ok
#help-0
wind cloak
What do you mean exactly
wind cloak
We can help you with specific questions
errant shuttle
oh
nocturne vapor
according to the answer key im supposed to find 6 different points, but i have no clue how
nocturne vapor
here is the answer key
i have no idea how they got 6 as a y coordinate when a is negative and c is negative
and then they also got -6 as a y coordinate
lone heartBOT
@nocturne vapor Has your question been resolved?
vocal dirge
my best bet would be that these are answers on a multiple choice question, and the correct answer is (-4, -10)
you have to realize that y=f(x) is the parent function of y=-2f(2(x+5))-4, and it is also given that (2,3) is a point on the graph of f(x), therefore f(2)=3
to find the corresponding coordinate, you basically just have to make sure that the term inside of the function would be equal to 2, so that you have the f(2) to work with. there's only 1 number that satisfies this, since 2(x+5)=2 has only one solution, which is -4
with the x coordinate being -4, you just plug that in the equation and get y=-2f(2)-4, or since f(2)=3. y=-10
so the point you're looking for would just be (-4,-10)
@nocturne vapor
nocturne vapor
my best bet would be that these are answers on a multiple choice question, and t...
it's on a multiple choice question
rocky atlas
I guess then bill said the answer with process as well
vocal dirge
oh so yeah, the answer is just (-4, -10) as explained
vocal dirge
wrong answers for the multiple choice
rocky atlas
Didn't you say it's MCQ
rocky atlas
Is that 3 the answer
nocturne vapor
yes that is the answer for question 3
rocky atlas
So does this mean that this question has multiple answers
vocal dirge
there's probably an error somewhere
i looked up that exact exercise and found this guy asking for help
he also claims there are multiple points, but the verified answer only states the 1 point
vocal dirge
yeah...besides, it wouldnt even make sense for (2,3) to be one of those points after all that shifting the function
nw
nocturne vapor
.close
rich cargo
what algorithm is used to numerically invert very large matrices?
mortal trellis
well usually you dont need an inverse
rich cargo
well usually you dont need an inverse
lol what
rich cargo
well usually you dont need an inverse
maybe iuse wrong terminlogy?
i want this
matrix M
inverse of it
M^-1
rich cargo
usually when you handle large matrices you rarely need the inverse
often sparse matrices that can fill up 64gb+ ram require inversion too
multi physics simulations etc.
in my case not but still i would like to know
mortal trellis
are you sure that you need inversion or do you just want to solve Mx=b
thats a huge difference
lone heartBOT
@rich cargo Has your question been resolved?
rich cargo
are you sure that you need inversion or do you just want to solve Mx=b
well i want to solve Mx=b for x
so i need x = ( M^-1 * b )
?
mortal trellis
no
rich cargo
mortal trellis
there are lots of algorithms which solve Mx=b without finding M^-1
rich cargo
like which ones 
mortal trellis
all of numerical linear algebra essentially
QR decomposition
LU decomposition
or lots of iterative solvers which give you an approximation
mortal trellis
LU decomposition
(this is essentially row reducing. not good for sparse matrices tho)
jacobi method, gauss seidel, minres, conjugated gradient, ...
rich cargo
QR decomposition
then how do those yield the x vector at any point then?
or is it like x = QR b or smth
mortal trellis
QR was actually not a good thing to mention cause its not used for big sparse matrices
this iterative solvers take an approximation x0 of x
and then using some steps achieve a better approximation x1
then x2
and so on
until its close enough
rich cargo
hmm probably thats rather difficult to do
btw the sparse was only a example
of where big matrices exist
in my specific case its not
its one of those idk
a0*x0^0 + a1*x0^1 = y0
a0*x1^0 + a1*x1^1 = y1
and matrix ends up being those x-terms
X a = y
and i must solve for a vector
if that makes sense
mortal trellis
how big is your matrix
tardy stag
row reduction should do that pretty quickly
lone heartBOT
mortal trellis
.reopen
lone heartBOT
✅
tardy stag
bots have feelings too
mortal trellis
or in more fancy terms, compute LU decomposition
rich cargo
im not sure ihad this in school
probably not?
will it be hard to implement
my programming skills are somewhat limited 💀
and using libraries that have LU decomposition probably not possible
mortal trellis
what language are you using
these algorithms have been implemented very efficiently in a lot of languages
mortal trellis
why not
mortal trellis
but even just coding row reduction by hand is probably enough
I mean you are clearly not looking for speed anyway
rich cargo
yes im just looking for a algorithm i can implement
i found something where i would need to calculate alot of determinants and traverse through matrix recursively until i reach matrix sub sizes of 3x3 or smth
but im not sure i can do it
vale crag
i found something where i would need to calculate alot of determinants and trave...
if you're talking about cramer's, just don't
rich cargo
and since this is probably some O(n!) algorihtm anyway ithought i might ask for smth simpler xD
vale crag
yeah
rich cargo
if you're talking about cramer's, just don't
let me see the name
vale crag
this method of solving systems is usually called cramer's rule
rich cargo
Step 1: Matrix of Minors
Step 2: Matrix of Cofactors
Step 3: Adjugate (also called Adjoint)
Step 4: Multiply by 1/Determinant
this?
vale crag
but yeah if you want to code it yourself, row-reduction (or gaussian elimination they're pretty much the same thing anyway) is a good start
vale crag
Step 1: Matrix of Minors
Step 2: Matrix of Cofactors
Step 3: Adjugate (also call...
That's just straight up inverting your matrix
but yeah O(n!)
you better not use it
mortal trellis
even for 100x100 its not fine
vale crag
even 100x100 it's mega huge prolly
mortal trellis
its just a theoretical linear algebra tool
vale crag
row-reduction is just O(n^3)
rich cargo
oh wait
mortal trellis
completely useless for any practical applications
rich cargo
if its 100x100 theres no way i dont get recursion problems anyway
you are right
is there a good book on this
rich cargo
just take a numerical analysis book
pwell probably thers a gazillion arent there?
vale crag
yeah there's a lot of them
I mostly know books not in english tho
so I can't really help on that
maybe @mortal trellis has some recommendations
mortal trellis
no, sry. dont know books either
vale crag
pwell probably thers a gazillion arent there?
they pretty much cover all the same material anyway
so you're prolly fine just picking one
rich cargo
well yes but when i asked for C++ books, there were alot of recommended and not recommended ones for exmaple
id assume its similar everyhwere
ty !
mortal trellis
we have a books channel
vale crag
yeah true
mortal trellis
maybe there is some stuff there
naive valley
if you're looking specifically for a book on matrix computation, try Watkins "Fundamentals of Matrix Computations"
very accessible
vale crag
Applied Numerical Linear Algebra (Demmel)
Familiarity with basic linear algebra
Demmel’s book is one of the gold standards for this area. It presents a very good perspective towards NLA and is very comprehensive. It covers both theoretical aspects and practical considerations in considerable detail, for each algorithm it discusses. It also has good problems. All in all, an excellent book to both learn from and refer to. Additionally, Demmel has notes for his NLA class that he teaches with his book, so there are additional online resources if one so desires.
that's from #books
mortal trellis
"matrix computations" by golub is a classic I just remembered
that said I've never read it
naive valley
Demmel is more advanced than the Watkins I mentioned above, fyi
There's also Trefethen and Bau which is quite good
rich cargo
**Applied Numerical Linear Algebra** (Demmel)
*Familiarity with basic linear alg...
i only find a book from 1997
probably the math didnt change tho, so its fine
rich cargo
ok ty guys
lone heartBOT
Available help channel!
cloud zodiac
Can someone explain 11 to me please
lone heartBOT
Can someone explain 11 to me please
naive valley
there may be a short cut that i'm not seeing, but i would start by working out what is the actual arithmetic progression
i.e. express it as an + b, where a and b are constants and n is the term number
you have three unknowns (a,b,k) and three equations, so you should be able to solve for a and b (and k)
cloud zodiac
Would I subtract 5k-10 by 2k to find the difference?
naive valley
yea, that would give you one of the unknowns in terms of k (which one?)
cloud zodiac
It would give me
3k-10
Then would I do the same
7k-14
And
5k -10
To get
2k-4?
@naive valley ?
naive valley
i would write out the three equations
if the general term looks like an+b, then term 2 would be 2a + b, and set that equal to 2k
similarly term 3 gives the equation 3a + b = 5k - 10
what's the third equation?
cloud zodiac
7k-14?
naive valley
yep
so that's a system of three equations and three unknowns:
2a + b = 2k
3a + b = 5k - 10
4a + b = 7k - 14
you can solve that for a,b,k
(you really only care about a and b)
then you'll have a general formula for the n'th term
namely an + b
cloud zodiac
Oh
naive valley
like i said, maybe there is a faster way to do this, but that's what comes to my mind
naive valley
sure
cloud zodiac
Maybe I’m being dumb but how would I know if it wouldn’t be like an^2 + b
Also I’m assuming it only work for arithmetic sequences?
@naive valley
naive valley
Also I’m assuming it only work for arithmetic sequences?
arithmetic sequence means it's of the form an+b
that's the definition of arithmetic sequence
if they hadn't said that, then you're right, it could be of any form and you wouldn't have enough information to solve this
odd night
WHAT THE SQUARE ROOT OF 705
lone heartBOT
WHAT THE SQUARE ROOT OF 705
worn fox
why are we shouting
vale crag
,w square root 705
vale crag
wtf
naive valley
i suppose it's not a mean number but it's not a nice one either
worn fox
deffo for a roblox quiz
naive valley
$\sqrt{705} e^0$, what a useful representation, haha
ocean sealBOT
Bungo
sour dove
$\sqrt{705} e^0$, what a useful representation, haha
you never know when you'll need to be as explicit as possible 😉
torn elk
sour dove
$(\sqrt{e})^{ln(47) + ln(3) + ln(5)}$
ocean sealBOT
mellowdramallama
naive valley
@odd night i hope you appreciate these beautiful results!
torn elk
That's correct, I just wanted to write as the definition of a^y
must be a troll or a bad coincidence xD
worn fox
they've been playing roblox for 10 hours its almost definitely a quiz world
torn elk
lol, didn't notice that
lone heartBOT
@odd night Has your question been resolved?
granite kelp
so are we listing all the solutions of the sqrt of 705
naive valley
so are we listing all the solutions of the sqrt of 705
yes!
dude needs it for his roblox quiz apparently
the more convoluted the solution, the better
omg i can't wait to help someone who talks like that
vale crag
,w factor 705
ocean sealBOT
worn fox
god forbid we do anything other with our time than help you
tacit arch
Don't spread your butthurt into other people's channels
granite kelp
,w factor 705
granite kelp
thought you might need the divisors
rose sigil
joke on you, i'm only on my 4th time retaking complex analysis
granite kelp
thats another one of the solutions
lone heartBOT
@odd night Has your question been resolved?
worn fox
@odd night do you have any more questions to ask or are you going to keep clicking the cross without saying anything
granite kelp
think shes choosing the second option
lone heartBOT
@odd night Has your question been resolved?
marble swift
I don't understand where the extra "n" comes from....
lone heartBOT
I don't understand where the extra "n" comes from....
placid zinc
The 5n distributing into that term
marble swift
oh wait...
if i write it other way i around i understand i think....
if i write (5n)(4m^2-3n) it makes sense....
don't ask me why.
ah, ok. ype.
yep. thanks
.closed
.close
torn tapir
a positive integer is equal to the sum of the squares of its 4 smallest positive divisors. what is the largest prime that divides this positive integer?
torn tapir
so i have that 1 is a guaranteed number as a factor, and the other 3 numbers squared should be prime so avoid duplicate factors?
someone correct me if im wrong
i just need a quick insight into the problem and i should be able to solve it
lone heartBOT
@torn tapir Has your question been resolved?
trim sorrel
so i have that 1 is a guaranteed number as a factor, and the other 3 numbers squ...
Wdym by this?
torn tapir
so its really 3 factors squared + 1
becuase 1 is always a smallest factor
of any number
torn tapir
idk
thats like a prediction
im not sure how im supposed to approach the other numbers
torn tapir
<@&286206848099549185>
torn tapir
um
i have no thought process
the only other things i can think of is that all them added together should be odd
if even theres a lot of composite factors
gray isle
the only other things i can think of is that all them added together should be odd
how did you reach that conclusion
torn tapir
so it would be 1^2 + x^2 + y^2 + z^2
which would add to an even number if all were odd
torn tapir
> the only other things i can think of is that all them added together should be...
i kinda assumed
but i just realised it doesnt work lol
which means 2 is now a factor of the bunch
gray isle
yes
torn tapir
1^2 + 2^2 + x^2 + y^2
or
5 + x^2 + y^2
the number should have more than like say 5 factors
im guessing as well
because the second largest factor squared either is gonna be more than or equals to the number (lets assume n)
where to now...?
gray isle
still thinking
I suppose the next step would be to consider whether 4 would be a factor
lone heartBOT
@torn tapir Has your question been resolved?
lone heartBOT
@torn tapir Has your question been resolved?
torn tapir
.close
brave junco
Hello, having some issues with a DE problem
lone heartBOT
Hello, having some issues with a DE problem
brave junco
pastel cradle
@proud shoal
brave junco
solving this by the undetermined coefficients method, for finding the particular solution, I've gotten (Ax^2+Bx+C)sin(2x)+(Dx^2+Ex+F)cos(2x)
and the 2nd derivative for that is
So my question is how should I handle the x^2-9 part of the right side of the initial problem when plugging these into y''-9y?
because wouldn't that need to incorporate both the A constants in front of x^2's as well as the lone constants such as 2E, 2C, A, and C?
neon scarab
Yours should be 3x but not 2x right(?)
brave junco
*srry used 2x instead of 3x
yeah, i'll put that in corrected in a sec
(Ax^2+Bx+C)sin(3x)+(Dx^2+Ex+F)cos(3x)
y and y''
scratch that
y'' ^
neon scarab
You may solve by comparing coefficients.
brave junco
so -8Ax^2 - 6E - 9C + 2A + C = x^2 - 9 right?
but isn't that too many variables for a system of equations?
neon scarab
You have 6 variables, so you should be able to set up 6 equations.
Or you may solve some coefficients by substitutions.
Say 3x = 0, 3x = pi, which can make either the sin or cos goes 0.
brave junco
wouldn't all the terms being multiplied to the cos have to settle to 0 to make it disapear from the equation?
neon scarab
Say in this pic, 3x = 0 means sin 3x = 0, which will make the left part of the expression (the long bracket multiplied by the sin thing) immediately goes away.
rigid smelt
shouldn't assuming the particular solution is of the form (Ax^2+Bx+C)sin(3x) be better, this should get rid of 3 of the undetermined coefficients
as well as making everything looks cleaner
neon scarab
You have 6 variables, so you should be able to set up 6 equations.
Actually by various substitution values, you should be able to set up enough equations to solve for the coefficients.
brave junco
shouldn't assuming the particular solution is of the form (Ax^2+Bx+C)sin(3x) be ...
is that something which can always be applied based on whether the given function only has a sin or cos in it? (referring to the (x^2-9)sin(3x))
rigid smelt
no just an assumption based on the fact that this ode only involves y'', it wouldnt work if there were a y'
neon scarab
neon scarab
shouldn't assuming the particular solution is of the form (Ax^2+Bx+C)sin(3x) be ...
I think both sin and cos are needed 🤔
rigid smelt
that is just a general guide for finding the particular solution, you dont have to explicitly follow it
any particular solution that solves this ode will work
for example, in y''-y=sinx, the solution y=-2sinx will work, plugging this into the ode gives 2sinx-sinx=sinx
and that solution can be found thru the assumption the particular solution is of the form Asinx
brave junco
gotcha, is it ok if I leave this channel open while I work throught the algebra rq?
rigid smelt
yes, it's your channel, you can keep it open until you need no further help
lone heartBOT
@brave junco Has your question been resolved?
brave junco
uhh, I think I'm in another pickle
so if 9y is plugged in
you get (9Ax^2)sin(3x) for that term
bit in y'' you have (-9Ax^2)sin(3x)
resulting in (0Ax^2)sin(3x) right?
so how can you have that be equal to (x^2-9)sin(3x) on the right side
if all the (x^2)sin(3x) terms are cancelled out
rigid smelt
well if it doesnt work then too bad, you'll just have to go back to the first assumption
so 6 undetermined coefficients are what you are going to have deal with here
brave junco
err yeah, but I think that still has the issue right?
(-9Ax^2-12Dx-9Bx-6E-9C+2A)sin(3x)-(9Dx^2+9Ex-21Ax+9F-2D-6B)cos(3x) + 9(Ax^2+Bx+C)sin(3x)+9(Dx^2+Ex+F)cos(3x)
->
(0Ax^2)sin(3x) + (other stuff) = (x^2-9)sin(3x), right?
rigid smelt
lemme see, if you assumed (...)sin(3x)+(...)cos(3x), you are going to end up with (...)sin(3x) + (...)cos(3x)=(x^2-9)sin(3x), which implies that the expression multiplied to cos(3x) is 0 and the other would be equal to x^2-9. For the expression that is equal to 0, any terms multiplied to x has to be equal to 0 as well as the constants. So that should give you 3 equations. For the other expression, the conefficients multiplied to x^2 has to be equal to 1, the constant is equal to -9 and the one multiplied to x is 0. So that's another 3
should be 6 equations
i havent exactly did the algebra but that should be what you are having
brave junco
yeppers, one sec ill put up the equation with y and y'' subbed in
(0Ax^2-12Dx+0Bx-6E+0C+2A)sin(3x)+(0Dx^2-0Ex+12Ax+0F-2D-6B)cos(3x) = (x^2-9)sin(3x)
(-12Dx-6E+2A)sin(3x)+(12Ax-2D-6B)cos(3x) = (x^2-9)sin(3x)
so we have
-12DX = 0 -> D = 0
-6E + 2A = - 9 -> E = -9/6
12Ax = 0 - > A = 0
-2D - 6B = 0 -> B = 0
rigid smelt
right, let's slow down a bit, can i have some context on the two equations you sent above?
the x^2 terms shouldnt cancel out
brave junco
sure,
rigid smelt
actually nvm, got it
hmmm, still sth is off, i feel like im missing some signs doing this in my head
,w second derivative of (ax^2+bx+c)sin(3x)+(dx^2+ex+f)cos(3x)
ocean sealBOT
rigid smelt
ok yeah, the x^2 shouldn't cancel out at all
the result you should be getting is
(-18ax^2-(18b+12d)x+2a-18c-6e)sin(3x) + (-18dx^2+(12a-18e)x+6b+2d-18f)cos(3x)=(x^2-9)sin(3x)
jeez, this is headaching to look at
you might have forgot a minus sign somewhere
rigid smelt
sure, did you use it to calculate your derivative?
rigid smelt
yeah it got the same thing, you likely forgot a minus sign when subtracting that with 9*(the assumption made)
brave junco
wait for real?
rigid smelt
the first term when subtracting with 9(ax^2+bx+c)sin(3x) does not cancel out the x^2
brave junco
oh my god I'm a massive goober
gotcha
thanks
so we get
(-18ax^2-(18b+12d)x+2a-18c-6e)sin(3x) + (-18dx^2+(12a-18e)x+6b+2d-18f)cos(3x)=(x^2-9)sin(3x)
-18Ax^2 = x^2 -> A = -1/18
-18Dx^2 = 0 -> D = 0
-18Bx - 12Dx = 0 -> -18Bx = 0 -> B = 0
12Ax - 18Ex = 0 -> E = -1/27
2A-18C-6E = 9 -> -2/18 + 6/27 -9 = 18C -> C = -40/81
rigid smelt
looks good
hmm and apparently we could have gotten away with this by using (ax^2+bx+c)sin(3x)+(dx+e)cos(3x)
maybe if a bit of analysis were done, could have reached that assumption, the assumption (...)sin(3x) couldnt work simply because it exists a term (2ax+b)cos(3x). so to deal with that, we can add (dx+e)cos(3x) to make sure that both a and b are not 0
red raptor
Could someone explain to me how 2^x+1 - 2^x = 2^x
little hazel
tardy stag
can also just try a bunch of examples idk, depends on how comfortable you are with algebra whether an algebraic proof is convincing
vale wigeon
Could someone explain to me how 2^x+1 - 2^x = 2^x
do you know exponent laws
red raptor
do you know exponent laws
I'm able to do any of these with plus sign but when it's this specific thing idk how
red raptor
So 2^1(2^x)
vale wigeon
thanks for not answering my question...
red raptor
Well there is a few what do you want me to say
vale wigeon
i want you to say at least one exponent law properly in its general form
you should be able to recall them that way
not just be limited to applying them in specific circumstances
red raptor
red raptor
i want you to say at least one exponent law properly in its general form
All of these
vale wigeon
ok that'll do
and you have already said earlier that 2^(x+1) = 2 * 2^x.
that was you applying the first of this list to 2^(x+1)
red raptor
Yeah
So then for thr -2^x I leave I like that right nothing I can really do
Or (2)^x?
red raptor
ok that'll do
I tried making the 2^x = y but then it gives a wrong answer
vale wigeon
what "wrong answer" are you getting, exactly
vale wigeon
So then for thr -2^x I leave I like that right nothing I can really do
yes, the -2^x can't really have anything done with it on its own
vale wigeon
why is it not right?
red raptor
I'm a fk dummy
Nvm
So wait then
Cause this was half the equation I was getting confused on
It's 2^8=2x+1 -2x
So the answer is 8?
vale wigeon
did you mean 2^8 = 2^(x+1) - 2^x
red raptor
Yee
vale wigeon
yes the answer is x=8
lone heartBOT
@red raptor Has your question been resolved?
vivid dome
Hmm I thought i and k evaluate to 0
lone heartBOT
Hmm I thought i and k evaluate to 0
vivid dome
but yeah that wouldnt make sense
tacit arch
Have you learned cross product
vale wigeon
@vivid dome how did you end up at the zero vector?
vivid dome
I dont know why but this is part of "review" for my multi class. But no, I'm googling how to do the linalg questions.
ocean sealBOT
kenny.xie
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
vivid dome
so thats from wikipedia
vale wigeon
how did you end up at 0i+0j+0k
did you just guess or was there some work that went into it
vivid dome
i and k are 0 so that would make it 0i+0j+0k?
tacit arch
i and k are 0 so that would make it 0i+0j+0k?
That's not a unit vector
vivid dome
yeah i know. i dont know how to make i and k nonzero
vale wigeon
i and k are 0 so that would make it 0i+0j+0k?
i and k are i and k, not zeros...
you're still not telling us how you got any of this
like
what did you do
did you try to add a and b?
did you try to subtract them?
vivid dome
I assumed i and k were 0 but I misread
vale wigeon
or what
...
you assumed that the letters i and k somehow stood for the zero vector
thats pretty wrong
vale wigeon
so you don't have anything written on paper, do you?
a good starting point would be to take the cross product of a and b, and to do it properly.
vivid dome
so you don't have anything written on paper, do you?
no this is what I wrote down previously
vale wigeon
bruh
vivid dome
vale wigeon
at no point did you decide to tell me you did try to take the cross product.
except you took the cross product of 9j and 17j
and not of i+9j+k and i+17j+k
vivid dome
I would say im tired because its late but I dont know what I'm thinking right now
vale wigeon
are you on a deadline
tacit arch
no this is what I wrote down previously
This would have been important to show in the beginning
vale wigeon
i+9j+k = 1 i + 9j + 1 k
vale wigeon
also you did not assume "i and k were zero"
you assumed that the coefficients on i and k were zero
when they are 1
vivid dome
So this is what I have now but it says <-8,0,8> is wrong
vivid dome
ah ok thx
vivid dome
Those are probably the problems here
So would it be <1,0,-1>?
trim sorrel
Do u know the definition of a unit vector?
vivid dome
length is 1?
trim sorrel
Right
trim sorrel
So would it be <1,0,-1>?
Whats the magnitude of this vector?
vivid dome
root2
trim sorrel
Right
vivid dome
.close
lone heartBOT
lone heartBOT
Available help channel!
wicked radish
lone heartBOT
wicked radish
how do we do this
slender gull
x + a being a factor of the two equations gives you a root of the equations.
Do you know what that root's going to be?
wicked radish
sum of roots or product of roots?
slender gull
neither.
It gives you one exact root.
Which is going to be a root of both the equations.
wicked radish
idk
slender gull
Do you know what a factor is?
wicked radish
ik
slender gull
What is it?
slender gull
...
wicked radish
^_^
wicked radish
ye
slender gull
Do you have the "words" to explain that?
wicked radish
the one which you have in your profile is a root
wicked radish
oh wait
slender gull
I'm talking about roots of an algebraic expression.
slender gull
I don't know what you know!
wicked radish
i just can't seem to express it in words
slender gull
You might have an entirely wrong idea for all I know, and it won't lead us anywhere.
wicked radish
no
trust me
and proceed
i'll try my best to understand
ik what are roots and factors very well
slender gull
Then you don't quite know either.
Regardless, rather than being a pain about it. I'll tell you how they relate.
A factor of any expression is another expression that leaves no remainder when the original expression is divided by the factor.
A root is a number, that makes the expression zero, when we use it in the variable's place in the expression.
(x-1)(x-2),
1,2 are roots of this expression. As when you put x = 1, you have 0*(-1), similarly for x = 2.
however, (x-1) and (x-2) are factors.
Get it?
wicked radish
however, (x-1) and (x-2) are factors.
till here i'm clear
slender gull
Sweet.
Now, note when x = 1 being a root => (x-1) is a factor automatically.
x=2 being a root means (x-2) is a factor.
Yes?
wicked radish
yes
wicked radish
x=-a
slender gull
Yes.
And x= -a is a root of both the expressions in your problem.
Meaning if you ise x = -a in any of the equations, they'll yield zero.
...?
lone heartBOT
@wicked radish Has your question been resolved?
civic willow
To find the equation of a line passing through J, that Bisects angle HJK, I first found the midpoint of HK which would be the bisector as well I'm pretty confident, and noted that JP would be the direction vector. SO therefore, the equation I got was J + t(JP). Fair enough I thought.. But then, part 2 of the question asks where line J intersects HK, which is that point P i used earlier. Did I do it correctly, and just in a dumb way? Or, did I do it totally wrong.
gleaming ridge
interesting idea!
but the bisector of some angle does not always pass through the midpoint of its opposite side
Like this for example
Q's bisector clearly does not pass through the midpoint of PR
civic willow
Ahhh, so do I instead need to let P = (x,y,z) and note that JP dot HK = 0?
wait but I'll have a 3 variable equation hmmm
gleaming ridge
I have nooo idea how to find the equation 
civic willow
HMMMM hm
well wait
P would not
LMAO
P would not have x,y,z
it would have the parametric equations of HK
ok let me try it
I really like your style of prompting me to find the answer btw greatly appreciated
yeah okay lol it worked thanks 🙂
.close
tranquil tinsel
$A=\frac{1}{2}abtan\theta$
lone heartBOT
$A=\frac{1}{2}abtan\theta$
gray isle
what's this supposed to be for?
gray isle
still an equation with no context
where's this coming up and what's your question about it
lone heartBOT
@tranquil tinsel Has your question been resolved?
cinder sundial
lone heartBOT
cinder sundial
I have been confused by this question
That the term f(14) would not exists with the restrictions of x being smaller than 6.
indigo lotus
That the term f(14) would not exists with the restrictions of x being smaller th...
f(x + 6) = f(x)
cinder sundial
That is only f(r) which r is smaller than 12 is available
cinder sundial
f(x + 6) = f(x)
X has to be smaller than 6
vale wigeon
f(x+6) = f(x) applies for all x (even though it is not otherwise stated)
so the domain of f is not [0,6) or [0,12)
you can apply the property f(x+6) = f(x) as much as you want
indigo lotus
yup
cinder sundial
I see
cinder sundial
you can apply the property f(x+6) = f(x) as much as you want
I should consider f(x+6) as another function like g(x) and its domain for the x is Real numbers with no further restriction
Am I right?
indigo lotus
yes
cinder sundial
you are making it more complicated than necessary
I don’t get it, aren’t that true?
vale wigeon
isn't*
cinder sundial
Thank you
vale wigeon
it is true but you do not need to consider f(x+6) as a function in its own right. and you do not need to give it a new name
cinder sundial
then what is f(x+6)
its a function other than the f(x)
since it has different domain than the f(x)
i think it's much clearer to view this question in the way that f(x+6) is another function than the f(x)
im still using it...
the channel
alpine sable
im still using it...
oh sorry
cinder sundial
so f(14) = f(8)
i found it impossible to get the f(14); since we dont actually know what f(x+6) stands for. If you put x=8 on both side of the expression f(x+6) = f(x) you would get f(14)=f(8). However, 8 is not in the domain of x for f(x). Therefore i think it is either i comprehend it the wrong way or theres another way of doing it
i wanna know if my comprehension is right that i cant get f(14) by putting x=8 to the expression
<@&286206848099549185>
vale wigeon
However, 8 is not in the domain of x for f(x).
it's understandable you were confused by how the problem wrote the function, but the domain of f is not just [0,6).
cinder sundial
it's understandable you were confused by how the problem wrote the function, but...
vale wigeon
yes, i see that loud and clear.
cinder sundial
f(x) is only defined in the domain
vale wigeon
yes, normally it would imply that f can only take numbers from 0 to 6 as inputs.
but you're also told f(x+6) = f(x)
if you're able to apply it once to conclude f(14) = f(8), you're able to apply it twice to conclude f(8) = f(2).
cinder sundial
if you're able to apply it once to conclude f(14) = f(8), you're able to apply i...
excuse me, but apply what?
vale wigeon
f(x+6) = f(x)
cinder sundial
do u agree the f(8) in the expression of f(14) = f(8) is not defined?
queen oxide
辅导费
cinder sundial
but the f(8) is actually the result after putting x=8 into f(x)
vale wigeon
yes, it is.
again, i think you are confusing yourself because of how confusing the problem statement is.
a better way to write what they're going for would be:
indigo lotus
@cinder sundial they never restricted you on putting x > 6. They just stated the definition of f(x) if 0 ≤ x ≤ 6
vale wigeon
Let $f: \bR \to \bR$ be a function such that:
\begin{itemize}
\item $f(x) = x^2 + 3$ for $0 \leq x < 2$
\item $f(x) = 9-x$ for $2 \leq x < 6$
\item $f(x+6) = f(x)$ for all $x$.
\end{itemize}
Find $f(14) + f(f(13))$.
ocean sealBOT
Ann (glomed)
vale wigeon
that's what the problem wanted to say
cinder sundial
ohhh
so the output of f(x) outside the domain of 0 to 6 is actually unknown
that we need to use the given expression as the bridge to find out
tardy stag
yall writing diagonally is really hard
cinder sundial
yall writing diagonally is really hard
so you are saying when x>6 for f(x) would be just x-6
tardy stag
no, look where the arrow is going
vale wigeon
i think you might also want to make a case for when x < 0
cinder sundial
im not sure why does the arrow came back to the beginning
vale wigeon
also x > 6 should be x ≥ 6
indigo lotus
so you are saying when x>6 for f(x) would be just x-6
no its saying that replace x by (x - 6)
cinder sundial
so the output of f(x) outside the domain of 0 to 6 is actually **unknown**
^ is that correct
cinder sundial
that we need to use the given expression as the bridge to find out
^
indigo lotus
im not sure why does the arrow came back to the beginning
it means to repeat the process until 0 ≤ x ≤ 6
tardy stag
^ is that correct
I mean it's not unknown, it's perfectly well defined
it's just not in that list of cases
tardy stag
i think you might also want to make a case for when x < 0
I decided unilaterally that $f: \R_+ \to \R_+$, definitely didn't just forget about that
ocean sealBOT
Hayley
indigo lotus
that f(x + 6) = f(x)
take f(22) for example. how would you proceed based on the given info
cinder sundial
that mean x+6=22 that is x=16
indigo lotus
yes how would you compute f(16) now
indigo lotus
f(16) = f(10 +6), no?
cinder sundial
what is f(16+6)
tardy stag
we're trying to figure that out
cinder sundial
f(16)= f(10+6)=f(10)
indigo lotus
yup
indigo lotus
then f(10) = f(4+6)=f(4)
no
cinder sundial
fixex
cinder sundial
its f(994+6) but it gonna take you a lot effort to get through the repetitive process to gain its value
a lot of calculation for f(1000)
indigo lotus
can you see that we are repeatidly subtracting 6?
cinder sundial
it seems like, yes
indigo lotus
f(1000) = f(994) = f(988) = f(982) ...
cinder sundial
but can we do it at once? from f(1000) to f(4)
cinder sundial
you mean 1000/6
cinder sundial
i cant figure
indigo lotus
or you can just keep on subtracting 6 but yeah repetitive subtraction is same as division
tardy stag
you could subtract 12 at a time to make it slightly faster
cinder sundial
i dont know why you put 1000/6 as the input in that expression could get f(4)
indigo lotus
i dont know why you put 1000/6 as the input in that expression could get f(4)
we are not putting 1000/6 as the input
cinder sundial
you could subtract 12 at a time to make it slightly faster
im not sure how to do it
cinder sundial
f(22)=f(16+6)
f(16+6)=f(16)
f(16)= f(10+6)=f(10)
then f(10) = f(4+6)=f(4)
and so on....
indigo lotus
how many times did you subtract 6?
cinder sundial
its kinda like chain reection like chemestry imo
cinder sundial
how many times did you subtract 6?
3
indigo lotus
22 = 6*3 + 4 right?
cinder sundial
yes
cinder sundial
yes we do
indigo lotus
okay so how would you calculate f(45)
cinder sundial
let it be blurred, its sth like f(45+6r)=f(45)
cinder sundial
okay so how would you calculate f(45)
im not sure
indigo lotus
proceed the same way you did for f(22)
indigo lotus
yes
cinder sundial
that is f(45)=f(39)=f(45/6)
indigo lotus
yup
you can subtract 6 seven times from 45 to reach at f(3)
or just divide 45 by 6 and take the remainder
both of the methods would work
cinder sundial
im shocked by the repetitive feature of f(x+6)=f(x)
im never seen sth like this before
lets be more general
is there a name for this?
i would name it "chain reaction" tho
indigo lotus
i guess its called recursion
cinder sundial
and f(x+r)=f(x) that is the general term of "chain reaction"
indigo lotus
This is the definition of a recusrive function
or wait i might be wrong
not sure about the terminology sorry
you can ask someone else maybe
cinder sundial
i wanna know what is the characteristic of this chain reaction tho
like sth i can recognize at the first sight
cinder sundial
and f(x+r)=f(x) that is the general term of "chain reaction"
is that true
or theres actually any other form
indigo lotus
you can say that its a periodic function
for 0 ≤ x ≤ 6
lets say it outputs particular values
the function will output the same values for
6 ≤ x ≤ 12
12 ≤ x ≤ 18
and so on
you can say it has a period of 6
do you know about sine function
or any trignometric function
cinder sundial
yes i do know about sine function and its kinda like waves
the graph looks like a wave with the period of 2 pi
indigo lotus
indigo lotus
the graph looks like a wave with the period of 2 pi
yup
so the function mentioned in the question would have a period of 6
cinder sundial
how do i put the fucntion on desmos? i wonder how it looks like
indigo lotus
cinder sundial
thats it actually looks
with that expression
f(x+6)=f(x)
its not being drawn in a very accurate way
but i think its clear that the function is periodic
indigo lotus
yes
tardy stag
this is how you do piecewise in desmos in one line, unfortunately it does not support recursive functions
cinder sundial
it does looks like a actually wave like the one we see in the sea
indigo lotus
yes haha
https://www.desmos.com/calculator/ictmqlpi8r
the link to the plot if you wanna play with it
cinder sundial
https://www.desmos.com/calculator/ictmqlpi8r
the link to the plot if you wanna p...
thank you so much, it would be helpful ,for me, to taking a note about recursive functions
median oar
(hey look it's not stats)
lone heartBOT
(hey look it's not stats)
median oar
is it common to write this kind of "vector"
on the last line
to write some v in V, by [v]_B
and does that mean that entries of a matrix need only be from a ring and not a field?
median oar
(that last line coming from the fact that you can add and multiply in rings, but no division of elements, and it seems here that each component of the vector v might be a matrix itself, so you'd have components being able to add and multiply?)
maybe that's gibberish actually
median oar
no?
ok for example, let V be R^2
and W be M_2x2
those are 2 vector spaces
im sure it's easy to find some f such that f: V x W -> F
and say both are over the field of reals
then f will go to reals
how does it make sense to multiply a matrix A, by [w]_C
wouldn't that w look like
thorn tapir
to write some **v** in V, by [**v**]_B
Atleast that's a notation I have seen in Hoffman-Kunze
B is the corresponding basis
which determines the coordinates of v
vale crag
nah
the coord vector is just the coordinates
so your vector would just be (w1, w2, w3, w4)
median oar
so we first map the basis vectors "C" of W to F^n?
vale crag
assuming you're representing the matrix ((w1, w2), (w3, w4)) in std basis
vale crag
so we first map the basis vectors "C" of W to F^n?
it's always what you do when you study matrices as elements of a vector space
if you wanna check linear independence of a set of 2x2 matrices
you end up with a 4x4 system right
median oar
i only first encountered this in my assignment that every half assed so idk
idk if that's what you always do because i've rarely ever done it
median oar
if you wanna check linear independence of a set of 2x2 matrices
what does this mean
if a set of 2x2 matrices are linearly independent?
oh just make it F^4
i see what you mean
vale crag
like the only thing we care about when looking at M_2x2 as a vector space is the addition and the scalar multiplication
forget about everything else
median oar
so when analysing any finite dimensional vector space, we first transport it to the world of F^n for ease of algebra
median oar
and that's the [v]_C notation
median oar
they're all isomorphic to some F^n anyway
yes i had a stint looking at how that all works
vale crag
you can transport your "vector" to F^n
vale crag
same stuff with polynomials for example
median oar
yeah right
median oar
x+1?
vale crag
then the representation of v in basis B would be [v]_B = (1, 0, 1)
v = 1*x^2 + 0*x + 1*(x+1)
median oar
so when we look at a vector of a vector space being described by $\vec v=\sum_{i=1}^na_iv_i$
ocean sealBOT
frosst
median oar
where $B = {v_1, v_2, ..., v_n}$
ocean sealBOT
frosst
median oar
we have that $[\vec v]_B = (a_1, a_2, ..., a_n)$
ocean sealBOT
frosst
vale crag
yeah
ofc this kind of mapping V -> F^n is not unique
take any basis B, it will give you a valid isomorphism between the two
vale crag
yea
dusk granite
HELP
alpine sable
Is there an intuitive way of understanding u substitution?, i understand what they're doing and how to apply it, but it seems kind of like some arbitrary method i just have to memorize
alpine sable
.close
heady void
So how did we get to the bottom equation
lone heartBOT
So how did we get to the bottom equation
vale wigeon
aint that just the same as the top equation but with the units written in??
and with the value of V'(t) plugged in as well
heady void
oh wait
sorry
wrong segment
it's just letting the equation be in terms of r'(t) ?
mellow grail
mhm
mellow grail
it's just letting the equation be in terms of r'(t) ?
they divided both sides by $4\pi[r(t)]^2cm^2$
ocean sealBOT
itzkraken.
heady void
.close
rotund briar
lone heartBOT
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
1
ruby current
try completing the square in the denominator
rotund briar
can you help?
ruby current
you need help completing the square?
rotund briar
i need help solving the question completely
lone heartBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
ruby current
you can rewrite the denominator in the form (x-a)² + b, and the make the substitution u = x - a
that’s the gist of the solution
errant dagger
lone heartBOT
worn fox
If you think its not a subspace, which subspace axiom do you think it doesn't satisfy?
errant dagger
If you think its *not* a subspace, which subspace axiom do you think it doesn't ...
its outside the dimension
cus of x^3
worn fox
What do you mean by that
errant dagger
What do you mean by that
dw im dumb
this is a practice test btw
i can prove if needed
I dont understand how to do this one
like how would i prove this
that it is indeed closed
worn fox
Well what does closed mean
errant dagger
Well what does closed mean
that it satisfies S or lies in the set S
worn fox
What is "it"?
errant dagger
x sry
worn fox
That doesn't make a lot of sense, x is already the name for something in S
Check your notes for what closed under vector addition means
errant dagger
i tried but i wasnt reallyyy sure
all i got was
x1 + y1 - 2(x2 + y2) + 3(x3 + y3)
but
yeah i have no idea
(x1 -2x2 + 3x3) + (y1 - 2y2 +3y3)
both sums are >= 0 therefore
sum is >= 0?
worn fox
Yeah thats exactly it
errant dagger
icic
i have another question if thats ok
similar style
this is an example of one that is not closed
is there a way to try and generally prove this
or just search for counter examples if it doesnt look easily provable
worn fox
Counter example is best really
You can start to try and prove it, and see what starts to not work, which might give you a hint of a counter example
Trying to prove it might tell you what vectors it would be true for, and then you pick a counter example that isn't one of them
lone heartBOT
@errant dagger Has your question been resolved?
alpine sable
lone heartBOT
alpine sable
Hello guys can anyone teach me how to take common like this
I don't know what this is called
subtle birch
$\left(x-3\right)^{40}=\left(x\left(1-\frac{3}{x}\right)\right)^{40}=x^{40}\cdot\left(1-\frac{3}{x}\right)^{40}$
ocean sealBOT
beard420
subtle birch
mainly use of converse of rule 5
alpine sable
**beard420**
Ooooo my dumb mind never thought like this