aPlatypus
#help-28
glossy valveBOT
kindred grove
yea
untold vessel
so we have L(...)
where we write it as L(z(...)) because z is a function found in the image where LeakyReLU(z) is described
so we get the derivative we use multivariable chain rule
kindred grove
yeah I mean the L(z(...)) is just a hint to help you notice that you have a composition of 2 functions
untold vessel
if I was sitting an exam, whats a good rule to remember that we have a composition of 2 functions?
from what I have understood, we get a gradiant:
(dL/dZ * dZ/dw0, dL/dZ * dZ/dw1, dL/dZ * dZ/dw2, dL/dZ * dZ/dw3)
kindred grove
yeah that's the whole gradient
untold vessel
but how come its w0, w1, w2 instead of x1, x2, x3 ?
kindred grove
cause of reasons
untold vessel
hha
kindred grove
you're doing a neural networks class I guess
untold vessel
yep
kindred grove
to change the parameters of your neural network, you'll want to compute gradients wrt to parameters of your model (w0, w1, w2, ...) not the inputs
that's why we don't really care about gradients wrt to the inputs to the model
untold vessel
right, okay
x1 = 1, x2 = 2, x3 = 3 and weights are given as:
w0 = -10, w1=w2=w3=1
I have to calculate the gradient based on these
But I don't know how to compute the value of dL/dZ
kindred grove
what did you get for dL/dz ?
kindred grove
yeah z is -4 here
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hazy drum
could somebody help me with this? i’m not sure where to start
proper hawk
Do implicit differentiation
primal stag
can you rotate pls
proper hawk
what help do you need
hazy drum
primal stag
just subsitute i guess try to solve x.d
hazy drum
do i just implicitly differentiate the x^3y=asinnx?
primal stag
try and check if u really get 0
hazy drum
because there’s an a in there and idk where it would go
proper hawk
do i just implicitly differentiate the x^3y=asinnx?
try it first
I'm sure a will cancel eventually
hazy drum
i did
hazy drum
this is what i have
idk how to differentiate acos(nx)n
that’s three things multiplied together
i probably did it wrong
proper hawk
$\frac{d}{dx}a\sin{nx}$
glossy valveBOT
WhereWolf(ping if needed)
glossy valveBOT
WhereWolf(ping if needed)
hazy drum
ohh
how do we know it’s a constant though?
couldn’t it have x’s in it
since it’s a variable it can have anything
or am i wrong
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snow crest
please correct this <3
snow crest
<@&286206848099549185>
queen crater
I don't know the notation, does S stand for side and A for angle?
snow crest
I don't know the notation, does S stand for side and A for angle?
Yes.
SSS = Side-Side-Side
SAS = Side-Angle-Side
ASA = Angle-Side-Angle
AAS = Angle-Angle-Side
queen crater
Then the first one only has two pairs of equal sides.
torn jolt
first one i guess is SAS
snow crest
here is the blank
torn jolt
vertical angles are congruent
queen crater
The third one only has two pairs of equal angles and no pair of equal sides
And the fifth one looks odd
snow crest
this look good?
snow crest
The third one only has two pairs of equal angles and no pair of equal sides
hmmm so AAS?
teal hearth
No
queen crater
Wait I actually got confused by the figure
It's only one pair of equal angles and one pair of equal sides
Hint: all of these acronyms have three letters because you need three pieces of information in order to show that two triangles are congruent
snow crest
I got it!
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torn jolt
help pls. i dont understand why- (the highlighted part)
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help pls. i dont understand why- (the highlighted part)
willow sedge
well, let the AP be a1,a2,a3,a4 with common difference d
The general term would be a+(n-1)d
Now reverse the AP
a_n,a_(n-1)... would be another AP with common difference equal to (-d)
So, the general term would be a_n+(n-1)(-d)
Here a_n is represented by l
So, the general term is given by l-(n-1)d
@torn jolt
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torn jolt
<@456226577798135808>
is this correct?
willow sedge
yes
steady swallow
Please help me... We are with 5 people which 3 have too work at a location.. We have 2 people extra that can rotate Tru out the whole day.. We work from 1800-2300 and 2400-0500
Thats 10 hours in total.. We want too give everyone the same amount of break via rotation from the 2 extra persons.. Can someone please help me?? I asked chat gpt but it can't give a good answer
steady swallow
Please help
onyx glen
here is what i understand right now:
- you have a location that needs to be staffed by 3 people at all times
- every day, there are two shifts: 18:00-23:00 and 00:00-05:00
- you have 5 people that can work on the location
- you want to design a schedule that gives everybody the same amount of rest
@steady swallow is this correct?
onyx glen
that's why i am asking op to confirm
neon basin
Some friendly advice: don't ask ChatGPT anything other than inquiry questions
Especially not anything math-related.
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steady swallow
Hey sorry
I am working At the momemt
So didn't have time too respond
So in total we work 12 hours
1 hour is break
The 23-24 is normal break
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steady swallow
here is what i understand right now:
- you have a location that needs to be sta...
Hey Ann the only thing that's wrong with this is that there are no 2 shifts
steady swallow
The one hour break is a normal break that everyone needs too have.. Today we are just lucky having 2 people extra that can rotate
gritty rose
can you pick a channel to close
gritty rose
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elder karma
hi there! i have a pretty weird one, i need to know the number of times id need to divide a very big number by a quite big number to get close to 0, because dividing it once barely changes it visually
elder karma
the big number is 17.8 nonillion digits long btw
7.3e+17860623038301067137151847182024564090627624290867126830058
&
5.0005e+81
viral jasper
Define close to 0
viral jasper
Use base 10 logarithms
elder karma
what might that be?
viral jasper
You've not learned logarithms yet?
elder karma
nah
viral jasper
Then who gave you this problem? What grade is this for?
fast peak
roughly speaking, each time you divide by an 80 digit number you lose 80 digits. so you need to check how often 80 fits into 17.8 nonillion
elder karma
this is my own personal project, finding out the number of Minecraft world permutations. now ive found it, and im trying to show the scale of it by dividing it by the number of atoms in the universe a ** number of times
elder karma
roughly speaking, each time you divide by an 80 digit number you lose 80 digits....
wait so i just divide number of digits of big number by number of digits of smaller number?
to get the digits of number of times id need to divide it by to get it small
fast peak
yes
elder karma
ok, so 2.2e+56 is the result i got, would that be the number of times, or the number of digits of the number of times?
fast peak
the number of times
runic spruce
I used this formula and subbed in the respective things, but wanted to ask if I need to divide by Q at any point
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ivory moon
How do I prove that sqrt(m) - sqrt(3) is irrational for all natural number m with m≠3?
ivory moon
I tried induction but got stuck on the inductive step
ivory moon
Ahh no latex
Okay I got m is (a^2 +2absqrt(3) +3b^2)/b^2
tranquil cave
the numerator is an irrational number
ivory moon
Ah yes but m is a natural number so it’s a contradiction
Thanks for ur help
I might need more help so I’ll leave it here for now
tranquil cave
And in the case when m is 3 a is 0 so thats why its working (i just tell this cause i was looking for a good minit where did we used the m=3 case)
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astral lintel
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astral lintel
<@&286206848099549185>
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jade totem
Is my solution right ?
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frail quarry
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frail quarry
i want to ask why the answer is 2/3
since i used u sub to solve the problem using 3+x^2 for my u
and then finding the new x values due to using u sub
hot herald
show your work for usub
frail quarry
show your work for usub
when i do this method this is the work but the answer is wrong
i forgot to add a power of 2 to the first fraction in the computation but it still gives 0 cause of the limit being applied
hot herald
you're mixing up how you're applying your bounds,
if you're updating the bounds, those bounds are for your new variable of integration
there's no longer any need to bring back x into this
if you were to bring back x, you'd use those original bounds for x
those bounds are u= 3+t^2 and u=3
and you can use those directly
if using
you'd go with x=0 to x=t→inf
frail quarry
those bounds are u= 3+t^2 and u=3
and you can use those directly
so we woudlnt apply the orginal u sub back into the problem after completion
hot herald
that more for integrals without bounds where you want to express the result in terms of the original variable
if you're updating the bounds as you go, that isn't needed
you "can" still do it with the proper notation if you really want
and you'd need to be careful with what bounds you use
frail quarry
and you'd need to be careful with what bounds you use
i see
well thank i apperciate thats something i didnt know
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lucid vessel
how do I find all the complex roots of x^5 + 1
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how do I find all the complex roots of x^5 + 1
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vague flame
I'm trying to make a power series using the first four nonzero terms of this function, but it does not work when starting at k=0, only at k=1
vague flame
rocky vale
you can write odd numbers as 2k+1 instead of 2k-1
That way k=0 gives 1
rather than k=1 giving 1
vague flame
i'm trying this
thank you
rocky vale
no problem 👍
vague flame
so actually it's changing every minus sign into a plus?
rocky vale
no, not the (-1)^{k+1} and 4^{k-1}
those will have to change
but it won't just be the opposite sign
vague flame
i'm confused on how to change the alternator to work at k=0
vague flame
thank you
i was overthinking it
rocky vale
it happens lol no problem
vague flame
is there a formula for this or do I just rationalize it because i've just been rationalizing it
vague flame
like i've just been looking at the numbers and surmising what the function should be like
it's a lengthy process for me
rocky vale
yeah, a taylor series centered at $x=a$ is $\sum_{k=0}^{\infty} \frac{f^{(n)}(a) \cdot (x-a)^n}{n!}$
glossy valveBOT
tatpoj
vague flame
this makes it easier i think 😭
rocky vale
I didn't check your work on finding the individual terms but that matches them
oh
🎉
vague flame
this makes me feel happy! (not an exclamation mark that is happy factorial)
vague flame
thank you 🫡 i will try to find the interval of convergence now
not in this chat though i think i got it
rocky vale
sounds good, good luck mate
torn jolt
A rotation Ro, and a symmetry whose axis passes through the center of rotation O.
A symmetry whose axis passes through the center of rotation O and has rotated by an angle of -a/2.
torn jolt
Why it has rotated -a/2?
rocky vale
Can you post the original question? It's not really clear what you're trying to do here
torn jolt
Oh, it's not a question
like, it's not a problem
I am trying to understand why in my module says that it as rotated -a/2
torn jolt
Yes, well it's alpha I think but what I am asking is why it has rotated -a/2 mm
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eternal drum
dis how
eternal drum
i need write equation for the problem
ivory moon
Do you know the formula relating time distance and speed
ivory moon
What’s the formula for distance in terms of speed and time
ivory moon
Well so distance is the same
What is distance in terms of time if we let the time it takes to go to grand mom be t
eternal drum
speed x t?
ivory moon
Right
eternal drum
which speed tho
ivory moon
What is t
eternal drum
3 hrs
ivory moon
No how did I define t
ivory moon
Yes
So now we can form an equation for the first part of the journey
distance = 4t
Now consider when he comes back
What is the time it takes for he to come back if the total time is 3 hrs
ivory moon
If it takes t hrs to go from house to grand mom
And it takes 3 hrs to go from house to grand mom and back
How long does it take to get back
eternal drum
3-t?
eternal drum
distance = 6(3-t)?
eternal drum
so the equation would be 4t = 6(3-t)?
ivory moon
Right
eternal drum
ok
wait is it 12x + 30 = 4x if x is time
i feel like its wrong
i give up
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calm trail
$$\int _{-1}^1\int {y^2-1}^{\sqrt{1-y^2}}\int 0^{xy+1}dzdxdy:::$$ I've computed to $$\int{-1}^{1} -\frac{1}{2}y^5 + \frac{1}{2}y^3 - y^2 + \sqrt{1-y^2}dy:.$$ But somehow I am getting the wrong answer: $$-3y^6 + 2y^3 - y^3/3 + \frac{1}{2}(\arcsin \frac{1}{y} + y\sqrt{1-y^2})\mid{-1}^1 = -2/3 + \pi/2.$$ How am I messing up the integral
glossy valveBOT
Zander
astral belfry
**Zander**
doesnt the integrand here have another term? +1?
astral belfry
🙂
calm trail
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peak grotto
A car is accelerating at a fixed rate .2 per second and is acted upon by friction which is .9 times the velocity. At what velocity does the acceleration force equal the friction force? Thank you.
onyx glen
no units
"acceleration force"
this is suspicious as shit from a physics/mechanics standpoint
!xy
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Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
peak grotto
@onyx glen can you explain your objection?
it's just a simple system I'm trying to understand.
The acceleration is being added, and the friction is being multiplied, so at a certain velocity the forces equal eachother.
onyx glen
fixed rate .2 per second
i don't understand what units this 0.2 is meant to be in
did you maybe mean 0.2 m/s^2...?
and also friction is a force (measured in newtons) while velocity is, well, a velocity (measured in m/s) and one of them cannot equal 0.9 times the other
peak grotto
I don't think it needs to be squared.
onyx glen
m/s^2, also known as (m/s)/s, is the SI unit for acceleration.
peak grotto
I didn't think velocity is a force, because it remains constant unless acted upon by friction or acceleration.
onyx glen
velocity is in fact not a force, yes.
but friction is.
a velocity and a force can't be equal, it makes no physical sense for them to be so.
peak grotto
i didn't ask that.
onyx glen
and is acted upon by friction which is .9 times the velocity.
peak grotto
at what velocity does acceleration force equal friction force.
onyx glen
there isn't any such thing as an "acceleration force" either!
peak grotto
you're honestly not being helpful.
onyx glen
and again, you're posting things here that make zero sense from the viewpoint of classical mechanics
and i am trying to make sense of them
onyx glen
and that is why i asked you to post the original problem exactly as it was given to you, with all context
ok whatever
peak grotto
The problem wasn't given to me.
I'll explain the problem again, for the next person. @onyx glen have a nice day.
hot herald
The problem wasn't given to me.
wdym
where did the question come from?
the issue as explained is that
terms are improperly used
units are absent
and as a result what you've presented makes no sense, so no-one will be able to help as it's very unclear whats being asked
peak grotto
A car is accelerating at a fixed rate of .2 units per second.
In other words at 1 second the car is traveling .2 per second.
At 2 seconds the car is traveling at .4 per second.
The car is acted upon by a friction force, which is the current speed multiplied by .9.
In other words, in each second the car's current speed is multiplied by .9, thus reducing the car's speed.
In the case of the 'acceleration' each second the speed is being added to, and the amount never changes.
In the case of the 'friction' each second the speed is being reduced by a quantity that is multiplicative rather than additive.
Thus there is a speed at which the force slowing the speed of the car is equal to the force accelerating the car.
What is that speed.
If you don't like the question, please just move on, rather than criticizing me for asking it, thanks.
hot herald
where did the question come from?
peak grotto
From me. Please go elsewhere if you don't want to help me with my question. thanks.
hot herald
people want to help. but can't if the question doesn't make sense. doesn't matter how many people you try asking
peak grotto
<@&268886789983436800>
pseudo frigate
What's going on
peak grotto
I just want help with my quesiton. If you don't like the question please move on.
I'm just trying to ask a question, and people are just criticizing me for asking it.
onyx glen
we are not criticizing you for asking the question, we are criticizing the question + trying to make sense of it.
peak grotto
I just want to wait for someone who is willing to explore my question rather than trying to rebuff people who dont' think it's a valid question.
pseudo frigate
Well your question is a little strange as it's missing a few things, like units. Do you have a picture of the problem?
peak grotto
I could draw a picture.
pseudo frigate
If the helpee doesn't want help from specific users, probably best to leave them alone, though
fast peak
they are trying to "explore" your question. which in this case means asking things about it
because as stated it doesnt make much sense
peak grotto
Well explain how it could be improved then, rather than just saying my question is invalid.
Because actually it does make sense.
hot herald
From me. Please go elsewhere if you don't want to help me with my question. than...
when attempting to make up a problem, components may end up getting ill formed.
was there a similar problem from your book/resource with better wording
peak grotto
It's a simple system.
peak grotto
when attempting to make up a problem, components may end up getting ill formed. ...
please move on. thanks for trying. I'd rather talk to someone who is helpful.
fast peak
in terms of the numbers you are imagining maybe. but not in the context into which you put the numbers. because in that context numbers have units. and you cant for example compare 3 m/s with 3 newton
peak grotto
A car is accelerating at a fixed rate of .2 units per second.
In other words at 1 second the car is traveling .2 per second.
At 2 seconds the car is traveling at .4 per second.
The car is acted upon by a friction force, which is the current speed multiplied by .9.
In other words, in each second the car's current speed is multiplied by .9, thus reducing the car's speed.
In the case of the 'acceleration' each second the speed is being added to, and the amount never changes.
In the case of the 'friction' each second the speed is being reduced by a quantity that is multiplicative rather than additive.
Thus there is a speed at which the force slowing the speed of the car is equal to the force accelerating the car.
Let me know if you need any clarification on any particular aspect of the question. thanks.
fossil condor
You keep telling people off when they're trying to ask for clarification
Not the best way to get help
peak grotto
No one has asked for clarification they just are telling me the question is invalid.
A car is accelerating at a fixed rate of .2 units per second.
In other words at 1 second the car is traveling .2 per second.
At 2 seconds the car is traveling at .4 per second.
The car is acted upon by a friction force, which is the current speed multiplied by .9.
In other words, in each second the car's current speed is multiplied by .9, thus reducing the car's speed.
In the case of the 'acceleration' each second the speed is being added to, and the amount never changes.
In the case of the 'friction' each second the speed is being reduced by a quantity that is multiplicative rather than additive.
Thus there is a speed at which the force slowing the speed of the car is equal to the force accelerating the car.
what is that speed?
Let me know if you need any clarification on any particular aspect of the question. thanks.
olive saffron
There is some degree of pedantry occurring here, to be fair
The point that people are making is that acceleration is not itself a force
Nevertheless
An acceleration is proportional to a force via Newton's second law
peak grotto
The particulars of using 'physics' forces can be done away with if it's not helpful for the question. it's really a simple math question about an additive and a multiplicative change and where they intersect.
I don't really know what math to use to find the intersection.
olive saffron
Yes, the mathematics is simple
The equations you are looking for are F = ma and the equation for the type of friction you're working with
Though based on how you have worded the question it seems you're working with F = 0.9v for your friction force
And a = 0.2
Since you don't have mass you can't use F = ma (and do not need it here)
My bad
peak grotto
yeah or the mass can just be 1
olive saffron
This works also
Yeah no you need some value for mass
If m = 1 then F = ma = a = 0.2
So you have F = 0.2 and F = 0.9v
Using this you can solve for the velocity v at which these forces become equal
If you want to solve for time you will need v = u + at
This should be everything you need
peak grotto
ok, so if I solve for V, I get f/v = 0.9 right?
but I admit i don't really know how to interpret that.
ma/v = 0.9
1*.2/v = 0.9
.2/v = 0.9
.2=0.9v
.2/0.9 = v
I'm not sure if i'm doing that right or it's getting me anywhere.
olive saffron
This is the correct answer, yes
v = 2/9
You have taken a long-winded approach but a correct one nevertheless
Much easier is to observe that F = 0.2 and F = 0.9v implies that 0.2 = 0.9v
Then v = 2/9 follows immediately
peak grotto
hmm, thank you. Do you know how I can graph this so that I can see the intersection? That would make it make a lot more sense to me.
olive saffron
Depends what you want to visualise
Force over time? Force over velocity?
If you want it over time then you need to use v = u + at, where u is the initial velocity
Assuming u is 0
You end up with F = 0.2 and F = 0.9*0.2*t = 0.18t
You can plug these into something like desmos to see the result
In this case the car will stop moving very quickly since the friction is very high relative to the acceleration
peak grotto
yes a more appropriate value would be .995 for the 'car', I used .9 for convenience.
As to the graph, I am imagining a graph which shows the intersection of the acceleration and the friction. But then It gets hard for me to decide which common elements should be on each axis to show their intersection.
If I imagine the velocity increasing over time, then V and T are on the axis, but if I imagine the friction increasing with the velocity then F and V are on the axis.
I don't know if that means that the graph is 3d, or perhaps I can isolate them properly. I don't know. What do you think?
olive saffron
Friction is proportional to velocity is proportional to time
This whole relationship is linear
peak grotto
yeah.
olive saffron
If you plot F = 0.2 and F = 0.18t you find find these intersect
peak grotto
on what graph? v/t v/f?
olive saffron
Force over time
This is implicit from writing F = 0.18t
A graph, generally, is going to correspond to an equation y = f(x) for some function f
In this case we have simply relabelled y to F, x to t, and the function f is "multiply by 0.18"
This force over time graph is, imo, the most sensible way to represent the scenario
Interpreting a graph with velocity as the x axis can be difficult and can get very difficult indeed if your acceleration is non-constant
peak grotto
But what about that the velocity increases over time, and the friction increases over velocity?
Does this mean the two quantities can't be expressed on the same graph, but then I recall you say friction is proportional to velocity/time.
Since you have been entertaining this question, can you show me the equations in desmos that would plot this intersection?
If i can see the intersection I will be satisfied, and then spend some time working with the numbers as you worked out here. I havent' done that yet, as i'm slow lol.
it seems to make sense.
olive saffron
I'm on a phone so do not have desmos to hand, but all you need to do is type F = 0.2 and F = 0.18t and it'll do the plotting for you
As for whether they can be plotted on the same graph
olive saffron
Yes
I'm pretty sure desmos will interpret the variables correctly
To be safe you could type y = 0.2 and y = 0.18x
You could plot a force-time-velocity graph in 3d but it is not going to be very enlightening
And again, rather difficult to interpret
I don't know what it looks like
peak grotto
thank you. time isn't really important to what I'm trying to calculate, i'm interested in the relationship between the forces and the velocity.
olive saffron
I see
Well then plot F = 0.2 and F = 0.9v
Or y = 0.2 and y = 0.9x
Again you will see the intersection happens almost immediately since the friction is enormous
peak grotto
yeah, it's good. i haven't yet fully gotten the transformation done on the equations to bring the velocity = .2 time, and friction = .9 velocity equations into alignment yet, but it seems to make sense.
olive saffron
Well the friction = 0.9 velocity thing is just what you wrote above
Well, kinda
You actually wrote two different things
The first was that the friction force is 0.9 * velocity
The second was that each second the car's velocity is multiplied by 0.9
peak grotto
yes.
olive saffron
The latter doesn't really correspond to anything you might call friction
peak grotto
the second one is what i'm working with.
peak grotto
I apologize for the confusion created by calling that force 'friction' when it is not in alignment with classical friction.
It is a toy representation of a friction like force.
olive saffron
Hmm
peak grotto
But that is the equation that I am trying to learn to solve. the second one, in which the velocity is being added to over time, and multiplied by a value less than 1 over time.
Thank you for being so thorough in reviewing my original questions.
olive saffron
Mathematically this is no longer straightforward
peak grotto
however, at least they are both enacting their changes over time.
peak grotto
it might be 3d.
olive saffron
I mean in terms of writing an equation
peak grotto
let me see,
olive saffron
Unfortunately it is somewhat ambiguous as written since there are a few equations you could write that would fit this description
peak grotto
v = .2 * t
and
v = v - (v * .995)
olive saffron
My initial thought based on taking what you wrote verbatim would give you a recurrence relation
Right, so you do want a recurrence relation
Hmm
peak grotto
the second equation also involves time, but i'm not sure how to put it in.
olive saffron
As written it is somewhat nonsensical
But I see what you're trying to do
The equation you want here is uhh
Let me think
peak grotto
v = (v - (v * .995)) * t
olive saffron
v = .2 * t
and
v = v - (v * .995)
This second equation should be written umm
$$v(t+1) = 0.995v(t)$$
glossy valveBOT
rat v2.3.1
olive saffron
Thankfully this has a straightforward solution; namely, $$v(t) = 0.995^{1-t}$$
glossy valveBOT
rat v2.3.1
olive saffron
But you also have an acceleration so probably we want a differential equation
peak grotto
you have a negative exponent there, so does the relationship end up not being linear?
that's what i intuitively thought when I imagined the graph
because as the speed increases the friction increase is greater at each step.
and by 'friction' i mean this toy friction force, which is a multiplicative change to the current velocity over time, of course.
olive saffron
Correct, the relationship is exponential
peak grotto
But you also have an acceleration so probably we want a differential equation
oh i read different, but you said differential.
yeah i suspected this might end up being calculus.
i haven't done calculus yet.
or physics, just algebra.
peak grotto
coool. that makes me happy.
olive saffron
You are applying a nonlinear process to your velocity so classical kinematics does not apply any longer
peak grotto
i'm honestly stoked. lol.
olive saffron
Well no, classical kinematics still applies
peak grotto
i encountered a calculus problem in the wild. thank you for helping me interpret it.
olive saffron
But the usual SUVAT stuff doesn't
Technically all of kinematics is calculus
Velocity and acceleration are derivatives
This is usually hidden, however
Since everything is nice and polynomial
$F = ma$ is more generally stated as $F = m\frac{\dd^2 x}{\dd t^2},$ for example
glossy valveBOT
rat v2.3.1
olive saffron
Kinematics is a good place to start with learning calculus, honestly
Solving kinematic problems was, after all, the primary reason why Newton and Leibniz developed the tools for calculus
peak grotto
ok, i will take this opportunity to spend more time with calculus.
But before I dive in, can you help me to phrase my problem in proper calculus terms? I'd be very grateful. You've helped me a lot already though, thanks a ton.
olive saffron
I do not have the time right now but sure I can do this later
peak grotto
Ok, no worries. I probably have enough here to piece it together. Thanks again very much for going into this question with me. I'll definitely be interested to see your formulation of the problem if you come back to it later, but regardless. Thanks.
uncut nebula
How can i prove that a measure integral of a simple function is always non negative?
uncut nebula
by the definition c_i may not be positive right?
gritty rose
How can i prove that a measure integral of a simple function is always non negat...
!original
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How can i prove that a measure integral of a simple function is always non negat...
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
uncut nebula
this is the problem
gritty rose
oh ok
rapid rain
we also need your lesson's definition of a simple function
gritty rose
do you know what a simple function is
gritty rose
by the definition c_i may not be positive right?
why is this notation drastically different
gritty rose
from this one
uncut nebula
we also need your lesson's definition of a simple function
the normal
gritty rose
wow look at that
uncut nebula
linear combinations of indicator functions
gritty rose
did you read the definition of the coefficients
rapid rain
the normal
uhuh and the alphas are taken in...?
uncut nebula
but thats for the $s$ which is the simple function
glossy valveBOT
Rootsyl
uncut nebula
i need to show that by definition of a measure integral the result would be positive
uncut nebula
ok xD
gritty rose
or s
uncut nebula
now i see that c_i is the same with the simple function
i thought they were different for some reason
.close
hallow moth
hi can anyone help me wuth this question ?? it's true identities and I hv no idea how to solve it. ty
hallow moth
trig identities**
glossy valveBOT
primal stag
question 6 ?
primal stag
in order to find the maximum value of f(theta) then u need to derive f(theta)
hallow moth
how do I derive it??
hallow moth
wait how do u get this
primal stag
so its sqrt(2) xd
shrewd hamlet
Max value of the sin function will be 1, and min of cos function will be -1, find where sin(2theta) = 1 and cos(2theta) = -1
shrewd hamlet
U right bro mb
primal stag
u can also derive x.d
hallow moth
this is the ans from my tecaher
#6
primal stag
ye its the same but in radian'em
hallow moth
ok I think I got it thx
pseudo cape
Let $a = p_1^{a_1}p_2^{a_2}\cdots p_n^{a_n}$ and $b = p_1^{b_1}p_2^{b_2}\cdots p_n^{b_n}$, where $p_i$ are primes. \ Prove that $\gcd(a, b) = p_1^{\min(a_1, b_1)}p_2^{\min(a_2, b_2)} \cdots p_n^{\min(a_n, b_n)}$.
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pseudo cape
<@332561465405865985>
Oh, alright. Could you give an outline?
limber sandal
If a A is divisible by C then for any p prime number
[A]p>=[C]p []p its like the highest power of p that divides this number
pseudo cape
Can we also make this follow from Bezout's lemma or Euclid's lemma, for example?
I should move this to #elementary-number-theory, probably
.close
wild linden
This mgiht make no sense but:
a) for a span in Rn that forms a line, all the vectors are a scalar multiple of each other right? (so in a ref form matrix it's rank 1)
b) For a span in Rn that forms a plane it would form a matrix thats rank 2?
c) And for a span that that is Rn the matrix is rank n?
limpid moat
a is true
b is true
yes ...rank M =dim(Im(M)) so if Im(M)=R^n then rank M =n
so c is true
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rapid rain
so a and b are true no matter what value of n in Rn?
yes, provided that you CAN form a line or a plane in R^n.
If you can form a line in R^n, then n >= 1
If you can form a plane in R^n, then n >= 2
rugged narwhal
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rugged narwhal
<@&286206848099549185>
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@rugged narwhal Has your question been resolved?
rugged narwhal
<@&286206848099549185>
gritty rose
Wouldn’t the 2nd option become sin (-2+k(5/n))dx?
There should be no k and n variable in the integral
Write out carefully what the Riemann sum is and then do a shift if you need in order to match the bounds of the second option
rugged narwhal
Write out carefully what the Riemann sum is and then do a shift if you need in o...
I just learned this so what shift are you referring too
rugged narwhal
There should be no k and n variable in the integral
Isn’t x equal to -1 + k(5/n)?
gritty rose
I just learned this so what shift are you referring too
To -1 and 4 since I assume you understand the first expression
gritty rose
Wouldn’t the 2nd option become sin (-2+k(5/n))dx?
Where does -2 come from then?
rugged narwhal
To -1 and 4 since I assume you understand the first expression
It’s the same distance though
rugged narwhal
That's x_k
So why does the site bring this up?
gritty rose
Yes, 5-0 = 4 -(-1)
gritty rose
So why does the site bring this up?
Bring what up
rugged narwhal
Bring what up
A different x
gritty rose
Have you learned Riemann sums yet?
rugged narwhal
Have you learned Riemann sums yet?
Like I said I just started
gritty rose
Look at the definition of what a Riemann sum is
There's an x_k variable
Or maybe some other index like x_i
It comes from that definition
rugged narwhal
Look at the definition of what a Riemann sum is
So does I correspond to the left riemann sum and II to the right riemann sum?
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gritty rose
You're taking the limit as n goes to infinity so they're the same
The index starting point tells you which to use
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torn jolt
What about it?
hot herald
definition of tan
torn jolt
Tan(a+b)
=sin(a+b)/cos(a+b)
Now dérive sin a +b
Do a sketch
See what it means
Use cosine or sine laws to find what you’re looking for
Also read question 7
Also it’s easier to do cos(x-y)
First
Then change it to cos(x-(-y))
To get cos(x+y)
vapid ivy
the difference is literally just change - to +
torn jolt
Then sin (x-y) is just cos(pi/2 -x+y)
Let x and y be coordinates on a unit circle
Let cos x, sin x be in the 2nd quadrant
And cosy, sin y in the first
Coordinates on a cartestian plane
Same thing
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@torn jolt Has your question been resolved?
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@torn jolt Has your question been resolved?
fiery copper
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fiery copper
So when answering this question
I would get a LOT of constants like A, B, C, D and so on
Do I have to solve for all of them?
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@fiery copper Has your question been resolved?
glossy valveBOT
gritty rose
that's (r^2-9)^2, so you can probably do some order reduction
fiery copper
But here the particular solution has a bunch of constants
gritty rose
oh those constants you mean
fiery copper
Yeah
Like what do we do with them
Just simply add y_p to y_h here to get the answer?
kindred sable
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kindred sable
having a complete mindbank help bles
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kindred sable
<@&286206848099549185>
naive pewter
<@&286206848099549185>
Juzzie
unkempt thorn
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Available help channel!
stiff swan
How am I supposed to know the bounds for Z
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How am I supposed to know the bounds for Z
onyx glen
wdym bounds
stiff swan
lets say f(x) = x/2 for 0 <= x <= 2 and 0 otherwise
onyx glen
but in practice you cut off the parts where the stuff under the integral sign is 0
onyx glen
0 to x, you mean? for the CDF?
onyx glen
well formally it is -infty to x
stiff swan
yeah i know
onyx glen
it's just that your density function is zero below x=0
stiff swan
yeah
onyx glen
anyway it's a bit hairy to do this here
stiff swan
but for here, how do I know bounds for z?
ok
so I just gotta do it the improper integral way?
onyx glen
.... i guess, yes
stiff swan
okay
stiff swan
.... i guess, yes
so f(z) will no longer be a piecewise function?
onyx glen
sorry, i don't have the energy to untangle your somewhat sloppy notation rn
stiff swan
what notation
stiff swan
bruh
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vital torrent
does anyone know how to do this
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does anyone know how to do this
vital torrent
the asnwer should be 60 m/s
next sail
does anyone know how to do this
use suvat equations
proper hawk
umm it's already given though
proper hawk
does anyone know how to do this
the height is 180, which symbol represent heght?
vital torrent
x
vital torrent
how would u do it during differentiation
proper hawk
you can find the maximum height using differentiation
say x = f(t)
then when f'(t) = 0, it reaches maximum height
vital torrent
oh i just found out what to do
vital torrent
plug in 180 for x
i did this
and rearranged for b
then differentiated it
and got b'=5 -180t^-2
then i did 0=5-180t^-2
t= plusminus 6
-6 not possible since time cant be negative
so t=6
then plug into original equaiton 180=b(6) -5(6)^2
b=60
the wording was confusing me
i throught they were asking at a specific height of 180 to find the speed
i didnt realsie that 180 was the max height and they were asking for b
thx
.close
modest rain
is this even true?
full forumBOT
is this even true?
modest rain
where does the other L go from the nominator?
fast peak
cancels out with one of the Ls from the numerator
modest rain
yea i know
torn jolt
isn't it right here
torn jolt
i don't really get what you mean - the L^2 cancels with the factor of L on the denominator, leaving only one L on top
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modest rain
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modest rain
@torn jolt and can the 3+alpha be canceled out?
torn jolt
Is the fraction: $\frac{6L\frac{6 + 5\alpha}{3 + \alpha}L}{6L\frac{6 + 5\alpha}{3 + \alpha}L}$ or am i misreading it
glossy valveBOT
kitty boy!
torn jolt
or are they separate
modest rain
no
modest rain
the bottom is an addition of 6l
glossy valveBOT
kitty boy!
torn jolt
ahh i see
$\frac{6L\frac{6 + 5\alpha}{3 + \alpha}L}{6L+\frac{6 + 5\alpha}{3 + \alpha}L}$
glossy valveBOT
kitty boy!
torn jolt
$\frac{6L\frac{6 + 5\alpha}{3 + \alpha}L}{6L + \frac{6 + 5\alpha}{3 + \alpha}L} = \frac{36L^2 + 30\alpha L^2}{(3+\alpha)(6L + \frac{6L + 5\alpha L}{3 + \alpha})}$
glossy valveBOT
kitty boy!
modest rain
you multiplied ith with 3+aplha
torn jolt
yeah i did
torn jolt
well, i actually put all the Ls on the top on the numerator of the fraction in the numerator (recursion 
) and then moved the denominator down to the bottom
it's a weird fraction but i think my verdict is that you can't cancel the alphas, unless there's some more shifting around you do
torn jolt
have fun in japan lol
thanks 
modest rain
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strange condor
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obsidian shadow
how do i resolve this?
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how do i resolve this?
misty briar
worn matrix
!occupied
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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
merry holly
Mb
glossy valveBOT
🐱!Yajat! 【Catfan1398】🐱
worn matrix
and use this
golden tartan
golden tartan
help is desperatly needed
wise wyvern
Have you tried anything?
golden tartan
i hoesntly dont know where to start
it would be great if someone could help me on a call
golden tartan
sorry what
ive tried labeling the pulleys va and vb
but im still confused
i know its dependent motion
golden tartan
ok
wise wyvern
Well you realise net force on a pulley here in this case (which I'm going to assume massless) is 0
golden tartan
ye
worn matrix
,, 2a_A=a_B
glossy valveBOT
🐱!Yajat! 【Catfan1398】🐱
worn matrix
i dont know if this would help but this is the relation b/w the acc
yea ik
that is not a answer here
still mb
wise wyvern
The constraint between acceleration, velocity and displacement is same as long as there's only translatory motion.
Anyways, net work done by tension is also zero.
@golden tartan
$\sum T_i x_i$ = 0
glossy valveBOT
! What the hell am I doing here?
wise wyvern
Differentiate to get what's being asked.
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hot lark
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hot lark
Isnt the solution to this a line with a random but set angle?
hot lark
It asks the solution to that
And it says a circle or line etc
For what i thinked
This is like x = x , y
So every x is true
Y can be whatever but it will be one
So it should be a line the result
With a random angle
minor crater
if we have a+bi as our form
when does a^2 + b^2 = a-bi?
hot lark
My friend says the result is a line on the real axis tho im not sure why or if its true
hot lark
if we have a+bi as our form
when does a^2 + b^2 = a-bi?
Let me think
minor crater
what are the possible values for b?
proper hawk
sqrt(a^2 + b^2) = a-bi
hot lark
sqrt(a^2 + b^2) = a-bi
Wtf why
minor crater
because |z|
proper hawk
|a+bi|=sqrt(a^2+b^2)
hot lark
A value
minor crater
what value
proper hawk
|z| = sqrt(a^2+b^2) and z*=a-bi
hot lark
minor crater
wut
glossy valveBOT
hot lark
minor crater
where did you get a+b
hot lark
sqrt(a^2 + b^2) is real
.
minor crater
$|z| = \sqrt{a^2 + b^2} = a-bi = \overline{z}$
glossy valveBOT
artemetra
minor crater
is this clear
glossy valveBOT
artemetra
hot lark
Yeah so?
hot lark
**artemetra**
This is what I wrote
hot lark
Here
glossy valveBOT
artemetra
hot lark
Why
minor crater
you can't do that
hot lark
Okok bro
glossy valveBOT
artemetra
