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would R^n be R^5?
also im having trouble translating over what i should be doing. i tried rewriting in terms of the linear dependence definition, but i couldnt figure out how to use it
saying the dimension of the span is 4 is saying an output vector of Ax=b is in R4 where A represents a matrix whose columns are the vectors right?
how can i translate the dimSpan equaling 4 over to something usable in proving this?
nvm figured it out
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Here why will the power of x^n from the first bracket not be considered
you will have to send the actual question for anyone of us to be able to help you
i see that you did this an hour ago and didnt send the original, if you want help you will have to send the original and respond
Yes sorry
Question 6 here
@tardy lark
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,rotate
have you guys learnt the multinomial theorem/expansion?
Yes
instead of going the way that website , ill try to explain it with multinomial
xd_senBugha
n can be anything <=22, with that it is impossible to find a combination of integer values of t and u which satisfies the second equation in this
you can try if you want and check for yourself
so 253*4 is 1012, which fixes u = 4 and we've fixed t to be 0
10 - 4 = k
which is 6
and then just put it in the multinomial thing
Makes sense
!done
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Hello, I'm looking for help converting a point on a 2d arbitrary convex quadrilateral into a normalized coordinate within that quadrilateral. I've been doing some googling, and am confused whether this would be a projection mapping, or bilinear interpolation, or some other technique and I'm not really sure where exactly to begin. Here's an example of the inputs I have. And I would love to get the output as a normalized point showing the green points normalized position within the quadrilateral.
@finite gale Has your question been resolved?
you can go in one direction and map the unit square $[0,1]\times[0,1]$ to your polygon via $$(s,t)\mapsto (1-s)(1-t)A_{00}+(1-s)tA_{01}+s(1-t)A_{10}+stA_{11}$$ where $A_{ij}$ is representative of the point $(i,j)$ in the unit square
Flip
I'm not sure if that's immediately helpful in normalizing your points in the quadrilateral
It looks like it might be... still just trying to process it. Aren't the function inputs in this case (s, t) actually the normalized point already though?
yeah, so we'd really just like for the function to have a known inverse
what I did was interpolate two interpolations
you can map the bottom line of the unit square (the one that has the points $(0,0)$ and $(1,0)$) to the line containing $A_{00}$ and $A_{10}$ by sending $(s,0)$ to $(1-s)A_{00}+sA_{10}$ and $(s,1)$ to $(1-s)A_{01}+sA_{11}$ respectively
Flip
for shorthand we can call these $f_0(s)$ and $f_1(s)$ respectively; then we just interpolate between them and send $$(s,t)\mapsto (1-t)f_0(s)+tf_1(s)$$ which is equivalent to the above thingy by some algebra
Flip
so that's another way to see how that's working
Hmmm, I'm not sure I quite understand this yet. Just playing around with some examples though to see if I can make it click. Thank you for the answer though!
so I think we're hoping that a system of equations in $(x,y)$ of the form $$\left{\begin{array}{rcl} a_{11}xy+a_{12}x+a_{13}y+a_{14}&=&0\ a_{21}xy+a_{22}x+a_{23}y+a_{24}&=&0\end{array}\right}$$ has a unique solution
Flip
the two equations correspond to the two coordinates being equal, and the coefficients come from banging the numbers together (the constant terms carry the data of the point in the polygon that we're trying to reverse the mapping for)
this desmos link is interactive
in case you hadn't played with that yet lol
you can click and drag points, and play with the s and t sliders
lol, yea I am right now. I think it's starting to clarify a bit better. Thank you!
nice
Sweet, yea I think this was exactly what I was looking for actually. Thanks @rare girder! I appreciate the help.
cool, no problem! thanks for sharing
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. Help
are there any helpers nobody helping me
Please don't occupy multiple help channels.
first off no more than one channel
second, noadvert, wait your turn
again.
For sure bro
this is not your channel
It is
oh fantastic
It’s not? Or am I messed up
it is
For sure
i think i'm lagging
.something never counts as opening the channel
;-;
ah
This shit stressing me out
Won’t open if you have a dot at the start
all g
This math
Lmk if any one can help I been waiting for 25
Dam can no one help😭
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What exactly do you need help with
4th one maybe
I need help solving the rest and thank you all help is appreciated
The X,Y coordinate and the intervals
Solve f(x)=-4 is asking for what x is f(x)=-4
You can read it from the graph
f(x)=y
Yes
y = -4 is neither the case at x=-2 nor at x=8
Do you know how to read x and y from a graph?
I do
This domain and range stuff just gets tricky with me
I have to now get x and y coordinates
This isn’t domain or range
Yeah you have to find the points where the y values are -4
For sure I will try that now
And determine their x values
Fuck bro
I’m so confused 😭
Could you maybe help on a easier question I can skip for now
Look at the graph. Where is y = -4
You should
Y=-4 looks like it’s on 0.5
What do you mean
Are there other x for which f(x)=y=-4? Or is 0.5 the only one?
What about x=4
Indeed
Right where my mouse is
For which problem exactly ?
Becuase the next 2 are finding
Y and X intercept coordinates
Oh it’s not 2?
Why would it be 2
Or -2?
Why would it be -2
How would u convert 0.5 into a point
(0.5,f(0.5))
The system tells me this is not a decimal or integer value
I’m saying now that we have the solutions how I would enter it for F(x)=-4
There’s another solution
Fuck man I must be slow this is killing me
So we need to solve for another solution
I think it doesn’t need the ()
Correct your right
It didn’t
Now I need Y intercept coordinates and X
It looks like it crosses the graph at 7?
And -1.5
Any other helpers ?
<@&286206848099549185>
For any other helpers wanting to help am I here so far
@copper flare Has your question been resolved?
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pls help me settle my math related debate with one of my friend
he belives x raised to power dx is totally integratable and have a phsyical significance as well>
while i tried objecting that we cant just randomly create integrals with highscool level knowlegde as after further studies we'll see how complexities in these function increase due to their physical significance and graphical significance.
thus he propose that it doesnt matter if he is a highscool student
he knows how integration works hence he can go ff creating his own throrems and stuff
tl:dr does x raised to power dx exists graphically? is tehre a physical significance? is it even integratable? and im not asking x^dx dx just x^dx?
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its controversial
but apparently can be interpreted
but the more imp ques is
is this the way to do it?
the right way?
claiming that just using highschool knwoledge to come up with stuff is totaly alright?
meanwhile i belive u need to have a higher level of knowledge , at least a degree in that field before u even try to "come up with stuff" in that respective field. what they teach in highscool is pretty bland and incomplete isnt it. before proposing any random integral shouldnt u worry about the higher level of complexities
no it cant
@tranquil pine Has your question been resolved?
for question 1 and 2 it really depends on who discovered the idea and how they use it
for the third question it really depends cuz he maybe found/used some knowledge outside of high school + even still most of the newly discovered maths are not used or we may did not find a use for it yet
exactly
im lacking words how i could explain it to him
nah
in 3rd ques he basically wanna say that highscool knoweledge is enough to start coming up with your theories if u have enough creativity in your mind
i belive u cant do sht with creativity as long as all u have is incomplete knowledge
just tell him the notation is really weird
Yeah high school knowledge of a single topic isn’t enough I get you
but highschool knowledge covered a lot of math but I won’t think it’s enough to prove somethings in calculus
"imagine a rectangle with length l breadth b
l^b * b is defined
in the rectangle dx approaches to zero
to x^dx approaches to 1
integral approximately equal to x"
his words not mine
he keeps being fixated over this
exactly
if only he could understand that
ngl same logic I got in my mind when I saw that notation but where did that other b come from ?
like the integral is x^dx
not x^dx dx
exactly
if it wouldve been dx dx it wouldve made more sense comparitively
yea
and only thing that comes to mind when find the integral of x^dx is using ln at this point
The more I think about it the more it doesn’t make sense
lnFx = int of lnxdx
tho are we treating dx as a variable?
and if yes why?
while doing our normal int we treat dx as just a notation of a very small change of rate
hre as we are using the properties of ln it must either be a constant or a variable?
can we change the nature of dx?
and isnt lnxdx just product of two functions now
lnx and dx
bcz now dx is a variable
as per what weve assumed
Yea fr the more I think about it doesn’t make sense
idk if the there is a logic he tries to imply like solving a differential equation using log_x(n)
or something
he havent done differential equations yes
so thats out of the way
anyway
thanx for clearing it out for me
have a great day ahead
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sin(32) + sin(28) = cos(2)
how to prove this identity
i'm assuming that these angles are in degrees?
yes
okay
try writing sin(32) as sin(30+2) and sin(28) as sin(30-2)
(The reason is that we know the value of sin and cos at 30 degrees, so we're hoping for some sort of simplification)
yes
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Pls explain this question steps
@steep ivy Has your question been resolved?
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Potentially stupid question but i was under the impression that this graph is supposed to approach -9?
Well it’s exponential with a base > 1 so it has to blow up
It approaches -9 in the negative limit
wdym "blow up"
Goes to Infinity for large x
Wait im confused why is there even an asymptote there
An exponential function a^x goes to 0 as x approaches -infinity, so here in principle the 1/3 * 3^(x+2) goes to 0 for very small x and what remains is the -9 as the limit
Then where would the asymptote be if it goes to 0?
That wouldnt be an asymptote if its defined there no?
i kinda dont get what you mean...
Like for 1/x
there is an asymptote there since 0 can not equal 1
OH WAIT
is it just saying the equation will never reach zero?
Since a^x is never 0?
erm
a^x is never 0 correct
but for very small x you a^x -> 0
so the limit would be 0-9 = -9
here
But i thought a limit would usually indicate what it APPROACHES but never actually gets to
yes
it will never touch the asymptote y = -9
that's why it's an asymptote
a line it approaches but never reaches
Thats what i thought too
but like
then why is it showing me that 💀
bacc (unhelpful)
Well I figured it was -9 but i was just REALLY confused why it was showing me that as an asymptote
i thought i had my definition of an asymptote wrong
correcto
I hope you see why im confused now
because even on the graph you jsut sent its still rounding and telling me that they should coincide
it also does this when i plug in -25
One thing that it shows you though is how fast the exponential decays when you go left. In only 20 ish units to the left Desmos considers that the answer is so close to 0 that it is 0
try to zoom
you will always find a small gap
Which is not true tho correct?
between the asympote and the function
I see that which is honestly even more mind boggling
especially because it's strictly increasing
i feel like if i zoomed it would self correct
computers
they are limited regarding memory and stuff, so at some point they stop calculating and round
its interesting that it stops so early though
for this one, why is it at -8?
they fall down as they grow fast
I thoguht it would be at positive 8
how do you come up with -8?
The y intercept
approach 8?
Now you have an exponential function whose base is smaller 1 so that goes to 0 for large x
what about the other term if x=0
it would become positive
it would be -2(8)?
which is -16
oh wait it would be uhm
ah
-8
ur better at explaining it to my teacher
She has a strong russian accent and she has a tendancy to dumb stuff down
and its to a point where its just
wtf
well teacher and tutor are again two different things but yea
Idk sometimes i wish i could have a more personal experience with teachers so that they could better understand my level
Like im starting to research calc jsut because im really interested in it and i asked her a question and she practically bit my head off for it
I mean i get it
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Trying to make sure my answer is right
I got 204
I did For n = 9:
a_9 = 12 + (9 - 1) * 24 = 12 + 8 * 24 = 12 + 192 = 204 feet
@green onyx Has your question been resolved?
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I don’t think that’s the correct answer 🥲
I got 200 , not 204
204 is correct
12+24(n-1)
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How do i find the y intercepts? I have the asymptotes idk if that helps
For b)
Y intercept means when the graph intercepts y axis
So make x = 0?
Yup
No worries
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Could someone check if this is right?
how do you propose doing partial fractions on 4
ah
could this be then done by trig?
cus of x^2?
but that's also not part of the identities
trig can work yeah
since we could get it to look like 1/(1+x^2) we can pull out something tan related
i see no real issue with them
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One message removed from a suspended account.
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the derivative of 1/sin5 x - the derivative of cos3 x, then I would use chain rule
did you did something similar?
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Lock in
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power rule and chain rule
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huhhh
yuh
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yes
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💯💯
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🙏🏻🙏🏻
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propus knows best
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yea
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🤔🤔
Rewrite in good notation
(sin(x))^-5
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Ye what’s the inside function
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Btw just write it like -5(sinx)^-6
So you can easily see the inside function
No
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Yesssszzz!!!
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😭😭🙏🙏🙏😹😹
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💀💀💀
I mean you just dropped the negative
It was -5
But atleast if it was positive 5
You would be right
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he used division though
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Just do the problem lil bro😹🙏
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❤️❤️
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🔥🔥
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Closed by @amber portal
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what's a general form of a polynomial in P_4(F)
I feel this isn't even a vector space in the first place
$(x-6)^2+6;-(x-6)^2+6$
A dense set
The set of a all polyonomials of degree less than or equal to 4
in the case of 12
we have deg 0
yes
so is that not in the set?
neither of those satisfy p(6) = 0
^^
find a basisss
A dense set
1 is not 0 at 6
This is a basis as every linealy independent list of the right length can act as a basis for a given vector space
hmm
but 0 cannot be part of the basis
you don't know the dimension of U, so that argument cant work
you do know that a LI set of length 5 will span all of P_4(R) though
A dense set
but U is not P_4(R)
so the dimension of this is at most 5
It's a proper subspace
yeah
So the dimension has to be strictly smaller
U has dimension <= 4
A dense set
what is your proof
Well, we know for sure that the space contains (x-6), (x-6)^2,(x-6)^3,(x-6)^4
and that all linear combinations of them must also belong in the space by definition
the part you should explain is why this set is LI
it shouldnt be a very long explanation but you should briefly justify it
Using the definition of linear independence, $a(x-6)+b(x-6)^2+c(x-6)^3+d(x-6)^4=0$, it's trivial to verify that all the coefficents must be zero
A dense set
yes
and can you explain it?
okay
so you should at least say that, rather than saying "its trivial to ..."
because nobody wants to read that
I mean that's eveyr math textbook
yeah and thats bad writing
Though there is a lemma I've used here that I'm yet to prove
The only time things are actually trivial are when you don't feel the need to mention the fact that they are trivial
theres that, and theres also using "this is trivial" to hide the fact that you don't actually understand something
every time you claim something is trivial, you should be able to provide a proof immediately
So, let's consider polynomials of at most degree 4 which have at least one root at 6.
Would this not be (x-6)(ax^3 + bx^2 + cx + d)? @warm python
@warm python Has your question been resolved?
Yes
But(x-6) (ax^3 +bx^2+cx+d) can be written as a linear combination of $(x-6),(x-6)^2,(x-6)^3,(x-6)^4$
A dense set
Compile Error! Click the
reaction for more information.
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yep
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$\frac{ds}{dt} = -\sin(\theta) \frac{d\theta}{dt}$
knief
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no
because theta is a variable and we’re differentiating dt
theta changes as a function of t
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you’re multiplying by the derivative of theta dt
which is dtheta / dt
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same way derivative of s dt is ds/dt
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they told you
dtheta/dt is 5
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for theta yes
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oh brother
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it’s not a fraction sir
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dtheta doesn’t have a singular value
neither does dt
it’s notation
perhaps you’d prefer
$s’ = -\sin(\theta) \theta^{\prime}$
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knief
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he sent me the problem in a dm
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oh
Alex
LaTeX source sent via direct message.
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💀
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^
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^
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it’s notation to denote an operation
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hence why i wrote this
derivative
it’s not a fraction
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with a numerator value and denominator value
there is no ds
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there’s ds/dt
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think of it like dy/dx
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we don’t treat dy and dx as independent quantities
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sin(3pi/2) = -1
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we get the other -1 from derivative of cos
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yes
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^
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no
$s’ = -\sin(\frac{3\pi}{2}) \cdot 5$
knief
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bro has eyebags
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Can someone explain to me how the (1/5) got the to top part and became 5?
Forgot about this part of mahts....
The 1/5 in denominator went to numerator actually, so
1/(1/a) = a
To understand it you can multiply a in numerator and denominator
(1×a)/(1/a)×a
a will be cancelled in denominator and you will be left with
a/1 = a```
Oh ok
So anytime there is a fraction with 1 as the denominator I can just move the fraction up and discard the numerator right?
Yeah yeah a/1 is nothing but a
Yes if you have a fraction in denominator, you can reverse and bring it in numerator, for an example
1/(a/b) can be written as b/a
Ok! Thanks!
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That is mean lol
what can i say, i put a lot of effort into making sure the helpees get the help they need

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not sure if i should ask here, but any good sources for induction questions that are like this level?
or if you have any questions around that i would appreciate if you send it here
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same question i asked, i tried turning this equation into the definition of deveritive, like (sin(ln(e+h)) - sin(ln(e)))/(ln(e+h) - (ln(e))) = ddx sin(ln(x)), where x = e, but then after differentiate it, it is cos1 x 1/e ? how is this cos 1
Please don't occupy multiple help channels.
please dont use lhopital rule
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?
idk hwo how to do
the problem is with the derivative
you're supposed to differentiate with respect to the ln{x}
so dy/d(lnx)?
yup
can i know why?
since the denominator is ln{x+h}-ln{x}
and from the definition of derivative that thing is what you differentiate with respect to
so denominator can only be a variable like u = lnx, anything that isnt cannot? like ln sin etc
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A dense set
for (b)
Ah
Who is blake?
A dense set
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✅
Any progress?
So The set of polynomials less than or equal to degree 4, except for second degree polynomials
so like $(x-6)^2 \notin U$
A dense set
so a possible list of linearly independent vector would be ${1,x ,(x-6)^3, (x-6)^4}$
A dense set
hmm, you sure?
@warm python sorry I was asleep when you replied originally
I don't think your basis is going to work
hmm?
And I don't think it's equivalent to my suggestion
,w is (x-6),(x-6)^2,(x-6)^3, (x-6)^4 linearly independent
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Which was (x-6)(ax^3 + bx^2 + cx + d)
You've been including constants in it, or am I wrong?
(x-6)^2 when differentiated once is 2(x-6), differentiated again is 2
x^3 + 2x^2 + 3x + 2 also has a non-zero 2nd derivative at x = 6
sure
So we have p''(6) = 0 implying that we have a 2nd degree polynomial of the form (x-6)(ax + b) = ax^2 + (b-6a) x - 6b
Then integrating twice will get the original form of the polynomial
,w x^3+2x^2+3x+2
true\
nvm, i got it
seems fine to me
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As not every polynomial is such tha $p"(6)=0$ , it follows that $U \subseteq P_4(\mathbb{R})$. It thus follows $dim(U) \leq 5$. We now prove that no polynomial of degree $2$ belongs to this vector space. A polynomial of degree $2$ is of the general form $p(x)= ax^2+bx+c; a,b,c \in \R$, it then follows that $p"(x)=2a$, which is non-zero for $a \neq 0$. From this it follows that $dim(U) \leq 4$. We now attempt to find as many linearly independent vectors in $U$ as possible. One possible list is ${1,x,(x-6)^3,(x-6)^4}$. From which it follows that $dim(U) \geq 4$. From this we can conclude that $dim(U)=4$
A dense set
Is this fine?
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need help in solving unit direction vector n
vector
yes
i tried
made it more confusing
is it 'illegal' to compare like this
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A small uniform ball of radius r rolls without slipping down from the top of a sphere of radius R. Find the angular velocity of the ball at the moment it breaks off the sphere. The intial velocity of the ball is negligible.
Moment of inertia of a solid sphere is 2/5 MR^2
my guess is we need to assume that the large sphere is stationary.
Initially, yes
otherwise the answer would appear to be dependent on the masses of the two balls?
but eventually it'll move, right?
okay, let's make two cases
so if the large ball moves, let's say it has a mass of M and the small ball has a mass of m.
well, if you can just assign masses, then you probably should just do 2.
so angular momentum and linear momentum are conserved. So we know that the angular momentum at any time t will be equal to zero, and similarly for the linear momentum in the x direction.
in the x direction
oh alright
angular momentum is conserved when external torque is zero
there's external torque due to gravity (mgr)
well not initially
hmm... good point.
but when it's begun to fall, it has been started
so the point where the two spheres separate seems to be dependent on gravity. We can treat this as the acceleration due to gravity is less than the acceleration required to move a point along a sphere of radius r+R
I think
but my wife is home, so I gtg.

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Isn't it easier with energy
oh right, energy is conserved here
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i was going to respond then i realized i have no idea how to do that LOL
i believe it should be -4, as well as some other columns not looking correct
for example 3 - 2*0 should be 3 not 1, in the column 1 left from the last column
It should be -4 like epi said
do you know why?
0 - 2(2)
yes!
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is this assuming g(w) would be sin(w)? if so it would be A right?
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help plss
Have U learnt about alternate angle theorum or whatever it called
yeah
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with 2ab, is it 2xaxb or 2xa + 2xb?
2*a*b
so the first option?
2ab means 2 * a * b yeah
also should really use astericks instead of x's
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this question doesn’t make much sense right?
shouldn’t it be “if at least one of the numbers is a 1”, no?

"one of the numbers is a 1" is an assumption, like in a conditional probability sense
so it's like
output 3 random numbers
yeah that makes sense but shouldn’t it be “at least one of the numbers is a 1”
i did the conditional probability and got 6/27
but assumed this in the process