S & T is quadrant 2 &3
#help-13
abstract lotus
remote lantern
u go around the quadrants basicall
until u get to that refence angle
since I did a full circle
it would now be 2pi - refernce angle
brazen wigeon
no, it would be 2pi + reference angle
remote lantern
hmm
i think its minus because the refernce angle is before 2pi
so from a full circle u are reducing the ref angle
its just like the first one
first one is pi - ref
brazen wigeon
so from a full circle u are reducing the ref angle
nope, you go around the circle, then, continue along a little more until you reach the reference
remote lantern
180 - ref angle
the second one is + because its now 180 + ref angle
right but we are now back in quadrant 2 right
so it would be - again
brazen wigeon
huh
abstract lotus
if u look at ur blue line, u made a full circle and kept going
brazen wigeon
idk how to even explain it
brazen wigeon
if u look at ur blue line, u made a full circle and kept going
but he does
remote lantern
yes thats why that one is +
because it extends beyond pi which is 180
so the ref angle is added to the 180
its 180 + ref angle for the 3rd quadrant
abstract lotus
im talking about the 3rd one, you made a full circle and went right back to the same angle as the first drawing 2pi/3
remote lantern
right so thats why i made it 2pi
so to my understanding its 2pi = full circle - ref angle
brazen wigeon
so to my understanding its 2pi = full circle - ref angle
nope
if it was subtraction, you would have gone backwards after completing the full circle
but after completing the circle, you continued on, so you add a value to 2pi
abstract lotus
brazen wigeon
5pi/3 = 2pi - pi/3
your angle: 2pi + 2pi/3
remote lantern
but like isnt that ref angle taken away from the full circle
ohh
i see now
im so dumb
i already drew a full circle
i just slowly drew it on paint
and just realied
i had already made a full 360
abstract lotus
YES
remote lantern
fuck
ok
so lemme just be clear
so i made the full circle
so that angle is now added to it?
brazen wigeon
so that angle is now added to it?
yes
because after makoing the full circle, you continued along
abstract lotus
u ended up back at the first angle
remote lantern
yea say we do this
oksay we kept going
so say i do a full circle
would this
be
2pi + pi - ref angle
?
brazen wigeon
2pi + pi - ref angle
what is your reference angle again
remote lantern
pi/3
brazen wigeon
ok then yes
what you said over text is correct, but it doesn't match the image you sent. just an fyi
remote lantern
like no the initial rotation
brazen wigeon
youre correct
remote lantern
its the second go around the circle
m ok
hm
i see now
thanks a lot guys
imma try and do more practice
il open another box if i have questions
thanks again
both of u
cedar kilnBOT
@remote lantern Has your question been resolved?
peak prism
Hey guys, help with the following?
$ f(x,y) = 8 - \frac{x^2}{15} - \frac{y^2}{8}$ I need to get from the point A = (0,2) to B = (2,0).
denote x(t) = t, y(t) = 2-t
calculate $\frac{df}{dn}(x(t),y(t)$ where df/dn is the directed derivative
What I did was calculating the gardient of f so
$\frac{df}{dx} = -\frac{-2x}{15}$
$\frac{df}{dy} = -\frac{-y}{4}$
then $\nabla (f(x(t),y(t)) = (-\frac{2t}{15}, \frac{t-2}{4})$
now they ask what is the max value of this directed derivative and for which t i get it.
so since I'm going from the point (0,2) I get (0,0) * some vector which is also 0, so it doesn't depend on t. probably ii'm doing something wrong, so would appreciate any help
wraith daggerBOT
mtr123
cedar kilnBOT
@peak prism Has your question been resolved?
fallen moat
Hey guys, help with the following?
$ f(x,y) = 8 - \frac{x^2}{15} - \frac{y^2}{...
i am totally unfamiliar with directional directional derivatives so it took me quite a bit of time to get myself together with this question
First thing:
is the "max value of this directed derivative" means the max norm of this directional derivative?
if yes, i bet you still need to find t so that sqrt((-2t/15)²+((t-2)/4)²) is maxed?
cedar kilnBOT
@peak prism Has your question been resolved?
pulsar carbon
How to Interpret the following graph in detail:
I am novice in Algebra I need to find domain and range, Does this graph represent a function and a one-one function. Why/Why not?
Also sorry for bad english I am not good at it
cedar kilnBOT
How to Interpret the following graph in detail:
I am novice in Algebra I need t...
@pulsar carbon Has your question been resolved?
fallen moat
if it's a curve on a graph that needs to be check, we can use a very useful tool:
every vertical line on the graph will only pass through the curve at most once
if it's a yes, then it's a function
sorry, typo
pulsar carbon
Okay if it passes twice its not function
and one-one function? also domain and range?
fallen moat
Okay if it passes twice its not function
e.g. a circle will not be a function
fallen moat
and one-one function? also domain and range?
one-to-one function, there's another tool:
every horizontal line on the graph will only pass through the curve at most once
if it's a yes, then it's an one-to-one function
domain is the interval(s) that function is on x,
e.g.
pulsar carbon
I understand So this graph is not a one to one fusntion as passes twice at (0,5)
fallen moat
I understand So this graph is not a one to one fusntion as passes twice at (0,5)
yea, it pass through twice on the horizontal line pass through (0,5)
fallen moat
the x values between the blue arrows would be the domain
fallen moat
and one-one function? also domain and range?
however we don't know if this curve will extend forever, so you'll have to be careful if it does or not
pulsar carbon
the x values between the blue arrows would be the domain
-1 and 2 I see
fallen moat
-1 and 2 I see
nope, if just seen from the drawn cruve, it will be -1.something and 2.something instead of strict -1 and 2
so it really depends if there's an equation of the curve is given or just a graph
pulsar carbon
nope, if just seen from the drawn cruve, it will be -1.something and 2.something...
We can assume its strict -1 and 2 .something can be ignored and Range?
pulsar carbon
so it really depends if there's an equation of the curve is given or just a grap...
Just this image sadly
fallen moat
We can assume its strict -1 and 2 .something can be ignored and Range?
range will be the same method, but with y-values
fallen moat
We can assume its strict -1 and 2 .something can be ignored and Range?
e.g. for this ,we can write
-1≤x≤2
or in interval notation
[-1,2]
brb
pulsar carbon
e.g. for this ,we can write
-1≤x≤2
or in interval notation
[-1,2]
I understand
But for y 10 and - 10 would be it??
fallen moat
I understand
But for y 10 and - 10 would be it??
sorry, i had my eye check just now
pulsar carbon
Ohhh Eye check you at a clinich or hospital or something?
fallen moat
I understand
But for y 10 and - 10 would be it??
range is for y values,
so we can write
-10≤y≤10
or in interval notation
[-10,10]
pulsar carbon
sorry, i had my eye check just now
sorry for disturbing man
fallen moat
Ohhh Eye check you at a clinich or hospital or something?
I was walked out of eye hospital lol
fallen moat
sorry for disturbing man
no worries, i was waiting for the eye check just now, that's why I'm free to help XD
pulsar carbon
Thanks man
fallen moat
Thanks man
Cheers!
fallen moat
Take care man
tyty
pulsar carbon
I pray for you speedy recovery
fallen moat
good luck with your maths~
cedar kilnBOT
cedar kilnBOT
Available help channel!
tender badge
Can someone help me with this question. I don’t know how to continue it from where I got up to
cedar kilnBOT
Can someone help me with this question. I don’t know how to continue it from whe...
@tender badge Has your question been resolved?
lavish lily
Can someone help me with this question. I don’t know how to continue it from whe...
first 3^(k+1) > 3k^3
then you have to prove that 3k^3 > (k+1)^3 for k greater than or equal to 4
tender badge
where is the 3k^3
tender badge
oh so basically multiplying a 3 to both side
tender badge
yes thanks
cedar kilnBOT
@tender badge Has your question been resolved?
civic coral
cedar kilnBOT
civic coral
can someone check it?
hot crag
yes
you made ONE small mistake
you forgot that the sqrt function NEVER gives negative values
so -2/3 cannot be the sqrt of y^2 - 5
so no solutions exist
@civic coral
civic coral
shit
civic coral
you forgot that the sqrt function NEVER gives negative values
why
hot crag
sighs
Please Correct My Grammar, I have enough experience to know not to argue with you.
I cannot explain that to you.
royal loom
another way to see it might help
any y that is less than sqrt(5)
is not even possible
and both sides of the equation grow at a linear rate
you can think of the right hand side getting 'turned on' at y=sqrt(5)
at which point it will take value 3
but the other function will take value sqrt(5)
civic coral
is not even possible
why
royal loom
and then from there on out, they will grow at the same pace
civic coral
Can't take the square root of a negative value
why
royal loom
it isn't defined as a real number?
hot crag
sighs
Austin what have you gotten urself into
royal loom
What do you think sqrt(-6) should be?
civic coral
is it because when y<sqrt5, there will be some imaginary numbers in our equation, which is not allowed.
royal loom
What two real numbers, when multiplied together, will be -6?
hot crag
he knows complex numbers Austin
royal loom
It doesn't seem like it
royal loom
is it because when y<sqrt5, there will be some imaginary numbers in our equation...
yes
when |y|<sqrt(5)
that is
hot crag
ur giving me PTSD, my first PCMG channel was him claiming to be an expert on complex numbers and also saying that e^(i times pi) = e^180i
royal loom
idk about saying it "isn't allowed" but it certainly will make finding a solution impossible
royal loom
but why, couldn't we have some imaginary number in our equation.
think about it this way
if you plug in |y|<sqrt(5)
the right hand side, will have an imaginary number
hot crag
@civic coral if i may. the sqrt function never gives negative results since it wouldn't pass the vertical line test that way, and would cease to be a function
royal loom
that you understand yes?
hot crag
imma let @royal loom finish first
royal loom
but the left hand side will just be |y| which is an entirely real number
and you need them to be equal
a real number, will not equal an imaginary number
so even if you were considering complex solutions (which you aren't)
civic coral
I see
royal loom
there aren't any of those either
hot crag
,w y = 3 - sqrt(y^2 - 5)
wraith daggerBOT
royal loom
You're doing that to yourself
hot crag
@civic coral the solutions if the sqrt function gave negative results
royal loom
graphically the red is the right hand side of the equation, and the blue is the left hand side
the interval where the red takes on no values, is when |y|<sqrt(5)
outside of that interval, it grows at the same rate as y
but it starts at a different value
they'll never be equal
because it is like two parallel lines
that will never intersect
But you probably need to review what it means to be a function
as Ren is saying above, one of (if not the only) requirements of being a function is that for every 1 input, you only get 1 output
if sqrt(9) for example was 3 and -3, this would fail to be a function
when your problems involve the square root, they are using it as a function
and because of this, the value will always be positive
because that's what we decided
civic coral
im overwhelmed, i need some time to digest it
hot crag
!done
cedar kilnBOT
If you are done with this channel, please mark your problem as solved by typing .close
civic coral
a real number, will not equal an imaginary number
is it possible to tell that the equation has no solution at its original form
civic coral
as I just described you can visualize it graphically
I cannot visualize them graphically, i have no idea about their graphs.
royal loom
you don't know how to graph y=x?
civic coral
I do
royal loom
that's the left hand side
the right hand side is nearly as simple
and you don't even actually need to graph it
all you need to realize is that they will grow at the same pace
and start at different values
Anyways, that's hardly a method to solve the question, I was just trying to help your understanding
you should definitely just solve this algebraically
civic coral
and you don't even actually need to graph it
I cannot tell what it looks like, the graph of the right hand side.
There's some sqrt involves which i do not know its property.
royal loom
Okay then ignore it
royal loom
Anyways, that's hardly a method to solve the question, I was just trying to help...
Just solve it algebraically
civic coral
Okay then ignore it
but you do, i wonder if there's some property/ characteristics about the graph of a sqrt
royal loom
,w plot y=sqrt(x)
wraith daggerBOT
royal loom
they look basically like this
civic coral
I see
it is y^2 = x but the upper part
and the lower part didn't exist, because this is a coordinate plane
I have no questions now
civic coral
because it is like two parallel lines
Do you know it at the first sign when looking at the equation, or after put it on desmos.
im so curious is it pure talent to visualize the graph of an singular equation by looking at it.
and it is an equation which involves some umsimplifiable sqrt in it
hot crag
@civic coral ru done
civic coral
no.
hot crag
sighs
cedar kilnBOT
hot crag
shakes head
cedar kilnBOT
Available help channel!
bold ermine
cedar kilnBOT
cedar kilnBOT
@bold ermine Has your question been resolved?
cedar kilnBOT
@bold ermine Has your question been resolved?
bold ermine
.close
cedar kilnBOT
cedar kilnBOT
✅
worldly chasm
Sorry, I just saw this if you are.
bold ermine
I think I've figured it out
cedar kilnBOT
bold ermine
are you able to help me with this one?
cedar kilnBOT
Available help channel!
bold ermine
cedar kilnBOT
flint plinth
!status
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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
@bold ermine Has your question been resolved?
cedar kilnBOT
@bold ermine Has your question been resolved?
bold ermine
.close
honest torrent
C
honest torrent
Use the product rule to figure out the number of ways can we place the
two snakes. Explain your reasoning. The answer must be for a general n.
idk how to start this
cedar kilnBOT
@honest torrent Has your question been resolved?
worldly chasm
@honest torrent if the two snakes are confined to a single row, then they cannot interact with each other, so it's like you have one row, one snake, and then you just combine together two possible rows from that. Right?
worldly chasm
Sorry, I mean "if the two snakes are each confined to a single (different) row"
honest torrent
yes
honest torrent
what does that mean
worldly chasm
That the head is bigger than the tail?
honest torrent
"head must be greater than tail is still required"
worldly chasm
I means that if we number the squares from 1 to 2n, the head sits on a larger number than the tail
So the head is on 13 and the tail is on 10 is valid
But the head is on 13 and the tail is on 15 is not
worldly chasm
It's because you're probably not familiar with the game
The numbering works like this:
16 15 14 13 12 11 10 9
1 2 3 4 5 6 7 8
oh lmfao i didnt know
worldly chasm
The board is normally 10x10
And the numbers move back and forth like this all of the way up
worldly chasm
The size of the snake doesn't matter
For a single row single snake, there's (n-1) places the head can go
And then if it goes at n, then there's (n-1) places the tail can go, if the head goes to n-1 then there's n-2 places the tail can go, and so on.
So it's sum from i=1 to n-1 of i
honest torrent
but the head always has to be greater
worldly chasm
Yes and if the head is on the first tile there's no room for the tail
worldly chasm
Let's do n=3
The head can go on tiles 2 and 3
(n-1)
If the head is on tile 2, the tail can only go on tile 1.
But if the head is on tile 3, then the tail can go on either tile 2 or tile 1.
honest torrent
but where would the tail go if its at the end of the row
worldly chasm
I'm talking about a 1x3 grid, single snake, single row
worldly chasm
The three possible snakes are drawn on this image
The head is always right of the tail
honest torrent
i thought it cant go on tile 1 or that dont matter
worldly chasm
The tail can, the head cannot
honest torrent
The head is always right of the tail
isnt it only on the left
worldly chasm
In this image larger numbers are farther to the right
worldly chasm
This is an actual board
It depends on the way the board is made
That's why I wrote numbers on the squares
honest torrent
true so it would alternate
got it
there is n-1 ways to place a head on the row
@worldly chasm
worldly chasm
Yes
worldly chasm
For a single row single snake, there's (n-1) places the head can go
Reread from here
honest torrent
oooooo ok ty
honest torrent
so n(n+1)/2
primal ridge
please help https://hips.hearstapps.com/hmg-prod/images/math2-1569526052.jpg?crop=1xw:1xh;center,top&resize=980:*
worldly chasm
so n(n+1)/2
Sorry, that's for 1 to n
We're doing 1 to n-1
!occupied @primal ridge
cedar kilnBOT
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
worldly chasm
(also fr you're asking about goldbach's conjecture?)
cedar kilnBOT
@honest torrent Has your question been resolved?
sick ledge
can someoine explain
cedar kilnBOT
can someoine explain
sick ledge
why this
cedar kilnBOT
@sick ledge Has your question been resolved?
manic roost
I've been trying to do this problem for about an hour now, any ideas on how I can prove:
wraith daggerBOT
Neo
manic roost
I have tried using limit ratio tests
Induction
but I need to check if it is true, which it is true, and then prove it.
cedar kilnBOT
@manic roost Has your question been resolved?
minor terrace
I have to do this equation which includes permutations and variation, though i dont know how, since we did much easier problems besides this
cedar kilnBOT
I have to do this equation which includes permutations and variation, though i d...
@minor terrace Has your question been resolved?
cedar kilnBOT
@minor terrace Has your question been resolved?
cedar kilnBOT
@minor terrace Has your question been resolved?
wheat ocean
you just have to use these formulas: $$A_n^k = \frac{n!}{(n - k)!}$$ and $$P_n = n!$$
wraith daggerBOT
Alisia
dense pulsar
please help me with this: Give an example of partition N into three subsets.
dense pulsar
i mean {1}, {2}, {3,4,5,...} would be boring
found this solution but i don't want to copy everything
cedar kilnBOT
@dense pulsar Has your question been resolved?
dense pulsar
how do i start the solution from scratch?
mossy cave
I’m a bit confused by your question. There are lots of different partitions of N you just need to choose one. Are you struggling to find another partition into infinite sets like the example solution?
dense pulsar
yes, i want to use just N in my solution. the one shown involves Z
mossy cave
Do you see why this solution uses Z instead of N? Which numbers are missing if you replace Z with N?
Also, do you know modular arithmetic?
dense pulsar
Also, do you know modular arithmetic?
nope, haven't gotten to that yet
mossy cave
Ok, well in that case, you can just think about the reminders of the number when dividing by 3. You can then partition N using that (it should be the same as these partitions)
mossy cave
Do you see why this solution uses Z instead of N? Which numbers are missing if y...
Using this you can just union the missing numbers to each set (again should give the same answer)
dense pulsar
wait, can I do this?
S = {A_1, A_2, A_3}
A_1 = {x is element of N : x=3k for some k is element of N}
A_2 = {x is element of N : x=3k-1 for some k is element of N}
A_3 = {x is element of N : x=3k-2 for some k is element of N}
mossy cave
Yep.
Another partition (completely different and somewhat random) is:
All natural numbers with an odd number of digits;
all natural numbers with a multiple of 4 number of digits;
All natural numbers with a multiple of 2 but not a multiple of 4 number of digits
dense pulsar
ooh okay
mossy cave
There are a lot of options for partitions
dense pulsar
Yep.
Another partition (completely different and somewhat random) is:
All nat...
how would this look like in set builder notation?
mossy cave
Good question, it sort of depends. One option is (which is the simplest) is:
{n in N: n has an odd number of digits}
Alternative option: ${n \in \mathbb{N} : (n/10^m) \in [1,10] , \text{if m is odd}}$
wraith daggerBOT
Owen
cedar kilnBOT
cedar kilnBOT
Available help channel!
acoustic sigil
Calculate the mass of the arc defined by the equation y^3=x^2, 0<=y<=3, density distribution function = x
acoustic sigil
I know i need to do a line integral
But do i need to transform the equation of the arc before finding dl, or just leave it like it îs?
cedar kilnBOT
@acoustic sigil Has your question been resolved?
unborn pond
hi can we do this
cedar kilnBOT
hi can we do this
unborn pond
oblique stirrup
#8
bitter peak
you gotta find the total number of general surgery physicians first
bitter peak
so 25,000 general surgery physician
no that is general practice
unborn pond
oh i see
bitter peak
look for general surgery and find that total
unborn pond
yes 36000
bitter peak
id say 37000 cause it looks like 35000+2000 to me
unborn pond
so 3700*.24?
bitter peak
then look at the pie chart, the percentage for 35-44 is 24% so you must find 24% of that number
bitter peak
so 3700*.24?
yeah 37000*0.24
bitter peak
sure
unborn pond
so female general surgery physicians in under 35 category represented 8.5 percent of all general surgery physicians
so under 35 is 30%
bitter peak
so we know the total of general surgery physicians is 37,000, so we need to find how many are in under 35
unborn pond
yes
which is 37,000*30
but wait
arent we fdoing the number of female general surgery physician
not all?
bitter peak
yes, but it is 8.5% of how many are under 35, not all ages
bitter peak
you should do 37000*0.3
bitter peak
oh you mean 11,100
bitter peak
wait let me think one second
this question is worded a bit weirdly for me so give me one sec to think
unborn pond
okk
bitter peak
okay, i think i get it now. you need to find 8.5% of 37,000 to find how many of the under 35s are female
because 8.5% of the total number of them are females under 35
unborn pond
yes
wait
i thought it was saying that the number of female general surgery is arouns 2000 and 8/5 of that is all the general surgery physicians
bitter peak
in the question, they say that for the total of general surgery physicians, only 8.5% of them are females under 35
so you have to find out how many females under 35 there are using that information
bitter peak
yes, that is females and males
unborn pond
and with that, we know there's 11,100 people in general surgery
and then we would multiply by 8.5%
unborn pond
oh yes you're right
bitter peak
but females under 35 are only 8.5% of all ages all together
unborn pond
oh i see
bitter peak
so to find the number of females under 35 we have to find 8.5% of 37,000
bitter peak
so now we know that is how many females are under 35 out of 11,100 who are under 35
so the males are the rest
of the 11,100
unborn pond
wait im confuse
bitter peak
what's wrong?
unborn pond
give me a second to write this doqn and conceptulize
bitter peak
okay
unborn pond
ok so here is where im confused
so we did this
37,000 * .30
we got 11,100
which is the people under the age of 35
and then we did this
37,000 *.85= 3145
that i sall the people of all ages
we do not know how many people are under the age of 35% for females
we know of all age
unless i multiply 3145 by 30%?
bitter peak
we do not know how many people are under the age of 35% for females
it is 3145
because in the question, 8.5% is how many females are under 35 in all of the 37,000
so we used that to find 3145 which is how many females are under 35
unborn pond
so what we did was
for both male and female was 37000*.30
and then for just female under 35
we did 37,000* 8.5%
and then to find male
we can so 11,1000 -3145?
bitter peak
yes
sonic zephyr
Need help with questions 1.1a and 1.1b. What do i do with the 2 on the right for 1.1a, to get the brackets alone. How do I get the bases the same in 1.1b? Or do I need to do something else there
alpine rune
what is log(a) + log(b)?
alpine rune
happens lol nw
sonic zephyr
But I am not sure about the second one🤔
sonic zephyr
Hmmm
mild sundial
$a^x-y=a^x/a^y$
wraith daggerBOT
斯韋裡
inland ocean
$a^{x-y}=\frac{a^x}{a^y}$
mild sundial
ty
wraith daggerBOT
ColdTee
mild sundial
**ColdTee**
i think this can help
sonic zephyr
I am very confused
mild sundial
I am very confused
use the above exponent rule written
sonic zephyr
5^x times 5^-1 ?
alpine rune
yes
mild sundial
5^x times 5^-1 ?
no
sonic zephyr
mild sundial
oh yeah
mild sundial
nvm did not notice that -1
alpine rune
ok
strong yacht
.open
mighty shuttle
cedar kilnBOT
Available help channel!
open nest
Pre algebra simple
cedar kilnBOT
Pre algebra simple
open nest
this probably soudns dumb but i dont think i have the correct answer for a), this is just highschool begining algebra.
cedar kilnBOT
Available help channel!
earnest hound
Hi, Question 6 part b. For some reason I always get a mental block with these questions and would appreciate somebody explaining the steps. For part a vector AB = -5+12i.
cedar kilnBOT
@earnest hound Has your question been resolved?
cedar kilnBOT
@earnest hound Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
austere hull
is this correct?
cedar kilnBOT
is this correct?
void sand
looks good to me
cedar kilnBOT
void sand
its sometimes tiring to show all these steps
indeed
cedar kilnBOT
Available help channel!
full rover
Why is sin(2x)/2cos(x) considered to be equal to sin(x)? I know the proof that it comes from, but shouldn't it have a discontinuity at every maximum or minimum?
crimson sedge
Why is sin(2x)/2cos(x) considered to be equal to sin(x)? I know the proof that i...
its a removeable discontinuity
full rover
its a removeable discontinuity
how so?
crimson sedge
you can remove it by simplifying
full rover
pardon me being dumb
how
crimson sedge
pardon me being dumb
how
the same way you prove sin(2x)/2cosx = sinx
take the limit as x->pi/2
hollow trail
they're equal for every point that is not a minimum or maximum. you can call those discontinuities "removeable" because if we define the function to be 1 or -1 at those points and sin(2x)/2cos(x) elsewhere it would be a continuous function equal to sin(x) always
full rover
they're equal for every point that is not a minimum or maximum. you can call tho...
in those situations, should i just be using what multiple of pi/2 the angle is to get the minimum or maximum values
too lazy to do the extra step lmfao
hollow trail
is there a specific problem you're trying to solve here?
full rover
not really. im just asking: in the scenario that i'm solving for the sine of a half angle, and that angle happens to be a minimum or maximum, what should I do to know whether its 1 or -1
cedar kilnBOT
Available help channel!
fossil sleet
cedar kilnBOT
digital cliff
how can you have a length of -5?
fossil sleet
that's what i was thinking as well, but then how do i get to C?
digital cliff
if i model the center of the base of the triangle as the origin
saying A has coordinate (0,a) C has (3.4,0)
the info implies that a/-3.4=-5/4, a/3.4=5/4, then a=4.25
height is 4.25
,calc 4.250.56.8
wraith daggerBOT
Result:
14.45
how do i know when to change the sign of the slope? i mean, for a/-3.4 = -5/4, but for the other one, it was positive...
digital cliff
i just multiplied both sides by -1
cedar kilnBOT
cedar kilnBOT
Available help channel!
verbal mango
i am looking at the answer key my teacher made for the study guide he gave us, and i was wondering how he got this answer?
ancient lodge
find the reference angle and then note that sin is negative in the 3rd and 4th quadrants
so you're finding the angles in those quadrants w/ said reference angle
ancient lodge
$\arcsin(0.43)$
wraith daggerBOT
Civil Service Pigeon
cedar kilnBOT
cedar kilnBOT
Available help channel!
royal loom
cedar kilnBOT
royal loom
Working on the forward direction of part (a)
I thought I had something going but I'm getting stuck
Let tk be the partial sums of the product
then log(tk)=log(1+a0)+log(1+a1)+log(1+a2)+...+log(1+ak)
log(tk)=log((1+a0)(1+a1)....(1+ak))
we want to show limit tk k->infinity converges
if it does then lim log(tk) = log lim tk
so we now consider the limit as k-> infinity of the inside of the log
(1+a0)(1+a1)....(1+ak)
I thought to use here the absolute convergence of an
since it converges absolutely, there will be some index m where all the terms ai afterwards are |ai|<1
so since this product is
1+a0+a1+.......+a0a1a2a3....ak
we can get a bound on the a0a1...ak
since eventually adding more terms to the product will just make it smaller
and likewise a bound on any of the mixed product terms that have terms with index > m
and the rest will be finite
so I thought we could show convergence somehow using like the monotone sequence theorem or something
but then I remembered that an need not be increasing/decreasin
and I gave up
slate lintel
I see no Taylor's theorem also go to bed
royal loom
yah I thought I could do it without taylors theorem
Need to solve something first
I didn't have any ideas about taylors theorem
maybe
if tk=(1+a0)(1+a1)...(1+ak)
then
logtk=log((1+a0)(1+a1)...(1+ak))
logtk = log(1+a0)+log(1+a1)+...log(1+ak)
then using taylor's theorem on log(1+x) somehow
then we're summing a finite amount of infinite sums
crazy buisness
r u still here or am I going crazy
I might be
Okay but
an>-1
so
1+ai > 0
but for any ai > 0
we are outside this radius of convergence
mighty drift
then we're summing a finite amount of infinite sums
so make it finite
Find a relevant bound that lets you discard the later terms
royal loom
so make it finite
Find a relevant bound that lets you discard the later terms
working on that
I wish I could say there are only finitely many ai > 0
but that's not true
mighty drift
no, finitely many terms for the log expansion
royal loom
because like 1/n -> 0 will have infinitely many > 0 but still conv to 0
royal loom
no, finitely many terms for the log expansion
But I have many log expressions
I have
log(1+a0)+log(1+a1)+...log(1+ak)
for any ai<0 we can apply this taylor series
but for the rest we can't
mighty drift
for any ai<0 we can apply this taylor series
log(1 + x) <= x
log(1-x) >= -x
so either way, |log(1+ai)| <= |ai|
mighty drift
well, the Taylor series is a complicated way of proving this
royal loom
how can we prove log(1+x) <= x
and
log(1-x)>=-x
mighty drift
log(1 + x) <= x
log(1-x) >= -x
always look at convexity arguments
mighty drift
don't worry it's concave here
royal loom
sacred grail
so either way, |log(1+ai)| <= |ai|
,w plot |log(1+x)| and |x| for -0.9<x<1
wraith daggerBOT
sacred grail
its just not true lmao
mighty drift
right
vast pike
$d/dx[x-log(1+x)]=1-1/1+x\le 0$ for x>0
wraith daggerBOT
everg
mighty drift
or is it ?
vast pike
and x=log(1+x) for x=0
sacred grail
or is it ?
log on desmos is log_10 
mighty drift
doesn't matter, just a multiplicative constant
sacred grail
the inequality doesnt hold globally
sacred grail
especially not near -1
mighty drift
the inequality doesnt hold globally
yeah that's actually bothersome for the reciprocal
royal loom
guys look isn't it fine
sacred grail
no
royal loom
sacred grail
royal loom
why not 
sacred grail
you need to compare their absolute values
royal loom
I'd rather not
sacred grail
youd need to state the inequality as |log(1+x)| <= kx near x=0 for some k>1
mighty drift
but what can I do about positive ai
for that my argument works
sacred grail
the point is that because you have absolute convergence
royal loom
for that my argument works
ooh
sacred grail
the terms eventually will be small enough that the inequality kicks in
royal loom
I don't know what you're talking about
wraith daggerBOT
sacred grail
so eventually $\abs {a_i} < \epsilon$ for all $i > N$
wraith daggerBOT
sacred grail
youd need to state the inequality as |log(1+x)| <= kx near x=0 for some k>1
this is good enough to apply this inequality
sacred grail
you gotta interpret "apply" as use a bit of thinking not just use the inequality as is
royal loom
Yea I mean like
1+ai -> 1
since ai -> 0
so
if you're approaching log(1)
it will get small
near 0
but so will kx when x is near 0
so I'm not sure how to counterbalanace that
sacred grail
the point of this question is like
because you have absolute convergence
you know that the terms in the series will become very small eventually
and since log(1+x) and x are asymptotically equivalent near 0
this forces the sums to either both converge or both diverge
royal loom
I can buy that
sacred grail
the fact that you can write |log(1+x)| <= kx near x=0 is quantitatively stating this fact
royal loom
the fact that you can write |log(1+x)| <= kx near x=0 is quantitatively stating ...
I'll try this again tomorrow
thank you
sacred grail
sacred grail
i think taylor series makes it much easier
sacred grail
taylors theorem gives you the value of k you should use
royal loom
oh maybe I use it on
|log(1+x)| = | taylor series thing |
since x near 0 will be -1 < x <= 1
series triangle inequality
<= sum |(-1)^(n+1) x^n / n|
<= | x^n/n|
and x is our a_i ?
vast pike
lets use the taylor expation of log.. $log(1+x)=x+f(x)$ with $\lim_{x\to 0} |f(x)|/x=0 $ then $\sum_n |log(1+a_n)|=\sum |a_n+f(a_n)|<\sum |a_n|+\sum |f(a_n)|$ the first sum converges by assumption, the second one for comparison test $\lim_{n\to \infty} |f(a_n)|/a_n=0 $ (since $a_n\to 0$)
royal loom
which goes to 0
sacred grail
vast pike
wraith daggerBOT
everg
sacred grail
royal loom
<= | x^n/n|
<= |x|
right?
since as n gets large, a_i will go to 0
and dividing by something bigger
just drop the power
drop the denominator
sacred grail
i think that approximation is sus
royal loom
damn
sacred grail
<= | x^n/n|
especially this step
royal loom
oh
sacred grail
but as mateo was saying earlier
you shouldnt be using an infinite series here
you want a finite sum
taylors theorem isnt quite taylor series
taylors theorem tells you how to approximate by finite sums + remainder
royal loom
Oh okay
we did last HW
this problem
so I figured it was probably to use here
but I guess no
sacred grail
yeah the remainder is what taylors theorem tells you
that lets you prove convergence for the whole series
but you want to explicitly use the remainder here
sacred grail
something along the lines of [ \log(1 + x) = x g(x) ]
for some function $g$ which is continuous in a neighbourhood of $0$
wraith daggerBOT
royal loom
yes
sacred grail
,, \log(1 + x) = x \f {f'(\xi)} {1!}, \quad f(x) = \log(1 + x)
wraith daggerBOT
sacred grail
this is a degree 0 taylor polynomial + remainder
royal loom
Gotcha
Okay thanks again
I shouldnt open help channels this late
because I want to solve and feel bad for leaving
but its 3:30 am so I should go to bed
sacred grail
old fjord
hi!! really quick question how can you tell what formula a parabola is by the information given
subtle hinge
!originall
cedar kilnBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
subtle hinge
!status
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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
Basically I found how to do it and find the rule, but I had to look at the answer to figure out what i needed to find and what formula
example for the image i didnt know it was y = ax^2 + bx
just wondering how youd figure that out?
<@&286206848099549185>
livid hound
the ideal approach would depend on what you're given
here since you are given the x-intercepts (and another point)
its recommended that you start with factored form
and then use the additional given point to determine the leading coefficient/scaling factor
livid hound
for this, you could start with
vertex form or factored form
doesn't make a big difference
takes practice to recognise what works well
just do problems
and try approaching it in different ways
old fjord
okay its just so difficult to determine which one it is
pastel vault
okay its just so difficult to determine which one it is
It can be both
old fjord
so for this one what would the format be?
pastel vault
I'd say use vertex form for this tho
old fjord
just so i can get an idea of what they look like
pastel vault
so for this one what would the format be?
Factored form
Since you aren't given the vertex there
livid hound
the ideal approach would depend on what you're given
consdier the questions:
do you have the vertex
do you have the x-intercerpts
pastel vault
No k(x - a)(x - b)
old fjord
yeah
pastel vault
No k(x - a)(x - b)
This tells you the roots
old fjord
and that is to find k
pastel vault
ye
old fjord
yeah and then what next
livid hound
try not the use a,b there as those conflcit with the a,b,c in general form
pastel vault
k controls the vertical stretch
livid hound
and then use the additional given point to determine the leading coefficient/scaling factor
pastel vault
try not the use a,b there as those conflcit with the a,b,c in general form
Yeah r1 and r2 then
old fjord
> and then use the additional given point to determine the leading coefficient/s...
what would I do for that
livid hound
sub in the point
old fjord
into what?
livid hound
your factored form equation
livid hound
can you first identify the values of r_1 and r_2?
old fjord
1 and 3
livid hound
that'll give you
y = a(x-1)(x-3)
then can you identify the coordinates, x,y values of the other given point?
livid hound
forget about a for a sec
look at the graph you're given
you're given 3 points there, two of which are the x-interecepts
what are the coordinates of the third point?
old fjord
(0,3)
old fjord
so sub in the other points
and use substitiution?
ah
ok and why do we need the x and y though
livid hound
you want to sub in the x and y coordiates of your point
i.e. what is the x-coordinate of (0,3)
what is the y-coordinate of (0,3)
livid hound
yes
livid hound
no
we're not finding x and y
you're using the x and y coordiantes of the third point (0,3) to find a
old fjord
dont we already have a?
livid hound
do you?
old fjord
yes
livid hound
what is your value of a and how did you get it
old fjord
we subbed in the roots and the points
livid hound
upon getting the value of a
use that here
y = a(x-1)(x-3)
and you'll have your equation
you can then expand or w/e to get it in the desired form
or just leave it as is
old fjord
yes but the answer is in the format y = ax^2 + bx + c
livid hound
ok, so if they want general form then expand it out
livid hound
if they don't tell you what form they want, then you can use whatever's most convenient
and they can't mark you wrong for it
livid hound
well books only display on of the forms of the answer
old fjord
but if the parabola is lowered or higher does that mean there is bx
livid hound
unless they specifically ask for general form
leaving answers in factored of vertex form is completely fine
whether there is a bx term depends on whether the parabola is symmetric about the y-axis
old fjord
ah
okay thank you so much i will keep working towards understanding this all confusing stuff
loll
hot crag
!done
cedar kilnBOT
If you are done with this channel, please mark your problem as solved by typing .close
old fjord
.close