
differential equations) these steps were performed outside the
Mathematica sessions by hand!
--
michael.hartl at uni-konstanz.de
```
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Cody
# Problem 2121. Find offset of given matrix element from first matrix element
Solution 1043705
Submitted on 3 Nov 2016 by Justin Ng
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
m=[11 2 34; 40 51 6; 87 8 109] element = 51; offset = 5; assert(isequal(FindOffset(m,element),offset));
m = 11 2 34 40 51 6 87 8 109
2 Pass
m=reshape([1:10],5,2); element = 9; offset = 9; assert(isequal(FindOffset(m,element),offset));
3 Pass
m=eye(7); element = 0; offset = 2; assert(isequal(FindOffset(m,element),offset));
4 Pass
m=[10 20 30 40; 50 60 70 80;]; element = 56; offset = 0; assert(isequal(FindOffset(m,element),offset));
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# SIMPLE ELEVATOR
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## Jayson Balasulia
on 20 October 2014
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#### Transcript of SIMPLE ELEVATOR
SIMPLE ELEVATOR
FEATURES AND LIMITATIONS
FEATURES
The system uses mini dynamo motor for lifting up and down the elevator box.
The device is operated with 9 volts Power Supply.
The design project uses LED as indicator for designating floor levels.
The simple elevator uses push buttons for choosing desired floors.
LIMITATIONS
The project focuses only the vertical movement of elevator car.
The design project has a limited number of elevator control system.
CONCLUSION
The team finally accomplished the Simple Elevator after all the months of planning, research, testing and all the hard works and the infinite value of determination. The system can successfully move vertically to serve the user.
The researcher aimed to construct a logic design circuits through elevator system, to provide effective solution of transporting goods and to move easily between the floors of the building was achieved.
Hence the objectives of the design project have been successfully obtained. The whole process of the design was successfully functioning.
SCHEMATIC DIAGRAM
Project Objective
GENERAL OBJECTIVE:
To design and construct a “Simple Elevator” on or before October 22, 2014.
SPECIFIC OBJECTIVE:
1. To design and construct a logic design circuit that utilizes AND and NOT Gates.
2. To be able to analyze the operation of logic components and switching functions of logic gates.
3. To construct a device that provides accessibility solution for moving vertically between floors of a building.
Introduction
Logic gates are the basis of digital electronics. It performs basic logical functions and the fundamental building blocks of digital integrated circuits. Each chip usually has about 4 logic gates. Most logic gates take an input of two binary values, and output a single value of a 1 or 0. Real computers don't use these kinds of gates, because they take far too much space. Some circuits may have only a few logic gates, while others, such as microprocessors, may have millions of them.
There are seven types of logic gates, three main basic types of digital logic gate, the AND Gate, the OR Gate and the NOT Gate. By combining thousands or millions of logic gates, it is possible to perform highly complex operations. The maximum number of logic gates on an integrated circuit is determined by the size of the chip divided by the size of the logic gates.
SIGNIFICANCE OF THE DESIGN PROJECT
With the population growing the need for housing is increasing with the lack of livable area, housing as increased into multi level units. But these multi-level structures would have basically unusable if it weren’t for another technological innovation that came along around the same time, thus our design project ’Simple Elevator’ is built and developed.
An elevator system is an easy concept to understand. In its simplest form there is a single elevator, moving vertically for serving users an all floor levels. It control the flow of foot traffic between various floors of buildings, It allow disabled persons to access between level floors.
Our design project is also beneficial for transporting heavy items and office equipment between various levels of the building. Most importantly, it enhances the quality of our lives.
CIRCUIT DESCRIPTION AND ANALYSIS
Basically an elevator system consists of power supply, IC logic gates 7404 and 7408, LEDs, relays, motor, push button and resistor. Push button is connected on the Vcc , when it is pressed it activates the relay and gives signal to the AND gate that runs the LEDs and the motor. When the other button is pushed it goes the same way but now, an inverter is activated. The input signal of the inverter is connected on Vcc stands for logic 1, therefore the output signal is 0. The output of the inverter is connected on the AND gate of the other circuit. When the inverter gives the 0 signal on the AND Gate then it will produce a 0 output that is connected on relay. If the AND Gate gives the 0 signal to the relay, it turns the relay off. If the relay is turned off it also turned the LED and the motor of the other circuit off.
BLOCK DIAGRAM
SUMMARY OF FINDINGS
Through observations, planning and researches, the group came up with the idea of a “Simple Elevator.”
Having tested the device, reflections and conclusions have been formulated. In connecting together AND and NOT gates and others electronic components we constructed a logic circuit of an elevator system. This simple elevator is not only for controlling the flow of foot traffic between various floors of buildings, not only for accessibility of physically disabled person, transporting large things but also for creating more usable space. Elevators helped us a lot to use our spaces more effectively. If you compare with stairs you would see elevators take less space than a stair.
The whole project uses the relay an electrical device in which a small change in current or voltage controls the switching on or off of circuit.
Overall, the importance of the project is to attain effectively the objectives that had been created. Then correct layouts of prototype complete the design project.
Full transcript
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Explore BrainMass
# Solutions to First-Order Ordinary Differential Equations
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Question 1: (1 points)
Solve the following equations for given initial condition.
Question 2: (1 points)
Integrate the following equations and write a solution for given initial condition.
Question 3: (1 points)
Integrate the following equations and write a nontrivial solution.
Question 4: (1 points)
Integrate the following equations and write a nontrivial solution.
Question 5: (1 points)
Integrate the following equations and write a nontrivial solution.
Question 6: (1 points)
Integrate the following equations and write a nontrivial solution.
Question 7: (1 points)
Integrate the following equations and write a solution for given initial condition.
Question 8: (1 points)
Integrate the following equations and write a nontrivial solution.
Question 9: (1 points)
Solve the following equation
Question 10: (1 points)
Make the change of variable for the homogenous differential equation
and find , where
Question 11: (1 points)
Solve the following equation
Question 12: (1 points)
Find the general solution of the linear differential equation (use C to denote a constant)
.
Question 13: (1 points)
Find the solution of the linear differential equation which satisfies the initial condition :
.
Question 14: (1 points)
Find the solution of the following differential equation:
.
Question 15: (1 points)
Solve the following equation for given initial condition
https://brainmass.com/math/calculus-and-analysis/solutions-first-order-ordinary-differential-equations-446611
#### Solution Preview
Question 1: (1 points)
Solve the following equations for given initial condition.
This is a separable differential equation. We write it as
.
Integrating both sides we obtain
for some constant c. Plugging in the initial condition we obtain
whence . Thus the solution is , or
The implicit form of this equation is .
Question 2: (1 points)
Integrate the following equations and write a solution for given initial condition.
Dividing both sides by we obtain
which is separable, so we write it as
.
Integrating both sides we obtain
.
Exponentiating both sides we now obtain
.
Plugging in the initial condition we obtain
.
Thus the solution is , or .
Question 3: (1 points)
Integrate the following equations and write a nontrivial solution.
This is another separable differential equation. Separating variable and integrating both sides we obtain
.
To evaluate the integral on the right (with the minus sign), we make the substitution , whence and the integral becomes
.
Similarly, the integral on the left becomes
.
Thus we have
whence
,
which yields a nontrivial solution for any constant c such that . For instance, when we have the nontrivial solution .
Question 4: (1 points)
Integrate the following equations and write a nontrivial solution.
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# Documentation
### This is machine translation
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# peaks
Example function of two variables
## Syntax
```Z = peaks; Z = peaks(n); Z = peaks(V); Z = peaks(X,Y); peaks(...) [X,Y,Z] = peaks(...); ```
## Description
`peaks` is a function of two variables, obtained by translating and scaling Gaussian distributions, which is useful for demonstrating `mesh`, `surf`, `pcolor`, `contour`, and so on.
`Z = peaks;` returns a 49-by-49 matrix.
`Z = peaks(n);` returns an `n`-by-`n` matrix.
`Z = peaks(V);` returns an `n`-by-`n` matrix, where ```n = length(V)```.
`Z = peaks(X,Y);` evaluates `peaks` at the given `X` and `Y` (which must be the same size) and returns a matrix the same size.
`peaks(...)` (with no output argument) plots the peaks function with `surf`. Use any of the input argument combinations in the previous syntaxes.
`[X,Y,Z] = peaks(...);` returns two additional matrices, `X` and `Y`, for parametric plots, for example, `surf(X,Y,Z,del2(Z))`. If not given as input, the underlying matrices `X` and `Y` are
```[X,Y] = meshgrid(V,V) ```
where `V` is a given vector, or `V` is a vector of length `n` with elements equally spaced from -3 to 3. If no input argument is given, the default `n` is 49.
## Examples
collapse all
Create a 5-by-5 matrix of peaks and display the surface.
```figure peaks(5);```
``` z = 3*(1-x).^2.*exp(-(x.^2) - (y+1).^2) ... - 10*(x/5 - x.^3 - y.^5).*exp(-x.^2-y.^2) ... - 1/3*exp(-(x+1).^2 - y.^2) ```
## See Also
#### Introduced before R2006a
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### Find the budget to be allocated now for this improvement
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##### Reference no: EM131027209
Recently Scott spent \$200,000 for landscaping on his home. Every 10 years, it need to be improved with the cost of \$100,000. Find the budget (P) to be allocated now for this improvement every 10 years, forever (n=infinity) if i=10%/year.
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# Cell centres
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June 4, 2003, 02:41 Cell centres #1 Tom Guest Posts: n/a Hi, what is the most accurate method to calculate the centre of a general polyhedral cell?
June 4, 2003, 05:28 Re: Cell centres #2 John Guest Posts: n/a Divide it into tetrahedrals and calculate the individual tetra center: the polyhedral center is sumup(ViCi)/sumup(Vi), where Vi is volume of each tetra, and Ci is the center of each tetra.
June 4, 2003, 08:56 Re: Cell centres #3 Tom Guest Posts: n/a Thank you. Will do, but how do I calculate the centre of a tet?
June 4, 2003, 09:04 Re: Cell centres #4 Tom Guest Posts: n/a ...and the volume of a tet for that matter? thank you
June 4, 2003, 12:28 Re: Cell centres #5 Ananda Himansu Guest Posts: n/a the centroid of a tet (or any simplex, the tet being a 3D simplex) is the arithmetic average of [the position vectors representing] its vertices. note that such a statement is not in general true for polytopes other than simplices, which is why you have to decompose a general polyhedron into tets. pick any vertex of the tet. label it "O". label, in any order, the other three vertices "A", "B", "C". denote the vector from O to A by the label "a", from O to B by "b", similarly "c". the volume of the tet is one-sixth of the magnitude of the scalar or "box" product of the three vectors a, b, c. that is, volume(OABC)=|a.(bxc)|/6. (the analogous formula for a triangle embedded in 3D is that the area of triangle OAB is half the magnitude of the cross-product of a and b, that is, area(OAB)=||axb||/2.) the decomposition of general polyhedra into the union of tetrahedra with plane faces is not in general exact. for example, if a polyhedron has a quadrilateral for a side, the vertices of the quad need not lie in a plane. then the decomposition of the quad into two or more tets is only an approximation. these issues have been noted in old papers, and are probably discussed in any CFD text that explains unstructured meshes. also probably covered previously in this forum (do a search).
June 4, 2003, 14:54 Re: Cell centres #6 john Guest Posts: n/a Another means is to calculate the determinate of the following expression: V=(1/6)*|x1 y1 z1 1 | |x2 y2 z2 1 | |x3 y3 z3 1 | |x4 y4 z4 1 | where the x, y & z's are the coordinates of the respective points defining the tetrahedron. Note that the volume will always be positive, even though the value of the determinate might be negative depending upon the ordering of the points.
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# Lady Gaga Tests Borderline Positive for Lupus
Article
Lady Gaga announced that she recently tested borderline positive for lupus. Dr. Pullen talks about lupus tests, false positives, and diagnostic uncertainty.
This article originally appeared online at DrPullen.com, part of the HCPLive network.
Lady Gaga and lupus. Is this just another publicity stunt? After all Lady Gaga is an all-star at self-promotion. In a Larry King live full-hour interview, Lady Gaga stated that she tested borderline positive for lupus. To the average listener this may sound a bit goofy. After all, lupus is a serious rheumatologic disease that is often associated with arthritis, skin rashes, kidney problems, extreme fatigue and sometimes neurologic problems. What is a borderline positive test for lupus anyway? In fact this is a common scenario for several reasons. First is that lupus is not a simple diagnosis to make. Secondly, the commonly done screening tests for systemic lupus erythematosus (lupus) is a very non-specific test. The anti-nuclear antibody (ANA) test is often ordered as part of a screening evaluation for rheumatologic symptoms. In a general medical setting 35% of patients may have an ANA test positive at a titer of 1:40. The false positive rate at a higher titer of 1:160 closer to 5%. I have no idea what Lady Gaga’s titer was, but calling the test borderline implies a low titer. In any test the chances of a true positive test as opposed to a false positive test is dependent on the chances that the individual has the condition being tested. If given the overall collection of symptoms, family history and physical findings a patient is felt to have a 10% chance of having lupus, and has a titer of 1:40 then the test can be a true positive or a false positive test. The odds of each of these are:
False Positive = .9 x .35 = 31.5% -- prevalence of non-disease x false positive rateTrue Positive = .1 x .65 = 6.5% -- prevalence of disease x true positive rateRatio of false positive tests to true positive tests = 5.1 : 1
If a patient has a vague rheumatologic set of symptoms that are unlikely to be lupus, then the chances of a false positive test may be much higher than the chances of a true positive test as shown in the hypothetical situation above.
If the titer was >1:160 then the false positive rate drops to 5%, but still in a patient where the clinical presentation suggests a 10% chance of disease the test is far from conclusive:
False positive = .9 x .05 = 4.5% -- prevalence of non-disease x false positive rateTrue positive = .1 x .95 = 9.5% -- prevalence of disease x true positive rateRatio of true positive tests to false positive tests = 2.1 : 1
There are additional rheumatologic tests, specifically anti-smooth muscle DNA and others that are more specific for lupus, and can help define these scenarios, but often a patient is left concluding that their test for lupus was “borderline.”
The other issue with lupus is that the diagnosis is not made solely based on a blood test. To make a diagnosis of lupus a patient must have at least 4 of a list of 11 diagnostic criteria, only one of which is a positive ANA titer, and only one other is another of the more specific blood tests for lupus. If a patient does not have physical findings suggestive of lupus the diagnosis cannot be conclusively made.
So Lady Gaga may or may not have lupus, but she appears to join many others in feeling that she has one or more blood tests that suggest possible lupus, but lack other lab tests or physical findings to make a diagnosis of lupus. She is certainly not a common person, but she is in a common predicament.
Ed Pullen, MD, is a board-certified family physician practicing in Puyallup, WA. He blogs at DrPullen.com — A Medical Bog for the Informed Patient.
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trig identites
posted by .
since sin2x+√3cos2x=2sin(2x+60) find the general solution for the equation sin2x+√3cosx=0
thank you
• trig identites -
since sin(n*180) = 0,
2x+60 = n*180
x = n*90 - 30
Similar Questions
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2. math (trig)
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find the general solution for the equation: sin2x+ square root 3cosx =0
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## Learn to Use ITensor
main / classes / autompo
# AutoMPO
AutoMPO is a very powerful system for translating sums of local operators into an MPO (or IQMPO) tensor network. The notation for AutoMPO input is designed to be as close as possible to pencil-and-paper quantum mechanics notation.
The code for AutoMPO is located in the files "itensor/mps/autompo.h" and "itensor/mps/autompo.cc".
# Synopsis
// Make a chain of N spin 1/2's with QN information
auto Nx = 12, Ny = 6;
auto N = Nx*Ny;
auto sites = SpinHalf(N,{"ConserveQNs=",true});
//
// Use AutoMPO to create the
// next-neighbor Heisenberg model
//
auto ampo = AutoMPO(sites);
for(int j = 1; j < N; ++j)
{
ampo += 0.5,"S+",j,"S-",j+1;
ampo += 0.5,"S-",j,"S+",j+1;
ampo += "Sz",j,"Sz",j+1;
}
//Convert the AutoMPO object to an MPO
auto H = toMPO(ampo);
//....
//
// Create a model with further-range interactions
// capturing a 2D lattice (with a 1D ordering of sites)
//
auto ampo2D = AutoMPO(sites);
auto lattice = squareLattice(Nx,Ny);
for(auto b : lattice)
{
ampo2D += 0.5,"S+",b.s1,"S-",b.s2;
ampo2D += 0.5,"S-",b.s1,"S+",b.s2;
ampo2D += "Sz",b.s1,"Sz",b.s2;
}
auto H2D = toMPO(ampo2D);
//....
//
// Create the 1D Hubbard model
//
auto sitesElec = Electron(N);
auto ampoHub = AutoMPO(sitesElec);
for(int i = 1; i <= N; ++i)
{
ampoHub += U,"Nupdn",i;
}
for(int b = 1; b < N; ++b)
{
ampoHub += -t,"Cdagup",b,"Cup",b+1;
ampoHub += -t,"Cdagup",b+1,"Cup",b;
ampoHub += -t,"Cdagdn",b,"Cdn",b+1;
ampoHub += -t,"Cdagdn",b+1,"Cdn",b;
}
auto H = toMPO(ampoHub);
//....
//
// Create a spin model with four-site terms
//
auto ampo4 = AutoMPO(sites);
for(auto i : range1(N-4))
{
ampo4 += "Sz",i,"Sz",i+1,"Sz",i+2,"Sz",i+4;
}
for(auto i : range1(N-1))
{
ampo4 += 0.5,"S+",i,"S-",i+1;
ampo4 += 0.5,"S-",i,"S+",i+1;
}
auto H4 = toMPO(ampo4);
## AutoMPO Interface
• AutoMPO(SiteSet sites)
Construct an AutoMPO object.
• AutoMPO += op
The += operator of an AutoMPO object adds a operator product into the sum of operators represented by the AutoMPO.
Examples of valid input:
ampo += "Sz",i;
ampo += "Sz",i,"Sz",j;
ampo += 0.2,"Sz",i,"Sz",j;
ampo += 0.5,"S+",i,"S-",j;
ampo += 0.5,"S+",i,"S-",j,"S+",k,"S-",l;
The operator products on the right-hand side of the += operator begin with an optional real- or complex-valued coefficient, then continue with a comma separated list of string-integer pairs. The coefficient can be either a numeric literal or a variable of type Real (double) or Cplx (std::complex<double>).
The string-integer pairs, such as "Sz",i, represent an operator $S^z_i$ .
Which strings are acceptable as operator names are determined by the SiteSet used to construct the AutoMPO. All operator names must be valid input to the .op method of the SiteSet (which is used to construct local site operators as tensors).
• .sites() -> SiteSet const&
Retrieve the SiteSet used to construct the AutoMPO.
## Converting an AutoMPO to an MPO
### toMPO function
Call the toMPO function to create an MPO from an AutoMPO. You can pass various named arguments to control which backend is used to process the AutoMPO and to control the accuracy of this process.
Examples:
auto H1 = toMPO(ampo);
auto H2 = toMPO(ampo,{"Exact=",true});
Named arguments recognized:
• "Exact" — boolean (default: false). Set whether to use the 'exact' backend or the approximate backend. The approximate backend can handle operators acting on more than two sites, and compresses the resulting MPO by doing a series of singular value decompositions (SVDs). Normally these SVDs are very accurate but could become an issue if an important term has an unusually small coefficient. The exact backend makes no approximations whatsoever, but is limited to at most two-site operators and can sometimes result in larger MPO bond dimensions than the approximate backend.
• "MaxDim" — integer (default: 5000). Set the maximum bond dimension of the resulting MPO when using the approximate backend. If Exact=true, this has no effect.
• "Cutoff" — real. (default: 1E-13). Set the SVD truncation error cutoff used by the approximate backend. If Exact=true, this has no effect.
### toExpH function
The toExpH function converts an AutoMPO representing a sum of operators $H$ into an MPO which approximates the operator $e^{-t H}$ for a small time step t, making an error of order $t^2$ per time step. The time step t can be real or complex.
The method used to do the approximate exponentiation is based on the following article: Phys. Rev. B 91, 165112 (arxiv:1407.1832) using the $W^{I}$ method, and has the advantage that unlike naive approaches for exponentiating an MPO, the time-step error per site is independent of system size.
Note that the true amount of error per step, or the quality of the results can depend very highly on how short- or long-range the interactions are in the input Hamiltonian. It is recommended to test your results against a different method, such as Trotter gates, global Krylov, or TDVP to confirm that your results are accurate and controlled.
Examples:
auto ampo = AutoMPO(sites);
...
Real tau = 0.1;
//Real time evolution
auto expiH = toExpH(ampo,tau*Cplx_i);
//...
//Imaginary time evolution
auto expH = toExpH(ampo,tau);
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# 6th Grade Mathematics (TEKS) Practice
« Back to Texas Middle School
Discover the most effective and comprehensive online solution for curriculum mastery, high-stakes testing, and assessment in . Our 6th Grade Mathematics (TEKS) curriculum and test review is aligned to the most current standards.
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• Questions 3,630
• Vocabulary Terms 229
• Instructional Videos 81
### Test Standards
1. (6.2.A) Classify numbers
2. (6.2.B) Opposites and absolute values
3. (6.2.C) Order rationals
4. (6.2.D) Order in contexts
5. (6.2.E) Division
6. (6.4.C) Ratios as comparisons
7. (6.4.D) Rates as comparisons
8. (6.4.E) Rates and percents
9. (6.4.F) Benchmark fractions
10. (6.4.G) Fractions, decimals, percents
11. (6.5.C) Use fractions, decimals, percents
12. (6.7.A) Numerical expressions
13. (6.7.B) Expressions and equations
14. (6.7.C) Equivalent expressions
15. (6.7.D) Simplify expressions
1. (6.3.A) Reciprocals
2. (6.3.B) Fraction multiplication
3. (6.3.C) Integer operations
4. (6.3.D) Operate with integers
5. (6.3.E) Multiply and divide rationals
6. (6.4.A) Compare rules
7. (6.4.B) Apply reasoning
8. (6.5.A) Ratios and rates
9. (6.5.B) Solve problems
10. (6.6.A) Independent and dependent quantities
11. (6.6.B) Write an equation
12. (6.6.C) Represent a situation
13. (6.9.A) Write equations and inequalities
14. (6.9.B) Solutions on number lines
15. (6.9.C) Problems from equations
16. (6.10.A) Model and solve
17. (6.10.B) Check solutions
1. (6.4.H) Convert units
2. (6.8.A) Triangles
3. (6.8.B) Area formulas
4. (6.8.C) Area equations
5. (6.8.D) Determine solutions
6. (6.11.A) Graph points
1. (6.12.A) Represent data
2. (6.12.B) Describe data
3. (6.12.C) Summarize numeric data
4. (6.12.D) Summarize categorical data
5. (6.13.A) Interpret data graphs
6. (6.13.B) Data variability
7. (6.14.A) Compare features
8. (6.14.B) Debit and credit cards
9. (6.14.C) Check register
10. (6.14.E) Credit reports
11. (6.14.F) Value of credit reports
12. (6.14.G) Paying for college
13. (6.14.H) Annual salaries
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Cody
# Problem 44446. Add a vector to a matrix
Solution 1695481
Submitted on 18 Dec 2018 by Athi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
2 Pass
mat = ones(2); v = 1:3; comb_correct = [1 1 inf; 1 1 inf; 1 2 3]; assert(isequal(addVecToMat(mat, v), comb_correct))
3 Pass
mat = ones(3, 4); v = [2 3]; comb_correct = [ones(3,4); 2 3 -inf -inf]; assert(isequal(addVecToMat(mat, v), comb_correct))
4 Pass
mat = ones(2); v = [2 3]; comb_correct = [1 1; 1 1; 2 3]; assert(isequal(addVecToMat(mat, v), comb_correct))
5 Pass
mat = ones(3); v = 1:6; comb_correct = [repmat([ones(1,3) inf(1,3)], 3, 1); 1:6]; assert(isequal(addVecToMat(mat, v), comb_correct))
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# Graphing linear systems
A system of linear equation comprises two or more linear equations. The solution of a linear system is the ordered pair that is a solution to all equations in the system.
One way of solving a linear system is by graphing. The solution to the system will then be in the point in which the two equations intersect.
Example:
Solve the following system of linear equations
$\\ \left\{\begin{matrix} y=2x+4\\ y=3x+2 \end{matrix}\right. \\$
The two lines appear to intersect in (2, 8)
It's a good idea to always check your graphical solution algebraically by substituting x and y in your equations with the ordered pair
$\\ \underline{y=2x+4} \\ {\color{green} 8}\overset{?}{=} 2\cdot {\color{green} 2}+4 \\8=8 \\\\ \underline{y=3x+2} \\ {\color{green} 8}\overset{?}{=}3\cdot {\color{green} 2}+2 \\8=8 \\$
A linear system that has exactly one solution is called a consistent independent system. Consistent means that the lines intersect and independent means that the lines are distinct.
Linear systems composes of parallel lines that have the same slope but different y-intersect do not have a solution since the lines won't intersect. Linear systems without a solution are called inconsistent systems.
Linear systems composed of lines that have the same slope and the y-intercept are said to be consistent dependent systems. Consistent dependent systems have infinitely many solutions since the lines coincide.
Video lesson: Solve the linear system graphically
$\\\left\{\begin{matrix} 2y - 4x = 2 \\ y = -x + 4\\ \end{matrix}\right.$
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# S'COOL Lesson: Comparing the Amount of Rainfall in Different Geographic Areas
In this lesson, students construct a rain gauge, collect and graph precipitation data, specifically the amount of rainfall at a locality, then compare their findings with other students' data.The resource includes teaching notes, a vocabulary list... (View More)
# S'COOL Lesson: Visual Opacity
In this lesson bridging art and science, students build understanding of the terms translucent, opaque, and transparent, as they apply to cloud descriptions, and create a collage using materials matching these characteristics, as well as a... (View More)
# Can you Show that the Temperature of Air has an Effect on its Weight and its Direction of Vertical Movement?
This activity has two purposes: challenge the learner to develop a procedure for investigating a research question and to learn more about factors affecting the dynamics of air in motion. It demonstrates that warm air and cold air differ in weight... (View More)
# How Clouds Form-Understanding the Basic Principles of Precipitation
The purpose of this investigation is to understand the change that takes place when water condenses from a gas to a liquid, and how a change in pressure affects this transformation. Materials needed for the experiment include a large (2L) soda... (View More)
# Design Challenge: What Factors Determine the Comfort of Air?
The purpose of this investigation is to understand how the amount of water vapor in the air at various temperatures affects the way the human body responds. This is an important basic concept for understanding why one might feel either comfortable... (View More)
# Does Air Have Weight?
This experimental activity is designed to develop an understanding that air has mass. Students conduct an investigation and observe the change in the position of a bar balancing a balloon inflated with air on one end and a uninflated balloon on the... (View More)
# Barometer Basics
This experimental activity is designed to develop a basic understanding of the interrelationship between temperature and pressure and the structure of a device made to examine this relationship. Resources needed to conduct this activity include two... (View More)
# Constructing a Barometer
This experimental activity is designed to develop a basic understanding of the relationship between temperature and pressure and that a barometer can be constructed to detect this relationship. Resources needed to build a simple barometer include a... (View More)
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http://math.stackexchange.com/questions/134340/lebesgue-decomposition-theorem-and-fundamental-theorem-of-calculus
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# Lebesgue decomposition theorem and fundamental theorem of calculus
On page 42 of 'Evans-Gariepy, Measure theory and fine properties of functions' it's stated and proved this theorem: let $\mu$ and $\nu$ be Radon measures on $\mathbb{R}^n$. Then $\nu=\nu_{ac}+\nu_s$ where the first is absolutely continuous respect to $\mu$ and the second is singular respect $\mu$.
Furthermore the derivative $D_{\mu}\nu=D_{\mu}\nu_{ac}$ and $D_{\mu}\nu_s=0 \; \mu-a.e$. Up to this everything is ok and proved, but then is stated a consequence not proved (because it seems obvious) but I don't understand: $\nu(A)=\int_{A}D_{\mu}\nu_{ac} d\mu+\nu_s(A)$. Why is there the term $\nu_s(A)$?
I thought this: $\nu(A)=\int_{A}D_{\mu}\nu_{} d\mu =\int_{A}D_{\mu}\nu_{ac} d\mu+\int_{A}D_{\mu}\nu_{s}d\mu=\int_{A}D_{\mu}\nu_{ac} d\mu$
-
You almost have the answer already. $\nu(A) = \nu_{ac}(A) + \nu_{s}(A)$, and $\nu_{ac}(A) = \int_A D_\mu \nu_{ac} d\mu$. You wrote $\nu_s(A) = \int_A D_\mu \nu_s d\mu$, but this is not in general true, because the fundamental theorem of calculus (equality) doesn't hold for $\nu_s$ singular with respect to $\mu$. You can only say that $\int_A D_\mu \nu_s d\mu \leq \nu_s(A)$. – Nicholas Stull Apr 20 '12 at 12:40
Ok, thanks i got it – balestrav Apr 20 '12 at 12:44
No problem. I wasn't sure if I had explained it well, but this was the argument that came to mind. – Nicholas Stull Apr 20 '12 at 12:46
Think about $\nu_{ac}$ the Lebesgue measure and $\nu_s$ the counting measure.
For the computation $\int_{A} d nu = \int_A d \nu_s + \int_A d \nu_s = \int_A D_\mu \nu_{ac} d \mu + \nu_s(A)$, but
In general $$\int_A d \nu_s \neq \int_A D_\mu \nu_{s} d \mu = 0.$$
-
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https://classroom.synonym.com/classroom-activities-teach-segregation-7966273.html
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Classroom Activities to Teach About Segregation
The effects of segregation can still be felt throughout society. Teachers can help children understand segregation through numerous classroom activities. However, take care when performing some of these activities, especially with younger children. Leave enough time after the activity to explain the exercise in depth and allow children to share their feelings.
1The Sneetches
For younger children, try an activity based on "The Sneetches" by Dr. Seuss. Using the class list, mark every third name. Read the marked names from the list aloud and announce that, from now on, these are the "good" students. They will have special privileges, such as lining up first or having five additional minutes of recess. Place a dot on their foreheads using a washable marker. Allow the situation to play out for half the day, with the "good" children favored in all activities. Then gather the class and explain that it was an experiment. Read aloud from "The Sneetches" and explain the randomness of picking the "good" children and compare it to the arbitrary way people decide one race or ethnic group is superior. Lead a long class discussion to soothe any hurt feelings.
2Segregation Tic-Tac-Toe
Draw a grid containing 64 numbered squares on a large piece of paper and tape it to the floor. Place 25 pennies and 30 paper clips randomly on the grid. Have the students gather around to watch. Explain that the grid represents a neighborhood and the paper clips and pennies symbolize families. Each penny wants to live near -- touching -- at least two other pennies, while each paper clip wants most of its neighbors to be paper clips. Have students suggest moves to make, such as "Move the penny on square 2 to square 24." Move each piece according to the students' directions. As students try to make every piece happy, they start to create a pattern of extreme segregation. Take the pennies and paper clips from the board and tell the students to try to create a more integrated "neighborhood." Move the pennies and paper clips according to their instructions. Discuss how an individual's social race preferences can lead to segregation.
3Eye Color
This exercise is more suitable for older children. Like the Sneetches experiment, the eye color exercise involves giving certain students special privileges but in this instance, the students are not told why the teacher favors one student over the other. Choose a particular eye color and begin treating students with that feature as superior; for example, only answer their questions or let them line up first. Never mention directly why some students are treated better than others. Observe that many students may unconsciously begin to assume their new roles by either acting superior or resenting those students who have been singled out. The next day, gather the class, explain the exercise and allow all the students to share their reactions. Leave plenty of time to discuss and air any grievances.
4Frogs and Snakes
Have children put on a play with the theme of segregation. Download the play "Why Frogs and Snakes Never Play Together" -- available from websites such as Kidsinco -- and assign children to the roles. This story of a forbidden friendship teaches that segregation is the result of learned behavior. Afterward, tell the children to write their own ending in which the snake and frog remain friends. This activity is suitable for young children and teaches a lesson about the unfairness of segregation.
Sarah Wallman has been writing in newspapers since 2004 in publications such as the "Spectator." Her primary interests are health, religion, politics and psychology. She has earned two bachelor's degrees, and has a passion for providing information on health and fitness.
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https://www.bartleby.com/solution-answer/chapter-1-problem-9rcc-precalculus-mathematics-for-calculus-6th-edition-6th-edition/9780840068071/63d69449-500b-4af9-9105-fe5a469a7011
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# The factoring of difference of squares.
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
#### Solutions
Chapter 1, Problem 9RCC
(a)
To determine
## To find: The factoring of difference of squares.
Expert Solution
### Explanation of Solution
Given:
(ab)2.
Concept used:
The y - coordinates of points :
(ab)2
(ab)2=a2+b22ab(ab)(ab)=a2+b22aba2+b2abab=a2+b22aba2+b22ab=a2+b22ab
(b)
To determine
Expert Solution
### Explanation of Solution
Given:
(ab)3.
Concept used:
The y - coordinates of points:
(ab)3
(ab)3=(ab)(ab)(ab)(ab)3=(a2+b2abab)(ab)(ab)3=(a2+b22ab)(ab)(ab)3=(a3a2b2a2b+2ab2+ab2b3)(ab)3=(a33a2b+3ab2b3)
(c)
To determine
Expert Solution
### Explanation of Solution
Given:
(a+b)3.
Concept used:
The y - coordinates of points:
(a+b)3
(a+b)3=(a+b)(a+b)(a+b)(a+b)3=(a2+b2+ab+ab)(a+b)(a+b)3=(a2+b2+2ab)(a+b)(a+b)3=(a3+a2b+2a2b+2ab2+ab2+b3)(a+b)3=(a3+3a2b+3ab2+b3)
### Have a homework question?
Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!
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https://justaaa.com/physics/263098-in-a-pool-game-the-cue-ball-which-has-an-initial
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Question
# In a pool game, the cue ball, which has an initial speed of 4.0 m/s, make...
In a pool game, the cue ball, which has an initial speed of 4.0 m/s, make an elastic collision with the eight ball, which is initially at rest. After the collision, the eight ball moves at an angle of 35° to the original direction of the cue ball.
(a) Find the direction of motion of the cue ball after the collision.
° (from the original line of motion)
(b) Find the speed of each ball. Assume that the balls have equal mass.
m/s (cue ball)
m/s (eight ball)
#### Earn Coins
Coins can be redeemed for fabulous gifts.
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| 2.515625
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CC-MAIN-2024-33
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https://lists.blender.org/pipermail/bf-extensions-cvs/2020-January/009318.html
| 1,591,004,902,000,000,000
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# [Bf-extensions-cvs] [57203ac3] blender-v2.82-release: ant_landscape: fix invalid identity checks
Wed Jan 29 03:41:36 CET 2020
```Commit: 57203ac3f2983c2689f66f8642d12441171ea32e
Author: Campbell Barton
Date: Wed Jan 29 13:40:59 2020 +1100
Branches: blender-v2.82-release
https://developer.blender.org/rBA57203ac3f2983c2689f66f8642d12441171ea32e
ant_landscape: fix invalid identity checks
===================================================================
M ant_landscape/ant_noise.py
===================================================================
diff --git a/ant_landscape/ant_noise.py b/ant_landscape/ant_noise.py
index d9f13483..e9a090f1 100644
--- a/ant_landscape/ant_noise.py
+++ b/ant_landscape/ant_noise.py
@@ -109,34 +109,34 @@ def no_bias(a):
def shapes(x, y, z, shape=0):
p = pi
- if shape is 1:
+ if shape == 1:
# ring
x = x * p
y = y * p
s = cos(x**2 + y**2) / (x**2 + y**2 + 0.5)
- elif shape is 2:
+ elif shape == 2:
# swirl
x = x * p
y = y * p
s = ((x * sin(x * x + y * y) + y * cos(x * x + y * y)) / (x**2 + y**2 + 0.5))
- elif shape is 3:
+ elif shape == 3:
# bumps
x = x * p
y = y * p
z = z * p
s = 1 - ((cos(x * p) + cos(y * p) + cos(z * p)) - 0.5)
- elif shape is 4:
+ elif shape == 4:
# wave
x = x * p * 2
y = y * p * 2
s = sin(x + sin(y))
- elif shape is 5:
+ elif shape == 5:
s = (z * p)
- elif shape is 6:
+ elif shape == 6:
s = (y * p)
- elif shape is 7:
+ elif shape == 7:
s = (x * p)
else:
@@ -154,24 +154,24 @@ def marble_noise(x, y, z, origin, size, shape, bias, sharpnes, turb, depth, hard
z += origin[2]
value = s + turb * turbulence_vector((x, y, z), depth, hard, noise_basis=basis)[1]
- if bias is 1:
+ if bias == 1:
value = cos_bias(value)
- elif bias is 2:
+ elif bias == 2:
value = tri_bias(value)
- elif bias is 3:
+ elif bias == 3:
value = saw_bias(value)
else:
value = sin_bias(value)
- if sharpnes is 1:
+ if sharpnes == 1:
value = 1.0 - sharp(value)
- elif sharpnes is 2:
+ elif sharpnes == 2:
value = 1.0 - sharper(value)
- elif sharpnes is 3:
+ elif sharpnes == 3:
value = soft(value)
- elif sharpnes is 4:
+ elif sharpnes == 4:
value = sharp(value)
- elif sharpnes is 5:
+ elif sharpnes == 5:
value = sharper(value)
else:
value = 1.0 - soft(value)
@@ -557,7 +557,7 @@ def noise_gen(coords, props):
x, y, z = coords
# Origin
- if rseed is 0:
+ if rseed == 0:
origin = x_offset, y_offset, z_offset
origin_x = x_offset
origin_y = y_offset
```
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https://www.freelancer.pl/project-directory/C-5-19
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# Katalog Projektów : Calculate and display order total on web form - Calculate Beats-Per-Minute for Songs
### Projekty, które zaczynają się od
Calculate and display order total on web form Calculate and export from Access to Excel CALCULATE and LOGIN Calculate and pivot Calculate and Plot Blocking Probabilities for a Circuit-Switched System Calculate and Plot Blocking Probabilities for a Circuit-Switched System - repost Calculate and Plot Blocking Probabilities for a Circuit-Switched System - repostt Calculate and Plot Blocking Probabilities for a Circuit-Switched System - repostt - repost Calculate and Plot Blocking Probabilities for a Circuit-Switched System - repostt - repost 2 Calculate and Plot Blocking Probabilities for a Circuit-Switched System - repostt - repostt calculate and visualise volume of soil needed to be displaced - open to bidding calculate and visualise volume of soil needed to be displaced . Calculate angle between two coordinates sets in Python Calculate angle in x, y and z axis using Arduino + sensor Calculate angle in x, y and z axis using Arduino + sensor -- 2 Calculate Annual Leave Remaining + Send Notification Email calculate app in Java
Calculate Arabic Numerology then Convert to Words Calculate Arbitrage Betting Calculate Area and Collect a text vaues from DWG-file calculate area of ellipse in excel or ANSYS software Calculate area of land from polygon coordinates in Excel Spreadsheet Calculate ascendant and zodiac Calculate ascendant and zodiac - repost calculate average number of goals/points for hockey/basketball/football from given odds Calculate average, highest, lowest, and standard deviation in Visual Basic Calculate Averages of test scores using arrays Calculate Avg. & Stand Dev. calculate azimuth and elevation(repost) Calculate Azimuth angle and Elevation angle for normal vector of plane (latitude, longitude, elevation) Calculate basic mathletics Calculate bearing / heading from point A to point b if given latitude and longatude for 2 points Calculate Beats-Per-Minute (on Mac) Calculate Beats-Per-Minute for Songs
| 416
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| 2.859375
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CC-MAIN-2017-22
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http://www.numbersaplenty.com/24643
| 1,590,428,940,000,000,000
|
text/html
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| 195,992,045
| 3,686
|
Search a number
24643 = 191297
BaseRepresentation
bin110000001000011
31020210201
412001003
51242033
6310031
7131563
oct60103
936721
1024643
1117573
1212317
13b2a8
148da3
15747d
hex6043
24643 has 4 divisors (see below), whose sum is σ = 25960. Its totient is φ = 23328.
The previous prime is 24631. The next prime is 24659. The reversal of 24643 is 34642.
Subtracting 24643 from its reverse (34642), we obtain a palindrome (9999).
It can be divided in two parts, 246 and 43, that added together give a square (289 = 172).
It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 34642 = 217321.
It is a 5-Lehmer number, since φ(24643) divides (24643-1)5.
It is a cyclic number.
It is not a de Polignac number, because 24643 - 25 = 24611 is a prime.
It is a Harshad number since it is a multiple of its sum of digits (19), and also a Moran number because the ratio is a prime number: 1297 = 24643 / (2 + 4 + 6 + 4 + 3).
It is an Ulam number.
It is a d-powerful number, because it can be written as 213 + 4 + 62 + 47 + 33 .
It is a Duffinian number.
24643 is a lucky number.
It is a self number, because there is not a number n which added to its sum of digits gives 24643.
It is not an unprimeable number, because it can be changed into a prime (24623) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (5) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 630 + ... + 667.
It is an arithmetic number, because the mean of its divisors is an integer number (6490).
224643 is an apocalyptic number.
24643 is a deficient number, since it is larger than the sum of its proper divisors (1317).
24643 is a wasteful number, since it uses less digits than its factorization.
24643 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1316.
The product of its digits is 576, while the sum is 19.
The square root of 24643 is about 156.9808905568. The cubic root of 24643 is about 29.1003263190.
The spelling of 24643 in words is "twenty-four thousand, six hundred forty-three".
Divisors: 1 19 1297 24643
| 679
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http://mathv.chapman.edu/~jipsen/structures/doku.php/lattice-ordered_rings?rev=1280443597&do=diff
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Differences
This shows you the differences between two versions of the page.
lattice-ordered_rings [2010/07/29 15:46] (current)
Line 1: Line 1:
+=====Lattice-ordered rings=====
+Abbreviation: **LRng**
+====Definition====
+A \emph{lattice-ordered ring} (or $\ell$\emph{-ring}) is a structure $\mathbf{L}=\langle L,\vee,\wedge,+,-,0,\cdot\rangle$ such that
+
+
+$\langle L,\vee,\wedge\rangle$ is a [[lattice]]
+
+
+$\langle L,+,-,0,\cdot\rangle$ is a [[ring]]
+
+
+$+$ is order-preserving: $x\leq y\Longrightarrow x+z\leq y+z$
+
+
+${\uparrow}0$ is closed under $\cdot$: $0\leq x,y\Longrightarrow 0\leq x\cdot y$
+
+
+Remark:
+
+====Definition====
+==Morphisms==
+Let $\mathbf{L}$ and $\mathbf{M}$ be $\ell$-rings. A morphism from $\mathbf{L}$ to $\mathbf{M}$ is a function $f:L\rightarrow M$ that is a
+homomorphism: $f(x\vee y)=f(x)\vee f(y)$, $f(x\wedge y)=f(x)\wedge f(y)$, $f(x\cdot y)=f(x)\cdot f(y)$, $f(x+y)=f(x)+f(y)$.
+====Examples====
+
+
+====Basic results====
+The lattice reducts of lattice-ordered rings are [[distributive lattices]].
+
+====Properties====
+^[[Classtype]] |variety |
+^[[Equational theory]] | |
+^[[Quasiequational theory]] | |
+^[[First-order theory]] | |
+^[[Congruence distributive]] |yes, see [[lattices]] |
+^[[Congruence extension property]] | |
+^[[Congruence n-permutable]] |yes, $n=2$, see [[groups]] |
+^[[Congruence regular]] |yes, see [[groups]] |
+^[[Congruence uniform]] |yes, see [[groups]] |
+
+^[[Definable principal congruences]] | |
+^[[Equationally def. pr. cong.]] | |
+^[[Amalgamation property]] | |
+^[[Strong amalgamation property]] | |
+^[[Epimorphisms are surjective]] | |
+
+====Finite members====
+
+$\begin{array}{lr} +None +\end{array}$
+
+====Subclasses====
+[[Commutative lattice-ordered rings]]
+
+====Superclasses====
+[[Abelian lattice-ordered groups]]
+
+
+====References====
+
+[(Ln19xx>
+)]
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https://cstheory.stackexchange.com/questions/14079/shortest-cycle-with-a-specific-number-of-vertices
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# Shortest cycle with a specific number of vertices
Given an undirected graph with n nodes, I need to find the shortest cycle of involving exactly n/2 vertices (i.e. keeping the distance traveled by the cycle to a minimum). Some nodes cannot directly connect; they're in specific areas and I can only travel from one area to another, not within one area.
I'm just asking in general here, what kind of techniques can I use to get the best (shortest) possible cycle of n/2 nodes? So far I've written a basic substitution optimizer and just a greedy Depth First Search. I'm wondering what would be the best approach for me to start on now?
• It seems you are using "length/distance" in two different ways. If I understand correctly, I would say that you are looking for "the shortest cycle that involves exactly 40 vertices." Oct 25, 2012 at 2:15
• That's exactly what I mean, thank you, I'll edit it now. Oct 25, 2012 at 15:39
• This sounds like a generalization of TSP Oct 25, 2012 at 16:22
• You should make the question general, 80 and 40 should be replaced by n and n/2. Are you sure there exists a cycle of length at least n/2? Because in general finding the longest cycle is a very hard question. Nov 15, 2012 at 16:28
• Note that if you don't require the cycle to be simple (the cycle can repeat edges and vertices, and you count edges with multiplicity), then you can solve this with fairly standard dynamic-programming algorithms. Nov 26, 2012 at 4:37
• Constant factor for k-TSP applies only for metric graphs. Otherwise the problem is not approximable via TSP. For metric k-TSP there is a $2$-approximation due to Naveen Garg dl.acm.org/citation.cfm?id=1060650. For asymmetric distances that satisfy triangle inequality, an $O(\log^3 k)$ is known from cs.illinois.edu/homes/chekuri/papers/orienteering-journal.pdf and an $O(\log^2 n \log k/\log \log n)$ result from researcher.watson.ibm.com/researcher/files/us-viswanath/…. Oct 26, 2012 at 15:49
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TOPICS
# Wheat and Chessboard Problem
Let one grain of wheat be placed on the first square of a chessboard, two on the second, four on the third, eight on the fourth, etc. How many grains total are placed on an chessboard? Since this is a geometric series, the answer for squares is
a Mersenne number. Plugging in then gives .
Mersenne Number
## Explore with Wolfram|Alpha
More things to try:
## References
Pappas, T. "The Wheat & Chessboard." The Joy of Mathematics. San Carlos, CA: Wide World Publ./Tetra, p. 17, 1989.Steinhaus, H. Mathematical Snapshots, 3rd ed. New York: Dover, pp. 23-24, 1999.
## Referenced on Wolfram|Alpha
Wheat and Chessboard Problem
## Cite this as:
Weisstein, Eric W. "Wheat and Chessboard Problem." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/WheatandChessboardProblem.html
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# Projection
## 1 Definitions
### 1.1 Distance between point and set
The distance d from point \mathbf{y} \in \mathbb{R}^n to closed set S \subset \mathbb{R}^n:
d(\mathbf{y}, S, \| \cdot \|) = \inf\{\|x - y\| \mid x \in S \}
### 1.2 Projection of a point on set
Projection of a point \mathbf{y} \in \mathbb{R}^n on set S \subseteq \mathbb{R}^n is a point \pi_S(\mathbf{y}) \in S:
\| \pi_S(\mathbf{y}) - \mathbf{y}\| \le \|\mathbf{x} - \mathbf{y}\|, \forall \mathbf{x} \in S
• if a set is open, and a point is beyond this set, then its projection on this set does not exist.
• if a point is in set, then its projection is the point itself
• \pi_S(\mathbf{y}) = \underset{\mathbf{x}}{\operatorname{argmin}} \|\mathbf{x}-\mathbf{y}\|
• Let S \subseteq \mathbb{R}^n - convex closed set. Let the point \mathbf{y} \in \mathbb{R}^n и \mathbf{\pi} \in S. Then if for all \mathbf{x} \in S the inequality holds:
\langle \pi -\mathbf{y}, \mathbf{x} - \pi\rangle \ge 0,
then \pi is the projection of the point \mathbf{y} on S, so \pi_S (\mathbf{y}) = \pi.
• Let S \subseteq \mathbb{R}^n - affine set. Let we have points \mathbf{y} \in \mathbb{R}^n and \mathbf{\pi} \in S. Then \pi is a projection of point \mathbf{y} on S, so \pi_S (\mathbf{y}) = \pi if and only if for all \mathbf{x} \in S the inequality holds:
\langle \pi -\mathbf{y}, \mathbf{x} - \pi\rangle = 0
• Sufficient conditions of existence of a projection. If S \subseteq \mathbb{R}^n - closed set, then the projection on set S exists for any point.
• Sufficient conditions of uniqueness of a projection. If S \subseteq \mathbb{R}^n - closed convex set, then the projection on set S is unique for any point.
Example
Find \pi_S (y) = \pi, if S = \{x \in \mathbb{R}^n \mid \|x - x_0\| \le R \}, y \notin S
• Build a hypothesis from the figure: \pi = x_0 + R \cdot \frac{y - x_0}{\|y - x_0\|}
• Check the inequality for a convex closed set: (\pi - y)^T(x - \pi) \ge 0
\left( x_0 - y + R \frac{y - x_0}{\|y - x_0\|} \right)^T\left( x - x_0 - R \frac{y - x_0}{\|y - x_0\|} \right) =
\left( \frac{(y - x_0)(R - \|y - x_0\|)}{\|y - x_0\|} \right)^T\left( \frac{(x-x_0)\|y-x_0\|-R(y - x_0)}{\|y - x_0\|} \right) =
\frac{R - \|y - x_0\|}{\|y - x_0\|^2} \left(y - x_0 \right)^T\left( \left(x-x_0\right)\|y-x_0\|-R\left(y - x_0\right) \right) =
\frac{R - \|y - x_0\|}{\|y - x_0\|} \left( \left(y - x_0 \right)^T\left( x-x_0\right)-R\|y - x_0\| \right) =
\left(R - \|y - x_0\| \right) \left( \frac{(y - x_0 )^T( x-x_0)}{\|y - x_0\|}-R \right)
• The first factor is negative for point selection y. The second factor is also negative, which follows from the Cauchy-Bunyakovsky inequality:
(y - x_0 )^T( x-x_0) \le \|y - x_0\|\|x-x_0\|
\frac{(y - x_0 )^T( x-x_0)}{\|y - x_0\|} - R \le \frac{\|y - x_0\|\|x-x_0\|}{\|y - x_0\|} - R = \|x - x_0\| - R \le 0
Example
Find \pi_S (y) = \pi, if S = \{x \in \mathbb{R}^n \mid c^T x = b \}, y \notin S.
• Build a hypothesis from the figure: \pi = y + \alpha c. Coefficient \alpha is chosen so that \pi \in S: c^T \pi = b, so:
c^T (y + \alpha c) = b
c^Ty + \alpha c^T c = b
c^Ty = b - \alpha c^T c
• Check the inequality for a convex closed set: (\pi - y)^T(x - \pi) \ge 0
(y + \alpha c - y)^T(x - y - \alpha c) =
\alpha c^T(x - y - \alpha c) =
\alpha (c^Tx) - \alpha (c^T y) - \alpha^2 (c^Tc) =
\alpha b - \alpha (b - \alpha c^T c) - \alpha^2 c^Tc =
\alpha b - \alpha b + \alpha^2 c^T c - \alpha^2 c^Tc = 0 \ge 0
Example
Find \pi_S (y) = \pi, if S = \{x \in \mathbb{R}^n \mid Ax = b, A \in \mathbb{R}^{m \times n}, b \in \mathbb{R}^{m} \}, y \notin S.
• Build a hypothesis from the figure: \pi = y + \sum\limits_{i=1}^m\alpha_i A_i = y + A^T \alpha. Coefficient \alpha is chosen so that \pi \in S: A \pi = b, so:
A(y + A^T\alpha) = b
Ay = b - A A^T\alpha
• Check the inequality for a convex closed set: (\pi - y)^T(x - \pi) \ge 0
(y + A^T\alpha - y)^T(x - y - A^T\alpha) =
\alpha^T A(x - y - A^T\alpha) =
\alpha^T (Ax) - \alpha^T (A y) - \alpha^T (AA^T \alpha) =
\alpha^T b - \alpha^T (b - A A^T\alpha) - \alpha^T AA^T \alpha =
\alpha^T b - \alpha^T b + \alpha^T AA^T \alpha - \alpha^T AA^T \alpha = 0 \ge 0
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# Class 8 RD Sharma Solutions – Chapter 13 Profit Loss Discount And Value Added Tax – Exercise 13.2 | Set 1
### Question 1: Find the S.P. if M.P. = Rs 1300 and Discount = 10%
Solution:
Using Formula S.P = (M.P * (100 – Discount%))/100
Hence, S.P = Rs (1300 * (100 – 10))/100
= Rs (1300 * 90)/100
= Rs 1170
### Question 2: Find the S.P. if M.P. = Rs 500 and Discount = 15%
Solution:
Using Formula, S.P = (M.P * (100 – Discount%))/100
Hence, S.P = Rs (500 * (100 – 15))/100
= Rs ((500 * 85))/100
= Rs 425
### Question 3: Find the M.P. if S.P. = Rs 1222 and Discount = 6%
Solution:
Using Formula, M.P = (S.P * 100)/(100 – Discount%)
Hence, M.P = Rs (1222 * 100)/(100 – 6)
= Rs (1222 * 100)/94
= Rs 1300
### Question 4: Find the M.P. if S.P. = Rs 495 and Discount = 1%
Solution:
Using Formula, M.P = (S.P * 100)/(100 – Discount%)
Hence, M.P = Rs (495 * 100)/(100 – 1)
= Rs (495 * 100)/99
= Rs 500
### Question 5: Find discount in percent when M.P. = Rs 900 and S.P = Rs 873
Solution :
Using Formula, Discount% = ((M.P – S.P) * 100)/M.P
Hence, Discount% = (((900 – 873) * 100)/900)%
= ((27 * 100)/900)%
= 3%
### Question 6: Find discount in percent when M.P. = Rs 500 and S.P. = Rs 425
Solution:
Using Formula, Discount% = ((M.P – S.P) * 100)/M.P
Hence, Discount% = (((500 – 425) * 100)/500)%
= ((75 * 100)/500)%
= 15%
### Question 7: A shop selling sewing machines offers 3% discount on all cash purchases. What cash amount does a customer pay for a sewing machine (correct up to two decimal places) the price of which is marked as Rs 650.
Solution:
Here, M.P of sewing machine = Rs 650
Discount% = 3%
To find, S.P of sewing machine.
Using Formula, S.P = (M.P * (100 – Discount%))/100
Hence, S.P of sewing machine = Rs(650 * (100 – 3))/100
= Rs (650 * 97)/100
= Rs 630.50
### Question 8: The marked price of a ceiling fan is Rs 720. During offseason, it is sold for Rs 684. Determine the discount percentage.
Solution:
Here, M.P of ceiling fan = Rs 720
S.P of ceiling fan = Rs 684
To find, Discount%
Using Formula, Discount% = ((M.P – S.P) * 100)/M.P
Hence, Discount% = (((720 – 684) * 100)/720)%
= ((36 * 100)/720)%
= 5%
### Question 9: On the eve of Gandhi Jayanti a saree is sold for Rs 720 after allowing 20% discount. What is its marked price?
Solution:
Here, S.P of saree = Rs 720
Discount% = 20%
To find M.P of saree
Using Formula, M.P = (S.P * 100)/(100 – Discount%)
Hence, M.P of saree = Rs (720 * 100)/(100 – 20)
= Rs (720 * 100)/80
= Rs 900
### Question 10: After allowing a discount of 7.50% on the marked price, an article is sold for Rs 555. Find its marked price.
Solution:
Here, S.P of article = Rs 555
Discount% = 7.50%
Using Formula, M.P = (S.P * 100)/(100 – Discount%)
Hence, M.P of article = Rs (555 * 100)/(100 – 7.50)
= Rs (555 * 100)/92.50
= Rs 600
### Question 11: A shopkeeper allows his customers 10% off on the marked price of goods and still gets a profit of 25%. What is the actual cost to him of an article marked Rs 250?
Solution:
Here, M.P of goods = Rs 250
Discount% given to customer = 10%
Profit% earned by shopkeeper = 25%
To find C.P of goods.
Using Formula, S.P = (M.P * (100 – Discount%))/100
Hence, S.P of goods = Rs (250*(100 – 10))/100
= Rs (250 * 90)/100
= Rs 225
Using Formula, C.P = (S.P * 100)/(100 + Profit%)
Hence, C.P of goods = Rs (225 * 100)/(100 + 25)
= Rs (225 * 100)/125
= Rs 180
### Question 12: A shopkeeper allows 20% off on the marked price of goods and still gets a profit of 25%. What is the actual cost to him of an article marked Rs 500?
Solution:
Here, M.P of goods = Rs 500
Discount% given to customer = 20%
Profit% earned by shopkeeper = 25%
To find C.P of goods.
Using Formula, S.P = (M.P * (100 – Discount%))/100
Hence, S.P of goods = Rs (500 * (100 – 20))/100
= Rs (500 * 80)/100
= Rs 400
Using Formula, C.P = (S.P * 100)/(100 + Profit%)
Hence, C.P of goods = Rs (400 * 100)/(100 + 25)
= Rs (400 * 100)/125
= Rs 320
### Question 13: A tradesman marks his goods at such a price after allowing a discount of 15%, he makes a profit of 20%. What is the marked price of an article whose cost price is Rs 170?
Solution:
Here, C.P of the article = Rs 170
Discount% = 15%
Profit% earned by tradesman = 20%
To find M.P of the article.
Using Formula, S.P = (C.P * (100 + Profit%))/100
Hence, S.P of the article = Rs (170 * (100 + 20))/100
= Rs (170 * 120)/100
= Rs 204
Using Formula, M.P = (S.P * 100)/(100 – Discount%)
Hence, M.P of the article = Rs (204 * 100)/(100 – 15)
= Rs (204 * 100)/85
= Rs 240
### Chapter 13 Profit Loss Discount And Value Added Tax – Exercise 13.2 | Set 2
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# If y=(ax+b)/(cx+d) and if (a+d)=0 then show that (y-x)y"=2y(1+y')
Giridharan
13 Points
3 years ago
y=(ax+b)/(cx+d) and a+d=0
a+d=0 ⇒d=-a
Hence y=(ax+b)/(cx-a)
⇒y(cx-a)=(ax+b)
Differentiating w.r.t.x
cy+(cx-a)y1= a––––(1)
Differentiating again w.r.t.x
2cy1+(cx-a)y2=0–––(2)
⇒a = 2cy1/y2 + cx–––(3)
(1) ⇒ cy +(cx-a)y1 = a
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we get 2y1(1+y1)=y2(y-x)
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# Hot water of mass 0.35kg at 80degrees celsius is poured into a copper cup which is initially at 25 degrees celsius. The mass of the cup is 0.80kg. Assume that no energy is lost to the surroundings.Give - Specific Heat Capacity of water = 4200Jkg-1C-1 Specific Heat Capacity of Copper = 400Jkg-1C-1 Question) Find the initial Temperature of water.Thanks,Vishnu
Rohith Gandhi
24 Points
14 years ago
Dear Vishnu,
since no energy is lost to surroundings
heat lost by water = heat gained by copper cup
let final common temperature of the mixture = T
therefore,
0.35X4200X(80-T) = 0.8X400X(T-25)
117600-1470T = 320T-8000
=> 117600+8000 = 1470T+320T
125600=1790T
=>T = 125600/1790 = 70.16 degree C
final common temperature of mixture = 70.16 degree C
Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We
are all IITians and here to help you in your IIT JEE preparation.
All the best Vishnu !!!
Regards,
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# CS203 Lecture 6 - PowerPoint PPT Presentation
CS203 Lecture 6. John Hurley Cal State LA.
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CS203 Lecture 6
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## CS203 Lecture 6
John Hurley
Cal State LA
Suppose two algorithms perform the same task such as search (linear search vs. binary search) and sorting (selection sort vs. insertion sort). Which one is better? One possible approach to answer this question is to implement these algorithms in Java and run the programs to get execution time. But there are two problems for this approach:
• First, there are many tasks running concurrently on a computer. The execution time of a particular program is dependent on the system load.
• Second, the execution time is dependent on specific input. Consider linear search and binary search for example. If an element to be searched happens to be the first in the list, linear search will find the element quicker than binary search.
### Execution Time
It is very difficult to compare algorithms by measuring their execution time. To overcome these problems, a theoretical approach was developed to analyze algorithms independent of computers and specific input.
This approach approximates the effect of a change on the size of the input. In this way, you can see how fast an algorithm’s execution time increases as the input size increases, so you can compare two algorithms by examining their growth rates.
### Growth Rate
Consider linear search. The linear search algorithm compares the key with the elements in the array sequentially until the key is found or the array is exhausted. If the key is not in the array, it requires n comparisons for an array of size n. If the key is in the array, it requires n/2 comparisons on average.
The algorithm’s execution time is proportional to the size of the array. If you double the size of the array, you will expect the number of comparisons to double. The algorithm grows at a linear rate. The growth rate has an order of magnitude of n. Computer scientists use the Big O notation to abbreviate for “order of magnitude.” Using this notation, the complexity of the linear search algorithm is O(n), pronounced as “order of n.”
### Big O Notation
For the same input size, an algorithm’s execution time may vary, depending on the input. An input that results in the shortest execution time is called the best-case input and an input that results in the longest execution time is called the worst-case input.
Best-case and worst-case are not representative, but worst-case analysis is very useful. You can show that the algorithm will never be slower than the worst-case. An average-case analysis attempts to determine the average amount of time among all possible input of the same size.
Average-case analysis is ideal, but difficult to perform, because it is hard to determine the relative probabilities and distributions of various input instances for many problems. Worst-case analysis is easier to obtain and is thus common. So, the analysis is generally conducted for the worst-case.
### Best, Worst, and Average
The linear search algorithm requires n comparisons in the worst-case and n/2 comparisons in the average-case. Using the Big O notation, both cases require O(n) time. The multiplicative constant (1/2) can be omitted. Algorithm analysis is focused on growth rate. The multiplicative constants have no impact on growth rates. The growth rate for n/2 or 100n is the same as n, i.e., O(n) = O(n/2) = O(100n).
### Ignore Multiplicative Constants
Consider the algorithm for finding the maximum number in an array of n elements. If n is 2, it takes one comparison to find the maximum number. If n is 3, it takes two comparisons to find the maximum number. In general, it takes n-1comparisons to find maximum number in a list of n elements.
Algorithm analysis is concerned with large input size. If the input size is small, there is no significance to estimate an algorithm’s efficiency. As n grows larger, the n part in the expression n-1 dominates the complexity. The Big O notation allows you to ignore the non-dominating part (e.g., -1 in the expression n-1) and highlight the important part (e.g., n in the expression n-1). So, the complexity of this algorithm is O(n).
constant time
executed
n times
### Repetition: Simple Loops
for (i = 1; i <= n; i++) {
k = k + 5;
}
Time Complexity
T(n) = (a constant c) * n = cn = O(n)
Ignore multiplicative constants (e.g., “c”).
constant time
inner loop
executed
n times
executed
n times
### Repetition: Nested Loops
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
k = k + i + j;
}
}
Time Complexity
T(n) = (a constant c) * n * n = cn2 = O(n2)
Ignore multiplicative constants (e.g., “c”).
constant time
inner loop
executed
i times
executed
n times
### Repetition: Nested Loops
for (i = 1; i <= n; i++) {
for (j = 1; j <= i; j++) {
k = k + i + j;
}
}
Time Complexity
T(n) = c + 2c + 3c + 4c + … + nc = cn(n+1)/2 = (c/2)n2 + (c/2)n = O(n2)
Ignore non-dominating terms
Ignore multiplicative constants
constant time
inner loop
executed
20 times
executed
n times
### Repetition: Nested Loops
for (i = 1; i <= n; i++) {
for (j = 1; j <= 20; j++) {
k = k + i + j;
}
}
Time Complexity
T(n) = 20 * c * n = O(n)
Ignore multiplicative constants (e.g., 20*c)
inner loop
executed
20 times
executed
n times
executed
10 times
### Sequence
for (j = 1; j <= 10; j++) {
k = k + 4;
}
for (i = 1; i <= n; i++) {
for (j = 1; j <= 20; j++) {
k = k + i + j;
}
}
Time Complexity
T(n) = c *10 + 20 * c * n = O(n)
O(n)
Let n be
list.size().
Executed
n times.
### Selection
if (list.contains(e)) {
System.out.println(e);
}
else
for (Object t: list) {
System.out.println(t);
}
Time Complexity
T(n) = test time + worst-case (if, else)
= O(n) + O(n)
= O(n)
The Big O notation estimates the execution time of an algorithm in relation to the input size. If the time is not related to the input size, the algorithm is said to take constant time with the notation O(1). For example, a method that retrieves an element at a given index in an array takes constant time, because it does not grow as the size of the array increases.
### Constant Time
In the last couple of weeks, we have covered various data structures that are implemented in the Java Collections Framework. As a working programmer, you may often use these, and most of the rest of the time, you will use various libraries that supply alternatives. You will not often have to implement the data structures yourself.
However, in order to understand how these structures work at a lower level, we will cover implementation this week.
### Implementation of Data Structures
A list stores data in sequential order. For example, a list of students, a list of available rooms, a list of cities, and a list of books, etc. can be stored using lists. The common operations on a list are usually the following:
·Retrieve an element from this list.
·Insert a new element to this list.
·Delete an element from this list.
·Find how many elements are in this list.
·Find if an element is in this list.
·Find if this list is empty.
### Lists
There are two common ways to implement a list in Java.
• Using arrays. One is to use an array to store the elements. The array is dynamically created. If the capacity of the array is exceeded, create a new larger array and copy all the elements from the current array to the new array.
• Using a linked list. The other approach is to use a linked structure. A linked structure consists of nodes. Each node is dynamically created to hold an element. All the nodes are linked together to form a list.
### Two Ways to Implement Lists
Sample Code is linked from course page
### Two Ways to Implement Lists
For convenience, let’s name these two classes: MyArrayList and MyLinkedList. These two classes have common operations, but different data fields. The common operations can be generalized in an interface or an abstract class. A good strategy is to combine the virtues of interfaces and abstract classes by providing both interface and abstract class in the design so the user can use either the interface or the abstract class whichever is convenient. Such an abstract class is known as a convenience class.
MyList
### MyList Interface and MyAbstractList Class
MyAbstractList
Once an array is created, its size cannot be changed. Nevertheless, you can still use array to implement dynamic data structures. The trick is to create a new larger array to replace the current array if the current array cannot hold new elements in the list.
Initially, an array, say data of Object[] type, is created with a default size. When inserting a new element into the array, first ensure there is enough room in the array. If not, create a new array twice the size of the current one. Copy the elements from the current array to the new array. The new array now becomes the current array.
### Array List
Before inserting a new element at a specified index, shift all the elements after the index to the right and increase the list size by 1.
### Insertion
To remove an element at a specified index, shift all the elements after the index to the left by one position and decrease the list size by 1.
### Implementing MyArrayList
Since MyArrayList is implemented using an array, the methods get(int index) and set(int index, Object o) for accessing and modifying an element through an index and the add(Object o) for adding an element at the end of the list are efficient.
However, the methods add(int index, Object o) and remove(int index) are inefficient because it requires shifting potentially a large number of elements. You can use a linked structure to implement a list to improve efficiency for adding and removing an element anywhere in a list.
A linked list consists of nodes. Each node contains an element, and each node is linked to its next neighbor. Thus a node can be defined as a class, as follows:
class Node<E> {
E element;
Node next;
public Node(E o) {
element = o;
}
}
### Nodes in Linked Lists
The variable head refers to the first node in the list, and the variable tail refers to the last node in the list. If the list is empty, both are null. For example, you can create three nodes to store three strings in a list, as follows:
Step 1: Declare head and tail:
### Adding Three Nodes
Step 2: Create the first node and insert it to the list:
### Adding Three Nodes, cont.
Step 3: Create the second node and insert it to the list:
### Adding Three Nodes, cont.
Step 4: Create the third node and insert it to the list:
### Adding Three Nodes, cont.
Each node contains the element and a data field named next that points to the next element. If the node is the last in the list, its pointer data field next contains the value null. You can use this property to detect the last node. For example, you may write the following loop to traverse all the nodes in the list.
Node<E> current = head;
while (current != null) {
System.out.println(current.element);
current = current.next;
}
### Traversing All Elements in the List
Run
public void addFirst(E o) {
Node<E> newNode = new Node<E>(o);
size++;
if (tail == null)
}
### Implementing addFirst(E o)
public void addLast(E o) {
if (tail == null) {
head = tail = new Node<E>(element);
}
else {
tail.next = new Node(element);
tail = tail.next;
}
size++;
}
### Implementing addLast(E o)
public void add(int index, E o) {
if (index == 0) addFirst(o);
else if (index >= size) addLast(o);
else {
Node<E> current = head;
for (int i = 1; i < index; i++)
current = current.next;
Node<E> temp = current.next;
current.next = new Node<E>(o);
(current.next).next = temp;
size++;
}
}
### Implementing add(int index, E o)
public E removeFirst() {
if (size == 0) return null;
else {
Node<E> temp = head;
size--;
if (head == null) tail = null;
return temp.element;
}
}
### Implementing removeFirst()
public E removeLast() {
if (size == 0) return null;
else if (size == 1)
{
Node<E> temp = head;
head = tail = null;
size = 0;
return temp.element;
}
else
{
Node<E> current = head;
for (int i = 0; i < size - 2; i++)
current = current.next;
Node temp = tail;
tail = current;
tail.next = null;
size--;
return temp.element;
}
}
### Implementing removeLast()
public E remove(int index) {
if (index < 0 || index >= size) return null;
else if (index == 0) return removeFirst();
else if (index == size - 1) return removeLast();
else {
Node<E> previous = head;
for (int i = 1; i < index; i++) {
previous = previous.next;
}
Node<E> current = previous.next;
previous.next = current.next;
size--;
return current.element;
}
}
### Time Complexity for ArrayList and LinkedList
• A circular, singly linked list is like a singly linked list, except that the pointer of the last node points back to the first node.
### Circular Linked Lists
• A doubly linkedlist contains the nodes with two pointers. One points to the next node and the other points to the previous node. These two pointers are conveniently called a forward pointer and a backward pointer. So, a doubly linked list can be traversed forward and backward.
### Doubly Linked Lists
• A circular, doubly linked list is doubly linked list, except that the forward pointer of the last node points to the first node and the backward pointer of the first pointer points to the last node.
### Circular Doubly Linked Lists
A stack can be viewed as a special type of list, where the elements are accessed, inserted, and deleted only from the end, called the top, of the stack.
### Stacks
A queue represents a waiting list. A queue can be viewed as a special type of list, where the elements are inserted into the end (tail) of the queue, and are accessed and deleted from the beginning (head) of the queue.
### Queues
• Using an array list to implement Stack
• Use a linked list to implement Queue
Since the insertion and deletion operations on a stack are made only at the end of the stack, using an array list to implement a stack is more efficient than a linked list. Since deletions are made at the beginning of the list, it is more efficient to implement a queue using a linked list than an array list. This section implements a stack class using an array list and a queue using a linked list.
### Implementing Stacks and Queues
There are two ways to design the stack and queue classes:
• Using inheritance: You can define the stack class by extending the array list class, and the queue class by extending the linked list class.
• Using composition: You can define an array list as a data field in the stack class, and a linked list as a data field in the queue class.
### Design of the Stack and Queue Classes
Both designs are fine, but using composition is better because it enables you to define a complete new stack class and queue class without inheriting the unnecessary and inappropriate methods from the array list and linked list.
GenericStack
GenericQueue
### MyStack and MyQueue
A regular queue is a first-in and first-out data structure. Elements are appended to the end of the queue and are removed from the beginning of the queue. In a priority queue, elements are assigned with priorities. When accessing elements, the element with the highest priority is removed first. A priority queue has a largest-in, first-out behavior. For example, the emergency room in a hospital assigns patients with priority numbers; the patient with the highest priority is treated first.
### Priority Queue
The term "Heap" is used in several different ways. The Priority Queue example code uses a heap that adheres to this definition:
A heap is a binary tree (one in which each node has at most two children) that with these two properties:
Order property:
If A is a parent node of B then the key of node A is ordered with respect to the key of node B with the same ordering applying across the heap. Either the keys of parent nodes are always greater than or equal to those of the children and the highest key is in the root node (max heap) or the keys of parent nodes are less than or equal to those of the children and the lowest key is in the root node (min heap).
Shape property:
1- All leaves are either at depth d or d-1 (for some value d).
2- All of the leaves at depth d-1 are to the right of the leaves at depth d.
3- (a) There is at most 1 node with just 1 child. (b) Any such child is the left child of its parent, and (c) it is the rightmost leaf at depth d.
### Heap
For a node at position i, its left child is at position 2i+1 and its right child is at position 2i+2, and its parent is (i-1)/2. For example, the node for element 39 is at position 4, so its left child (element 14) is at 9 (2*4+1), its right child (element 33) is at 10 (2*4+2), and its parent (element 42) is at 1 ((4-1)/2).
### Representing a Heap
Adding 3, 5, 1, 19, 11, and 22 to a heap, initially empty
### Adding Elements to the Heap
Adding 88 to the heap
### Rebuild the heap after adding a new node
Removing root 62 from the heap
Move 9 to root
Swap 9 with 59
Swap 9 with 44
Swap 9 with 30
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# Homework Help: Given a series of output, how to determine function?
1. Apr 12, 2014
### iamsmooth
1. The problem statement, all variables and given/known data
I am a dumb programmer trying to figure out the relationship for a sequence of output. I can't seem to figure it out by guessing, so I assume there's a way to mathematically work this out.
Anyways I wrote a program to do discrete event simulation and I have the following numbers on a given input x that represents the % of time the system spends in this state. The numbers are not exactly what they should be because it's simulated, but there's a huge sample so it should be very close.
x = 1, f(x) = 0
x = 2, f(x) = 0.063
x = 3, f(x) = 0.156
x = 4, f(x) = 0.26
x = 5, f(x) = 0.366
x = 6, f(x) = 0.467
2. Relevant equations
This may have something to do with the fraction 1/4.
My system has events that spend 25% of the time doing one thing, then 75% of the time doing another.
3. The attempt at a solution
I tried guessing and checking, the only way I know how. I figured 0.25^0 = 1, 0.25^1 = 0.25, 0.25^2 = 0.063, then 0.25^3 = 0.015625 which is close, but one decimal off and finally 0.25^4 = 0.00390625 which isn't even close anymore to the above.
I also tried dividing each subsequent output with the last output to try and get some sort of pattern and it seems sporadic. I'm so confused >< Please point me in the right direction!
2. Apr 12, 2014
### Simon Bridge
Nope - there is no way to uniquely determine the generating function from a series of discrete samples.
You have to start out with some idea about what the function could be, then test that idea by doing some sort of fitness test.
I can fit your data really well to a quartic or a quintic polynomial.
... it sounds more like a stochastic process. If it is statistical - involving some randomness - then f(x) is not going to be smooth.
Presumably you know the generating algorithm?
Last edited: Apr 12, 2014
3. Apr 12, 2014
### iamsmooth
The algorithm is that each event will perform an action A followed by action B. Action A is performed for an exponentially distributed amount of time with a mean of 10 minutes, and B is performed for an exponentially distributed amount of time for 30 minutes.
So what I'm trying to figure out is the percentage of time the system has 2 or more events performing action A simultaneously. So when there is 1 event, the answer is 0%, because it can't spend anytime with the system having more than 2 events doing activity A. 2 events = 0.063, 3 events = 0.156, 4 events = 0.26. Eventually it approaches 100% when you have a lot of events, because it's going to be very rare that only 0 or 1 events are performing action A.
So basically that's my problem. I simulated the output and received this, and I don't know how to analyze what it means. It definitely has to do with the 0.25 but I'm really unsure because my math skills are VERY rusty (I'm planning on taking some time over the next few months to relearn some of the university-level math I learned a while back).
4. Apr 12, 2014
### Simon Bridge
What makes you so sure that 0.25 has anything to do with it?
Using least-squares fit for polynomial order 4, the coefficient vector is:
c = [-6.6667e-05 1.4167e-03 -1.3000e-02 6.3583e-02 -5.5933e-02 4.0000e-03]
$f(x)=c(1)x^4+c(2)x^3+c(3)x^2+c(4)x+c(5)$
But I don't understand your description of the generating algorithm.
What do you mean by an "event"?
How does the "event" decide which action to perform?
5. Apr 12, 2014
### iamsmooth
Well, I am simulating events. Each event starts out doing action B for an amount of time that's exponentially distributed with a mean of 30 minutes, to get this number I do: -(actionTime) * log(randomNumber)
where actionTime is the amount of time the event is spent performing action and randomNumber is just a randomly generated floating point number from 0-1.
So if I say there are 2 events. They both start out performing event B from time 0 for -(30) * log(randomNumber) minutes, then depending on the specific number they rolled, they switch to action A for -(10) * log(randomNumber) minutes. I simulate a million minutes and receive the sequence of results for input which represents the number of events I want to start the system with. Hopefully that helps?
6. Apr 12, 2014
### haruspex
I think that what you are calling events others might call processes. You start x processes at once. Each process independently is in state A for an exponentially distributed random time, then in state B for an exponentially distributed random time, then ... returns state A, picking a new random time, ad nauseam?
I assume the two times are independent. For all processes, the mean sojourn in state A is 10 minutes, in state B 30 minutes.
And you want to know the probability that at any instant there are two or more processes in state A. Is that it?
Assuming the system has been running long enough that it has ceased to matter whether all started in the same state, I don't see that the distribution of time periods matters. Doesn't it just come down to each independently being in state A with probability 1/4? That gives an excellent fit to your data.
7. Apr 12, 2014
### Simon Bridge
How does that relate to the values of x - is x the number of "events"?
i.e.
We could consider an event to be a fisherman, action A is "fish" and action B is "cut bait" and x is the number of fishermen in the boat. Each starts out cutting bait, and when that's done, they start fishing for a bit - alternating fish and cut-bait?
So actionTime is 30 for event B and 10 for event A.
Is randomNumber taken from a rand() function - i.e. equally likely to be any real number between 0 and 1?
I'm still a bit confused:
You said each starts out doing action B for a time that is exponentially distributed with a mean of 30mins.
You then tell me that this time is determined by: Time = actionTime*log(randomNumber)
But then you say that actionTime is the time spent doing the action.
... or is it the mean-time doing the action?
You have a reference that this is actually the mean?
You have been running the simulation to gather statistics on what exactly?
i.e. perhaps you want to know how much time out of a set period (1 million minutes, or a day, or whatever) that the fishermen spend actually fishing ... you would need that to determine how many fishermen you need to hire to get a certain size catch in the time period.
See how making things concrete helps think about it?
Keeping the goals a secret will just make life harder for you.
BTW: I agree with haruspex (above)
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## Quiz by Hussain
### Our brand new solo games combine with your quiz, on the same screen
Correct quiz answers unlock more play!
16 questions
• Q1
In a Common-Emitter configuration, an emitter resistor is used for:
Collector Bias
AC Signal Bypass
Stabilization
Higher Gain
60s
• Q2
What is the output signal form the circuit shown below
Same input signal
Amplifying and inverted signal
Only amplifying signal
Only amplifying signal
60s
• Q3
From the given transistor amplifier circuit shown below Calculate the voltage of R2 (if IB = 0)
8.3 V
1.6 V
1.8 V
0.8 V
60s
• Q4
Amplifier signal without phase shaft you must connect common emitter amplifier
Three stage
Single stage
Non above
Two stage
30s
• Q5
What is the amplifier configuration of the circuit shown below?
Common Emitter Amplifier
Common Base Amplifier
Inverting Amplifier
Common Emitter Amplifier
60s
• Q6
When a silicon transistor is properly biased, what is VBE for a C-E configuration?
0.5 V
0.7 V
0.79 V
0.3 V
60s
• Q7
The function of the speaker in an audio amplifier circuit is to:
Convert pressure signal into sound
Convert sound signal into electric
Convert electric signal into light
Convert electric signal into sound
60s
• Q8
The CE amplifier below has an input of 20mV and an Av of 60, calculate the output voltage.
None of the above
1.2 V
12 V
0.12 V
60s
• Q9
For the multistage amplifier shown in the figure below, calculate the Voltage output Vo/p
100 V
10 V
1 V
1000 V
60s
• Q10
Which reading below is for emitter terminal
None
0.617
Both
0.605
60s
• Q11
From the transistor figure below which part number is the emittier
3
none
2
1
60s
• Q12
calculate the gain of inverting Amplifier
10
20
100
40
60s
• Q13
This is an example of the output swing for a class ________ amplifier.
A
C
B
AB
60s
• Q14
A transistor amplifier has a voltage gain of 100. If the input voltage is 75 mV, the output voltage is:
15 V
7.5V
1.33V
13.3 V
60s
• Q15
the current gain of an amplifier which has an input signal of 100mA and output signal 1.5 A is
15
0.5
0.015
1.5
60s
Teachers give this quiz to your class
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https://electronics.stackexchange.com/questions/683354/colpitts-oscillator-has-negative-feedback-why-does-it-oscillate
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Colpitts oscillator has negative feedback, why does it oscillate?
The following is a picture of colpitts oscillator and CE.
Barkhausen stability criterion said positive feedback is needed in order to oscillate. However, Common-Emitter has 180 degree phase shift. Thus it's negative feeback. How can this oscillate?
• How about you assume that there are other components in the circuit that provide an additional 180 degree phase shift at a particular frequency.
– Nedd
Sep 30, 2023 at 8:03
• Don't you think that just saying CE is 180 degrees and declaring this as negative feedback is a smidge over simplistic? Don't you recognize that there can be frequency dependent responses by external components (and parasitics) that may alter such a simplistic view? Sep 30, 2023 at 8:03
• @Nedd I am not sure which components can provide additional 180 degree shift. Please point out those components
– kile
Sep 30, 2023 at 8:13
• @periblepsis Can you specify those external components?
– kile
Sep 30, 2023 at 8:14
• Consider collector voltage is going up so inductor starts charging via C2 to ground, but C2 voltage rise is delayed. Another words, L+C2 makes a phase shift in Vcollect vs. Vc2 point of view. During opposite cycle the inductor current flows backward through C1 so C2 is charged opposite. Sep 30, 2023 at 9:06
Colpitts oscillator has negative feedback, why does it oscillate?
No, it has positive feedback. It's just that you didn't analyse sufficiently.
It oscillates because $$\C_2\$$ and $$\L\$$ provide about 150° of phase shift AND, the collector resistor $$\R_C\$$ and $$\C_1\$$ add another 30°. Meaning there is another 180° of phase shift and, this makes it oscillate.
Maybe these words and images (from my basic website) might help a bit: -
To analyse the circuit and derive the oscillation frequency formula, the collector node cannot be assumed to be unloaded. This mistake is made by many websites that attempt to describe functionality; important subtleties are glossed-over and, the formulas are presented matter-of-fact with no attempt to reveal the workings. Some websites misleadingly describe the operation by referring to a tank circuit (where C1 and C2 are in series) but, because C1 and C2 are grounded, they form a filter not only with inductor L1, but with resistor R4.
R4 makes this a 3rd order filter and only a 3rd order filter can produce sufficient phase shift for the circuit to oscillate. R4 is that important!
So, R4, C1, C2 and L1 form a 3rd order low-pass filter that produces 180° phase-shift whilst still providing sufficient signal gain to initiate and maintain oscillation. A 2nd order analysis is insufficient to deliver the theory. Bear that in mind. This is a proper analysis.
So, we "recognize" that Q1's collector is a current source in parallel with R4 and that this is equivalent to a voltage source in series with R4. Hence, R4 and C1 form a low-pass filter that introduces a lagging phase angle at the collector. Below is the bode plot when we apply a signal generator at the circuit input (C3) and C1 is connected to the collector (L and C2 are not connected yet): -
Normally, a CE amplifier delivers constant gain across a wide frequency range (with a phase angle of 180°) but, with C1 loading the collector, the phase shift is tending towards +90° at high frequencies. The gain is also reducing at high frequencies. The faint lines in the diagram above show the natural gain of the CE amplifier (12.6 dB). This gain is due to the ratio of R4 to R1 i.e. 2000 ÷ 470 = 4.255 = 12.6 dB.
Then, if L and C2 are added to R4 and C1 we get this bode plot after inductor L1: -
Note that the amplitude peak at about 2 MHz corresponds with a 0° phase angle. When the feedback loop is closed, this is where the circuit will oscillate. The 0° phase shift is a result of the collector resistor (R4) feeding C1, L and C2 i.e. this is a 3rd order filter. The gain peak at around 2 MHz is 12.6 dB because, at the oscillation frequency, when C1 equals C2, the 3rd order filter has unity gain.
I won't repeat the full version from my website.
• But Common Emitter amplifier has 180 degree phase shift.
– kile
Sep 30, 2023 at 9:23
• @kile the feedback signal gets shifted a further 180 degrees by the components mentioned in my answer. Sep 30, 2023 at 9:24
• What's collector RC? Do you mean $R_c$ and $C_o$?
– kile
Sep 30, 2023 at 9:25
• I actually said collector resistor RC but I can see it may be confusing. Sep 30, 2023 at 9:27
• Can you point it collector resistor RC out in this diagram?
– kile
Sep 30, 2023 at 9:31
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Residue questions
brianhawaiian
Sorry I don't have equation editor, for some reason every time I install it on Microsoft Word it never appears...
1. The problem statement, all variables and given/known data
Calculate the residue at each isolated singularity in the complex plane
e^(1/z)
2. Relevant equations
#1 Simple pole at z0 then,
Res[f(z), z0] = lim (z - z0)(f(z)) as z goes to z0.
#2 Double pole at z0 then,
Res[f(z), z0] = lim d/dz [(z - z0)^2*f(z)] as z goes to z0.
#3 If f(z) and g(z) are analytic at z0, and if g(z) has a simple zero at z0 then
Res[f(z)/g(z), z0] = f(z0)/g'(z0)
#4 If g(z) is analytic and has a simple zero at z0, then
Res[1/g(z), z0] = 1/g'(z0).
3. The attempt at a solution
The problem occurs when z = 0 so looking at
Res[e^(1/z), 0], Using #1, #3, #4 don't help the problem. So using #2
lim as z goes to z0 [d/dz z^2 * e^(1/z)], there's still a problem... I'm completely lost at this point.
1. The problem statement, all variables and given/known data
Evaluate the following integral, using the residue theorem
Integral |z| = 1 (sin(z)/z^2)dz
2. Relevant equations
See Above
3. The attempt at a solution
How would I start this? z = 0 gives a problem, would I take the integral first and then evaluate?
latentcorpse
z=0 isn't a pole for this function.
write out the laurent series. it has all negative powers. thus z=0 is an essential isolated singularity.
knowing the laurent series, how do you find the residue?
Count Iblis
New Edit: I agree with Latentcorpse (he replied whe I was still typing the message)
You know that exp(z) is an analytical function everywhere on the complex plane. This means that you can replace z by 1/z in the series expansion without any problem. It will yield a Laurent series that converges everywhere (except at z = 0). You can then read-off the residue at z = 0.
The Laurent expansion contains arbitrarily large negative powers of z, we then say that the function has an "essential singularity" at z = 0.
You can still compute the residue at this essential singularity like in case of ordinary poles, by considering the mapping z ---> 1/z
The residue is 1/(2 pi i) times a contour integral that encircles the singularity in an anti-clockwise way. If then perform a change of variables z -->1/z, the integral changes to:
1/(2 pi i) exp(z) (-1/z^2) dz
which is now traversed clockwise. Changing it to anticlockwise will yield a minus sign. This means that the residue at zero of exp(1/z) is the same as the residue at zero of the function
exp(z)/z^2
This function has an ordinary pole (a "double pole") at zero.
brianhawaiian
Thank you, appreciate the comments... I'm completely lost in Complex Analysis (First Grad Level class, never took it as an undergrad)
latentcorpse
read off the $c_{-1}$ coefficient i.e. the coefficient of the $z^{-1}$ term in the laurent series for $\frac{e^z}{z^2}$ and you'll get the residue is 1. hopefully.
brianhawaiian
read off the $c_{-1}$ coefficient i.e. the coefficient of the $z^{-1}$ term in the laurent series for $\frac{e^z}{z^2}$ and you'll get the residue is 1. hopefully.
Yeah, I figured it out thanks to you guys, much appreciated, I have another question up on the page, was wondering if you could maybe help with that one? If not it's okay, Thanks again!
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### Author Topic: Building a Terrain Object3D from scratch (Read 3472 times)
#### paulscode
• double
• Posts: 863
##### Building a Terrain Object3D from scratch
« on: October 18, 2008, 04:59:08 pm »
I have a couple more questions related to building a terrain Object3D.
My first question is why does building an Object3D increase its vertice count? Building the following square Object3D increases its vertice count from 4 to 12
Code: [Select]
` Object3D terrain = new Object3D( 2 ); SimpleVector bottomLeft = new SimpleVector( 0, 0, 0 ); SimpleVector bottomRight = new SimpleVector( 5, 0, 0 ); SimpleVector topRight = new SimpleVector( 5, 0, 5 ); SimpleVector topLeft = new SimpleVector( 0, 0, 5 ); int triangle1 = terrain.addTriangle( topLeft, 0, 0, bottomLeft, 0, 1, bottomRight, 1, 1 ); int triangle2 = terrain.addTriangle( bottomRight, 1, 1, topRight, 1, 0, topLeft, 0, 0 ); System.out.println( "Unique Vertice count before build: " + terrain.getMesh().getUniqueVertexCount() ); System.out.println( " Triangles: " + terrain.getMesh().getTriangleCount() ); terrain.build(); System.out.println( "Unique Vertice count after build: " + terrain.getMesh().getUniqueVertexCount() ); System.out.println( " Triangles: " + terrain.getMesh().getTriangleCount() );`Strangely, the number of triangles remains 2 even though the number of vertices triples. What is causing this, and how do I prevent it from happening?
My second question is what is the best way to determine a polyID given a triangle number. I don't suppose they are the same, are they?
My last question is about OcTree's. The JavaDoc states that the best maxPoly value to use for an OcTree depends on the object and its usage. Unfortunately, I don't know enough to know what the best value to use for my situation might be. What would you recommend for terrain objects that are 32X32 grids with two triangles per square?
#### EgonOlsen
• Posts: 11899
##### Re: Building a Terrain Object3D from scratch
« Reply #1 on: October 18, 2008, 10:04:45 pm »
Strangely, the number of triangles remains 2 even though the number of vertices triples. What is causing this, and how do I prevent it from happening?
You can't...and you shouldn't. build() always increases the vertex count by 8. Those are a vertices for the object's bounding box that are stored in the same structure. That's all.
My second question is what is the best way to determine a polyID given a triangle number. I don't suppose they are the same, are they?
If you are just adding triangles and are not building triangle strips afterwards on that object, the polyID should be in the same order in which you've added the polygons. Apart from that, there is no good way to find out which polygon gets which id.
My last question is about OcTree's. The JavaDoc states that the best maxPoly value to use for an OcTree depends on the object and its usage. Unfortunately, I don't know enough to know what the best value to use for my situation might be. What would you recommend for terrain objects that are 32X32 grids with two triangles per square?
That's pretty hard to tell. Just give it a try and see what performs best in your case. Don't forget to enable the OcTree for collision detection or otherwise, only the rendering will benefit from it.
#### paulscode
• double
• Posts: 863
##### Re: Building a Terrain Object3D from scratch
« Reply #2 on: October 18, 2008, 10:52:26 pm »
You can't...and you shouldn't. build() always increases the vertex count by 8. Those are a vertices for the object's bounding box that are stored in the same structure. That's all.
I just finished some testing before I read your post, and discovered it was always 8 more, not 3 times as many. I hadn't thought of the bounding box (DUH!). Yeh, that is pretty obvious now.
If you are just adding triangles and are not building triangle strips afterwards on that object, the polyID should be in the same order in which you've added the polygons. Apart from that, there is no good way to find out which polygon gets which id.
Ok, that should be simple enough to work with. I ought to be able to get an offest value like this (as long as all Object3Ds are created on the same thread):
Code: [Select]
`int polyIdOffset = terrain.getPolygonManager().getMaxPolygonID() - terrain.getMesh().getTriangleCount();`Then just add that number to a triangle number to get its polyID.
That's pretty hard to tell. Just give it a try and see what performs best in your case. Don't forget to enable the OcTree for collision detection or otherwise, only the rendering will benefit from it.
Ok, I suppose I will make a FPS application, start with maxPoly = 2, then randomly try higher values and see what works best.
#### paulscode
• double
• Posts: 863
##### Re: Building a Terrain Object3D from scratch
« Reply #3 on: October 19, 2008, 02:04:50 am »
I'm having a little trouble getting calcMinDistance to work. Not sure if I made my OcTree incorrectly or something.
The code I used to generate the terrain Object3D:
Code: [Select]
` public void createTerrain() { int x, y; terrain = new Object3D( 2048 ); // Create the terrain's vertices: SimpleVector[][] vertices = new SimpleVector[33][33]; for( y = 0; y < 33; y++ ) { for( x = 0; x < 33; x++ ) { vertices[x][y] = new SimpleVector( x * 5, 0, y * 5 ); } } // Create the terrain's polys, set their UV coordinates, // and set their initial texture to grey: for( y = 0; y < 32; y++ ) { for( x = 0; x < 32; x++ ) { terrain.addTriangle( vertices[x][y+1], 0, 0, vertices[x][y], 0, 1, vertices[x+1][y], 1, 1, TextureManager.getInstance().getTextureID( "Grey" ) ); terrain.addTriangle( vertices[x+1][y], 1, 1, vertices[x+1][y+1], 1, 0, vertices[x][y+1], 0, 0, TextureManager.getInstance().getTextureID( "Grey" ) ); } } terrain.build(); OcTree ocTree = new OcTree( terrain.getMesh(), 2048, OcTree.MODE_OPTIMIZED ); ocTree.setCollisionUse( OcTree.COLLISION_USE ); terrain.setOcTree( ocTree ); }`
The camera orientation:
Code: [Select]
` camera = world.getCamera(); camera.setPosition( -80, -160, -80 ); camera.lookAt( new SimpleVector( 80, 0, 80 ) );`
And the calcMinDistance part:
Code: [Select]
` SimpleVector position = new SimpleVector( camera.getPosition() ); SimpleVector direction = new SimpleVector( camera.getDirection() ).normalize(); float distance = world.calcMinDistance( position, direction, 10000 ); if( distance == Object3D.COLLISION_NONE ) outputString = "No Collision"; else outputString = "Collision - Distance = " + distance;`
I also posted a link to the actual applet:
http://www.paulscode.com/source/calcMinDistanceTest/
« Last Edit: October 19, 2008, 04:39:36 am by paulscode »
#### paulscode
• double
• Posts: 863
##### Re: Building a Terrain Object3D from scratch
« Reply #4 on: October 19, 2008, 06:25:37 am »
I'm not so sure the problem is related to how I created the terrain any more. I've tried loading a 3DS instead. The camera is looking directly at the object, but plugging the camera's position and direction vectors into calcMinDistance still returns Object3D.COLLISION_NONE. I must somehow be using the method incorrectly or something.
#### paulscode
• double
• Posts: 863
##### Re: Building a Terrain Object3D from scratch
« Reply #5 on: October 19, 2008, 06:31:47 am »
OMG, I am dumb! I forgot to make the terrain object a collision listener!! Ok, it works great now, ROFL!
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Your research can change the world
More on impact ›
# Frontiers in Applied Mathematicsand Statistics
## ORIGINAL RESEARCH article
Front. Appl. Math. Stat., 28 April 2021 | https://doi.org/10.3389/fams.2021.638996
# Random-Walk, Agent-Level Pandemic Simulation (RAW-ALPS) for Analyzing Effects of Different Lockdown Measures
• Department of Statistics, Florida State University, Tallahassee, FL, Unites States
This article develops an agent-level stochastic simulation model, termed RAW-ALPS, for simulating the spread of an epidemic in a community. The mechanism of transmission is agent-to-agent contact, using parameters reported for the COVID-19 pandemic. When unconstrained, the agents follow independent random walks and catch infections due to physical proximity with infected agents. Under lockdown, an infected agent can only infect a coinhabitant, leading to a reduction in the spread. The main goal of the RAW-ALPS simulation is to help quantify the effects of preventive measures—timing and durations of lockdowns—on infections, fatalities, and recoveries. The model helps measure changes in infection rates and casualties due to the imposition and maintenance of restrictive measures. It considers three types of lockdowns: 1) whole population (except the essential workers), 2) only the infected agents, and 3) only the symptomatic agents. The results show that the most effective use of lockdown measures is when all infected agents, including both symptomatic and asymptomatic, are quarantined, while the uninfected agents are allowed to move freely. This result calls for regular and extensive testing of a population to isolate and restrict all infected agents.
## 1 Introduction
There is a great interest in stochastic modeling and analysis of medical, economical, and epidemiological data resulting from the ongoing COVID-19 pandemic [1]. Until a large amount of infection, treatment, vaccination, containment, and recovery data from this pandemic become available, the community will have to rely primarily on simulation models to help assess situations and to evaluate countermeasures [2]. Naturally, simulation systems that follow precise mathematical and statistical models will play an important role in understanding this dynamic and complex situation [3]. There have been a large number of models proposed in the past literature, relating to the spread of epidemics through human contact or otherwise. They can be broadly categorized in two main classes (a more detailed taxonomy of simulation models can be found in [4]):
1. Population-Level, Deterministic, Dynamic Models: A large number of epidemiological models have focused on coarse, population-level summaries, that is, counts of infected (I), susceptible (S), removed or recovered (R), etc., whose evolutions are governed by deterministic differential equations. The original model of this type is the Susceptible-Infected-Removed (SIR) model [5], proposed by Kermack and McKendrick in 1927, that uses ordinary differential equations to model a constrained growth of the counts in those three categories. Since then, researchers have developed several advancements and variations of this model, including the SIRD model [6] given by
The scalar parameters β, γ, and µ together control the dynamics of infections, recovery, and mortality. The zero-sum condition $dS\mathrm{/}dt\mathrm{+}dI\mathrm{/}dt\mathrm{+}dR\mathrm{/}dt\mathrm{+}dD\mathrm{/}dt\mathrm{=}0$ ensures constancy of the community size: $S\mathrm{\left(}t\mathrm{\right)+}I\mathrm{\left(}t\mathrm{\right)+}R\mathrm{\left(}t\mathrm{\right)+}D\mathrm{\left(}t\mathrm{\right)}=N$. Several studies have applied these general models to different epidemics, such as Ebola and SARS [7]. A recent article [8] extends this model to include hospitalizations and different strata of infections, and estimates these quantities from the Wuhan COVID-19 pandemic.
While they provide useful population-level summaries, these models do not generally focus on capturing spatial dynamics. Specifically, they do not explicitly model agent dynamics as residents move around in a community or across communities. Also, these models often provide deterministic outcomes, with no mechanism to incorporate randomness in the model. Several recent simulation models, focusing directly on the COVID-19 illness, also rely on such coarser community-level models [9].
2. Agent-Level Modeling: While dynamical evolutions of population variables are simple and effective, especially for the overall assessment, they do not take into account any social dynamics, human behavior, societal restrictions, and complexities of human interactions explicitly. The models that study these human-level factors and variables, while tracking disease at an individual level, are called agent-level models [10]. Here one models the mobility, health status, and interactions of individual subjects (agents) to construct an overall population-level picture in a bottom-up way. The advantages of agent-based models are that they are more detailed and one can vary the parameters of lockdown measures, such as social distancing, at a granular level to infer overall outcomes. Furthermore, these models have built-in stochastic components for agent motions, interactions, infections, and recoveries, thus enabling a more realistic simulation environment. Agent-based models have been discussed in several articles, including [4, 1113] and so on. The importance of simulation-based analysis of epidemic spread is emphasized in Ref. [14] but with a focus on infection models within a host. Ref. [12] constructed a detailed agent-based model for spread of infectious diseases, taking into account population demographics and other social conditions, but they did not consider countermeasures such as lockdowns in their simulations. A broad organization of different agent-based simulation methods has been presented in Ref. [4]. A recent article [15] studies the socioeconomic impact of social distancing using agent-based models. Although there are numerous other articles on the topic of agent-based simulations for simulating the spread of infections, we have only listed the most relevant ones.
In this article, we develop a mathematical simulation model, named RAW-ALPS, to simulate the spread of an infectious disease, such as COVID-19, in a confined community and to study the influence of some external interventions on outcomes. Since RAW-ALPS is purely a simulation model, the underlying assumptions and choices of statistical distributions for random quantities become critical in its success. On the one hand, it is important to capture the intricacies of the observed phenomena as closely as possible, using sophisticated modeling tools. On the other hand, it is desirable to keep the model efficient and tractable (for individual laptops) by using simplifying assumptions. One can, of course, relax these assumptions and obtain more and more realistic models as desired, albeit with increased computational complexity.
Simulation Setup: We assume a closed community with infection initiated by a single infected agent at the beginning. The infections are transmitted through physical exposure (proximity) of mobile susceptible agents to mobile infected agents, as shown in the middle panel of Figure 1. When unconstrained, the agents follow a smooth random-walk motion, independent of other agents (see the left panel of Figure 1). When instructed to enter lockdown, an agent moves toward his/her allocated household unit and stays put until the restrictions are imposed. The households are arranged, so that they are equally spaced at the grid points of a square domain (see the right panel of Figure 1).
FIGURE 1
FIGURE 1. Left: random walk model for agent motions. Middle: proximity-based spread of infections from infected to susceptible agents. Right: layout of households on a square grid in the center of the domain.
An agent’s health situation follows the chart shown on the left side of Figure 2. Susceptible agents become infected when exposed to infected agents, with a certain probability. The infected agents go through a period of sickness, with two eventual outcomes—full recovery for most and death for a small fraction. That is, one starts as noninfected or susceptible, potentially gets infected with a certain probability, and later recovers or dies according to their event probabilities. Those with nonfatal infections are further labeled as symptomatic or asymptomatic agents. Naturally, agents with fatal outcomes are labeled as symptomatic. This labeling allows for the selective imposition of lockdown measures on a subset of the population. Once recovered, we assume that the agents can no longer be infected, as suggested by the CDC FAQ [16]. The social dynamical model used here is based on a fixed domicile, that is, each agent has a fixed housing unit. Under unrestricted conditions, that is, no lockdown, the agents are free to move over the full domain using a simple motion model. These motions are independent across agents and encourage smooth paths. Under lockdown conditions, the required agents head directly to their housing units and generally stay there during that period. The remaining agents, including a small fraction of agents, termed essential workers, are allowed to move freely under the restrictions. The infected agents under lockdown can only infect other susceptible agents living in the same household and not the general public, as shown on the right side of Figure 2. Similarly, mobile infected agents can only infect other mobile agents but not those under the lockdown. Three types of lockdowns are considered: 1) lockdown of the full population or LD1, 2) lockdown of infected agents or LD2, and 3) lockdown of symptomatic agents or LD3.
FIGURE 2
FIGURE 2. Left: flow chart for evolution of infection dynamics for the population. Right: scheme for infection of susceptible agents from infected agents.
The main highlights and contributions of this article are as follows: 1) agent-level transmission of infections and thus, a more granular analysis than (population-level) deterministic dynamical models, 2) random-walk motions of agents when unconstrained and restricted to households when under lockdowns, 3) different classifications of infections: fatal and nonfatal, with the latter being either symptomatic or asymptomatic, 4) selective lockdowns for different strata of the population, and 5) statistical quantifications of gains resulting from lockdown restrictions and their timings on infection rates.
The rest of this article is as follows. Random-Walk, Agent-Level Pandemic Simulation (RAW-ALPS) develops the proposed RAW-ALPS model, specifying the underlying assumptions and motivating model choices. It also discusses choices of model parameters and provides comparisons with the SIRD model. Exemplar Outcomes and Computational Cost presents some illustrative examples and discusses the computational complexity of RAW-ALPS, while Analyzing Effects of Lockdown Measures develops the use of RAW-ALPS in understanding the influences of countermeasures. The article ends by discussing model limitations and suggesting some future directions.
## 2 Random-Walk, Agent-Level Pandemic Simulation (RAW-ALPS)
In this section, we develop our simulation model for agent-level interactions and the spread of the infections. For this, we consider a population in a predetermined geographical domain. In terms of the model design, there are competing requirements for such a simulation to be useful. On the one hand, we want to capture detailed properties of agents and their pertinent environments, to render a realistic scenario for pandemic evolution. On the other hand, we want to keep model complexity reasonably low, to utilize it for analyzing variable conditions and countermeasures. Also, to obtain statistical summaries of pandemic conditions under different scenarios, we want to run a large number of simulations and compute averages. This process also requires keeping the overall model simple, from a computational perspective, to allow for multiple runs of RAW-ALPS.
### 2.1 Simplifying Assumptions
The overall setting of the simulation model is as follows. We assume that the community is located in a square geographical region D with h household units arranged centrally on a uniformly spaced square grid in D (see the right panel of Figure 1). We assume that there is a fixed number, say N, of total agents in the community (including all classifications—uninfected, infected, dead, etc.) and their health status updates every unit interval (e.g., 1 h), indexed by variable t. The unconstrained agents can traverse freely through all parts of D but are largely restricted to their home units under restrictions. Next, we specify some simplifying assumptions:
Independent Agents: An agent’s movements and infection status are independent of those of other agents. The actual infection event is, of course, dependent on one being in close proximity to an infected carrier (within a certain distance, say ≈ 6 feet) for a certain exposure time. But the probability of an agent being infected is independent of such events for other agents.
Full Mobility in Absence of Restrictive Measures: As mentioned above, we assume that each agent is fully mobile and moves across the domain freely when no restrictive measures have been imposed. In other words, there is no effect of an agent’s age, gender, or health on his/her mobility. Also, we do not impose any day/night schedules on the motions. Some articles, including [17], provide two- or three-state models where the agents transition between some stable states (home, workplace, shopping, etc.) in a predetermined manner. To implement such detailed models requires careful considerations about the daily movement patterns of agents, and that increases model complexity.
Homestay During Restrictive Measures: We assume that most agents comply with instructions and stay at home at all times during the lockdown conditions. Only a small percentage (set as a parameter ${\rho }_{0}$) of the population, representing essential workers, are allowed to move freely, but a large majority stays at home.
Sealed Region Boundaries: In order to avoid complications resulting from a transportation model in the system, we assume that there is no transfer of agents into and out of the region D. The region is modeled to have reflecting boundaries to ensure that all citizens stay within the region. The only way the population of D can be changed is through death.
Fixed Domicile: The whole community is divided into a certain number of living units (households or buildings). These units are placed in square blocks with uniform spacing. Each agent has a fixed domicile at one of the units. During a lockdown period, the agents proceed to and stay at home with a high compliance level. We assume that all agents within a domicile unit are exposed to each other, that is, they are in close proximity and can potentially infect others.
No Reinfection: We assume that once a person has recovered from the disease, he/she cannot be infected again for the remaining observation period. While this is an important unresolved issue for the current COVID-19 infections, it has been a valid assumption for the past coronavirus infections and it remains the current CDC guideline [16].
Single Patient Ground Zero: The infection is introduced in the population using a single carrier, termed patient ground zero at time t = 0. This patient is selected randomly from the population and the time t is measured relative to this initial event.
Constant Immunity Level: The probability of infection of agents, under the exposure conditions, remains the same over time. We do not assume any increase or decrease in an agent’s immunity level over time. Also, we do not assign any age or ethnicity to the agents and all agents are assumed to have equal immunity levels.
### 2.2 Model Specifications
There are several parts of the model that require individual specifications. These parts include modeling the movements of each agent (with or without restrictions in place), the mechanisms of transmitting infections from agent to agent, and the processes of recovery and fatality for infected agents. A full listing of the model parameters and some typical values are given in Table A1 in the Appendix.
Motion Model: The movement of an agent follows a simple random-walk model where the instantaneous velocity ${v}_{i}\left(t\right)$ is a weighted sum of three components: 1) velocity at the previous time, that is, ${v}_{i}\left(t\mathrm{-}1\right)$, 2) a directed component guiding them to their home, $\left({h}_{i}\mathrm{-}{x}_{i}\left(t\mathrm{-}1\right)\right)$, where ${x}_{i}\left(t\right)$ is the agent location at time t, and 3) an independent Gaussian increment $\sigma {w}_{i}\left(t\right)$, ${w}_{i}\left(t\right)\sim \mathcal{N}\left(0,1\right)$. Note that the motions of different agents are kept independent of each other. The location ${h}_{i}\mathrm{\in }D$ denotes the home unit (or stable state) of the ${i}^{th}$ agent.
Using mathematical notation, the model for instantaneous position $xi(t)$ and velocity $vi(t)$ of the $ith$ agent are given by
Here $α∈ℝ+$ determines how fast one moves toward their home and µ quantifies the degree to which one follows the directive to stay home. When $μ=0$, an agent reaches home and stays there, except for a random perturbation $wi$. However, if $μ=0.5$, then a significant fraction of motion represents continuity in velocity, irrespective of the home location. The value $μ=1$ implies that either there is no lockdown or the agent does not comply with the directive.
Reflecting Boundaries: When a subject reaches the boundary of the domain D, the motion is reflected, and the motion continues in the opposite direction.
Figure 3 shows examples of random agent motions under different simulation conditions. The leftmost panel shows a situation with no lockdown and the agents are moving freely according to Eq. 2. The middle panel shows the case where the restrictions are imposed on Day 10 and stay in place after that. The last panel shows the case where a lockdown is imposed on Day 10 and then lifted on Day 20. The blue curves represent free motion before a lockdown, the red denotes motion toward home during a lockdown, and the green represents free motion again after the lockdown is lifted.
Lockdown Model: Once the lockdown starts, at time say ${T}_{0}$, agents are directed toward their homes and asked to stay there. We assume that $\rho \mathrm{%}$ of the subjects follow this directive, while the others $\left(\left(100\mathrm{-}\rho \right)\text{%}\right)$ follow a different motion model. The variable ρ changes over time according to
where ${\rho }_{0}$ is set to be a large number: 98% or 99%. We note that the people who do not follow restrictions follow the prescribed motion model Eq. 2 with $\mu \mathrm{=}1$. We study three types of lockdowns in this article: 1) lockdown of the full population, termed LD1, 2) lockdown of infected agents, termed LD2, and 3) lockdown of symptomatic agents, termed LD3.
Exposure–Infection Model: Infection of a susceptible agent depends on the level of exposure to an infected agent, irrespective of the infected agent being symptomatic or not. Recall that mobile infected agents can only infect other mobile agents, while agents in a lockdown can only infect agents in the same household. To create the conditions for the spread of infection:
• The physical distance between the susceptible agent and the infected agent during exposure should be less than ${r}_{0}$.
• The exposure time in terms of the time units is at least ${\tau }_{0}$. In this study, we use the cumulative time exposure for an agent over the whole history, rather than just the recent history.
• Under these conditions, the probability of being infected at each time t is an independent Bernoulli random variable with probability ${p}_{I}$.
RecoveryDeath Model: Once a subject is infected, we randomly assign a label or infection type immediately. Either an infected agent is going to recover (nonfatal type, or NFT) or the person is eventually going to die (fatal type, or FT). The probability of having a fatal type, given that a person is infected, is ${p}_{F}$. An agent with the NFT label is further classified as either Symptomatic (S) or Asymptomatic (AS) with a certain probability ${p}_{S}$. All agents with the FT label are classified as symptomatic.
Recovery: An agent with a nonfatal type (NFT) is sick for a period of ${T}_{R}$ days. After this period, the person can recover at any time, according to a Bernoulli random variable with probability ${p}_{R}$ in each time step.
Fatality: An agent with a fatal type (FT) is sick for a period of ${T}_{D}$ days. After this period, the person can die at any time, independently according to a Bernoulli random variable with probability ${p}_{D}$.
FIGURE 3
FIGURE 3. Sample agent motions under different conditions. Blue curves denote unrestricted movements, red curves denote movement toward home units during a lockdown, and green curves denote movement after lockdown.
### 2.3 Chosen Parameter Values
A complete listing of the ALPS model parameters is provided in Appendix Table A1. In this section, we motivate the values chosen for those parameters in these simulations. We justify these choices from the current reports of the COVID-19 pandemic. As argued in [18], a realistic choice of parameters is very important in establishing the validity of simulation models.
• Population Density: We use a square domain D of size $2\mathrm{\ast }2$ miles2 or $\mathrm{5\ast 5}$ miles2 for a community with a population of N agents. The value of N changes in different experiments. The values of $N\mathrm{=}900$ and $D\mathrm{=}{\left[0,2\right]}^{2}$ represent a population density of 225 people/mile2. The community contains h living units (buildings) with a domicile of $N/h$ people per unit. In case $N/h$ is high ($\approx 100$), a unit represents a tall building in metropolitan areas, but when $N/h$ is small ($\approx 5$), a unit represents a single-family home in a suburban area.
The time unit for updating configurations is 1 h and the occurrence of major events is specified in days. For example, the lockdown can start on Day 1 and end on Day 30.
• Agent Speeds: The standard deviation for accelerations, denoted by σ, in agent mobility are approximately 1–5 feet/h (fph). Through integration over time, this results in agent speeds up to 1,000 fph. We assume that ${\rho }_{0}\mathrm{=}0.9\mathrm{-}0.98$, that is, 90–98% of the people follow the restriction directives.
• Infection Rates: The physical distance between agents to catch infection should be at most ${r}_{0}\approx 6\text{ }\text{ft}$ and the exposure time should be at least ${\tau }_{0}\mathrm{=}5$ time units (hours). The probability ${p}_{I}$ of getting infected, under the right exposure conditions, is set at 5% at each time unit (hour) independently. There is no current reference literature on selecting this value since it is difficult to measure precise exposure events for people who have tested positive for COVID-19. While contact tracing [19] is being developed to ascertain infection rates, there are no public data currently available to measure this infection rate. This value leads to overall infection rates that are similar to national and international infection rates [20].
• Fatality Rate: Once infected, the probability of having a fatal outcome is set at 5–10%, according to the mortality rate listed by the CDC [21]. The period of recovery for agents with nonfatal outcomes starts at 7 days. The probability of reaching full recovery for those agents is ${p}_{R}\mathrm{=}0.001$ at each subsequent time unit (hour). Similarly, for the agents with fatal outcomes, the period of being infected is set to be 7 days and after that, the probability of death at each time unit (hour) is set to be ${p}_{D}=0.1$.
• Asymptomatic Infected Agents: The percentage of infected agents who remain asymptomatic is set to be in the range of 15–40% [22, 23]).
Figure 1 provides a pseudocode for the RAW-ALPS algorithm.
ALGORITHM 1
Algorithm 1. RAW-ALPS Pseudocode
### 2.4. Model Validation
Although RAW-ALPS is perhaps too simple model to capture the intricate dynamics of an actual active society, it does provide an efficient tool for analyzing effects of countermeasures during the spread of a pandemic. Before it can be used in practice, there is an important need to validate it in some way.
As described in Ref. [4], there are several ways to validate a simulation model. One is to use real data (an observed census of infections over time) in a community to estimate model parameters, followed by a statistical model testing. While such data may emerge for COVID-19 in the future (especially with the deployment of tracking apps in many countries), there are currently no such agent-level data available for COVID-19. The other approach for validation is to consider coarse population-level variables and their dynamics and compare them against established models such as SIR and its variations. We will perform validation of RAW-ALPS in two ways.
##### 2.4.1. Qualitative Comparisons with SIR Model
As the first comparison, we study shapes of infection curves resulting from the RAW-ALPS model and compare them qualitatively to the shapes resulting from the SIR model. Figure 4 shows plots of the evolutions of global infection counts (susceptible, infected, and recovered) in a community under the well-known SIR model (on the left) and the proposed ALPS model (on the right). In the ALPS model, the counts for recovered and fatalities are kept separate, while in the SIR model, these two categories are combined. One can see a remarkable similarity in the shapes of the corresponding curves, and this provides a certain validation to the RAW-ALPS model. In fact, given the dynamical models of agent-level mobility and infections, one can derive the parameters of the population-level differential equations used in the SIRD model. We pursue this topic in the next section.
FIGURE 4
FIGURE 4. Evolution of population-level infection measurements under a typical SIR model (left) (source: Wikipedia) and the RAW-ALPS model (right). Visually, the RAW-ALPS model can generate infection curves with shapes similar to those of the SIR model.
##### 2.4.2 Estimation of SIRD Parameters
In this section, we use data simulated from RAW-ALPS to fit the SIRD model given in Eq. 1 and use estimated SIRD model parameters to provide interpretations. Rearranging equations in the SIRD model (Eq. 1), we get the instantaneous values of the parameters:
$μ(t)=D′(t)I(t),γ(t)=R′(t)I(t),β(t)=−NS′(t)I(t)S(t)=NI′(t)+γI(t)+μI(t)I(t)S(t) .$
Let $1f(t)>0$ denote the domain over which a function $f(t)$ is strictly positive. We define the estimators of overall SIRD parameters to be
$μ^=mean(D′(t)I(t)1I(t)>0) ,γ^=mean(R′(t)I(t)1I(t)>0),β^=N2mean(1I(t)S(t)>0I(t)S(t)(I′(t)+γI(t)+μI(t)−S′(t))).$
Since in our simulations we have a small population size (N), the number of dead agents becomes a constant soon after any changes are complete, making $D′(t)=0$ for most of the study. To focus on the mortality rate in the active period, we modify the estimator of µ to be$μ^=mean(D′(t)I(t)1D′(t)I(t)>0)$.
Figure 5 shows an example of this estimation using data simulated from RAW-ALPS for an unrestricted situation. The left plot shows the evolution of $S/I$ functions under the chosen simulation parameters. The infection reaches its peak around Day 26 and there are no uninfected agents left after that. For these data, the estimated values of parameters are $β^=0.036281$, $γ^=0.000860$, and $μ^=0.001005$ (at hourly unit time). We clarify that the simulation is updated at an hour unit, but the final estimates of these rates are converted into daily time unit to compare with published values. The basic reproduction number for this simulation comes out to be $R0=β^γ^=42.1703$.
FIGURE 5
FIGURE 5. Estimation of SIRD parameters from data simulated by RAW-ALPS for an unconstrained spread. The pandemic curves generated by the SIRD model with estimated parameters match the RAW-ALPS output.
Figure 6 shows when the restrictions are imposed on Day 5. The left plot shows the evolution of $S/I$ functions—the infection reaches its peak around Day 20 even though there is a significant portion of uninfected agents. For these data, the estimated values of parameters are $β^=0.002860$, $γ^=0.001005$, and $μ^=0.002459$ (at hourly unit time). The basic reproduction number for this simulation comes out to be $R0=β^γ^=2.8458$.
FIGURE 6
FIGURE 6. Estimation of SIRD parameters from data simulated by RAW-ALPS for the case when the restrictions are imposed on Day 5. Once again, the pandemic curves generated by the SIRD model with estimated parameters match the RAW-ALPS output.
Compared with the previous example, we see that the infection rate comes down significantly, from $β^=0.036281$ to $β^=0.002860$, when the restrictions are imposed. Similarly, the reproduction number also comes down from 42.1703 to 2.8458.
## 3 Exemplar Outcomes and Computational Cost
We illustrate the use of the RAW-ALPS model by presenting some sample outcomes under some typical scenarios. Furthermore, we discuss the computational cost of running RAW-ALPS on a laptop computer.
### 3.1 Examples From RAW-ALPS
We start by showing sample outputs of RAW-ALPS under some interesting settings. In these examples, we use a relatively small number of agents ($N=1500$) and household units ($h=289$), with $T=50$ days, in order to improve visibility of displays.
1. Example 1No Restrictive Measures: Figure 7 shows a sequence of temporal snapshots representing the community at different times over the observation period. In this example, the population is fully mobile over the observation period and no social distancing restrictions are imposed. The snapshots show the situations on Day 8, 17, 33, and 42. The corresponding time evolutions of global count measures $\left[S\mathrm{\left(}t\mathrm{\right)},I\mathrm{\left(}t\mathrm{\right)},D\mathrm{\left(}t\mathrm{\right),}\text{ }\text{and}\text{ }R\mathrm{\left(}t\mathrm{\right)}\right]$ are shown in the bottom right panel. The infection starts to spread rapidly around Day 5 and reaches a peak infection level of 78% around Day 17. Then, the recovery starts and continues until very few infected people are left. In this simulation, the number of fatalities is found to be 6%.
2. Example 2Early Restrictions: In the second example, a lockdown of all the infected agents is introduced on Day 5 and these measures stay in place after that. The results are shown in Figure 8. Once again we show snapshots for situations on Day 8, 17, 33, and 42. The corresponding time evolutions of global count measures are shown in the bottom right panel. As the summary shows, an early restriction of lockdown is very effective in controlling the spread of the infection and the peak infection rate is minuscule at 3%.
3. Example 3Early Restrictions but Removed Later: In the next example, a full lockdown or LD1 is introduced on Day 2 but removed on Day 30. The results are shown in Figure 9. In early snapshots, the population is under a lockdown and the spread is minimal. By Day 33, the lockdown is released, and the population is fully mobile. Consequently, the infection begins to spread again, and by Day 42, the infection reaches its peak value.
4. Example 4Delay in Imposing Restrictions: In the last example, with results shown in Figure 10, a full lockdown (LD1) is introduced very late (on Day 10). In this simulation run, the infection rate reaches a peak value of 80% despite a lockdown. This is because the infection had already spread extensively in the population by the time the lockdown starts.
FIGURE 7
FIGURE 7. Example 1: model outputs at different times under no restrictive measures. Blue dots are susceptible agents, red dots are infected agents, green dots are recovered agents, and purple circles denote fatalities. The pandemic curves in the bottom right exhibit typical behavior of the RAW-ALPS model at the population level.
FIGURE 8
FIGURE 8. Example 2: displays of model outputs when a lockdown of infected people (LD2) is imposed on Day 5. The pandemic curves in the bottom right show a drastic reduction in infection due to the lockdown.
FIGURE 9
FIGURE 9. Example 3: model outputs when a full lockdown or LD1 is imposed on Day 2 and removed on Day 30. As seen in the plot of the resulting pandemic curves, the infection is controlled at first by the lockdown, but it spreads fast after the lockdown is lifted.
FIGURE 10
FIGURE 10. Example 4: model outputs when a full lockdown is imposed on Day 10. The pandemic curves show that the infection has spread widely in the community by the time a lockdown is imposed.
### 3.2. Computational Complexity
The main computational cost in the simulation comes from the need to update the following variables at each time t:
• locations of all agent according to their independent motion model;
• pairwise distance matrix between Susceptible agents and Infected agents; and
• infection status of each agent according to the infection dynamics.
In general, we obtain some efficiency by performing matrix operations, rather than using “for loops” for these updates across agents. Additionally, we increase speed by maintaining a list of neighborhoods for each agent and checking interactions only between the neighbors at each t.
Since the computational efficiency of the simulator is of vital importance, we study the computational cost of running RAW-ALPS for different variable sizes. In these experiments, we note the time taken by RAW-ALPS code on a MacBook Pro laptop with an Intel 2.8 GHz Core i7 processor and 16 GB memory. In Figure 11 we plot average run times (using five runs in a setting) of the code for different values of N, h, and T. Recall that h is the number of household units, N is the number of agents, and T is the period length. From these results, we see that the computational cost is linear in T, which makes sense. The computational cost is superlinear in N, keeping other variables fixed. This is because an increase in N represents a higher density and increased infection rates, thus requiring additional computations for tracking infected agents. Interestingly, for the range of parameter values studied here, the computational cost does not grow with an increase in h. As an aside, we note that for $N=5000$ and $h=1521$, the run time for $T=100$ days is approximately 330 s and for $T=150$ days is approximately 484 s.
FIGURE 11
FIGURE 11. Plots of run times vs. simulation parameters T, N, and h. The algorithmic cost is linear in T but superlinear in N.
## 4 Analyzing Effects of Lockdown Measures
There are several ways to utilize this model for prediction, planning, and decision-making. We illustrate some of these ideas using examples.
### 4.1 Timing of Imposition of Full Lockdown
First, we study the effect of time of a full lockdown on the epidemic infections. In the following simulations, we have used $N=1000$ agents with $h=361$ households in the scene domain $[0,2]2$ miles. For the infection parameters, we use a contact radius $r0=0.02$ miles, contact period $τ0=5 h$, and $pI=0.1$. The motion parameters are: $σ=0.0002 mph$, $μ=0.02$, and $α=0.1$. In each setting, we run the code 30 times and collect the simulation outputs.
Figure 12 shows examples of RAW-ALPS outputs when we impose a full lockdown on the community but at different times. From top to bottom, the plots show lockdowns starting on Day 2, Day 10, and Day 30, respectively, with the last row showing results for no lockdown. Once the restrictions are imposed, they are not removed in these examples. As expected, the best results are obtained for the earliest imposition of restrictions. In the case of no lockdown, the peak infection rate in the population ranges from 60 to 80%, which is very high for a community. The fraction of fatalities ranges from 6 to 8%, and the fraction of community that is never infected is zero in all runs. In case the restrictions are imposed on Day 2, with all other parameters held the same, there is a remarkable improvement in the situation. The peak infection goes down to 2–3%, the fatalities decrease to 0.1–0.2%, and the fraction of uninfected goes up to 98–99%. Thus, an early imposition of full lockdown measures helps significantly reduce infection in the community.
FIGURE 12
FIGURE 12. Results from RAW-ALPS runs for a full lockdown (LD1) starting at different times. Going from top to bottom, we impose lockdowns later and later. The effect of a lockdown diminishes significantly if the start of the lockdown is delayed.
### 4.2 Timing of Removal of Restrictions
In the next set of simulations, we study the effects of lifting restrictions and thus re-allowing full mobility in the community. Some sample results are shown in Figure 13. Each plot shows the evolution for a different end time $T1$, while the start time is kept fixed at $T0=5$. As these plots indicate, the gains made by early imposition of restrictions are nullified when the restrictions are lifted too soon. In the case of early lifting of restrictions, the full population gets infected eventually. Since we do not assume any change in the immunity levels of the agents over time, the results from early lifting of restrictions are quite similar to those from not imposing any restrictions in the first place. The results appear to be the same, just shifted in time.
FIGURE 13
FIGURE 13. Results from RAW-ALPS runs for a full lockdown (LD1) ending at different times. Going from top to bottom, we keep the lockdowns for longer periods. A longer imposition of lockdown helps in reducing the spread of the epidemic.
### 4.3 Statistical Summaries
In the next set of experiments, we compute average values of some variables of interest using 30 runs of RAW-ALPS. In the first result, we study three variables—number of deaths, number of agents remaining uninfected, and the peak infection rate—using $N=1000$ agents living in a community of $h=361$ households, observed over $[0,100]$ days. We vary the start time $T0$ (start day of restrictions) from 2 to 30 and then to 100 and study the resulting outcomes. (The value of $T0=100$ implies that the restrictions are never imposed in that setting.)
Figure 14 shows box plots of these three variables against $T0$, with each row representing a different type of lockdown. The first column shows the median percentage of fatalities in the population for different lockdown types. The second column shows the fraction of the uninfected population and the last column shows the peak infection rates. It can be seen that the results are very similar for a full lockdown (LD1) and a lockdown of infected agents (LD2) but are quite different for a lockdown of only the symptomatic agents (LD3). We find that this last type of lockdown is ineffective in containing the spread of the infection, except for when $T0=2$. These results suggest that a lockdown of only the infected agents (LD2) can be an effective strategy in controlling the epidemic. Of course, given the asymptomatic nature of the disease, it is not possible to ascertain the infection status precisely. One can only estimate this status for a large population using accurate and extensive testing schemes.
FIGURE 14
FIGURE 14. Statistical summaries of infection variables obtained using 30 runs of RAW-ALPS, plotted against the starting day of the lockdown. In all three lockdown types, a delay in imposing lockdown causes an increase in infections and casualties.
In Figure 15, we study the impact of changing $T1$ while $T0=1$ is kept fixed (and other experimental conditions being the same as in the last experiment). The three panels show the fractions of deaths (left), the number of uninfected (middle), and the peak infection rates (right). Each curve in the panel corresponds to a different lockdown type: full LD or LD1, infection LD or LD2, and symptomatic LD or LD3. The results show that for LD1 and LD2, a delay in the lifting of the restrictions is effective in controlling the epidemic. In contrast, the imposition of LD3 is not as effective. Interestingly, these results show that LD2 is more effective than LD1. This may be because in full lockdown, the chances of susceptible agents coinhabiting with the infected agents increase and this, in turn, increases the infection rates.
FIGURE 15
FIGURE 15. Statistical summaries of infection variables obtained using 30 runs of ALPS, plotted against $T1$, the reopening day. In all three types of lockdowns, a delay in reopening after a lockdown helps control the number of infections and deaths.
## 5 Discussion and Conclusion
This article develops an agent-based simulation model, called RAW-ALPS, for modeling the spread of an infectious disease in a closed community. Several simplifying yet reasonable assumptions make RAW-ALPS efficient and effective for statistical analysis. The model is validated at a population level by comparing it with the popular SIR model in epidemiology.
The results from RAW-ALPS show that a lockdown of only the infected agents (LD2) is the most effective kind. However, this includes both symptomatic and asymptomatic agents, with the latter ones not being easy to detect. This calls for regular and extensive testing of the population to isolate and restrict all infected agents while allowing for free movements of all uninfected agents. Furthermore, these results indicate that 1) early imposition of lockdown measures (right after the first infection) significantly reduces infection rates and fatalities; 2) lifting of lockdown measures recommences the spread of the disease, and the infections eventually reach the same level as that of the unrestricted community; and 3) in the absence of any extraneous solutions (a medical treatment/cure, a weakening mutation of the virus, or natural development of agent immunity), the only viable option for preventing large infections is the judicious use of lockdown measures.
The strengths and limitations of the RAW-ALPS model are the following. It provides efficient yet comprehensive modeling of the spread of infections in a self-contained community, using simple model assumptions. The model can prove very useful in evaluating costs and effects of imposing different types of social lockdown measures in a society. In the current version, the initial placement of agents is set to be normally distributed with means given by their home units and fixed variance. This variance is kept large to allow for near arbitrary placements of agents in the community. In practice, however, agents typically follow semirigid daily schedules of being at work, performing chores, or being at home. Thus, at the time of imposition of a lockdown, the agents can be better placed in the scenes according to their regular schedules, rather than being placed arbitrarily.
In terms of future directions, there are many ways to develop this simulation model to capture more realistic scenarios. 1) It is possible to model multiple, interactive communities instead of a single isolated community. 2) One can include typical daily schedules for agents in the simulations. A typical agent may leave home in the morning, spend time at work during the day, and return home in the evening. 3) It is possible to provide age demographics to the community and assign immunity to agents according to their demographic labels [24]. 4) As more data becomes available in the future, one can change immunity levels of agents over time according to the spread and seasons. 5) In practice, when an agent is infected, he/she goes through different stages of the disease, associated with varying degrees of mobility [13]. One can introduce an additional variable to track these stages of infections in the model and change agent mobility accordingly. 6) One can incorporate super spreader events in the model to help capture these mechanisms of transmission. 7) Finally, one can use the output of RAW-ALPS, in conjunction with techniques for the analysis of epidemic curves [2527], to further adapt simulation parameters to a given community or region.
## Data Availability Statement
The raw data supporting the conclusion of this article will be made available by the authors, without undue reservation.
## Author Contributions
All authors listed have made a substantial, direct, and intellectual contribution to the work and approved it for publication.
## Funding
This research was supported in part by the grants NSF DMS-1621787 and NSF DMS-1953087.
## Conflict of Interest
The author declares that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
## Appendix: listing of alps parameters
Table A1 provides a listing of all the parameters one can adjust in ALPS to achieve different scenarios. It also provides some typical values used in the experiments presented in the article.
TABLE A1
TABLE A1. Listing of parameters associated with different model components of ALPS.
## References
1. Bertozzi, AL, Franco, E, Mohler, G, Short, MB, and Sledge, D. The challenges of modeling and forecasting the spread of COVID-19. Proc Natl Acad Sci (2020). 117:16732–16738. doi:10.1073/pnas.2006520117
2. Adam, D. Special report: the simulations driving the world’s response to COVID-19. Nature (2020). 580 (7803):316–8. doi:10.1038/d41586-020-01003-6
3. Chao, DL, Halloran, ME, Obenchain, VJ, and Longini, JIM. FluTE, a publicly available stochastic influenza epidemic simulation model. Plos Comput Biol (2010). 6:e1000656. doi:10.1371/journal.pcbi.1000656
4. Hunter, E, Mac Namee, B, and Kelleher, JD. A taxonomy for agent-based models in human infectious disease epidemiology. J Artif Societies Soc Simulation (2017). 20. doi:10.18564/jasss.3414
5. Kermack, WO, and McKendrick, AG. A contribution to the mathematical theory of epidemics. Proc R Soc A (1927). 115:700–721. doi:10.1098/rspa.1927.0118
6. Osemwinyen, AC, and Diakhaby, A. Mathematical modelling of the transmission dynamics of ebola virus. Appl Comput Math (2015). 4:313–320. doi:10.11648/j.acm.20150404.19
7. Timpka, T, Eriksson, H, Gursky, EA, Nyce, JM, Morin, M, and Jenvald, J. Population-based simulations of influenza pandemics: validity and significance for public health policy. Bull World Health Organ (2009). 87:305–311. doi:10.2471/BLT.07.050203
8. Hao, X, Cheng, S, Wu, D, Wu, T, Lin, X, and Wang, C. Reconstruction of the full transmission dynamics of COVID-19 in wuhan. Nature (2020). 584:420–424. doi:10.1038/s41586-020-2554-8
9. Verity, R, and Okell, LC, Dorigatti, I, Winskill, P, Whittaker, C, and Imai, N. Estimates of the severity of coronavirus disease 2019: a model-based analysis. Lancet Infect Dis (2020). 20:P669–P677. doi:10.1016/S1473-3099(20)30243-7
10. Gilbert, NG. Agent-based models. California, US: Sage Publications (2008).
CrossRef Full Text
11. Epstein, JM, and Axtell, RL. Growing artificial societies Social science from the bottom up. Cambridge, MA: MIT Press (1996).
12. Hunter, E, Namee, BM, and Kelleher, J. An open-data-driven agent-based model to simulate infectious disease outbreaks. PLoS ONE (2018). 13:e0208775. doi:10.1371/journal.pone.0208775
13. Perez, L, and Dragicevic, S. An agent-based approach for modeling dynamics of contagious disease spread. Int J Health Geographics (2009). 08:50. doi:10.1186/1476-072X-8-50
14. Nguyen, V, Mikolajczyk, R, and Hernandez-Vargas, E. High-resolution epidemic simulation using within-host infection and contact data. BMC Public Health (2018). 18. doi:10.1101/133421
15. Silva, PC, Batista, PV, Lima, HS, Alves, MA, Guimaraes, FG, and Silva, RC. COVID-ABS: an agent-based model of COVID-19 epidemic to simulate health and economic effects of social distancing interventions. Chaos Solitons Fractals (2020). 139:110088. doi:10.1016/j.chaos.2020.110088
16.Center for Disease Control. Clinical questions about COVID-19: questions and answers (2020). Available at:https://www.cdc.gov/coronavirus/2019-ncov/hcp/faq.html
17. Dunham, JB. An agent-based spatially explicit epidemiological model in MASON. J Artif Soc Soc Simul (2005). 9:1–3.
18. Gurdasani, D, and Ziauddeen, H. On fallability of smulations models in informing pandemic responses. Lancet Glob Health (2020). 8:E776–E777. doi:10.1016/S2214-109X(20)30219-9
19.Center for Disease Control. Contact tracing resources: resources for conducting contact tracing to stop the spread of COVID-19 (2020). Available at:https://www.cdc.gov/coronavirus/2019-ncov/php/open-america/contact-tracing-resources.html
20.Covid Volunteer Team. The COVID tracking project (2020). Available at: https://covidtracking.com/
21.Center for Disease Control. Coronavirus disease 2019 (COVID-19) (2020c). Available at: https://www.cdc.gov/coronavirus/2019-ncov/covid-data/covidview/index.html
22. Byambasuren, O, Cardona, M, Bell, K, Clark, J, McLaws, M-L, and Glasziou, P. Estimating the extent of asymptomatic COVID-19 and its potential for community transmission: systematic review and meta-analysis. medRxiv (2020). 10.1101/2020.05.10.20097543
23.Science News. Whole-town study reveals more than 40% of COVID-19 infections had no symptoms (2020). Available from: https://www.sciencedaily.com/releases/2020/06/200630103557.htm (Accessed September 2020).
24. Chang, SL, Harding, N, Zachreson, C, Cliff, OM, and Prokopenko, M. Modelling transmission and control of the COVID-19 pandemic in Australia. arXiv (2020).
25. Srivastava, A, and Chowell, G. Understanding spatial heterogeneity of COVID-19 pandemic using shape analysis qof growth rate curves. medRxiv (2020). doi:10.1101/2020.05.25.20112433
26. Srivastava, A, and Klassen, E. Functional and shape data analysis. New York, US: Springer (2016).
CrossRef Full Text
27. Srivastava, A, Klassen, E, Joshi, SH, and Jermyn, IH. Shape analysis of elastic curves in Euclidean spaces. IEEE Trans Pattern Anal Mach Intell (2011). 33:1415–1428. doi:10.1109/TPAMI.2010.184
Keywords: COVID-19 simulations, random-walk models, lockdown measures, agent-level models, SIR model
Citation: Srivastava A (2021) Random-Walk, Agent-Level Pandemic Simulation (RAW-ALPS) for Analyzing Effects of Different Lockdown Measures. Front. Appl. Math. Stat. 7:638996. doi: 10.3389/fams.2021.638996
Received: 08 December 2020; Accepted: 16 February 2021;
Published: 28 April 2021.
Edited by:
Waleed Isa Al Mannai, New York Institute of Technology Bahrain, Bahrain
Reviewed by:
Shuo Chen, University of Maryland, Baltimore, United States
Halim Zeghdoudi, University of Annaba, Algeria
Copyright © 2021 Srivastava. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
*Correspondence: Anuj Srivastava, anuj@stat.fsu.edu
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# How Do I Create a Cash Flow Model?
Osmand Vitez
A cash flow model helps owners and managers calculate cash expected from certain business activities. These models are often different for each situation, as few activities in business are exactly the same. To create a cash flow model, steps include selecting the time frame to measure, using a formula to estimate cash inflows and outflows, deciding whether or not to use a discounted cash flow model and comparing positive cash flows to previous projects. These steps are similar for each model situation.
Measureable time frames are a must when measuring cash flow. In a project management type of system, owners and managers will often break down models to each project and specific time lines within each project. For example, a cash flow model is common in construction companies. Individuals in these firms must complete projects on a timeline; this provides an opportunity for measuring cash flow on a particular basis. Using a specific time frame helps companies create a deeper analysis for measuring the company’s overall cash flow process.
Formulas for estimating cash flows will most certainly vary. A common practice for this process is to estimate the total revenues a company will receive, and then deduct the expenses necessary to earn these revenues. For example, if each member in a sales force delivers an average of \$25,000 US Dollars (USD) monthly, then a company would compare how much it costs to hire, train and pay for this new team member. The remaining figure is then the cash flow left to the company for paying expenses and rewarding owners or shareholders with a profit.
Using a discounted cash flow model helps estimate future cash flows for multiple years. The value of these estimates needs to be based on current dollar values for accurate comparison to money spent on new projects. Owners and manages will estimate cash flows and discount them back to the current dollar amount using a discount rate, such as the cost of capital related to borrowed funds. This process is common because it gives a quantitative analysis to the model that can eliminate the subjectivity of human error.
Business models allow for comparison between one or more projects. Although each project is different, comparing the effectiveness of models can help the company refine their business modeling process. Comparisons also help companies discover which projects result in the highest cash returns. Owners and managers can also help the company diversify by taking on projects that have low risk and low cash returns, which can offset riskier projects with questionable cash returns.
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# CHAPTER
1
Flat Slabs
1.1 INTRODUCTION
Common practice of design and construction is to support the slabs by beams and support the beams by columns. This may be called as beam-slab construction. The beams reduce the available net clear ceiling height. Hence in warehouses, offices and public halls some times beams are avoided and slabs are directly supported by columns. This types of construction is aesthetically appealing also. These slabs which are directly supported by columns are called Flat Slabs. Fig. 1.1 shows a typical flat slab.
d 2
Critical section for shear
Fig. 1.1
A typical flat slab (without drop and column head)
The column head is some times widened so as to reduce the punching shear in the slab. The widened portions are called column heads. The column heads may be provided with any angle from the consideration of architecture but for the design, concrete in the portion at 45º on either side of vertical only is considered as effective for the design [Ref. Fig. 1.2].
Critical section for shear d 2
90° Concrete in this area is neglected for calculation
Fig. 1.2 Slab without drop and column with column head
2
Moments in the slabs are more near the column. Hence the slab is thickened near the columns by providing the drops as shown in Fig. 1.3. Sometimes the drops are called as capital of the column. Thus we have the following types of flat slabs:
d 2 d 2 Critical section for shear
Critical section for shear
Fig. 1.3 Slab with drop and column without column head
(i) (ii) (iii) (iv)
Slabs Slabs Slabs Slabs
without drop and column head (Fig. 1.1). without drop and column with column head (Fig. 1.2). with drop and column without column head (Fig. 1.3). with drop and column head as shown in Fig. 1.4.
d 2
Critical section for shear 45° 45°
Fig. 1.4
Slab with drop and column with column head
The portion of flat slab that is bound on each of its four sides by centre lines of adjacent columns is called a panel. The panel shown in Fig. 1.5 has size L1 ´ L2. A panel may be divided into column strips and middle strips. Column Strip means a design strip having a width of 0.25L1 or 0.25L2, whichever is less. The remaining middle portion which is bound by the column strips is called middle strip. Fig. 1.5 shows the division of flat slab panel into column and middle strips in the direction y.
if Fe 415 or Fe 500 steel is used If drops are not provided or size of drops do not satisfy the specification 1. 1. The drops when provided shall be rectangular in plan.3 Thickness of Flat Slab From the consideration of deflection control IS 456-2000 specifies minimum thickness in terms of span to effective depth ratio.5 Panels. if HYSD bars are used It is also specified that in no case. if mild steel is used.e. 1. then the ratio shall not exceed 0. = 32 ´ 0. .8. For exterior panels.2 and 1. and have a length in each direction not less than one third of the panel in that direction.2.1 PROPORTIONING OF FLAT SLABS Drops IS 456-2000 [Clause 31..9 = 28.2. shall be considered for design purpose as shown in Figs. then the maximum value of ratio of larger span to thickness shall be = 40. 1. that portion of the column head which lies within the largest right circular cone or pyramid entirely within the outlines of the column and the column head.2. if mild steel is used = 32.9 = 36.2. = 40 ´ 0. If drop as specified in 1. the width of drops at right angles to the non continuous edge and measured from the centre-line of the columns shall be equal to one half of the width of drop for interior panels.1.2] gives the following guidelines for proportioning.9 times the value specified above i.1 is provided.2 Column Heads Where column heads are provided. 1. the thickness of flat slab shall be less than 125 mm. column strips and middle strips is y-direction 1.Flat Slabs L2a C of panel A L2b C of panel B 3 L1 y o Column strip x Middle strip Column strip Middle strip Column strip L2a 4 L but < 1 4 L2b 4 but < L1 4 Fig.2 1. For this purpose larger span is to be considered.2.4.
Total Design Moment The absolute sum of the positive and negative moment in each direction is given by M0 = Where. the following clauses are to be carefully noted: (a) Circular supports shall be treated as square supports having the same area i. squares of size 0. L1 and L2. (c) The successive span length in each direction shall not differ by more than one-third of longer span. (b) The panels shall be rectangular and the ratio of the longer span to the shorter span within a panel shall not be greater than 2.e.C. L2 shall + L2b . 1.886D. M0 = Total moment W = Design load on the area L2 ´ Ln Ln = Clear span extending from face to face of columns.4 THE DIRECT DESIGN METHOD This method has the limitation that it can be used only if the following conditions are fulfilled: (a) There shall be minimum of three continuous spans in each directions. capitals. and L2 = Length of span transverse to L1 In taking the values of Ln. (e) The end span must be shorter but not greater than the interior span. (d) The design live load shall not exceed three times the design dead load. brackets or walls but not less than 0.65 L1 L1 = Length of span in the direction of M0. (f) It shall be permissible to offset columns a maximum of 10 percent of the span in the direction of the offset not withstanding the provision in (b).4 Advanced R. In Fig. Design 1..C. 2 (c) When the span adjacent and parallel to an edge is being considered. WL n 8 be taken as the average of the transverse spans.3 DETERMINATION OF BENDING MOMENT AND SHEAR FORCE For this IS 456-2000 permits use of any one of the following two methods: (a) The Direct Design Method (b) The Equivalent Frame Method 1. the distance from the edge to the centre-line of the panel shall be substituted for L2.5 it is given by bL 2a g Distribution of Bending Moment in to ve and +ve Moments The total design moment M0 in a panel is to be distributed into –ve moment and +ve moment as specified below: . (b) When the transverse span of the panel on either side of the centre line of support varies.
35 M0 = 0. The moment in the middle strip shall be the difference between panel and the column strip moments. Distribution of Bending Moments Across the Panel Width The +ve and –ve moments found are to be distributed across the column strip in a panel as shown in Table 1. Table 1.65 OPM MN1 + a1 PQ c 0 ∑K ∑K c s Kc = Sum of the flexural stiffness of the columns meeting at the joint.65 M0 0.Flat Slabs 5 In an interior span Negative Design Moment Positive Design Moment In an end span Interior negative design moment 0. LM 0.1 S. expressed as moment per unit rotation. a b c Distribution of Moments Across the Panel Width in a Column Strip Distributed Moment Per cent of Total Moment 100 75 60 Negative BM at the exterior support Negative BM at the interior support Positive bending moment .75 - Positive design moment LM MN LM MN .28 M0 1 1+ ac OP PQ OP PQ = 0.1. 010 M0 1 1+ ac 0.63 - Exterior negative design moment = where ac is the ratio of flexural stiffness at the exterior columns to the flexural stiffness of the slab at a joint taken in the direction moments are being determined and is given by ac = Where. and Ks = Flexural stiffness of the slab. No.
1.4. The critical section for shear shall be at a distance Critical section d/2 Support section column / column head (a ) Support section d/2 d/2 Critical section (b) Fig. Critical section Support section d/2 Support section d/2 Critical section d/2 d/2 d/2 (a) (b) Fig. 1.C.7 In case of columns near the free edge of a slab.C. The shape of the critical section in plan is similar to the support immediately below the slab as shown in Fig. Free edge Critical section Free corner d/2 d/2 (a) Corner column (b) Critical section Fig. These critical sections are shown in Figs. 31. 1. Shear Force d from the periphery of the column/capital drop 2 panel. 1. the critical section shall be taken as shown in Fig. 1.6 Advanced R. 1. the critical section shall be taken as indicated in Fig. Hence if drops are provided there are two critical sections near columns.4.7. Design Moments in Columns In this type of constructions column moments are to be modified as suggested in IS 456–2000 [Clause No.8.1 to 1. 1.6 For columns sections with re-entrant angles.5].8 .6.
c 0. EDGE OF DROP e b 150 mm 24BAR DIA OR 300 mm min. Clear span .15 l max 75 mm max.20 ln c 0.22 ln * Bent bars at exterior supports may be used if a general analysis is made.Flat Slabs 9 Minimum percentage of steel at section 50 Remainder 50 Remainder 50 Type of bars Strip WITHOUT DROP PANEL WITH DROP PANEL Straight bars d b 75 mm max d b e b e b Column strip 24 BAR DIA OR 300 mm min. EDGE OF DROP c 75 mm max. 150 mm c c Straight bars 100 Middle Strip 50 Remainder 50 150 mm c (ALL BARS) Bent bars* Remainder 50 Remainder 150 mm f f 150 mm 75 mm max.30 ln e 0.ln Face of support interior support D C 75 mm max. Note.125lmax d b Bent bars* Remainder 50 Remainder d b b e g g o.14 ln b 0. D C Clear span .33 ln f 0.15 l max 150 mm c a a c C (ALL BARS) 150 mm 0.20 ln g 0.ln D Face of support C interior support [NO SLAB CONTINUITY] [CONTINUITY PROVED] [NO SLAB CONTINUITY] Bar Length From Face of Support Minimum Length Maximum Length d 0. 1. ALL BARS 150 mm min.9 Minimum bend joint locations and extensions for reinforcement in flat slabs . Fig.24 ln Exterior support Mark Length a 0. D is the diameter of the column and the dimension of the rectangular column in the direction under consideration.
166 kN \ tv = 675 500 368166 × 1000 .58 Checking the thickness selected: Since Fe 415 steel is used.40 = 92.675 ´ 0.3125 ´ 106 Nmm = 211.85 26.25 fck = 0.5 = 4.44 kNm Distribution of moment into column strips and middle strip: Column Strip in kNm –ve moment +ve moment 0.65 ´ 189.35 ´ 189.44 = 39. Check for Shear d from the column face. Mu lim = 0.675 = 368.5 m \ Total design load in a panel W = 15 L2 Ln = 15 ´ 5 ´ 4.5 ´ = 189.118 N/mm2 .75 ´ 123. Design Ln = 5 – 0.25 20 = 1. Hence periphery of critical 2 section around a column is square of a size = 500 + d = 500 + 175 = 675 mm The critical section for shear is at a distance Shear to be resisted by the critical section V = 15 ´ 5 ´ 5 – 15 ´ 0.138 ´ 20 ´ 2500 ´ 1752 = 211.86 Middle Strip in kNm 30.35 ´ 189.84 kNm 8 8 = 0.5 = 337.5 = 337. = 0.C.5 kN Moments Panel Moment Panel –ve moment M0 = WL n 4.779 N/mm2 4 × 675 × 175 500 675 ks = 1 + bc subject to maximum of 1.84 = 66.3125 kNm Hence singly reinforced section can be designed i.10 Advanced R.55 0..5 ´ 5000 = 2500 mm \ Mu lim = 0.84 = 123.40 kNm Panel +ve moment = 0. thickness provided is satisfactory from the consideration of bending moment.e.138 fck b d2 Width of column strip = 0. bc = \ safe in shear since tv < tc L1 5 = =1 L2 5 ks = 1 tc = 0.60 ´ 66.C.84 = 0.
spacing required is s= p 4 ´ 102 ´ 2500 = 301. i.87 ´ 415 ´ Ast ´ 175 1 − i. For +ve moment in column strip: Mu = 39.3 OP Q LM N OP Q 415 Ast × 2500 × 175 20 OP Q A st 2 – 21084.55 ´ 106 = 0.3 = 0 Ast = 1583.3 = 0 Ast = 651 mm2 Using 10 mm bars.74 Provide 12 mm bars at 175 mm c/c.86 kNm \ 39.6 mm < 2 ´ thickness of slab 651 Hence provide 10 mm bars at 300 mm c/c.87 f y A st d 1 − LM N A st f y bd fck = 0.3Ast + 1464. .86 ´ 21084. provide similar reinforcement in other direction also.86 ´ 106 = 0. 1464. Hence using 12 mm bars..74 mm2 This is to be provided in a column strip of width 2500 mm.e.78 = Ast 1 − LM N A st 21084.3 OP Q LM N 415 A st × 2500 × 175 20 OP Q Ast2 – 21084.86 = Ast 1 − or \ LM N A st 21084. Since span is same in both directions.. spacing required is given by s= p 4 ´ 12 2 ´ 2500 = 178 mm 1583.87 ´ 415 ´ Ast ´ 175 1 − 630.55 kNm 92. Provide 10 mm diameter bars at 300 mm c/c in the middle strip to take up –ve and +ve moments.3 Ast + 630.e.Flat Slabs 11 Reinforcement For –ve moment in column strip: Mu = 92.78 ´ 21084.
1.5 mm 32 Provide 190 mm thickness.C. Solution : Thickness : Since Fe 415 steel is used and drop is provided.300 c\c Cover -25 500 500 Section through column strip 3000 10 . Let the cover be 30 mm \ Thickness of flat slab = \ Overall thickness D = 220 mm Let the drop be 50 mm.C. Provide suitable drop. 1.10 Reinforcement details [all dimension in mm units] Example 1.300 c/c 3000 500 section through middle strip 500 Fig.12 Advanced R. Design Reinforcement Details It is as shown in Fig.2: Design an interior panel of a flat slab with panel size 6 ´ 6 m supported by columns of size 500 ´ 500 mm. Take live load as 4 kN/m2.10 Column Strip 12-175 c/c Middle Strip Column strip Column strip 12-175 c/c 10-300 c/c 10-300 c/c 5000 Sign convention Top reinforcement Bottom reinforcement Column Strip Middle Strip 5000 5000 5000 12-175 c/c 5000 5000 200 10 . Hence at column head. Use M20 concrete and Fe 415 steel. maximum span to thickness ratio permitted is 32 6000 = 187. d = 240 mm and D = 270 mm .
= = 400 kNm 8 8 \ Total negative moment = 0.928 ´ 106 Nmm Thus Mu lim > Mu.4 ´ 140 = 56 kNm Width of column strip 2 = width of middle strip = 3000 mm = 476.6 ´ 140 = 84 kNm Middle Strip 0.35 ´ 400 = 140 kNm The above moments are to be distributed into column strip and middle strip M0 = Column Strip –ve moment +ve moment 0.25 ´ 260 = 65 kNm 0.75 kN/m2 = 1.5 = 581.27 ´ 1 ´ 1 ´ 25 = 6. . Hence thickness selected is sufficient.928 kNm Mu lim = 0.00 kN/m2 = 11.00 kN/m2 = 4.625 kN Design Total Moment Total moment Width of column strip = width of middle strip = 3000 mm.625 ´ 6 ´ 5.75 kN/m2 Finishing load Live load Total load \ Design (factored) load Clear span \ Design load = 1.625 kN/m2 Ln = 6 – 0.75 ´ 260 = 195 kNm 0. W0 Ln 581625 ´ 55 .5 = 5. \ Loads For the purpose of design let us take self-weight as that due to thickness at column strip \ Self-weight = 0.138 ´ 20 ´ 3000 ´ 2402 = 476.5 ´ 11. Check for Shear The critical section is at a distance .Flat Slabs 13 Size of Drop It should not be less than 1 ´6m = 2 m 3 Let us provide 3 m ´ 3 m drop so that the width of drop is equal to that of column head.75 = 17.5 m W0 = Wu ´ L2 ´ Ln = 17.138 fck b d = 0.65 ´ 400 = 260 kNm Total positive moment = 0.
118 N/mm2 Design shear stress permitted = 1.87 ´ 415 ´ Ast ´ 240 1 − 2250. Design d 240 = = 120 mm from the face of column 2 2 \ It is a square of size = 500 + 240 = 740 mm V = Total load – load on 0.38 = Ast 1 − LM N A st 34698.8 = 0 Ast = 2419 mm2 in 3000 mm width .880 N/mm2 4 × 740 × 240 740 k s = 1 + bc subject to maximum of 1 bc = 500 L1 =1 L2 ks = 1 500 740 tc = 0.38 ´ 34698. Shear strength may be checked at distance Reinforcement (a) For –ve moment in column strip Mu = 195 kNm Thickness \ d = 240 mm Mu = 0. 2 LM N fy A st × b × d fck 195 ´ 106 = 0. It is quite safe since drop size is large.740 ´ 0.87 fy Ast d 1 − 120 500 120 d from drop.C.118 N/mm2 > tv Hence the slab is safe in shear without shear reinforcement also.625 ´ 6 ´ 6 – 17.740 = 624.25 20 = 1.625 ´ 0.740 area = 17.C.740 ´ 0.14 Advanced R.8 OP Q LM N OP Q 415 A st × 3000 × 240 20 OP Q A st 2 – 34698.849 kN \ Nominal shear Shear strength where where \ = tv = = ks tc 624.489 × 1000 = 0.8 Ast + 2250.
. Thickness = 190 mm Provide 10 mm bars at 230 mm c/c in this portion also. spacing required is s= p 4 ´ 12 2 ´ 3000 = 140. 1.9 OP Q LM N 415 A st × 3000 × 190 20 OP Q 2 Ast – 27469.26 mm 2419 Provide 12 mm bars at 140 mm c/c (b) For +ve moment in column strip Mu = 84 kNm = 84 ´ 106 Nmm.9 OP Q LM N 415 A st × 3000 × 240 20 OP Q Ast = 1285 mm2 Provide 10 mm bars at 180 mm c/c (c) For –ve moment in middle strip: Mu = 65 kNm. Thickness d = 190 mm 84 ´ 106 = 0.11.9 = 0 Ast = 983 mm2 in 3000 mm width Using 10 mm bars s= p 4 ´ 102 ´ 3000 = 239.Flat Slabs 15 Using 12 mm bars. Thickness = 190 mm 65 ´ 106 = 0. provide similar reinforcement in both directions.87 ´ 415 ´ Ast ´ 190 1 − 947. The details of reinforcement are shown in Fig.5 = Ast 1 − \ Using 10 mm bars s= p 4 ´ 102 ´ 3000 = 183 mm 1285 LM N A st 27469.7 mm 983 Provide 10 mm bars at 230 mm c/c (d) For +ve moment in middle strip Mu = 56 kNm.9 Ast + 947. Since span is same in both direction.5 ´ 27469.5 = Ast 1 − LM N Ast 27469.87 ´ 415 ´ Ast ´ 190 1 − 1224.
625 ´ 6 ´ 4.67 = 493.C.33 = 4.67 m W0 = 17.C. 1.67 = 288. if columns are of 500 mm diameter.33 m Ln = 6 – 1.52 = a2 4 a = 1. Design Column strip =Dorp width Column strip =Dorp width Middle strip 12–140 c/c Column strip =Dorp width 10–180c/c 10–230 c/c 6000 =Dorp width Middle strip 12–230c/c 6000 Column strip 10–180c/c 6000 6000 12 @ 140 6000 6000 10 @ 180 c/c 190 240 500 10 @ 230 Cover .3: Design the interior panel of the flat slab in example 1.85 kN M0 = Wo Ln 493.85 × 4.25L = 0.3 kNm = 8 8 .5 m It’s equivalent square has side ‘a’ where \ π × 1.11 Reinforcement details Example 1. providing a suitable column head.25 ´ 6 = 1.16 Advanced R.30 500 Section through column strip 10 @ 230 c/c 190 240 500 500 Fig. Solution: Let the diameter of column head be = 0.2.
87 ´ 415 ´ Ast ´ 240 1 − 1622 = Ast 1 − LM N A st 34698.928 ´ 106 Nmm > Mu Hence thickness selected is sufficient.85 kNm 0.3 = 100.138 ´ 20 ´ 3000 ´ 2402 = 476. \ d = 240 mm 140.55 kNm. 4 OP Q Maximum shear permitted = ks × 0.740 p Shear on this section 1500 120 V = 17.625 6 ´ 6 \ tv = 592. Check for Shear The critical section is at a distance d 240 = = 120 mm from the face of column head 2 2 Diameter of critical section = 1500 + 240 =1740 mm = 1.55 kNm 0.8 OP Q LM N 415 Ast × 3000 × 240 20 OP Q .4 ´ 100.55 ´ 106 = 0.9 kNm The distribution of above moment into column strip and middle strips are as given below: Column Strip –ve moment +ve moment 0. there is no need to provide shear reinforcement Design of Reinforcement (a) For –ve moment in column strip Mu = 140.45 N/mm2 π × 1740 × 240 LM N p ´ 174 2 = 592.3 = 187.59 × 1000 = 0.60 ´ 100.4 = 46.36 kNm Width of column strip = width of middle strip = 3000 mm \ Mu lim = 0.65 ´ 288.9 = 60.75 ´ 187.59 kN .4 = 140.4 kNm Total +ve moment = 0.138 fck bd2 = 0.54 kNm Middle Strip 0.740 m Perimeter of critical section = p D = 1.118 N/mm2 Since ks works out to be 1 Since maximum shear permitted in concrete is more than nominal shear tv.Flat Slabs 17 \ Total –ve moment = 0.35 ´ 288.25 ´ 187.9 = 40.25 20 = 1.
The size of the columns supporting the system is 500 ´ 500 mm and floor to floor height is 4. Calculate design moments in interior and exterior panels at column and middle strips in both directions.C. s= (b) For the +ve moment in column strip Mu = 60.18 Advanced R.87 ´ 415 ´ Ast ´ 190 1 − 882.54 kNm. It is to be designed using M20 grade concrete and Fe 415 steel.9 = 0 Ast = 913 mm2 LM N OP Q LM N 415 A st × 3000 × 190 20 OP Q Using 10 mm bars π 4 × 10 2 × 3000 = 258 mm 913 Provide 10 mm bars at 250 mm c/c.5 m. π 4 × 12 2 × 3000 = 199 mm 1705 Provide 12 mm bars at 190 mm c/c.9 Ast + 882.9Ast + 683 ´ 27469. π 4 × 10 2 × 3000 = 336 mm 701 Provide 10 mm bars at 300 mm c/c. It has to carry a live load of 4 kN/m2 and a finishing load of 1 kN/m2.9 2 Ast – 27469.85 kNm.9 2 Ast – 27469.8 Ast + 1622 ´ 34698. d = 190 mm 46.51 = Ast 1 − A st 27469. Example 1. provide similar reinforcement in both directions.87 ´ 415 ´ Ast ´ 190 1 − 683 = Ast 1 − A st 27469. d = 190 mm 60. As span is same in both directions.4: A flat slab system consists of 5 m ´ 6 m panels and is without drop and column head.85 ´ 106 = 0.9 = 0 Ast = 701 mm2 LM N OP Q LM N 415 A st × 3000 × 190 20 OP Q Using 10 mm bars. Reinforcement detail may be shown as was done in previous problem. s= (c) For –ve moment in middle strip: Mu = 46.51 ´ 27469.54 ´ 106 = 0.8 = 0 Ast = 1705 mm2 Using 12 mm bars.C. . s= (d) Provide 10 mm bars at 300 mm c/c for +ve moment in middle strip also. Design Ast2 – 34698.
5 ´ L2Ln = 16. width of middle strip = 6 – 2.5 m Hence.5 m Width of middle strip = 5 – 2.5 = 2.5 = 453.25 m on either side INTERIOR PANELS Moments Along Longer Size L1 = 6 m \ Load on panel L2 = 5 m Ln = 6 – 0.25 m on either side of column centre line \Total width of column strip = 1.9 m Ln = 5.5 = 5.9 = 28.25 ´ 2 = 2.5 m subject to minimum of 0.5 m Along Width L1 = 5 m \Total width of column strip = 2.5 m W0 = 16.25 L1 or L2 whichever is less.25 ´ 5 = 1.5 = 3.5 kN/m2 Panel Dimensions Along length L1 = 6 m Width of column strip and L2 = 5 m = 0.24 ´ 1 ´ 1 ´ 25 = 6 kN/m2 Finishing weight = 1 kN/m2 Live load = 4 kN/m2 Total = 11 kN/m2 Wu = 1.Flat Slabs 19 Solution: Thickness: Since Fe 415 steel is used and no drops are provided. longer span to depth ratio is not more than 32 ´ 0. 6000 = 208 28.8 d= Let us select Loads Self weight 0.5 ´ 11 = 16.25 ´ 5 = 1.65 ´ L1 = 3.8 d = 210 mm and D = 240 mm = 0.75 kN .5 m L2 = 6 m Width of column strip = 0.5 ´ 5 ´ 5.
5 m.59 kN-m 8 8 Appropriation of Moment: Total –ve moment = 0.51 kNm Middle Strip 202.95 = 202.88 kN-m Total +ve moment = 250.18 = 65.77 = 109.59 – 162. Design M0 = Appropriation of Moment .65 ´ 250.71 kN-m \ Moments in column strip and middle strip are as shown below: Column Strip –ve moment +ve moment 0. effective length of column = 1.75 ´ 162.112 m Size of column = 500 ´ 500 mm 1 × 500 4 mm 4 12 Moment of inertia of column = .65 ´ 311.77 kNm \ Total +ve moment = 311.75 – 152.06 kNm 0.88 = 87. Hence as per Table 28 in IS 456-2000.16 kNm 0.60 ´ 87.51 = 43.60 ´ 109.2 ´ length = 1. W0 Ln 45375 ´ 55 = = 311. = = 250.24 = 4.5 .26 = 5.5 kN M0 = W0 Ln 4455 ´ 4.18 kNm Hence moment in column strip and middle strip along longer direction in interior panels are as given below: Column Strip –ve moment +ve moment 0.88 = 40.40 ´ 87.26 m The building is not restrained from lateral sway.95 kNm 8 8 Total –ve moment = 0.5 ´ 6 ´ 4.5 – 0.72 kNm 0.5 = 4.C.95 – 202.67 kNm Along Width L1 = 5 m Panel load Panel moment L2 = 6 m and Ln = 5 – 0. = W0 = 16. .25 ´ 162.2 ´ 4.59 = 162.20 Advanced R.08 kNm EXTERIOR PANELS Length of column = 4.18 – 65.75 = 152.69 kNm 109.88 = 122.63 kNm Middle Strip 0.75 ´ 202.C.5 = 445.71 = 52.06 = 50.71 = 35.
43 = 53.75 7 Dead load \ Relative stiffness ratio is ac = kc1 + kc2 2 × 1018844 = 1.72 – 120.25 = 161.678 = 1+ .123 = 960000 ks .25 – 161.66 Middle Strip 215.463 Positive moment coefficient = 0.66 = 57.95 = 144.72 0.69 ´ 311.Flat Slabs 21 \ kc = I 1 500 4 = 101844 mm4 = × L 12 5112 LONGER SPAN DIRECTION Moment of inertia of beam Is = Moment of inertia of slab 1 = × 6000 × 240 3 12 = L2 = 5000 mm kc = Its length \ Is 1 6000 ´ 2403 = 1382400 mm4 = ´ 5000 12 5000 4 Live load = < 0.82 120.63 – α Total moment M0 = 311.75 – 0.690 α 0.25 0.72 = 120. 1474 ac Hence various moment coefficients are: a = 1+ Interior –ve moment coefficient = 0.77 Shorter Span Direction \ \ ks = 1 5000 × 2403 = 96000 × 12 6000 k + kc2 2 × 1018844 ac = c1 = 2.387 α 0.463 ´ 31.72 = 0 144.95 kNm \ Appropriation of moments in kNm is as given below: Exterior –ve moment coefficient = Total Interior –ve Exterior –ve + Moment 0.43 – 86.60 ´ 144.387 ´ 311.43 = 86.95 = 215.43 1.95 = 120.28 = 0.65 = 0.1 = 0.43 Column Strip 0.00 ´ 120.72 0.75 ´ 215.474 = 1382400 ks \ 1 1 = 1.
06 –122.43 –152.66 2 57.06 2 –122.76 2 –161. 1.25 2.76 – 110.5 m –122.72 120.471 0.76 = 128.65 Exterior –ve moment coefficient = = 0. Design 1 = 1.09 In the exterior panel in each column strips half the above values will act.59 = 110.72 Middle Strip –53.00 ´ 110.1 0.16 Middle Strip –50.16 86.90 – 128.5 2.471 αc 0.66 65.16 2 –128.1 Interior –ve moment coefficient = 0.25 – 66.67 44.5 2.5 Fig.18 86.67 35.82 52.43 –50.51 + 2 2 –50.5 m –42.440 = 0.442 ´ 250.682 ´ 250.5 m –40.44 ´ 250.72 44.43 –15.09 65.471 Total moment M0 = 250.63 Col Strip –161. 1.16 2 43. These moments are shown in Fig.63 − Positive moment coefficient = 0.16 Middle Strip 170.76 –42.5 2.82 2.63 – α 1.69 52.76 = 0 110.16 2 –128.28 0.51 3.06 –122.72 –152.76 0.25 m 120.12 .18 66.25 Column Strip 0.63 2 66.72 110.69 66.65 0.59 kNm \ Appropriation of moments in shorter span exterior panel in kNm is as given below: \ a1 = 1 + Total Interior –ve Exterior -ve + Moment 0.63 2 –152.471 0.76 1.18 = 42.90 0.06 2 –122.72 1.63 Col Strip –152.06 –110.16 –152.16 2.682 = 0.72 –40.16 –110.C.16 –128.77 52.25 = 66.77 35.08 86.75 a 1.09 –42.06 2 –110.51 + 2 2 43.18 1.82 –4 66.28 = 0.C.75 – = 0.72 –40.12 Col Strip –120.76 0.75 ´ 170.66 2 57.16 = 44.72 –53.59 = 110.16 2 –161.08 65.51 3.5 m –122.72 –53.442 = α 1.76 = 110.18 2 86.69 52.59 = 170.22 Advanced R.60 ´ 110.66 65.
Use M20 concrete and Fe 415 steel. Sketch the reinforcement details by showing cross sections (i) at column strip (ii) at middle strip. 3. Use M20 concrete and Fe 415 steel. Calculate the design moments at various strips in the interior and exterior panels. For the flat slab system of size 6 m ´ 6 m provide suitable drop and fix up overall dimensions. Floor to floor distance is 3. The floor system is supported by columns of size 500 mm ´ 500 mm. Design the typical interior panel of a flat slab floor of size 5 m ´ 5 m with suitable drop to support a live load of 4 kN/m2.6 m. Design the exterior panel of a flat slab of size 6 m ´ 6 m with suitable drop to support a live load of 5 kN/m2.6 m. The floor is supported by columns of size 450 mm ´ 450 mm.Flat Slabs 23 REVIEW QUESTIONS 1. 2. Give the plan of the floor system showing these design moments. the floor height being 3. The floor system is supported by columns of size 500 mm ´ 500 mm. .
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https://www.storyofmathematics.com/a-piece-of-wire-10-m-long-is-cut-into-two-pieces/
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# A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is a maximum?
This question aims to find the total area enclosed by a wire when it is cut down into two piecesThis question uses the concept of the area of a rectangle and an equilateral triangle. The area of a triangle is mathematically equal to:
$Area \space of \space triangle \space = \space \frac{Base \space \times \space Height}{2}$
Whereas the area of a rectangle is mathematically equal to:
$Area \space of \space rectangle \space = \space Width \space \times \space Length$
Let $x$ be the amount to be clipped from the square.
The sum remaining for such an equilateral triangle would be $10 – x$.
We know that the square length is:
$= \space \frac{x}{4}$
Now the square area is:
$= \space (\frac{x}{4})^2$
$= \space \frac{x^2}{16}$
The area of an equilateral triangle is:
$= \space \frac{\sqrt 3}{4} a^2$
Where $a$ is the triangle length.
Thus:
$= \space \frac{10 – x}{3}$
$= \space \frac{\sqrt 3}{4} (\frac{10 – x}{3})^2$
$= \space \frac{\sqrt 3(10-x)^2}{36}$
Now the total area is:
$A(x) \space = \space \frac{x^2}{16} \space + \space \frac{\sqrt 3(10-x)^2}{36}$
Now differentiating $A'(x) = 0$
$= \space \frac{x}{8} \space – \space {\sqrt 3(10 – x)}{18} \space = \space 0$
$\frac{x}{8} \space =\space {\sqrt 3(10 – x)}{18}$
By cross multiplication, we get:
$18x \space = \space 8 \sqrt(3) (10 – x)$
$18x \space = \space 80 \sqrt(3) \space – \space 8 \sqrt(3x)$
$(18 \space + \space 8 \sqrt(3) x) = \space 80 \sqrt(3)$
By simplifying, we get:
$x \space = \space 4.35$
The value of $x = 4.35$ is where we can obtain the maximum area enclosed by this wire.
## Example
A 20 m long piece of wire is divided into two parts. Both pieces are bent, with one becoming a square and the other an equilateral triangle. And how would the wire be spliced to ensure that the covered area is as large as possible?
Let $x$ be the amount to be clipped from the square.
The sum remaining for such an equilateral triangle would be $20 – x$.
We know that the square length is:
$= \space \frac{x}{4}$
Now the square area is:
$= \space (\frac{x}{4})^2$
$= \space \frac{x^2}{16}$
The area of an equilateral triangle is:
$= \space \frac{\sqrt 3}{4} a^2$
Where $a$ is the triangle length.
Thus:
$= \space \frac{10 – x}{3}$
$= \space \frac{\sqrt 3}{4} (\frac{20 – x}{3})^2$
$= \space \frac{\sqrt 3(20-x)^2}{36}$
Now the total area is:
$A(x) \space = \space \frac{x^2}{16} \space + \space \frac{\sqrt 3(20-x)^2}{36}$
Now differentiating $A'(x) = 0$
$= \space \frac{x}{8} \space – \space {\sqrt 3(20 – x)}{18} \space = \space 0$
$\frac{x}{8} \space =\space {\sqrt 3(20 – x)}{18}$
By cross multiplication, we get:
$18x \space = \space 8 \sqrt(3) (20 – x)$
$18x \space = \space 160 \sqrt(3) \space – \space 8 \sqrt(3x)$
$(18 \space + \space 8 \sqrt(3) x) = \space 160 \sqrt(3)$
By simplifying, we get:
$x \space = \space 8.699$
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https://byjus.com/question-answer/an-object-is-projected-with-a-velocity-of-20-text-m-s-making-an-angle/
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Question
# An object is projected with a velocity of 20 m/s making an angle of 45∘ with horizontal. The equation for the trajectory is h=Ax−Bx2 where h is height, x is horizontal distance, A and B are constants. The ratio A : B has value of (g=10 ms−2)
A
1:5
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B
5:1
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C
1:40
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D
40:1
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Solution
## The correct option is D 40:1Standard equation of projectile motion y=xtanθ−gx22u2cos2θ Comparing with given equation A=tanθ and B=g2u2cos2θ So AB=tanθ×2u2cos2θg=u2sin2θg=40010=40 (As θ=45∘,u=20 m/s,g=10 m/s2)
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# Posts by Baby_Banana
Total # Posts: 61
1. ### Algebra
A basket contains the following pieces of fruit: 3 apples, 2 oranges, 2 bananas, 2 pears, and 5 peaches. Jack picks a fruit at random and does not replace it. Then Bethany picks a fruit at random. What is the probability that Jack gets a peach and Bethany gets an orange? I'...
2. ### Algebra
11. A basket contains the following pieces of fruit: three apples to oranges two pairs two bananas and five peaches. Jack pics of fruit at random and does not replace it then, Bethany pics of fruit at random. what is the probability that Jack gets a peach and Bethany gets an ...
3. ### Science
What process occurs after a forest fire wipes out an entire ecosystem? What is this process called? What are the steps of this process after a natural disaster, such as a forest fire, occurs? I don't quite understand what it means, my father and I think it's talking ...
4. ### Algebra
Which of the following expressions is true? A)4^3 x 4^4 = 4^12 B)5^2 x 5^3 > 5^5 C)3^2 x 3^5 = 3^10 D)5^2 x 5^4 < 5^8 Which of the following expressions is true? a)2^4 x 2^3 = 2^12 b)3^3 x 3^6 > 3^8 c)4^2 x 4^2 > 4^4 d)5^2 x 5^2 = 5^10 Which one is correct? 6^2 x 6...
5. ### Algebra
Which of the following expressions is true? A)4^3 x 4^4 = 4^12 B)5^2 x 5^3 > 5^5 C)3^2 x 3^5 = 3^10 D)5^2 x 5^4 < 5^8 Which of the following expressions is true? a)2^4 x 2^3 = 2^12 b)3^3 x 3^6 > 3^8 c)4^2 x 4^2 > 4^4 d)5^2 x 5^2 = 5^10 I don't get how to do ...
6. ### Algebra
I'm kind of stuck on these two, so if you could help I'll be very grateful (m^2-m+3) + (m-1) A. m^2-m-2 B. m^2+2 *My answer* C. m^2-2 D. m^2+m+2 (4x^2 - 2x - 1) - (-3x^3 + 2) A. -3x^3 + 4x^2 - 2x + 1 *My answer* B. -3x^3 + 4x^2 - 2x - 3 C. 3x^3 + 4x^2 - 2x + 1 D. 3x^3...
7. ### Algebra
Ohhhh, so it would be 13.7! Thank you very much
8. ### Algebra
So it would be 12.2?
9. ### Algebra
3.6 divided by 2 1.8 3.14x1.8x1.8 10.2 10.2 divided by 2 5.1 1/2 4.8x5.1 2 divided by 24.5 12.3 This is my work
10. ### Algebra
Find the area of the irregular figure (use 3.14 for pi). SHow your work. Round to the nearest tenth. Triangle with a semi-circle on it. The triangles height is 4.8ft and the circle's diameter is 3.6ft. Every time I do my work I keep getting 12.3 and I have no clue if I'...
Thank you!
12. ### History
Why did John Wilkes Booth assassinate President Lincoln? A. Booth thought Lincoln was too easy on Southerners. B. Booth had lost is a series of slavery debates with Lincoln. C. Booth was angry about the Confederate defeat. D. Booth was a Confederate official. I chose A, but if...
13. ### History
"With malice toward none, with charity for all...let us strive...to bind up the nation's wounds...to do all which may achieve a just and a lasting peace among ourselves and with all nations." - Abraham Lincoln, Second Inaugural Address Based on the lesson and ...
14. ### Science
Thank you so much! But just to check, Tropical matches more with my statement, so would that be the answer?
15. ### Science
Name ONE of the air masses that can be found in North America (remember to use both the temperature and humidity descriptors). I searched everywhere, even looked back at my lesson, and I still can't find out, if you can help me or send me a link, I will be very grateful
16. ### History
So it would be A? I read both of them and they talk about the Louisiana Purchase
17. ### History
1. What is a reason that the Missouri Compromise did not have lasting effects? A. It only applied to the lands of the Luisiana Purchase B. It only applied to the lands of the Mexican Cession C. It relied on popular sovereignty D. It limited the number of states that could be ...
18. ### Algebra
Ooooo, okay. When I did the math I got 3,362.57, is that correct or did I do something wrong?
19. ### Algebra
Hi, can someone please walk me through this? I forgot how to do this and I lost my notes about it from last year Audrey deposited \$2,800 in a savings account that pays 2.65% interest compounded annually. What is the total value of the account after 7 years?
Thank you!
21. ### Language Arts
President Lincoln’s Declaration of Emancipation, January 1, 1863" by Frances E. W. Harper 1 It shall flash through coming ages, 2 It shall light the distant years; 3 And eyes now dim with sorrow 4 Shall be brighter through their tears. 5 It shall flush the mountain ...
22. ### History
I do agree with you, but I'm not a professional nor do I know much about history for I'm still learning it myself
23. ### Science
Oooh, I see, so I just do math? When I did A+B-LiquidProduct, I got 13.8. Thank you very much for helping me!
24. ### Science
A student mixed two clear liquids together in a beaker. A gas and a new liquid formed. The gas escaped, so the student was unable to measure its mass. She guessed that its mass was no more than 10.0 [LW1] grams. Her data is shown in the table below. Mass (g) liquid reactant A ...
Thank you!
26. ### History
Where did the Mexican-American War take place? I want to say near the Rio Grande
27. ### Science
A researcher wants to experiment with an element that reacts like phosphorus (P) but has a greater atomic mass. Which element should the researcher select for the experiment? A. Nitrogen (N) B. Sulfur (S) C. Arsenic (As) D. Silicon (Si) I choose Arsenic, because it's in ...
28. ### Algebra
Oh, forgot it was called expression, thanks for telling me that. And thank you for helping me out <3
29. ### Algebra
Ok, this equation is confusing me and I don't know how to do it 5/8 - (1/4) ^2 I know that 5/8-1/4 is 3/8, but what else do I do?
30. ### Social Studies
I changed it and said it was C
31. ### Social Studies
Ohhhh, so it would be C, since they're on the pacific coast and they were farmers and Oregon offers nice fertile soil
32. ### Social Studies
Why were settlers attracted to the Oregon Country? Select all that apply. A. The Oregon Country offered an abundance of open fields for farming. B. The Oregon Country was dryer than the East Coast and Pacific coast. C. The Oregon Country offered fertile soil along the Pacific ...
33. ### Algebra
Find the area of the irregular shape. Round to the nearest tenth. There's a semi-circle on top of a rectangle with 8 as the base and 5 as the height. A. 90.2 sq cm B. 241 sq cm C. 65.1 sq cm D. 72.3 sq cm Find the area of blue portion of the irregular shape. Round your ...
34. ### social studies
I can, seeming as I have nothing else to do and helping is fun
No problem
36. ### social studies
I agree with you!
37. ### Alegbra
Thank you very very much!
38. ### Alegbra
Ok, I'm good on that one, but what about 6? Can you like give me hints on how to solve it? I don't really want to know the answer until I give up
39. ### Alegbra
Can you help me with 3 and 6? When I do number 3 I do x/5+6=10 -6 -6 x/5=4 x5 x5 x=20
40. ### Algebra
1. –9x – 5 = –95 a) 17 b) 11 c) 10 d) –10 2. x/4 – 5 = –8 a) –27 b) –12 c) –7 d) 12 3. x/5 + 6 = 10 a) 44 b) 30 c) 20 d) –20 4. –2(m – 30) = –6m a) –15 b) –13 c) –8 d) 8 5. 3.75x + 3.7 = 1.7 + 1....
41. ### History
I can't really find anything there to help me. The writing that I was given says: Compare and contrast the views of the Federalist Party and the Democratic Republican Party on at least three different historical issues. I can't find anything to really help me with it
42. ### History
Can someone help me or send me a good link to find more information about the Federalist Party and the Democratic Republican Party, preferably ones that state their differences and similarities? I can't really find any help on other websites
43. ### Algebra
But how did you get 40 and 5x? 4x+4=9x-36 9x and 4x are positives so wouldn't it be addition ? Same thing with -36 and +4, they are negative and positive so I was thinking I had to subtract them
44. ### Algebra
4x + 4 = 9x – 36 I need help on how to solve this
45. ### Algebra
With the origin as the center of dilation, rectangle PQRS will be dilated by a scale factor of 6x24 to form rectangle P’Q’R’S’ . 220x219 What will be the length of 23x10 ? A 3 units B 2 units C 27 units D 18 units I chose D, I read and watched a video on ...
46. ### Algebra
Thank you, Ms.Sue!
47. ### Algebra
1. 7r – 7 = 2r + 18 (1 point) r = –5 r = 5 r = 2.2 r = 1.2 2. 2x + 12 = 18 – x (1 point) x = 3 x = 10 x = 6 x = 2 3. 8x – 3 = 15x + 18 (1 point) x = –3 x = 3 x = 2.1 x = .9 4. 6y – 6 = 4y + 16 (1 point) y = 2.2 y = –2.2 y = 11 y = 5 5. 3(x &#...
48. ### Algebra
Thank you, Ms.Sue and Bob! I never really understood how to solve these types of equations
49. ### Algebra
I need help on how to solve equations like this (y-5)/3 =1 I don't quite understand it
50. ### Social Studies
Thank you so much, Reed and Writeacher!
51. ### Social Studies
It is a local issue, so I can A as well. That means it's C! Because C says " Alisha is participating in democracy by speaking out on local issues "
52. ### Social Studies
No, you're not missing something. But I know it wouldn't D or B, because in B it's says she's trying to change a law but she isn't
53. ### Social Studies
A local park is always littered with trash. Every time Alisha takes her dog to the park, she can't help but notice that trash is everywhere. She knows that this is a hazard to the environment. When she looks around, Alisha realizes that the park only contains a few ...
54. ### Algebra
Ohhhh, I get it now, a little, but nothing like studying that can help me! Thank you very much !
55. ### Algebra
I know that when it says 'is' it's suppose to be an equal sign, but the rest is still kind of confusing. How did you know that when it says "the difference of a number and 10" is (x-10)?
56. ### Algebra
Which equation matches the statement "The product of 3 and the difference of a number and 10 is 15?" A. 3 + (x-10)=15 B. 3x-10=15 (My answer) C. 3 (x-10) = 15 D. 3 x/10 = 15 Can you please check to see if I'm right? If I'm not explain, please and thank you
57. ### Algebra
So it would be D? Because on A when I did .56x134 I got 75.04 and 75 is one of the answers and on B I got 15.06 and some other numbers, but rounded it would be 15.1. So the answer would be D
58. ### Algebra
9.During the basketball season, Diane took 134 shots and made about 56% of them. a. How many shots did Diane make? b. The team made a total of 498 shots. What percent of the team’s made shots did Diane make? (1 point) 24 shots; 15.1% 753 shots; 27% 240 shots; 27% 75 shots...
59. ### Spanish
No problem :D
60. ### Spanish
It's A. An Asado is a traditional celebration where Chilean families come together to celebrate whatever they want.
61. ### Language Arts
I actually took the test just a minute ago The answers are 1.A 2.D 3.C 4.D 5.B 6.C 7.D 8.A 9.A 10.D 11.C 12.A 13.A 14.B 15.B 16.B There you go, I promise with all my heart this is 100% correct
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# What is the Sandbox?
This "Sandbox" is a place where Code Golf users can get feedback on prospective challenges they wish to post to the main page. This is useful because writing a clear and fully specified challenge on the first try can be difficult. There is a much better chance of your challenge being well received if you post it in the Sandbox first.
See the Sandbox FAQ for more information on how to use the Sandbox.
## Get the Sandbox Viewer to view the sandbox more easily
To add an inline tag to a proposal use shortcut link syntax with a prefix: [tag:king-of-the-hill]
# Kaos Pendulum and Einstein's Revenge
The Double Pendulum system is a relatively simple system with quite complex behavior that's highly sensitive to initial conditions (i.e., "chaotic").
Doctor Kaos has designed a particularly fiendish double pendulum, using his diabolical machine "Einstein's Revenge". Einstein's Revenge is a device which couples matter and energy, causing the mass of the first pendulum (connected to the anchor point) to be a function of the energy of the second: m1 = sqrt(PE2**2 + KE2**2), with the potential energy measured from the rest position of the pendulum.
For this challenge, you must write a program or function to simulate a basic Kaos Pendulum, where the second pendulum has unit mass and both pendulums are unit length. It must accept the starting position of the Pendulum - two real numbers, theta1 and theta2 - as input or command-line arguments, and one positive integer equal to the number of seconds the pendulum will be active.
Your program must output the value of theta1 and theta2 for each second, either printing to stdout or returning as a list, or something similar.
## Notes
I don't know if this should be a golf, a popularity contest, or if there should be some scoring method for some other type of challenge.
# Blind (deterministic) Jenga
The idea for this came up in the chat room, under the broader topic of how to make a human dexterity game into a software KotH challenge. Here's the basic idea, which I don't think works, but I'd love to get feedback to make it better.
The jenga tower has three blocks per layer. Layers alternate between N/S aligned blocks and E/W aligned blocks.
There may be just two players, or more than two.
Each turn a player selects a block location to probe. If there's no block there, the player gets to go again. If there's a block there, it is removed from the stack. If this causes the tower to fall, the player loses. Otherwise, the block is placed in the next safe position at the top of the tower, unless all available positions are unsafe in which case the player loses. If the player has not lost, it is now the next player's turn. Continue until some player loses.
I think this won't work well because it will boil down to a relatively random competition between bots that start with edge pieces and bots that start with center pieces. I don't want to run the contest just to find out that that's the case. Maybe putting four blocks per layer would make that a non-issue. I'm open to other suggestions on how to make this challenge work.
A idea. Maybe I'll do it a long time later...
You have these options each turn (they need better names and a plausible background):
• Meet k If the bot k also did meet you, and nobody injects, both of your scores are increased by 1. Otherwise nothing happens. But if k is yourself, this is invalid and will be just like doing nothing.
• Invite k The bot k gets a notification that you are doing this, and in the next round, your output is ignored and you must do Meet k.
• Inspect k You will get these informations at the beginning of next round:
• What the bot k was doing.
• The score change of bot k.
• Who was inviting bot k.
• Who was trying to meet bot k. (This is not known by bot k if not inspecting himself and his score is not changed.)
• Who was inspecting bot k. (This is not known by bot k if not inspecting himself.)
• Who was injecting bot k (and another bot). (This is not known by bot k if not inspecting himself and his score is not changed.)
• Inject j k If j is meeting k and k is meeting j, your score is increased by 10 and their scores are both decreased by 1 instead. If one of them is pretending doing that, and the other is pretending or really doing that, your score doesn't change. Otherwise, your score is decreased by 10. They'll not know who is injecting unless they are inspecting you or self-inspecting instead.
• DoNothing Nothing happens.
Possible other options:
• Pretend action: action is any action other than Pretend. It's the same as doing nothing, but inspectors will be told that you are doing that action. The score change that the inspectors will see is calculated as if you really did that.
• Multiply: For each turn until the next time you meet some bot each other, the points you, the other bot and all the bots injecting you get are doubled (or incremented, I'm not sure).
Everybody can see only their own score, bots inviting them and the result from Inspect.
• I think that Pretend is a great action, and should definitely be included. – Nathan Merrill May 22 '15 at 19:30
# Is this a red-black tree?
A red-black tree is a binary search tree where each node has an additional 'colour' property, which can be black or red. The root and leaf nodes must be black, each red node must have black children, and the path from any node to any child leaves must have the same number of black nodes. For instance:
2B
| \
1R 3B
| |
~ ~
Is invalid because of a black violation (the path from the root to each leaf has a different number of blacks).
Your task is to determine whether a given tree is a valid Red-Black tree.
## Input
The input will be a string matching [ a-zA-Z]. A space represents a leaf, a lowercase character a black node, and an uppercase character a red node. The value of a node is the position in the alphabet: a is 0, b is 1, C is 2, and so on. The children of the node in position n are in positions 2n+1 and 2n+2. If the string ends before the position you're looking for, that node is a leaf. For instance, baD cf E corresponds to:
1,B
/ \
0,B 3,R
/ \
2,B 5,B
/
4,R
## Output
Your program should output something other than whitespace on STDOUT if and only if the input represents a valid tree.
## Simple test cases
True False
" " "" (Root black violation)
"a" "A" (Root black violation)
"dBeac" "dBea" (Black violation)
"dAE B" (Red violation)
Question about extrapolating data from an incomplete tennis scoreboard
Introduction
It is a fine day at the Stack Exchange tennis club. The players have just finished playing a grand tennis tournament in a round robin style (every player plays every other player once). The final results were about to be announced when suddenly, the scoreboard explodes! It is a total disaster - now nobody knows what the final scores were! Luckily, a piece of the scoreboard is still intact. Can you write a program to figure out the rest of the scoreboard from only a small part of it?
If there are k players, our scoreboard would have had k*3 entries, indicating each player's wins, losses, and draws. It is presented in a list of space separated comma separated tuples. For instance, in a game with 4 players here is a possible final scoreboard:
0,2,1 3,0,0 1,1,1 1,2,0
In this example the first player did not win anything, lost twice, and drew once; the second player won all three of their matches; the third player won once, lost once, and drew once; and the fourth player won once and lost twice.
Input Description
As input, you are given a scoreboard similar to the one above, except some of the numbers have been replaced by ?. The ? indicate the numbers that were unrecoverable after the explosion. For instance:
?,?,1 0,?,0 ?,?,?
Output Description
Output the scoreboard, with the ?s replaced with the actual scores. The input scoreboard will always have enough information for you to deduce the final scoreboard.
For example, consider the above input. We know that the second player has 0 wins and 0 draws, therefore, they must have lost both of their games. The first player has a single draw, and since we know the first player beat the second player, that draw must be with the third player. Thus the output is:
1,0,1 0,2,0 1,0,1
Sample inputs and outputs
TODO: Sample inputs and outputs of higher k
Sandbox Questions
• I have an alternate formulation for this question where instead of an incomplete scoreboard, the complete scoreboard is given and the program has to generate a table of which player beat which player. Which challenge do you think would be more interesting?
Sudoku Swapping Shenanigans
Honestly, there's not many shenanigans in this challenge but I wanted that sweet tautogram title.
Introduction
Imagine you're on a train, and there is a Sudoku grid that's already been entirely filled in left on the seat. We'll represent this grid as a series of 81 comma separated integers from 1-9 on a single line. Each cell in the grid can be numbered as follows:
As with all Sudoku grids, there will be exactly 9 of each number present in the input.
The Sudoku grid will not be completed correctly. It will have the right number of each number, but they will be positioned wrong. Your program's task is to swap these numbers to solve the Sudoku puzzle correctly.
(Since your stop is next, you want to make this program solve the Sudoku puzzle as fast as possible.)
or:
(Since you're writing this program on a napkin, you want to make it use as little bytes as possible.)
Input Description
Input consists of 81 positive integers that are comma separated. Some of them will be prefixed with an X which means those are the ones that are already placed and you can't swap them.
Example input goes here.
Output Description
Output the swaps required to solve the grid in the fewest number of swaps. You should have a line for each swap, consisting of two numbers in the form:
a,b
where a and b are both different and in the range 0-80.
Here's an example output (and I haven't actually written the example input yet!)
80,0
8,9
14,15
So that would mean that to solve the grid we need to swap the number in position 80 with the number in position 0, the number in position 8 with the number in position 9, etc.
Sandbox Questions
• Are we trying to solve it in the fewest number of swaps, or just transform it into any solved sudoku grid (meaning I could swap every input into a single predetermined pattern)? – Geobits May 27 '15 at 16:11
• @Geobits Fewest number of steps. – absinthe May 27 '15 at 22:03
• Do you guarantee that the original clues give a unique solution? If so, there's a lot of overlap with existing Sudoku-solver questions, and fastest-code would need a very large and well-chosen set of test cases to be confident that it's not overly sensitive to the order in which heuristics are applied. The interesting (IMO) part of the question reduces to "Decompose a non-simple graph into cycles", and I think there may be better settings to present it. – Peter Taylor May 30 '15 at 13:40
Code golf challenge: Write a program or a function that solves the following problem. Normal code-golf rules apply.
Given a set of subnets, give the smallest possible network these can belong to. You should consider the network- and broadcast addresses in your calculations.
Example input:
128.208.0.0/18
128.208.128.0/17
128.208.96.0/19
Example output:
128.208.0.0/16
You might want to read up on IP prefixes before trying this.
# Flash Cards Viewer
In this challenge, your goal is to produce a flash card viewer. Since this is a Code Golf challenge, it won't be a particularly featureful flash card viewer - in fact, it'll be the bare minimum a flash card viewer can be. We'll implement the following features:
1. Reading a list of flash cards from a file;
2. Quizzing the user on either "side" of the flash cards;
3. Reporting to the user which ones they got incorrect.
Input Description
As feature number 1 suggests, for this challenge you must read input from a file (unfortunately, your language cannot participate if it does not support reading from a file). The file shall be called cards.txt and will be formatted as a simple list of values separated by a pipe character |. For instance, a "Family Names in Chinese" flash card deck might look like this:
Mother|妈妈
Father|爸爸
Elder Sister|姐姐
Younger Sister|妹妹
Older Brother|哥哥
Younger Brother|弟弟
Output Description
First, your program should ask the user which side of the cards they want to revise, using a user input function such as prompt() or raw_input(). An input of 0 indicates the user wants to revise the left side of the cards, and an input of 1 indicates the user wants to revise the right side of the cards. Use the message "Choose side:" when asking the user.
After that, the program will pick out random entries from the opposite side of the user's selection, and prompt the user to type in the entry's corresponding value. After the user has gone through the entire list, output the pairs that he or she got incorrect.
Here is an example of what a possible session might look like. What the user inputted in this example is signified with >.
Choose side:
>0
>Younger Sister
>Mother
>Father
>Younger Brother
>Older Brother
Older Brother|哥哥
Younger Brother|弟弟
# Fibonacci Box Packing Factory
### Introduction
The Electronic Goods Company is a company that produces electronic item of various sizes. They need to package their items in boxes so that they can be shipped off to the store. There are some restrictions regarding what item can be stored in what box:
1. Each box can only contain a single item, and each item can only be contained in one box;
2. The volume of the item cannot exceed the volume of the box;
3. If the volume of the item is exactly the volume of the box, the item can be placed in the box without any padding material required;
4. If the volume of the item is less than the volume of the box, then padding material equal to the difference of volumes must be placed in the box to prevent the item breaking.
For example, an item that had a volume of 2m3 could be placed in a box of volume 2m3 without any extra padding material, but if the same item was placed in a box of volume 5m3, we'd need to add padding material of 3m3.
The supplier company is called the Fibonacci Box Company, which supplies The Electronics Goods Company with N boxes and P metres squared of filling material. Each box's volume is determined by the Fibonacci sequence (so the first box's volume is 1, the second is also 1, the third is 2, the fourth is 3, the fifth is 5, etc.).
Your program goal is to figure out the minimum amount of padding material required to pack all of the electronic items.
### Input Description
Input is a space separated list of integers that indicate the volume in square metres of the electronic items.
//TODO: Example inputs and outputs
### Output Description
Output the minimum value of P required to fill all the boxes.
//TODO: Example inputs and outputs
• I'm assuming that items can't be split between multiple boxes, and that a box can't contain multiple items? – Nathan Merrill Jun 1 '15 at 4:03
• @NathanMerrill Yep. Rule 1 states only one item can be placed in each box. I'll edit for clarity. – absinthe Jun 1 '15 at 4:05
• A couple example would help understand the challenge. You could also add some explanation to them. E.g. 1 5 8 7 => 6 (N = 7, boxes = 1 1 2 3 5 8 13, used boxes = 1 5 8 13) if I understand it correctly. Volume is cubic metres, m3. – randomra Jun 1 '15 at 4:43
# NP Cops and Robbers
This is an idea I've had for a while, and I really want to get it to work, but there are some large hurdles.
Hurdle 1: I need to pick a puzzle, preferably one that is NP-Complete. I think lots of Nikoli puzzles are good candidates, and I am leaning towards Light Up
# Cops:
Cops will write a program to generate a puzzle. The puzzle has a maximum size of WxH. The cops must be able to generate the puzzle in under 1 second. Puzzles must be deterministic, but randomness is allowed as long as I can set the seed. The puzzles generated must have only 1 solution. The cop will post a sample puzzle in his post (for the robbers to use)
Your score is the shortest time any of the robbers solved your puzzle. Highest score wins.
# Robbers:
Robbers write programs to solve the puzzles. Robbers are not allowed to target the specific puzzles, but are allowed to target the specific algorithms used to generate the puzzle. If the cop were to change the sample puzzle posted, the robber should still be able to solve the new puzzle in a similar amount of time.
If you solve a puzzle faster than any other robber, your score increases by the time taken to solve that puzzle. Highest score wins.
Hurdle 2: How to measure time taken. I don't want to run all of the cops/robbers, especially if this becomes popular. I also don't want to use the GOLF framework, as that would take a really long time to run. So, possible I could have the person run som CPU-intesive algorithm on their computer, measure how long it takes, and scale all of their answers according to that time.
# Create a basic spell checker
In this challenge, you have to implement a basic spell checker. Your program will use the first input as a source.
## Specifications
• You will receive two inputs, one is a sentence/paragraph you use as a source for your spell checker, and the other includes wrong words (they can be correct too) which you have to correct using the first input.
• Possible mistakes: Missing one letter (e.g. helo), one extra letter (e.g. heello), substitution of one character (e.g. hilp instead of help), transposition of two adjacent characters (e.g. hlep instead of help).
• In other words, the mistake and the original word have to have a Damerau–Levenshtein distance of one.
• You can separate the inputs in any reasonable format. For example, instead of using a newline, you can use a pipe sign to separate them. (|) However, watch out so you don't use something like letters because they are used in the 1st input.
• It's possible for a word to have neither an identical word nor a word with 1 distance from the first input. You will have to leave them unchanged.
• You don't need to do anything with punctuation. That is, I won't test words that are followed by a punctuation. (E.g. From the first example, I won't ask the correction of 'blod')
• Challenge is case insensitive, both the input and output. Meaning that if 'The' is used in the 1st input and the 2nd input is 'teh' you can change it to 'The', 'teh', 'teH', and similar cases.
• If a word is in the first input, but also has a distance of 1 with one of the other words, do not correct it.
• If there are two words with the same distance, you can output either of them. (E.g. 'en' can change into 'on' and 'in' if the words are both in the first input.) You can also output both of them, but you'll have to use a slash between them. (E.g. 'en' can be corrected into 'in/on')
• You can assume words are just sequences of letters separated by spaces.
• You can use STDIN or the closest alternative in your language.
• Standard loophole rules apply.
• This is , so the shortest code wins.
## Examples
Input:
A criminal strain ran in his blood, which, instead of being modified, was increased and rendered infinitely more dangerous by his extraordinary mental powers.
increasd mnetal
Output:
increased mental
Input:
The fact is that upon his entrance I had instantly recognized the extreme personal danger in which I lay.
I recognizi dnger en
Output:
I recognize danger in
• There are several definitions of "distance". Since the most common one is probably Levenshtein, you might want to specifically say that original and mistake have to have Damerau–Levenshtein distance of one. Other questions: what is a word? Only sequences of letters? What about "doesn't" or "up-to-date"? (It's probably simplest to rule those out.) What do we output if there are two different words with distance one? E.g. does en become in or on if both are in the first string? – Martin Ender Jun 3 '15 at 20:33
• Is the input case insensitive? If so, is case of the output arbitrary too or do we have to preserve the input capitalisation? And are we guaranteed that every word in the second string has a word in the first string that is no more than distance 1 away? – Martin Ender Jun 3 '15 at 20:34
• @MartinBüttner I clarified it in the bullet point above it. Only transposition of adjacent letters, one missing letter, one extra letter and a wrong letter. I guess that's Damerau-Levenshtein distance. Regarding the two words problem, I guess you could output either of them. I'm not really sure about this. I'm open for ideas. – JNV Jun 4 '15 at 4:49
• @MartinBüttner Good point. It's case insensitive, and the output can be in any case. (e.g. If the first input includes 'The' and there's 'teh' in the second input, it should output 'the' or 'THE' or 'tHE' or ...). No, you're not guaranteed that there's a word with a distance of 1 in the first input. Take a look at the example #2, 'I' was not changed. Also, in your previous post, you asked what counts as a word. I already said it in the post 'You can assume all words in the 2nd input are separated by spaces.' I'll edit it for clarity. – JNV Jun 4 '15 at 4:56
• I was asking for distance "no more than 1 away". That is, could there be a word that has neither a word with distance 1 nor an identical word in the first input? "You can assume words are just sequences of letters separated by spaces." Yet in the first input, there is punctuation. What sort of punctuation should we expect? The rest looks good. – Martin Ender Jun 4 '15 at 12:43
• @MartinBüttner Yes, it's possible for a word to have neither an identical word nor a word with 1 distance from the 1st input. You can ignore the words that are followed by a punctuation, like blood in the 1st input. They won't be in the test cases. (E.g. I won't ask for the correction of 'blod') Thank you for your help. – JNV Jun 4 '15 at 13:10
• As for the 'lay' in the first example, I'll fix it now, thanks. – JNV Jun 4 '15 at 13:19
• Right, but the question is what do we do with words that have neither an identical word nor a word with 1 distance from the 1st input. Remove them? Leave them changed? Also, that second point might be worth mentioning in the spec. – Martin Ender Jun 4 '15 at 13:19
• @MartinBüttner Leave them unchanged. Thank you for your support. – JNV Jun 4 '15 at 13:26
# Multiply with restricted operations: Best lower bound
Multiply with restricted operations is a challenge to multiply two variables using as few as possible of four allowed operations: addition, reciprocal, negation, and variable assignment. The current best is 22 operations.
Your goal is to prove a lower bound L, a number for which you demonstrate that there are no solutions with fewer that L operations. The highest L wins.
You must explain why your code demonstrates the lower bound, and I must be able verify your code by running it. There is a time limit of 30 minutes. The natural approach is to search the space of possible solutions, perhaps shrinking it via mathematical arguments, but any method is allowed. Purely mathematical proofs without code are also valid.
Your code is limited to 30 minutes and 3GB of memory on my machine (Windows 7 with a Intel Core i5-460M processor). I may be satisfied with someone else running your code to confirm with a conservative adjustment for different machine speeds.
Your code needs to run on easily-available free software. Please include instructions on how to run your code.
The winner will be the highest lower-bound proven by [date]. Tiebreak is fastest runtime on my computer. I will give a bounty of 500 rep to the winner.
If your lower bound matches a solution, narrowing down the optimum to a single value, I will award you an additional bounty of 300 rep.
For Sandbox:
• I haven't done a fastest code before. Anything I'm missing? Any machine specs I should add?
• Maybe you should require the program to be able to prove lower bounds for arbitrary expressions, rather than only a*b. Otherwise, it will be impossible to tell if a brute-forcing program is doing anything correctly, for any bound too low to find an answer. – feersum Jun 8 '15 at 17:35
• @feersum I'd really like to let people optimize for just multiplication since that's the function I care about. Is it really impractical to verify that code is doing the search correctly if the user is required to give a detailed explanation of their method and why it works? I trust people to be honest as to what the code does. – xnor Jun 12 '15 at 17:49
• I do believe it is impractical. How can you really tell that someone is not randomly twiddling bits and then printing 19? It seems that (innocently) bugged answers are more likely than correct ones in an environment where there is no testing. – feersum Jun 12 '15 at 18:01
# Date-A-List!
### Introduction
The year is 20XX. Lists of integers are now members of society with a full set of rights. Like humans, lists like to have dates and get married. In this challenge, we'll produce a program "Date-A-List!" that will get as many lists as possible on happy and compatible dates. Lists tend to prefer using programs that use as little bytes as possible, so this program will be golfed.
Lists are compatible (that is, you can assign them to a date together) if they share 2 or more integers in the same order. For instance, the lists (1 2 3) and (2 3 5) would be compatible. The lists (1 2 3) and (3 2 5) would not be compatible because they do not share 2 or more integers that are in the same order.
Lists are also always monogamous -- they will never go on a date with more than one list. For instance, consider the lists (1 2 3), (2 3 5) and (5 1 2) -- although the first list is compatible with both of the others, it may only date one of them.
### Input Description
Input is a list of lists in brackets, space separated:
(1 2 3) (2 3 5) (5 1 2)
That input would indicate three lists to be assigned to dates: [1, 2, 3], [2, 3, 5], and [5, 1, 2].
### Output Description
If n is the maximum number of dates possible, output n lines, each line consisting of two space separated lists which are to be matched up. For instance, on the example input there are two possible outputs:
(1 2 3) (2 3 5)
or:
(1 2 3) (5 1 2)
[TODO: More complex inputs and outputs]
• It would be nice to include the word "matching" somewhere in the text, for searchability. It would also be nice to have the basic examples use slightly more variety: at present they're all 3-element lists which overlap on adjacent pairs, but neither of those are stated as constraints. How about using (1 2 3) and (2 5 4 3) as the example of compatibility? – Peter Taylor Jun 6 '15 at 13:43
# 5-a-side Toroidal Bot Soccer
## Randomised teams
This is a team game. Rather than being assigned permanently to one team or the other, each bot will play in a number of games, each for a different randomly composed team of 5 players ("5-a-side"), and score a point for each game in which its team wins. This means every game requires team work, but there is still a single overall winner after all games are played.
## Rules
There are no rules for the players (no referee, no penalties, no off-side rule). The movement of the ball defines the score and the players can do as they see fit.
### Scoring
There are no goal posts. If the ball moves off the right hand edge of the field, it reappears at the left hand edge of the field and team 1 scores a goal. Similarly team 2 score if it moves off the left hand edge and reappears at the right. The score will be represented by a single integer, that is increased by one when team 1 scores, and decreased by one when team 2 scores. At the end of a game, team 1 wins if the score is positive, team 2 wins if the score is negative, and a zero score is a draw/tie.
Movement over the top and bottom edges has no effect on the score.
The game lasts for 2000 time steps (each bot provides 2000 moves). If there is no winner by the end of the game it is extended by up to a further 1000 moves, with the game ending if either team scores.
## Physics
Physics is the main obstacle to real world toroidal soccer, and will not be respected in this game. The physics have been simplified as much as possible to hopefully allow games to be viewed live.
This is a non-contact sport. Bots pass straight through other bots (of either team) with no interaction. Each bot can only interact with the ball.
The playing field is a continuous rectangle of width 100 and height 50. The bots and the ball have radius 1. Bots and ball can move freely in any direction without meeting a boundary - the edges of the field wrap. The ball will rebound from any bot if their circumferences overlap.
### Turning and acceleration
Due to the use of simplified physics this is quick to explain:
A bot has a facing angle and a velocity. A constant acceleration applies in the direction specified by the facing angle. Drag is a deceleration proportional to the velocity (an acceleration in the opposite direction to the velocity, proportional to the size of the velocity). This means a bot that does not turn will accelerate to a maximum velocity where the drag matches the acceleration.
A bot can turn any angle instantaneously, but the velocity (and hence direction) will only change gradually. For example, a bot changing direction by 180 degrees will continue travelling backwards while it slows down to zero velocity, and then accelerate in its new direction.
The ball has no acceleration of its own, so other than collisions, its only change in velocity is due to drag. There is no spin.
### Collisions
Only the ball can be involved in a collision with a bot - bots pass straight through each other.
Although the radius of both bots and ball will be displayed visually as 1, I believe the results are the same if collision calculations are based on the bots having radius 2 and the ball being a point.
To keep calculation simple, the ball will be tested for collision with each bot by considering the bot to be a circle moving with constant velocity and the ball to be a point moving with constant velocity (that is, the acceleration will occur at instants, rather than continuously over time). Since this is also how the motion will be modelled generally, the collisions should be consistent with the motion of objects in the game.
This allows the exact point of collision to be calculated so the rebound can occur with no overlap.
## Communication
There is no communication between bots. Each bot communicates solely with the controller.
The communication method will depend on the controller type (language agnostic/specific). Players will either be functions/objects in a specific language, or separate programs that communicate through STDIN/STDOUT.
Each step all bots will be supplied with the same information, and will provide a facing angle which will be a float in the range [0, 360).
The information supplied to bots will be as follows.
• Team direction (1 for team 1 or -1 for team 2)
• Current score (positive if team 1 is winning, negative if team 2 is winning)
• Facing angle and velocity of itself
• Facing angle and velocity of 4 team mates
• Facing angle and velocity of 5 opponents
Facing angle will be given as a float. Velocity will be given as x and y components, so two floats.
The information will therefore be received as two integers followed by 15 floats.
# Sandbox questions
• Are any terms not familiar that would be worth linking or further explaining?
• Are there any further simplifications that could be made, without detracting from the game?
• Are any of the simplifications too much? Am I overlooking some way in which the game could become trivial?
• Is it correct to model the bots as radius 2 and the ball as a point? Does this give identical results to modelling the bots as radius 1 and the ball as radius 1?
• Are there any problems likely to arise from basing collision detection on constant velocity bots and ball? (Acceleration being applied instantaneously each step, rather than spread out over continuous time.)
• How long does each game last for? I like your idea of randomising the teams. How many bots are in each team? – euanjt Jun 16 '15 at 7:55
• @TheE The "5-a-side" in the title refers to 5 players per team - I wasn't sure how widely known that was so I'll edit that in. – trichoplax Jun 16 '15 at 12:41
• @TheE I'm not sure how long each game should last yet - I'll have to test with some example players once the controller is written. For now I'll say 2000 steps in order to have something that people can comment on. I guess it will depend on what acceleration and drag figures I settle on too - as that affects how far each bot can move per time step. – trichoplax Jun 16 '15 at 12:45
• If a player of the right team stays on the left part of the field, he could counter score every goal it's team take, right? – Katenkyo Jun 16 '15 at 13:26
• Yeah I didn't read the title :) oops – euanjt Jun 16 '15 at 14:00
• @TheE no problem - and thanks for the other feedback :) – trichoplax Jun 16 '15 at 15:17
• @Katenkyo all the players can move freely over the edges of the field, so if the left team can get the ball past the right team's players then it shouldn't matter which side of the field they are on. The left team can reappear at the left side of the field and keep kicking the ball further right. – trichoplax Jun 16 '15 at 16:06
• @Katenkyo also all players can see all other players, so a bot from team 1 kicking the ball past the right hand edge of the field can see any team 2 bots waiting at the left hand edge of the field. It can therefore aim to avoid them so they cannot simply kick it back. – trichoplax Jun 16 '15 at 16:08
# Build a Rorschach Generator
2 possible challenges:
## 1.
Help a poor struggling game dev write some code to quickly and efficiently generate a Rorschach inkblot.
Using your language of choice, generate a Rorschach inkblot. You may use any method of generating a random seed.
The generated image must be in black and white and reflect down the middle of the image. This is a code golf so the shortest submission will win.
OR
## 2.
Help a poor struggling game dev write some code to quickly and efficiently generate a Rorschach inkblot.
Given 2 input images (a source image and an image of a template inkblot), convert the first into the Rorschach inkblot of the second without changing the colours of the image.
You may assume that the images are the same size, however you may not display the original image unaltered. Anywhere in the template where the RGB values of the pixel are (255,255,255), you may not display the original image. The produced image must be vertically mirrored, so you will have 2 copies of the source image distorted and reflected in the result.
Then I'd provide source images some of which would probably be shamelessly stolen from either the Voroni Map or the Mona Lisa Colour Palette questions because those questions are awesome.
My thoughts:
I'm essentially hoping to get the source image in a twisted demented fragmented form, similar to if you gave a small child red cordial then the source image and a blur tool and said go nuts. I feel like my second challenge has potential, but I feel like I need to improve the challenge description and detail exactly what I want to be produced, but I'm not sure how to do that without a wall of text and without unnecessarily restricting the challenge.
• Your suggestions aren't code-golf. Code-golf means that the shortest code in byte/char wins. First of all, you should determine what will be the winning criterion. It will be hard to determine some if you want to keep it as a code-golf AND a pop-con. The first one is more designed to be a pop-con, as submissions can be creative. The second one is too restrictive to be a pop-con : for the same images, ouput will always be the same. It would go as a code-golf. Be aware that image-processing is language retrictive by essence, and that could be badly welcomed as a code-golf :) – Katenkyo Jun 23 '15 at 14:31
• Not necessarily a duplicate, but something to look over to make sure yours differs sufficiently Make a Rorschach image – trichoplax Jun 23 '15 at 16:48
# [Insert] Nerd Sniping Pattern (Series)
Having learned from Prime Nerd Sniping Pattern that hoping an optimal solution will not be spotted too quickly is not a good idea, I'd like to judge how much interest there is in a series of similar contests that have been demonstrated to not have an achievable optimal solution, allowing long term open ended competition.
Each one would need work to demonstrate this lack of an achievable optimal solution, and I'll put that work in if there is suffficient interest here.
For example, there could be a Fibonacci Nerd Sniping Pattern, Factorial Nerd Sniping Pattern, and so on. Each one would need to have a different method of defining the scoring pattern, so that different optimisation techniques and algorithms would be required for each one. This is what would ensure they are distinguished from each other as separate challenges in a series, rather than near duplicates.
# What I've learned from my mistake with the primes
The prime scoring pattern had a checkerboard optimal solution because all of the scoring pixels were on opposite coloured checkerboard squares to the pixel being scored. I need to avoid this in any future scoring patterns, and more generally avoid any pattern that divides the image into two regions for scoring (where all of one region are scored by combinations of pixels from the other). Intuitively, there should be plenty of mixing.
As I think about it more, I'll add ideas here on things to rule out and things to ensure I have before considering a pattern scoring rule ready for posting.
# Build a Mahjong AI
Mahjong is a traditional Chinese gambling game played throughout Asia in which four players draw and discard tiles in order to try to complete a hand of 14 tiles.
In this problem, a specification for a simplified version of Mahjong called "PPCG Mahjong" was given. Your task in this problem is to build a program that will play a version defined here called "Full PPCG Mahjong" (hereafter known as the FPM rules).
Your program will be an online program, taking input as it plays the game. (that does not necessarily mean it needs networking capabilities, but it means that it will output in reaction to ongoing input, not just all input at once.)
## Tiles
There are 34 types of tiles in FPM.
1. The “bǐng”/“pin” (餅) tiles, also known as the circle tiles / dot tiles. These will be represented in text with a number from 1 to 9 and then the letter b, as in 1b, 2b, up to 9b.
2. The “suǒ”/“sou” (索) tiles, also known as the bamboo tiles. These will be represented in text with a number from 1 to 9 and then the letter s, as in 1s, 2s, up to 9s.
3. The “wàn”/“man” (萬) tiles, also known as the character tiles. These will be represented in text with a number from 1 to 9 and then the letter w, as in 1w, 2w, up to 9w.
The three tile types totalling 27 tiles above are called the number tiles.
1. The “fēng”/”fuu” (風) tiles, also known as the wind tiles. There is one wind tile for each cardinal direction. These will be represented in text with the letter of the wind's cardinal direction in capitals, repeated twice, as in: EE, SS, WW, or NN.
2. The “yuán”/“gen” (元) tiles, also known as the dragon tiles. There is one red, one green, and one white dragon, represented as ZZ (for zhōng 中/center), FF (for 發/fortune), and BB (for bái 白/white) respectively.
The two tile types above are called the honour tiles. There are four of each type of tile, for a total of 136 tiles.
## Objective
The goal of Mahjong is to form a complete hand, which consists of four sets and a pair.
A set is one of:
• A sequence of number tiles. (e.g. 5w 6w 7w or 3s 4s 5s)
• A triplet of any tile (number or honour tile). (e.g. 3b 3b 3b or EE EE EE)
• A quadruplet of any tile. (e.g. 3b 3b 3b 3b or EE EE EE EE)
A complete hand will have 14 tiles if there are no quadruplets present; each quadruplet adds one tile to the hand's size, up to a possible 18.
## Gameplay
Mahjong is a game played in several hands (局, ).
At the beginning of each hand, each player receives a text input of START [Wind] [Round number]. Each player is then given an input of 13 tiles, represented as a set of 13 space-separated strings (e.g. 1b 9s BB 3w 6w 4b ZZ 3w 5w 2s 6b NN NN) and assigned a different seat wind depending on where they "sit" at the virtual table (given as a single letter out of NESW).
Starting with the player at position East and moving clockwise around the compass, each player takes a turn in which:
• The player whose turn it is will be given a tile (自摸牌, zìmōpái) from the "wall" (the collection of tiles that have not been drawn or dealt yet) as a two-character input.
• The player can take actions during the standby phase:
• If the player has four of the same tile in his hand, it can call kong (槓, gàng) and draw an extra tile. Those tiles must stay as a quadruplet from then on and cannot be used for anything else. The program does this by outputting kong [quadruplet tile] as text.
• If the player previously called pung on a triplet and has a fourth of the tile, it can call kong as well.
• If the player's hand is complete with this drawn tile, it may declare hu in which case the hand ends.
• The player will then produce an output of whatever tile it decides to discard. The tile that was discarded will be propagated as output to the other players in the form [seat] [discarded tile] (e.g. E 8w). The players can then call that tile if it will complete a set:
• The player immediately after the discarding player may call chi (吃, chī) by outputting chi as text during the call phase, to complete a sequence.
• Any other player can call pung (碰, pèng) to complete a triplet, similarly by outputting pung as text. Pung overrides chi when it happens.
• Any player can also call kong (槓, gàng) to complete a quadruplet, also by outputting text, in which case they are given an extra tile as text input.
• Any player can also call hu (胡, ) to complete their hand regardless of whether they are completing a sequence, triplet, or pair. Hu overrides both pung/kong and chi.
• If a player does not want to call tiles, it must output pass.
The turn immediately goes to the player who called the tile, and they skip their draw phase, discarding another tile immediately. If nobody called a tile, the next player draws one. At any point if a player calls, the other three players will receive an input indicating that the player in that seat made a call and what was called.
• For chi, the input will be in the form of [seat] chi [other two tiles] (e.g. S chi 6w 7w).
• For pung and kong, the input will be in the form of [seat] pung/kong. This is because all the tiles are identical to the called tile anyway.
• For hu, the input will be in the form of [seat] hu [all other tiles in hand].
In the above three cases, the tile being called is always the previously discarded tile.
• For a kong during the standby phase, the input will be in the form of [seat] kong [kong tile], because the kong tile is not implied to be the previously discarded tile.
• For a hu during the standby phase, the input will be in the form of [seat] hu [tile drawn]
Once a tile is called, the set that it makes is locked in place and cannot change for the rest of the hand. So if two tiles would make use of a tile in a called set, they may not use that tile.
A hand ends when either somebody calls hu validly or there are no more tiles to draw. (There is no dead wall in this game.) At that point, all players will receive an input of END to signal that the round has ended, and a summary of their points.
If somebody other than the dealer won, then the deal rotates. If the dealer or nobody won, then the deal stays the same. After four deal rotations, the prevailing wind also rotates.
A game of FPM consists of 16 such deal-rotations (rounds), such that each player gets to play one round with each combination of seat wind and table wind. Each player starts with a bank of 500 points. At the end of 16 rounds, the player with the most points is the winner of the game.
## Example gameplay interaction
> START HAND EAST 1
> 1b 4b 6b 2s 9s 3w 3w 5w 6w NN NN ZZ BB
> E // The player is East (dealer)
> NN // East drew the north-wind first.
BB // East decides to discard the white-dragon.
> S 3w // Nobody called it, and South discards a 3-character.
pung // East wants to pung the 3w tile.
ZZ // East discards the red-dragon
> S pung // South pungs the red-dragon.
pass // East has no use for this tile, so it passes.
> W kong 2s // West declares a standby kong on 2-bamboo.
> W 3b // West discards a 3-circle.
pass // East passes again.
> N chi 2b 4b // North chi-s it with a 2b and 4b.
> N 7s // North discards a 7-bamboo.
pass
> 2w // East's hand: 1b 4b 6b 2s 9s 2w 5w 6w NN NN NN -- 3w 3w 3w
2w // East doesn't need the 2w, so he discards it.
> S hu 2b 3b 4b 7s 8s 9s 2w 6w 7w 8w ZZ ZZ ZZ
// South wins on that tile. The 2w completed his pair.
> END
> 496 506 499 499 // Their current point counts. South's hand was worth
// 2 points, so East pays 4 while the others pay 1.
> START EAST 2
> [13 more tiles]
> N // The deal rotated, and this player is now North.
## Brackets
Your program can participate in one of three levels of gameplay:
• Level 1, where every hand is worth exactly 8 points, and the only thing that matters is completing hands.
• Level 2, where every hand is worth the number of points in the points table section below, but any hand can still win.
• Level 3, where every hand is worth the number of points in the points table, but a hand must have at least 8 points total with that system to win.
(TBD)
## Penalties
The following are invalid moves:
• Failing to produce an output when required.
• Discarding a tile that is not in your hand, or when it is not your turn.
• Calling chi on any player other than the one immediately before you, or calling any tile that does not actually complete a set in your hand.
• Calling hu when your hand is not complete, or when your hand value does not meet the minimum point requirement in level 3 gameplay.
If your program makes an invalid move, the move will be rejected, your program will pay 20 points to each of the three other players, and for the rest of that round, it will be a "forced tsumogiri" player, which automatically discards every tile it draws and never calls anything.
## Special Rules
There are no special rules such as dora or furiten in FPM. Any person may complete a hand at any time as long as it is a valid complete hand.
• Sounds fun :) I hope there'll be enough entries to actually play a game though – Sp3000 Jun 24 '15 at 4:34
# Team Gun Battle
This is an individual competition, where the goal is to be the last player surviving. Fighting always takes place between two teams. At the beginning, every player is randomly assigned to one of two teams and placed randomly on the board.*
Each turn, a player can move north, south, east or west or stay. In addition to moving, they can also shoot a bullet north, south, east, or west. Note that players won't know about other players' movements when they move, but the shoot() method is called after all moving has taken place. The player can choose to shoot a light bullet, which travels 4 squares per frame and does 1 damage, or a heavy bullet, which travels 2 squares per frame and does 3 damage. The bullets do move on the same step they were shot in, so if a target is less than 4 units away, you can shoot a light bullet without any chance of the target dodging.
Every player has a starting health of 30 and 30 bullets to shoot.
After 1000** or so steps, the round will end. Whichever team has more players alive will be considered the winners. Ties will be broken by whichever team has the highest total health.*** Every player not on the winning team will die, and the winning team will be split up into two new teams. This means that half of those who were originally on your side are now your enemies. Another 1000 steps are executed and another winning team is chosen. This process repeats until there is only one player left.
This has some interesting strategic consequences. For example, if your team is winning, it is in your interest to start shooting players on your own team because they may not expect it and you will have to fight them eventually. Similarly, you don't want to waste all of your bullets on the first few rounds because your bullets and your health is not replenished after each round. I like trichoplax's suggestion where ammunition and health could be slightly restored after each round. You want to get your teammates to do as much work as possible before killing them.
The controller almost ready. This is the spec:
public class Player1 extends Player {
public Direction move(World w, int stepsLeft) {
// stepsLeft = # of steps left in round
// List of players: w.getPlayers()
// List of bullets: w.getBullets()
Player p1 = w.getPlayers().get(0);
// Check health: p1.getHealth()
// Check # of bullets left: p1.getBullets()
// Check team: p1.isOnSameTeam(this)
// Check position: p1.getPosition()
return Direction.NORTH; // always move north
}
public Bullet shoot(World w, int stepsLeft) {
// similar to above
return new Bullet(Direction.NORTH, true); // heavy bullet shot north
}
}
*I think the board should be a square with side length 3*(# of players). Hopefully this would make to not too sparsely populated.
**Exact number to be determined. It may be based on the number of players (higher # of players = more steps until the end of the round). 1000 seems like a good number.
*** I'm working on a system that's somewhat more fair. Currently, if there was originally an odd number of players, they were split with one random team getting an extra player. Obviously, that team has a significant advantage under this system. Perhaps it could be percentage of surviving players?
• It sounds like it would be easy to forever avoid opponents by just staying off their orthogonal lines. What happens if it never gets down to one player for that reason (or everyone runs out of bullets)? – Geobits Jun 25 '15 at 16:37
• @Geobits If everyone runs out of bullets, then I have two options I'm considering: a) Nothing happens, and when the 1000 frames ends at the end of the round, one team is randomly chosen to proceed (that is the default when teams tie exactly; if one team has higher total health than the other, then that team wins) b) everyone's bullets are replenished. Which is better? – soktinpk Jun 25 '15 at 17:06
• Would it help to give everyone some extra ammunition at the start of each round, but also let them carry over whatever ammunition they saved from the last round? – trichoplax Jun 25 '15 at 17:21
• If you were feeling mean you could make the arena wrap and have bullets not stop until they hit someone. So dodged bullets will keep looping across the arena, slowly filling it up until there's no way for everyone to avoid them. – trichoplax Jun 25 '15 at 17:22
• @trichoplax I like your first idea. The second seems like a little harsh since you have a lot less control of who your bullets hit. – soktinpk Jun 25 '15 at 17:30
• I did think it might be a little too mean :) – trichoplax Jun 25 '15 at 17:31
• If everyone runs out of bullets, then there is obviously a knife fight. – feersum Jun 25 '15 at 21:40
A few minutes ago, I posted this question as a popularity-contest which was quickly put on hold as too broad. Specifically, the "there are no restrictions on what the programs/functions do" was apparently not quite right. A commenter also suggested that an empty program would be a possible answer which I knew, but didn't necessarily want to restrict as I think the popularity contest would have taken care of that. I'd be interested in any suggestions people have to make the question more appropriate for PPCG.
Also, I plan on asking with the tags popularity-contest and restricted-source. Does that seem right?
Here is the question:
Inspired by the bonus in this question
Write some code that can be run in multiple directions. In each direction, the code must be a complete program or a function. The code must be written in one language (so for example, you cannot have a python program that is also a whitespace program when run backwards).
### Explanation of "multiple directions"
abcd efgh
ijklm
nop
Then backwards is
hgfe dcba
mlkji
pon
And sideways is
a n
bio
cjp
dk
l
em
f
g
h
### Trivial example
Brainfuck
.
.+++++++++
+++++++++
+++++++++
+++++++++
+++++++++
+++++++++
+++++++++
+++++++++
+++++++++.
.
### Notes
• there are no restrictions on what the functions/programs do
• your code must be able to run in at least two directions, but more is better
• abuse of comments is frowned upon but allowed
• Welcome to the sandbox! You deleted your post before I could add another comment, so here's what I wanted to say: What makes an answer "good" in this question? If we go by the criteria "your code must be able to run in at least two directions, but more is better", which the empty program satisfies, that seems a little contradictory. Thus I think it's good to place some sort of restriction on the programs. I think source-layout is better than restricted-source - take a look at the questions there and see what you think :) – Sp3000 Jun 27 '15 at 6:17
• Also, as is even if the question wasn't closed as too broad, it'd probably be closed as a duplicate of this question, which you might want to also take a look at. – Sp3000 Jun 27 '15 at 6:18
• Thanks, I'll look through those questions. And I think you're right. That other question is pretty similar. I might just let it go. – sudo rm -rf slash Jun 27 '15 at 13:40
### Allongiator code-golf
There are a lot of acronyms on the internet -- ROTFL, RTFM, NSA, laser, taser, and so on. Your job is to write code that, given an acronym, can figure out what that acronym might stand for. You'll be given an alphabetically-sorted list of words that you can use, to be provided in the format of your choice.
Note that if you're given an actual acronym, you don't have to provide what it really stands for. As an example, given ROTFL, you could output "rolling on the floor laughing", or you could output "rearrange orange thrice free language".
### Input
1. A dictionary of all of the words you're allowed to use. This is guaranteed to have at least one word for each letter; your program is allowed to do anything it wants if it encounters a letter that it can't find a word for.
• Not necessarily in alphabetical order
• All in lowercase
2. A string of lowercase letters, with no whitespace, punctuation, or non-letter characters.
They can be given in any format you want. For example, the acronym could be the first command line argument and the dictionary the rest of them, or they could be arguments to a function, or given via standard input, or whatever.
The dictionary contents can be found [here].
### Output
1. A string containing a space-delimited list of words that form the acronym given as input, from the dictionary given.
• No trailing whitespace except for a single optional newline
• The words must be chosen (pseudo-)randomly from the dictionary -- that is, if I run it twice with identical inputs, I should get different outputs1.
• It doesn't have to make sense or be plausible or grammatically correct. Any words that start with the right letters will do; however, bonus points2 to anyone who manages that.
Since this is , the shortest answer wins.
All the standard prohibitions apply here, though with the note that accessing the dictionary from the web is acceptable, so long as it's the unmodified dictionary and nothing else.
1: Barring practically impossible situations.
2: That is, a browser cookie
Sandbox (foot)notes:
This may very well be a duplicate. I looked around as much as I could, but I don't even know what this is called, so I didn't find much. Apologies if it is.
Would they be acronyms or initialisms? I'm not sure.
I'll add the actual dictionary once I've found a nice one that isn't massive.
• A quick search for acronyms turned up this – Sp3000 Jun 27 '15 at 6:31
• @Sp3000 ...I even upvoted that question and I forgot about it. Should I delete this post? – Fund Monica's Lawsuit Jun 27 '15 at 6:32
• I'm not sure tbh... might be good to get a second opinion, since I'm not sure whether that one being a popcon changes anything – Sp3000 Jun 27 '15 at 6:33
• @Sp3000 Alrighty. I'll leave this up. (Thanks for replying so quickly, by the way) – Fund Monica's Lawsuit Jun 27 '15 at 6:51
• Eww, CR..... =) – Faraz Masroor Jun 28 '15 at 3:42
• @FarazMasroor Nooo my secret is revealed (I recognize you from that meta question! Hi!) – Fund Monica's Lawsuit Jun 28 '15 at 3:42
## Introduction
You're fed up. You've tried to sell your car but failed, not knowing whether your price is too high, too low or people are just too picky. Then a programmer decides to take a look at your car and makes you an offer using a program that he had written for judging a car's worth.
## Challenge
Given a car's make, model and mileage, you have to find how much a car is worth using the method below. The shortest program to do so wins.
If the make or model of the car is incorrect, you must output "Error: [input make/model] is invalid".
## Method
First of all, you should scrape all of the prices and mileages of all of the listings on AutoTrader.com of the specified car. Next, plot these on a graph (this graph is not needed to be shown) of mileage against prices.
On the graph you should draw a line of best fit. Now you can read the price on the line of best fit for the specified mileage of the car. Now output this with a dollar sign before the price.
## Example
None of these prices are correct output
Ford Fiesta, 200 miles
Input: ford, fiesta, 200
Output: $15000 Subaru Outback, 0 miles Input: subaru, outback, 0 Output:$27000
## Example script
Here's my example test script in Python 2.7 (uses autotrader.co.uk though, not .com):
from BeautifulSoup import BeautifulSoup
import urllib2, sys, re, pylab
model, make, mileage, postcode = sys.argv[1:]
mileage = int(mileage)
soup = BeautifulSoup(page)
results = soup.findAll("div", {"class":"search-result__content"})
cars = []
for j in results:
pricetag = j.findAll(text=re.compile(r"£.*"))
price = int(pricetag[0].replace(',','').replace('£',''))
mileages = j.findAll(text=re.compile(r".+ miles"))
mileagecar = int(mileages[1].replace(" miles","").replace(",",""))
cars.append([mileagecar, price])
x = [k[0] for k in cars]
y = [i[1] for i in cars]
poly = pylab.polyfit(x, y, 2)
print("£"+str(int(pylab.poly1d(poly)(mileage)+0.5)))
• I believe scraping autotrader.com violates their visitor agreement. Section 5 seems to imply that it's illegal as well (though that may well depend on the jurisdiction). – Geobits Jul 1 '15 at 20:00
• @Geobits Damn it haha... I'll look for some kind of API for another online car sales site... – Beta Decay Jul 1 '15 at 20:01
• I'm also not sure where the pounds/dollars difference comes in. Both of those cars are available in the US, for example, but output is in pounds. I don't see anything in the input to say which to use. – Geobits Jul 1 '15 at 20:05
• @Geobits It was based on whether the program used AutoTrader**.com** or .co.uk – Beta Decay Jul 1 '15 at 20:07
• Well... I mean I got that. But it's a code golf; why would you use the longer one? I assumed there had to be some reason you made the choice, not just because you felt patriotic ;) – Geobits Jul 1 '15 at 20:09
• For instance, what if I decide to use the US site, and there are no matches for a particular make/model (even if there may be on the UK site)? Is my output "wrong"? – Geobits Jul 1 '15 at 20:10
• @Geobits I see... Well I'll limit it to an American site when/if I find one – Beta Decay Jul 1 '15 at 20:14
• Doesn't really matter where the site is from, really, as long as it's one site. Or there's something in the input to determine that. If the input was, say ford fiesta 200 dollars (or us/uk, gbp/usd, etc) then that part wouldn't be an issue. – Geobits Jul 1 '15 at 20:15
# Meta-Manufactoria
Manufactoria is a pretty great programming game. However, instead of writing machines to solve the problems, your goal is to write a program that creates the machines for you. If you are already familiar with Manufactoria, then for the next section, you only need to read the italicized phrases.
# The Rules of Manufactoria
In this version of Manufactoria, you are placed in a 9x9 world. The input generator is in the center top square (4, 0), and the output acceptor is in the center bottom square (4, 8). When input is generated, it moves immediately down.
Input is represented by a robot that has a queue of colors. Output will be the same robot, but the queue may contain a different sequence of colors. Robots must always eventually end up in the output square. There are 2 colors available, Red and Blue.
Each tick, the robot moves 1 square. The direction the robot moves, and any modifications to the queue is determined by the square it is on. There are 3 types of squares:
1. Movement square. This square either moves the robot North, East, South, or West and does not affect the queue. You cannot have two movement squares on the same tile
2. Choice square. This square moves the robot based on the top color in the queue. The general Choice square moves the robot east if red, west if blue, south if empty, and never north. This square can be rotated or reflected to change the directions traveled.
3. Writer square. This square writes a color to the back of the queue, and moves the robot in a given direction.
# STDIO and Scoring
Your program will be passed two lists of the same length, one containing the inputs, and the other containing the outputs. Each item of each list contains only the R and B characters, representing the input queue.
Your output is the board that solves the Manufactoria puzzle. Each square is represented by two characters. The first character represents the square type, the second represents the direction. The square types can be:
• Movement
• Choice
• Red writer
• Blue writer.
• .. Empty square
• II Input
• OO Output
The directions are: - North - East - South - West.
The direction listed for a choice tile represents the direction traveled if the queue is empty. If the direction is lowercase, the tile is reflected (which swaps the colors).
The program which solves more puzzles than any other program wins. In the event of a tie, the tiebreaker is the program that produces the most efficient solutions. Efficiency is measured by the number of times the robot moves across all of the solutions.
Your program should be a general solver. I may add puzzles to the ones listed below at any time.
# Sample Puzzles
The solutions given below don't necessarily have to do what the title indicates. Your program's solution also doesn't have to match the ones listed below. It only needs to solve the inputs given. The format below is
[Inputs] [Outputs]
Solution Map
Example 1: Don't do anything
["","R","B","RRBR","BRRBRB"] ["","R","B","RRBR","BRRBRB"]
........II........
........MS........
........MS........
........MS........
........MS........
........MS........
........MS........
........MS........
........OO........
Example 2: Print second R and anything after:
["","R","B","RR","BRB","RBRBBB"] ["","","","R","","RBBB"]
........II........
..MSCSMWCSMW......
..RSMEMEMS........
BECsMEMEMS........
........MS........
........MS........
........MS........
........MS........
........OO........
Example 3: Reverse String (This doesn't reverse any string, it only works for the examples listed)
["","BBR","B","RRB","BRRRB","BBRB"] ["","RBB","B","BRR","BRRRB","BRBB"]
........IICsMERS..
......CECSCEMSBS..
......RSMSCEMSMW..
......REMSCEMS....
........MSRSMS....
........MSRSMS....
........MSRSMS....
........MSBWMW....
........OO........
Example 4: Only include Rs
["R","","BBR","BRR","B","RBRBRBBR","BBBBBBRRR","RBBBRBRBBB","RRRR"] ["R","","R","RR","","RRRR","RRR","RRR","RRRR"]
........II........
........MEMEMEMS..
MSCSMWCSMWCSMWCSMW
REREMERERSMW..MS..
........MSMWMWMW..
........MS........
........MS........
........MS........
........OO........
• Sounds extremely difficult. – feersum Jul 7 '15 at 3:07
• I enjoyed playing that game...Hard indeed. :-) – Spikatrix Jul 7 '15 at 12:39
(Need a title.)
Write two programs (or functions) in the same language for these two tasks:
1. Given a list of integers, split at every non-negative integer. If there are two or more consecutive non-negative integers, your program should preserve the empty list between them.
2. Given a list of lists of negative integers, concatenate the lists and insert non-negative integers between them. The inserted integer after the nth list (zero-based) should be n.
Alternatively, you can increment every integer by one for both tasks. So you split at positive integers, accept non-positive integers in task 2 and the inserted integers start at 1.
The input for task 1 and the output for task 2 should have the same format, with the exception that one can be the string representation of the other in your programming language. The same goes for the output for task 1 and the input for task 2.
Your score is the total length of your two programs times (30 + their Levenshtein distance). Lowest score wins.
Input: [-5 7 8 -5 2 -6 -3 -3 8]
Output: [[-5] [] [-5] [-6 -3 -3] []]
Input: [[-5] [] [-5] [-6 -3 -3] []]
Output: [-5 0 1 -5 2 -6 -3 -3 3]
• if(1>2){code1}else{code2}. There's no need to ever have the edit distance any greater than 1 or 2. – John Dvorak Jul 12 '15 at 6:03
• better yet: if(input[0] instanceof Array<int>){code1}else{code2}. Haskell might be able to pull this off even more cleanly because the typeclasses are resolved at compile time, meaning that even empty arrays have the proper type. – John Dvorak Jul 12 '15 at 6:06
• The first task description is extremely unclear. What does it mean to split? How can one preserve something that didn't exist in the first place? – Peter Taylor Jul 12 '15 at 16:55
• @JanDvorak If the answer is less than 30 bytes, it is beneficial to remove a byte while increase the edit distance by one. But maybe I should change it to the length plus the edit distance. – jimmy23013 Jul 12 '15 at 19:48
• @PeterTaylor It just works like the CJam and GolfScript /. I'll try to reword it later if I get the scoring method working. – jimmy23013 Jul 12 '15 at 19:50
# Return of the 5318008
## Introduction
A week or so ago, I posted the challenge 5318008, with a massive reaction. Now, I want you to do the same again but with musical chords.
## Challenge
Given a word list, you must output a list of words which can be formed using musical chords. However, the word must also sound good too.
The chords which you may use are:
A B C D E F G
These are all the major chords without the flat/sharp chords.
A word will sound good if all of its letters are within the same key. All the chord progressions which you must concern are listed below:
• A major: A D E G
• B minor: D G A
• D major: D G A
• E major: E A B D
Other progressions such as G or C major could not make any word because all of the major chords (which were not sharps or flats) were consonants.
For my tests, I used the UNIX wordlist, gathered by typing:
ln -s /usr/dict/words w.txt
Or alternatively, get it here.
There are some uppercase letters in the dictionary, so make all of the letters in all of the words in the dictionary lowercase.
## Winning
The shortest code to output a list of words wins.
• What about A minor? – Nathan Merrill Jul 13 '15 at 17:03
• @NathanMerrill No vowels in Am, just C F and G – Beta Decay Jul 13 '15 at 17:05
• I don't understand then. A minor includes "a" and "e" – Nathan Merrill Jul 13 '15 at 17:09
• What about case sensitivity? Also I'm expecting this to end up as a regex challenge, if I've understood the problem correctly – Sp3000 Jul 13 '15 at 17:10
• @Sp3000 Edited to say that all words should be converted to lowercase – Beta Decay Jul 13 '15 at 17:13
• @Nathan Sorry, I was talking about chords instead of notes :/ – Beta Decay Jul 13 '15 at 17:14
• So I'm to assume that A would only be included if A major is a subset of the key? – Nathan Merrill Jul 13 '15 at 17:18
• @Nathan Uhh yes – Beta Decay Jul 13 '15 at 17:22
• Ok. Also, B minor and D major include the same notes, so the notes listed should be the same. D major shouldn't have the C, as it has a C#. – Nathan Merrill Jul 13 '15 at 17:24
• I don't know much about music theory, but from what little I do know I find this question very confusing. A B C D E F G doesn't look like a list of chords, but like the scale of C major. If by the "key of C major" you mean the tonic chord of the scale of C major, that would be C E G, which is not all consonants. I also promise to downvote any question which allows people from one specified nation to answer a different question to people from the rest of the world. – Peter Taylor Jul 13 '15 at 19:25
• @PeterTaylor Refer to this: The chord chart...lists all the common traids and four note extended chords belonging to the key of C major. – Beta Decay Jul 13 '15 at 19:35
• Yes, but that doesn't clear anything up. Unless... are you sure you're not confusing chords with chord progressions? – Peter Taylor Jul 13 '15 at 20:55
• @PeterTaylor Yes, that's exactly what I meant. I didn't know the word for it – Beta Decay Jul 13 '15 at 20:58
• You are definitely confusing notes and chords here. B minor and D major both contain the notes B C# D E F# G A, The possible chords are Bm Em F#m Dmaj Gmaj Amaj and C#dim, with F#m and Em often subsituted for Fmaj and Emaj. One valid interepretation is that both keys contain all these chords, another is that the minor is just B E (F#) and the major is just D G A. Under no circumstance can Bm be considered D G A. You are also misusing the word progression. This refers to the sequence of chords in a particular song and has nothing to do with the chords available in a particular key. – Level River St Aug 8 '15 at 12:32
• A major contains A B C# D E F# G# so the only natural notes are ABDE,which is what you have written for E major. E major contains one additional sharp E F# G# A B C# D# so the only natural notes are E A B. – Level River St Aug 8 '15 at 12:35
# Break out of the digital world!
Note: The tag is just something I knocked up quickly, if anyone has any better ideas please let me know.
## Challenge
You must write a morse code to ASCII translator using an external device (such as a button or the spacebar on the keyboard).
### Dots
A button press is a dot when it has been pressed for 0.5 seconds or longer and released.
### Dashes
A button press is a dash when it has been pressed for 1 second or longer and released.
### Exit
A button press is an exit when it has been pressed for 5 seconds or more. This should end the program and display the result.
### Space
A space is when there has been no button input for 1 second. This should move on to the next letter.
Your code should support every character on the following:
http://zunkworks.com/images/bluetoothmorse/morsecode1.gif
## Winning
The shortest code wins. You may write this in any language that allows input from an external source. For example, Arduino or Python (on a Raspberry Pi).
• Does this actually have to be able take input from hardware? Can you allow feeding in simulating data? I think a standard format like a sequence of numbers measuring current/voltage every millisecond would make it more accessible. – xnor Nov 18 '14 at 1:35
• While I do own an arduino, I see absolutely no reason why this challenge should be limited to using an external switch. Is there anything wrong with using the space bar or a mouse button? If it HAS to be an external switch, how about I rip open my mouse and connect an old doorbell push across one of the mouse button contacts? – Level River St Nov 27 '14 at 20:45
# VSEPR Strikes Back!
(Note: I know that the VSEPR method fails sometimes, and that there are exceptional molecules. This is addressed in the scoring system, and otherwise the challenge is about the molecules which conform.)
Most people who have taken an introductory chemistry course know about molecules, and (probably) the VSEPR theory of chemical bonding. Basically, the theory predicts the shape of a molecule given three main properties: the central atom A, the number of atoms bonded to the central atom X, and the number of lone electron pairs E on the central atom. Applying the VSEPR theory is simply a matter of finding the values of X and E (A always has an implied value of 1, as we'll see). This is called the AXE method. For example, a molecule which has 1 lone pair and 3 atoms bonded to the main atom is AX3E1, which is the trigonal pyramidal configuration.
## The Challenge
Your job is, given a covalent chemical compound, to output the geometrical shape of the molecule inputted. But wait! exclaim the exasperated programmers, you can't expect us to input all of the molecular data for every element! Of course we can, but I'm not feeling particularly masochistic today, so born was the scoring system below.
The input is any molecule, such as CO2 or HgCl2, and the output should be the name of the shape and NOT the AXE form. Ions should have their charge put in parenthesis directly after the ion, such as CO3(2-) or NH4(1+). 1- and - (and 1+ and +) are all acceptable.
## About the central atom A
In most cases, the central atom will be apparent. It usually has no subscript in the chemical formula: the C in CO2 is the central atom, for example. In a few cases, though, you might face compounds like ethelyne (C2O4), in which no clear atom is the central one. In this case, it is worth noting that such compounds are usually symmetrical, and considering any carbon to be the central atom will do.
## Scoring System
The base score is the number of bytes the program takes.
1. There is a minimum of 6 elements to be implemented for input; every 5 extra you add multiplies your score by 0.9.
2. Multiply your score by 0.8 if you can make your code work for ions.
3. Multiply your score by 0.9 if you can detect ionic compounds (VSEPR only works for covalent compounds) and reject them accordingly.
4. Multiply your score by 0.8 if you also output the bond angles.
5. There are certain classes of molecules which VSEPR fails to predict correctly, due to various reasons. (Check the first link in the post for the section on the exceptions.) Multiply your code by 0.9 per class if it returns those correctly.
## Test Cases
The parts in brackets are optional bonuses from above.
CO2: CO2 -> linear [180*]
HgCl2: HgCl2 -> linear [180*]
H2O: H2O -> bent [104.5*]
[CO32-: CO3(2-) -> trigonal planar 120*]
BrF: BrF3 -> T-shaped [90*]
[NaCl: NaCl -> Ionic compound]
XeF4: XeF4 -> square planar [90*]
Notes:
- The bonding patterns are not valid when the central atom is a transition metal (so don't do that.).
- Standard loopholes apply.
# One transparent colour sprite
## Overview
This challenge will only involve a hex string as input and a hex string as output, but will be explained in terms of computer graphics.
Given a fixed size background image, a smaller fixed size sprite image and its location, place the sprite on the background image allowing the background to show through for one specified colour of the sprite image.
## Details
• The background image will always be 32 by 18 pixels.
• The sprite image will always be 8 by 8 pixels.
• The colours will always be in the range 0 to 3 inclusive.
• There will always be exactly one transparent colour, from 0 to 3 inclusive.
• The location (x, y) of the sprite may be outside or partially outside the background image.
## Input
Since all inputs will always be the same size, there will be no separators, just a single string of hex digits.
To avoid having to define negative inputs, the top left pixel of the background image will be (128, 128), increasing left and down, and x and y will be defined by 2 hex digits each, giving a range of 0 to 255.
The input hex string will be made up of:
• Background pixels, 2 per hex digit, a string of 32 * 18 * 0.5 = 288
• Sprite pixels, 2 per hex digit, a string of 8 * 8 * 0.5 = 32
• Transparent colour, a single hex digit (the most significant half of the hex digit is unused)
• Sprite x and y, 2 hex digits each, a string of 2 * 2 = 4
The input will therefore be a string of 288 + 32 + 1 + 4 = 325 hex digits with no separators.
## Output
The output is the finished 32 by 18 pixel image, as a string of 288 hex digits with no separators.
## Format
### Pixel order
Each image is ordered in English reading order, left to right then top to bottom. That is, the pixels are listed in rows. The first pixel listed will be the top left. This applies to both the background image and the sprite image.
### Colour encoding
Pixel colours are encoded 2 per hex digit, most significant first.
For example, the hex digit B corresponds to binary 1011. This means the first pixel has value 10 = 2 and the second pixel has value 11 = 3.
### Transparent colour
Since the single transparent colour is encoded in one hex digit, which normally holds 2 colours, the first 2 bits (most significant) are ignored and the last 2 bits (least significant) are used to represent the transparent colour.
For example, the hex digit D corresponds to binary 1101. The 11 is ignored and the 01 is used to indicate colour 1.
Although the first two binary digits are always ignored, you may not assume they are always 00. They may take any value and all hex inputs of the correct length should produce a valid output.
### Coordinates
x and y are encoded as 2 hex digits each, most significant first.
For example, the hex string 2A corresponds to binary 00101010, and decimal 42
### Case sensitivity
Your code may accept input as either upper case or lower case hex digits, or both. If it only accepts one or the other case, it must also output in the same case.
You may not choose an arbitrary 16 characters to represent hexadecimal. You must use either (or both) of the following as input:
0123456789abcdef
0123456789ABCDEF
You must choose one or other (not both/mixed) to use as output. That is, you may accept inconsistent inputs if you wish to, but the output must be either always lower case or always upper case.
# Sandbox thoughts
• I'm considering adding example inputs and outputs and maybe a snippet to verify the output of arbitrary inputs for testing.
# Live Tennis Scoreboard
In this challenge, we're going to create a tennis scoreboard that could (theoretically) be used in an actual tennis match. The tennis scoreboard will display the score while listening on STDIN. As new results come in on STDIN, the tennis scoreboard will update itself accordingly.
### Description of Input
When the scoreboard is first instantiated, it will be provided input for two strings -- these are the names of the competitors. This can be via command line arguments, user input functions, etc.
After that, the scoreboard will recieve input via STDIN. These correspond to events that the scoreboard should reflect in the score.
• 0
The first player has scored a point.
• 1
The second player has scored a point.
• U
Undo the previous point and revert to the previous state of the scoreboard.
The scoreboard must handle games, deuces, etc. as described in the next section.
### Terse Description of Tennis Scoring System
This section here describes the exact rule set to use.
A match is between two players. To win a match, a player must win three sets. To win a set, a player must win at least six games and have a two-game lead over their opponent. To win a game, a player must win at least four points and have a two-point lead over their opponent.
The number of sets and games won are displayed as is. Points are displayed as follows in order of priority:
• 1 point = 15
• 2 points = 30
• 3 points = 40
• Number of points won are equal to the opponent = 40
• Number of points won are one greater than the opponent = Ad
Tiebreaks are optional. Implementing tiebreaks grants you -40 to your score.
### Scoreboard Format
During the first set and first game:
Name | x
Name | x
where x is the point display for each player.
During the first set:
Name a | x
Name a | x
where a is the number of games won by each player.
During the second set:
Name b a | x
Name b a | x
where b is the result of the first set.
During the third set:
Name c b a | x
Name c b a | x
where c is the result of the second set.
etc. for fourth and fifth sets.
Example:
Federer 5 6 6 3 2 | 0
Murray 7 4 4 6 2 | 15
• A bit of a far stretch, but it would be cool if you could point the programs at URL which gave the scores for an actual tennis match – Beta Decay Jul 27 '15 at 8:04
# Punch buggy?
As some of you might know, "punch buggy" is a children's(?) game that involves spotting Volkswagen Beetles and punishing other players. For reference, the car on the left is a punch buggy, while the car on the right is not:
Your task is: given an image, determine whether there is a punch buggy shown in it. Entries will be scored by percent correct among all n (TBD) test images. If there is a tie, it will be broken by run time on my computer (specs will be posted on main).
Test images may contain just about anything, and come in various sizes. Most will be automobiles of various types, but I will make sure they have a maximum of one auto in them (so if you see a non-matching car, you don't need to look for more). I won't use toy cars or heavily modified cars to try to fool you. Images will be pulled from Google Image Search, and will be of varying quality (from ad-quality to candid in-the-yard).
This is not a Kolmogorov complexity challenge. The point is not to recognize specific image files, but to recognize a Beetle in any picture. If it looks like someone is cheating this, I will change to a different test set and re-score.
### Spec
Input is the name of an image file in JPG or PNG format (should I restrict this to one?).
Output is Punch Buggy! if a punch buggy is present in the picture, and any other non-empty string if not.
Programs are run once per image. I will write a simple test controller for this using STDIO, so programs must be runnable from the command line, using a language that is freely available on Ubuntu.
You are allowed to read/write other files, but only in a folder with the relative path /yourProgramName. You are allowed to read files whose names are passed as input via STDIN.
### Sandbox
Yes, this seems a bit difficult to me, too. I don't see that as a problem. Is there anything that isn't clear about the task or spec? Other notes/questions/suggestions?
If it needs fluff (I'm not sure yet), it will probably be something like "help my robot become more human via children's games".
• Personally, I don't think this is that hard. You just look for a large grouping of yellow pixels around the centre of the image. I think you should limit it to a PNG. – Beta Decay Jul 29 '15 at 13:21
• @BetaDecay That would only work if all punch buggies are yellow and all other images/cars are not. The images shown here are only representative samples. The test cases will include various cars of various colors at various angles in various lighting conditions. – Geobits Jul 29 '15 at 13:27
• Ohh I thought Punch Buggy was when you have a yellow car, not just a Beetle – Beta Decay Jul 29 '15 at 13:30
• Ah. That's the Yellow Car Game. Similar, but yea, that would be a different difficulty level. – Geobits Jul 29 '15 at 13:32
• I afraid even if you people try to recognize buggies their results will come from randomness and overlearning on the provided images. Recognizing a car at all would be a difficult task itself. – randomra Jul 30 '15 at 11:49
• If you want to avoid this being a kolmorogov challenge, would it help to provide a much larger training set and test on a small subset of this that is not announced in advance? That does introduce a deadline excluding future submissions though. – trichoplax Aug 1 '15 at 18:30
• @trichoplax Not necessarily, because you could always test the new submissions using the subset (which is still unannounced) – Beta Decay Aug 5 '15 at 11:48
• @BetaDecay good point. That makes this sort of challenge seem much more promising. – trichoplax Aug 5 '15 at 12:27
• @trichoplax Though it could still be a kolmogorov complexity because a program could be programmed to recognise all pictures in the album, and choosing a subset wouldn't stop that program from getting full marks – Beta Decay Aug 5 '15 at 12:37
Question about symmetry in ascii grid (temp title)
Synopsis: Given a grid of ASCII characters, determine if, and in what ways, the grid is symmetrical.
Some characters are symmetrical to others. For example:
• d is symmetrical vertically to b
• W is symmetrical horizontally to M
• O is symmetrical both ways to O
• ! is symmetrical horizontally to i (takes some imagination!)
• ^ is symmetrical horizontally to v
• b is rotationally symmetrical 180 degrees to q (that is, if you rotate the first character 180 degrees, you get the second character)
• (space) is symmetrical vertically, horizontally, and rotationally to
In this challenge, there are only three types of symmetry that need to be considered: vertical, horizontal, and rotational by 180 degrees. You can find a full list of which characters are considered symmetrical at the bottom of this post.
If we arrange some characters into an n by m rectangle (where n and m are both even integers), then the rectangle may be symmetrical. Here is an example:
M^^^^MM^^M
OOOOOOOOOO
OOOOOOOOOO
WvvvvWWvvW
It is symmetrical horizontally.
Here's another example.
dddbbb
dddbbb
dddbbb
ddOObb
It is symmetrical vertically.
Here's one more example:
++++++
+p++q+
++++++
++++++
+b++d+
++++++
This one has three ways it is symmetrical -- horizontally, vertically, and rotationally.
Write a program which will, when given a rectangle of characters as input, output the number of ways that rectangle is symmetrical. For the three inputs I've shown above, the outputs would be 1, 1, and 3 respectively.
[Todo: list of characters]
## Sandbox Questions
1. One thing I've been considering is allowing entrants to access the list of which characters are symmetrical (in some sort of CSV format) in a file. This would mean that entrants wouldn't have to encode the symmetries themselves. On one hand, this probably saves some frustration at having to code in all the possible symmetries. On the other, this reduces the complexity of the challenge as the aspect of encoding the symmetries with minimal characters is removed. I'd like your thoughts on this.
• Yet more symmetries: all of []{}() are vertically symmetric to themselves. | belongs in "Both". lower-case c belongs in vertical. Maybe also lower-case a depending on the font your going by (apparently not the one used on SE). Likewise t might belong in horizontal symmetry. We also forgot about digits: 08 both ways, 3 vertical. – Martin Ender Sep 1 '14 at 11:58
• What do you think about adding rotational symmetry (by 180 degrees) to the mix? – Martin Ender Sep 1 '14 at 11:59
• Like 6 and 9? I think rotational symmetry would do well as a seperate question actually. – absinthe Sep 1 '14 at 12:08
• Yes, like 6 and 9. If a new question for that is based on the same concept, it would probably be too similar to not be a duplicate (at least the solution I have in mind would be rather easily adapted to do rotational symmetry, I think). Adding it here, might open some possibilities for interesting compression/golfing. – Martin Ender Sep 1 '14 at 12:11
• You've convinced me. I'll edit in the rotational part when I have time. – absinthe Sep 1 '14 at 12:14
• Since there will probably be combinations of all three symmetries on different characters, I'd recommend only having one list for each symmetry (with some pairs in multiple lists). Here are the rotational pairs I can think of: d/p, b/q, (/), [/], {/}, </>, %/%, ///, \/\, ,/,, S/S, s/s, z/z, Z/Z, W/M, !/i, ^/v, $/$, as well as all those which are symmetric both ways to themselves. – Martin Ender Sep 1 '14 at 12:22
• Some of the characters are not visible (at least on my screen) so would benefit from being followed by a description. I think they are just spaces but markdown appears to fail to display them even in backticks. – trichoplax Sep 1 '14 at 14:09
• Displaying a single space in an inline code tag has to be done with an actual <code> tag rather than with backticks. codegolf.stackexchange.com/questions/5194/… is a good starting point for rotational symmetries. Some digits also have linear symmetries. – Peter Taylor Sep 1 '14 at 14:39
• Would you also consider J and L as symmetrical (also requires a little imagination...)? – WallyWest Sep 2 '14 at 22:41
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Search by Topic
Resources tagged with Working systematically similar to Exploring Wild & Wonderful Number Patterns:
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There are 340 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
Six Is the Sum
Stage: 2 Challenge Level:
What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
Stage: 2 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
Rabbits in the Pen
Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
Mystery Matrix
Stage: 2 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
A Mixed-up Clock
Stage: 2 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
Stage: 2 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
Zargon Glasses
Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
All Seated
Stage: 2 Challenge Level:
Look carefully at the numbers. What do you notice? Can you make another square using the numbers 1 to 16, that displays the same properties?
How Much Did it Cost?
Stage: 2 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
A-magical Number Maze
Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
Bean Bags for Bernard's Bag
Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
Seven Square Numbers
Stage: 2 Challenge Level:
Add the sum of the squares of four numbers between 10 and 20 to the sum of the squares of three numbers less than 6 to make the square of another, larger, number.
Route Product
Stage: 2 Challenge Level:
Find the product of the numbers on the routes from A to B. Which route has the smallest product? Which the largest?
Page Numbers
Stage: 2 Short Challenge Level:
Exactly 195 digits have been used to number the pages in a book. How many pages does the book have?
A Right Charlie
Stage: 2 Challenge Level:
Can you use this information to work out Charlie's house number?
Two Primes Make One Square
Stage: 2 Challenge Level:
Can you make square numbers by adding two prime numbers together?
Sealed Solution
Stage: 2 Challenge Level:
Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes?
It Figures
Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
Pouring the Punch Drink
Stage: 2 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
The Pied Piper of Hamelin
Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
Polo Square
Stage: 2 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
Finding Fifteen
Stage: 2 Challenge Level:
Tim had nine cards each with a different number from 1 to 9 on it. How could he have put them into three piles so that the total in each pile was 15?
Forgot the Numbers
Stage: 2 Challenge Level:
On my calculator I divided one whole number by another whole number and got the answer 3.125. If the numbers are both under 50, what are they?
Rolling That Cube
Stage: 1 and 2 Challenge Level:
My cube has inky marks on each face. Can you find the route it has taken? What does each face look like?
Journeys in Numberland
Stage: 2 Challenge Level:
Tom and Ben visited Numberland. Use the maps to work out the number of points each of their routes scores.
Shapes in a Grid
Stage: 2 Challenge Level:
Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column?
Sums and Differences 1
Stage: 2 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
Sums and Differences 2
Stage: 2 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
Build it Up
Stage: 2 Challenge Level:
Can you find all the ways to get 15 at the top of this triangle of numbers?
Build it up More
Stage: 2 Challenge Level:
This task follows on from Build it Up and takes the ideas into three dimensions!
The Dice Train
Stage: 2 Challenge Level:
This dice train has been made using specific rules. How many different trains can you make?
A Shapely Network
Stage: 2 Challenge Level:
Your challenge is to find the longest way through the network following this rule. You can start and finish anywhere, and with any shape, as long as you follow the correct order.
Fractions in a Box
Stage: 2 Challenge Level:
The discs for this game are kept in a flat square box with a square hole for each disc. Use the information to find out how many discs of each colour there are in the box.
Being Thoughtful - Primary Number
Stage: 1 and 2 Challenge Level:
Number problems at primary level that require careful consideration.
Dart Target
Stage: 2 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, invites you to explore the different combinations of scores that you might get on these dart boards.
Open Squares
Stage: 2 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, focuses on 'open squares'. What would the next five open squares look like?
The Puzzling Sweet Shop
Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
Two Egg Timers
Stage: 2 Challenge Level:
You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how?
Today's Date - 01/06/2009
Stage: 1 and 2 Challenge Level:
What do you notice about the date 03.06.09? Or 08.01.09? This challenge invites you to investigate some interesting dates yourself.
Prison Cells
Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
Twenty Divided Into Six
Stage: 2 Challenge Level:
Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?
Magazines
Stage: 2 Challenge Level:
Let's suppose that you are going to have a magazine which has 16 pages of A5 size. Can you find some different ways to make these pages? Investigate the pattern for each if you number the pages.
Arranging the Tables
Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
X Is 5 Squares
Stage: 2 Challenge Level:
Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?
Oh! Harry!
Stage: 2 Challenge Level:
A group of children are using measuring cylinders but they lose the labels. Can you help relabel them?
Dice and Spinner Numbers
Stage: 2 Challenge Level:
If you had any number of ordinary dice, what are the possible ways of making their totals 6? What would the product of the dice be each time?
On Target
Stage: 2 Challenge Level:
You have 5 darts and your target score is 44. How many different ways could you score 44?
Train Carriages
Stage: 1 and 2 Challenge Level:
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
How Old?
Stage: 2 Challenge Level:
Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information?
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# From the given statement, take the cube of the diameter mean d 3 then multiply by π and divide by 6 as: V ( d ) = d 3 π 6 ⇒ V ( d ) = π 6 d 3
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
#### Solutions
Chapter 2.1, Problem 68E
To determine
## From the given statement, take the cube of the diameter mean d3 then multiply by π and divide by 6 as: V(d)=d3π6⇒V(d)=π6d3
Expert Solution
The required values of the given statement are shown in table as below:
dV(d)10.52352 4.18873 14.13710010.5235
### Explanation of Solution
Given information: Given statement is that, take the cube of the diameter mean d3 then multiply by π and divide by 6 as:
Calculation:
Take and Plug d=1 in V(d)=π6d3
V(1)=π6(1)3 =0.5235
Similarly,
V(2)=π6(2)3 =4.1887
V(3)=π6(3)3 =14.1371V(0)=π6(0)3 =0V(1)=π6(1)3 =0.5235
Thus, the required values of the given statement are shown in table as below:
d V( d ) 1 0.5235 2 4.1887 3 14.1371 0 0 1 0.5235
Graph of the function V(d)=π6d3 as shown below:
### Have a homework question?
Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!
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http://www.instructables.com/id/How-to-Model-a-Simple-Spring-mass-damper-Dynamic-S/
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# How to Model a Simple Spring-Mass-Damper Dynamic System in Matlab
3,215
1
Published
## Introduction: How to Model a Simple Spring-Mass-Damper Dynamic System in Matlab
In the field of Mechanical Engineering, it is routine to model a physical dynamic system as a set of differential equations that will later be simulated using a computer. These systems may range from the suspension in a car to the most complex robotics. The aim of this Instructable is to explain the process of taking a state-space system and simulate the step response using Matlab. To do this, the mass-spring-damper system shown above will be used as an example. There is no training required to complete these steps, however to fully understand the procedure a background in mathematics is advised.
## Step 1: Understand the Terminology
In order to understand the steps to follow, one must first understand some of the terminology that will be used. First, a dynamic system is anything that changes with time. This could be a pot that heats up on a stove or a speaker that plays music when an audio signal is applied. The second thing to understand is that dynamic systems can be modeled using a branch of mathematics known as ordinary differential equations (ODE's). Knowing how to solve ODE's is not important to complete this series of steps, but it should be understood that they are a mathematical model. ODE's can be written in a variety of forms, however to simulate systems in computer software it is convenient to use something called the state-space form. State-space form is a way to organize ODE's such that there exist 4 matrices containing the relevant system properties. Lastly, by using a computer program called Matlab, one can simulate the system with respect to a variety of inputs. Matlab is a linear language that executes each command line at a time. The steps mentioned in this Instructable are set up in such a way that each new line of code will be entered into the next available line.
## Step 2: Prepare Matlab
Verify that Matlab is installed on the computer in use. If the software cannot be found, try searching for it using the bar at the bottom left of the screen. This searching method only applies to the Windows 10 operating system. Once Matlab is open, a new script should be created. This is done by clicking the "New Script" button located in the top left corner of the program. The cursor should be in the editor window and the coding process may begin. Matlab has a convenient feature of telling the user what mistakes have been made. This will happen in two ways. The first is a red line under an error. To learn the nature of the error, hover the mouse over the line. The second will be discovered upon operation. When the program runs, if it encounters an error it will type in red what the error is and the echo the line that the error occurs on.
## Step 3: Begin the Code
For easy debugging and viewing, type in the following lines of code.
clear;
clc;
This will clear all workspace variables and the output screen before each script execution. While not necessary, this small amount of code will allow a user to easily see any issues with the program and clean up the output. It is important to note that Matlab code is case sensitive.
## Step 4: Confirm That the System Is in State-Space Form
There will always be 4 separate matrices when modeling a system in state-space form. These will be noted as the A, B, C, or D matrix depending on the location in the equation. Before the state-space matrices will be defined in Matlab, it is good practice to confirm that all matrices are present in the information given. The state-space representation for the mass-spring-damper system is shown here.
## Step 5: Define the Constants.
In mass-spring-damper problems there are several numerical constants to note. The constant k is called the spring constant and refers to the rigidity of the spring. The constant b is known as a damping coefficient and is significant in that it helps model fluid resistance. M in this case simply represents the mass of the block. For this simulation we will assume k = 24, b = 8, m = 25. Typically these values would have units associated, but for simplicity they have been omitted. To define these in Matlab begin with the variable name, such as "k," and set it equal to the number associated with it. For example, "k = 24;" is a proper declaration.
## Step 6: Create the a Matrix
Now that it has been confirmed that the given form is correct and the system constants have been defined, it is time to begin defining the system in the code. To begin, declare a variable called "A". This is done by typing "A = " on a single line followed by the matrix. To code the matrix, open with a bracket and type each term in order from left to right, top to bottom. Each column should be separated by a space and each row by a semicolon. Once the information is entered, close the line with a closing bracket followed by another semicolon. The final result should look something like the picture above.
## Step 7: Define the B, C, and D Matrices
Once the A matrix is coded into the software it is time to define the other 3. The method for this is the same for step 6 with the only difference being that the name, numbers, and size of the matrices may be different. It is important to remember that new columns are separated by a space, and new rows are separated by a semicolon. The final result should look something like the code pictured above.
## Step 8: Create a State-Space System Variable
One reason that Matlab is so widely used for dynamic system simulation is that it has most of the hard code already programmed. Now that the state-space matrices have been defined, the entire system can be condensed into a single variable in one line of code. To create a system variable, define the variable like any other using the variable name and an equals sign. In this case, the name "sys" will be used to represent the system. Next, on the right hand of the equal sign type "ss(A,B,C,D);" which will reference a Matlab state-space function. By doing this, the program now recognizes "sys" as a system of ODE's that can be used to simulate a response. When entered into Matlab, this line of code should look like the picture above.
## Step 9: Give the System an Input
Now that the system has been modeled entirely in Matlab, it is time to see how the system responds to a certain input. In this case, a step input will be used. A step input is an instantaneous change to some constant value, and in this system would signify a force being added to a system at rest. Simply type in the line "step(sys)" to code this in Matlab.
## Step 10: Verify Consistency
It is important to check that the code entered is correct corresponding to the state-space equations given. Unlike many problems in engineering, even if the wrong lines of code have been entered the code may execute and display information. This is an issue due to the fact that not all information output will be correct. This step exists to force the programmer to walk through the steps a second time with the intent of comparing the code with the instructions looking for inconsistencies. To further check for errors a copy of the final code may be seen above.
## Step 11: Execute the Code
Once all steps have been completed and the quality of the code has been verified, the code may be executed. This is done by either pressing the F5 key on the keyboard while the cursor is inside the editor block or clicking the "Run Section" button located at the top right of the Editor Tab.
## Step 12: Observe the Results.
When the program execution completes a plot similar to the one shown above should appear. It is at this point that a visual analysis of the system may be done if desired. Should the code not execute or the plot looks different than expected, take another look at the code and compare it to the guide.
## Recommendations
• ### 3D CAM and CNC Class
582 Enrolled
• ### Clocks Contest
We have a be nice policy.
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Win a copy of Functional Reactive Programming this week in the Other Languages forum!
# How to convert one hundered fifteen thousand into 115000?
Greg Kearney
Greenhorn
Posts: 7
Are there methods that exist that will help convert word numbers into actual numbers?
Ex:
"one hundred thousand" = 100000
"two hunderd twenty five thousand three hundered fifteen" = 225315
Manish Hatwalne
Ranch Hand
Posts: 2596
Originally posted by Greg Kearney:
Are there methods that exist that will help convert word numbers into actual numbers?
Ex:
"one hundred thousand" = 100000
"two hunderd twenty five thousand three hundered fifteen" = 225315
NO! AFAIK.
You can however do this -
Write a method that would convert each digit of a number to its word (such as one, two, ..., zero) and vice versa. So you can get "one zero zero zero zero zero" for 100000, this should be fairly simple and straightforwrd.
What you say can be done, but you have to take care of many special cases, such as -
11 => eleven, 12 =>twelve (not ten one, ten two)
21 =>twenty one 22 => twenty two ...
Also,
1500 can be fifteen hundred or it can be one thousand and five hundred. I suggest you go for approach no 1 ("One zero zero...").
HTH,
- Manish
[This message has been edited by Manish Hatwalne (edited November 01, 2001).]
Dave Vick
Ranch Hand
Posts: 3244
Greg
Someone had a similar question the other day and it was suggested that they go look for check writing programs or source code.
------------------
Dave
Sun Certified Programmer for the Java� 2 Platform
Greg Kearney
Greenhorn
Posts: 7
I have missed led you, sorry. I am receiving a string of:
"One hundred fifty five thousand" and I need to return 155000.
Are there any methods that will assist in this effort?
Gregg Bolinger
Ranch Hand
Posts: 15304
6
Greg,
I think they answered that question. There are no exisitng methods that will convert any type of string, "One Hundred", or "One", "Two", etc, to a 100, 1, 2 etc.
There may be a third party vendor that has developed one though. Just search the web..
------------------
Happy Coding,
Gregg Bolinger
Argm Mastoi
Ranch Hand
Posts: 35
Greg u have to work a bit to create this method, and then u can offer others to utilize it too ;-)
Junilu Lacar
Bartender
Posts: 7778
62
Greg,
I suspect this to be a class assignment so all I'll do is give you some hints.
1. Use java.util.StringTokenizer to break down your input string into separate words
2. Use an appropriate class from the Collections framework (see java.util.*) to keep key words and their equivalent types/values
3. Example: one hundred fifty five thousand
inputtype operationresult
onevalue +1
hundredplace marker* 100
fifty value+150
fivevalue+155
thousandplace marker*155000
4. Use polymorphism to do the operations rather than if-then or switch-case statements.
HTH
------------------
Junilu Lacar
Sun Certified Programmer for the Java� 2 Platform
UBB doesn't handle tables very well
[This message has been edited by JUNILU LACAR (edited November 03, 2001).]
Junilu Lacar
Bartender
Posts: 7778
62
Just realized something too:
There are other things that you need to check besides those implied in my previous post. Take for example the input string of "two hundred twenty five thousand three hundred fifteen". If you don't consider other factors, you might end up with 22500315 instead of 225315.
------------------
Junilu Lacar
Sun Certified Programmer for the Java� 2 Platform
[This message has been edited by JUNILU LACAR (edited November 03, 2001).]
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# Mathematics (NTSE Stage-2 (National-Level) Scholastic-Aptitude & Language Comprehension): Questions 36 - 43 of 112
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## Question number: 36
» Mathematics » Number Systems
MCQ▾
Zero is
### Choices
Choice (4) Response
a.
a natural number
b.
a positive number
c.
a whole number
d.
a negative integer
## Question number: 37
» Mathematics » Basic Geometry
MCQ▾
### Question
ABCD is a trapezium in which AB|| CD, if AB = 7 cm, DC = 3 cm and the area of the trapezium is 18 cm2, then the area of BCD is
### Choices
Choice (4) Response
a.
b.
c.
d.
6
## Question number: 38
» Mathematics » Statistics
MCQ▾
### Question
A card is drawn at random from a pack of 52 playing cards. The probability that drawn card being a red king is
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 39
» Mathematics » Basic Geometry
MCQ▾
### Question
Volume of a cube is. The length of its diagonal is
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 40
» Mathematics » Mensuration
MCQ▾
### Question
The diagonal of a square field measures 50 m. The area of square field is
### Choices
Choice (4) Response
a.
1205 m2
b.
1250 m2
c.
1200 m2
d.
None of the above
## Question number: 41
» Mathematics » Quadratic Equations
MCQ▾
### Question
Sum of roots is -1 and sum of their reciprocal is , then equation is
### Choices
Choice (4) Response
a.
x2 + x - 6 = 0
b.
x2 - 6x + 1 = 0
c.
6x2 + x + 1 = 0
d.
x2 - x + 6 = 0
## Question number: 42
» Mathematics » Statistics
MCQ▾
### Question
The median of the following data is
2,4, 6,8, 10,12,14
### Choices
Choice (4) Response
a.
8
b.
9.5
c.
10
d.
6
## Question number: 43
» Mathematics » Arithmetic
MCQ▾
### Question
The total cost of 12 pens and 5 pencils is equal to Rs. 111. Also, the cost of 1 pencil is Rs. 5 less than the cost of 1 pen. What will be the cost of 8 pens and 9 pencils?
### Choices
Choice (4) Response
a.
Rs. 91
b.
Rs. 89
c.
Rs. 98
d.
Rs. 97
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# social studies
If a bow with a draw weight of 60 kilograms were to fire at someone 5 meters away but the user had to roll first before drawing back and firing two arrows. Would the target have enough time to dive for cover behind a rock?
1. 👍 0
2. 👎 0
3. 👁 157
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# Hierarchical Risk Parity: Introducing Graph Theory and Machine Learning in Portfolio Optimizer
In this short post, I will introduce the Hierarchical Risk Parity portfolio optimization algorithm, initially described by Marcos Lopez de Prado1, and recently implemented in Portfolio Optimizer.
I will not go into the details of this algorithm, though, but simply describe some of its general ideas together with their associated implementation tweaks in Portfolio Optimizer.
### Hierarchical risk parity algorithm overview
#### Step 1 - Hierarchical clustering of the assets
The first step of the Hierarchical Risk Parity algorithm consists in using a hierarchical clustering algorithm to group similar assets together based on their pairwise correlations.
In the original paper, a single linkage clustering algorithm is used1.
In Portfolio Optimizer, 4 hierarchical clustering algorithms are supported:
A good summary of the pros and cons of each of these algorithms can be found in the dissertation of Jochen Papenbrock2.
That being said, practical performances is what ultimately matters, and there seem to be no agreement in the literature or in the industry about which algorithm to choose.
For example, the performances of the Hierarchical Risk Parity algorithm are reported as deteriorating when single linkage is NOT used3, but Ward’s linkage is used in the computation of the institutional index FIVE Robust Multi-Asset Index
Anyway, as an illustration of this first step, the following assets correlation matrix taken from a Munich Re4 research note:
is transformed into the following hierarchical clustering tree, visualized as a dendrogram:
#### Step 2 - Quasi-diagonalization of the assets correlation matrix
The second step of the Hierarchical Risk Parity algorithm consists in reordering the rows and the columns of the assets correlation matrix according to the order of the leaves of the hierarchical clustering tree computed in the first step5.
Because similar assets are grouped together in the first step, this permutation of the assets correlation matrix tends to reveal a block diagonal structure, hence the name of this second step.
One possible problem with this step is that the order of the leaves of a hierarchical clustering tree is not unique.
Worse, for $n$ assets, there are actually $2^{n-1}$ such orders6!
In Portfolio Optimizer, 2 leaf ordering methods are supported:
• (Default) Leaf ordering following clusters heights, identical to what is implemented in the R function hclust
• Leaf ordering minimizing the distance between adjacent leaves6
The second method is the most appropriate w.r.t. the aim of this second step, but has a running time of one or two orders of magnitude higher than the computation of the hierarchical clustering tree itself, which makes it very costly!
For a small universe of assets ($n \leq 100$), though, the impact on performances is minimal.
As an illustration of this second step, the assets correlation matrix of Figure 1 becomes, with the default leaf ordering method:
and becomes, with the optimal leaf ordering method:
In both Figure 3 and Figure 4, the block diagonal structure hidden in the initial assets correlation matrix now clearly appears.
#### Step 3 - Recursive bisection and assets weights computation
The third step of the Hierarchical Risk Parity algorithm consists in recursively bisecting the reordered assets correlation matrix computed in the second step and computing assets weights in inverse proportion to both their variance and their clusters variance.
In the original paper, Marcos Lopez de Prado suggests that constraints on assets weights might be incorporated to this step1.
This is what is done in Portfolio Optimizer, which supports minimum and maximum assets weights constraints7 thanks to a variation of the method described in Pfitzinger and Katzke8.
### Hierarchical risk parity algorithm usage with Portfolio Optimizer
As a quick practical example of Portfolio Optimizer usage, I propose to reproduce the example provided as Python code in the original paper1.
For this, it suffices to call the Portfolio Optimizer endpoint /portfolio/optimization/hierarchical-risk-parity with the following input data:
This results in the following output, which matches with the expected assets weights9:
{
"assetsWeights": [0.06999366422693354, 0.07592150585740284, 0.10838947598985076, 0.19029103653662804, 0.09719886787191244, 0.10191545037929818, 0.06618867657198627, 0.09095933462021928, 0.07123881246563615, 0.1279031754801325]
}
### Last words
I hope that you will find this new portfolio optimization algorithm useful.
As usual, feel free to reach out if you need any algorithm to be implemented, or if you would like support in integrating Portfolio Optimizer!
1. Such a procedure is called matrix seriation
2. This is in contrast to some other closed-source implementations of the Hierarchical Risk Parity algorithm, in which such constraints are managed by a post-process.
3. Adaptations to the Python code provided in the original paper1 are required in order to execute it in Python 3, and the computed covariance matrix is not the same between Python 2 and Python 3!
Tags:
Updated:
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• #### Graphing Polynomial Functions - Problem 3
##### Math›Precalculus›Polynomial and Rational Functions
How we graph a factored quintic (5th degree polynomial) function.
• #### Find an Equation of the Polynomial Function - Problem 2
##### Math›Precalculus›Polynomial and Rational Functions
How we identify the equation of a quintic function when we are given the intercepts of its graph.
• #### Graphing Polynomial Functions with Repeated Factors - Problem 3
##### Math›Precalculus›Polynomial and Rational Functions
How we graph a quintic (5th degree polynomial) function with two repeated factors.
• #### Find an Equation of the Polynomial Function - Problem 3
##### Math›Precalculus›Polynomial and Rational Functions
How we identify the equation of a quintic function when we are given the x-intercepts and one other point on the graph.
• #### Finding Zeros of a Polynomial Function - Problem 3
##### Math›Precalculus›Polynomial and Rational Functions
How to find the zeros of a quintic (5th degree polynomial) function.
• #### Using the Conjugate Zeros Theorem - Problem 3
##### Math›Precalculus›Polynomial and Rational Functions
How we find the zeros of a quintic (5th degree polynomial) function when two imaginary zeros are known.
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# Decimal to Binary
In this page we will learn about decimal to binary conversion.Binary number is number with base 2 and decimal number is a number with base 10.This conversion is very important because in computers the data is stored in binary format.and the user enters the number in decimal format so system will convert decimal to binary and store it.aLet us see what is a binary number and decimal number and conversion of decimal to binary.
Is this topic Binary Division hard for you? Watch out for my coming posts.
Binary numbers:
Binaryrefers to characterized by or consisting of two parts or components. The word binary denotes two (sections or subjects or values). The binary value in mathematics is having only two values 0s and 1s. The number which is expressed only in 0s and 1s are called binary numbers. The binary numbers are represented as that of decimal numbers. The numbers are placed left or right of the point. The numbers may be greater than one or less than one.
Decimal numbers:
Decimal number is a number which is formed with combination of two parts.one part is called integer part and other part is calledfractional part.part which exist left of decimal point is called integer part and part right of decimal point is called fractional part.for example. 5.36
here 5="integer" part
36="fractional" part
## How to convert decimal to binary:
step 1:Divide given number starting from 2 as suitable.
step 2:write remainder on the right side of quotient.
step 3:divide untill quotient will be 0.
step 4:now write binary number starting from lower end of that divison.
step 5:now write given number including quotient of lower end.that should be starting point.
## Convert decimal to binary
Below are the problems based on how to convert decimal to binary -
Problem 1:
Convert 35 into binary number.
Solution:
The binary number can be calculated by using L division method
2 | 35
-------
2 | 17 -- 1
-------
2 | 8 -- 1
-------
2 | 4 --0
-------
2 | 2 -- 0
--------
1 -- 0
The binary number of 35 is 1000112.
Problem 2:
Convert 65 into binary number.
Solution:
2 | 65
-------
2 | 32 -- 1
-------
2 |16 -- 0
-------
2 | 8 --0
-------
2 | 4 -- 0
--------
2 | 2 -- 0="1....0
So binary number of 65 is 1000001
Problem 3:
Convert 142 into decimal numbers.
Solution:
2 | 142
-------
2 | 71 -- 0
-------
2 | 35 -- 1
-------
2 | 17 -- 1
-------
2 | 8 -- 1
--------
2 | 4 -- 0
---------
2 | 2 -- 0
----------
1 -- 0
Fromthe above conversion, we divide the given decimal number by the base (2) and get the remainder. Then the remainders all are considered from bottom to top to get the binary value of the decimal number.This way, wecan convert the decimal to binary.
The binary number of 142 is 100011102
I am planning to write more post on samacheer kalvi 10th maths book and Convert a Decimal to a Fraction. Keep checking my blog.
## Practice problem
Problem 1:
Convert the following decimal number into binary number: 56
Solution: 111000
Problem 2:
Convert the following decimal number into binary number: 32
Solution: 100000
Problem 3:
Convert the following decimal number into binary number: 78
Solution: 1001110
Problem 4:
Convert the following decimal number into binary number: 158
Solution: 10011110
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# Documentation
Mathlib.Order.Sublattice
# Sublattices #
This file defines sublattices.
## TODO #
Subsemilattices, if people care about them.
## Tags #
sublattice
structure Sublattice (α : Type u_2) [] :
Type u_2
A sublattice of a lattice is a set containing the suprema and infima of any of its elements.
• carrier : Set α
The underlying set of a sublattice. Do not use directly. Instead, use the coercion Sublattice α → Set α, which Lean should automatically insert for you in most cases.
• supClosed' : SupClosed s.carrier
• infClosed' : InfClosed s.carrier
Instances For
instance Sublattice.instSetLike {α : Type u_2} [] :
SetLike () α
Equations
• One or more equations did not get rendered due to their size.
@[inline, reducible]
abbrev Sublattice.ofIsSublattice {α : Type u_2} [] (s : Set α) (hs : ) :
Turn a set closed under supremum and infimum into a sublattice.
Equations
• = { carrier := s, supClosed' := (_ : ), infClosed' := (_ : ) }
Instances For
theorem Sublattice.coe_inj {α : Type u_2} [] {L : } {M : } :
L = M L = M
@[simp]
theorem Sublattice.supClosed {α : Type u_2} [] (L : ) :
@[simp]
theorem Sublattice.infClosed {α : Type u_2} [] (L : ) :
@[simp]
theorem Sublattice.isSublattice {α : Type u_2} [] (L : ) :
@[simp]
theorem Sublattice.mem_carrier {α : Type u_2} [] {L : } {a : α} :
a L.carrier a L
@[simp]
theorem Sublattice.mem_mk {α : Type u_2} [] {s : Set α} {a : α} (h_sup : ) (h_inf : ) :
a { carrier := s, supClosed' := h_sup, infClosed' := h_inf } a s
@[simp]
theorem Sublattice.coe_mk {α : Type u_2} [] {s : Set α} (h_sup : ) (h_inf : ) :
{ carrier := s, supClosed' := h_sup, infClosed' := h_inf } = s
@[simp]
theorem Sublattice.mk_le_mk {α : Type u_2} [] {s : Set α} {t : Set α} (hs_sup : ) (hs_inf : ) (ht_sup : ) (ht_inf : ) :
{ carrier := s, supClosed' := hs_sup, infClosed' := hs_inf } { carrier := t, supClosed' := ht_sup, infClosed' := ht_inf } s t
@[simp]
theorem Sublattice.mk_lt_mk {α : Type u_2} [] {s : Set α} {t : Set α} (hs_sup : ) (hs_inf : ) (ht_sup : ) (ht_inf : ) :
{ carrier := s, supClosed' := hs_sup, infClosed' := hs_inf } < { carrier := t, supClosed' := ht_sup, infClosed' := ht_inf } s t
def Sublattice.copy {α : Type u_2} [] (L : ) (s : Set α) (hs : s = L) :
Copy of a sublattice with a new carrier equal to the old one. Useful to fix definitional equalities.
Equations
• Sublattice.copy L s hs = { carrier := s, supClosed' := (_ : ), infClosed' := (_ : ) }
Instances For
@[simp]
theorem Sublattice.coe_copy {α : Type u_2} [] (L : ) (s : Set α) (hs : s = L) :
(Sublattice.copy L s hs) = s
theorem Sublattice.copy_eq {α : Type u_2} [] (L : ) (s : Set α) (hs : s = L) :
theorem Sublattice.ext {α : Type u_2} [] {L : } {M : } :
(∀ (a : α), a L a M)L = M
Two sublattices are equal if they have the same elements.
instance Sublattice.instSupCoe {α : Type u_2} [] {L : } :
Sup L
A sublattice of a lattice inherits a supremum.
Equations
• Sublattice.instSupCoe = { sup := fun (a b : L) => { val := a b, property := (_ : a b L) } }
instance Sublattice.instInfCoe {α : Type u_2} [] {L : } :
Inf L
A sublattice of a lattice inherits an infimum.
Equations
• Sublattice.instInfCoe = { inf := fun (a b : L) => { val := a b, property := (_ : a b L) } }
@[simp]
theorem Sublattice.coe_sup {α : Type u_2} [] {L : } (a : L) (b : L) :
(a b) = a b
@[simp]
theorem Sublattice.coe_inf {α : Type u_2} [] {L : } (a : L) (b : L) :
(a b) = a b
@[simp]
theorem Sublattice.mk_sup_mk {α : Type u_2} [] {L : } (a : α) (b : α) (ha : a L) (hb : b L) :
{ val := a, property := ha } { val := b, property := hb } = { val := a b, property := (_ : a b L) }
@[simp]
theorem Sublattice.mk_inf_mk {α : Type u_2} [] {L : } (a : α) (b : α) (ha : a L) (hb : b L) :
{ val := a, property := ha } { val := b, property := hb } = { val := a b, property := (_ : a b L) }
instance Sublattice.instLatticeCoe {α : Type u_2} [] (L : ) :
Lattice L
A sublattice of a lattice inherits a lattice structure.
Equations
• One or more equations did not get rendered due to their size.
instance Sublattice.instDistribLatticeCoe {α : Type u_5} [] (L : ) :
A sublattice of a distributive lattice inherits a distributive lattice structure.
Equations
• One or more equations did not get rendered due to their size.
def Sublattice.subtype {α : Type u_2} [] (L : ) :
LatticeHom (L) α
The natural lattice hom from a sublattice to the original lattice.
Equations
• One or more equations did not get rendered due to their size.
Instances For
@[simp]
theorem Sublattice.coe_subtype {α : Type u_2} [] (L : ) :
= Subtype.val
theorem Sublattice.subtype_apply {α : Type u_2} [] (L : ) (a : L) :
a = a
theorem Sublattice.subtype_injective {α : Type u_2} [] (L : ) :
def Sublattice.inclusion {α : Type u_2} [] {L : } {M : } (h : L M) :
LatticeHom L M
The inclusion homomorphism from a sublattice L to a bigger sublattice M.
Equations
• One or more equations did not get rendered due to their size.
Instances For
@[simp]
theorem Sublattice.coe_inclusion {α : Type u_2} [] {L : } {M : } (h : L M) :
theorem Sublattice.inclusion_apply {α : Type u_2} [] {L : } {M : } (h : L M) (a : L) :
a =
theorem Sublattice.inclusion_injective {α : Type u_2} [] {L : } {M : } (h : L M) :
@[simp]
theorem Sublattice.inclusion_rfl {α : Type u_2} [] (L : ) :
@[simp]
theorem Sublattice.subtype_comp_inclusion {α : Type u_2} [] {L : } {M : } (h : L M) :
instance Sublattice.instTop {α : Type u_2} [] :
Top ()
The maximum sublattice of a lattice.
Equations
• Sublattice.instTop = { top := { carrier := Set.univ, supClosed' := (_ : SupClosed Set.univ), infClosed' := (_ : InfClosed Set.univ) } }
instance Sublattice.instBot {α : Type u_2} [] :
Bot ()
The empty sublattice of a lattice.
Equations
• Sublattice.instBot = { bot := { carrier := , supClosed' := (_ : ), infClosed' := (_ : ) } }
instance Sublattice.instInf {α : Type u_2} [] :
Inf ()
The inf of two sublattices is their intersection.
Equations
• Sublattice.instInf = { inf := fun (L M : ) => { carrier := L M, supClosed' := (_ : SupClosed (L M)), infClosed' := (_ : InfClosed (L M)) } }
instance Sublattice.instInfSet {α : Type u_2} [] :
The inf of sublattices is their intersection.
Equations
• One or more equations did not get rendered due to their size.
instance Sublattice.instInhabited {α : Type u_2} [] :
Equations
• Sublattice.instInhabited = { default := }
def Sublattice.topEquiv {α : Type u_2} [] :
≃o α
The top sublattice is isomorphic to the lattice.
This is the sublattice version of Equiv.Set.univ α.
Equations
• Sublattice.topEquiv = { toEquiv := , map_rel_iff' := (_ : ∀ {a b : }, () a () b () a () b) }
Instances For
@[simp]
theorem Sublattice.coe_top {α : Type u_2} [] :
= Set.univ
@[simp]
theorem Sublattice.coe_bot {α : Type u_2} [] :
=
@[simp]
theorem Sublattice.coe_inf' {α : Type u_2} [] (L : ) (M : ) :
(L M) = L M
@[simp]
theorem Sublattice.coe_sInf {α : Type u_2} [] (S : Set ()) :
(sInf S) = ⋂ L ∈ S, L
@[simp]
theorem Sublattice.coe_iInf {ι : Sort u_1} {α : Type u_2} [] (f : ι) :
(⨅ (i : ι), f i) = ⋂ (i : ι), (f i)
@[simp]
theorem Sublattice.coe_eq_univ {α : Type u_2} [] {L : } :
L = Set.univ L =
@[simp]
theorem Sublattice.coe_eq_empty {α : Type u_2} [] {L : } :
L = L =
@[simp]
theorem Sublattice.not_mem_bot {α : Type u_2} [] (a : α) :
a
@[simp]
theorem Sublattice.mem_top {α : Type u_2} [] (a : α) :
@[simp]
theorem Sublattice.mem_inf {α : Type u_2} [] {L : } {M : } {a : α} :
a L M a L a M
@[simp]
theorem Sublattice.mem_sInf {α : Type u_2} [] {a : α} {S : Set ()} :
a sInf S LS, a L
@[simp]
theorem Sublattice.mem_iInf {ι : Sort u_1} {α : Type u_2} [] {a : α} {f : ι} :
a ⨅ (i : ι), f i ∀ (i : ι), a f i
instance Sublattice.instCompleteLattice {α : Type u_2} [] :
Sublattices of a lattice form a complete lattice.
Equations
• One or more equations did not get rendered due to their size.
theorem Sublattice.subsingleton_iff {α : Type u_2} [] :
instance Sublattice.instUniqueSublattice {α : Type u_2} [] [] :
Equations
• Sublattice.instUniqueSublattice = { toInhabited := Sublattice.instInhabited, uniq := (_ : ∀ (x : ), x = default) }
def Sublattice.comap {α : Type u_2} {β : Type u_3} [] [] (f : ) (L : ) :
The preimage of a sublattice along a lattice homomorphism.
Equations
Instances For
@[simp]
theorem Sublattice.coe_comap {α : Type u_2} {β : Type u_3} [] [] (L : ) (f : ) :
() = f ⁻¹' L
@[simp]
theorem Sublattice.mem_comap {α : Type u_2} {β : Type u_3} [] [] {f : } {a : α} {L : } :
a f a L
theorem Sublattice.comap_mono {α : Type u_2} {β : Type u_3} [] [] {f : } :
@[simp]
theorem Sublattice.comap_id {α : Type u_2} [] (L : ) :
= L
@[simp]
theorem Sublattice.comap_comap {α : Type u_2} {β : Type u_3} {γ : Type u_4} [] [] [] (L : ) (g : ) (f : ) :
=
def Sublattice.map {α : Type u_2} {β : Type u_3} [] [] (f : ) (L : ) :
The image of a sublattice along a monoid homomorphism is a sublattice.
Equations
Instances For
@[simp]
theorem Sublattice.coe_map {α : Type u_2} {β : Type u_3} [] [] (f : ) (L : ) :
() = f '' L
@[simp]
theorem Sublattice.mem_map {α : Type u_2} {β : Type u_3} [] [] {L : } {f : } {b : β} :
b ∃ a ∈ L, f a = b
theorem Sublattice.mem_map_of_mem {α : Type u_2} {β : Type u_3} [] [] {L : } (f : ) {a : α} :
a Lf a
theorem Sublattice.apply_coe_mem_map {α : Type u_2} {β : Type u_3} [] [] {L : } (f : ) (a : L) :
f a
theorem Sublattice.map_mono {α : Type u_2} {β : Type u_3} [] [] {f : } :
@[simp]
theorem Sublattice.map_id {α : Type u_2} [] {L : } :
= L
@[simp]
theorem Sublattice.map_map {α : Type u_2} {β : Type u_3} {γ : Type u_4} [] [] [] {L : } (g : ) (f : ) :
theorem Sublattice.mem_map_equiv {α : Type u_2} {β : Type u_3} [] [] {L : } {f : α ≃o β} {a : β} :
a Sublattice.map { toSupHom := { toFun := f, map_sup' := (_ : ∀ (a b : α), f (a b) = f a f b) }, map_inf' := (_ : ∀ (a b : α), f (a b) = f a f b) } L () a L
theorem Sublattice.apply_mem_map_iff {α : Type u_2} {β : Type u_3} [] [] {L : } {f : } {a : α} (hf : ) :
f a a L
theorem Sublattice.map_equiv_eq_comap_symm {α : Type u_2} {β : Type u_3} [] [] (f : α ≃o β) (L : ) :
Sublattice.map { toSupHom := { toFun := f, map_sup' := (_ : ∀ (a b : α), f (a b) = f a f b) }, map_inf' := (_ : ∀ (a b : α), f (a b) = f a f b) } L = Sublattice.comap { toSupHom := { toFun := (), map_sup' := (_ : ∀ (a b : β), () (a b) = () a () b) }, map_inf' := (_ : ∀ (a b : β), () (a b) = () a () b) } L
theorem Sublattice.comap_equiv_eq_map_symm {α : Type u_2} {β : Type u_3} [] [] (f : β ≃o α) (L : ) :
Sublattice.comap { toSupHom := { toFun := f, map_sup' := (_ : ∀ (a b : β), f (a b) = f a f b) }, map_inf' := (_ : ∀ (a b : β), f (a b) = f a f b) } L = Sublattice.map { toSupHom := { toFun := (), map_sup' := (_ : ∀ (a b : α), () (a b) = () a () b) }, map_inf' := (_ : ∀ (a b : α), () (a b) = () a () b) } L
theorem Sublattice.map_symm_eq_iff_eq_map {α : Type u_2} {β : Type u_3} [] [] {L : } {M : } {e : β ≃o α} :
Sublattice.map { toSupHom := { toFun := (), map_sup' := (_ : ∀ (a b : α), () (a b) = () a () b) }, map_inf' := (_ : ∀ (a b : α), () (a b) = () a () b) } L = M L = Sublattice.map { toSupHom := { toFun := e, map_sup' := (_ : ∀ (a b : β), e (a b) = e a e b) }, map_inf' := (_ : ∀ (a b : β), e (a b) = e a e b) } M
theorem Sublattice.map_le_iff_le_comap {α : Type u_2} {β : Type u_3} [] [] {L : } {f : } {M : } :
M L
theorem Sublattice.gc_map_comap {α : Type u_2} {β : Type u_3} [] [] (f : ) :
@[simp]
theorem Sublattice.map_bot {α : Type u_2} {β : Type u_3} [] [] (f : ) :
theorem Sublattice.map_sup {α : Type u_2} {β : Type u_3} [] [] (f : ) (L : ) (M : ) :
theorem Sublattice.map_iSup {ι : Sort u_1} {α : Type u_2} {β : Type u_3} [] [] (f : ) (L : ι) :
Sublattice.map f (⨆ (i : ι), L i) = ⨆ (i : ι), Sublattice.map f (L i)
@[simp]
theorem Sublattice.comap_top {α : Type u_2} {β : Type u_3} [] [] (f : ) :
theorem Sublattice.comap_inf {α : Type u_2} {β : Type u_3} [] [] (L : ) (M : ) (f : ) :
theorem Sublattice.comap_iInf {ι : Sort u_1} {α : Type u_2} {β : Type u_3} [] [] (f : ) (s : ι) :
Sublattice.comap f (iInf s) = ⨅ (i : ι), Sublattice.comap f (s i)
theorem Sublattice.map_inf_le {α : Type u_2} {β : Type u_3} [] [] (L : ) (M : ) (f : ) :
theorem Sublattice.le_comap_sup {α : Type u_2} {β : Type u_3} [] [] (L : ) (M : ) (f : ) :
theorem Sublattice.le_comap_iSup {ι : Sort u_1} {α : Type u_2} {β : Type u_3} [] [] (f : ) (L : ι) :
⨆ (i : ι), Sublattice.comap f (L i) Sublattice.comap f (⨆ (i : ι), L i)
theorem Sublattice.map_inf {α : Type u_2} {β : Type u_3} [] [] (L : ) (M : ) (f : ) (hf : ) :
theorem Sublattice.map_top {α : Type u_2} {β : Type u_3} [] [] (f : ) (h : ) :
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# Owl Themed Multiplication Facts Cards
Subjects
Resource Types
Common Core Standards
Product Rating
4.0
File Type
PDF (Acrobat) Document File
12.07 MB | 59 pages
### PRODUCT DESCRIPTION
Click below to find the multiplication and division bundled set.
"Owl" Help with Your Multiplication and Division Facts {But First, Let Me Take a Selfie!}
Click below to find a matching set of just the division cards.
"Owl" Help with Your Division Facts {But First, Let Me Take a Selfie!}
***A complete set with less color is included at the end of the file!***
I wanted to keep these multiplication flash cards simple so the students could really focus on the facts they need to learn. Included is one of each basic fact card, 2 through 9, just 36 facts to learn in all! I explain to the students these are the facts they need to focus on since they all know their 0’s and 1’s, and I’ve taught the Commutative Property of Multiplication as well. I plan to have students practice with each other or with parent volunteers daily to help improve math fact fluency.
The Known Basic Facts list can be used to simply highlight the facts the student knows, and the student can continue to study the unhighlighted ones. Students can work in pairs or independently, using the Answer Sheet. Camera answer cards are included if you would like to make this a matching game.
It is so important for the students to know their facts with automaticity for success with the Common Core Math Standards! Simply print, cut, and laminate for years of use!
***I have also included a set of 0, 1, 10, 11, and 12 with matching camera cards if you would like to use those as well.
Total Pages
59
Included
Teaching Duration
1 Year
4.0
Overall Quality:
4.0
Accuracy:
4.0
Practicality:
4.0
Thoroughness:
4.0
Creativity:
4.0
Clarity:
4.0
Total:
17 ratings
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iProject: #production
For this part, it is really hard because I could really find the right time to actually start making it. I didn’t really decide on what certain thing that I was going to do, so I just randomly chose them. Β To the QUESTIONS…
What are some of the successes that you have had in producing your project?
The successes that I have probably produced are when I bought the things and also making the items. I just simply bought the stuff online, but I couldn’t find some eyes for it. I never actually got to get the eyes, but I just figured that I could just use black wool to do for the eyes. The making is pretty simple really, so it isn’t really that hard to do.
What have been some challenges that you have had to overcome during the process?
The challenges that I have had to overcome are probably when I could find time to actually do it. My parents thought that my project was useless with actual work, so I had to try and find some time to actually do it without them knowing. Also, videoing was another problem too. If I ever find time, I always just remember that I need a camera and so then I couldn’t do it… cause I didn’t have a camera with me at those times… π
How did you overcome the challenges?
I actually never got to overcome my challenges, I only just could secretly do it without them knowing, which was really hard. This is also why my project wouldn’t have that many things to show on my Expo day.
This project has been fun, but yet hard to catch up to… πππππ
Reflective Post: Term 4 Week 8
MATHS:
Maths Groups- In Maths Groups, we have been focusing on this new topic called ‘My Life In Numbers’. It is where we have to first think of anything in our life that has numbers in it like our age and stuff like that. After that, we had to choose our own ideas to use to make an equation out of it. We can choose what to use like indices, BODMAS, anything pretty much to make the number that we chose. A few of my ideas are how wide my feet are and how long my pencil case is compared to Astrid’s.
Project Maths- In Project Maths, we have been continuing the project Create your own Garden, something like that. I still am buying stuff, but I just need to buy some plants and some animals too. I then noticed that I still had not placed in all of the objects into my drawn out backyard. I then had to search up everything again and place them in my yard.Β
ENGLISH:
In English, I have finished off Toothpaste my piece and now I just need to get some feedback off it. It is really hard to find some people to read it because it is 2 pages long… and usually we only use one page. I have so far gotten feedback from Astrid and she said that it was really good. All I need is to have two more people to give me some feedback and then I can see if I can improve on it or not.
INQUIRY:
In Inquiry, we have finished off our Personal Project film and my group has already rendered it. We also got to look at everyone else’s work and it was very interesting on what the other groups had chosen to do . Our’s weren’t up there because Bailey was away on that day and she had the film on her laptop. I found that the other film were quite inspiring and a few were actually quite helpful for next year’s students who were going to move up to Year 6.
It’s nearly the end of the year already!!!!!π
Reflective Post: Term 4 Week 7
MATHS:
Project Maths- In Project Maths, we have started a new project which had to do with our backyard and re- decorating it. We had a limit of \$10, 000 to spend, so we couldn’t spend over that. So far, I haven’t used much money because I have been trying to save some up. I actually don’t know why I want to save it up, but it’s just me and how I think how to use money. I have so far only bought stools and sheds and a few more plants to start off my backyard.Β
ENGLISH:
For English, I have finished publishing Blue and I have pinned it in the Twilight Zone. You can’t really read it because I made it quite faint. I am now continuing my piece Toothpaste and I have also conferenced it with Mr Henderson. This piece is one of my longest pieces that I have every made and it is one and a half pages long. He helped me with the tenses and how to end the story quickly because I had no idea on how to get it to finish quickly.
INQUIRY:
In Inquiry, my group have finished our Stop Motion recording and are now just needing to record the voice overs. The challenges that came by were when we had to restart the first scene because the camera got moved. The lighting changed a lot a bit in the video, but that was okay because we didn’t really want to keep on retaking the whole scenes over and over again. The voice overs are going well and it is very fun to hear what we have to say and what tone we have to use.
This week has been a really fun week and also quite busy too!ππππππππ
Reflective Post: Term 4 Week 6
(In this post, there won’t be much to say since there was swimming for the whole week, so this post may be set up differently. Thank you for your patience.)
Inquiry:
In Inquiry, we have been continuing our topic on the stop motion animation film. Since we have already finished our shotlist, my group had to make the background and the characters. I finished making the Red Ghost Guy and also I have made the cart where White Ghost Guy’s remains will go in when the baby finds it. Bailey has made a raibow and CC has made some background parts!
ENGLISH:
In English, I have started a story called Extinction, where Extinction is an actual alive thing which had to help of humans to get rid of the dodos. I still have to continue Radio and 7 other seeds and also continue publishing my first piece Blue.
PROBLEM SOLVING BOOKS HAVE BEEN HANDED IN TO THE TEACHER TO MARK.
Swimming was really fun and tiring, especially when we had to do lots of laps. Today, we got to do some fun relays! It was really fun and really interesting!
Reflective Post: Term 4 Week 5
MATHS:
PROJECT MATHS– In Project Maths, we have been continuing our topic Algebra. Today, we started this sheet of paper with some more Algebra questions. The way that I did it was like this:
4x-10=3x-5
4x-3x= 10-5
x=5
(I made that one up by myself.π) Miss Torney had a different way of doing it, but I didn’t really understand it as much as the one that I use right now.
PROBLEM SOLVING– In Problem Solving, we shared our answers with others. I noticed that for one of the questions, I actually interpreted it in a really different way. I actually didn’t really get what the question really wanted, so I just did what I thought it wanted. My other answer was okay and there were also other answers to it too. The next THREE (Yes, THREE!) questions that we got were quite different. One of them were these two people talking about ages, the second one had letters in it, and the third one had cubes and dates in it.
Next week, we are going to hand in our booklets for submission.π£Β
ENGLISH:
In Writer’s Workshop, I have started publishing my piece Blue! I have only done the title though. I have also continued my seed Radio and No Title. Other than those, I also went to a Masterclass Guru Session with Mr Panckridge! The topic for English that he was talking about was how to think up of ideas. He told us what he did as an author and how he could get ideas. After that, we had to think up of our own creative ideas. He also said that we had to always have a twist in our story to make it interesting.Β
INQUIRY:
In Inquiry, since we had finished our Zoo Film, we were given the last project that we had to do for the rest of the year. It is called a Personal Project. We get into groups or by ourselves and then we choose what we want to do that we did this year. For example, my group, Bailey, CC and I are doing a stop motion film with a message in it. Our message was Kindness. We have already brainstormed our ideas and also have already made the story up. We all are looking forward to starting the whole project process!π
I am really enjoying the last term of my Primary School life!!!!!π
iProject: #planning
All through these weeks, I have been thinking about my iProject a lot. I have decided that if I have time, then I would add in another part to it, which involves zips. Anyhow, back to more QUESTIONS…
What have you needed to do to prepare for your project?
I have needed to buy the things that I would be needing to use and also to prepare what I have decided to make and what I am not making. I also need to prepare the times of when I can do it and when I can’t. That is the tricky bit because I can’t really find a spot other than the weekends. I will just have to confirm it with my parents.
What types of shots have you included in your shotlist? Why?
The shot that I am using the most is ‘Close Up’ because I have to make sure that the people watching are seeing what I am going to explain and what I am doing. The reason why I don’t want to put in my face is because I just want the audience to watch what I am doing with my hands and not what my face is like.
What is the next step for you?
The next step for me is to MAKE MAKE MAKE and also video while I am making it. I feel like I haven’t done enough for the past few projects and I just feel like this one has to be the best out of the last two ones. I also find that when I start making, I just keep on going and going. That’s the good part about this project.Ummm…πΆπΆπΆπΆ
It’s a bit blurry…π
I find that this project will probablyΒ be the most fun out of the other two!!!πππ
Zoo Conservation Film
At the start of the project, we had to choose an animal from the SOS10 website. SOS stands for Save Our Species and there are 10 animals to choose from. I chose the Baw Baw Frog because I had never heard of it before and it also sounded quite unique. When I researched about it, I was quite shocked about how many there were left in the world. So, then I decided to officially choose the Baw Baw Frog.Β
When I looked up some Baw Baw Frog websites, I found out that the Baw Baw Frog wasn’t like any ordinary frog. There were lots of interesting facts about what this amazing frog could do. I took down all the facts that I could find about it and put them all in the script, along with my own words. Finding a P.T.C part in the script was a bit hard for me because I didn’t know which bit needed a P.T.C. At the end, I managed to find one bit for the P.T.C to fit in. This was only the start.π
After that, we had an excursion to the zoo. We were meant to bring our scripts, but sadly, I FORGOT THE SCRIPT.Β My P.T.C sentence was easy to remember, so that was okay. I was put with the Helmeted Honey eaters group and the Philippines Crocodile group. While we were walking, we got lost and we didn’t know where we were going. We missed out on all of the presentations that we needed to go to. I managed to get some footage, but the frog wasn’t doing anything interesting It was just staying still on a branch the whole time. I didn’t do my Peace To Camera there because the cage that it was in was a bit small, and you wouldn’t really see the small frog in the video.π
When I put down all the footage that I used, I found that I didn’t have enough. So, I went on the internet and borrowed some off it. I also found that I had quite a long script for the film. When I did my voice over, I sort of had to rush through the whole script to be able to fit it in and not let it overtake the film of the Baw Baw Frog. I found it hard for some time, but then I got used to it and managed to get the V.O done. A big challenge was that I wasn’t there for the First Film Feedback time so I kind of had to continue on with my film without any feedback from five people.π
Today was when we had to share our Final Film and get some feedback from anyone. Here are some feedback that I got from other students…
This film was soo good. The music complemented the message perfectly and you really made me care. The footage you used also had a positive impact on the information you were giving me. The voice overs were also really clear This is one of the best films i have seen yet-Bailey G
Your voice over was really good and your footage and imagery where really great. I loved how you conveyed their message and the music really suited your topic and message. –Mikayla
That was great, it really made me care about the frog. you did a great job it was really sad and the music really complemented your message. well done-Tilly
That film was so heart breaking! I had no idea what so ever that these frogs are different to ordinary frogs!! I can’t believe the number of frogs there are left in the world! I have a question for you. Do all of the Baw Baw frogs live in zoos for the little amount left of them there are, or are there still some in the wild that people haven’t found yet??-Astrid
I liked Astrid’s question.πΒ Here’s the film that I made:
Reflective Post: Term 4 Week 4
MATHS:
Problem Solving- In Problem Solving, 6A had a debating session before French and we had to check with the others to see if we got the same or similar answers. Lots of people got the answers that I got for both questions (well, the second one had I think a few answers to it).
Project Maths- In Project Maths, we have been doing Algebra and this Ladder Problem. Here is what we have been doing in these topics.
Algebra: In Algebra, we did the sheet of paper which was where we had to find out each of the letter’s number, and then find out what words they would do make near the bottom of the page. To work out what number each letter were, we had to work out these algebra questions so then we could actually solve them. At the start, we were given two answers so then we could start. At the end, I managed to complete the alphabet, but I didn’t get to finish making the words.
ENGLISH:
In English, I can now PUBLISH my Blue piece!!!πΒ Well I could of ages ago, but i wasn’t sure if I was allowed to. I was 95% sure. Now, I have started a piece called ‘Grandpa’s cap’ and am continuing ‘Radio.’ Β I thought of doing a seed called Radio because when I was in the car, I was just pretty much listening to the radio and suddenly, it came up in my mind to do a piece about it. Also, I am still doing the seed ‘1 Hour later,’ just not as much.
INQUIRY:
In Inquiry, we finished off our zoo films for our endangered animals. We had a feedback day, but I was away for that. I did have some challenges that were blocking my way, but I managed to get my way through. The challenges were how to put the sources on because it wasn’t fitting in to the screen. Some of the sources had a few letters that couldn’t fit in the screen properly. The other challenge was that I didn’t know how to do a green screen. But, at the end, Bailey taught me how to and that problem was solved!
This week was a challenging yet fun week to get through!!!ππππππππππππΆ
Reflective Post: Term 4 Week 3
Maths:
Problem Solving-Β In Problem Solving, we had a few Malaysian Students that were over and also were seeing how our school worked. While we were doing our things, they were listening to what we were doing. For the first question, I may have calculated something wrong, because I got a tiny bit smaller answer than the real answer. Whoops!π
Β For my second answer, there were several ways to get the the same total number, which was 27. Next, we got TWO more questions to work on along the week. One was with a coin and the other was with this pentagon shape and you have to put in these numbers and they can’t be one different away.
Project Maths- In Project Maths, we continued the subject Algebra. One lesson, we went on our laptops and worked on the website that Miss Torney sent to us. Then, we had to record down what question we were doing and how we worked it out. The other lesson was that we could choose out of an easier sheet or a harder sheet to work on. I chose the easier sheet to see if I could do it and believe it or not, I couldn’t work it out on paper. It just didn’t work! I had to work it in my head so then it would work out. After that, I did the harder sheet and I never got to finish it…π’π
English:
In Writing, we all had to do two poems in 30 minutes with Mr Wade, one about an apple and the other about a random thing that we want to choose. I chose a banana. the banana piece had rhyming in it and rhythm, but some bits didn’t make any sense. I gave it to Miss Torney, and she also said to change the sentences which didn’t fit in with the whole thing.
I also continued with my other seeds that were said in the previous post like Radio.
Inquiry:
In Inquiry, we continued on with our animal project and there was a big problem that was really affecting my film. I didn’t have enough space to get into it! I had to pretty much delete a whole bunch of things to get back on it. After only adding a few more videos, it started not working again! And again it said that there wasn’t enough space! I yet had to delete more things again. In the film, I am up to doing the voice overs and then adding in the peace to camera shot. Then I would be all done.
What a nice week!!!π
Reflective Posts: Term 4 Week 2
MATHS:
PROBLEM SOLVING- Β Today we got our books back from the teachers and started a new section. We were given two questions (again.) and we had to write them down. These ones were probably some easy ones to warm up our brains to get ready to do other harder ones. Since this is only the start for this section, there isn’t much to say in this.Β
PROJECT MATHS-Β In Project Maths, we started the topic Algebra. The first thing that we did with Algebra was this topic called The Function Machine. It was where you put in a random number in, then it would come out with this way that changes it to a different number. Later on, we partnered up and made up our own to give to other people to figure out. Another topic that we did was where we had to answer questions like this: 19+4= 22+1 (This answer is true…) And then we had to say whether or not it was true or false. Then we again did our own to show to other people and they had to figure out of it was true or false.
ENGLISH:
In Writing, I did a one-on-one conference with Miss Torney and conferenced my piece ‘Blue…’ and she said that she was really happy with it. I had to improve it when I last conferenced it with Mr Henderson, he gave me lots of suggestions to improve it. I then gave it to Mr Henderson to look at. Other than improving some words to make it better, he was quite happy with it too. So, once I change those things and conference it with my group, THEN I WILL BE ABLE TO PUBLISH IT!!!
I also continued my seed 1 Hour Later seed and also the seed Radio.
INQUIRY:
In Inquiry, we went to the Melbourne Zoo to take some photos and videos of our chosen animals. We met so many interesting animals that were so cool. When I got to the Baw Baw Frog, I couldn’t see it because it was camouflaged in a branch. I managed to take videos and photos of it and I had to change my script because of the new information that I got.
I am really looking forward to allllll of these things!π
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https://www.danielvonrhein.nl/r/apply/
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# Apply
``````# Apply function (paste pre) to every element of a list
lapply(c(1:22,'X','Y'), function(pre,post) paste(pre,post, sep =''), pre="chr")
# the same as lapply but output is vector of characters
sapply(c(1:22,'X','Y'), function(pre,post) paste(pre,post, sep =''), pre="chr")
# Get specific list elements from list with arrays
lapply(tt, `[`, i =c(1:3))
# Apply custom function to data set
findRegionIndex <-function (chromosome, position, regions){
return (regions[chr == chromosome & start >= position & stop <= position])
}
apply(df.vaf,1,findRegionIndex, region=df.cov)
``````
# Lapply
``````# Apply function to vector resulting in lists with different length
freqs=list("999", "99,9,0" , "99,93,0")
lapply(freqs, 'str_split', pattern = ',' )
out = sapply(freqs, 'str_split', pattern = ',' )
lapply(lapply(out, as.numeric), sum)
``````
# Map and mapply
Both map and mapply apply element-wise function to lists of parameters. Map is a wrapper for mapply which tries to simplify output
``````# Apply custom function with two parameters as input to every element of a vector
isCT <- function(x, y){
return (x == "C" & y == 'T')
}
mapply(isCT,c("A","C"),c("T", "T"))
``````
# Do call
Do call calls a function with a certain name and feeds it with a
``````
# Apply paste function row-wise to data.table columns
do.call(paste, dt[,2:9])
do.call(str_remove_all, list(string = dt\$text, pattern = ' NA')
``````
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https://askworksheet.com/waves-worksheet-2-amplitude/
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# Waves Worksheet #2 Amplitude
Amplitude height of wave. Measured from the equilibrium position to the top of a crest or the bottom of a trough see vertical arrow wavelength length of a single wave cycle horizontal arrow.
Worksheet With 3 Experiments On Sound Waves That Incorporates The Australian Curriculum For Year 1 Science Sound Science Teaching Sound Sound Waves
### Label each part in the space provided.
Waves worksheet #2 amplitude. The music ends at piano quietly with a sweet melody. The access is granted simply by clicking on the link and downloading it. These exercises refer to the three waves shown below.
Amplitude connection juan is playing the piano. Frequency of waves that pass a point in a given. It begins with an identification of the wave parts then a deeper look at amplitude and frequency.
Key for waves unit 2 worksheet 5 l. Waves intro phet interactive simulations. C how are the wavelengths of the dashed and dotted waves related.
The music ends at piano quietly with a sweet melody. The illustration below shows a series of transverse waves. The music starts of at meso forte medium high volume.
Amplitude wavelength frequency and period 6 exercise 1. X cm 2 y cm 1 2 1 2 0 a what is the amplitude of the solid wave. To get the worksheet click here.
It then crescendos into forte loud and juan plays dramatically. This is a middle school worksheet for waves. Some of the worksheets for this concept are wave speed equation practice problems name key period speed frequency wavelength fourth grade science waves sound and waves work sound waves wave waves lesson physical science wave theory and sound.
Which wave represents the music at the beginning. Amplitude frequency wavelength answers physcial science displaying top 8 worksheets found for this concept. The music starts of at meso forte medium high volume.
It then crescendos into forte loud and juan plays dramatically. To get the answers click here. A metric ruler is needed for this activity to take measurements of several waves one full wave cycle wave train two or more waves amplitude measures the energy of a transverse wave is measured from the equilibrium position to the top of a crest or the bottom of a trough.
Measure distance between 2 identical points on consecutive waves. B what is the wavelength of the solid wave. Which wave represents the music at the beginning wave 6.
I get a lot of people emailing me requesting access. Amplitude connection juan is playing the piano.
Sound Digital Pdf Included Activities Classroom Activities Teaching Science
Https Encrypted Tbn0 Gstatic Com Images Q Tbn 3aand9gctwk9 Uu037nof5pfclu6mc0cdvjcfqjdqgvw Usqp Cau
Elements Compounds And Mixtures Worksheet Crossword Puzzle In 2020 Compounds And Mixtures Elements Compounds And Mixtures Crossword Puzzle
Https Encrypted Tbn0 Gstatic Com Images Q Tbn 3aand9gcrku8f Gkun6hbpkxdgggzd3m5grhlopkebhg Usqp Cau
Five Senses The Sense Of Hearing How The Body Works Biology For Kids Learning Science Cool Science Experiments
Timbre 2 Electronics Basics Physics Sound Waves
Pin By Monica Durapau On Chemistry Anchor Charts Ultrasound Physics Waves Physical Science
Waves Worksheet 2 Answers Pdf Google Drive Worksheets Worksheet Template Answer Keys
What Visibility Results There If A Their Electric Field Vectors Are Para Electric Field Electricity Light Wave
Calculo De Areas Sombreadas
Thedifference Wavelengths Low High Frequencies Wissenschaftler
Teachers Notebook Ellis Island The Unit Little Learners
Google Image Result For Https D1uvxqwmcz8fl1 Cloudfront Net Tes Resources 11455701 8ba35f41 106b 4d05 A732 In 2020 Parts Of A Flower Parts Of A Plant Amazing Flowers
Light And The Modern Atom Physical Science Chemistry Science
Waves Worksheet 2 Answers Pdf Google Drive Worksheets Worksheet Template Answer Keys
Hearing And Sound By Two Tiny Tables Teachers Pay Teachers In 2020 1st Grade Worksheets English Worksheets For Kids Hearing Sounds
Teacher S Pet Sound And Hearing Posters Free Classroom Display Resource Eyfs Ks1 Ks2 Sounds Hear Sound Science 4th Grade Science Fourth Grade Science
These Waves Represent The Different Frequency Levels One Going At A Low Speed One Medium And One High This Also Shows How Freque Waves Teaching Frequencies
Four Seasons Science Video For Kids Kindergaten 1st 2nd Grade In 2020 Science Videos For Kids Science Videos Science Experiments Kids
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https://www.jiskha.com/questions/120943/How-would-you-say-I-went-to-a-family-reunion-in-French-Would-it-be-Je-suis-allee
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# French
How would you say, "I went to a family reunion" in French?
Would it be "Je suis allee a la reunion de ma famille"?
(accents are missing)
Thanks
1. π 0
2. π 0
3. π 45
1. Please see your later post, which I saw first. If you would like to know how to make accent marks with the computer, I'll need to know:
1. Do you have a PC or MAC
2. Do you have Windows or not.
Sra (aka Mme)
1. π 0
2. π 0
posted by SraJMcGin
## Similar Questions
1. ### FRENCH REPOST
How would you say, "I went to a family reunion" in French? Would it be "Je suis allee a la reunion de ma famille"? (accents are missing) Thanks
asked by Anonymous on September 3, 2008
2. ### Math
Q. Hectors extended family was gathering a reunion. There were 8 sets of aunts and uncles. There were twice as many cousins as uncles. There were two grand parents,and than there was Hector's family of five. How many people were
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6. ### algeBRA WORD PROBLEM
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7. ### math
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8. ### Math
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9. ### English (Reunion by John Cheever)
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10. ### French
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https://www.vcalc.com/wiki/MichaelBartmess/Centroid+of+a+Beam+Section
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# Centroid of a Beam Section
Not Reviewed
bary =
Tags:
Rating
ID
MichaelBartmess.Centroid of a Beam Section
UUID
c374e102-ec14-11e6-9770-bc764e2038f2
The Centroid of a Beam Section calculator computes the y-centroid (y being vertical axis of an I-beam's cross-section).
INSTRUCTIONS: Choose units and enter the following:
• (D1) Top width
• (W1) Top thickness
• (W2) Vertical thickness
• (D2) Interior height
• (D3) Bottom width
• (W3) Bottom thickness
y-centroid (y): The calculator return the centroid in millimeters. However, this can be automatically converted to compatible units via the pull-down menu.
#### The Math / Science
The Centroid of a Beam Section calculator computes the y-centroid (y being vertical axis of an I-beam's cross-section). The centroid is the center of mass of the beam's cross-sectional sections. The centroid is used to compute the moment of inertia of the beam, which is used in analysis of the beam's bending stress under weight, beam shear, and beam deflection or other analysis questions.
We split the cross-section into into three segments, each segment having a nice rectangular symmetry. We then calculate the area and y-centroid of each of the three segments and compute the entire centroid as:
bary = sum(A_i*y_i)/(sum A_i)
The three segments are shown in the figure below, where A_i = w_i *D_i:
And the y-centroids of the segments are given as:
• y_1 = w_3 +D_2 + w_1/2
• y_2 = w_3 +D_2/2
• y_3 = w_3/2
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https://netlib.org/lapack/explore-html/df/d8b/group__trsyl_gaeff794eab56241190831454fae7dd6d6.html
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LAPACK 3.12.0 LAPACK: Linear Algebra PACKage
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## ◆ ctrsyl()
subroutine ctrsyl ( character trana, character tranb, integer isgn, integer m, integer n, complex, dimension( lda, * ) a, integer lda, complex, dimension( ldb, * ) b, integer ldb, complex, dimension( ldc, * ) c, integer ldc, real scale, integer info )
CTRSYL
Purpose:
``` CTRSYL solves the complex Sylvester matrix equation:
op(A)*X + X*op(B) = scale*C or
op(A)*X - X*op(B) = scale*C,
where op(A) = A or A**H, and A and B are both upper triangular. A is
M-by-M and B is N-by-N; the right hand side C and the solution X are
M-by-N; and scale is an output scale factor, set <= 1 to avoid
overflow in X.```
Parameters
[in] TRANA ``` TRANA is CHARACTER*1 Specifies the option op(A): = 'N': op(A) = A (No transpose) = 'C': op(A) = A**H (Conjugate transpose)``` [in] TRANB ``` TRANB is CHARACTER*1 Specifies the option op(B): = 'N': op(B) = B (No transpose) = 'C': op(B) = B**H (Conjugate transpose)``` [in] ISGN ``` ISGN is INTEGER Specifies the sign in the equation: = +1: solve op(A)*X + X*op(B) = scale*C = -1: solve op(A)*X - X*op(B) = scale*C``` [in] M ``` M is INTEGER The order of the matrix A, and the number of rows in the matrices X and C. M >= 0.``` [in] N ``` N is INTEGER The order of the matrix B, and the number of columns in the matrices X and C. N >= 0.``` [in] A ``` A is COMPLEX array, dimension (LDA,M) The upper triangular matrix A.``` [in] LDA ``` LDA is INTEGER The leading dimension of the array A. LDA >= max(1,M).``` [in] B ``` B is COMPLEX array, dimension (LDB,N) The upper triangular matrix B.``` [in] LDB ``` LDB is INTEGER The leading dimension of the array B. LDB >= max(1,N).``` [in,out] C ``` C is COMPLEX array, dimension (LDC,N) On entry, the M-by-N right hand side matrix C. On exit, C is overwritten by the solution matrix X.``` [in] LDC ``` LDC is INTEGER The leading dimension of the array C. LDC >= max(1,M)``` [out] SCALE ``` SCALE is REAL The scale factor, scale, set <= 1 to avoid overflow in X.``` [out] INFO ``` INFO is INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value = 1: A and B have common or very close eigenvalues; perturbed values were used to solve the equation (but the matrices A and B are unchanged).```
Definition at line 155 of file ctrsyl.f.
157*
158* -- LAPACK computational routine --
159* -- LAPACK is a software package provided by Univ. of Tennessee, --
160* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
161*
162* .. Scalar Arguments ..
163 CHARACTER TRANA, TRANB
164 INTEGER INFO, ISGN, LDA, LDB, LDC, M, N
165 REAL SCALE
166* ..
167* .. Array Arguments ..
168 COMPLEX A( LDA, * ), B( LDB, * ), C( LDC, * )
169* ..
170*
171* =====================================================================
172*
173* .. Parameters ..
174 REAL ONE
175 parameter( one = 1.0e+0 )
176* ..
177* .. Local Scalars ..
178 LOGICAL NOTRNA, NOTRNB
179 INTEGER J, K, L
180 REAL BIGNUM, DA11, DB, EPS, SCALOC, SGN, SMIN,
181 \$ SMLNUM
182 COMPLEX A11, SUML, SUMR, VEC, X11
183* ..
184* .. Local Arrays ..
185 REAL DUM( 1 )
186* ..
187* .. External Functions ..
188 LOGICAL LSAME
189 REAL CLANGE, SLAMCH
191 EXTERNAL lsame, clange, slamch, cdotc, cdotu, cladiv
192* ..
193* .. External Subroutines ..
194 EXTERNAL csscal, xerbla
195* ..
196* .. Intrinsic Functions ..
197 INTRINSIC abs, aimag, cmplx, conjg, max, min, real
198* ..
199* .. Executable Statements ..
200*
201* Decode and Test input parameters
202*
203 notrna = lsame( trana, 'N' )
204 notrnb = lsame( tranb, 'N' )
205*
206 info = 0
207 IF( .NOT.notrna .AND. .NOT.lsame( trana, 'C' ) ) THEN
208 info = -1
209 ELSE IF( .NOT.notrnb .AND. .NOT.lsame( tranb, 'C' ) ) THEN
210 info = -2
211 ELSE IF( isgn.NE.1 .AND. isgn.NE.-1 ) THEN
212 info = -3
213 ELSE IF( m.LT.0 ) THEN
214 info = -4
215 ELSE IF( n.LT.0 ) THEN
216 info = -5
217 ELSE IF( lda.LT.max( 1, m ) ) THEN
218 info = -7
219 ELSE IF( ldb.LT.max( 1, n ) ) THEN
220 info = -9
221 ELSE IF( ldc.LT.max( 1, m ) ) THEN
222 info = -11
223 END IF
224 IF( info.NE.0 ) THEN
225 CALL xerbla( 'CTRSYL', -info )
226 RETURN
227 END IF
228*
229* Quick return if possible
230*
231 scale = one
232 IF( m.EQ.0 .OR. n.EQ.0 )
233 \$ RETURN
234*
235* Set constants to control overflow
236*
237 eps = slamch( 'P' )
238 smlnum = slamch( 'S' )
239 bignum = one / smlnum
240 smlnum = smlnum*real( m*n ) / eps
241 bignum = one / smlnum
242 smin = max( smlnum, eps*clange( 'M', m, m, a, lda, dum ),
243 \$ eps*clange( 'M', n, n, b, ldb, dum ) )
244 sgn = isgn
245*
246 IF( notrna .AND. notrnb ) THEN
247*
248* Solve A*X + ISGN*X*B = scale*C.
249*
250* The (K,L)th block of X is determined starting from
251* bottom-left corner column by column by
252*
253* A(K,K)*X(K,L) + ISGN*X(K,L)*B(L,L) = C(K,L) - R(K,L)
254*
255* Where
256* M L-1
257* R(K,L) = SUM [A(K,I)*X(I,L)] +ISGN*SUM [X(K,J)*B(J,L)].
258* I=K+1 J=1
259*
260 DO 30 l = 1, n
261 DO 20 k = m, 1, -1
262*
263 suml = cdotu( m-k, a( k, min( k+1, m ) ), lda,
264 \$ c( min( k+1, m ), l ), 1 )
265 sumr = cdotu( l-1, c( k, 1 ), ldc, b( 1, l ), 1 )
266 vec = c( k, l ) - ( suml+sgn*sumr )
267*
268 scaloc = one
269 a11 = a( k, k ) + sgn*b( l, l )
270 da11 = abs( real( a11 ) ) + abs( aimag( a11 ) )
271 IF( da11.LE.smin ) THEN
272 a11 = smin
273 da11 = smin
274 info = 1
275 END IF
276 db = abs( real( vec ) ) + abs( aimag( vec ) )
277 IF( da11.LT.one .AND. db.GT.one ) THEN
278 IF( db.GT.bignum*da11 )
279 \$ scaloc = one / db
280 END IF
281 x11 = cladiv( vec*cmplx( scaloc ), a11 )
282*
283 IF( scaloc.NE.one ) THEN
284 DO 10 j = 1, n
285 CALL csscal( m, scaloc, c( 1, j ), 1 )
286 10 CONTINUE
287 scale = scale*scaloc
288 END IF
289 c( k, l ) = x11
290*
291 20 CONTINUE
292 30 CONTINUE
293*
294 ELSE IF( .NOT.notrna .AND. notrnb ) THEN
295*
296* Solve A**H *X + ISGN*X*B = scale*C.
297*
298* The (K,L)th block of X is determined starting from
299* upper-left corner column by column by
300*
301* A**H(K,K)*X(K,L) + ISGN*X(K,L)*B(L,L) = C(K,L) - R(K,L)
302*
303* Where
304* K-1 L-1
305* R(K,L) = SUM [A**H(I,K)*X(I,L)] + ISGN*SUM [X(K,J)*B(J,L)]
306* I=1 J=1
307*
308 DO 60 l = 1, n
309 DO 50 k = 1, m
310*
311 suml = cdotc( k-1, a( 1, k ), 1, c( 1, l ), 1 )
312 sumr = cdotu( l-1, c( k, 1 ), ldc, b( 1, l ), 1 )
313 vec = c( k, l ) - ( suml+sgn*sumr )
314*
315 scaloc = one
316 a11 = conjg( a( k, k ) ) + sgn*b( l, l )
317 da11 = abs( real( a11 ) ) + abs( aimag( a11 ) )
318 IF( da11.LE.smin ) THEN
319 a11 = smin
320 da11 = smin
321 info = 1
322 END IF
323 db = abs( real( vec ) ) + abs( aimag( vec ) )
324 IF( da11.LT.one .AND. db.GT.one ) THEN
325 IF( db.GT.bignum*da11 )
326 \$ scaloc = one / db
327 END IF
328*
329 x11 = cladiv( vec*cmplx( scaloc ), a11 )
330*
331 IF( scaloc.NE.one ) THEN
332 DO 40 j = 1, n
333 CALL csscal( m, scaloc, c( 1, j ), 1 )
334 40 CONTINUE
335 scale = scale*scaloc
336 END IF
337 c( k, l ) = x11
338*
339 50 CONTINUE
340 60 CONTINUE
341*
342 ELSE IF( .NOT.notrna .AND. .NOT.notrnb ) THEN
343*
344* Solve A**H*X + ISGN*X*B**H = C.
345*
346* The (K,L)th block of X is determined starting from
347* upper-right corner column by column by
348*
349* A**H(K,K)*X(K,L) + ISGN*X(K,L)*B**H(L,L) = C(K,L) - R(K,L)
350*
351* Where
352* K-1
353* R(K,L) = SUM [A**H(I,K)*X(I,L)] +
354* I=1
355* N
356* ISGN*SUM [X(K,J)*B**H(L,J)].
357* J=L+1
358*
359 DO 90 l = n, 1, -1
360 DO 80 k = 1, m
361*
362 suml = cdotc( k-1, a( 1, k ), 1, c( 1, l ), 1 )
363 sumr = cdotc( n-l, c( k, min( l+1, n ) ), ldc,
364 \$ b( l, min( l+1, n ) ), ldb )
365 vec = c( k, l ) - ( suml+sgn*conjg( sumr ) )
366*
367 scaloc = one
368 a11 = conjg( a( k, k )+sgn*b( l, l ) )
369 da11 = abs( real( a11 ) ) + abs( aimag( a11 ) )
370 IF( da11.LE.smin ) THEN
371 a11 = smin
372 da11 = smin
373 info = 1
374 END IF
375 db = abs( real( vec ) ) + abs( aimag( vec ) )
376 IF( da11.LT.one .AND. db.GT.one ) THEN
377 IF( db.GT.bignum*da11 )
378 \$ scaloc = one / db
379 END IF
380*
381 x11 = cladiv( vec*cmplx( scaloc ), a11 )
382*
383 IF( scaloc.NE.one ) THEN
384 DO 70 j = 1, n
385 CALL csscal( m, scaloc, c( 1, j ), 1 )
386 70 CONTINUE
387 scale = scale*scaloc
388 END IF
389 c( k, l ) = x11
390*
391 80 CONTINUE
392 90 CONTINUE
393*
394 ELSE IF( notrna .AND. .NOT.notrnb ) THEN
395*
396* Solve A*X + ISGN*X*B**H = C.
397*
398* The (K,L)th block of X is determined starting from
399* bottom-left corner column by column by
400*
401* A(K,K)*X(K,L) + ISGN*X(K,L)*B**H(L,L) = C(K,L) - R(K,L)
402*
403* Where
404* M N
405* R(K,L) = SUM [A(K,I)*X(I,L)] + ISGN*SUM [X(K,J)*B**H(L,J)]
406* I=K+1 J=L+1
407*
408 DO 120 l = n, 1, -1
409 DO 110 k = m, 1, -1
410*
411 suml = cdotu( m-k, a( k, min( k+1, m ) ), lda,
412 \$ c( min( k+1, m ), l ), 1 )
413 sumr = cdotc( n-l, c( k, min( l+1, n ) ), ldc,
414 \$ b( l, min( l+1, n ) ), ldb )
415 vec = c( k, l ) - ( suml+sgn*conjg( sumr ) )
416*
417 scaloc = one
418 a11 = a( k, k ) + sgn*conjg( b( l, l ) )
419 da11 = abs( real( a11 ) ) + abs( aimag( a11 ) )
420 IF( da11.LE.smin ) THEN
421 a11 = smin
422 da11 = smin
423 info = 1
424 END IF
425 db = abs( real( vec ) ) + abs( aimag( vec ) )
426 IF( da11.LT.one .AND. db.GT.one ) THEN
427 IF( db.GT.bignum*da11 )
428 \$ scaloc = one / db
429 END IF
430*
431 x11 = cladiv( vec*cmplx( scaloc ), a11 )
432*
433 IF( scaloc.NE.one ) THEN
434 DO 100 j = 1, n
435 CALL csscal( m, scaloc, c( 1, j ), 1 )
436 100 CONTINUE
437 scale = scale*scaloc
438 END IF
439 c( k, l ) = x11
440*
441 110 CONTINUE
442 120 CONTINUE
443*
444 END IF
445*
446 RETURN
447*
448* End of CTRSYL
449*
subroutine xerbla(srname, info)
Definition cblat2.f:3285
complex function cdotu(n, cx, incx, cy, incy)
CDOTU
Definition cdotu.f:83
complex function cdotc(n, cx, incx, cy, incy)
CDOTC
Definition cdotc.f:83
CLADIV performs complex division in real arithmetic, avoiding unnecessary overflow.
real function slamch(cmach)
SLAMCH
Definition slamch.f:68
real function clange(norm, m, n, a, lda, work)
CLANGE returns the value of the 1-norm, Frobenius norm, infinity-norm, or the largest absolute value ...
Definition clange.f:115
logical function lsame(ca, cb)
LSAME
Definition lsame.f:48
subroutine csscal(n, sa, cx, incx)
CSSCAL
Definition csscal.f:78
Here is the call graph for this function:
Here is the caller graph for this function:
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http://math.stackexchange.com/questions/71196/does-this-matrix-have-a-name
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Does this matrix have a name?
If $L$ is a lower triangular matrix of ones, does the following matrix have a special name?
$$A = \left(\begin{matrix}L & -L \\ -L & L \end{matrix}\right)$$
-
It might be $\pmatrix{1 &-1\\ -1 &1}\otimes L$. ($\otimes$ denotes the Kronecker product) – user13838 Oct 9 '11 at 19:25
What does this have to do with graph theory? – anon Oct 9 '11 at 22:19
well I had it written as part of a solution to an LP formulation of a flow network, so perhaps graph theorists could have a better idea on what it's called. – István Oct 10 '11 at 8:56
1 Answer
It's a special type of "block matrix". Or, as user13838 points out, it can be described as a Kronecker product (or direct product).
(Or, you could just call it $A$.)
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https://www.bartleby.com/essay/Heat-Transfer-Radiation-Lab-Report-F3B62J4KRZYS
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# Heat Transfer Radiation Lab Report
3378 Words14 Pages
Module : Heat Transfer – Free Convection and Radiation Laboratory Date : 22nd March 2012 CONTENTS INTRODUCTION 3 AIMS & OBJECTIVES 3 Objectives 3 To investigate Free Convection and Radiation 3 Theory 3 EXPERIMENT 3 Apparatus Used 3 Procedure 4 RESULTS, CALCULATIONS, OBSERVATIONS & CONCLUSIONS 5 Observations During Tests 5 Table 1 5 Table 2 5 Calculations 6 Calculating Power (Watts) 6 Calculating Heat Transfer Emissivity (Ɛ) 6 Emisssivity of a black body 6 Calculating Q rad 6 Calculating Q rad 6 Calculating Q conv 7 Equation for Free Convection 7 Percentage values calculation 7 Absolute Pressure calculation 7 Graph of Pressure Against Temp Difference 8 Conclusions 8 Conclusion…show more content…
| |TEL –TV (K) |(Mb)^1/4 |W |W |% |% | WM^-2K^-1 | |144 |2^1/4 = 1.19 |4.87 |1.14 |81 |19 |2.57 WM^-2K^-1 | |133 |16^1/4 = 2 |4.31 |1.66 |72 |28 |4.06 WM^-2K^-1 | |123 |61^1/4 = 2.79 |3.81 |2.13 |64 |36 |5.64 WM^-2K^-1 | |111 |203^1/4 = 3.77 |3.25 |2.71 |55 |45 |7.95 WM^-2K^-1 | |97 |503^1/4 = 4.73 |2.68 |3.24 |45 |55 |10.88 WM^-2K^-1 | |87 |1018^1/4 = 3.22 |2.27 |3.65 |38 |62 |13.66 WM^-2K^-1 | Table 2 Calculations Heat losses in the connecting leads Q = (0.94 x Volts x Amperes) in watts Calculating Power (Watts) Power = Volts x Amperes (Watts) Power = 8.21volts x 0.779 amps = 6.39 (W) x Heat loses Power = 6.39 (W) x 0.94 = 6.01 Watts Heat Transfer = 0.94 x 8.21 x 0.779 = 6.01 watts Calculating Heat Transfer Emissivity (Ɛ) Emisssivity of a black body ( copper ) = 1 If Ɛ = >1 Use Ɛ = 0.97 to calculate Q rad Ɛ = Q rad Joules or Watts A x σ x
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https://northcornwallnt.org/and-pdf/348-graphs-and-their-uses-by-oystein-ore-pdf-301-501.php
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# Graphs And Their Uses By Oystein Ore Pdf
File Name: graphs and their uses by oystein ore .zip
Size: 10308Kb
Published: 19.04.2021
Most of the volumes in the New Mathematical Library cover topics not usually included in the high school curricu- lum; they vary in difficulty, and, even within a single book, some parts require a greater degree of concentration than others. Thus, while you need little technical knowledge to understand most of these books, you will have to make an intellectual effort.
## The domination number of round digraphs
Oystein Ore was born in Norway in Ore came to the United States in to a position at Yale, where he spent his entire career, retiring in Ore had a great interest in writing accessibly about mathematics. Both of his MAA books were written for the general reader and were intended to provide a stimulating taste of their respective subjects. Invitation to Number Theory comprises eight brief chapters that introduce the reader to the elementary beginnings of number theory. Chapter 2 introduces the notion of prime numbers, the sieve of Eratosthenes, and Fermat and Mersenne primes.
## Øystein Ore
What is combinatorics? Notations and conventions. These notes which are work in progress and will remain so for the foreseeable fu-ture are meant as an introduction to combinatorics the mathematical disciplinethat studies finite sets roughly speaking. When finished, they will cover topicssuch as binomial coefficients, the principles of enumeration, permutations, parti-tions and graphs. The emphasis falls on enumerative combinatorics, meaning the artof computing sizes of finite sets counting , and graph theory. I have tried to keep the presentation as self-contained and elementary as possible.
## Graphs and their Uses
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. I am looking for fun, interesting mathematics textbooks which would make good studious holiday gifts for advanced mathematics undergraduates or beginning graduate students.
Search this site. Aimery de Goyon PDF. Alien Albion PDF.
### Combinatorics: afirstencounter - Math User Home dgrinber/comb/cafe.pdf[Aigner07]Martin Aigner, A...
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The concept of the domination number plays an important role in both theory and applications of digraphs. In this paper, the domination number of round digraphs is characterized completely. The domination theory of graphs was derived from a board game in ancient India. In , Ore formally gave the definitions of the dominating set and the domination number in [ 1 ]. Due to the universality of its applications to both theoretical and practical problems, domination has become one of the important research topics in graph theory. A summary of most important results and applications can be found in [ 2 ].
In Oystein Ore wrote this classic volume, which was published in the New Mathematical Library Series. This elegant book has provided students and.
Ore graduated from the University of Oslo in , with a Cand. In , the University of Oslo awarded him the Ph. In , he was appointed research assistant at the University of Oslo. In , Yale hired Ore as an assistant professor of mathematics, promoted him to associate professor in , then to full professor in In , he became a Sterling Professor Yale's highest academic rank , a position he held until he retired in
Let G be a Hamiltonian graph. In this article, we study simple graphs without loops or parallel edges. For terminology and notations not defined here we refer the reader to [ 1 ]. Let G be a graph with n vertices. A subgraph of G induced by a subset X is denoted by G [ X ].
Она чувствовала, что здесь что-то не то, но не могла сообразить, что. Она достаточно хорошо знала Танкадо и знала, что он боготворил простоту. Его доказательства, его программы всегда отличали кристальная ясность и законченность. Необходимость убрать пробелы показалась ей странной. Это была мелочь, но все же изъян, отсутствие чистоты - не этого она ожидала от Танкадо, наносящего свой коронный удар.
Чьи-то стальные руки прижали его лицо к стеклу. Панк попытался высвободиться и повернуться. - Эдуардо. Это ты, приятель? - Он почувствовал, как рука незнакомца проскользнула к его бумажнику, чуть ослабив хватку.
Вопрос насколько. уступил место другому - с какой целью?. У Хейла не было мотивов для вторжения в ее компьютер.
Не поддается. Сьюзан не могла поверить, что это сказал человек, двадцать семь лет работавший с шифрами. - Не поддается, сэр? - с трудом произнесла. - А как же принцип Бергофского.
Перед глазами возник текст: PRIMEDIFFERENCEBETWEEN ELEMENTSRESPONSIBLE FORHIROSHIMAANDNAGASAKI - Введите пробелы, - приказала Сьюзан. - Нам предстоит решить одну задачку.
Сьюзан, - начал он, - этого не должно было случиться. - Он провел рукой по своим коротко стриженным волосам. - Я кое о чем тебе не рассказал. Иной раз человек в моем положении… - Он замялся, словно принимая трудное решение.
Сьюзан хмыкнула. Этот фонд, всемирная коалиция пользователей компьютеров, развернул мощное движение в защиту гражданских свобод, прежде всего свободы слова в Интернете, разъясняя людям реальности и опасности жизни в электронном мире. Фонд постоянно выступал против того, что именовалось им оруэлловскими средствами подслушивания, имеющимися в распоряжении правительственных агентств, прежде всего АНБ. Этот фонд был для Стратмора постоянной головной болью. - Не вижу ничего нового, - сказала Сьюзан.
Массажистка быстро убрала руки из-под полотенца. В дверях появилась телефонистка и поклонилась: - Почтенный господин. - Слушаю.
И все был подписаны одинаково: Любовь без воска. Она просила его открыть скрытый смысл этих слов, но Дэвид отказывался и только улыбался: Из нас двоих ты криптограф.
Она не шевельнулась. - Ты волнуешься о Дэвиде. Ее верхняя губа чуть дрогнула. Стратмор подошел еще ближе.
С годами она приобрела гибкость и грацию. У нее была высокая стройная фигура с пышной грудью и по-юношески плоским животом. Дэвид шутил, что она может стать первой моделью для рекламы купальников, имеющей докторскую степень по прикладной математике и теории чисел. Через несколько месяцев оба начали подозревать, что обрели нечто такое, что может продлиться всю жизнь.
У ее ног лежало тело Хейла. Прошло еще несколько минут. Она пыталась не думать о Дэвиде, но безуспешно. С каждым завыванием сирены слова Хейла эхом отдавались в ее мозгу: Я сожалею о Дэвиде Беккере.
В этой встрече было что-то нереальное - нечто, заставившее снова напрячься все его нервные клетки. Он поймал себя на том, что непроизвольно пятится от незнакомцев. Тот, что был пониже ростом, смерил его холодным взглядом.
You may use these HTML tags and attributes: ```<a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong> ```
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# NavList:
## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Message:αβγ
Message:abc
Add Images & Files Posting Code: Name: Email:
Re: Position from altitude and azimuth.
From: Robin Stuart
Date: 2020 May 21, 08:47 -0700
It may be a bit late now but in order to determine whether a isoazimuth is a small circle I decided to construct a counter example. The plan was to obtain 4 points lying on an isoazimuth. Choosing 3 of those points calculate the centre of the small circle on which they fall. Then choose a different combination of 3 out of the 4 points and calculate the centre on the circle that those fall on. If the centres are different then an isoazimuth is not a small circle.
Circles great and small are conveniently handled analytically by stereographic projection onto the complex plane as described here. Circles on the sphere map to circles on the plane. Moreover since the Littrow projection is a simple analytic function of the stereographic projection this offers a consistent approach for the treatment of equal altitude circles and isoazimuths. In an earlier post by Bill Lionheart provided a link to a useful article which stated "If the complex z-plane is the Littrow map and the w-plane is the polar stereographic map, then z = w +1/w". After some time wrestling with this and trying to make it behave I realized there is a typo. It should read "z = w -1/w" and that makes all the difference.
In any case here's the counter example generated by appealing to the properties of the Littrow projection
Assume the GP of the body being observed is at 16°46.8'N 45°51.1'E. An observer at any of the points
P1 = 10.306896°N 55.375875°E
P2 = 0.248451°S 70.71412°E
P3 = 25.433371°S 95.262638°E
P4 = 48.38593°S 102.746085°E
The points P1, P2 and P4 all lie on a small circle with centre or pole at
c124 = 37.001705°S 28.211937°E
this can be checked calculating their distances from the centre and seeing that they are all the same.
The points P1, P2 and P3 all lie on a small circle with centre or pole at
c123 = 43.700701°S 18.268654°E
These centres are not the same hence isoazimuths are not in general small circles.
Under stereographic projection onto the complex plane isoazimuths appear to satisfy 3rd order equations. It may still be possible to calculate the intersection of 2 isoazimuths or an isoazimuth with a small circle. This would be analogous to the algebraic solution of the double altitude problem discussed here.
Robin Stuart
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Compound Restricted Choice Problem
Yesterday, I opened 1NT and played it there. In dummy I found K87x of spades, in my hand were 96xx. That left the defenders with AQJTx. Obviously, I wanted to win two spade tricks. On the first spade, I led low toward dummy, LHO played the small spade, I ducked, and RHO won the Ten. My first thought is that RHO might have held one, two, or all three of the missing minor honors (QJT), with relative likelihood of 1 : 1/2 : 1/3, meaning that he was 6-5 to have played the Ten because it was the only one of those cards he'd held. That meant in turn that, from his relevant holdings (those on which I could actually win two tricks - AT, QT, JT, QJT), the AT combination was 55% of those holdings, meaning that I should duck the King again on the second spade trick. Of course, I'd get to see another card from LHO before I had to decide.
When I regained the lead, I played another small spade toward dummy, and LHO played the Jack. Her relevant holdings were then (an original) AJx or QJx, with the AJx twice as likely as the QJx, meaning that I should rise with the King.
But wait, thinking back to the first trick, the matching holdings from RHO had then been AT or QT, with the AT being twice as likely, telling me to duck again.
On the other hand, it might not have been apparent to RHO that his QT were actually equals when he had to choose one, so maybe his choice was restricted to the Ten either way, in other words restricted choice did not apply to the first trick, so rising with the King is indicated (by LHO's play).
But what if RHO had played the Jack to the first trick, which would have been clearly equal to either the Ten or Queen, now restricted choice would again have applied to him, so maybe I should duck after all.
Finally, with both opponents having been potentially restricted in their choices of which minor honor to play from two, maybe the two choices cancel each other out, and it is a tossup whether to rise with the King, or duck.
What do you think?, and this time I really don't know.
Maybe I should have played bingo. Of course, I don't understand that game either, in the sense that I don't understand how anyone over the age of seven could possibly find it interesting.
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# Get instant live expert help on I need help with excel absolute reference multiple cells
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I need to be able to copy this formula (=SUMIFS(data!\$E:\$E,data!\$C:\$C, \$A\$3, data!\$A:\$A, \$A\$1, data!\$H:\$H, \$D\$2)) to multiple cells while the last absolute value increases by 1 column to make it absolute for the new cell.
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Starting in cell B5 in \$-PFSalesSum create formulas ro populate the cells in column B by linking to the appropiate source cells in PFQ1. Hint: After creating the first formula, edit the entry in cell B5 to use a relative reference to the source cell (instead of an absolute reference) so you can copy and paste the formula in cell B5 to the range B6:B9.
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# Recall
You have to be Odd, to be number one. Copy from one, it’s plagiarism; copy from three or more, it’s research.
American actress Ilka Chase enjoyed success as a novelist. In one encounter, an anonymous actress said to her: “I enjoyed reading your book. Who wrote it for you?” …to which Chase replied: “Darling, I’m so glad that you liked it. Who read it to you?”
Definition. A function f is a rule, relationship, or correspondence that assigns to each element of one set (x ∈ D), called the domain, exactly one element of a second set, called the range (y ∈ E).
The pair (x, y) is denoted as y = f(x). Typically, the sets D and E will be both the set of real numbers, ℝ.
An exponential function is a mathematical function of the form f(x) = b·ax, where the independent variable is the exponent, and a and b are constants. a is called the base of the function and it should be a positive real number (a>0).
Examples are 2x, 7x, and (14)x. There are three kinds of exponential functions depending on whether a > 1, a = 1, 0 < a < 1 -1.a.-
1. If a > 1, it can be observed that as the exponent increases, the curve get steeper. As x increases, f(x) heads to infinity, $\lim_{x \to ∞} a^x = ∞$. As x decreases, f(x) heads to zero, $\lim_{x \to -∞} a^x = 0$. It is strictly increasing and has a horizontal asymptote along the x-axis.
It is important to realize that as x approaches negative infinity, the results become very small but never actually attain zero, e.g., 2-5 ≈ 0.03125, 2-15 ≈ 0.00003052. Besides, the base of an exponential function determines the rate of growth or decay. For a > 1, the larger the base, the faster the function grows.
2. 0 < a < 1. As x increases, f(x) heads to zero, $\lim_{x \to ∞} a^x = 0$. As x decreases, f(x) heads to ∞, $\lim_{x \to -∞} a^x = ∞$. It is strictly decreasing and has a horizontal asymptote along the x-axis.
3. When a = 1, the graph is a horizontal line, y = 1.
Loosely speaking, the inverse function of a function f is a function that undoes the operation of the original function. If f is a function that maps each element x in its domain to a unique element y in its codomain, then the inverse function, denoted as $f^{-1}$, maps each y back to its corresponding x.
For a given function f, its inverse g(=f-1) is a function that reverses its result (g reverses the input and output of the original function), g(f(x))=x. An example is f(x)=$\sqrt{x},$ (x ≥ 0) g(f(x))=x ↭ g(x)= x2, the square function (x2) is the inverse of the square root function ($\sqrt{x}$).
# Logarithm function
In mathematics, the logarithm is the inverse function to exponentiation. We call the inverse of ax the logarithmic function with base a, that is, logax=y ↔ ay=x, that means that the logarithm of a number x to the base a is the exponent to which a must be raised to produce x, e.g., log4(64) = 3 ↭ 43 = 64, log2(16) = 4 ↭ 24 = 16, log8(512) = 3 ↭ 83 = 512, but log2(-3) is undefined.
The logarithm function is a type of mathematical function which is helpful in many areas of science, such as finance, biology, and physics. For example, investors use logarithms to calculate compound interest. Various solutions to applied mathematical issues can be done by using logarithms, such as radioactive decay and nuclear waste, the elimination of medicines from the body, etc.
Although the base of a logarithm can be any number, the most common bases used are 10 and e. The logarithm with base e is the natural logarithm. The natural logarithm is the inverse of ex, that is, lnx=y ↔ ey=x where e is the Eulers Number, a mathematical constant approximately equal to 2.71828.
The logarithm’s domain consists of real positive numbers. Its range is ℝ (figure i and ii). x-intercept: (1, 0), y-intercept: none. It is one-to-one and has a vertical asymptote along the y-axis at x = 0.
1. If a > 1, it can be observed that as the value of the argument increases, the logarithm’s value increases as well. As x increases, f(x) heads to infinity, $\lim_{x \to ∞} log_a(x) = ∞$. As x decreases, f(x) heads to zero, $\lim_{x \to 0+} log_a(x) = 0$. It is strictly increasing and has a vertical asymptote along the y-axis at x = 0.
The bigger the logarithm base, the graph approaches the asymptote of x = 0 quicker (Figure iii and iv).
2. If 0 < a < 1, it can be observed that as the value of the argument increases, the logarithm’s value decreases. As x increases, f(x) heads to minus infinity, $\lim_{x \to ∞} log_a(x) = -∞$. As x decreases, f(x) heads to zero, $\lim_{x \to 0+} log_a(x) = 0$. It is strictly decreasing and has a vertical asymptote along the y-axis at x = 0.
# Logarithmic Properties
1. Product Rule. loga(xy)=loga(x) + loga(y).
2. Quotient Rule. loga(xy) = loga(x) - loga(y).
3. Power Rule. loga(xn) = n·loga(x).
4. loga(1) = 0 ↭ a0 = 1.
5. loga(a) = 1 ↭ a1 = a.
# Derivate of the Logarithmic Function
If x > 0, y = ln(x) ⇒ ey = x ⇒[Differentiating both sides of this equation. Recall exponential function (ex)’=ex] $e^y\frac{dy}{dx} = 1 ⇒ \frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}.$
If x > 0, a ≠ 1, y = loga(x) ⇒ ay = x ⇒ ln(ay) = ln(x) ⇒[Power Rule. loga(xn) = n·loga(x).] y·ln(a) = ln(x) ⇒ y = $\frac{ln(x)}{ln(a)}⇒$[1ln(a) is a constant] $\frac{dy}{dy} = \frac{1}{xln(a)}$.
Examples:
• f(x) = $ln(\frac{x}{x+1})$
f(x) = $ln(\frac{x}{x+1}) = ln(x)-ln(x+1)$
f’(x) = $\frac{1}{x}-\frac{1}{x+1} = \frac{x+1}{x(x+1)} -\frac{x}{x(x+1)} = \frac{1}{x(x+1)}$
• f(x) = $ln(\frac{x^2·sin(x)}{2x+1})$
f(x) = $ln(\frac{x^2·sin(x)}{2x+1}) = ln(x^2·sin(x))-ln(2x+1) = ln(x^2)+ln(sin(x))-ln(2x+1)= 2ln(x) +ln(sin(x))-ln(2x+1)$
f’(x) = $\frac{2}{x} + \frac{cos(x)}{sin(x)} -\frac{2}{2x+1} = \frac{2}{x} + cot(x) -\frac{2}{2x+1}$
• f(x) = xx.
y = xx ⇒ ln(y) = ln(xx) ⇒ ln(y) = xln(x) ⇒ $\frac{d}{dx} ln(y) = \frac{d}{dx} xln(x) ⇒ \frac{y’}{y} = ln(x)+1 ⇒ y’ = y(ln(x)+1) = x^x(ln(x)+1).$
# Change of Base of Logarithm
Let loga(x) be the logarithm to base a of x. Then, $log_b(x)=\frac{log_a(x)}{log_a(b)}$.
Proof.
Let y = logb(x) ↭ by = x and z = loga(x) ↭ az = x.
Then, z = loga(x) = $log_a(b^y) = y·log_a(b) ⇒ log_b(x) = y = \frac{z}{log_a(b)} = \frac{log_a(x)}{log_a(b)}∎$
Examples:
• Let’s convert $\log_2(9)$ to base e (natural logarithm), $log_2(9)=\frac{ln(9)}{ln(2)}$= [$ln(9) \approx 2.1972, ln(2) \approx 0.6931$] ≈ 3.1699 ≈ 3.17.
• $log_3(81)=\frac{log_{10}(81)}{log_{10}(3)}$= [$log_{10}(81) \approx 1.9082, log_{10}(3) \approx 0.4771$] ≈ 4.
# Solved examples
• Simplify $log_8(\frac{1}{12})+log_8(\frac{12}{5})$
$log_8(\frac{1}{12})+log_8(\frac{12}{5})$ =[Product Rule. loga(xy)=loga(x) + loga(y)] $log_8(\frac{1}{12}·\frac{12}{5}) = log_8(\frac{1}{5}).$
• Solve ln(4x-3) = 7.
ln(4x -3) = 7 ⇒[Apply ex to both sides] $e^{ln(4x-3)} = e^7 ⇒ 4x -3 = e^7 ⇒ x = \frac{e^7+3}{4}.$
• Solve log4(x2-2x) = log4(5x-12).
log4(x2-2x) = log4(5x-12) ⇒[a = 4 > 1, logarithm is strictly increasing⇒ one-to-one] x2-2x = 5x-12 ⇒ x2 -7x + 12 = 0 ⇒ (x-3)(x-4) = 0 ⇒ x = 3 or 4.
It is essential to make sure to plug these numbers into the original equation so we don’t end up taking logarithms of zero or negative numbers.
log4(32-2·3) = log4(3) = log4(5·3-12) and log4(42-2·4) = log4(8) = log4(5·4-12).
• Solve the logarithmic equation $log_3(\sqrt{-7x+1})-7 = -4$
$log_3(\sqrt{-7x+1})-7 = -4$⇒[Isolate the Logarithmic Term] $log_3(\sqrt{-7x+1}) = 3$⇒[Convert to Exponential Form] $3^{log_3(\sqrt{-7x+1})}=3^3 = 27 ⇒ \sqrt{-7x+1} = 27$ ⇒[Square both sides of the equation to solve for x] -7x+1 = 272 = 729 ⇒ -7x = 728 ⇒ $x = \frac{-728}{7}$ = -104.
Besides, $log_3(\sqrt{-7·-104+1})-7 = -4↭log_3(\sqrt{729})-7 = -4 ↭ log_3(27)-7 = -4 ↭ 3-7 = -4$ and obviously $log_3(\sqrt{729})$ is a valid number.
• Solve the logarithmic equation $log_3(2x+1)+log_3(x+8)=3$
$log_3(2x+1)+log_3(x+8)=3$⇒[Product Rule. loga(xy)=loga(x) + loga(y)] $log_3((2x+1)(x+8)) = 3 ⇒$[By definition of logarithms] (2x+1)(x+8) = 33 = 27.
Expand and rearrange the equation, 2x2 +17x +8 = 27. It has two solutions x = -192, x = 1.
It is essential to make sure to plug these numbers into the original equation so we don’t end up taking logarithms of zero or negative numbers.
1. $log_3(2\frac{-19}{2}+1)= log_3(-18)$ ⊥ Since logarithms of negative numbers are undefined, this is not a valid solution.
2. $log_3(2·1+1)+log_3(1+8)=3↭log_3(3)+log_3(9)=3↭1+2=3$, so 1 is a valid solution.
# Bibliography
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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# Figure 1 the new marginalized bayes net 42 the
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Figure 1: The new marginalized Bayes net 4.2 The procedure is as follows: 1) Remove node X i 2) For each parent of X i , create an edge to each child of X i 3) Number each child of X i from k = 1 . . . n where n is the number of children of X i . For each k , create an edge from child X k to children X k + 1 . . . X n . Also, create an edge from each parent of X k to the children X k +1 . . . X n . (If we don’t add these extra edges we create an independency not in the original graph). We can see why this works by actually performing the marginization. Consider 3-node subgraphs: 7
X W Y where we want to remove W: w P ( X, W, Y ) = w P ( X ) P ( W | X ) P ( Y | W ) = P ( X ) w P ( W | X ) P ( Y | W, X ) = P ( X ) w P ( Y, W | X ) = P ( X ) P ( Y | X ) which implies that X and Y are not necessarily independent, meaning we need an edge between X and Y . For X W Y where we want to remove W: w P ( X, W, Y ) = w P ( W ) P ( X | W ) P ( Y | W ) = w P ( W, X ) P ( Y | W, X ) = w P ( Y, W, X ) = P ( X ) P ( Y | X ) Again, this implies that X and Y are not necessarily independent, meaning we need an edge between X and Y . For X W Y where we want to remove W: w P ( X, W, Y ) = w P ( X ) P ( Y ) P ( W | X, Y ) = P ( X ) P ( Y ) w P ( W | X, Y ) = P ( X ) P ( Y )(1) = P ( X ) P ( Y ) X and Y are marginally independent before marginalizing W , and remain so after marginal- ization. One other thing to consider is the case N X W Y . If we condition on X , then N is not necessarily independent of Y . But after we marginalize W we have N X Y . In this case, if we condition on X then N is independent of Y . Therefore, we need an edge between N and Y otherwise we introduce an extra independence not in the original graph. 8
5 Bayesian Network Inference 5.1 Given that r 2 and r 3 have been observed the nodes that are d-connected to (influenced by) r 1 are the following: u 1 .Age, u 1 .Gender, b 1 .Genre, b 2 .Genre, u 2 .Age, u 2 .Gender, r 4 , r 5 . Aside : In the na¨ ıve Bayes model one rating is independent of another rating given Θ NB . In the elaborate model two ratings can be dependent given Θ EL . In this question, knowing r 1 influences r 4 and r 5 , even if Θ EL is given to you. This is the basic motivation for template models in relational learning. 5.2 The answers to the three na¨ ıve Bayes queries are 0 . 7716, 0 . 3719, and 0 . 5815. The answers to the three elaborate model queries are 0 . 2778, 0 . 1800, and 0 . 6654.
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Quiz 3
# Quiz 3 - Math 3A(44351 Quiz 3 Problem 1 Let A1 = Show that...
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Math 3A (44351) Quiz 3 Problem 1. Let A 1 = 1 0 3 4 , A 2 = 4 0 - 2 3 , A 3 = 0 0 1 2 Show that T = 7 0 0 1 Span( A 1 , A 2 , A 3 ) . Do { A 1 , A 2 , A 3 } span all of R 2 x 2 ? Explain. Solution: 3 A 1 + A 2 - 7 A 3 = 3 0 9 12 + 4 0 - 2 3 - 0 0 7 14 = 7 0 0 1 and so T is a linear combination of A 1 , A 2 , A 3 , and thus T Span( A 1 , A 2 , A 3 ). Since all three vectors have a top-right component of 0, it is clear that the vector 0 1 0 0 , which is in R 2 x 2 , is not in Span( A 1 , A 2 , A 3 ), so no, they do not span all of R 2 x 2 Problem 2. Suppose that { v 1 , v 2 , . . . , v k } , with k > 2, is a set of lin- early independent vectors in R n . Prove that the set { v 2 , . . . , v k } must also be linearly independent. Solution: Suppose that c 2 v 2 + . . . + c k v k = 0 for some set of constants
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# Need help with probability interpretation for my assignment?
## Need help with probability interpretation for my assignment?
Need help with probability interpretation for my assignment? Please update my paper. A: Somewhat related: First of all, if you have a vector $\bm{\phi}$, you have to notice that $\overline{\phi}$ is not a vector. If you would like to know whether a $\bm{\phi}$ is a vector, the probability of the probability of applying a vector to a probability distribution will depend on the value of $\overline{\bm{\phi}}$ in $\{ \phi, \phi + \bm{\alpha}, \phi + \bm{\alpha + \beta}\}$ and the parameters $\bm{\alpha}$ and $\bm{\beta}$. But in this example, $yh_1=\phi_1h_1=\phi_2h_1=\phi_2h_2=0$ which indicates that the probability density function (PDF) of the probability distribution of $h_1, h_2$, $\phi$ and $\phi_1$, and $\phi_2$ are different, since a vector of PDFs is always a vector. The result that Huxley’s value theorem says that $$P=\frac{n!}{n-1}$$ should be computed as $$P=\frac{m!}{n!n-1}$$ So if there is a $1/\sqrt{n}$-dependence, and if $\bm\phi$ is a vector of $\phi$ or vector of $\phi_1, \phi_2$, then $\overline{\phi}$ or the $1/\sqrt{n}$-dependence should be $\overline{\phi} + \bm\gamma$ but unfortunately we cannot compute $\overline{\phi}$ themselves. Need help with probability interpretation for my assignment? I’m not too familiar with using probabilistic statistics. If I want to find values of the other variables that can take the value that I know the value of the least frequent variables, I’d need to know the probabilities of each variable. Any ideas? A: The Bayes method (which measures a probability density to mean) is generally used for estimating a probability density function (PDF), as $$f(x)=\rho(x)e^{-\frac{1}{2}\log{(1-x)}}$$ is often used, as it is well suited because of its simplicity, and the fact that $f$ is function of its dimension (variables like $x$). If $f$ is not smooth, then there is always a first pass necessary to measure the PDFs, that are thus much less informative over time. Need help with probability interpretation for my assignment? I need to have a single-member estimator in my model which estimates the probability distribution of a candidate model. I need this estimator to be self-consistent in terms of real-world parameters and not just being assumed to be linear in parameters. I need a single parameter estimator for estimating the posterior distribution of a binary candidate model, and ideally a two parameter estimator for the general case and only including parameterizable inputs (e.g. fitting a likelihood function of the form $L$ if there are known parameters). I’m struggling to be clear, but so far I’ve had no success because I can’t seem to achieve what the problem state of knowledge should have done. I have a probability-based estimator that says I can estimate the probability distribution of a possible mixture of possible associations of genes and genes-besides explaining the need for a second estimator for the probability distribution of resource true association. Here’s a rough example of what I’ve done and how it’s going to work: define weight_parameters(value) float how much weight_parameters = require(“r=type(value”)); if r==2 then weight_parameters := float(3) //…and the two parameters can be set to 0 N = 100; // 2 for a weight function (meaning, 2 different input values ) where N can be some number of different plausible or probable values or, for more general reasons, just a random hypothesis value.
## Pay Someone To Take Online Test
N = 10; // 10 to be called from my use-case. define 1/(1/(N/2)) @(xs) { d(x) := (xj / N) + (-2 *(xs+1)/2) + (xs + 1 + 1 * f(x)) * d(x); } define 2/(1/(N/2)) @(x) { x : int(1) := (x / 2) – ((n+1)/2) : int(2)*n as 2 useful reference n/2 } @(x: int(1/2)) // For example 1 == 1/(2)=(1/2). define 3/(1/(N/2)) @(i) { i : int(1) ~= -n: int(2)*i e = -n //…and the two parameters can be set to 0 define 3/(N/2)) @(j) { j : int(1/2)*j as 2 * n-1 define 3/(n-1)) @(k) { k : int(1/2)*k as 2 * k define 3/(n-1)) Check This Out (k-3) << do define 3/(A)/(B) { true **: int(1) as n as 0 } define 3/(B) { true **: true as 0 } define 3/(A)/(B+1)/(B)+1 / (A + 1) / (A + 1 + 2 * n - 1/2) as (A + 1 + 2 * n - 1)/(B + 2) / (A + 1 + 2 * n - 1)/(B + 1) define 4/(D)/(E/A) { (x : int(1) - i * x) for (i : int(2)) } /** ### Parameters: ** @returns The value of (x) that was passed to nn. ** / // -- For simplicity, we consider either n or 2 depending on the value ** @item (n < 5) - 1 is the number of possibilities for the factor *n* = 5; ** @item (n : 2) - 3 is the number of possibilities for the factor *n* = 10; ** @
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Grams To Kilograms
99.3 g to kg99.3 Grams to Kilograms
g
=
kg
How to convert 99.3 grams to kilograms?
99.3 g * 0.001 kg = 0.0993 kg 1 g
A common question is How many gram in 99.3 kilogram? And the answer is 99300.0 g in 99.3 kg. Likewise the question how many kilogram in 99.3 gram has the answer of 0.0993 kg in 99.3 g.
How much are 99.3 grams in kilograms?
99.3 grams equal 0.0993 kilograms (99.3g = 0.0993kg). Converting 99.3 g to kg is easy. Simply use our calculator above, or apply the formula to change the length 99.3 g to kg.
Convert 99.3 g to common mass
UnitMass
Microgram99300000.0 µg
Milligram99300.0 mg
Gram99.3 g
Ounce3.5027044216 oz
Pound0.2189190263 lbs
Kilogram0.0993 kg
Stone0.0156370733 st
US ton0.0001094595 ton
Tonne9.93e-05 t
Imperial ton9.77317e-05 Long tons
What is 99.3 grams in kg?
To convert 99.3 g to kg multiply the mass in grams by 0.001. The 99.3 g in kg formula is [kg] = 99.3 * 0.001. Thus, for 99.3 grams in kilogram we get 0.0993 kg.
Alternative spelling
99.3 Gram to Kilograms, 99.3 Gram in Kilograms, 99.3 Gram to Kilogram, 99.3 Gram in Kilogram, 99.3 Grams to kg, 99.3 Grams in kg, 99.3 g to Kilogram, 99.3 g in Kilogram, 99.3 g to Kilograms, 99.3 g in Kilograms, 99.3 Grams to Kilogram, 99.3 Grams in Kilogram, 99.3 Gram to kg, 99.3 Gram in kg
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https://cs.stackexchange.com/questions/113080/whats-an-upper-bound-for-this-recurrence-so-i-can-take-advantage-of-the-master
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# What's an upper bound for this recurrence so I can take advantage of the Master Theorem?
Let
$$T(N) = \begin{cases}1 & \text{if } N = 1\\ T(\varphi(N)) + 2T(\sqrt{N}) + \lg(\varphi(N))^3 & \text{otherwise} \end{cases}$$
where $$\varphi(N)$$ is Euler's totient function. My objective is to find an upper bound so that I can apply the Master Theorem and find a closed-form formula.
In order to make this answer self-contained, I will repeat the first part of my answer to your other similar question.
First, show that for $$N>2$$, $$\phi(\phi(N)) < N/2$$. This can be done as such:
Let $$N = \prod_{i=1}^rp_i^{k_i}$$ be the prime factorisation of $$N$$ ($$p_i$$ prime, $$k_i>0$$)
• Suppose $$N$$ is even. Then $$\phi(N) = N\prod_{i=1}^r(1-\frac{1}{p_i}) \leq N(1-\frac{1}{2}) \leq N/2.$$ Thus $$\phi(\phi(N)) < N/2$$.
• Suppose $$N$$ is odd and $$N > 1$$. Then $$\phi(N) = \prod_{i=1}^r (p_i-1)p_i^{k_i-1}$$ is even and smaller than $$N$$. By the previous result $$\phi(\phi(N)) < N/2$$.
So we get the desired result.
Now, let $$k>2$$ be some fixed integer, and suppose $$N\geq k^2$$. You can write: $$T(N) = T(\phi(N)) + 2T(\sqrt{N})+\lg(\phi(N))^3$$ $$T(N) = T(\phi(\phi(N))) + 2T(\sqrt{\phi(N)})+\lg(\phi(\phi(N)))^3 + 2T(\sqrt{N})+\lg(\phi(N))^3$$ $$T(N) \leq T(N/2) +4T(N/k) + 2\lg(n)^3.$$ Where we used that $$N\geq k^2 \implies T(\sqrt{\phi(N)}) \leq T(\sqrt{N}) \leq T(N/k)$$ in the last step.
(Note that I am implicitly assuming $$T(n)$$ is a monotonically increasing function. One could get rid of that assumption by working on $$S(n) = max(T(n),S(n-1))$$, at the cost of making the argument more tedious and harder to follow)
Now we can apply the Akra-Bazzi method (a generalisation of the master theorem) with:
• $$k = 2$$
• $$a_1 = 1$$, $$b_1 = \frac12$$, $$a_2 = 4$$, $$b_2 = \frac1k$$
• $$g(x) = 2\lg(x)^3$$
First we need to find $$p$$ such that $$a_1b_1^p + a_2b_2^p = 1$$, that is: $$(\frac12)^p+4(\frac1k)^p = 1 \;\;\; (*)$$ Let's call $$p_0$$ the solution to $$(*)$$.
Then, we need to plug $$p = p_0$$ into the equation: $$T(N) \in O(N^p(1+\int_1^N\frac{g(u)}{u^p+1}du))$$ To get:
$$T(N) \in O(N^{p_0}(1+\frac{6(N^{p_0}-1)}{p_0^4N^{p_0}\ln(2)^3})) \in O(N^{p_0})$$ (I just trusted WolframAlpha and did some simplifications for the integral, I'm too clumsy/lazy to attempt it myself if I don't have to)
Notice that in $$(*)$$ we can make $$p_0$$ arbitrarily close to $$0$$ by picking a big enough $$k$$. In other words, for any $$\epsilon > 0$$ there exists some $$k>0$$ such that $$p_0 < \epsilon$$. So let's pick any $$\epsilon > 0$$ and set $$k$$ such that $$p_0 < \epsilon$$. Plugging that into the last result we get:
$$T(N) \in O(N^\epsilon)$$
And this is true for any $$\epsilon>0$$.
In other words, $$T(N)$$ grows slower than any polynomial function in $$N$$.
I suspect it is asymptotically not much bigger than $$\lg(\phi(N))^4$$, as the square root term will vanish pretty fast, but I can't prove anything better than what I showed.
Also note that you could use the master theorem by setting $$k=2$$ and working on the upper bound $$T(N) \leq 5T(N/2) + 2\lg(N)^3$$, but this would of course lead to a much weaker result.
• +1 for phi(phi(N)) < N/2 alone. Aug 27 '19 at 19:27
What is $$\varphi(p)$$ when $$p$$ is prime? Why is this the worst case?
Note that the master theorem is not going to be applicable. It handles subdivision into subproblems whose size is a fixed proportion of the original problem's size: $$2T(\sqrt N)$$ doesn't fit this model.
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https://www.utwente.nl/en/education/bachelor/programmes/applied-mathematics/study-programme/second-year/
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Second year of Applied Mathematics
During the second year of the Bachelor’s Applied Mathematics your knowledge of mathematics and its application potential will grow. The assignments and projects you work on become more realistic and complex.
• Year 2Year 2
• Statistics & Analysis
• Dynamical Systems
• Discrete Structures & efficient Algorithms
• Modelling & Analysis of stochastic Processes
Module 5: statistics and analysis
In this module you will learn to look for linear relationships in statistical data, using an advanced software package, SPSS. In the Mathematical statistics part of the module, you will learn much of the underlying theory. You will continue to develop your knowledge of basic mathematics in Analysis, Part 2 (having done Part 1 in module 2 of the first year). Also, you will learn in this module how to reflect on different cultures within and outside the field of mathematics. For example, the different approaches mathematicians take and consider to develop a theory, or how the same behaviour can be interpreted differently in different cultures. In this module you will follow courses like Mathematical Statistics, Analysis II, Prooflab Revisited: Diversity in Cultures.
Module 6: dynamical systems
Mathematical models often involve differential equations. Such models are called dynamic, because they show the relationship between the variables and their changes, whereas so-called static models only represent the variables. In this module you will study dynamic models from the perspective of differential equations, mathematical system theory and numerical aspects, making use of MATLAB software. In the team project you will create a model for a dynamic process of the body, such as walking or running or jumping. The challenge here is to capture in a model the combination of stability and movement that characterises that movement. In this module you will follow courses like Ordinary Differential Equations, Systems Theory, Numerical Mathematics.
Module 7: discrete structures and EFFICIENT ALGORITHMS
You will be studying so-called discrete structures. As opposed to continuous structures, such as a collection of real numbers, these structures are about finite or countably infinite sets and variations thereon. The focus is on making calculations within such sets, such as finding the shortest path between a certain number of points and the links between those points. The computability and complexity of these calculations are important factors. You will use elements from abstract algebra, such as groups, rings and fields, as well as finite automata and Turing machines. This module is joint with Technical Computer Science. In this module you will follow courses like Algorithmic Discrete Mathematics, Algebra, Implementation Project on Graph Isomorphism.
Module 8: Modelling and analysis of stochastic processes
As far as mathematics is concerned, this module is about Markov chains. These are stochastic models, or models in which outcomes depend on chance, and in which you go from state to state according to a given probability distribution, regardless of where you were in the past. These models find all sorts of applications, for example, in modelling queues. They also play a vital role in modelling and simulating complex stochastic systems. In your multidisciplinary project group, together with Civil Engineering and Industrial Engineering and Management students, you will learn how to use such models to make good decisions in practical situations in which chance plays an important role, for example, logistics, a hospital environment or traffic. In this module you will follow courses like Stochastic Models, Markov Chains, Multidisciplinary Project.
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# To guide 4 k grade 12 curriculum
. 2018-8-30 · CURRICULUM OVERVIEW FOR K – 9 MATHEMATICS Page 4 NEW BRUNSWICK MATHEMATICS GRADE 9 CURRICULUM GUIDE ASSESSMENT Ongoing, interactive assessment (formative assessment) is essential to effective teaching and learning., 2016-8-29 · Grades K-12 Curriculum Guide. Curriculum Guide Contributors: Susan R. Adams, Grade 4 - Selected Elementary Social Studies Standards with Crispus Attucks Historical Inquiry Grade 9 - 12.
2019-1-10 · K to 12 Curriculum Guide MATHEMATICS (Grade 1 to Grade 10) GRADE 4 The learner demonstrates understanding and appreciation of key concepts and skills involving numbers and number sense (whole numbers up to 100 000, multiplication and division of whole numbers, order of operations, factors and multiples, addition and subtraction of fractions 2019-1-10 · K to 12 Curriculum Guide MATHEMATICS (Grade 1 to Grade 10) GRADE 4 The learner demonstrates understanding and appreciation of key concepts and skills involving numbers and number sense (whole numbers up to 100 000, multiplication and division of whole numbers, order of operations, factors and multiples, addition and subtraction of fractions
2019-7-10 · Grade 4 English Curriculum & Pacing Guide . Teacher Notes ACPS will utilize a theme approach to integrate the English strands of reading, writing, research, and communication. To the extent possible, a variety of genres should be utilized during a unit of study. Best practices include: meaningful expression . 12 2019-1-10 · K to 12 Curriculum Guide SCIENCE (Grade 3 to Grade 10) K to 12 BASIC EDUCATION CURRICULUM At the end of Grade 4, learners can investigate changes in some observable properties of materials when mixed with other materials or when force is applied on them. They can identify materials that do not decay and use this knowledge to help minimize
2018-10-30 · MATHEMATICS GRADE 4 CURRICULUM GUIDE 2014 1 INTRODUCTION Background INTRODUCTION The Mathematics curriculum guides for Newfoundland and Labrador have been derived from The Common Curriculum Framework for K-9 Mathematics: Western and Northern Canadian Protocol, January 2008. These guides incorporate the conceptual framework for Kindergarten 2018-8-30 · CURRICULUM OVERVIEW FOR K – 9 MATHEMATICS Page 4 NEW BRUNSWICK MATHEMATICS GRADE 9 CURRICULUM GUIDE ASSESSMENT Ongoing, interactive assessment (formative assessment) is essential to effective teaching and learning.
2018-8-30 · CURRICULUM OVERVIEW FOR K – 9 MATHEMATICS Page 4 NEW BRUNSWICK MATHEMATICS GRADE 9 CURRICULUM GUIDE ASSESSMENT Ongoing, interactive assessment (formative assessment) is essential to effective teaching and learning. 2018-10-30 · MATHEMATICS GRADE 4 CURRICULUM GUIDE 2014 1 INTRODUCTION Background INTRODUCTION The Mathematics curriculum guides for Newfoundland and Labrador have been derived from The Common Curriculum Framework for K-9 Mathematics: Western and Northern Canadian Protocol, January 2008. These guides incorporate the conceptual framework for Kindergarten
2014-11-24 · GRADE 4 SCIENCE CURRICULUM GUIDE STANDARDS REFERENCE Elkhart Community Schools Page 12 Grade 4 Science Curriculum September 2002 Standards Referenced SCIENTIFIC THINKING Standard 2: Students use a variety of skills and techniques when attempting to answer questions and solve problems. They describe 2019-1-10 · K to 12 Curriculum Guide SCIENCE (Grade 3 to Grade 10) K to 12 BASIC EDUCATION CURRICULUM At the end of Grade 4, learners can investigate changes in some observable properties of materials when mixed with other materials or when force is applied on them. They can identify materials that do not decay and use this knowledge to help minimize
2019-1-10 · K to 12 Curriculum Guide MATHEMATICS (Grade 1 to Grade 10) GRADE 4 The learner demonstrates understanding and appreciation of key concepts and skills involving numbers and number sense (whole numbers up to 100 000, multiplication and division of whole numbers, order of operations, factors and multiples, addition and subtraction of fractions 2019-7-10 · Grade 4 English Curriculum & Pacing Guide . Teacher Notes ACPS will utilize a theme approach to integrate the English strands of reading, writing, research, and communication. To the extent possible, a variety of genres should be utilized during a unit of study. Best practices include: meaningful expression . 12
2018-10-30 · MATHEMATICS GRADE 4 CURRICULUM GUIDE 2014 1 INTRODUCTION Background INTRODUCTION The Mathematics curriculum guides for Newfoundland and Labrador have been derived from The Common Curriculum Framework for K-9 Mathematics: Western and Northern Canadian Protocol, January 2008. These guides incorporate the conceptual framework for Kindergarten 2019-1-10 · K to 12 Curriculum Guide SCIENCE (Grade 3 to Grade 10) K to 12 BASIC EDUCATION CURRICULUM At the end of Grade 4, learners can investigate changes in some observable properties of materials when mixed with other materials or when force is applied on them. They can identify materials that do not decay and use this knowledge to help minimize
. 2016-11-8 · K to 12 Curriculum Guide ART (Grade 1 to Grade 10) Grade 4 Through the formal introduction of elements, the learner can identify the basic knowledge and skills in music and art, towards self-development, the K to 12 Arts Curriculum Guide May 2016 Page 12of 102 . K to 12 Arts Curriculum Guide May 2016 Page of 102, 2019-1-10 · K to 12 Curriculum Guide SCIENCE (Grade 3 to Grade 10) K to 12 BASIC EDUCATION CURRICULUM At the end of Grade 4, learners can investigate changes in some observable properties of materials when mixed with other materials or when force is applied on them. They can identify materials that do not decay and use this knowledge to help minimize.
. 2019-1-10 · K to 12 Curriculum Guide SCIENCE (Grade 3 to Grade 10) K to 12 BASIC EDUCATION CURRICULUM At the end of Grade 4, learners can investigate changes in some observable properties of materials when mixed with other materials or when force is applied on them. They can identify materials that do not decay and use this knowledge to help minimize 2018-10-30 · MATHEMATICS GRADE 4 CURRICULUM GUIDE 2014 1 INTRODUCTION Background INTRODUCTION The Mathematics curriculum guides for Newfoundland and Labrador have been derived from The Common Curriculum Framework for K-9 Mathematics: Western and Northern Canadian Protocol, January 2008. These guides incorporate the conceptual framework for Kindergarten.
2016-8-29 · Grades K-12 Curriculum Guide. Curriculum Guide Contributors: Susan R. Adams, Grade 4 - Selected Elementary Social Studies Standards with Crispus Attucks Historical Inquiry Grade 9 - 12 2019-1-10 · K to 12 Curriculum Guide MATHEMATICS (Grade 1 to Grade 10) GRADE 4 The learner demonstrates understanding and appreciation of key concepts and skills involving numbers and number sense (whole numbers up to 100 000, multiplication and division of whole numbers, order of operations, factors and multiples, addition and subtraction of fractions
2014-11-24 · GRADE 4 SCIENCE CURRICULUM GUIDE STANDARDS REFERENCE Elkhart Community Schools Page 12 Grade 4 Science Curriculum September 2002 Standards Referenced SCIENTIFIC THINKING Standard 2: Students use a variety of skills and techniques when attempting to answer questions and solve problems. They describe 2019-1-10 · K to 12 Curriculum Guide MATHEMATICS (Grade 1 to Grade 10) GRADE 4 The learner demonstrates understanding and appreciation of key concepts and skills involving numbers and number sense (whole numbers up to 100 000, multiplication and division of whole numbers, order of operations, factors and multiples, addition and subtraction of fractions
2016-11-8 · K to 12 Curriculum Guide ART (Grade 1 to Grade 10) Grade 4 Through the formal introduction of elements, the learner can identify the basic knowledge and skills in music and art, towards self-development, the K to 12 Arts Curriculum Guide May 2016 Page 12of 102 . K to 12 Arts Curriculum Guide May 2016 Page of 102 2018-8-30 · CURRICULUM OVERVIEW FOR K – 9 MATHEMATICS Page 4 NEW BRUNSWICK MATHEMATICS GRADE 9 CURRICULUM GUIDE ASSESSMENT Ongoing, interactive assessment (formative assessment) is essential to effective teaching and learning.
2019-1-10 · K to 12 Curriculum Guide SCIENCE (Grade 3 to Grade 10) K to 12 BASIC EDUCATION CURRICULUM At the end of Grade 4, learners can investigate changes in some observable properties of materials when mixed with other materials or when force is applied on them. They can identify materials that do not decay and use this knowledge to help minimize 2019-1-10 · K to 12 Curriculum Guide SCIENCE (Grade 3 to Grade 10) K to 12 BASIC EDUCATION CURRICULUM At the end of Grade 4, learners can investigate changes in some observable properties of materials when mixed with other materials or when force is applied on them. They can identify materials that do not decay and use this knowledge to help minimize
2018-8-30 · CURRICULUM OVERVIEW FOR K – 9 MATHEMATICS Page 4 NEW BRUNSWICK MATHEMATICS GRADE 9 CURRICULUM GUIDE ASSESSMENT Ongoing, interactive assessment (formative assessment) is essential to effective teaching and learning. 2019-1-10 · K to 12 Curriculum Guide MATHEMATICS (Grade 1 to Grade 10) GRADE 4 The learner demonstrates understanding and appreciation of key concepts and skills involving numbers and number sense (whole numbers up to 100 000, multiplication and division of whole numbers, order of operations, factors and multiples, addition and subtraction of fractions
2019-7-10 · Grade 4 English Curriculum & Pacing Guide . Teacher Notes ACPS will utilize a theme approach to integrate the English strands of reading, writing, research, and communication. To the extent possible, a variety of genres should be utilized during a unit of study. Best practices include: meaningful expression . 12 2018-10-30 · MATHEMATICS GRADE 4 CURRICULUM GUIDE 2014 1 INTRODUCTION Background INTRODUCTION The Mathematics curriculum guides for Newfoundland and Labrador have been derived from The Common Curriculum Framework for K-9 Mathematics: Western and Northern Canadian Protocol, January 2008. These guides incorporate the conceptual framework for Kindergarten
2019-1-10 · K to 12 Curriculum Guide MATHEMATICS (Grade 1 to Grade 10) GRADE 4 The learner demonstrates understanding and appreciation of key concepts and skills involving numbers and number sense (whole numbers up to 100 000, multiplication and division of whole numbers, order of operations, factors and multiples, addition and subtraction of fractions 2016-11-8 · K to 12 Curriculum Guide ART (Grade 1 to Grade 10) Grade 4 Through the formal introduction of elements, the learner can identify the basic knowledge and skills in music and art, towards self-development, the K to 12 Arts Curriculum Guide May 2016 Page 12of 102 . K to 12 Arts Curriculum Guide May 2016 Page of 102
2016-8-29 · Grades K-12 Curriculum Guide. Curriculum Guide Contributors: Susan R. Adams, Grade 4 - Selected Elementary Social Studies Standards with Crispus Attucks Historical Inquiry Grade 9 - 12 2016-8-29 · Grades K-12 Curriculum Guide. Curriculum Guide Contributors: Susan R. Adams, Grade 4 - Selected Elementary Social Studies Standards with Crispus Attucks Historical Inquiry Grade 9 - 12
2019-1-10 · K to 12 Curriculum Guide MATHEMATICS (Grade 1 to Grade 10) GRADE 4 The learner demonstrates understanding and appreciation of key concepts and skills involving numbers and number sense (whole numbers up to 100 000, multiplication and division of whole numbers, order of operations, factors and multiples, addition and subtraction of fractions 2019-1-10 · K to 12 Curriculum Guide SCIENCE (Grade 3 to Grade 10) K to 12 BASIC EDUCATION CURRICULUM At the end of Grade 4, learners can investigate changes in some observable properties of materials when mixed with other materials or when force is applied on them. They can identify materials that do not decay and use this knowledge to help minimize
2018-10-30 · MATHEMATICS GRADE 4 CURRICULUM GUIDE 2014 1 INTRODUCTION Background INTRODUCTION The Mathematics curriculum guides for Newfoundland and Labrador have been derived from The Common Curriculum Framework for K-9 Mathematics: Western and Northern Canadian Protocol, January 2008. These guides incorporate the conceptual framework for Kindergarten 2018-8-30 · CURRICULUM OVERVIEW FOR K – 9 MATHEMATICS Page 4 NEW BRUNSWICK MATHEMATICS GRADE 9 CURRICULUM GUIDE ASSESSMENT Ongoing, interactive assessment (formative assessment) is essential to effective teaching and learning.
2016-8-29 · Grades K-12 Curriculum Guide. Curriculum Guide Contributors: Susan R. Adams, Grade 4 - Selected Elementary Social Studies Standards with Crispus Attucks Historical Inquiry Grade 9 - 12 2019-1-10 · K to 12 Curriculum Guide SCIENCE (Grade 3 to Grade 10) K to 12 BASIC EDUCATION CURRICULUM At the end of Grade 4, learners can investigate changes in some observable properties of materials when mixed with other materials or when force is applied on them. They can identify materials that do not decay and use this knowledge to help minimize
2018-10-30 · MATHEMATICS GRADE 4 CURRICULUM GUIDE 2014 1 INTRODUCTION Background INTRODUCTION The Mathematics curriculum guides for Newfoundland and Labrador have been derived from The Common Curriculum Framework for K-9 Mathematics: Western and Northern Canadian Protocol, January 2008. These guides incorporate the conceptual framework for Kindergarten 2018-8-30 · CURRICULUM OVERVIEW FOR K – 9 MATHEMATICS Page 4 NEW BRUNSWICK MATHEMATICS GRADE 9 CURRICULUM GUIDE ASSESSMENT Ongoing, interactive assessment (formative assessment) is essential to effective teaching and learning.
. 2018-8-30 · curriculum overview for k – 9 mathematics page 4 new brunswick mathematics grade 9 curriculum guide assessment ongoing, interactive assessment (formative assessment) is essential to effective teaching and learning., 2014-11-24 · grade 4 science curriculum guide standards reference elkhart community schools page 12 grade 4 science curriculum september 2002 standards referenced scientific thinking standard 2: students use a variety of skills and techniques when attempting to answer questions and solve problems. they describe).
. 2018-8-30 · curriculum overview for k – 9 mathematics page 4 new brunswick mathematics grade 9 curriculum guide assessment ongoing, interactive assessment (formative assessment) is essential to effective teaching and learning., 2014-11-24 · grade 4 science curriculum guide standards reference elkhart community schools page 12 grade 4 science curriculum september 2002 standards referenced scientific thinking standard 2: students use a variety of skills and techniques when attempting to answer questions and solve problems. they describe).
. 2018-8-30 · curriculum overview for k – 9 mathematics page 4 new brunswick mathematics grade 9 curriculum guide assessment ongoing, interactive assessment (formative assessment) is essential to effective teaching and learning., 2016-8-29 · grades k-12 curriculum guide. curriculum guide contributors: susan r. adams, grade 4 - selected elementary social studies standards with crispus attucks historical inquiry grade 9 - 12).
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https://hubpages.com/health/forum/318593/how-do-you-measure-your-happiness
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How do you measure your happiness?
Jump to Last Post 1-8 of 8 discussions (17 posts)
1. 71
m abdullah javedposted 6 years ago
How do you measure your happiness?
2. 60
Wissam Qawasmehposted 6 years ago
By the distance between me and people i love
1. 71
m abdullah javedposted 6 years agoin reply to this
that's great, how do feel then?
2. 60
Wissam Qawasmehposted 6 years agoin reply to this
Feel happier as they closer. Plus, it's hard to measure a feeling since the human is complicated
3. 71
m abdullah javedposted 6 years agoin reply to this
Excellent Wissam, Jazakallahu Khair.
3. 61
alexatomasposted 6 years ago
It's difficult to be rational about happiness, but if we could have a mathematical approach to it, it should be a something like this:
how your life looks at the moment/ how you would like your life to be
The "dream life" part of the equation is always 100%, meaning full happiness.
The hardest part to measure would be the how your life looks at the moment. We would need to separate each component of life and give it a percentage of how much it really means for full happiness. Then we would have to score on how close we are from being fully happy.
Imagine that someone is happy by having a job as CEO of Google and having a big family. But in real life, they are at Google's first entry job and they are single. Imagining that google as 10 levels of employees and entry job is the first level and CEO is the 10th level, and imagining that being single equals to the first level of relationships out of 10, their happiness would be:
(1/10 + 1/10) / 100% = 20%
Does that make sense?
1. 71
m abdullah javedposted 6 years agoin reply to this
Very logical approach towards happiness, thanks for a profound reply, Alexa.
4. 72
connorjposted 6 years ago
Conn says, well he really types/texts it... When in happiness embrace the moments, do not measure it unless one wants to possibly end it...
1. 71
m abdullah javedposted 6 years agoin reply to this
That's wonderful way of looking at happiness, but you know Conn... If you start realising the happiness along with its quantity then you might strive for more.
5. 88
Austinstarposted 6 years ago
I measure happiness by contentment. Also, lack of physical pain.
I worked to earn a living for over 40 years and thought I would be so happy when I retired. Instead, I am full of pain and unable to physically enjoy many of the things that I used to enjoy.
So now, my happiness is measured by having a "good" day, as opposed to a bad day.
A good day is when I feel somewhat pain free and I can enjoy my reading and writing and my family.
I have learned to have "happy moments". Those moments in time like when I eat an ice cream bar and it is so delicious that it makes me smile.
When I read a wonderful book that takes me away to a fantasy that I enjoy.
When my pets do something so cute that I have to say, "Awwww".
When my husband does the housework for me or brings me a cup of coffee or tea.
When the weather is good enough to sit outside in my swing or my hammock.
These are "happy moments" and I treasure them.
1. 71
m abdullah javedposted 6 years agoin reply to this
Austin your views about contentment are commendable, true without it even happiness lack the glamour. Thanks.
6. 61
thelonelypoet02posted 6 years ago
Happiness shouldn't be measure. Appreciation is what really matters it give us the quality living that anything in this world cannot give.
1. 71
m abdullah javedposted 6 years agoin reply to this
That's true. Thanks.
7. 51
Traci Michelleposted 6 years ago
Happiness is never what we think it is. But then it is also exactly what we decide it is.
I'm sitting here at 0535 (5:35 in the morning) finishing a shift at work. If you asked what would make me happy, my prompt reply would be, "my bed." But once I get home, I'll be reluctant to go there. To miss the day, as I have most days for the past 7 years.
That's the situation if I allow my thoughts and emotions to flow without consciousness. My normal operational mode. However, I've given some thought to experimenting with choice. I should choose my emotional flavor at any given time, or at the very least be conscious of it.
So I suppose I measure happiness according to the amounts in which I choose to have it...
1. 71
m abdullah javedposted 6 years agoin reply to this
Excellent Traci, you have rightly described your perception of happiness. Thanks.
8. 60
moisesbebeposted 5 years ago
No sé si la mido o no, pero la reconozco cada vez que mira mi pareja y cuando mi mascota (gato) se me sube encima y me ronronea.
Un saludo
1. 71
m abdullah javedposted 5 years agoin reply to this
I can't understand this language Toni, will you please translate it. Thanks in advance.
working
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### Unit 4: Addition, Subtraction & Measurement
Go HERE for Dreambox learning journey!
### Unit 3: Fractions & Decimals
Click image above for full description. Click image above for full description.
### Unit 2: Multi-Digit Multiplication & Early Division
Click image above for full description. Click image above for full description.
### Understanding Fractions
Regions and Sets Fractions and Division Estimating Fractional Amounts Equivalent Fractions Fractions in Simplest Form Improper Fractions & Mixed Numbers Comparing Fractions Ordering Fractions Writing to Explain
### Polygons
Points, Lines, and Planes Line Segments, Rays, and Angles Measuring Angles Polygons Triangles Quadrilaterals Make & Test Generalizations
### Dividing by 1-Digit Divisors
Using Mental Math to Divide Estimating Quotients Dividing with Remainders Connecting Models and Symbols Dividing 2-Digit by 1-Digit Numbers Dividing 3-Digit by 1-Digit Numbers Deciding where to start dividing Factors Prime and Composite Numbers Problem Solving: Multiple-Step Problems At home practice ideas clickg4-division-home-ideas
### Multiplying by 2-Digit Numbers
Using Mental Math to Multiply 2-digit Numbers Estimating Products Arrays and an Expanded Algorithm Multiplying 2-Digit Numbers by Multiples of Ten Multiplying 2-Digit by 2-Digit Numbers Special Cases Problem Solving: Two-Question Problems Need some ideas for at home? Click g4-mult-2-digit-topic.
### Patterns and Expressions
Variables and Expressions Addition and Subtraction Expressions Multiplication and Division Expressions Problem-Solving: Use Objects and Reasoning Ideas for at home? Click on the link below: g4-topic-6
### Division Meanings and Facts
Want to know how to help at home? Click on the link below: g4-topic-4
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## 4th std Maths-2nd Semester-Lesson-10-Addition And Subtraction Of Money-Ex-10.1-(Video-01).
4th std Maths-2nd Semester-Lesson-10-Addition And Subtraction Of Money-Ex-10.1-(Video-01).
## 4th std Maths-2nd Semester-Lesson-1-Addition And Subtraction Of Money-(Video).
4th std Maths-2nd Semester-Lesson-1-Addition And Subtraction Of Money-(Video).
## 4th std Maths-Lesson-10-Addition And Subtraction Of Money-(Video).
4th std Maths-Lesson-10-Addition And Subtraction Of Money-(Video).
## Class 4th Maths (Addition and Subtraction of money) Exercise 10.3 – (11.01.2021)
Class 4th Maths (Addition and Subtraction of money) Exercise 10.3 – (11.01.2021) Note: Write neatly in your classwork. Solve the remaining problems yourself. After each problem draw the line
## Class 4th Maths (Addition and Subtraction of money) Exercise 10.2 (28.12.2020)
Class 4th Maths (Addition and Subtraction of money) Exercise 10.2 (28.12.2020) Note: Write neatly in your class work Solve the remaining problems yourself After each problem draw the line
## Class 4th Maths Chapter-10 addition and substraction of money (2nd Semester)-14.12.2020
Class 4th Maths Chapter-10 addition and substraction of money Note: Write the notes in second semester class work Write neatly After each problem draw the line
## Class 4th Maths Chapter-9 Decimals – (12.11.2020)
Class 4th Maths Chapter-9 Decimals Note: Draw the diagrams neatly using scale, pencil and compass.
## Class 4th Maths Activity-5 Fraction – (06.11.2020)
Class 4th Maths Activity-5 Fraction Note: Do the activity neatly.
## Class 4th Maths Chapter-9 (Fractions and decimals) – (30.10.2020)
Class 4th Maths Chapter-9 (Fractions and decimals) Note: Draw the figures neatly and colour. Write these sums in homework book.
## Class 4th Maths Chapter-8 (Mental Arithmetic) – (23.10.2020)
Class 4th Maths Chapter-8 (Mental Arithmetic) – (23.10.2020) Note: Write these sums in homework book neatly.
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### জাহাঙ্গীরনগর বিশ্ববিদ্যালয়
#### E ইউনিট
12
16
24
32
0
1
2
3
##### 3. In a room all except 18 of the people are above 50 years of age . If 15 of the people din the room are under 50 years of age. How many people are in the room?
27
30
33
can't be determined
648
1800
2700
2400
25%
20%
30%
15%
62
72
69
9
11
27
29
16
18
21
24
1.2
4.5
12
80
##### 10. The cost of a pen is twice that of a pencil . If you buy x pens for TK. y how many pencils can you buy with Tk. (x+y) ?
$\frac{4{x}^{2}y}{x+y}$
42
72
57
57
0
1
2
3
15 miles
12 miles
13 miles
14 miles
Tk. 525
Tk. 600
TK. 775
Tk. 1400
TK. 1600
Tk. 2200
Tk. 2400
1500
500
2000
2500
12
15
22
24
2.3 kg
4.5 kg
3.9 kg
none
1
2
3
infinite number
4
2
0
1
0
1 or -1
2 or -2
None
1.5%
66.67%
133.33%
150%
30
45
50
70
Always
only if x <0
only if x>o
only if x=0
7
9
12
15
6.5
6
5.5
5.25
9
36
400
81
13
23
26
39
3
4
5
8
##### 30. In a cash box, there are equal numbers of 10 paisa, 25 paisa and 50 coins. If it contains TK. 170 . then what is then number of 10 paisa coin?
100
200
220
150
RightAns: 0 | WrongAns: 0 | Result: 0/0
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# Gross primary production and increase in biomass
This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!
Please help me calculate GPP, Increase in biomass, photosynthetic efficiency, etc., by explaining how you arrived at the final answer and showing all of the equations or formulas involved in each calculation. Kindly use the correct scientific units throughout the calculation, as well as in the final answer. Please use scientific notation and an appropriate number of significant figures in the final answers. Thank you.
In an ecological investigation of an area of European grassland, the energy values of organisms in the ecosystem were found. The ecologist summarized the results of the work thus:
The amount of solar energy reaching the grassland was 1.91 x 10^6 kJ m^-2 y^-1. The grasses of the area lost 36 x 10^2 kJ m^-2 y^-1 in respiration and showed a net primary productivity (NPP) of 20.4 x 10^3kJ m^-2 y-1.
1. What is the gross primary productivity (GPP) for this area? Please show all of your workings.
2. Calculate the photosynthetic efficiency, i.e., the percentage of solar energy reaching the vegetation that is converted to GPP.
3. Most of the NPP of the producers is lost by death or other pathways, but three particular groups of animals feeding on the grass were also studied:
a. field mice , which ingested 107 kJ m^-2y^-1 and lost 105 kJ m^-2y^-1 in respiration and faeces;
b. grasshoppers, which ingested 444 kJ m^-2y^-1 and lost 374.3 kJ m^-2y^-1 in respiration and faeces;
c. small birds, which ate grass seeds with an energy value of 60 kJ m^-2y^-1 and lost 59.2 kJ m^-2y^-1 in respiration and faeces;
CALCULATE the increase in biomass of each of the three groups: field mice, grasshoppers and small birds.
d. The energy value of the grasshopper population that is consumed by spiders is 0.8 kJ m^-2y^-1. In this habitat, grasshoppers do not survive beyond the end of the year. CALCULATE the percentage of the total energy value by grasshoppers lost by death and other pathways.
e. Some important groups of organisms found in each ecosystem have been omitted from this account . STATE what they are and their functions.
f. Please state the trophic level of each group of organism in this food web.
https://brainmass.com/biology/food-webs/gross-primary-production-and-increase-in-biomass-110746
#### Solution Preview
1. Gross Primary Production, GPP, is the total amount of CO2 that is fixed by the plant in photosynthesis.
Respiration, R, is the amount of CO2 that is lost from an organism or system from metabolic activity. Respiration can be further divided into components that reflect the source of the CO2.
Net Primary Production, NPP, is the net amount of primary production after the costs of plant respiration are included (NPP = GPP - R).
Therefore GPP = NPP + R = (20.4 x 10^3kJ m^-2 y-1)+ (36 x 10^2 kJ m^-2 y^-1) = 24.0 x 10^3kJ m^-3 y-1
2. Calculate the photosynthetic efficiency, i.e., the percentage of solar energy reaching the vegetation that is converted to GPP.
The photosynthetic efficiency is the fraction of light energy converted into other forms of ...
#### Solution Summary
Calculation of Gross Primary Production based on solar energy, respiration rates, and net primary productivity. Calculation of increase in biomass of field mice, grasshoppers and small birds.
\$2.19
## Biology and Society
Chapter 2: Biology and Society, Question 14, p.34-
? One solution to the problem of acid precipitation caused by emissions from power plants is to use nuclear power to produce electricity. The proponents of nuclear power contend that it is the only way that the United States can increase its energy production while reducing air pollution, because nuclear power plants emit little or no acid-precipitation-causing pollutants. What are some of the benefits of nuclear power? What are the possible costs and dangers? Do you think we ought to increase our use of nuclear power to generate electricity? Why or why not? If a new power plant were to be built near your home, would you prefer it to be a coal-burning or nuclear plant? Why?
Chapter 3: Biology and Society, Question 15, p.53-
? Each year, industrial chemists develop and test thousands of new organic compounds for use as pesticides, such as insecticides, fungicides, and weed killers. In what ways are these chemicals useful and important to us? In what ways can they be harmful? Is your general opinion of pesticides positive or negative? What influences have shaped your feelings about these chemicals?
Chapter 4: Biology and Society, Question 12, p.71-
? Doctors at a university medical center removed John Moore's spleen, which is standard treatment for his type of leukemia. The disease did not recur. Researchers kept the spleen cells alive in a nutrient medium. They found that some of the cells produced a blood protein that showed promise as a treatment for cancer and AIDS. The researchers patented the cells. The U. S. Supreme Court ruled against Moore, stating that his lawsuit "threatens to destroy the economic incentive to conduct important medical research." Moore argued that the ruling left patients "vulnerable to exploitation at the hands of the state." Do you think Moore was treated fairly? Is there anything else you would like to know about this case that might help decide?
Chapter 5: Biology and Society, Question 14, p.87-
? Lead acts as an enzyme inhibitor, and it can interfere with the development of the nervous system. One manufacturer of lead-acid batteries instituted a "fetal protection policy" that banned female employees of childbearing age from working in areas where they might be exposed to high levels of lead. Under the policy, women were involuntarily transferred to lower-paying jobs in lower-risk areas. A group of employees challenged the policy in court, claiming that it deprived women of job opportunities available to men. The U.S. Supreme Court ruled the policy illegal. Nonetheless, many people are uncomfortable about the "right" to work in an unsafe environment. What rights and responsibilities of employers, employees, and government agencies are in conflict in this situation? Whose responsibility should it be to determine what makes a safe environment and who should it be to determine what makes a safe environment and who should or should not work there? What criteria should be used to decide?
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Map Index Random Help Topics
## Topic: trees
topics > computer science > data > Group: data structures
Group:
sequences
Topic:
b-trees
Topic:
graphs
Topic:
hierarchical structures
Topic:
sparse arrays
Topic:
structured programming
Topic:
suffix trie and suffix array
Topic:
syntax analysis
#### Summary
In computer science, a tree is a hierarchical data structure of nodes and leaves. There is one root node. A node may have one or more children. Leaves are nodes without children. There may be a fixed or variable number of children per node.
Trees are useful data structures. Types of trees include binary tree, B-tree, syntax tree, suffix trie, height-balanced tree, and concatenation tree. (cbb 6/06)
Subtopic: tree transformation
Quote: transform tree structures by transforming a linear representation of the tree; use string transformations [»chriC8_1966, OK] QuoteRef: chriC8_1966 ;;570 Ambit for manipulating trees-- follow up
Subtopic: tree traversal
QuoteRef: morrJH8_1972 ;;760 treewalk special-range with node: @@ and leaf: expr2 specific, not understood treewalking
Subtopic: subtree
Quote: with hierarchical organization; users want to operate on subtrees [»akscRM5_1984]
Subtopic: trees for data communication
Quote: the Connection Machine uses trees and butterflies for collecting, combining, and spreading information, e.g., sets; need two-way pointers [»hillWD_1985] Quote: a butterfly structure avoids the exponential behavior of trees by keeping the levels constant sized; omega network, perfect shuffle, FFT [»hillWD_1985]
Subtopic: suffix tree or trie
Quote: compress assembly code by identifying common substrings via suffix trees; 7% compression on average [»frasCW6_1984] Quote: Bonsai trees compress large trie structures to 1/3; encodes indices with 32 by 32 bit multiplication [»darrJJ3_1993] Quote: on-line construction of suffix tree in O(N) [»nelsMR8_1996]
Subtopic: concatenation tree
Quote: represent rope/cord as an ordered concatenation tree with flat strings for leaves and shared nodes with other ropes [»boehHJ12_1995] Quote: efficient operations on ropes/cords by storing the length of the sub-tree in each node Quote: enhancements to ropes/cords: rebalance tree, user-defined functions at leaf nodes, substring nodes
Subtopic: B-tree
Quote: each B-tree level increases size by 400x; at most two physical I/O per operation due to caching highest levels [»bayeR_2002]
Subtopic: tree access
Quote: access a tree structure by specifying a path or a matching pattern [»elsoM3_1970, OK] QuoteRef: kiebRB9_1973 ;;4.3 p(n) father of p(n) has composite type of p(n)->p(n) where p(n) is a pointer to a note (this forms a function returning type p(n)) QuoteRef: kiebRB9_1973 ;;%Earley: pointer in a data structure is an access path
Subtopic: tree representation
Quote: stores data as multi-level strings from character, to symbol, ..., to library [»mullAP_1964, OK] QuoteRef: mullAP_1964 ;;452 definition by n A n(3)(3)H(1)B(2)C(1)D(3) then can "insert the value string into another, or "join" two named strings together(retainnames) Quote: threaded binary tree for detection and correction of errors: link null-pointers in a special way [»taylDJ11_1980, OK]
Subtopic: implicit representation
Quote: can implicitly represent regularly structured trees and butterflies by address [»hillWD_1985] QuoteRef: sammJE_1969 ;;456 matched parens to show tree structure %!!
Related Topics
Group: sequences (7 topics, 97 quotes)
Topic: b-trees (16 items)
Topic: graphs (18 items)
Topic: hierarchical structures (46 items)
Topic: sparse arrays (6 items)
Topic: structured programming (27 items)
Topic: suffix trie and suffix array (20 items)
Topic: syntax analysis
(29 items)
Updated barberCB 3/05
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High School Dude
For my science homework, I have to calculate the size of a shadow on a screen.
The object is a vertical pencil 15cm high which is 30cm away from a small spotlight which is almost level with the pencil. The screen is 1m away from the spotlight.
Please, I need help. Is there some sort of formula for calulating the size of a shadow formed by a certain object aginst a screen.
rpenner
Think about triangles. Light spreads in straight lines, so the source of the light is in a straight line with all of shadows of the top of the pencil. Also, the source of the light is in a straight light with all of the shadows of the bottom of the pencil. So if the screen and the pencil are parallel, we have a situation of similar triangles. Therefore the problem reduces to one of units and ratios.
CODE
+---------+ X / 1 m (at screen)
| /
| /
| /
| /
+----+ 15 cm / 30cm (at pencil)
| /
| /
| /
|/
+ source
X = 1 m * 15 cm / 30 cm = 100 cm * 1/2 = 50 cm.
Image by photoshop, copyright 2006 by Triangle Technologies Inc.
PhysOrg scientific forums are totally dedicated to science, physics, and technology. Besides topical forums such as nanotechnology, quantum physics, silicon and III-V technology, applied physics, materials, space and others, you can also join our news and publications discussions. We also provide an off-topic forum category. If you need specific help on a scientific problem or have a question related to physics or technology, visit the PhysOrg Forums. Here you’ll find experts from various fields online every day.
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# National Defence Academy Examination (I) – NDA & NA Exam I 2017 Syllabus
1. National Defence Academy Examination (I) – NDA & NA Exam I 2017 SyllabusSyllabus – General Principles
Syllabus of the UPSC NDA 1 Examination 2017
Paper I
Mathematics
( Code No. 01 )
( Maximum marks – 300 )
1. Algebra : Concept of a set, operations on sets, Venn diagrams. De Morgan laws, Cartesian product, relation, equivalence relation.
Representation of real numbers on a line. Complex numbers – basic properties, modulus, argument, cube roots of unity. Binary system of numbers. Conversion of a number in decimal system to binary system and vice-versa.
Arithmetic, Geometric and Harmonic progressions. Quadratic equations with real coefficients. Solution of linear inequations of two variables by graphs. Permutation and Combination. Binomial theorem and its application. Logarithms and their applications.
2. Matrices and Determinants : Types of matrices, operations on matrices. Determinant of a matrix, basic properties of determinants. Adjoint and inverse of a square matrix, Applications – Solution of a system of linear equations in two or three unknowns by Cramer’s rule and by Matrix Method.
3. Trigonometry : Angles and their measures in degrees and in radians. Trigonometrical ratios. Trigonometric identities Sum and difference formulae. Multiple and Sub – multiple angles. Inverse trigonometric functions. Applications – Height and distance, properties of triangles.
4. Analytical Geometry of Two and Three Dimensions : Rectangular Cartesian Coordinate system. Distance formula. Equation of a line in various forms. Angle between two lines. Distance of a point from a line. Equation of a circle in standard and in general form. Standard forms of parabola, ellipse and hyperbola. Eccentricity and axis of a conic.
Point in a three dimensional space, distance between two points. Direction Cosines and direction ratios. Equation two points. Direction Cosines and direction ratios. Equation of a plane and a line in various forms. Angle between two lines and angle between two planes. Equation of a sphere.
5. Differential Calculus : Concept of a real valued function – domain, range and graph of a function. Composite functions, one to one, onto and inverse functions. Notion of limit, Standard limits – examples. Continuity of functions – examples, algebraic operations on continuous functions. Derivative of function at a point, geometrical and physical interpretation of a derivative – applications.
Derivatives of sum, product and quotient of functions, derivative of a function with respect of another function, derivative of a composite function. Second order derivatives. Increasing and decreasing functions. Application of derivatives in problems of maxima and minima.
6. Integral Calculus and Differential Equations : Integration as inverse of differentiation, integration by substitution and by parts, standard integrals involving algebraic expressions, trigonometric, exponential and hyperbolic functions. Evaluation of definite integrals – determination of areas of plane regions bounded by curves – applications.
Definition of order and degree of a differential equation, formation of a differential equation by examples. General and particular solution of a differential equations, solution of first order and first degree differential equations of various types – examples. Application in problems of growth and decay.
7. Vector Algebra : Vectors in two and three dimensions, magnitude and direction of a vector. Unit and null vectors, addition of vectors, scalar multiplication of a vector, scalar product or dot product of two vectors. Vector product or cross product of two vectors. Applications work done by a force and moment of a force, and in geometrical problems.
8. Statistics and Probability :
Statistics : Classification of data, Frequency distribution, cumulative frequency distribution – examples. Graphical representation – Histogram, Pie Chart, Frequency Polygon – examples. Measures of Central tendency – Mean, Median and Mode. Variance and standard deviation – determination and comparison. Correlation and regression.
Probability : Random experiment, outcomes and associated sample space, events, mutually exclusive and exhaustive events, impossible and certain events. Union and Intersection of events. Complementary, elementary and composite events. Definition of probability – classical and statistical – examples. Elementary theorems on probability – simple problems. Conditional probability, Bayes’ theorem – simple problems. Random variable as function on a sample space. Binomial distribution, examples of random experiments giving rise to Binominal distribution.
Paper II
General Ability Test
( Code No.02 )
( Maximum Marks – 600 )
Part A – English ( Maximum marks 200 )
The question paper in English will be designed to test the candidate’s understanding of English and workman like use of words. The syllabus covers various aspects like : Grammar and usage, vocabulary, comprehension and cohesion in extended text to test the candidate’s proficiency in English.
Part B – General Knowledge
( Maximum marks – 400 )
The question paper on General Knowledge will broadly cover the subjects : Physics, Chemistry, General Science, Social Studies, Geography and Current Events.
The syllabus given below is designed to indicate the scope of these subjects included in this paper. The topics mentioned are not to be regarded as exhaustive and questions on topics of similar nature not specifically mentioned in the syllabus may also be asked. Candidate’s answers are expected to show their knowledge and intelligent understanding of the subject.
Section ‘A’ ( Physics )
Physical Properties and States of Matter, Mass, Weight, Volume, Density and Specific Gravity, Principle of Archimedes, Pressure Barometer.
Motion of objects, Velocity and Acceleration, Newton’s Laws of Motion, Force and Momentum, Parallelogram of Forces, Stability and Equilibrium of bodies, Gravitation, elementary ideas of work, Power and Energy.
Effects of Heat, Measurement of temperature and heat, change of State and Latent Heat, Modes of transference of Heat. Sound waves and their properties, Simple musical instruments.
Rectilinear propagation of Light, Reflection and refraction. Spherical mirrors and Lenses. Human Eye.
Natural and Artificial Magnets, Properties of a Magnet, Earth as a Magnet.
Static and Current Electricity, conductors and Non-conductors, Ohm’s Law, Simple Electrical Circuits, Heating, Lighting and Magnetic effects of Current, Measurement of Electrical Power, Primary and Secondary Cells, Use of X – Rays.
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### 63 Responses to “National Defence Academy Examination (I) – NDA & NA Exam I 2017 Syllabus”
1. sushant January 5, 2012 at 12:16 pm # Reply
how 2 check application status????
2. mohd baqir January 8, 2012 at 12:16 pm # Reply
qualification
3. isaac gabriel M January 10, 2012 at 6:43 pm # Reply
sir,
I didn’t receive my submit card i have applied to nda online
i didn’t get any details of examination please help me my number is 9551032554
thanking you sir
yours faithfully
isaac
4. ravi prakash singh January 14, 2012 at 7:53 pm # Reply
what is pattern of attempting nda?
commerce students eligeble for nda or not.
please give me information on 08574900595.
thanks!
5. Ajit kumar mishra January 20, 2012 at 7:00 pm # Reply
1.How many time i can attempt examination of N D A ?
6. sushil kumar January 29, 2012 at 6:44 pm # Reply
sir i did not recieve my admit card i did not get any information about exam please help me .
7. mohit February 3, 2012 at 5:12 pm # Reply
which date come nda form on 2012
8. tarun February 16, 2012 at 1:40 pm # Reply
is it that much tuff to clear nda exam?
9. vikas vicky February 19, 2012 at 7:45 pm # Reply
i m in 11th can i m eligable for nda .
10. muana February 20, 2012 at 9:53 am # Reply
Hello Sir…
My self Sagar Singh from Varanasi.
I am student of 12th & my subject is Math & Commerce. Can i apply for the NDA examination?
Thanks…
11. vedant patil February 23, 2012 at 8:07 am # Reply
sir,
I have not received the details of examination, I have applied on-line for NDA on 31/10/2010.
Please update me, my number is 91-9503631602
Yours faithfully
Vedant Patil
12. kunal sharma February 28, 2012 at 9:02 pm # Reply
sir when will i recieve my admit card registration no 11208851379
13. balamurugan March 7, 2012 at 11:57 pm # Reply
hai i am ready to serve my country ,now my age 23 pls help me to join my no.7845367325
14. Dharamveer Goyal March 11, 2012 at 4:37 pm # Reply
Sir,
kindly sent me the name list of cities where the NDA exam is taken
and when will be the forms for NDA-2 for 2012 will available in post offices.
my no. is 9914720193.
15. Yugandhar Parashar March 13, 2012 at 12:16 am # Reply
hello sir my birthday is 09-04-1996 can i give nda exam.
• UpscExam March 13, 2012 at 11:38 am # Reply
Hi
Please see the details given below :
Only unmarried male candidates born not earlier than 2nd July 1994 and not later than 1st January, 1996 are eligible”. The date of birth accepted by the Union Public Service Commission is that entered in the Matriculation or Secondary School Leaving Certificate or in a certificate recognized by an Indian University as equivalent to Matriculation or in an extract from a Register of Matriculates maintained by a University which must be certified by the proper authority of the University or in the Higher Secondary or an equivalent examination certificates.
16. kishan ml March 16, 2012 at 10:47 am # Reply
where shall we get application form? how much it costs?
• UpscExam March 19, 2012 at 10:54 am # Reply
Hi
Application for NDA Exam 2012 will be available on this UPSC website : http://upsconline.nic.in/. The application for NDA Exam 2012 costs about Rs. 50/-.
17. priyesh shukla March 17, 2012 at 10:44 am # Reply
sir,, I have not got my admit card till now, i have applied NDA online. please help me, my number is 8097829402 thanking you sir
18. SUMIT singh March 18, 2012 at 11:56 am # Reply
sir i dont understand the syllabus please give some hints about the syllabus
19. Jagmohan March 22, 2012 at 10:36 am # Reply
Jai Hind sir mujhe abhi tak mera admitcard nahi mila hai please muje status mere mob no par batya thanku mob 9813831612
20. Mansi March 23, 2012 at 9:51 pm # Reply
what is the eligibility for appearing in nda exam? on clearing the exam, what are the various available fields for making a career in?
21. Saurabh March 24, 2012 at 4:04 pm # Reply
sushant :how 2 check application status????
• UpscExam March 26, 2012 at 10:49 am # Reply
E-Admit Card for NDA Exam is not yet issued, please wait…
22. Saurabh March 24, 2012 at 4:08 pm # Reply
Pls. note that i have not yet received my hall ticket for NDA -I examination which is to be held on 15th April 2012. I had submitted my form by on line my registration id is 11203730486. MayI request you to kindly help me or suggest me where should i contact for hall ticket and my center for the examination. Thanking you,
Saurabh
23. thoran March 29, 2012 at 8:56 am # Reply
sir,when the halltickets will be issued
• UpscExam March 29, 2012 at 10:27 am # Reply
Hi
Please wait and visit : http://upsc.gov.in/exams/exams.htm#Venue%20Information
• UpscExam March 29, 2012 at 11:45 am # Reply
Hi
The eligible candidates shall be issued an e-Admission Certificate three weeks before the commencement of the examination. The e-Admission Certificate will be made available in the UPSC website [www.upsc.gov.in] for downlading by candidates. No Admission Certificate will be sent by post.
If a candidate does not receive his / her e-Admission Certificate or any other communication regarding his / her candidature for the examination three weeks before the commencement of the examination, he / she should at once contact the Commission. Information in this regard can also be obtained from the Facilitation Counter located in the Commissions office either in person or over phone Nos. 011-23385271 / 011-23381125 / 011-23098543. In case no communication is received in the Commission\’s Office from the candidate regarding non-receipt of his / her e-Admission Certificate at least three weeks before the examination, he / she himself/herself will be solely responsible for non-receipt of his / her e-Admission Certificate. No candidate will ordinarily be allowed to take the examination unless he / she holds an e-Admission Certificate for the examination.
On the receipt of e-Admission Certificate, candidates should check it carefully and bring discrepencies / errors if any, to the notice of UPSC immediately. The mere fact that an e-Admission Certificate to the Examination has been issued to a candidate, will not imply that his / her candidature has been finally cleared by the Commission or that entries made by the candidate in his / her application for the examination have been accepted by the Commission as true and correct. Candidates may note that the Commission takes up the verification of eligibility conditions of a candidate, with reference to original documents, only after the candidate has qualified for interview for Personality Test on the results of the Written Examination. Unless candidature is formally confirmed by the Commission, it continues to be provisional.
The decision of the Commission as to the eligibility or otherwise of a candidate for admission to the Examination shall be final. Candidates should note that the name in the e-Admission Certificate in some cases, may be abbreviated due to technical reasons.
Download E-Admit Card for NDA Exam now through online : http://upsc.gov.in/exams/exams.htm#Venue%20Information
24. Harsh April 1, 2012 at 9:28 pm # Reply
Sir,
I have Not taken Maths in my +12. Can i give the exam of NDA??
Please let me know as soon as possible.
Yours faithfully,
Harsh Bhandari
25. karan singh bhati April 4, 2012 at 2:58 pm # Reply
26. AVIJIT CHOWDHURY April 5, 2012 at 2:17 pm # Reply
sir,
Can a Student be selected in NDA if he couldnt pass the 12th exam..?? plz Answer me..
27. Rahul kumar April 12, 2012 at 1:15 am # Reply
sir, after qualifying NDA(I) What will happen in future ???? Is there any type of engineering course happening or not ??????????
28. mayank April 16, 2012 at 8:28 pm # Reply
i have not received my admit card yet
29. sani April 26, 2012 at 2:31 pm # Reply
i didn’t get my admit card. i have applied it online. how to check application status.help me please
30. AMIR SOYEL May 7, 2012 at 10:32 pm # Reply
WHAT IS THE LANGUAGE OF NDA EXAM?
PLEASE SIR GIVE ME THE ANSWERE ITS URGENT
31. suri May 10, 2012 at 2:59 pm # Reply
sir when nda & na (1) 2012 results will be announced…….plz mail me the details to my email account.
suri
32. Aman Anand May 12, 2012 at 2:06 pm # Reply
hello sir,
my self Aman Anand from patna
i am student of class 12th & my subject is physics,chemistry and biology.can i apply for N.D.A.
thanks
JAI HIND
33. Ankur Ashutosh May 13, 2012 at 4:34 pm # Reply
Respected sir,
I want to know the answers of following questions
1.the gap provided between written test and SSB test ?
2.how will you inform about the place allotted for SSB ?
• UpscExam May 14, 2012 at 11:30 am # Reply
34. Ajay May 17, 2012 at 7:45 pm # Reply
Respected Sir
can u tell me ?
how i fill the nda form ?
and when is exam ?
i doing the +2 non-med can i fill the nda form ?
plzzzzzz sir help me my num is 9780918208
35. yadvender May 24, 2012 at 9:15 pm # Reply
i have opted for commerce stream in 12th class..my question is..can i still appear for NDA examinations??is there any eligibilty for such stream holders in NDA???
36. kartik suresh June 4, 2012 at 4:03 pm # Reply
sir,
I didn’t receive my submit card i have applied to nda online
i didn’t get any details of examination please help me my number is 9995546040
thanking you sir
yours faithfully
Kartik
37. sumit June 5, 2012 at 6:49 pm # Reply
i have completed the registration but dont have any idea that when i will the hall ticket and also about the where i m going to write the NDA exam..
38. aditya June 8, 2012 at 11:40 am # Reply
query about nda exam 21aug,2011?
39. vishal June 16, 2012 at 5:37 pm # Reply
jai hind
40. jatin rana June 18, 2012 at 3:28 pm # Reply
sir,
this is to kindly inforn that i jatin rana of 12th will be appearing for nda exams this year so kindly tell me from where do i get the sampel papers for my practice. I will be highly oblidged for your feedback.
Thanking u,
jatin rana
41. PRATIK June 22, 2012 at 5:22 pm # Reply
What about my admit card? As exam. on 21st AUGUST.
42. vaibhav negi July 5, 2012 at 12:29 pm # Reply
when will be the written results of nda1 2012 declared
43. barkat ali August 13, 2012 at 1:02 pm # Reply
sir i am the student of 12 my age is19 (183\1993) till now i want to ask that i am eligibale for the nda exam of 2012-2013 or not plzzzzzzzzzz inform me on my no (9990575505-8285833089) or email id jamshedmalikk@gmail.com
44. vaibhav August 13, 2012 at 8:53 pm # Reply
i have not got my admit card….plz tell when it will reach…only one weeek is remaining…..
45. Sagar November 25, 2012 at 8:31 am # Reply
Hello Sir…
My self Sagar Singh from Varanasi.
I am student of 12th & my subject is Math & Commerce. Can i apply for the NDA examination?
Thanks…
• UpscExam November 25, 2012 at 11:36 am # Reply
Hi
Pls see the details given below :
National Defence Academy NDA & NA Exam (I) Educational Qualifications :
1. for Army wing of National Defence Academy : 12th Class pass of the 10+2 pattern of School Education or equivalent examination conducted by a State Education Board or a University.
2. For Air Force and Naval Wings of National Defence Academy and for the 10+2 Cadet Entry Scheme at the Indian Naval Academy : 12th Class pass of the 10+2 pattern of School Education or equilvalent with Physics and Mathematics conducted by a State Education Board or a University.
Candidates who are appearing in the 12th Class under the 10+2 pattern of School Education or equivalent examination can also apply for NDA examination.
Regards.
46. Akash February 16, 2013 at 8:23 am # Reply
I have applied for NDA Exam-2013..At What time the Exam will commence on 14 April 2013 at Chennai? I understand there are Two papers, each 2.5 Hrs of duration..I have to book the Train Reservation…Kindly help in intimating the location and start, end time of Exams. Thank you.
47. shashank May 26, 2013 at 10:16 pm # Reply
hi entered wrong educational qualification in my part 1 registration and also completed my part 2 registration. can i change my educational qualificationn nw??
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