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#help-39
pearl pondBOT
pearl pondBOT
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quasi veldt
need help proving that abs((x^2)*(floor(x))) is less that or equal to 2abs(x) on the interval or (-1;1) - {0}
finite onyx
Resolve it
Into a piecewise function
Write what abs x square floor x will be from -1 to 0, at 0, 0 to 1 at 1 etc
And also same for the other hand side
And compare each piecewise
quasi veldt
our teacher said to use the fact that for x in (-1;1) -{0} x is less than or equal to 1/x
hot canyon
well, lim_{x \to 0} right?
so x is in [0, 1] and floor (x) = x in [0, 1]
but I'm not sure if what you're talking and what you've wrote are the same things
quasi veldt
nah dont mind the lim stuff
all of this is to prove that lim of that function is 0 in 0
using epsilon delta definition
but i cant seem to get to the inequality of abs ((x^2)*E(1/x)) less than or equal to 2abs(x)
with E(x) being the floor function
pearl pondBOT
@quasi veldt Has your question been resolved?
quasi veldt
Write what abs x square floor x will be from -1 to 0, at 0, 0 to 1 at 1 etc
pretty sure you cant do that cuz its floor (1/x)
<@&286206848099549185>
pearl pondBOT
@quasi veldt Has your question been resolved?
quasi veldt
.close
surreal dove
Below are the ages of the students from Section Masiyahin: 16, 12, 13, 14, 15, 20, 17, 16, 18, 15
how do u know if its a sample variance or population variance
pearl pondBOT
Below are the ages of the students from Section Masiyahin: 16, 12, 13, 14, 15, 2...
@surreal dove Has your question been resolved?
cobalt hinge
Below are the ages of the students from Section Masiyahin: 16, 12, 13, 14, 15, 2...
does it encompass the entire group, or only a portion?
pearl pondBOT
cobalt hinge
.reopen
pearl pondBOT
✅
cobalt hinge
because its only from that section?
what is the context?
is there something larger?
like, idk, a school?
pearl pondBOT
@surreal dove Has your question been resolved?
surreal dove
like, idk, a school?
yea
sharp smelt
Simplify $\neg(\neg P \lor Q) \lor ( P \land \neg R)$
pearl pondBOT
Simplify $\neg(\neg P \lor Q) \lor ( P \land \neg R)$
jolly parrotBOT
What a wonderful world!
sharp smelt
I was thinking $(P \land \neg Q) \lor ( P \land \neg R)$
jolly parrotBOT
What a wonderful world!
sharp smelt
I think I use the distributive property now?
finite onyx
Don't u think it'll just be P ^ (~Q v ~R)
sharp smelt
Yeah, $((P \land \neg Q) \lor P )\land ( P \land \neg Q) \lor(\neg R)$
jolly parrotBOT
What a wonderful world!
sharp smelt
And then the distributive property again
so $[(P \lor P) \land ( P \lor \neg Q)] \land ( (\neg R \lor P) \land ( \neg Q \lor \neg R)]$
jolly parrotBOT
What a wonderful world!
prisma elk
**What a wonderful world!**
this is the start?
sharp smelt
yeah
prisma elk
Don't u think it'll just be P ^ (~Q v ~R)
it cant be simplified anymore than this though
sharp smelt
Yeah, how do I get there though
jolly parrotBOT
Goëtia
prisma elk
$P \land (\neg Q \lor \neg R)$
jolly parrotBOT
Goëtia
prisma elk
@sharp smelt clear?
pearl pondBOT
sharp smelt
<@&268886789983436800>
pearl pondBOT
Available help channel!
blissful thunder
A stone is thrown so that it will hit a bird at the top of a pole. However, at the instant the stone is thrown, the bird flies away in a horizontal straight line at a speed of 10m/s. The stone reaches a height double that of the pole and, in its descent, hits the bird. Find the horizontal velocity of the stone.
plush sundial
i have an idea, but im not sure if it wud give the ans
write yr equations from h, instead of O
blissful thunder
tried it doesnt work well
pearl pondBOT
@blissful thunder Has your question been resolved?
blissful thunder
wont the horizontal velocity of the stone be 10m/s?
Nah the bird is 10m/s and it starts further than the person throwing it
so it has to go slightly faster to catch up
pearl pondBOT
@blissful thunder Has your question been resolved?
torpid schooner
How would i begin to do this problem?
pearl pondBOT
How would i begin to do this problem?
torpid schooner
i know i can do x^2 +1 - 1 but that doesnt do anythin here
west sapphire
what if you rearrange that as (x^2 - 1) + 1
torpid schooner
what if you rearrange that as (x^2 - 1) + 1
then split it? into two fractions
x^2-1 / x-1 and 1/x-1
torpid schooner
by factoring x^2-1 and canceling common factors
how would i simplify this? sorryfor asking
west sapphire
can you factor x^2 - 1?
west sapphire
x^2 - 1 has x-1 as a factor, whereas x^2 + 1 does not
pearl pondBOT
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Available help channel!
worn goblet
Integration of [e^x log(sin e^x)]/tan(e^x) dx
pearl pondBOT
Integration of [e^x log(sin e^x)]/tan(e^x) dx
worn goblet
It's obvious that it's f(x)/f'(x) but what to do with the e^x
safe prairie
ln or log?
worn goblet
ln or log?
Ln
bitter lodge
$$\int \frac{e^x \ln(\sin(e^x))}{\tan(e^x)} \dd{x}$$
jolly parrotBOT
King Leo
bitter lodge
Youre trying to differentiate the function?
worn goblet
Yeah I'll take ln() as t then get dt= ()dx
safe prairie
If we integrate ln() we get e^x/tane^x ooo
i got its reciprocal tho
worn goblet
Now put it in original integration we and get t dt
After power rule put ln()= t back in
That's how my maths teacher taught me
It's different from usub but if it works it works
worn goblet
t is ln() right
Yes
worn goblet
u is your x and du is your dx right
Yes
worn goblet
The maths teacher told is on lhs it'll be dx/dt just multiple both sides by dt to get ___dx = dt
safe prairie
move everything on lhs other than du to rhs
worn goblet
then make du the subject of ur last line
Instead of substituting as u I used t and differentiated wrt t
So dt
safe prairie
nono even if my x is x and my u is ln(), my last line would be dx=___du
doesnt rly matter but apparently this tripped you
worn goblet
nono even if my x is x and my u is ln(), my last line would be dx=___du
If I took ln()=u then yes but I took ln()=t and differentiated wrt t
worn goblet
doesnt rly matter but apparently this tripped you
Yeah I just started integration yesterday
safe prairie
you need du=___dt
worn goblet
you need du=___dt
I don't get it
safe prairie
so youd put the entire ___dt in place of du in ur main work
or you can use ___du=dt
if you can find the ___ in your main work
worn goblet
or you can use ___du=dt
How did u get u?
safe prairie
It's obvious that it's f(x)/f'(x) but what to do with the e^x
tbh i didnt see this until you pointed it out
safe prairie
How did u get u?
your u is my x
your t is my u
you just have wonky variable naming
i like my u and x
worn goblet
you just have wonky variable naming
Wonky 😭
worn goblet
i like my u and x
OK but I wanted to ask I more thing
If dy/dx = 1 can u write it as dy=dx
safe prairie
sure but in what situations
worn goblet
Like when I differentiated ln() wrt t
I got the differentiation and then dx/dt
And rhs is dt/dt = 1
So just make it d/dt(ln()) dx = dt
That works?
safe prairie
t is ln() right
and if you arent lying yes it works
worn goblet
lhs = d(ln())
Should be d/dt ln()
safe prairie
Does this work?
rhs is dy/dz
safe prairie
Yeah it's just me butchering x
are you having x u and t in the same problem
safe prairie
Does the dx numerator and dx denominator cancel out?
yes thats what chain rule says
dx can be seen as a number
worn goblet
are you having x u and t in the same problem
Like u use usub for me it's just sub and t is general variable used
safe prairie
value being infsmall * x
worn goblet
dx can be seen as a number
It's just rate of change of x so it should be no
safe prairie
Like u use usub for me it's just sub and t is general variable used
i dont rly understand the necessity to distinguish this because usub isnt a special form of sub, its just that u is how we name the subbed
"it's just sub and t is general variable used" im using u
worn goblet
i dont rly understand the necessity to distinguish this because usub isnt a spec...
Yeah and the variable u use to substitute is u and for me it's t but I can also use anything else
safe prairie
It's just rate of change of x so it should be no
rate of change of x means dx/dt, not dx
safe prairie
no need to be sorry different teachers have their conventions
worn goblet
rate of change of x means dx/dt, not dx
Oh yeah dx is infisimally small x and dx/dy is rate of change of x wrt y
safe prairie
Yeah it's just me butchering x
but whats in the ln()? is it x or u
safe prairie
did u appear on last line or is it bad handwriting
worn goblet
Yeah it's just me butchering x
worn goblet
did u appear on last line or is it bad handwriting
It was me butchering x
Thanks a lot
worn goblet
Does this work?
I thought it wasn't allowed to cancel dx in here because of some insta reels
safe prairie
d/dt ln() doesnt really work in a straightforward manner
worn goblet
d/dt ln() doesnt really work in a straightforward manner
Won't it be same as d/dx ln() but only multiplied by dx/dt
safe prairie
in line 3, you d/dxed rhs, so you should d/dx lhs and nothing more
worn goblet
Implicit functions
safe prairie
dont add dx/dt yet
worn goblet
in line 3, you d/dxed rhs, so you should d/dx lhs and nothing more
In line 3 I took dt from lhs denominator to rhs numerator
safe prairie
this is done by implicit d
worn goblet
this is done by implicit d
Yeah
safe prairie
In line 3 I took dt from lhs denominator to rhs numerator
if "diff wrt t" doesnt count as a line to you then i meant line 2
worn goblet
What'd I do wronf
worn goblet
in line 3, you d/dxed rhs, so you should d/dx lhs and nothing more
I d/dted rhs
And lhs both
worn goblet
this is done by implicit d
Like in your step 3 d/dx y = dy/dy dy/dx equating both lhs
safe prairie
one of my dy is a bottom, others at top
both your dts are bottoms
how
by line 3, my bottom dx became dy
in your line 2, both sides have dt as denominator and theres no reason to introduce dx/dt
worn goblet
in your line 2, both sides have dt as denominator and theres no reason to introd...
There is so that we can get ___dx = dt cause rhs = 1
worn goblet
There is so that we can get ___dx = dt cause rhs = 1
It would be ___ dx/dt=1
safe prairie
There is so that we can get ___dx = dt cause rhs = 1
ur doing what i did in line 3
worn goblet
ur doing what i did in line 3
Yess
worn goblet
all this is based on d/dt being actually d/dx in the lhs bracket right
Yeah I'm extremely mentally challenged
I'm going to perform blood eagle on myself because of that
pearl pondBOT
@worn goblet Has your question been resolved?
visual canyon
howsit look? first time graphing domain
pearl pondBOT
howsit look? first time graphing domain
cobalt hinge
howsit look? first time graphing domain
this works fine
$x\in R\setminus \qty{-4}\cup\qty{4}$
broooooo
MY FUCKING PREAMBLE
dude
why
visual canyon
MY FUCKING PREAMBLE
what happened?
sweet bone
that should be a union btw
cobalt hinge
x is in the reals and x is not equal to 4,-4 which is what your notation says
cobalt hinge
that should be a union btw
yes i know
visual canyon
that should be a union btw
union?
sweet bone
or just R\{4,-4}
jolly parrotBOT
;(
cobalt hinge
or just R\\{4,-4}
can ya help out with restoring my preamble
specifically restoring \bR and such
sweet bone
I don't have a preamble loaded on texit
you can probably find one in the helper channel
visual canyon
can ya help out with restoring my preamble
your don't ping is confusing, what time zone? Also, kind of a discord noob even though i've had it forever, is pinging straight up @ing someone? or just responding like this
cobalt hinge
your don't ping is confusing, what time zone? Also, kind of a discord noob even ...
both @ and responding, though you have the option to disable pings for replies
sweet bone
your don't ping is confusing, what time zone? Also, kind of a discord noob even ...
replies will ping unless you disable them here
cobalt hinge
your don't ping is confusing, what time zone? Also, kind of a discord noob even ...
yeah im figuring that out right now
visual canyon
replies will ping unless you disable them here
ohhhh
cobalt hinge
its just that i have maybe 10-15 pings whenever i go to sleep
which obviously is going to affect
bitter lodge
specifically restoring \bR and such
\newcommand{\bR}{\mathbb{R}}```silent mode is a godsend, but I get it. I'm a pretty light sleeper so even the buzz alerts can bother me sometimes
bitter lodge
how can i implement it
,preamble --add
sweet bone
,preamble
visual canyon
I think my question has been answered, but I'll leave the channel open until you figure out whatever is going on with your preamble
pearl pondBOT
pearl pondBOT
Available help channel!
rapid wagon
Why is this not a linear transformation
T(x,y,z)=(2x,-12xy+z,3y-12z,y+3x)
summer imp
xy has degree 2
But you can check that it doesn’t follow the properties of linear functions. In particular think of what T(2x, 2y, 2z) would look like
rapid wagon
xy has degree 2
I dont get how this is illegal
rapid wagon
But you can check that it doesn’t follow the properties of linear functions. In ...
What do you mean look like? Try to compute it?
summer imp
I dont get how this is illegal
Well linear and degree 2 don’t really go together
summer imp
What do you mean look like? Try to compute it?
Yeah. You can pretty much see instantly that the result on this term with xy in it will be a multiplication by 4, not by 2
And for linearity you need
T(2v)= 2T(V)
pearl pondBOT
pearl pondBOT
Available help channel!
wet latch
confused what to do
pearl pondBOT
confused what to do
wet latch
i got the exact same answer for a and b
plush bramble
Show?
wet latch
Show?
simplifying part a i got sigma from i =1 to n (3i^2-3i+1) which i then split into 3 different thingies sigma 3i^2-sigma 3i + sigma 1 (still dont know if im allowed to even do this) then i wrote it in terms of n by googling how to represent specific summations in terms of n(i dont know how do i^2 in terms of n) and it told me that sigma i^2 = n(n+1)(2n+1)/6 so i just plugged all those n formulas in and got 3n(n+1)(2n+1)/6-3n(n+1)/2+n then when i simplfied everything i ended up just getting that sigma i=1 to n of (i^3-(i-3)^3) = n^3
which in retrospec there was probably a easier way
im still confused what the question is asking me to do
plush bramble
Yea there's a much simpler way
Simplify each consecutive pair
Write out a few terms of the sum and see how they cancel
It's called a telescoping sum
wet latch
Write out a few terms of the sum and see how they cancel
alr
pearl pondBOT
@wet latch Has your question been resolved?
wet latch
.close
pearl pondBOT
pearl pondBOT
Available help channel!
errant depot
can the derivative exist here? its at a discontinuity
pearl pondBOT
can the derivative exist here? its at a discontinuity
sharp tree
Whats the definition of derivative?
errant depot
Whats the definition of derivative?
the formula for rate of change?
cause the doubt surfaced when i was trying to see if this is true
if f(x) on [a, b] >> R is injective then f is strictly monotone
and since a disconituity doesnt mean its not injective, a discontinuity wouild allow for a graph like this one, which is injective, and not monotone
west sapphire
differentiability at a point implies continuity at that point
west sapphire
cause the doubt surfaced when i was trying to see if this is true
`if f(x) on [a...
is this injective though? as you keep going to the right, won't the line with negative slope eventually cross the x axis?
pine wadi
Ev and I are in a call together, we were confused because desmos gave us that the derivative in f(1) = 1
errant depot
is this injective though? as you keep going to the right, won't the line with ne...
lets just say it doesnt for the sake of the example
like lets say b is at the edge of the picture
were never passing calc 1 💯
lime crest
like lets say b is at the edge of the picture
If [a,b] is where your function is defined, then yeah your function is indeed injective. Bungo was just imagining how it could keep decreasing the same way outside of the window you showed us
west sapphire
what was your concern about differentiability? a function doesn't have to be differentiable in order to be injective
pine wadi
can the derivative exist here? its at a discontinuity
I was confused by the fact that the derivativa exists in x = 1 in the case of this function
west sapphire
no it doesn't exist
lime crest
Indeed, the limit on the left side of f(x) when x goes to 1 isn't the value f(1)
pine wadi
no it doesn't exist
so desmos is getting it wrong or did we write something wrong?
west sapphire
so desmos is getting it wrong or did we write something wrong?
it's getting it wrong
the derivative is 1 everywhere except x=1
and doesn't exist at x=1
lime crest
They wrote the right derivative instead of the derivative
pine wadi
oh ok thanks that clears things up
pearl pondBOT
Available help channel!
vivid pilot
I dont understand how to do part 2 because the gcd is less than both numbers
vivid pilot
or maybe im just midunderstanding
the gcd of 65024 and 128397 is 127
so how would i make an equation
127 = a * 65204 + b * 128397
where a and b are integers
tulip ore
I dont understand how to do part 2 because the gcd is less than both numbers
did you learn anything about the extended eucliean algorithm or bezout's identity
vivid pilot
did you learn anything about the extended eucliean algorithm or bezout's identit...
oh yeha bezouts identity
ax+by=gcd(a,b)
idk how to find x and y though
tulip ore
you know how gcd(65024, 128397) = gcd(65024, 128397 - 65024)?
vivid pilot
you know how gcd(65024, 128397) = gcd(65024, 128397 - 65024)?
oh i didnt know that
tulip ore
you can think of it this way
theres a particular gcd of which 65024 and 128397 is a multiple of, right
vivid pilot
yeah
tulip ore
now if you add/subtract the numbers, the resulting numbers would still be a multiple of the gcd, right
vivid pilot
now if you add/subtract the numbers, the resulting numbers would still be a mult...
yeah
tulip ore
you know how gcd(65024, 128397) = gcd(65024, 128397 - 65024)?
so does this make sense now
vivid pilot
so does this make sense now
yeah, but how would we use it
tulip ore
now what we can do is
vivid pilot
like i still dont understand how i would make such equation as 127 = a * 65204 + b * 128397
without a and b being fractions
OHHH
one can be negative
right
tulip ore
integers are allowed to be negative, yes
Im heading in a direction to help you find a, b, you know that right
vivid pilot
right
tulip ore
even with that you took quite a bit to say that
tulip ore
you know how gcd(65024, 128397) = gcd(65024, 128397 - 65024)?
anyways what we can do is use this to follow along with the euclidean algorithm
vivid pilot
even with that you took quite a bit to say that
sorry i was doing smth
tulip ore
oh I see, now worries
the algorithm's first steps is essentially "subtract 128397 by 65024 until you cant anymore"
vivid pilot
right
tulip ore
the algorithm then says this result should have the same gcd as before
since youll be repeating the steps until you get that gcd
right
vivid pilot
yeah
tulip ore
so if we follow along with the euclidean algorithm, we would begin with:
gcd(65024, 128397)
= gcd(65024, 128397 - 65024)
= gcd(65024, 63373)
= gcd(65024 - 63373, 63373)
= gcd(1651, 63373)
if you look at the steps you did in (i), youll see the same numbers and the same calculations being used
does that check out
vivid pilot
so if we follow along with the euclidean algorithm, we would begin with:
gcd(650...
right yeah
vivid pilot
does that check out
yup
tulip ore
thats good, now say we continue along
gcd(1651, 63373)
what would the next step be in the algorithm
tulip ore
so if we follow along with the euclidean algorithm, we would begin with:
gcd(650...
keep in mind Im talking about these steps, your talking about the next next step right now
so you need to state what to subtract from 63373 to get 635
= gcd(1651, 63373)
= gcd(1651, 63373 - ?????)
= gcd(1651, 635)
vivid pilot
= gcd(1651, 63373)
= gcd(1651, 63373 - ?????)
= gcd(1651, 635)
= gcd(1651, 63373 - 38*1651)
tulip ore
= gcd(1651, 63373 - 38*1651)
yep
tracking down that its 38 (the quotient) is what will get us to finding x and y
we can use some interesting thinking to beef up the euclidean algorithm to tell us x and y
tulip ore
dont worry about the specific details, Ill lead you there
theres a particularly clean way to do this
first off, since 63373 = 128397 - 65024,
63373 is a linear combination of 128397 and 65024, right
vivid pilot
right
tulip ore
that extends to the new numbers we get: 1651, 635, etc. right
vivid pilot
yeah
tulip ore
so suppose we keep track of how many multiples, positive/negative, of 63024 and 128397 we are using
the first steps of this new euclidean algorithm begin with
gcd(63024, 128397)
then we do
gcd(63024, 128397 - 63024)
gcd(63024, 63373)
and we keep track that 63373 = 1 * 128297 + -1 * 63024
does that make sense that this will lead us to the GCD
vivid pilot
ahh i think i see
tulip ore
lets try on the next step
gcd(63024, 63373) and that 63373 = 1 * 128297 + -1 * 63024
now the next step would be?
vivid pilot
1651 = 1*65024 - 1 * 63373
tulip ore
however 63373 is itself a linear combination of 128397 and 63024
so ultimately 1651 = ?? * 63024 + ?? * 128397, what would the ?s be
vivid pilot
oh it would be
128397-2*65024
no wait thats -1651
ohh
its
(-1)*128397+1*65024=1651
right
since its 1*65024 - (1 * 128397 + -1 * 65024
alr i think i can finish this
tulip ore
thats great
now what you are looking at is the idea behind the "extended euclidean algorithm"
vivid pilot
ah i see
tulip ore
its a version that uses the quotients to also write down x and y
I dont have the exact version down but I can show you what the beginning steps would be
do you want to see it
tulip ore
np
we first have to mildly edit the euclidean algorithm to make it more "algorithmic" or "computery"
heres what we have before:
gcd(65024, 128397)
= gcd(65024, 128397 - 65024)
= gcd(65024, 63373)
= gcd(65024 - 63373, 63373)
= gcd(1651, 63373)
now heres what it looks like after:
128397
65024
63373
1651```the euclidean algorithm begins with two numbers
then it finds the remainder of larger / smaller and attaches it
so far two numbers have been attached
the next one would be:
128397
65024
63373
1651
635```since 63373 % 1651 = 635 (the % symbol means "remainder" in computer programming, its name is the "modulo" operator)
you with me so far?
vivid pilot
you with me so far?
right yeah
tulip ore
now the new version will attach the "amount" of 65024s and 128397s we are using:
65024 1 0
128397 0 1```then each new entry will do the division & remainder as usual
in this case, 128397 / 65024 = 1 remainder 63373
in other words, we obtained 63373 by subtracting 128397 - 65024
similarly we subtract 0 1 - 1 0 on the numbers on the right
65024 1 0
128397 0 1
63373 -1 1```the idea here is the numbers on the left keep track of the usual euclidean algorithm
and subtract as usual
vivid pilot
ohh
tulip ore
but the numbers on the right will keep track of the same information
we subtract rows with other rows here
instead of just one number with another number
the next row considers 65024 / 63373 = 1 remainder 1651
this method doesnt look organized on which rows its picking, had that mixed up at first
its picking the smallest two rows, forgot to say
65024 - 63373 = 1651
65024 1 0 - 63373 -1 1 = 1651 2 -1
65024 1 0
128397 0 1
63373 -1 1
1651 2 -1```you can verify and see that 2 * 65024 + -1 * 128397 = 1651
does this next step 1651 2 -1 make sense to you
vivid pilot
yeah
tulip ore
with this, you will get the gcd along with the x and y required to make that gcd
tulip ore
np
vivid pilot
this makes sense
tulip ore
this is the extended euclidean algorithm, its useful in some ways
pearl pondBOT
tulip ore
np
pearl pondBOT
Available help channel!
visual canyon
are these correct?
if so, should I include additional explanation?
visual canyon
.close
pearl pondBOT
pearl pondBOT
Available help channel!
hoary wharf
zero clue how to do 95
pearl pondBOT
zero clue how to do 95
dusky fossil
zero clue how to do 95
If you need a clue to start, you can start by rationalising 1+sinx/1-sinx first
spiral pivot
You can also try using the tan addition formula for the left
hoary wharf
by multiplying both parts by 1+sinx?
hoary wharf
You can also try using the tan addition formula for the left
cant use that
dusky fossil
by multiplying both parts by 1+sinx?
Yes
spiral pivot
What do you have access to?
spiral pivot
because you do need to use some sort of addition formula
hoary wharf
because you do need to use some sort of addition formula
nvm i have access to it i just forgor
one sec
if i have to only alter one side the whole time which should i
tan side or sin side
spiral pivot
It's pretty cool that tan(x/2 + pi/4) can be represented as the sum of two trig functions
hoary wharf
its tan^2 btw
dusky fossil
tan side or sin side
this is a proof question right? the sin side is much easier (in my opinion)
hoary wharf
this is a proof question right? the sin side is much easier (in my opinion)
yes its a proof questino
ill try sin side then
spiral pivot
Honestly, if you have to only change ONE side, I would definitely attack from tan
inverting a sum formula always strikes me as more difficult.
dusky fossil
ill try sin side then
Mhm try rationalising it and then substitute using a random variable
hoary wharf
got (1+2sinx+sin^2x)/cos^2x
spiral pivot
use the definition of the tan and sec and simplify a little
hoary wharf
ohhh like 1+sec^2x=tan^2x
spiral pivot
you should get to (sec^2 x + 2 sec x tan x + tan^2 x)
hoary wharf
*1+tan^2x=sec^2x
dusky fossil
got (1+2sinx+sin^2x)/cos^2x
yea which is (1+sinx)^2/cos^2x
pick a variable like v and substitute it using tan x/2
And then ig you would know the rest
spiral pivot
then you can factor to get (sec x + tan x)^2
and via the tan addition formula, this is exactly the same as the left.
hoary wharf
yea which is (1+sinx)^2/cos^2x
pick a variable like v and substitute it using t...
i dont get what u mean by picking a variable
what does that do
ik what picking a variable means but idk how that would help
spiral pivot
Which is, to me, significantly more mysterious than going the other way
dusky fossil
i dont get what u mean by picking a variable
substitution method
okay nvm ig they dont teach it in pre university in your country
hoary wharf
substitution method
oh ik what subsitution method is but do u mean substituting a value or a variable
idk how subbing a variable would help
hot canyon
Do you perhaps know Componendo dividendo?
hoary wharf
Do you perhaps know Componendo dividendo?
no, maybe ik it but not in that name
dusky fossil
idk how subbing a variable would help
Okay so like, if you pick a variable t = tan (x/2), you can then write sin x and cos x in terms of t.
SIn x becomes 2t/(1-t^2) and cosx becomes (1-t^2)/(1+t^2).
And then boom, it becomes muchhh easier to simplify the whole thing.
And when you are done with the simplification, replace it back with tan(x/2) and then your lhs would be equal to rhs
hot canyon
hoary wharf
never heard of that one
dusky fossil
but I get it, this might feel like a complicated thing
hoary wharf
but I get it, this might feel like a complicated thing
yea it does
hot canyon
never heard of that one
not very difficult to verify
hoary wharf
not very difficult to verify
idk how its relevant
my teacher defintely hasnt taught us that and idk if she'll allow me to use it
hot canyon
Well it's a basic math thing, much like multiplication n division
hoary wharf
Well it's a basic math thing, much like multiplication n division
havent heard of it ever oddly
hot canyon
but alr, if you do not want to try... here's an alternative
Write sin x = - cos (x + pi/2)
Start with right hand side
$= \frac{1 - \cos \left(x + \frac{\pi}{2} \right)}{1 + \cos \left(x + \frac{\pi}{2}\right) }$
hoary wharf
yea
hoary wharf
half angle?
hot canyon
double angle ? whichever it is :/
hoary wharf
double angle i think
jolly parrotBOT
hot canyon
well :/ since you know abt it, you can proceed from here.
hoary wharf
no the first graphic was right
hoary wharf
well :/ since you know abt it, you can proceed from here.
so 2y in that case is x+pi/2, so y is x/2 + pi/4
hot canyon
... is that a thing you should be asking ?
hoary wharf
And no, `sin x = -cos (x + pi/2)`
aka sin(pi/2 -x)
hoary wharf
... is that a thing you should be asking ?
for personal clarification
hot canyon
for personal clarification, use the double angle formula, conclude the result, and then clarify the steps ^^"
hoary wharf
ty
hoary wharf
And no, `sin x = -cos (x + pi/2)`
.close
pearl pondBOT
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Available help channel!
cobalt hinge
bored, late at night, why not
pearl pondBOT
bored, late at night, why not
cobalt hinge
!status
pearl pondBOT
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
1
hot canyon
wha 
hot canyon
You know a + b = 1, a² + b² = 2/3, I'm pretty sure you can figure a^4 + b^4
fierce totem
bored, late at night, why not
missed opportunity to set sin^4(theta) = a and cos^4(theta)=b
reef marlin
How-
gusty flume
Suppose the real-valued function f with f(x)=1 for all real x except 2, where f is undefined.
Can we define an integral over that function on [1,3] ? or any other interval including 2?
Intuitively, yes we can. The area of a rectangle with "no width" is zero, so a single points shouldn't really matter to the value of an integral.
But what do we need to formally define that? Or am I misunderstanding something and it is impossible?
cobalt hinge
Suppose the real-valued function f with f(x)=1 for all real x except 2, where f ...
depends on the type of discontinuity here.
and also, the behavior of the function in a neighborhood of x=2.
gusty flume
constant, since the function is 1 everywhere else
bitter lodge
depends on the type of discontinuity here.
It seems like just a hole
west sapphire
yea you can integrate it regardless of the value at x=2, even if it has no value there
hot canyon
A function f is Riemann integrable if and only if f is discontinuous on a set of measure zero.
bitter lodge
So you can do $\lim_{b \to 2^-} \int_1^b f(x) \dd{x} + \lim_{a \to 2^+} \int_a^3 f(x) \dd{x}$
jolly parrotBOT
King Leo
gusty flume
```A function f is Riemann integrable if and only if f is discontinuous on a set...
a point-set is of measure zero over the continuum. So we can integrate it?
cobalt hinge
It seems like just a hole
which is what i was thinking, so it can be integrable.
hot canyon
a point-set is of measure zero over the continuum. So we can integrate it?
Yes Lebesgue-measure of one element set is zero
gusty flume
with the intuitive assumption being correct and the integral being the same as if f(x)=1 everywhere?
hot canyon
You can take it as an assignment to prove the above
pearl pondBOT
gusty flume
.reopen
pearl pondBOT
✅
earnest hound
is anyone good at trigonometric functions of real numbers and the unit circle?
gusty flume
```A function f is Riemann integrable if and only if f is discontinuous on a set...
isn't f(x)=1/x a simple counterexample?
hot canyon
!help @earnest hound
pearl pondBOT
To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.
hot canyon
isn't f(x)=1/x a simple counterexample?
No. f is not riemann integrable in [0, infty)
earnest hound
!help <@1331689646744272936>
i sent you a message
hot canyon
but discontinuous only at x=0
what about boundedness
tranquil jacinth
ans 7/18
cobalt hinge
hm?
tranquil jacinth
use identity sin^4 theta + cos^4 theta
gusty flume
what about boundedness
not bounded around 0, yes. So
f is Riemann integrable if and only if f is bounded over its entire domain and the set of its discontinuities is of measure zero.
?
hot canyon
Let me pull the conditions for Riemann integrability rq
gusty flume
not bounded around 0, yes. So
```
f is Riemann integrable if and only if f is b...
But this is also wrong, counterexample being f(x)=1 except f(5)=+infinity
which can still be Riemann integrated in a way analogous to my original question.
hot canyon
not bounded around 0, yes. So
```
f is Riemann integrable if and only if f is b...
basically hm
gusty flume
So then what about 1/x?
hot canyon
But this is also wrong, counterexample being f(x)=1 except f(5)=+infinity
but the set of discontinuities still have measure 0, i.e., {5}
hot canyon
So then what about 1/x?
1/x is not bounded in (0, infty)
gusty flume
But this is also wrong, counterexample being f(x)=1 except f(5)=+infinity
neither is this.
hot canyon
huh? x = 5 is a removable discontinuity
also, {5} has measure 0.
we can integrate in any set [a, b] containing {5} exactly due to that reason
Also, f(5) = infty is not a valid function . infty is a concept not a value
so if you wish to plug f(5) = infty, the only way is to plug lim_{x to 5} f(x) = infty, which then gives you a function that is not bounded and in that case you should be able to see why it's non-integrable as well
the integral diverges
pearl pondBOT
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visual canyon
what are my next steps here to finding the inverse?
pearl pondBOT
what are my next steps here to finding the inverse?
gusty flume
well what's the inverse operation of squaring an expression?
cobalt hinge
what are my next steps here to finding the inverse?
sqrt both sides and consider a domain restriction, since y=(x+2)^2 is not one-to-one
jolly parrotBOT
;(
cobalt hinge
$\sqrt{(y+2)^2}=\sqrt{x}$
jolly parrotBOT
;(
visual canyon
sqrt of x-2 - sqrt of x-2
cobalt hinge
sqrt of x-2 - sqrt of x-2
where are you getting these numbers from
cobalt hinge
sqrt of x-2 - sqrt of x-2
that would be 0
cobalt hinge
mayb, but i cannot assume
plush moss
Fair
fierce totem
what are my next steps here to finding the inverse?
there is one method that I find really satisfying (work specifically for all quadratics)
Hint: $x^2 + 4x + 4 - y = 0$
jolly parrotBOT
TargetVN
visual canyon
Idk maybe he typoed
kind of, i meant y=sqrt(x)-2, y-sqrt(x)-2
plush moss
kind of, i meant y=sqrt(x)-2, y-sqrt(x)-2
What's the use of the latter expression anyway
You completed your mission alr with y=sqrt(x)-2
fierce totem
no one asked but imma just throw this here:
cobalt hinge
kind of, i meant y=sqrt(x)-2, y-sqrt(x)-2
what is y-sqrt(x)-2
fierce totem
The inverse function of $y = ax^2 + bx + c$ is the root of $ax^2 + bx + c - y = 0$.
jolly parrotBOT
TargetVN
visual canyon
ok how is this
visual canyon
~~no one asked but imma just throw this here:~~
i have something but it's not your fun way 
cobalt hinge
ok how is this
👍
fierce totem
ok how is this
nice
limpid lily
That's unfortunately not enough.
visual canyon
whaaaat noooooo
limpid lily
For example, if f(x) = x^2 and g(x) = x/2, you'd have f(2) = 4 and g(4) = 2, but they're not inverse functions, even though you got back the 2 you started with.
cobalt hinge
this one is my fun way, i just think it's kind of neat
$f$ and $g$ are inverses iff $f(g(x))=x$
jolly parrotBOT
;(
limpid lily
One second, mistaken example.
cobalt hinge
so you have to prove that $f(g(x))=x$ which basically means to subsitute $g(x)$ into the x-input for f and see if it equals $x$.
limpid lily
Fixed.
jolly parrotBOT
;(
limpid lily
But f(5) = 25, and g(25) = 12.5, so you don't get back 5.
You need to prove that it works for all xs, not just one.
cobalt hinge
visual canyon
oh this is so sad
plush moss
this one is my fun way, i just think it's kind of neat
This just proves for one x value
f(g(x)) =4(x/4)=x
Since f(g(x)) = x the functions are inverses
I mean it's not far off from what you do it's just proves it for all x
cobalt hinge
oh this is so sad
indeed
plush moss
I mean it's not far off from what you do it's just proves it for all x
Which I think is honestly cooler
midnight haven
Physics question
pearl pondBOT
Physics question
midnight haven
the answer is 800m
Since this is destructive interference, the path difference should be a half-integer multiple of wavelength
however since the wave is undergoing a phase change after reflection
destructive interference should occur if the path difference is an integer multiple
this means that
RH - MH = 1600 (1)
the smallest integer multiple is just 1
the answer is 800m apparently
vestal tapir
i don't know sorry
midnight haven
no worries
I understand how 800 is an answer
however can't the answer just be anything
only RH - MH should be 1600?
vestal tapir
it should be 400
midnight haven
The answer is 800m
vestal tapir
i don;t understand how 800 is the answer
midnight haven
according to the answer booklet
800 can be one of the answers according to me
if we assume that RM and MH are equal in length
vestal tapir
unless it doesn't work like sound?
midnight haven
Destructive interference should work in all kinds of waves
including radio waves hopefully
Anyways thanks for your help @vestal tapir
If anyone else can help me get at the answer of 800m, feel free to ping me. I would be on another tab
hot canyon
If anyone else can help me get at the answer of 800m, feel free to ping me. I wo...
Idts it's 800m
isn't the shortest possible distance between house n mountain 400m?
midnight haven
yeah that does actually make sense
400m makes more sense
but this is what the book says
I am still not sure why its 2d?
2d = n lambda
hot canyon
for destructive interference, the path difference should be (n + 1/2) lambda
vestal tapir
sounds wrong yeah
hot canyon
for destructive interference, the path difference should be (n + 1/2) lambda
n for smallest n, 2d = lambda/2 gives d = 400
ohh right mb
midnight haven
for destructive interference, the path difference should be (n + 1/2) lambda
however the wave coming from the mountain has a phase change of pi
therefore its essentially the formula of constructive interference
hot canyon
there's a phase change of pi, so as the second wave reflected off the mountain, we need to add the phase change of pi to the path difference due to reflection
the equation would be
2d + (pi * lambda)/2pi = (n + 1/2) lambda
vestal tapir
alright
hot canyon
right lol ofc
midnight haven
$$2d + \frac{\pi \lambda}{2 \pi} = (n + \frac{1}{2}) \lambda$$
jolly parrotBOT
Edmund Cloudsley
hot canyon
here I'll show you why
midnight haven
this right?
ahhhhhhhh okay this makes sense
and the smallest n is 0
therefore
,w solve 2d + 800 = 1/2 * (800)
,w solve 2d + 800 = 1/2 * (800)
jolly parrotBOT
jolly parrotBOT
midnight haven
nvm
hot canyon
@midnight haven you missed the lambda/2 here on left side
midnight haven
but lambda is 1600
hot canyon
Yeah there corrected
midnight haven
where did the 800 * 3 come from ?
hot canyon
2d + l/2 = (n + 1/2)l => 2d = nl => for n = 1, 2d = l
hot canyon
where did the 800 * 3 come from ?
1600 * (1 + 1/2) = 1600 * 3/2 = 800 * 3
midnight haven
yup
hot canyon
$2d + \frac{\lambda}{2} = n\lambda + \frac{\lambda}{2}$
jolly parrotBOT
hot canyon
right?
midnight haven
yuppadoodle
rotund osprey
x/(x^2+x)^3/2
pearl pondBOT
x/(x^2+x)^3/2
rotund osprey
integrate
please
i cannot figure this one out
can someone give a hint on a usub perhaps that i should try
😭
inland ivy
$\int\frac{x}{(x^2+x)^\frac{3}{2}}\dd{x}$
jolly parrotBOT
kheerii
inland ivy
is this your question?
rotund osprey
yes
inland ivy
the substitution u=sqrt(x) looks promising
after you split the denominator into two roots
jolly parrotBOT
neon
glacial sequoia
Substitute 1 + x^(-1)
rotund osprey
ahh
glacial sequoia
then it's just the power rule
rotund osprey
wait
how can u just factor out x^2
cos the whole thing is to the power of
3/2
can u just do that ?
ohhh wait the x on top
disspeared
okok gotchu
wait but then
wouldnt it be
x^1/2
on the bottom
glacial sequoia
we factor x^2 out
so what is outside would be (x^2)^(3/2) = x^3
x on top will cancel so only x^2 remains
rotund osprey
ohhhh
okok thanks
wait i have a questoin
cos i got the answer
but the books answers says
and i get that integrals can have differnet answers cos of the plus c
but then when i pluged it into desmos
they look different
and im not sure why
they;re very similar tho
glacial sequoia
did you substitute back in
inland ivy
they're the same
inland ivy
they shouldn't
rotund osprey
and i added the sliders so they could match up but u see
on the lhs
they're like
flipped
inland ivy
yeah
that's because of the way you handled the square root
it's not a perfect antiderivative
inland ivy
it's because of the domain of the integrand
inland ivy
it's only real for values < -1 or > 0
inland ivy
your answer will work in both domains
inland ivy
this answer then
$\frac{2}{\sqrt{1+\frac1{x}}}=\frac{2\sqrt{x}}{\sqrt{1+x}}=\frac{2x}{\sqrt{x^2+x}}$
jolly parrotBOT
kheerii
inland ivy
that's why your answers match for x>0
but for x<-1 you can't split the square roots like that
pearl pondBOT
@rotund osprey Has your question been resolved?
spare lark
<@&268886789983436800>
opal lantern
.close
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lucid hazel
pearl pondBOT
lucid hazel
i know that resistance in parallel circuits = 1/R1 + 1/R2 ...
but we have 7 resistors here
pearl pondBOT
@lucid hazel Has your question been resolved?
grave hazel
treat each set of three like its one resistor(bc those three are in series with eachother)
so for the top and bottom pathways, its a series of three so their resistance is 3R
the middle ones resistance is R
so 1/total resistance = 1/3R + 1/R + 1/3R
@lucid hazel
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midnight haven
What is the probability of getting a double 6 atleast once in 24 throws with 2 dice
midnight haven
Case1 :- 1st die thrown with no probability of getting a 6 i.e (5/6)^24 no condition on second die
case 2:- 2nd die thrown with no 6 i.e (5/6)^24 and no condition on 1st die
since there is a repition
whose probability is (5/6)^24 (5/6)^24
so the probability of getting a double 6 atleast once in 24 throws with 2 dice is
1- (5/6)^24 -(5/6)^24 + (5/6)^24 (5/6)^24
but its not matching with the answer key
where have i gone wrong?
vestal tapir
1- ((5/6)^24 + (5/6)^24 − (5/6)^24 (5/6)^24)
midnight haven
sorry for the bad terminology
vestal tapir
oh that's what you did mb
midnight haven
yeah
vestal tapir
lemme think
midnight haven
1- ((5/6)^24 + (5/6)^24 − (5/6)^24 (5/6)^24)
the value of this is 0.975
vestal tapir
yep
well yeah, it's just wrong logic
you;re saying either there's never 6 on the left, or never 6 on the right (or never 6 ever anywhere)
but that's too restrictive
could be 6s on both sides
midnight haven
could be 6s on both sides
But this is the condition itself right,not it's complement
vestal tapir
no that's still the complement
6 2
2 6
6 2
2 6
...
this is neither of your 2 cases
vestal tapir
it's (35/36)^24
pearl pondBOT
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Available help channel!
vast berry
how do i do this
pearl pondBOT
how do i do this
compact ridge
this isn't your channel
vast berry
this isn't your channel
oh hii
pearl pondBOT
@vast berry Has your question been resolved?
vast berry
<@&286206848099549185>
hot canyon
<@&286206848099549185>
What's the formula for SD?
vast berry
What's the formula for SD?
hot canyon
!show
pearl pondBOT
Show your work, and if possible, explain where you are stuck.
hot canyon
Should've tried something perhaps?
For example, plugging:
nSD² = sum [(x - x{bar}]² = sum [(x - q) + q - (14 + q)]² = sum [(x - q) - 14]²
Note, x{bar} = 14 + q
You should be fine if you just expand the RHS without touching (x - q)² and simplify
vast berry
For example, plugging:
```nSD² = sum [(x - x{bar}]² = sum [(x - q) + q - (14 + q...
could u perhaps write this down please?
this is a bit confusing sorry
pearl pondBOT
@vast berry Has your question been resolved?
pearl pondBOT
@vast berry Has your question been resolved?
pearl pondBOT
@vast berry Has your question been resolved?
peak jetty
could u perhaps write this down please?
it's just this line
$n\cdot\text{SD} = \sum\left(x - \bar{x}\right)^2 = \sum\bigl(x - (14-q)\bigr)^2 = \sum\bigl( (x-q) - 14\bigr)^2$
$\bar{x} = 14-q$ comes from
jolly parrotBOT
Typo
peak jetty
using this with the values given in your question
jolly parrotBOT
Typo
pearl pondBOT
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flint drum
got a exam tomorrow need help with this question asap
pearl pondBOT
got a exam tomorrow need help with this question asap
flint drum
so ik
all areas have to be positive
to add them all up
but theres a negative area
and see idk what to do
spare lark
Take abs value
cobalt hinge
but theres a negative area
Just make the function negative for whichever intervals the area is negative. It automatically makes the area positive.
cobalt hinge
i.e., turn $\int_1^3 f(x)dx$ into $\int_1^3 -f(x)dx$, since the function's area is negative on that interval (or, in fact, the function itself is negative).
jolly parrotBOT
;(
cobalt hinge
im to dumb for this
No you aren't! It just takes a bit of practice and intuition.
flint drum
**;(**
still no clue how to do it
it was a test i did like 3 weeks ago
got a high grade and this was one of the things i got wrong
and all my teacher wrote was it cant be negative
plush moss
got a exam tomorrow need help with this question asap
Oh lmao I was wondering why that last integral had bounds 3 to -4
cobalt hinge
and all my teacher wrote was it cant be negative
Yes, because it is area.
plush moss
Oh lmao I was wondering why that last integral had bounds 3 to -4
I think my eyes are not eyeing
cobalt hinge
You are calculating "area", not "net area".
Net area here is similar to finding the area "under the curve" for the function f.
However, "area" represents the total, meaning that we cannot have any negative areas, else the unit itself will not make sense.
flint drum
I think my eyes are not eyeing
is that a mistake?
plush moss
is that a mistake?
I saw wrongly
flint drum
However, "area" represents the total, meaning that we cannot have any negative a...
ok
thats very well said i suppose
i still havent got a clue on how to answer it
cobalt hinge
It is the standard definition
cobalt hinge
Take abs value
As stated here, this is the way to solve it.
Do you know what absolute value does?