hmm yeah i think i get it
#help-13
primal lodge
it's just that you get 1 minute to solve it and not sure i would think of all this in 1 minute lol
lunar lynx
It doesn't have to do with primes but it requires you to consider cases.
Try this: Find all ordered pairs of (x,y) such that they are solutions of xy - x - 2y - 8 = 0 and x and y are natural numbers.
lunar lynx
it's just that you get 1 minute to solve it and not sure i would think of all th...
Just try to get one solution and answer for x+y then. If there are multiple answers, they'll all have same x+y. So, you don't have to solve fully.
primal lodge
Just try to get one solution and answer for x+y then. If there are multiple answ...
mhm makes sense
i'll try to find something of that sort
primal lodge
Try this: Find all ordered pairs of (x,y) such that they are solutions of xy - x...
okay sure
primal lodge
Try this: Find all ordered pairs of (x,y) such that they are solutions of xy - x...
what is your definition of natural numbers?
includes 0 or not
lunar lynx
It doesn't include zero.
Btw, i came up with another straight forward way for you to get the answer for this question. It's a bit more nuanced though. Lol
also, it only helps you with finding x+y without finding the solution.
cedar kilnBOT
@primal lodge Has your question been resolved?
primal lodge
It doesn't include zero.
oh
honestly no clue lol
not even sure how to approach it
lunar lynx
It's okay. It requires a bit manipulation at first. Then, it should be easy.
Let me give you a hint.
primal lodge
sure
lunar lynx
You have to kind of factorise the expression i.e. write it in form of products. You'll have to add stuff on each side to make the factorisation possible.
lunar lynx
You have to kind of factorise the expression i.e. write it in form of products. ...
Because once it's in form of a * b = c, then you can consider cases as c will have limited factorisation form corresponding to LHS.
lunar lynx
Yes. You are almost there.
You see. If only you could somehow make term inside second bracket to be (y-1), you would be done.
primal lodge
You see. If only you could somehow make term inside second bracket to be (y-1), ...
but that wouldn't just be products?
x(y-1) - 2(y-1) + 10
(x-2)(y-1) + 10
lunar lynx
Yes. But that is equal to zero.
lunar lynx
x(y-1) - 2(y-1) + 10
Also, it should be -10. Check your calculation once.
primal lodge
yeah it should be -10
lunar lynx
Okay.
lunar lynx
So, now (x-2)(y-1) - 10 = 0
=> (x-2)(y-1) = 10
Now see possible ways of writing 10 as a * b .
lunar lynx
Important things to notice is that x and y and natural numbers. Therefore we can find answers. Otherwise it'll have infinitely many solutions.
Yes.
primal lodge
x-2 = 10 -> x = 12
x -2 = 1 -> x= 3
x -2 = 5 -> x = 7
x -2 = 2 -> x = 4
each has corresponding y values so 4 ordered pairs?
lunar lynx
```10*1
1 * 10
5*2
2*5```
Just compare all the a's in a * b with x-2 to get x's.
all the b's in a * b with y-1 to get y's.
lunar lynx
x-2 = 10 -> x = 12
x -2 = 1 -> x= 3
x -2 = 5 -> x = 7
x -2 = 2 -> x = 4
Yes. Bingo.
lunar lynx
each has corresponding y values so 4 ordered pairs?
yes.
primal lodge
hmm okay interesting lemme check what we did lol
sure makes sense @lunar lynx
i guess you gotta develop the intuition lol
to write out cases
lunar lynx
Yes. It might seem a bit tricky at first but slowly it becomes easy.
primal lodge
u have another question?
lunar lynx
u have another question?
Umm... I don't think so. Let me see if i can find one online.
Well, i found one. It might be interesting. Though it's easy i think.
x^2 - y^2 = 24 and xy = 35
find (x+y).
@primal lodge
primal lodge
Well, i found one. It might be interesting. Though it's easy i think.
x^2 - y^2...
wait i have to do something
but yes i think i could probably brute force this with algebra
lol
instincts is to factorize with difference of squares
lunar lynx
Alright.
lavish lily
are you still trying to solve this?
primal lodge
are you still trying to solve this?
not really, but if you have any insights i don't mind
primal lodge
how was i supposed to solve it?
xy = 35
so 7*5 or 5*7
in either case we have the abs value of the difference to be 2
(x-y)(x+y) = 24
6*4
12*2```
the only option that has a 2 is the third oneYes
primal lodge
that was how i was supposed to do it?
and yeah the difference can be 2 or -2
so sum could be 12 or -12 you're right
lunar lynx
Yeah. Pretty much. Actually, in actual question: x and y weren't integers. With integers, it's extremely simple in hindsight. Lol
Just keep practicing. You have general idea of how to go about solving. Practicing will make you better.
primal lodge
Yeah. Pretty much. Actually, in actual question: x and y weren't integers. With ...
lol
i think without the integers constraint i can bash algebra w/o using my brain
but for integer part i had to use my brain so ig that's hard
also i don't think this has
other solutions
that aren't integers
well besides complex solutions but that's just a reiteration of our real solutions
lunar lynx
Yeah.
primal lodge
Just keep practicing. You have general idea of how to go about solving. Practici...
no questions to find lol
lunar lynx
no questions to find lol
I mean... you'll occasionally come across such questions. Hopefully. Lol.
primal lodge
I mean... you'll occasionally come across such questions. Hopefully. Lol.
right but then i'll forget my current intuition
welp
anyway thanks
for the help
lunar lynx
right but then i'll forget my current intuition
You probably won't.
lunar lynx
anyway thanks
You are welcome.
cedar kilnBOT
cedar kilnBOT
Available help channel!
signal vault
cedar kilnBOT
signal vault
hi i dont know how to approach 3
H 3
cedar kilnBOT
@signal vault Has your question been resolved?
cedar kilnBOT
crimson sedge
.reopen
cedar kilnBOT
Available help channel!
crimson sedge
@signal vault
cedar kilnBOT
<@523846493761699842>
dreamy sleet
.close
jolly sable
f A and B are idempotent matrices of same order, then AB is idempotent matrix if and only if:
Options :
- AB = 0
- AB + BA = 0
- BA = 0
- AB = BA
jolly sable
$(AB)^2 =B^2A^2$
wraith daggerBOT
arjunn5589
jolly sable
=BA
If AB =BA then it can be idempotent i guess
Any other clearer ideas which explain?
tropic oxide
$(AB)^2 \neq B^2A^2$ in general though
wraith daggerBOT
Ann
jolly sable
Ohh it works in determinants only right?
tropic oxide
i don't know what "it" is.
i think that you confused this with something else that \textbf{does} work: $(AB)^T = B^TA^T$
but $(AB)^T = B^TA^T$ is irrelevant here
wraith daggerBOT
Ann
drowsy coral
How to do factorial (!)?
south tundra
This channel is occupied, open your own one
cedar kilnBOT
@jolly sable Has your question been resolved?
jolly sable
$(AB)^2 =A^2B^2$
wraith daggerBOT
arjunn5589
cedar cypress
(AB)²=ABAB
If this is equal to AB, then BA should be equal to AB because then we can replace the middle BA with AB,
Giving us (AB)²=AABB=A²B²=AB
jolly sable
wait a minute
cedar cypress
$(AB)^2 =A^2B^2$
We're talking about matrices here, This is true only if AB=BA
jolly sable
because of matrix multiplication is not commutative
cedar cypress
(AB)²=ABAB
If this is equal to AB, then BA should be equal to AB because then we...
Another way to look at it is that (AB)²=AB implies (AB)²=A²B² implies AB=BA
cedar cypress
because of matrix multiplication is not commutative
In general we can say that yea
But it's not impossible for it to be commutative
jolly sable
yes true
fallen plaza
How can I find x?
cedar kilnBOT
How can I find x?
gray blade
SOHCAHTOA
fallen plaza
How do I know which one it is
gray blade
well
fallen plaza
gray blade
is the side with 40m opposite, adjacent, or the hypotenuse to the angle with 30 degrees?
fallen plaza
Opposite to 30degrees
fallen plaza
Adjacent
gray blade
so which one uses opposite and adjacent sides?
fallen plaza
TOA
gray blade
u got it
wraith daggerBOT
kihei
wraith daggerBOT
kihei
gray blade
so now we need to get x by itself and can multiply both sides by x
fallen plaza
X= 40/ tan30?
gray blade
yep
fallen plaza
X=-6.2
gray blade
,calc 40 / (tan(30 deg))
wraith daggerBOT
Result:
69.282032302755
perhaps you used radians
gray blade
np
gray blade
yeah
fallen plaza
gray blade
maybe draw a diagram as it may be easier to visualize
it’s similar to the last problem
gray blade
yep
proud python
(3a/10) / (6a^2/5b)
proud python
(3a/10) / (6a^2/5b)
help to solve this
proud python
ok
fallen plaza
.close
rough rampart
can someone explain me what kind of series is this, which doesnt even have any respected pattern but these cool dude have made equation even.
rough rampart
2nd series
rough rampart
its not even the ap
cedar cypress
The differences are in an arithmetic progression
cedar cypress
its not even the ap
The differences are themselves in an AP
rough rampart
so that makes it an ap?\
cedar cypress
I didn't even explain how to get the nth term yet 😅
rough rampart
i got it
cedar cypress
Or can you do that on your own?
rough rampart
yea i can
cedar cypress
Alright
rough rampart
but just i couldnt see through the patterns
cedar cypress
It's fine
jolly sable
determinant of skew symmetric matrix even order would be?
cedar kilnBOT
determinant of skew symmetric matrix even order would be?
jolly sable
so here I am thinking that it will be det(a)=(-1)^n det(a^t)
the n is even so
det a =det a^t
crimson sedge
det a =det a^t
this property is true for any matrix
jolly sable
how can we say?
crimson sedge
so here I am thinking that it will be det(a)=(-1)^n det(a^t)
how did you get the factor of (-1)^n?
jolly sable
only symmatric matrix?
crimson sedge
could you show a screenshot of the original problem you are working on?
jolly sable
how did you get the factor of (-1)^n?
ahh sorry it is skew
crimson sedge
ah ok
jolly sable
ohh yes it is true for all because changes in row doesn't make difference in value of derminant
crimson sedge
changes in rows swap the sign of the determinant
crimson sedge
well yeah, an even number of swaps means a factor of (-1)^even
jolly sable
The skew symmetric determinant of even order is
both will be equal so zero?
$det (a) =det (a)^t$
wraith daggerBOT
arjunn5589
jolly sable
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
torn birch
(message.txt 1) this code takes accel data and gyro data to create a equation representing rotation from "reference vector (gravity)". Specificly the code in the Mahony_update( function (message.txt 2)
How do I either apply inverse rotation to a vector or get the gravity vector from this?
torn birch
did you reach discord character limit for messages
yes, the question is the message, the fist file is the whole code for context. the second file is the specific function that takes the:
accelerometer x y z force vectors
roll pitch and yaw gyro changes in angular velocity
time delta
dire geode
give that a read and see if it solves your problem
https://en.wikipedia.org/wiki/Active_and_passive_transformation
cedar kilnBOT
@torn birch Has your question been resolved?
torn birch
give that a read and see if it solves your problem
https://en.wikipedia.org/wiki...
So the original code that i was using did a passive transformation so the reference vector changed, I have used that it the code and would like to leave it the same for side by side testing.
This new code instead provides as quaternaion which is the equivalent of an active transformation when applied to a vector.
I would like to find the passive transformation (the transformed reference vector) that corresponds to the the provided active transformation quaternaion.
is the notation
for what i am calling the "inverse rotation"?
dire geode
of a quaternion it's conjugate?
cedar kilnBOT
torn birch
.reopen
cedar kilnBOT
✅
torn birch
https://math.stackexchange.com/questions/4281031/conjugate-of-unit-quaternion
"see screenshot (symbols didn't copy paste)"
so if I do a rotation of a vector(a) by the conjugate of the output quaternion of that Mahony_update function then i will have the vector corresponding to that vector(a) if the world space was rotated by the quaternion?
dire geode
idk what mahony update is
torn birch
idk what mahony update is
dire geode
yea i'm not reading that
torn birch
yea i'm not reading that
afaik it outputs an "active"? quaternion representing the absolute rotation of the inertial measurement unit
cedar kilnBOT
@torn birch Has your question been resolved?
torn birch
<@411595460965433347> Has your question been resolved?
I guess so I'll just have to give it a test
cedar kilnBOT
long arrow
start from the equation of the line
y = mx + b, where m < 0
try to make it dependent on one variable only (here: slope)
in other words make an effort to find b in terms of m, can you?
waxen basalt
in terms of m?
long arrow
yes
waxen basalt
hmm
long arrow
hint: use the point
waxen basalt
7-3m?
waxen basalt
wdym?
long arrow
A = (..., ...), B = (..., ...)
long arrow
yes
waxen basalt
m=-1 correct?
waxen basalt
hmm
waxen basalt
B=(7-3m,0)
long arrow
rather
(0, 7 - 3m)
but yes
now area of the triangle is
1/2 * (3m - 7)/m * (7 - 3m)
I guess you know why
long arrow
now we have function which represents area of the triangle ABC
in terms of the slope m only
and we know it's the smallest as possible
so you should find for which value of "m" that minimum occurs
do you know calculus?
waxen basalt
so you should find for which value of "m" that minimum occurs
using which equation?
long arrow
$$A(m)=\frac{1}{2} \cdot \frac{3m-7}{m} \cdot (7-3m)=-\frac{(3m-7)^2}{2m}$$
wraith daggerBOT
modus7591
long arrow
do you know how we in general look for extrema of the function
do you know derivatives?
that's optimalization
waxen basalt
well
heres what i got
i just want to double check to see as i can only submit once for this hw
$$up: (-\infty,-2-\sqrt{2}) and (-2+\sqrt{2}, \infty), down: (-2-\sqrt{2},-2+\sqrt{2})$$
wraith daggerBOT
arctic14.
long arrow
seems good
waxen basalt
x cords= $$-2-\sqrt{2},-2+\sqrt{2}$$
wraith daggerBOT
arctic14.
long arrow
yes
waxen basalt
okay last one
so for the left sum
its
(sin(0)+sin(1/8)+sin(2/8)+...+sin(7/8))*1/8
correct?
rounding each sin to three decimals btw
<@&286206848099549185>
sinful summit
$$
L = \sum_{i=0}^{7} f(i \cdot \Delta x) \cdot \Delta x
$$
where $f(x) = \sin(x)$, $\Delta x = \frac{1}{8}$, and $i$ ranges from $0$ to $7$.
$$
R = \sum_{i=1}^{8} f(i \cdot \Delta x) \cdot \Delta x
$$
where $f(x) = \sin(x)$, $\Delta x = \frac{1}{8}$, and $i$ ranges from $1$ to $8$.
wraith daggerBOT
adzetto
sinful summit
,w L=sum_{i=0}^{7} sin(i * 1/8) 1/8, R= sum_{i=1}^{8} sin(i * 1/8)1/8
wraith daggerBOT
cedar kilnBOT
waxen basalt
.close
cedar kilnBOT
Available help channel!
mellow zodiac
need help filling out this chart for momentum in physics
cedar kilnBOT
need help filling out this chart for momentum in physics
mellow zodiac
formula for momentum: p=mv
formula for kinetic energy: E(K)=1/2 mv^2
<@&286206848099549185>
cedar kilnBOT
@mellow zodiac Has your question been resolved?
sinful summit
need help filling out this chart for momentum in physics
write the momentum and kinetic energy conservation equations and solve for $v_1^\prime, v_2^\prime$
$$m_1 v_1 + m_2 v_2 = m_1 v_1^\prime + m_2 v_2^\prime$$
$$\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 (v_1^\prime)^2 + \frac{1 }{2} m_2 (v_2^\prime)^2$$
wraith daggerBOT
adzetto
robust drift
how do i write x E [5, 13] in form Ix - aI <= b? i missed this lesson
untold torrent
$x \in [5, 13]$
wraith daggerBOT
redstoneplayz09
untold torrent
do you know what $[5, 13]$ means
wraith daggerBOT
redstoneplayz09
untold torrent
or in general, what $[a, b]$ means?
wraith daggerBOT
redstoneplayz09
robust drift
yes x is equal to those numbers or any values inbetween
wraith daggerBOT
redstoneplayz09
untold torrent
is used to say if something is IN that set
robust drift
ahh okay
untold torrent
so if you say $2 \in [1, 3]$
wraith daggerBOT
redstoneplayz09
untold torrent
it means 2 is in the set [1, 3]
that is true because the set [1, 3] represents all numbers between 1 and 3 (including 1 and 3)
robust drift
gotcha
untold torrent
so we know x is a number between 5 and 13
now, |x - a| <= b
is the same as saying "the distance between x and a is AT MOST b"
so LESS THAN OR EQUAL TO b
now how can u classify the numbers between 5 and 13
by saying their distance to some number a, is at most some number b?
what numbers a and b would work here
robust drift
5 and 13?
untold torrent
|x - 5| <= 13?
does 3 satisfy this condition?
answer: ||yes it does||
does it belong in [5, 13]?
answer: ||no it doesn't||
robust drift
yes and no?
ayy 2/2
sorry what was the 3 for, just an example or is it part of the process?
untold torrent
just an example
to show a = 5 and b = 13 doesn't work
try doing the opposite
if I gave you
|x - 1| <= 3
what set of numbers would this correspond to?
hint: ||try imagining this on the number line||
robust drift
im sorry i dont see a connection rather than just trialling values to find the answer
untold torrent
hint: ||try imagining this on the number line||
look at this
robust drift
sorry i dont quite get the connection
untold torrent
what's the midpoint of the interval [5, 13]
robust drift
9
robust drift
4
untold torrent
and 5 and 9?
robust drift
4
robust drift
no
robust drift
im so sorry i know this is very basic but i cant figure it out could you just show me and maybe i can peice it together?
cedar kilnBOT
@robust drift Has your question been resolved?
untold torrent
alright
do u understand that all points in [5, 13] are distance of at most 4 from 9?
so for any x in [5, 13] the distance between 9 and x must be less than or equal to 4
pseudo merlin
How did they get 1.67
cedar kilnBOT
How did they get 1.67
pseudo merlin
dire geode
How did they get 1.67
did you solve for x when setting y=0?
pseudo merlin
Yes
pseudo merlin
I got 2 numbers
dire geode
I got 2 numbers
show your entire work?
dire geode
this?
pseudo merlin
Yes buf i didnt simplify final answer
dire geode
,calc (20 + sqrt(480)) / 2
pseudo merlin
Im in bed lol
wraith daggerBOT
Result:
20.954451150103
,calc (20 - sqrt(480)) / 2
wraith daggerBOT
Result:
-0.95445115010332
Yes
dire geode
,w roots -1/12 (x-10)^2 + 10
wraith daggerBOT
dire geode
,calc 2(5 + sqrt(30))
wraith daggerBOT
Result:
20.954451150103
,calc -2(sqrt(30) - 5)
wraith daggerBOT
Result:
-0.95445115010332
distant pecan
ive reached the point where i solved using the quadratic formula getting the roots 4 and -6
distant pecan
idk what to do to find the extranious solutions
slate lintel
try both of them in the equation and see which one doesn't work! but i think you might want to recheck your work because it looks to me like neither one satisfies the equation
distant pecan
bruh
muted bear
you forgot the 2 on the denominator i think
slate lintel
oh yeah
but you should be able to factor that so no need for quad formula
muted bear
called it lol
distant pecan
oh yeah
but you should be able to factor that so no need for quad formula
i already forgot howot factor
can u give me a rundown with this problem so i remember
slate lintel
um it'll take practice but you go from $x^2 + x - 6 = 0$ to $(x+3)(x-2) = 0$ which gives you your roots right away
wraith daggerBOT
Hayley
slate lintel
anyway
carry on with your roots for now
distant pecan
how did u get to (x+3)(x-2)
slate lintel
i looked for two numbers that multiplied to -6 and added to 1
distant pecan
added to one?
slate lintel
yes, i needed a positive and a negative number because i wanted them to multiply to -6
being able to factor stuff like that is a lot faster than setting up the quadratic formula, but it isn't always possible
distant pecan
so for this i just plug each x value in and see if it works?
slate lintel
yeah
jaunty mural
🤨
jaunty mural
so for this i just plug each x value in and see if it works?
Try what you said for yourself and have confidence in what you end up finding
If you want further clarification, show us your working.
slate lintel
to be clear the "extraneous" solutions are the ones that don't work
distant pecan
yeah
but to clarify if it says x=1 and i get x=3 after solving its extraneous
right?
doesnt work
jaunty mural
well can you select multiple checkboxes?
jaunty mural
most likely
distant pecan
or something
jaunty mural
kinda stupid of them to use the word 'extraneous' as well. what is this, an english class?
to me at least.
distant pecan
they haveanother typo in one of the questins
lol
"round ot teh nearest hundredth"
for this i just plug in and see which works ...also in this case what defines "works"
jaunty mural
There's another way to approach the question
That doesn't rely on brute forcing dumb multiple choice questions (quite frankly)
jaunty mural
Bringing it back to quadratics
Suppose we have a quadratic x^2 + bx + c
And I told you its roots are 1 and 2
(edited)
Do you know how to find b and c
distant pecan
no
jaunty mural
Well in factored form, do you know what the quadratic must look like
distant pecan
uhm
jaunty mural
for this i just plug in and see which works ...also in this case what defines "w...
tbh you should know this stuff pretty confidently before trying out these kinds of quetions
distant pecan
x(a-something)(c-something)
wraith daggerBOT
wraith daggerBOT
jaunty mural
this is factored form.
distant pecan
ohk
jaunty mural
alpha and beta are the roots.
distant pecan
ok
jaunty mural
Were you already aware or?
distant pecan
no
jaunty mural
🫤 You really do need to know this stuff first
distant pecan
idk what alpha is they never mentioned ir
jaunty mural
Like how are you learning math
alpha and beta are just some choice of letters by me
jaunty mural
$$a(x-p)(x-q)$$
wraith daggerBOT
jaunty mural
this if you like.
distant pecan
yeah ive seen that
jaunty mural
Either way, those 2 numbers are your roots.
distant pecan
yeah
jaunty mural
They are the roots because of the zero-product property
distant pecan
x-2 means x=2
wraith daggerBOT
jaunty mural
So this is your quadratic in factored form
wraith daggerBOT
jaunty mural
🤨 not sure what u mean
oh ig maybe thats one method...
you mean like plug in x = 1 and x = 2?
distant pecan
sure
jaunty mural
The other way is to remember this factored form and standard form are equal
You can multiply out the factored form.
$(x-1)(x-2) = x^2 - 3x + 2$
wraith daggerBOT
distant pecan
huh
jaunty mural
for this i just plug in and see which works ...also in this case what defines "w...
Going back to the original question, you can do exactly this method but for cubics
distant pecan
solved (x-1)(x-2) to get that
jaunty mural
expanded is the word
jaunty mural
same method?
Unfortunately theres a small twist that didnt exist in my example
distant pecan
but do the steps make sense
can u do a quick rundown of the steps and i can see
jaunty mural
ok
wraith daggerBOT
jaunty mural
Now the key point to remember is that these two forms are equal
In the example, I said 1 and 2 are the roots.
Also a = 1
So now we know
wraith daggerBOT
jaunty mural
We then expand the left hand side
distant pecan
ok but how does finding b and c matter
jaunty mural
It matters for finding the standard form of the polynomial
Which is what this question is after
distant pecan
ok sure
jaunty mural
The thing that made my example simpler was that a = 1
You can't assume that here.
So since this is a cubic
Your factored form will be
wraith daggerBOT
jaunty mural
And standard form is
wraith daggerBOT
jaunty mural
its a bit trickier to multiply that out, but is good practice
and well - not all questions in math are multiple choice.
distant pecan
what is a
jaunty mural
for this i just plug in and see which works ...also in this case what defines "w...
if you look at your multiple choices - you'll see it will either be 1 or -1
but you don't have to worry about this til later
jaunty mural
yep
Now leaving the a aside, multiply those 3 brackets out
(so multiply two brackets out first then multiply another)
wraith daggerBOT
jaunty mural
oof an error somewhere
distant pecan
BRUHHHHHHHh
jaunty mural
in the x term
distant pecan
and the 12
distant pecan
wraith daggerBOT
jaunty mural
This is what we have so far
now remember this needs to be equal to one of our 4 multiple choices
So now you compare this thing with each of them
the coefficient of x^3 tells you what a must be for that particular choice
jaunty mural
open another channel. i will be hopping off unfortunately but someone else will
distant pecan
ok
jaunty mural
distant pecan
.close
stark quest
no clue where to even begin here
cedar kilnBOT
no clue where to even begin here
stark quest
i think im confused by the question as well
what does it mean by two planes which are parallel to both planes?
cedar kilnBOT
@stark quest Has your question been resolved?
cedar kilnBOT
@stark quest Has your question been resolved?
stark quest
.close
stray vale
State the base function and then describe the transformations
$f(x)=-3\left ( 4 \right )^{x+1}$
wraith daggerBOT
deviousglxy
crimson sedge
Well you got something that's[a^x]
wraith daggerBOT
crimson sedge
Any idea of what that is
cedar kilnBOT
cedar kilnBOT
Available help channel!
jovial wing
Not entirely sure where to start, if someone could give me some suggestions I can get the ball rolling
cedar kilnBOT
Not entirely sure where to start, if someone could give me some suggestions I ca...
@jovial wing Has your question been resolved?
slate lintel
that's a fun one
slate lintel
Not entirely sure where to start, if someone could give me some suggestions I ca...
the fact that f has a turning point at (1, 1/2) tells you two things: that it goes through that point, and that it has a turning point there
can you translate those into equations?
I can't really read it but it looks to me like you've got at least one of those already there
jovial wing
yeah that is erased working out
give me 2 secs
That look right @slate lintel ? (This is cas enabled btw)
slate lintel
that will get you the right answer i think but I'm very curious where your f ' (1) expression came from
generally you'd find f ' (x) first and then substitute 1
jovial wing
my cas spat that out when I chucked in derivative at point 1
jovial wing
yeah suggesting the two equations helped a lot
cheers
wouldn't I need a third equation though?
or is there another way to show that the variables equal eachother
slate lintel
you can do it with just those two 
you just won't get their values
but you can show that they're all equal
cedar kilnBOT
Available help channel!
crimson sedge
how do I prove S is convex
cedar kilnBOT
how do I prove S is convex
crimson sedge
S = {x in R2 | x_1^2 + x_2^2 <= 1, x_2 > 0}
My first attempt:
To show that it is convex,
WTS: $\lambda x + (1-\lambda)y \in S$
wraith daggerBOT
matthewzz
crimson sedge
for all lambda in [0,1]
Step 1:
Since $x_1 >= 0$, and $y_1 >= 0$;
$\lambda x_1 + (1-\lambda)y_1 >= 0$, this is obviously true
wraith daggerBOT
matthewzz
crimson sedge
Step 2:
WTS: $(\lambda x_1 + (1-\lambda)y_1)^2 + (\lambda x_2 + (1-\lambda)y_2)^2 \leq 1$
wraith daggerBOT
matthewzz
crimson sedge
by FOIL:
$(\lambda x_1 + (1-\lambda)y_1)^2 + (\lambda x_2 + (1-\lambda)y_2)^2 = (\lambda x_1)^2 + ((1 - \lambda) y_1)^2 + (\lambda x_2)^2 + ((1 - \lambda) y_2)^2 + \lambda * (1-\lambda)(x_1y_1+x_2y_2)$
wraith daggerBOT
matthewzz
crimson sedge
$=\lambda^2 (x_1^2 + x_2^2) + (1 - \lambda)^2 (y_1^2 + y_2^2) + 2 \lambda * (1-\lambda)(x_1y_1+x_2y_2)$
wraith daggerBOT
matthewzz
crimson sedge
$<= \lambda^2 + (1-\lambda)^2 + 2 \lambda * (1-\lambda)(x_1y_1+x_2y_2)$
wraith daggerBOT
matthewzz
crimson sedge
im not sure where to go from here @crimson delta
crimson delta
the crucial inequality that you want is $2ab\leq a^2+b^2$
wraith daggerBOT
denascite
crimson sedge
how do i know that?
crimson sedge
a/b + b/a >= 0?
crimson delta
no
crimson sedge
$(a-b)^2 >= 0$
wraith daggerBOT
matthewzz
crimson sedge
$a^2 + b^2 - 2ab \geq 0$
wraith daggerBOT
matthewzz
crimson sedge
$a^2 + b^2 \geq 2ab$
crimson delta
yes
wraith daggerBOT
matthewzz
crimson sedge
**matthewzz**
so from here
$\leq \lambda^2 + (1-\lambda)^2 + 2 \lambda * (1-\lambda)(x_1y_1+x_2y_2)$
$\leq \lambda^2 + (1-\lambda)^2 + \lambda (1-\lambda) 2x_1y_1+2x_2y_2$
$\leq \lambda^2 + (1-\lambda)^2 + \lambda (1-\lambda) (x_1^2 + y_1^2 +x_2^2 + y_2^2)$.
crimson delta
well lets not ignore the brackets
hallow mango
hi
crimson delta
you got rid of some brackets
probably the space before the \leq
the bot doesnt seem to like spaces near the $
crimson sedge
$\leq \lambda^2 + (1-\lambda)^2 + 2 \lambda (1-\lambda)$
wraith daggerBOT
matthewzz
crimson sedge
$=(\lambda + 1-\lambda)^2$
wraith daggerBOT
matthewzz
crimson delta
yes
crimson delta
whatever those are
crimson delta
what do you want to ask about them?
crimson delta
well thats a much harder question
crimson sedge
from what ive learned $d := {x+d | x \in S \implies x+d \in S}$
wraith daggerBOT
matthewzz
crimson delta
d shouldnt be both a set and a vector
what you mean is that d is a feasible direction for x in S if there exists lambda with x+lambda d in S
"d is a feasible direction if you can go a (small) step in direction of d without leaving S"
crimson sedge
i thought it is a set of vectors?
crimson delta
you are using d to mean to different things
in the theorems d has to be a vector
otherwise the expressions written down dont make sense
crimson sedge
oh ok
can you help me walk through the feasible directions of S = {x in R2 | 0 <= x_1^2 - x_2} at (0,0)
we want to find D = set of feasible directions
$[0,0] + [d_1, d_2] \in S$
wraith daggerBOT
matthewzz
crimson delta
make a sketch of S
crimson sedge
yea its a parabola
crimson delta
its not
crimson sedge
with the bottom partshaded
crimson sedge
so i know intuitivly any feasible direction should probably be like some downward looking vector, but i want to try to work through it algebraicly
but how do i make this into a vector
crimson delta
what you mean is that d is a feasible direction for x in S if there exists lambd...
you are ignoring the lambda
you need lambda * [d1, d2] in S for at least some small lambda
however, what happens if you scale both d1 and d2 by lambda
no
crimson sedge
however, what happens if you scale both d1 and d2 by lambda
what do you mean?
wraith daggerBOT
matthewzz
crimson delta
you have the wrong inequality sign
the idea here is now that lambda^2 goes smaller much quicker than lambda
after dividing one lambda you get that d_2 <= lambda d_1 for all small lambda
but thats clearly not true
crimson sedge
wdym its not true?
crimson delta
assuming d_2 and d_1 positive
i.e. d goes "to the top right"
similar argument for top left
well d_2 is fixed but lambda d_1 isnt
so just pick lambda small enough
crimson delta
thats not even a vector
crimson sedge
oops i meant that
crimson delta
no the lambda is not part of the direction
crimson sedge
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
warped haven
n is a integer ( n > 5 ). Give a set X = { 1,2,3,... n }. Set X will have ability T if we can provide X into 2 sets A and B so that in any set A and B, in three abitary elements , multiple of two elements will be different with the other one ( in three elements ) . Prove with 7 ≤ n ≤41, set X always have ability T
warped haven
Sorry if my post is wrong about english grammar, help me please
cedar kilnBOT
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
i don't know where to begin
- I don't know where to begin
<@&286206848099549185>
cedar kilnBOT
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
indigo eagle
What does the question tell you to do?
warped haven
n is a integer ( n > 5 ). Give a set X = { 1,2,3,... n }. Set X will have abilit...
.
prove it
<@&286206848099549185>
cunning forum
it suffices to prove the assertion for n=41
for smaller n just ignore the "ghost" elements that aren't there
now how do you do that?
warped haven
for smaller n just ignore the "ghost" elements that aren't there
i dont unerstand what are u talking about
cunning forum
i dont unerstand what are u talking about
suppose we have proved it for n=41
ok, suppose
then we split it into two sets A and B that fulfil the condition
for n=40 for example we do the exact same thing except without the element {41}
for n=39, ignore {40},{41} and so on
warped haven
can you prove it clearly please
dire geode
n is a integer ( n > 5 ). Give a set X = { 1,2,3,... n }. Set X will have abilit...
What does ability T mean
cunning forum
What does ability T mean
it means it satisfies that condition
cunning forum
can you prove it clearly please
We claim it suffices to prove the statement for n=41.
Suppose we have proven it for n=41. Then we can split {1,2...41} into sets A and B such that neither A nor B have three elements with one being the product of the other two. Then for any n<41, we can still satisfy the condition by taking these two sets A and B and erasing all the elements above n, since by assumption these two sets fulfil the condition.
Hence it suffices to prove it for n=41
warped haven
okay, but how we can prove it for n=41.Can u help me
cunning forum
okay, but how we can prove it for n=41.Can u help me
It's not that difficult, could you try it yourself for a while? For example, when n=10 we could have
{1,2,3,4,5,7,9}
{6,8,10}
warped haven
It's not that difficult, could you try it yourself for a while? For example, whe...
prove it , not give an example. This way is too long
cunning forum
find the largest n could be
41
cunning forum
prove it , not give an example. This way is too long
Giving an example in this case would prove it as I have already shown
I can’t give you the example since that’s tantamount to an answer
cunning forum
Oh that was a typo sorry
I’ve seen this question before
Whatever I will construct the sets for you
{1,2,3,4,5,7,8,9,11,13,17,19,23,25,29,blah blah}
{6,10,12,14,15…} the rest
The first is from greedy algorithm
Second is the rest
It is clear that first set fulfils condition by construction, second one works because 6*10>41
vapid whale
cedar kilnBOT
vapid whale
i had a question about the vertex
according to my calculations, the vertex should be (1.5, 0.75)
but on desmos doesnt it show that vertex is (1,1) ?
south tundra
Well, why do you think the x-coordinate of the vertex is 1.5?
vapid whale
south tundra
t = 1.5 does not mean that x = 1.5 though
vapid whale
oh right
south tundra
So, try solving x + 1/x = 1.5
Hold on
I think after all you can't treat this as a usual quadratic function due to behaviour of x + 1/x
Better approach would be to complete the square I think
Yeah it seems to work, although you will need to notice a nuance
vapid whale
i know how to solve it, i was just confused about the vertex actually
but im curous if there's an alternate method
what did you have in mind?
south tundra
Ah, okay, you could solve for the vertex using derivatives if you are familiar with calculus
Hmm, but, wait, I think what I wrote right now could help as well, without calculus
Oh, yeah, nevermind
Completing the square is the way
Do you want me to explain why or you will try yourself?
vapid whale
my method is completely different, so yea i have no idea what u mean
south tundra
What did you use?
south tundra
Or are you talking about this?
vapid whale
yea i meant final solution, not vertex
south tundra
The issue is that t does not take any value
In fact t can't be equal to any number from (-2, 2) for real x
vapid whale
my method is completely different, so yea i have no idea what u mean
so basically, x + 1/x = t, so the domain of t will be this
vapid whale
so basically, x + 1/x = t, so the domain of t will be this
and then i can graph the f(x), but exclude the parts where t does not exist
south tundra
Hm? You mean the range of t?
vapid whale
uh range yes
south tundra
and then i can graph the f(x), but exclude the parts where t does not exist
Yeah that's the way
Simply consider the parabola t^2 - 3t + 3 for values t >= 2 and t <= -2
Do you have any other doubts/questions?
vapid whale
yea im just trying to solve it one sec
horrible drawing but ig something like this
south tundra
Everything looks good 
vapid whale
so it exists everywhere above 1 🤔
south tundra
Except the [ next to -infinity
vapid whale
Except the [ next to -infinity
oh yea lol i'll fix that
south tundra
so it exists everywhere above 1 🤔
Yup
vapid whale
yep got it
vapid whale
Completing the square is the way
what did you mean by this btw?
south tundra
At the time completing the square was the only thing I came up with so I thought it's the only way
Basically, after completing the square in $x + \frac1x$ I got [ f(x) = (( x + \frac1x ) - \frac32 )^2 + \frac34]
wraith daggerBOT
alonelybean
south tundra
So the problem was narrowed down to finding out the range of x + 1/x - 3/2
south tundra
**alonelybean**
Which is R\(-3.5; 0.5), so the smaller possible number of this square here is 1/4
Meaning the smaller possible value of f is 1/4 + 3/4
Which is 1
south tundra
But this approach is better
brittle wasp
Can some tell me why this is not correct?
cedar kilnBOT
Can some tell me why this is not correct?
dawn junco