not just the first
#help-28
fast peak
sturdy valve
ohk fine let me try that
im getting 1 again
tan(x+x^3/3) - sin(x-x^3/6) / x+x^3/3 -x + x^3/6
whole lily
tan(x+x^3/3) - sin(x-x^3/6) / x+x^3/3 -x + x^3/6
you can do it again for the terms in the numerator
sturdy valve
like tan y = y approximation or i need to use 2nd term again?
versed fable
you can do it again for the terms in the numerator
correct , you must be
whole lily
like tan y = y approximation or i need to use 2nd term again?
second or else you're missing out O(x^3)
sturdy valve
second or else you're missing out O(x^3)
so we need to get term with (x+x^3/3)^3??
..
whole lily
so we need to get term with (x+x^3/3)^3??
yep
although you can omit the higher O(x^5) terms
silk bridge
$\frac{(x+\frac{x^3}{3})+\frac{1}{3}(x+\frac{x^3}{3})^3+(x-\frac{x^3}{6})-\frac{1}{6}(x-\frac{x^3}{6})^3}{x+\frac{x^3}{3}-x+\frac{x^3}{6}}$
glossy valveBOT
scoob
silk bridge
this shit
glossy valveBOT
Roy
whole lily
because if you expand it out, the only cubic term will be x^3
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still current
give the rules of the following exponential functions in the form
still current
x is 0 no?
silk bridge
y intercept is - 3, substitue x=0
silk bridge
x is 0 no?
yes
still current
y intercept is - 3, substitue x=0
like -3 = a(c)^0 -2?
silk bridge
yes
silk bridge
no
silk bridge
😄
silk bridge
yes
silk bridge
(2,-11)
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silk bridge
<@1386023345547513916> mb for answering late
nw
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rigid glen
Had an exam and was wondering if my answer was right or wrong.
We knew that triangles EDC and ABC were equal and we had to prove if SC=KC
I basically said that since ABC=EDC with DE and AB as a base, and SC,KC were THE height of equal triangles, then SC=KC
(The intended solution was to compare right triangles)
Nevertheless was my answer one of the right ways to solve the exercise?
neon silo
Seems correct to me
Would appreaciate if you showed us how you wrote it in your exam tho just to be sure
rigid glen
Would appreaciate if you showed us how you wrote it in your exam tho just to be ...
Don't exactly remember and I am translatingso there is probably going to be a bit of paraphrasing.
Since ABC=DCE and CK is the height of ABC with base AB,SC is the height of DCE with base DE then CK=SC because they are corresponding heights of equal triangles
silk bridge
what language are you translating from
neon silo
Personally, this kinda reads more like stating a property of equal triangles more than proving it
Would be better if you said something like: since theyre equal, they have equal areas and AB=ED, therefore CK=SC
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rigid glen
Personally, this kinda reads more like stating a property of equal triangles mor...
Yeah but if it's a property isn't this case proven by default?
gusty flicker
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gusty flicker
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copper sable
the 3rd problem is simple:
Assume that the unit hypercube has vertices (±1/2,±1/2,±1/2,±1/2)
then take the 24cell as all permutations of (±1/2,±1/2,0,0)
gusty flicker
that 24cell is a valid rotation?
copper sable
yes it is a rotation
copper sable
uh consider the orthogonal basis of
(1/2,1/2,0,0)
(1/2,-1/2,0,0)
(0,0,1/2,1/2)
(0,0,1/2,-1/2)
under this basis it looks just like the 24-cell
copper sable
well consider the width i think
gusty flicker
it can go from (+1/2, +1/2, 0, 0) to (0, 0, -1/2, -1/2) but its not like obviously optimal
(or maybe im stupid, but i dont see how its optimal)
copper sable
that is the diameter;
the width is the (minimal) distance of pairs of parallel hyperplanes bounding it
fortunately, a 24-cell is regular and convex, very symmetric, all pairs of opposing surfaces are equivalent
copper sable
the 3rd problem is simple:
Assume that the unit hypercube has vertices (±1/2,±1/...
In this particular rotation, it is bound by hyperplanes x=1/2 and x=-1/2, having width 1
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gusty flicker
In this particular rotation, it is bound by hyperplanes x=1/2 and x=-1/2, having...
ah, i see
ok so problem 3 would be 24. gotcha.
for whoever will come: what about p1 and p2?
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robust slate
In the future, please show what you’ve done so far (and give any other relevant information) when asking for help. It gives us more context and prevents wasting time from explaining things unnecessarily. \ \
-
Recall that if two integers are congruent modulo the $t$-th primorial $P_t$, then their difference has at least $t$ different prime divisors. Consider what the pigeonhole principle tells you if $P_t<n$. \ \
-
Shoelace. The rest is just symbol pushing.
glossy valveBOT
Civil Service Pigeon
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split orbit
hi guys i might have a vague question but is there a possibility that i can learn this in 8 or so hours and how because i genuinely dont understand anything except the basics of integrals. i am capable enough of doing question 1 and question 5 till 1/2 sin (x^2) e^4 + c but i dont understand more about these integrals, this is a pratice test in the real test i need 7 out 10 points
neon basin
You can definitely learn quite a bit in 8 hours
If your professor / school has provided material, go through that
split orbit
we got some powerpoint slides but on the subject of practice questions or etc we got only this practice test and one from 3 years ago
neon basin
we got some powerpoint slides but on the subject of practice questions or etc we...
That's a bit lackluster
Well there's plenty of material online
I will let someone else recommend though, because I am not really familiar with it. And don't want to recommend something that I am not an expert in
split orbit
i understand thanks for the, tip though
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digital void
,tex
Hi there. The following is a proof our professor wrote today, about f's bijectivity. I am a lil bit confused because i think she rushed the proof a little too much. Specifically, in the injectivity proof, why did we just take $\frac{n_1}{2} = \frac{n_2}{2} $ and the respective for n odds like that? Afterwards, for surjectivity, where a is negative, why did we just multiply it with -1? to make it a natural? If so, whats a better way to explain these parts of the process?
\vspace{0.3cm}
Suppose we have the function $f: \mathbb{N} _0 \to\mathbb{Z} $, and:
$ f(n) = \begin{cases} k & n = 2k ,k\in\mathbb{N} _0 \ -k & n=2k-1 , k\in\mathbb{N} \end{cases} $
\
\vspace{0.3cm}
$f(n_1)=f(n_2) \implies \left( n_1,n_2 \ \text{ even, and} \ \frac{n_1}{2} =\frac{n_2}{2} \right) \lor
\
\left( n_1,n_2 \ \text{odd, and} \ -\frac{n_1+1}{2} =-\frac{n_2+1}{2} \right) $
\vspace{0.2cm}
so, in any case, we have:
$f(n_1)=f(n_2) \implies n_1=n_2 $
\vspace{0.2cm}
Now, for surjectivity:
\vspace{0.2cm}
$\text{Let} \ a\in\mathbb{Z} : \text{If} \ a\geq 0: \ f(2a) =a , 2a\in\mathbb{N} _0 $
\vspace{0.15cm}
$\text{If} \ a<0 : n= 2(-a) -1 = -2a-1 . \ f(-2a-1) = a $
\vspace{0.2cm}
So, in any case, $$\forall a\in\mathbb{Z} , \ \exists n\in\mathbb{N} _0 : \ f(n) = a \square \textbf{MEOW} $$
glossy valveBOT
fijokazż
queen crater
To prove injectivity, you need to prove that for any pair of inputs (n_1, n_2), if the outputs are the same then the inputs are as well
Let K = f(n_1) = f(n_2)
Either K is negative, or not
That means, given the definition of f, either both n_1 and n_2 are even, or both are odd
Well, if they're even, f(n_1) = n_1/2 and f(n_2) = n_2/2
digital void
oh okay that makes sense
queen crater
So n_1/2 = n_2/2, so n_1 = n_2
digital void
similarly for odds
queen crater
Yes
digital void
how bout the surjectivity part?
queen crater
You need to show that for any element a of the codomain (Z), there exists an element n in the domain such that f(n) = a
So you take any element (let a in Z)
If a >= 0, that can only come from an even n (again, definition of f)
If a < 0, that can only come from an odd n
Well, if a >= 0, n = 2a works
If a < 0, since the definition is "-k if n = 2k-1", you have -k = a so k = -a, so n = 2(-a)-1
digital void
oh lol
queen crater
In other words, if a < 0, n = -2a-1 works
digital void
mann if shed just written that id get it
yeah that makes sense
thank you for your time
.close
severe mist
Question 17 part a and b I don’t know where to start. I’m working on solving systems of equations by substitution.
prime pier
,rccw
glossy valveBOT
neon basin
Question 17 part a and b I don’t know where to start. I’m working on solving sys...
Any ideas?
or thoughts
severe mist
I’m not sure if I can get help on finding the equations for a j can solve b I just don’t k ow what to start with
neon basin
I’m not sure if I can get help on finding the equations for a j can solve b I ju...
uh
sorry I don't understand
spiral vigil
Question 17 part a and b I don’t know where to start. I’m working on solving sys...
first figure out how many mL of acid they need, and how many mL of water (you don't need systems of equations for that)
now, if they pour a ounces of solution A, how many ounces of acid will they get? how many ounces of water?
same for pouring b ounces of solution B
severe mist
Umm
outer lotus
Umm
What does it mean to have say 70% of something if the whole thing (100%) equals some number b?
severe mist
0.7?
outer lotus
0.7b
Right
So if the whole solution is b mL, then 70% of it is 0.7b mL
Given that we need a sample from each beaker, we can take a mL from solution A and b mL from solution B
And a+b must equal 5 mL
severe mist
I got that
outer lotus
In our target solution, the acid content is 65%
So 0.65 * 5 mL of acid
That means
0.7a + 0.2b = 3.25 mL of acid
a+b=5
severe mist
I think that makes sense
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somber tusk
!help
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stiff pier
faint pilot
hi can someone just help with this question
ye i gotchu
fallen light
$\frac{\binom{8}{3}}{\binom{14}{3}} = \frac{2}{13}$
glossy valveBOT
melon
brazen lark
yeah
craggy tapir
looks good to me
somber tusk
okay thanks
craggy tapir
okay thanks
remember to close the channel
somber tusk
.close
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✅ Original question: #help-28 message
somber tusk
$\frac{\binom{8}{3}}{\binom{14}{3}} = \frac{2}{13}$
hey could u help with this one
brazen lark
i think its right
somber tusk
i think its right
do you know someone who could check
fallen light
I got 7/10
seems right to me, P(blue) = 1 - P(not blue) which ends up giving 0.7 or 7/10
somber tusk
seems right to me, P(blue) = 1 - P(not blue) which ends up giving 0.7 or 7/10
some
other people are
getting 10/13
vivid stream
I got 7/10
Seems to be right
brazen lark
getting 10/13
that means there are 100% of getting a blue one, 30% for the non blue one
somber tusk
Applying the Rule to Your QuestionThe question states the ODDS in favor of NOT picking a blue sock are 30%.Convert Odds to a Ratio: 30% means $30$ to $100$, which simplifies to a ratio of 3:10.
brazen lark
that means there are 100% of getting a blue one, 30% for the non blue one
it doenst make sense
glossy valveBOT
k4sh
fallen light
Applying the Rule to Your QuestionThe question states the ODDS in favor of NOT p...
where does 100 come from? by definition of complement events we must have P(blue) + P(not blue) = 1
brazen lark
strictly speaking, that's correct.
fallen light
Applying the Rule to Your QuestionThe question states the ODDS in favor of NOT p...
oh wait i misread, yeah that checks out
somber tusk
so is it still 7/10
fallen light
should be
somber tusk
so not 10 / 13
fallen light
don't see why it would be
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dim trail
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dim trail
How did my teacher get the answer for #10
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glad knot
we're describing the position of the tip of the hand with respect to the floor
in centimeters
+200 is because the centre of the clock is 2 metres off the floor
2pi/60 is the frequency is because every second, the hand moves 1/60 of a full rotation (2pi)
dim trail
Ohh
glad knot
the coefficient of 4 is because sin and cos functions have an amplitude of 1, so it has to be scaled up because the hand is 4cm long
the first equation (with cos) describes the VERTICAL movement, assuming the origin of our coordinate plane is 2 m just below the centre of the clock
dim trail
Thanks bro
glad knot
no problem 👍
dim trail
.close
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dim schooner
Pls
stiff musk
,rccw
glossy valveBOT
dim schooner
But I got basic knowledge:))
stiff musk
fundamental theorem of calculus?
dim schooner
Yah
stiff musk
side note, why does it look like a kidnapping ransom note? 
dim schooner
Integrals
dim schooner
side note, why does it look like a kidnapping ransom note? <:opencry:58607861486...
Wat
Wym
dim schooner
Whats random note
coral ore
side note, why does it look like a kidnapping ransom note? <:opencry:58607861486...
because they took a picture and highlighted the texts to make it clear
rare pine
if you apply the "opposite" of a derivative on f'(x), it should give you f(x)
dim schooner
I'm not being kidnapped btw
dim schooner
because they took a picture and highlighted the texts to make it clear
Yah
Ya wanna see the original 1?
I translated it
Thats why it looks like that
Solved it
Solved it
Yah am so dum its kindergartenly ez
Hơw to end a help channel?
slate violet
Hơw to end a help channel?
type .close
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proven lynx
hello, what do i do if my z value is less than 1,00 in normal distribution
proven lynx
the table
slate violet
the table
your table should definitely have the values for 0 <= z < 1
what kind of table do you have?
proven lynx
your table should definitely have the values for 0 <= z < 1
i reallzed so
if its like z = 30
then ik its something like 0.99999…
bro i got an exam in an hour
slate violet
then ik its something like 0.99999…
yeah
the area of the tails is really small, so P(z < 30) = 1 - P(z > 30) is super, super close to 1
onyx glen
if its like z = 30
if |z| is greater than like 5 then you start running out of precision on most computers to calculate values this close to 0 or 1
onyx glen
e-198 
proven lynx
if |z| is greater than like *5* then you start running out of precision on most ...
yeah
brittle steeple
,calc e^(30^2/2)
glossy valveBOT
Result:
2.7071782767869e+195
oh damn
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proven lynx
oh damn
btw when i got a normal distribution its for example
atleast 33 and less than 77
so then id do
phi(77-mu+0.5/sigma) - phi(33-mu-0.5/sigma)
slate violet
<@&268886789983436800>
proven lynx
phi(77-mu+0.5/sigma) - phi(33-mu-0.5/sigma)
yeah this is correct but if i were to do this with the calculator F(n;p;k) function
then id have to do 32, why
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torn jolt
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✅ Original question: #help-28 message
onyx glen
when to use intersection and union for (i)
intersection means "all of these happen at once", union means "at least one will happen"
(ii) is a bit trickier; you might want a Venn diagram
torn jolt
intersection means "all of these happen at once", union means "at least one will...
ooh
so now i use intersection for (i)?
onyx glen
yes
torn jolt
(ii) is a bit trickier; you might want a Venn diagram
(ii) is nothing
ok thanks for the help
i can do ABC'+AB'C+A'BC right
for ii
wut
onyx glen
i can do ABC'+AB'C+A'BC right
and also ABC
and also you may want to fix your notation a bit
P(ABC') + P(AB'C) + P(A'BC) + P(ABC)
torn jolt
P(ABC') + P(AB'C) + P(A'BC) + P(ABC)
ye im sorry for that, but why P(ABC)?
its only either 2 right?
or does it mean atleast 2?
onyx glen
"any two hit" means at least 2.
if they wanted exactly 2 then they would have said exactly 2
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quasi star
The vietnamese text is just find all functions that satisfy
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The vietnamese text is just find all functions that satisfy
quasi star
For polynomials I was able to solve it by comparing the deg of both sides but for functions idk where to start beside just x=y which doesnt do much
sturdy valve
what do u have to do?
charred carbon
what do u have to do?
Find all functions satisfy the above
charred carbon
hmm how about first x=0
x,y>0
sturdy valve
oh shit
silk bridge
xf(y)^2 = x + y
quasi star
xf(y)^2 = x + y
How do you prove it's injective though?
fast peak
well it could be constant so thats a bummer
silk bridge
hm..
quasi star
well it could be constant so thats a bummer
I think that's the only solution
Well atleast for polynomials that's the only solution
fast peak
there are a lot more functions than just polynomials
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narrow mica
why did i get three equations like this
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why did i get three equations like this
narrow mica
i cant see any clear dominance
and obviously 5-3p = 1-3p doesnt have any solutions
so i feel like there must have been some sort of dominance that i missed
but with the column II it sometimes has benefits player A if row II is selected but has drawbacks for player A if row I is selected
same with III as it's still better than I in some circumstances and worse in others
idk its just confusing
idk what to do in this case
nvm
i think i sorted it
i think i just made 3 simultaneous equations v1, v2, v3
and found the min
wait no max
.solved
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quasi star
there are a lot more functions than just polynomials
Well I'm just saying
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Well I'm just saying
quasi star
This is how I solve it for polynomials
deg(f(x)) = a
a(2a+1) = a or 2a^2 = 0 or a = 0 so f(x) is a constant
But for f(x) being a function it's pretty hard to have a starting point
empty blaze
(you've accidentally claimed this channel)
cobalt summit
This is how I solve it for polynomials
deg(f(x)) = a
a(2a+1) = a or 2a^2 = 0 ...
do .close to close it
quasi star
(you've accidentally claimed this channel)
It's still unanswered so I think it's fine
cobalt summit
The channel
quasi star
The channel
Yeah but it's still unsolved
quasi star
Im prolly gonna make this a thread though cause this question is pretty hard
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glad marten
For the final question on here is my answer still correct even though it is not the one on the answer key
warped frost
,rccw
glossy valveBOT
warped frost
well, here you have a saving grace
you are asked for possible formulae
and as you can imagine with periodic functions, there are many possible representations
so the question becomes, "is it correct?" and, "does your teacher accept it?"
glad marten
Yea I just want to double check if it is correct
glad marten
so the question becomes, "is it correct?" and, "does your teacher accept it?"
My teacher did mention there are many possibilities
warped frost
indeed, when dealing with periodic functions
glad marten
So I don’t know why she would not accept it unless it wasn’t accurate
warped frost
I haven't really checked the function itself, so if it is right, I don't see a reason why your teacher would penalize you
glad marten
I’m just paranoid about it not being right
warped frost
do you want me to check it though
glad marten
If you could that would be great
glossy valveBOT
Hanako(x, y); ∂(fox)/∂x
warped frost
graphing your answer function gives me quite a whole other graph though
so is your first possible function
hint: note that both of your functions do not have a vertical shift. that's something you should look into, considering the graph given has a y-intercept of 2, not 0
warped frost
shifting your first function up by 2 fixes it
but shifting your second function up by 2 gives this shape, which looks to be exactly the other way round from your given graph
from here you should know what to do to fix the second one?
glad marten
I don’t really know how to fix it because I would say go right pi/4 units but I already put that
Or left
warped frost
I'll give you a hint
you don't have to shift it right or left
remember, I said that your graph is exactly the other way round
glad marten
Oh reflect
warped frost
you just need to add one symbol/sign
glad marten
Wait
I just realized
Nvm
I just don’t understand how that works because on the photo we are given reflecting over the x axis doesn’t fix it
It makes it worse
warped frost
you're not reflecting over the x-axis
you're reflecting about the midpoint
which you can do by...?
another hint: the provided answer is doing it
glad marten
Reflecting over the y axis? That still doesn’t work tho
Because we are still starting at the orgin if we reflect
warped frost
you're reflecting about the midpoint
^
warped frost
basically you're reflecting about the black line you drew in the graph
glad marten
I don’t see how this would make it cosine tho
warped frost
I never said it would make it cosine
glad marten
Or how to write that and not reflecting it across the x axis
warped frost
I said that your answer, when graphed, is exactly a reflected version of the given graph
which you can rectify with one fix
warped frost
compare these two graphs
glad marten
Yea they are different which is why I’m confused
warped frost
notice that your function has the graph going down through the y-intercept, while the graph given goes up through it
glad marten
On the graph we are given we have to shift left or right
warped frost
but otherwise the peaks are exactly where they should be
that is an indication that your graph is correctly shifted
but you just need to flip it vertically through the midline
for (co)sine graphs, you do this by multiplying the entire function by -1
glad marten
But flipping it won’t make a max be on the y axis
glad marten
for (co)sine graphs, you do this by **multiplying the entire function by -1**
But isn’t that reflecting over the x axis how do I write those differently
glad marten
Reflecting it doesn’t make it in the write slot
glad marten
not the whole function
That’s how we write reflecting over the x axis tho
glad marten
It does on there but not on the graph we are given which is why I’m confused
Yea but I don’t have the red graph by only reflecting J I would have to shift
glossy valveBOT
Hanako(x, y); ∂(fox)/∂x
glad marten
Yes
warped frost
what I'm saying, is that you stick a negative sign in front of the 3
and you're done
glad marten
But then the graph starts at the min not the max
warped frost
that's what I meant by the fact that you just need to add one sign to fix your function
doesn't matter, it is exactly the shape of your graph!
glad marten
But I don’t understand how that would be cosine because cosine has to start at the max on the y axis
warped frost
not necessarily!
glad marten
warped frost
a shifted cosine can start at its minimum at x = 0
do not fixate on where cosine and sine start their cycles
because it all depends on the shift itself
glad marten
But that’s what our teacher told us to do
warped frost
focus more on matching the shape of the graph
glad marten
Which is why I’m confused
glad marten
focus more on matching the shape of the graph
But we are moving the graph so how can we match it
We wouldn’t want to match it if we did not match it it would not be cosine
warped frost
do they not look the same?
glad marten
Yea but that’s sin
warped frost
glad marten
I would shift it pi/4 to the left to get cosine
But that’s our starting graph
So how would we know there was a reflection
distant void
you're trying to describe the graph in the picture, aren't you?
glad marten
I’m trying to move it to make it cosine
distant void
you already are using a cosine function
glad marten
Yea but cosine starts at the max at the y axis
distant void
play around with your function in Desmos or Geogebra to see for yourself
glossy valveBOT
distant void
,w plot cos(x - pi)
glossy valveBOT
glad marten
Now I would never know how to write a cosine g graph then because if it does not have to start at the max that’s the difference between cosine and sin
distant void
well see
a cosine graph... is just a sine graph shifted pi radians
this is why there are multiple possible representations to begin with!
glad marten
Yea that’s why I’m trying to shift it to make it cosine
because I wrote the sin graph correctly
distant void
yes, and your cosine graph is correct as well!
glad marten
but I don’t have a cosine graph
distant void
the fact that it starts at its max while the graph in the picture starts at its min is the hint telling you to flip the cosine graph upside down
by your logic then, the graph cannot be a cosine graph because it starts at its min
which makes no sense, because Hanako could reproduce the graph with a cosine function!
distant void
But that’s what our teacher told us to do
this is but a guide, I believe. Idk what they told you to do exactly, but presumably doing what they told you to may help you find sine/cosine representations faster. but those are not hard and fast rules.
glad marten
But I don’t understand why we are flipping upside down
because then the graph would start at the min
distant void
because your graph as written is the wrong way round
distant void
because then the graph would start at the min
don't get too fixated on where it starts
it is still represented by a cosine function
it's like saying a graph cannot be a sine graph because it's not y = 0 at x = 0
glad marten
but if I don’t focused on where it starts I won’t know how to write cosine or sin graphs
Because then there would be no difference between them
distant void
because there ARE no differences between them beyond an angle shift
that's the key point here
glad marten
So how would I know how to write equations because now I would have i have no idea
Like why would I not just reuse my sin equation and write cosine instead of sin but keep the numbers
distant void
**Hanako(x, y); ∂(fox)/∂x**
I mean, how did you come up with this function to start with?
glad marten
Because if I shift left pi/4 it starts at the max
distant void
use the same method to come up with it, but flip it if it needs to be flipped
distant void
I mean what I said. how did you first come up with the original second answer?
glad marten
I looked at the graph noticed the max was at pi/4 I wanted the max to be on the y axis so I shifted it
full whale
if you insist on not flipping it, then you can achieve the same thing by shifting the second answer pi along
glad marten
But that would also make it start at the min
Or not the min at zero which is sin
Because the period is pi
full whale
so do you mean to tell me that because the graph starts at the min, you cannot represent the function using the cosine function as the base function instead of the sine function?
glad marten
full whale
because if you say so, these two functions here would like to have a word
glad marten
so do you mean to tell me that because the graph starts at the min, you cannot r...
I can but why would I write extra stuff if one shift can fix it instead of reflecting and shifting
I’m just confused why the answer is not right
glad marten
Shifting it pi would make it not start at the max it would give you the same graph because the period is pin
full whale
your answer is not correct because your graph is in the wrong position. that's really all there is to it
glad marten
your answer is not correct because your graph is in the wrong position. that's r...
But I did not flip it
glad marten
But on the graph we are given if we move it and reflect it it results in the thing above
distant void
at this point you should probably head to Desmos or Geogebra and play around with the graphs
glad marten
I wouldent know how to get the original equation tho
full whale
start with cos(x), then slowly apply each transformation in turn
or if you want a base to work off of, the four functions we used for testing are in my latest screenshot
just copy them
glad marten
full whale
if the screenshot is unclear
glad marten
I’m just confused because it seems like we are going backwards
And if I am doing this wrong I don’t understand how I get others right
full whale
to me the goal is simple - match the shape of the graph given with any kind of transformation applied to the sine or cosine functions as necessary
glad marten
Yea
full whale
I don't care if the function ends up starting from the min or starting from y = 100 or something
if it matches the shape of the given graph, pass
glad marten
Like we are given this graph so why wouldent we just move it left
full whale
maybe you missed the point of my first message
also, you are not allowed to shift the given graph, if that is what you meant
that'd be changing the question
full whale
no? the question wants you to represent the graph with a function, not move the graph
what you're telling me is something like this
glad marten
now I’m confused
glad marten
How do we know how much to shift a graph if we don’t have a starting point
full whale
How do we know how much to shift a graph if we don’t have a starting point
you do have a starting point - cos(x), or sin(x)
from there, you can stretch, compress, shift, or flip as necessary
full whale
your answer is not correct because your graph is in the wrong position. that's r...
cosine is not at the right spot because otherwise we would not have this issue, right?
also, look at the graph a little bit more carefully
glad marten
Wait
full whale
notice that at x = 0, the graph is rising through the midpoint. only sine does that, meaning that the cosine graph has to be shifted anyway
glad marten
We are starting from cosine then we are making the graph on the picure?
full whale
yes
don't you always do that? you start with the base function, then you apply transformations as necessary
never start with some shifted function. that's how you get confused in all of this shifting madness like you're playing Tetris
glad marten
no I was just trying to make it fit the rules of cosine
from the given graph
I thought I was going from the sin to cosine
full whale
cosine may have rules, but starting at the max point is NOT a rule cosine has
full whale
I thought I was going from the sin to cosine
not necessary
I mean I suppose you can? you can start with sin(x) = cos(x - (pi/2)) and work from there
but I'd rather just start with cos(x) and apply any necessary shift myself
full whale
you could, if you accept that you may need to shift by a different amount, OR that you need to flip the graph
full whale
if you insist on not flipping it, then you can achieve the same thing by shiftin...
again, I will remind you, this was the first thing I said to you
glad marten
How would I do this now tho if I can’t use the graph to move sin
full whale
use the graph to move sin?
glad marten
But how do I know how much to move it
full whale
I don't get what you mean
but if you mean how much to move the base sin graph by to match this graph
glad marten
Wait could I think of it of making the given grown back to sin or cosine
full whale
well, sin starts at midpoint and goes up
full whale
Wait could I think of it of making the given grown back to sin or cosine
I would like to see what you mean before I say yes or no
full whale
well then, here's one way to do it that would work for most all (co)sine graphs
glad marten
because before I would look at the graph and see there is a x internet or midline intercept at pi/2 so I would translate it to the left but now I did that lash time it was wrong
So now I would think it’s the opposite but that’s wrong
full whale
- start with a bare sine/cosine graph.
- determine the period of the target graph. transform your graph to match the period of the target graph.
- determine the amplitude of the target graph. stretch your graph to match the amplitude of the target graph.
- determine the horizontal shift of the target graph. pick an interesting point, such as a peak close to the y-axis (usually a peak), and look at how far it is shifted compared to your graph. shift your graph using the minimum shift necessary to make the peak on your graph correspond exactly to the peak of the target graph.
- determine the vertical shift of the target graph. look at the y-intercept of the target graph, and shift your graph up/down to match.
and that's all you really need to do more or less
full whale
How would I do this now tho if I can’t use the graph to move sin
an example of step 4, since I see how that can be confusing.
the base sine graph has a positive peak at x = pi/2 (closest peak to y-axis). but here your peak is at x = pi, and the period of the graph is unchanged (2pi).
that means that you need to shift the base sine graph so that its positive peak is at x = pi, aka shifting by pi/2 units to the right.
of course, once you are convinced enough, you can use a transformed sine/cosine function to recover its equivalent cosine/sine representation, but I'd rather not confuse you more on that
glad marten
And then for this one how is the period 8 and not 9
glad marten
- ?
full whale
you've just answered your own question
glad marten
But why does the period not end with each peak
full whale
wdym end with each peak
glad marten
Everytime it repeats it’s a period
full whale
a period is when the graph returns to the same state
for example, at x = 0, the sine graph is at y = 0 heading upwards
the next time the sine graph is at y = 0 heading upwards, that would be the completion of one period
though peaks are easier to see, which is why we usually use peaks to determine period
glad marten
Oh wait I think I was including the first box when counting so I double counted
full whale
careful there.
glad marten
I think WHAG happened is last year when we did this my teacher told me something about how it is backwards what we think but that is not the way my teacher described it this year
It’s all coming back to me
Like he would tell us in this case when transiting left and right + would be moving right and not left like we thought
full forumBOT
@glad marten Has your question been resolved?
fallow ridge
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fallow ridge
how to do this question?
charred carbon
,tex .log rules
glossy valveBOT
Alexis_Fx <- sucks at Math
charred carbon
Hint: Base change rule and difference of squares
silk bridge
-# give me your preamble
fallow ridge
i know the rules. i am not able to solve this question as it requiires some extra steps which i am not able to think of
fallow ridge
-# give me your preamble
what?
full forumBOT
Show your work, and if possible, explain where you are stuck.
fallow ridge
i knew helpers would say that but if i don't know what to do what would i do?
i just applied the law of logs
which are -
log base A / log base B = log base B / log base A
Dont know what to do with it further
brittle steeple
cross multiply
undone radish
aops mentioned?
fallow ridge
log A^2 = log B^2
fallow ridge
aops mentioned?
it is AOPS v1
undone radish
it is AOPS v1
yess
undone radish
-# give me your preamble
its a package
brittle steeple
(log A)^2 = (log B)^2
this should be possible to extract more info out of now
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@fallow ridge Has your question been resolved?
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grizzled zodiac
Okay so I’m solving for the Vertex here, the star is the original problem. I numbered each of my steps down to help the helpers track my thought process along with what I did.
I got this question wrong slightly
The answer is -5, 3
I ended up with 5, -3
Where did I go wrong?
gritty rose
I ended up with -3, +5
how ?
grizzled zodiac
Sorry 5, -3
Uhm, you can see my work. Once I subtracted 3 from both sides I don’t really see what else I need to do to the equation and that was my end result
silk bridge
Okay so I’m solving for the Vertex here, the star is the original problem. I num...
Multiplying by 6 does not mean you multiply each and every thing , a single term is multiplied by 6
(x+5)^2/6 is a single term
6 *(x+5)^2/6 = (x+5)^2
grizzled zodiac
Okay so my goal with that multiplication thing was to remove the fractions, as I find fractions hard to work with. Typically what you do to one thing you have to do to everything. What makes this different?
Just trying to fully understand so I can make better notes
silk bridge
<@&268886789983436800>
grizzled zodiac
Thank you mods
silk bridge
Okay so my goal with that multiplication thing was to remove the fractions, as I...
eg treat (x+5)^2 as a
You have y =a/6 +3
Now simplify
tepid arrow
guys guys chill
silk bridge
guys guys chill
?
silk bridge
Ok
silk bridge
!nosols
full forumBOT
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.
silk bridge
Sorry 5, -3
As you can see your amswer is right but the step was wrong
tepid arrow
@grizzled zodiac in coordinate system the value of x is first and know as abscissa and y is known as ordinate (absicissa, ordinate)
grizzled zodiac
eg treat (x+5)^2 as a
You have y =a/6 +3
It helps me out to write it out, my mistake was interacting with the things in the parentheses
grizzled zodiac
<@586728689614454786> in coordinate system the value of x is first and know as a...
I get that,
grizzled zodiac
But I don’t understand what happened in between the 2nd and 3rd step at all
silk bridge
It helps me out to write it out, my mistake was interacting with the things in t...
When you multiply by 6 the 6 in denominator cancels out so no need to interact
grizzled zodiac
Ohhh
silk bridge
Even though if you were to interact you can't directly multiply anything because the paranthesis are squared
silk bridge
sorry im a new member pardon my silly mistake
-# im new too
tepid arrow
But I don’t understand what happened in between the 2nd and 3rd step at all
in the 2nd step i multiplied 6 to every term so the concepts is that 1/6multiplicatin6 would be one for the first term and in the second term 6*3 becomes 18 in the third steps i subtract the equation by 18 and through the distrubutive law the 6 is taken common from the term 6y-18 which becomes (y-3)6 and i suppose you know the restg
tepid arrow
bro what about understanding ????\
grizzled zodiac
Yeah, it’s easier for me to understand when I can like
There’s a difference between reading math written out and then like, seeing all the symbols and stuff properly
At least in my head
silk bridge
Yeah, it’s easier for me to understand when I can like
Experience helps in algebra
grizzled zodiac
Yes!
full forumBOT
If you are done with this channel, please mark your problem as solved by typing .close
grizzled zodiac
One sec I just wanna show the guy something
grizzled zodiac
bro what about understanding ????\
This is what I mean when I say I was writing it down to understand. I could follow the steps and in between them all write down what allowed me to progress
Now when I go and try to practice this concept, my notes will show me how to attack similar problems
Thank you
.close
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lime crescent
This problem is giving me some trouble
full forumBOT
This problem is giving me some trouble
lime crescent
I need to find the slope using this table
gritty rose
do you knwo the formula for $m_{PQ}$
glossy valveBOT
riemann
lime crescent
do you knwo the formula for $m_{PQ}$
No
vivid imp
what am i ment to do, straight line with a square?
full forumBOT
what am i ment to do, straight line with a square?
wary condor
what am i ment to do, straight line with a square?
a or b?
wary condor
assuming its a,
y = x^2 - 2x - 1 is the function
and its asking you for solutions where x^2 - 2x - 1 = 0
do you know what that means
vivid imp
No
wary condor
when is this true
wary condor
So i do alegvra
you dont need algebra
vivid imp
How does one make this true?
wary condor
you have the graph, follow the instructions
vivid imp
you have the graph, follow the instructions
Yes but how do i plot a straight line with a sqaured equation
wary condor
ok let me try to make it clearer
x^2 -2x - 1 = y
we're looking for,
x^2 - 2x - 1 =0
vivid imp
ok let me try to make it clearer
x^2 -2x - 1 = y
we're looking for,
x^2 - 2x - ...
3x
But yeah .
Nvm your explaining
Mb
vivid imp
ok let me try to make it clearer
x^2 -2x - 1 = y
we're looking for,
x^2 - 2x - ...
So
How
wary condor
what I've been trying to say is, it wants the solutions for when y = 0
vivid imp
what I've been trying to say is, it wants the solutions for when y = 0
Yea
Oh
So the turning points
vivid imp
2.4 and -0.4
wary condor
2.4 and -0.4
and what are these?
wary condor
estimating the solutions for x^2 - 3x - 2 = 0 is just like saying estimate solutions for when y = 0. and the solutions are where the graph intersects with y=0
y=0 is a line
a very speical line indeed
vivid imp
But its squared
wary condor
Yeah but i cant plot a squared
you're not gonna plot a squared function/line
vivid imp
you're not gonna plot a squared function/line
Ey
urban nebula
I think the question wants us to use the graph of x²-2x-1
To find the roots of the other polynomial
By adding a straight like
wary condor
oh my god!!!!
vivid imp
What straight line???
wary condor
I think the question wants us to use the graph of x²-2x-1
can you continue 
vivid imp
I didnt read the question properly
Np
delicate torrent
what am i ment to do, straight line with a square?
Where are yall at
silk bridge
By adding a straight like
^
vivid imp
Where are yall at
No where rn
silk bridge
Write the x^2 -3x equatiom as graph equation + something
Or subtract the graph equation from that
vivid imp
-x-1
delicate torrent
what am i ment to do, straight line with a square?
We need to add a straight line (aka in a form of $ax + b$) such that it will turn into $x^2 - 3x - 2 = 0$. In other words, we want: $x^2 - 2x - 1 = ax + b \Leftrightarrow x^2 - 3x - 2 = 0$
glossy valveBOT
1 divided by 0 equals Infinity
delicate torrent
**1 divided by 0 equals Infinity**
The 2 points that intersect the line and the parabola will have the x coordinates that are the roots of $x^2 - 3x - 2 = 0$
vivid imp
Right, so you have quadratic -x-1=0
Ive plotted it
glossy valveBOT
1 divided by 0 equals Infinity
delicate torrent
**1 divided by 0 equals Infinity**
Do you understand what im saying here
silk bridge
The line goes through the orginal graph line
As infinity said find points of intersection
silk bridge
Do you understand what im saying here
Change nick its too long
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peak knot
Can I get a simple definition of a determinant in matrix calculus?
gritty rose
Can I get a simple definition of a determinant in matrix calculus?
"simple" is relative. what definition did you learn
main vault
area enclosed by each unit vector i guess
peak knot
I've just been taught the cross product so none
gritty rose
I've just been taught the cross product so none
start here:
https://www.cuemath.com/algebra/determinants/
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neat thicket
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prime pier
you said part (c)? 🤨
neat thicket
you said part (c)? 🤨
B
neat thicket
can we see what you did?
Yea
I did
I assumed 1/10 was the true calue and then tried to calculate what values for S(3) would i need to go beyond 0. 95th percentile on the normal
And then once i got that calue i subtracted by 1 over 5 and then divided by Se
<@&286206848099549185>
prime pier
what i meant was, can we see your work, so we can see what you calculated exactly?
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@neat thicket Has your question been resolved?
Available help channel!
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hybrid abyss
this is implicit differentiation for derivatives help me understand plz
hybrid abyss
@boreal crane
hybrid abyss
i watched the tutorial but the tutorial has no exponent
do i need to apply a rule of some sort
boreal crane
Do you understand the power rule of $ax^{n}$?
glossy valveBOT
Flappy the Turd
hybrid abyss
yes
boreal crane
It also applies to y, except you'd also add $\frac{dy}{dx}$ to the term
Such that the derivative of $ay^{n}=any^{n-1}\frac{dy}{dx}$
glossy valveBOT
Flappy the Turd
boreal crane
Make sure you treat dy/dx as a separate, extra variable!
hybrid abyss
wait so the derivative of y is dy/dx?
boreal crane
Also remember the multiplication rule
derivative of $ax^{n}y^{n}$ is $a(x^{n})'y^{n}+a(y^{n})'x^{n}$
boreal crane
wait so the derivative of y is dy/dx?
yes, perfect!
glossy valveBOT
Flappy the Turd
boreal crane
slight adjustment dont mind it
boreal crane
yes it is!
hybrid abyss
now i do the second term which is y^3 or am i missing something else
boreal crane
you're safe to move it onto the next term ye
boreal crane
yes it is! now you want to go to the other side of the equation, = 10
boreal crane
yes indeed!
hybrid abyss
finally
wait
1 more question
uh maybe a couple more if you have the time
i did 𝒙𝟐 +𝒚𝟐 =𝟐𝟓
did i do it correctly
ik there must be something im missing out on this one
boreal crane
oh mb i vanished for a moment lemme check for you... it seems you did it all right!
hybrid abyss
do you have any knowledge to piecewise
boreal crane
oh. hmm... I'm not too confident on that
-# btw i think we can move back to #math-discussion or close this help channel and move to a new one if you're done with the initial question
-# use .close to close a help ticket
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dull flint
$$
\text{The numbers } 1,2,3,4,5 \text{ are written on five cards, one number on each card. Three cards are drawn in succession and at random from the deck. The resulting digits are written from left to right. The probability that the resulting three-digit number is even is}
$$
dull flint
$$
\text{The numbers } 1,2,3,4,5 \text{ are written on five cards, one number on each card.}
$$
$$
\text{Three cards are drawn in succession and at random from the deck.}
$$
$$
\text{The resulting digits are written from left to right.}
$$
$$
\text{The probability that the resulting three-digit number is even is}
$$
glossy valveBOT
BlackidoZΣ
dull flint
how to solve this
marsh spruce
you can reduce this to the probability that the last card is either 2 or 4
elder sandal
or bash all 60 permutations
marsh spruce
sotrue
elder sandal
or bash all 60 permutations
or maybe think of it alternatively
there's a total of 60 permutations, ABC
C can be any number from 1-5
this means that there should be equally many permutations for each ending digit
XX1, XX2,..., XX5
elder sandal
you can reduce this to the probability that the last card is either 2 or 4
and use this fact to get a quick solution
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@dull flint Has your question been resolved?
dull flint
$$
\begin{aligned}
&\text{A small pack of cards consists of three red cards, three blue cards,} \
&\text{three yellow cards, and three green cards.} \
&\text{The pack is shuffled thoroughly and the first four cards are turned face up.} \
&\text{The probability that there is exactly one card of each color is}
\end{aligned}
$$
dull flint
12c4 is our sample space but i don't have idea whether they are faced up in this permuation or not
but if we assume them face up that should work
how to do it now
glossy valveBOT
BlackidoZΣ
dull flint
okay
like when we draw 4 cards on shuffling then each of them must belong to R/G/B/Y respectively
little steppe
Hint: since we're looking for order, the first one could be any card (so you could assume without loss of generality that the first card is, say, red)
dry arch
i was thinking 3^4 / 12c4
little steppe
i think that's close but the numerator assumes cards in order
bc it doesn't matter what colour the first card is
cobalt summit
there are 12 total cards
and 1/4 chance of getting one of them
no i dont think that would work tha way
dull flint
i thikn Axe's answer is perfect here
glossy valveBOT
dull flint
yes
dull flint
i have one more question to ask
glossy valveBOT
little steppe
wait i'm not wrong i just didn't math correctly
little steppe
i have one more question to ask
go for it
dull flint
Lot A consisits of 3G and 2D articles. Lot B consists of 4G and 1D article. A new lot C is formed by taking 3 articles from A and 2 from B. The probability that an article chosen at random from C is decetive is
little steppe
what do G and D stand for
dull flint
good and defective i guess
little steppe
are we assuming no replacement
dull flint
yes ig
dry arch
,w (12*9*6*3)/(12*11*10*9)
yeah you can make the sample space ordered hands if you want, then the denominator is 12*11*10*9
little steppe
yeah you can make the sample space ordered hands if you want, then the denominat...
ye i cooked but my idea was half-baked
@dull flint i would think about turning this one into cases
case 1: A returns 0D, B returns 0D
case 2: A returns 1D, B returns 0D
... etc
bc then your ultimate D-Drawing probability is easy, and you can reuse your maths in multiple cases
dull flint
okay i will think it later i have to go now
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@dull flint Has your question been resolved?
bitter anchor
I'm working on some matrix calculus and getting a bit lost in gradients with transposes and the rules we're allowed to use.
I want to compute the simple gradient shown. This seems like a "linear" term so my instinct is to say this would be evaluated as "b^T*A" basicaally removing the linear term. I don't know if that's correct though, I've seen some solutions to other problems where you need to write the problem in the form "c^Tx" to obtain c^T.
atomic valve
when in doubt, start from first principles. what's the definition of a gradient?
bitter anchor
$$df = \nabla f \cdot d\bold{r}$$
glossy valveBOT
bennbatt
atomic valve
yea, it's the transpose of the total derivative
so how would you compute the total derivative as a limit
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@bitter anchor Has your question been resolved?
bitter anchor
I guess I'm not seeing where $$\nabla f = [ \frac{\partial f}{\partial x_1}, ... , \frac{\partial f}{\partial x_n}]^T$$ and calculating specific partial derivatives gets me to a generic solution for b^T*Ax.
glossy valveBOT
bennbatt
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
bitter anchor
b and A have no terms containing x_i
atomic valve
yes
bitter anchor
but the gradient is A^Tb not b^TA
atomic valve
ok i suppose you have your answer already
but yes the total derivative here is b^TA if you look at the definition of a total derivative
and since the gradient is the transpose it should be (b^TA)^T which is A^Tb
bitter anchor
Can you point me to the definitions you're referencing?
The help I'm looking for is understanding why and when we need to insert/manipulate an equation with transposes to get into the correct form
I'm trying to compute the Hessian of the regularized logistic regression objective function, this is just one of the terms and it caught me off guard
atomic valve
Let $\fn{f}{\bb{R}^n}{\bb{R}}$. The total derivative of $f$ at $\mathbf{x} \in \bb{R}^n$ is defined to be the linear transformation $Df$ such that
\[ \lim_{\mathbf{v}\to0}\frac{\|f(\mathbf{x}+\mathbf{v}) - f(\mathbf{x}) - Df(\mathbf{v})\|}{\|\mathbf{v}\|} = 0.\]
So since $b^TA$ is linear we have
\[ \frac{\|b^TA(\mathbf{x}+\mathbf{v})-b^TA\mathbf{x}-b^TA\mathbf{v}\|}{\|\mathbf{v}\|} = 0\]
already such that $b^TA$ is the total derivative.
bitter anchor
You're saying b^TA is the total derivative?
atomic valve
yeah
bitter anchor
And the total derivative and gradient are just transposes of one another?
atomic valve
if you think about it the tangent space of ℝⁿ at a point is just ℝⁿ while the tangent space of ℝ at a point is just ℝ so the total derivative has type ℝⁿ → ℝ
the gradient is a vector though, so you need to take a transpose
bitter anchor
Okay, I don't think I've seen the total derivative definition or it's relationship to the gradient
thanks
atomic valve
what did you mean by df = ∇f ∙ dr then?
full forumBOT
@bitter anchor Has your question been resolved?
night socket
How do i know the sign for each radius of curvature of each plano concave lens
