ik how to do derivative of 2x^2
#help-19
woeful rivet
copper quarry
yes once u do a few derivatives theyre all the same process
its just rinse and repeat
woeful rivet
whats In
copper quarry
natural log
woeful rivet
ohh kk
copper quarry
wait so youvbe done integration but not differentiation?
woeful rivet
so i dont need to study ahead for that
woeful rivet
wait so youvbe done integration but not differentiation?
i have
copper quarry
why
forest sky
well if you only have to use the power rule then of course it's easy
forest sky
the hard integrals come once you start on transcendental functions
copper quarry
wait i have done differentiation
so u know how to do the stuff in the image u sent
amber schooner
why not just learn in the proper order
forest sky
a transcendental function is something which isn't just powers of x, like trig, exponents, logs, etc
woeful rivet
and then im going to europe 😈😈
celest acorn
a transcendental function is something which isn't just powers of x, like trig, ...
the hard integrals come, when thine thing you are integrating are close to the derivative of a transcedenatl function
woeful rivet
a transcendental function is something which isn't just powers of x, like trig, ...
oh
ok
@copper quarry if im studying business
will they make me do maths
woeful rivet
a transcendental function is something which isn't just powers of x, like trig, ...
cuz i dont want to do this
copper quarry
<@722683408190406667> if im studying business
maybe some
woeful rivet
are u studying maths?
copper quarry
major in astrophys minor in math
woeful rivet
but its too hard for me 😢😢
copper quarry
sad life
woeful rivet
i just want to get this over with
copper quarry
whats the name of ur final hs exams overf there
woeful rivet
whats the name of ur final hs exams overf there
wace exam
copper quarry
such a wack name
frigid canopy
Knowing advanced maths is always useful
woeful rivet
Knowing advanced maths is always useful
how
frigid canopy
well, you can start off as an engineer and then eventually go into management for instance
woeful rivet
well, you can start off as an engineer and then eventually go into management fo...
omg
i was gonna do a double degree
in engineering and business
but i realise now..
i think im too dumb for it 😭😭😭
bcoz high school vectors is too hard for me
woeful rivet
such a wack name
hsc sounds worse
copper quarry
woeful rivet
wace is one syllable
celest acorn
ur doing accelerated math
what that mean?
copper quarry
what that mean?
accel math is when someone from a lower grade studies math from a higher grade
odd edgeBOT
@woeful rivet Has your question been resolved?
torn bloom
hi
torn bloom
a really quick question
Let K be a field and K[t]≤n−1 the set of polynomials in the unknown t from
Degree at most n − 1. Show that two polynomials p, q ∈ K[t]≤n−1 are equal if
p(βi) = q(βi) for pairwise different ¨ β1, . . . , βn ∈ K plated.
I made a linear equation system that has to be zero
then i have a vandermonde Matrix and with the different koeffizients
can i use the determinante instead of the vandermonde Matrix because i know that it cant be zero so the other Matrix with the koeffizients has to 0 in every line
<@&286206848099549185>
<@&286206848099549185>
.close
blazing mesa
a really quick question
lebron is that you?
odd edgeBOT
lebron is that you?
blazing mesa
.close
odd edgeBOT
odd edgeBOT
Available help channel!
tepid quarry
$4x^2 +8x +4 - (x+3)^2$
odd edgeBOT
$4x^2 +8x +4 - (x+3)^2$
clever fjordBOT
Simon James B
tepid quarry
i have no idea how to start factoring this
vale vapor
probably easiest to break up those parentheses first
warped glacier
4x^2 + 8x + 4 is also a multiple of a perfect square trinomial
tepid quarry
probably easiest to break up those parentheses first
what do you mean
vale vapor
the (x+3)^2
warped glacier
you don't actually need to expand (x + 3)^2 in this situation
but if you're unsure just try it
tepid quarry
the (x+3)^2
oh like use the formula on the binom?
warped glacier
yeah
mystic saffron
warped glacier
so like for this trinom what do we do with it
can you expand (x + 3)(x + 3) using FOIL
odd edgeBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
tepid quarry
can you expand (x + 3)(x + 3) using FOIL
x^2 +6x + 9?
warped glacier
x^2 +6x + 9?
yep!
mystic saffron
don't post the full solution
Pretty sure a question like this would be easier explained if the solution is sent and he asks about what he doesn't understand..
tepid quarry
but don't we have a binom so we have to use a^2 + 2ab + b^2 on it?
warped glacier
Pretty sure a question like this would be easier explained if the solution is se...
it's easy for you and me but you're not helping by giving them the answer
and also this is not the method we have been discussing
it's much more beneficial if they try by themselves
warped glacier
but don't we have a binom so we have to use a^2 + 2ab + b^2 on it?
yes, it's just FOIL
mystic saffron
First step plus 1 not plus 2 mb
warped glacier
(a + b)(a + b) = a^2 + ab + ba + b^2
tepid quarry
oh yea
warped glacier
so it's a foil shortcut for when we have a perfect square
tepid quarry
got it
but what if we want to solve this trinom first. I remember we had to factor somehow the second terms but i forgot how so we get 4 terms
warped glacier
but what if we want to solve this trinom first. I remember we had to factor some...
okay continuing your method, you get (4x^2 + 8x + 4) - (x^2 + 6x + 9)
and now we should expand this, collect like terms
tepid quarry
there was a trick for this right? make 8x = to somethings so that it also gives 4?
but i forgot
warped glacier
the trick you're talking about is here
recognise that 4x^2 + 8x + 4 = 4(x^2 + 2x + 1) = 2^2 * (x + 1)^2
= (2 * (x + 1))^2
then use the identity for a^2 - b^2 = (a + b)(a - b)
tepid quarry
we just factor a 4?
warped glacier
yeah
tepid quarry
$4x^2 + 8x + 4 = 4(x^2 + 2x + 1)$
clever fjordBOT
Simon James B
tepid quarry
and we write 4 as 2^2?
warped glacier
yeah
tepid quarry
$2^2(x^2 +2x +1)$?
clever fjordBOT
Simon James B
warped glacier
$2^2(x^2 +2x +1)$?
yep
tepid quarry
$2^2(x^2 +2x +1) - (x^2 +6x + 9)$
clever fjordBOT
Simon James B
tepid quarry
now what?
tepid quarry
yea i see it
warped glacier
x^2 + 2x + 1 = (x + 1)^2
x^2 + 6x + 9 = (x + 3)^2 like you did already
clever fjordBOT
Simon James B
warped glacier
yeah so $2^2 (x + 1)^2 - (x + 3)^2$
mystic saffron
Don't forget your 2^2 at the start
clever fjordBOT
south
tepid quarry
and we do a^2b^2 = (ab)^2 ?
warped glacier
and we do a^2b^2 = (ab)^2 ?
yes!
mystic saffron
Correct
tepid quarry
$2(x+1)^2 - (x+3)^2$ \newline (2x+2)^2 - (x+3)^2$
clever fjordBOT
Simon James B
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
warped glacier
**Simon James B**
Compile Error! Click the <:errors:504266556054962196> reaction...
correct except you mean 2^2
but yes that's correct
tepid quarry
$[2(x+1)]^2 - (x+3)^2$
clever fjordBOT
Simon James B
warped glacier
ah okay yes
tepid quarry
$(2x+2)^2 - (x+3)^2$
clever fjordBOT
Simon James B
tepid quarry
so difference of squares?
warped glacier
so difference of squares?
yes!
tepid quarry
$(2x+2 -x -3)(2x+2+x+3)$
clever fjordBOT
Simon James B
tepid quarry
$(x -1)(3x +5)$
clever fjordBOT
Simon James B
tepid quarry
I just can't see what to do at first and it is so annoying
warped glacier
ah yeah
okay so yeah keep practicing, don't be afraid to look at worked examples
but you clearly know your stuff
merry finch
I mean when I look at a question like this
warped glacier
I just can't see what to do at first and it is so annoying
also yeah as I suggested, there's no harm in expanding everything then factorising
merry finch
I just try stuff till it works
tepid quarry
I just try stuff till it works
that won't do it at my exam:(
merry finch
You haven’t done an exam where you look at the question and scratch your head as to how to even start it?
tepid quarry
our exams did not have such questions before so it will be my first
steep mantle
You haven’t done an exam where you look at the question and scratch your head as...
the geometry expereince
tepid quarry
until now it was all about using a formula or calculating stuff because like recently we learned about roots and stuff but now about oaprting with variables
merry finch
Well I’ll tell you now that using formulas and calculating stuff is not what maths is about
tepid quarry
operating*
merry finch
Maths is actually a very creative field
tepid quarry
I can't wait to move on to harder things thanks for the help i will go practice some more
merry finch
It’s all about finding interesting and creative ways to tackle complex problems
tepid quarry
.close
odd edgeBOT
warped glacier
until now it was all about using a formula or calculating stuff because like rec...
congrats you're levelling up in maths
but sounds like you're adjusting, having to use your brain more
that's going to be a common pattern the more maths you see and do
frosst is very very right
tepid quarry
congrats you're levelling up in maths
thanks for the support
odd edgeBOT
Available help channel!
near sparrow
odd edgeBOT
near sparrow
I tried lhop
if I assume cosx = 0 then I think get the right answer but I'm not sure if that's okay to do
near sparrow
if I assume cosx = 0 then I think get the right answer but I'm not sure if that'...
This is my reasoning for that
near sparrow
oh the log is base e
clever fjordBOT
Bonk
vale vapor
its not rigorous but: what is the highest power of x in the numerator and denominator?
near sparrow
3/2 in numerator and 5/2 in denominator
so the denominator grows faster
and it approaches 0?
is that right?
vale vapor
yeah the denominator is a higher power
so it dominates
if you were to divide everything by x^2, every other term becomes 0 but that term still goes to infinity
(well, not quite infinity because its times cos, but you get the point)
near sparrow
yep i got it thx
half sleet
Solve the inequality
In the answer, write the sum of all natural numbers that are NOT solutions to the inequality.
vale vapor
you need to find a range for which this inequality holds
half sleet
how do i do that
vale vapor
well, what i would firstly do is divide both sides by x-7
half sleet
x>-1
wild hull
correct
vale vapor
x>-1
not so fast
we indeed do get x+4>3
but if x<7 we divide by a negative number
and then we reverse the inequality
wild hull
good catch
tacit haven
well, what i would firstly do is divide both sides by x-7
isnt dividing by a variable usually not reccomended for inequalities? sorry
wild hull
ye but u can always make cases
half sleet
**but** if x<7 we divide by a negative number
you lost me here
(x-7) has -7 so we reverse the inequality sign?
vale vapor
isnt dividing by a variable usually not reccomended for inequalities? sorry
its alright to do, but you are usually just left with cases
wild hull
its because we don't know the variable
vale vapor
you lost me here
(x-7) has -7 so we reverse the inequality sign?
ill give an example, if x=4, then x-7=-3, which is negative
if you multiply an inequality by a negative number, the sign gets reversed
wild hull
you lost me here
(x-7) has -7 so we reverse the inequality sign?
we change the sign if u divide by a negative
5>4
Multiply by -1
-5<-4
half sleet
you lost me here
(x-7) has -7 so we reverse the inequality sign?
so basically yes?
wild hull
so basically yes?
if x<7 that implies that (x-7) would be negative
so u gotta change the inequality
wait sry
vale vapor
$(x+4)(x-7)>3(x-7) = \begin{cases}x+4>3 \text{ for } x-7>0\x+4<3 \text{ for } x-7<0\end{cases}$
clever fjordBOT
Bonk
half sleet
is there another way of solving this? for instance i don't catch that theres (x-7) on both sides
vale vapor
if x-7 is negative, then the sign gets reversed, if x-7 is positive, then its just regular
vale vapor
is there another way of solving this? for instance i don't catch that theres (x-...
wdym?
wild hull
how do u not see (x-7) is on both lhs and rhs
half sleet
wdym?
the way i solved it incorrectly is i started looking for the root i completely disregarded (x-7) on both sides
wild hull
root?
vale vapor
the way i solved it incorrectly is i started looking for the root i completely d...
well, you shouldnt do that
wild hull
this is an inequality not an equation
tacit haven
is there another way of solving this? for instance i don't catch that theres (x-...
you could expand and solve the quadratic inequality but this way is simpler
half sleet
well, you shouldnt do that
ait give me a second)
wild hull
@half sleet this method is much easier
wild hull
youll realise after practicing some questions
vale vapor
this is the easiest
mellow tundra
the way i solved it incorrectly is i started looking for the root i completely d...
you have to consider that x is not equal to 7
wild hull
$(x+4)(x-7)>3(x-7) = \begin{cases}x+4>3 \text{ for } x-7>0\\x+4<3 \text{ for } x...
Whatd happen if x=7?
vale vapor
Whatd happen if x=7?
then the inequality isnt true
quartz trellis
make it to the form ()>0
wild hull
then the inequality isnt true
right
vale vapor
because 0 >0 isnt true
vale vapor
right
also, in my cases you see x-7>0 and x-7<0, this doesnt include x=7 because then you divide by 0 and its not possible
ocean bramble
is there another way of solving this? for instance i don't catch that theres (x-...
yes, there's another way to solve this.
You'd first operate both sides, so you'd have:
x^2-3x-28 > 3x-21
Then you'd rearrange, by adding (-3x+21) to both sides:
x^2-6x-7 > 0
half sleet
alr im with u here
but then all i can do is find the solution right? I need intervals?
ocean bramble
i assume you know how to solve the equality x^2-6x-7 = 0?
half sleet
yes, there's another way to solve this.
You'd first operate both sides, so you'd...
yea ik give me a second ill solve this
tacit haven
yes, there's another way to solve this.
You'd first operate both sides, so you'd...
also, if you dont want to factorize you could do
(x-7)(x+4) >3(x-7)
(x-7)(x+4) - 3(x-7)>0
(x-7)(x+4-3)>0
ocean bramble
that's usually more complicated tbh
half sleet
i get -1 and 7
ocean bramble
also, if you dont want to factorize you could do
(x-7)(x+4) >3(x-7)
(x-7)(x+4) -...
remember that he said "if he doesnt notice the (x-7) on both sides", which i assume wouldnt notice in the next step either then
correct. Now, you know that the equality is true for x=-1 and x=7
tacit haven
remember that he said "if he doesnt notice the (x-7) on both sides", which i ass...
oh mb
ocean bramble
and you know that it's a parabola, so it will change signs on those two points
you can either go by the theory, the coefficient on x^2 is positive, thus the parabola will be positive-negative-positive
half sleet
aha hold on
ocean bramble
or you can select three points, one before -1, one between -1 and 7, and one after 7, and check their sign, and if they hold the inequality
this is the totally perfectly drawn parabola
half sleet
x is from (-infinity; to -1) and (7 to +infinity)
ocean bramble
and you can clearly see it's positive in the intervals you said
half sleet
alr so now I've got intervals
write the sum of all natural numbers that are NOT solutions to the inequality.
so from -1 to 7
vale vapor
then you take the complement
half sleet
28?
ocean bramble
correct
half sleet
you're a life saver
ocean bramble
so from -1 to 7
remember -1 is not a natural, so from 0 to 7
vale vapor
remember -1 is not a natural, so from 0 to 7
0 is also not natural
ocean bramble
**Bonk**
while this is correct, students for some reason struggle a lot with cancelling terms that might be negative on inequations
they usually dont respect the "if the thing is negative case 1, if the thing is positive case 2"
ocean bramble
~~0 is also not natural~~
depends on convention used. I consider it natural.
half sleet
how did we decide that x+4<3 it's specifically for x-7<0
i dont get how we got cases here at all
vale vapor
they usually dont respect the "if the thing is negative case 1, if the thing is ...
then they need to learn it
its much more elegant imo
ocean bramble
it really isnt, and it's usually significantly slower. You dont want to use that for the simpler ones
half sleet
then they need to learn it
as proven just now, not really
i got how to solve it using another method
ocean bramble
i dont get how we got cases here at all
okay, so lets compare with equations. On an equation, you know the following:
a = b <=> a*n = b*n, for n!=0
as in, if you multiply (or divide) both sides of the equation by non-zero, the equation is equivalent, correct?
half sleet
non zero?
ocean bramble
a number that is not zero
ocean bramble
if you multiply both sides by zero, it doesnt matter what you had before
half sleet
the same action happened to both so nothing should change
ocean bramble
and you cannot divide both sides by zero because that's heresy
okay
so to put an example, if x=1, then 2x=2. Obviously.
half sleet
yep
ocean bramble
now, for inequalities, that is not a rule
for inequalities, you have two, distinct, but similar, rules
a > b <=> a*n > b*n, for n>0
a > b <=> a*n < b*n, for n<0
now, if you multiply both sides by something positive, you keep the inequality side
if you multiply both sides by something negative, you flip the inequality
I'm gonna use the same example
x > 1
Multiplying by 2:
2x > 2
However, multiplying by -2 instead:
-2x < -2
half sleet
a > b <=> a\*n > b\*n, for n>0
a > b <=> a\*n < b\*n, for n<0
gonna be honest I'm not too familiar with the <=> sign
ocean bramble
if, and only if
it means that the thing on one side being true makes the thing on the other side true, AND the thing on one side being false means the thing on the other side also being false
so they are both true, or both false
half sleet
x > 1
Multiplying by 2:
2x > 2
However, multiplying by -2 instead:
-2x < -2
ok i get this but i dont rly get how to implement it
half sleet
yep
ocean bramble
now, let's multiply both sides by 2
6 > 2, obviously
however, if we multiply by -2 instead, we get:
-6 < -2
because the rule says that we need to alter the inequality direction
half sleet
yep
ocean bramble
if we do not, we'd get the wrong:
-6 > -2
which is not true
that's why for inequalities you distinguish the case of multiplying (or dividing) by a negative
half sleet
ah wait i think i got smt
so we have x+4>3 only if x-7>0
or x
no
wait
only if x-7>0
and x+4<3 only if x-7<0
i got this part now
but those aren't solutions they're cases
so im a bit lost again
ocean bramble
you'd need to solve each inequality, in each case
ocean bramble
so we have x+4>3 only if x-7>0
for example, this one.
x+4 > 3, with x-7 > 0
x+4 > 3 means that x+1 > 0
or in other words, x > -1
but since you're on the case x-7>0, you also need x >7
which means that both conditions must be true at the same time.
x > -1
x > 7
which obviously means that x > 7. So that's one part of your solution
half sleet
ah got it
cases seem a lot more complicated to me
im also very rusty on them since i havent used them since middle school or so and am going into my high school finals
im going to try solving it without any help and see if i get the correct answer)
I've got another question
Solve the inequality
Write the LARGEST solution to the inequality in your answer.
i got x<-25
so the largest would be -24?
or I guess -24,9 smt like that?
vale vapor
are they asking for an integer?
half sleet
Write the LARGEST solution to the inequality in your answer.
half sleet
then I guess an integer?
vale vapor
so then what is the largest integer such that x<-25?
half sleet
idk i wrote that it's incorrect
vale vapor
what was your answer?
half sleet
it said to write the largest possible asnwer
i tried
x<-25
-24
-24,9
none of them are correct
half sleet
oh shit i forgot -
vale vapor
yes
half sleet
honestly im out of options except for -26
yea it was -26
but itsn't that techncially wrong?
wooden gorge
Why is it wrong?
wooden gorge
,calc -(-26)/5
clever fjordBOT
Result:
5.2
what about -26,9
vale vapor
what about -26,9
but -26,9<-26
wooden gorge
Ooo you mean you have a problem with "largest solution". It doesn't say integers yes
vale vapor
-26 is bigger than -26,9
half sleet
Ooo you mean you have a problem with "largest solution". It doesn't say integers...
yes
half sleet
-26 is bigger than -26,9
yea i was wrong
wooden gorge
Hmm that's an issue yes, because we can take -25.0000000001 and there's still a larger solution
vale vapor
Hmm that's an issue yes, because we can take -25.0000000001 and there's still a ...
yeah exactly
half sleet
the answer the system accepted was -26
wooden gorge
Yes if the system designated it as an Integer Type Question
vale vapor
but it shouldve mentioned its an integer though
half sleet
.close
odd edgeBOT
odd edgeBOT
Available help channel!
whole anvil
I don't understand how to start tackling this problem
whole anvil
nimble blaze
simplify the nested fraction, factorise the x^3-8
woven pier
Wait, if x-> -2, then the denominator doesn't become 0, it becomes -16
woven pier
What the john
cold sage
<@&268886789983436800>
whole anvil
simplify the nested fraction, factorise the x^3-8
How do I simplify it?
woven pier
I already tried that and ended with 0
Ok, so the answer is 0?
nimble blaze
did you copy down the question correctly?
whole anvil
Alright
How do I simplify the numerator?
I don't see anything they have in common
Except being a fraction
Aah nvm
woven pier
Take the LCM
whole anvil
I see
woven pier
Ok
whole anvil
Take the LCM
Huh?
Nvm I don't get it
I tried to multiply the top and bottom
Nvm again
Idk where to go from here
woven pier
x^3 + 8 = (x+2)(x^2 + 4 - 2x)
So there is a common factor (x+2) in both the numeratora nd the denominator.
Cancel it, and now put x = -2
whole anvil
x^3 + 8 = (x+2)(x^2 + 4 - 2x)
How do I find this?
woven pier
x^3 + y^3 = (x+y)(x^2 + y^2 - xy)
This is an identity
whole anvil
But how do you like prove it
woven pier
uhm
that's another thing, but:
x^3 + y^3 = (x+y)(x^2 + y^2) - x^2 * y - x * y^2 = (x+y)(x^2 + y^2) - xy(x+y) = (x+y)(x^2 + y^2 - xy)
Similarly, x^3 - y^3 = (x-y)(x^2 + y^2 + xy)
hollow quarry
Hi I need to prove (using the bolzano weierstrass theorem) that a monotonic increasing sequence that has an upper bound converges. Now the BW theorem says that for a bounded sequence a_n, there always exists some subsequence a'_k that converges
hollow quarry
Here is what i tried :
So let a' be the limit of the subsequence, then we know that $\forall \varepsilon > 0 \ \exists k_0 \in N \ \forall k \geq k_0: | a'_k - a' | < \varepsilon$
but im not sure how i can use the fact that a_n is increasing here, i would appreciate any hints
clever fjordBOT
p1za
odd edgeBOT
@hollow quarry Has your question been resolved?
hollow quarry
<@&286206848099549185>
meager juniper
Consider such a subsequence $(a_{k_n})$, and consider all of the "skipped" elements of $(a_n)$ between $k_n$ and $k_{n+1}$. Could it be possible to violate the convergence condition with any of these elements?
clever fjordBOT
OmnipotentEntity
meager juniper
(Specifically, a convergent subsequence)
Also, were you instructed to use BW? I do not believe that you need it.
@hollow quarry
hollow quarry
yes i have to use it
hollow quarry
**OmnipotentEntity**
hm no i dont think so
dull blade
hm no i dont think so
"i don't think so " isn't an answer
hollow quarry
So you mean $| element - limit | < \varepsilon$ is still the case?
clever fjordBOT
p1za
meager juniper
What can we say about $a_m$ where $m > k_n$?
clever fjordBOT
OmnipotentEntity
meager juniper
Given that $(a_n)$ is monotonically increasing.
clever fjordBOT
OmnipotentEntity
dull blade
am> a(kn)
hollow quarry
…
meager juniper
@dull blade I'm glad you know but I'm asking p1za 😉
So if am > akn then what can we say about |am - a'|?
As compared to |akn - a'|
hollow quarry
Can you explain what you mean by k_n coz i think we had different notation in class
meager juniper
Ok, let k_n be a monotonically increasing sequence of integers.
$(a_{k_n})$ is a subsequence of $(a_n)$
clever fjordBOT
OmnipotentEntity
hollow quarry
So if am > akn then what can we say about |am - a'|?
that its smaller
so its also smaller than epsilon
so it converges?
meager juniper
Yes
hollow quarry
**OmnipotentEntity**
but for this both also have to be >= k_0 right
meager juniper
Not salty tho, more like good humored irony, thought the winky face conveyed that, but tone is difficult over text
hollow quarry
spoiling the solution wasnt neccessary in the first place, so now if you dont have anything to contribute to the convo please go to another channel or send something related to the problem, thx
meager juniper
So anyway, due to how limits are structured, you have an epsilon. So you choose an N such that for all n > N, |a_{k_n} - a'| < epsilon. Then you prove that for all m > k_n, |a_m - a'| < epsilon
This is how I would approach it anyway
hollow quarry
Yeah alright thanks, I really overcomplicated it for no reason 🤣
hollow quarry
I think they are trolling
@meager juniper do you mind helping me with another problem? Its about proving the convergence of two sequences, only if you have time ofc
$(1 - \frac{1}{n})^n$ and $(\frac{(n+1)!}{n!})^n \cdot \frac{1}{(n + 2)^n}$ that both converge to 1/e im pretty sure
clever fjordBOT
p1za
hollow quarry
yes, that is also what we used in class to approximate e and we also used $\sum_{k=0}^n \frac{1}{k!}$
clever fjordBOT
p1za
hollow quarry
so Im not sure if applying the basic epsilon definition makes sense here?
meager juniper
Well, can you show that this approaches that using the epsilon definition and the binomial theorem?
hollow quarry
or if it did make sense, im not sure how to approximate it correctly
meager juniper
Oh whoops, wrong message
Irrespective of whether your assertion is right or wrong, this is an extremely unpleasant way of conveying the situation. For the record, it was entirely unclear to me as well why you were answering. I just thought you were being overeager, which happens from time to time, and it no big deal.
meager juniper
Irrespective of whether your assertion is right or wrong, this is an extremely u...
This was intended for you @dull blade
The above was intended to p1za, my b
hollow quarry
Well, can you show that this approaches that using the epsilon definition and th...
$| \sum_{k=0}^{n} \binom{n}{k} \cdot (- \frac{1}{n})^{n - k} - \frac{1}{e} | < \varepsilon$
like this?
meager juniper
$\binom{n}{k}$
clever fjordBOT
OmnipotentEntity
hollow quarry
ah yeah forgot it
clever fjordBOT
p1za
meager juniper
So are you claiming this approaches 1/e or e?
hollow quarry
1/e
meager juniper
Ok cool, sorry, I got it in my head somehow that you were doing e.
hollow quarry
no problem
meager juniper
Ok, jfyi because $\binom{n}{k} = \binom{n}{n-k}$ you can change this to be slightly nicer
clever fjordBOT
OmnipotentEntity
hollow quarry
$| \sum_{k=0}^{n} \binom{n}{n - k} \cdot (- \frac{1}{n})^{n - k} - \frac{1}{e} | < \varepsilon$ or rewrite it using factorials?
clever fjordBOT
p1za
meager juniper
One more step for full niceness
In this sum you're effectively "counting down" from n to 0
hollow quarry
yeah
meager juniper
What if we change n-k to k and count up instead?
We hit everything the exact same way, yeah?
hollow quarry
wait do you mean counting down the sum index?
meager juniper
No, counting up on the sum index
We are essentially turning "a + b + c" into "c + b + a"
By commutativity, this is the same
hollow quarry
you mean the n-k in the exponent right
hollow quarry
but why are we reversing the first step
hollow quarry
**OmnipotentEntity**
this
meager juniper
We have to have like vs like
We're doing a substitution everywhere
Letting n-k -> k
hollow quarry
so $| \sum_{k=0}^{n} \binom{n}{k} \cdot (- \frac{1}{n})^{k} - \frac{1}{e} | < \varepsilon$
clever fjordBOT
p1za
meager juniper
Exactly
hollow quarry
okay yeah sure
meager juniper
Ok, now we have the partial sum definition of 1/e minus 1/e
Pretty straightforward from here
hollow quarry
we didnt do any partial sums i dont think so
meager juniper
One can rewrite using factorials if that makes it more recognizable
Brb going to drive back home, like 5 minutes
hollow quarry
sure np
maybe for some context, we approximated e in class using nested intervals, so maybe thats what my professor wants me to use but Im not sure
meager juniper
Your definition of isn't exp(z) = sum z^n / n! ?
hollow quarry
no i havent seen that
wise karma
so $| \sum_{k=0}^{n} \binom{n}{k} \cdot (- \frac{1}{n})^{k} - \frac{1}{e} | < \v...
can u write this in latex pls
1/2 [ (1+x)^(2^n) + (1-x)^(2^n) ] this whole is the numerator and this is the denominator n * 2^n * pi notation (from k=0 to n-1) [(1+x)^(2^k) + (1-x)^(2^k))]
meager juniper
!occupied
odd edgeBOT
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).
hollow quarry
i think #latex-help is better suited, sorry
what I thought was if we could rewrite our term in a way, so we get the sum in the picture as the denominator?
@meager juniper u got any idea?
meager juniper
(1/n)^k = n^(-k)
then you have 1/k! n^(-k) time n!/(n-k)! which is a term that vanishes as n gets large.
well, I guess it doesn't actually... hmm...
how does this work?
hollow quarry
You mean how it approximates e?
odd edgeBOT
@hollow quarry Has your question been resolved?
lapis widget
how to solve this by hand?
odd edgeBOT
how to solve this by hand?
wise karma
how to solve this by hand?
u^2 common
lapis widget
u^2 common
wdym
wise karma
take u^2 out
lapis widget
i can break out u^2 yes but how does that help
ember oak
It doesn't
wise karma
yep
hexed marten
Multiply by 6 and solve the cubic eqn
ember oak
I would transform to have integer coefficients, then use RRT
lapis widget
oh yeah rational root theorem, lemme look this up agane
ember oak
If no rational roots, then gg with that general cubic
quasi sparrow
,w solve x^2 - x^3 / 3 = 1/6
clever fjordBOT
wise karma
no rational
signal oar
$u^3 - 3u^2 + \frac12 = 0$
wise karma
ig
clever fjordBOT
Kepe
ember oak
no rational
!xy
odd edgeBOT
no rational
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.
wise karma
signal oar
$u^3 - 3u^2 + \frac12 = 0$
You can always use the cubic formula (or a variation of the steps used to derive it)
But it won't be very nice
lapis widget
Please show the original problem, **exactly as it was stated to you**, with the ...
The original problem is finding in part b here the median of a piecewise defined probability distribution function for a transformed variable and i arrive at the expression i gave here
Really messy, almost illegible but these are the pdf and cdf i arrive at
i know that the median will be between 1<u<2
quasi sparrow
lapis widget
<:sully:651816820122189834>
what does this mean haha, my handwriting being non-existent or the problem being weird?
odd edgeBOT
@lapis widget Has your question been resolved?
nimble garden
what does this mean haha, my handwriting being non-existent or the problem being...
What is the question you are stuck in?
lapis widget
What is the question you are stuck in?
Basically solving this by hand just using a normal calculator if needed
Its for finding the median u here for a given probability distribution function
ember oak
Basically solving this by hand just using a normal calculator if needed
Just use a calculator tbh
if this is for stats
lapis widget
Just use a calculator tbh
it is, not sure if my calculator can do this tho
ember oak
it is, not sure if my calculator can do this tho
online calculator
Generally, you can't do cubics by hand
If RRT fails, then your real solution is going to be very ugly
lapis widget
online calculator
it will be on exam with a normal non graphing calculator
i heard a rumor prof said mbe leaving the expression is enough
but hey wanted to be sure
ember oak
If you are expected to use the cubic formula in a stats exam, you are being taight the wrong things
ember oak
i heard a rumor prof said mbe leaving the expression is enough
Highly doubt for cubics. Most real solutions have to be written as a sum of cube roots of complex numbers. That is just nonsense
lapis widget
Highly doubt for cubics. Most real solutions have to be written as a sum of cube...
reasonable! I'll focus on other stuff then, this exam is massive anyways.....
thx guys!
.close
odd edgeBOT
odd edgeBOT
Available help channel!
heady ether
odd edgeBOT
olive fjord
so f"(x)=0
Will give you the point where it changes from concave down to concave up
correct?
heady ether
yes
olive fjord
So then the concave donward interval must include -3 as an endpoint
So it should be (-3, a) or (b, -3)
Now we only have an option that looks like this
heady ether
b ?
olive fjord
Yes
heady ether
1 and -1 but 1 double root
olive fjord
now it is already factored out so it's easy
olive fjord
1 and -1 but 1 double root
yes
heady ether
should ı take 1 or not
olive fjord
Do double roots not count?
I don't know
I think they should right? Or is there a rule about them?
heady ether
ıdk really
odd edgeBOT
@heady ether Has your question been resolved?
ocean bramble
are you allowed to use l'hôpital?
charred heart
are you allowed to use l'hôpital?
omgim so stupid
bro i thought 1/x as x-> inf was inf
heady ether
are you allowed to use l'hôpital?
ı dont know actually
ocean bramble
if you're allowed, i'd use that
heady ether
if you're allowed, i'd use that
can u solve ?
ocean bramble
i could, but the idea is that you have to do it yourself
hm, if you dont know lhopital rule i'm gonna need to know what rules you were given
odd edgeBOT
@heady ether Has your question been resolved?
rancid solar
Are Segments from a Point Equidistant to Two Points on Intersecting Lines Always Perpendicular? if so WHY ?
are lines AP and BP always perpendicular
vale vapor
imagine you draw a circle with center P and that goes through A and B
rancid solar
imagine you draw a circle with center P and that goes through A and B
oooooooooh! i got it
crude nacelle
Any tips to prove this?
I know that if x is in the left hand side:
Or x is in A or x is in B AND C
the rigth hand side
or x is in A or in B and C
But I don't know how to write this.
viscid flint
to show two sets are equal, the strategy is often to show that they're subsets of each other
so let $x \in A \cup (B \cap C)$ and show that $x \in (A\cup B)\cap(A\cup C)$
clever fjordBOT
hayley (measurable)
crude nacelle
to show two sets are equal, the strategy is often to show that they're subsets o...
So either $x \in A or x \in B and C$
clever fjordBOT
Guilhotina
odd edgeBOT
@crude nacelle Has your question been resolved?
weary pelican
So either $x \in A or x \in B and C$
yes that's a good start
if x is in A, is x in the other set? similarly in the other case
crude nacelle
yes that's a good start
Can I make the rigth hande size like this?. Since A is in both union, its the same as writing A U (B intersec C) ?
weary pelican
Can I make the rigth hande size like this?. Since A is in both union, its the sa...
I don't understand what you mean
weary pelican
Can I make the rigth hande size like this?. Since A is in both union, its the sa...
we are in fact trying to prove that the right hand size is this
so let's not jump the gun
crude nacelle
This is what I have:
until now
Can I assume that X is the only element in the left hand size ?
weary pelican
Can I assume that X is the only element in the left hand size ?
uh no
there are probably others
also don't confuse $subset\subset set$ with $element\in set$
clever fjordBOT
rafilou is not not born in 2003
weary pelican
oh are you asking
ok I think I get your question
'x' here is arbitrary
it's a random element of A U (B n C)
crude nacelle
also don't confuse $subset\subset set$ with $element\in set$
Yes, but if $ x \in A x \subset A$, right ?
weary pelican
Yes, but if $ x \in A x \subset A$, right ?
no
$x\in A$ is the same as ${x}\subseteq A$
clever fjordBOT
rafilou is not not born in 2003
crude nacelle
oooo you're right
weary pelican
a pencil might be in a box with a pencil and sharpener
but the pencil itself is not a "box"
it doesn't contain anything
crude nacelle
a pencil might be **in** a box with a pencil and sharpener
yess
weary pelican
alr
so
taking a random element x of A U (B n C)
you've shown if x is in A, then x is in the desired set
so what if x is in B n C?
crude nacelle
taking a random element x of A U (B n C)
How can I write that, so be clear to my professor ?
weary pelican
"Let $x\in ...$"
clever fjordBOT
rafilou is not not born in 2003
weary pelican
and if your professor is very much into being rigorous
they might ask you to put "logic connections" between statements
if you have something like:
"Statement 1
Statement 2"
and statement 2 is a consequence of statement 1
you should write
"Statement 1
then/thus Statement 2"
take this part for example
If your prof was really into logical words
you would write instead:
If $x\in A$ (replace the down arrow with this snippet)
then $x\in A\cup B$
and $x\in A\cup C$
Thus $x\in (A\cup B)\cap (A\cup C)$
clever fjordBOT
rafilou is not not born in 2003
weary pelican
hope this makes sense
crude nacelle
Is this right ?
crude nacelle
hope this makes sense
Yes, i get about the in part but don't understand how this prove that both are equal
weary pelican
Uh
Our objective for now was only to arrive to x in (A U B) n (A U C)
So you should have stopped at line 3 for both cases
Because that ends the proof of (first set) subset (second set)
Then and only then, we start showing the other inclusion
crude nacelle
**Then** and only then, we start showing the other inclusion
I dont get it
weary pelican
I dont get it
say we want to show $X = Y$
clever fjordBOT
rafilou is not not born in 2003
weary pelican
(both X and Y are sets)
that's the same as showing
$X\subseteq Y$ and $Y\subseteq X$
clever fjordBOT
rafilou is not not born in 2003
clever fjordBOT
rafilou is not not born in 2003
weary pelican
by picking an arbitrary $x\in X$
clever fjordBOT
rafilou is not not born in 2003
weary pelican
and show that $x\in Y$
clever fjordBOT
rafilou is not not born in 2003
clever fjordBOT
rafilou is not not born in 2003
weary pelican
which is done by picking an arbitrary $y\in Y$
clever fjordBOT
rafilou is not not born in 2003
weary pelican
and showing $y\in X$
clever fjordBOT
rafilou is not not born in 2003
weary pelican
**rafilou is not not born in 2003**
once you've shown this
the part of the proof where x is involved is DONE
you're not going to use the same x to show $x\in X$
clever fjordBOT
rafilou is not not born in 2003
weary pelican
since that's where we picked it from
weary pelican
**rafilou is not not born in 2003**
that's how we showed that
weary pelican
**rafilou is not not born in 2003**
and this
crude nacelle
that's how we showed that
Yes, but I dont understand why $x \in$ A U B n A U C implies that A∪(B∩C) ⊆ (A∪B)∩(A∪C)
clever fjordBOT
Guilhotina
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
weary pelican
Yes, but I dont understand why $x \in$ A U B n A U C implies that A∪(B∩C) ⊆ (A∪B...
you've taken a random x
in A∪(B∩C)
and you've shown that it must also be in (A∪B)∩(A∪C), all of the time
so no matter the element in the first set
the second set also has it
isn't that the definition of (first set) ⊆ (second set)?
crude nacelle
isn't that the definition of (first set) ⊆ (second set)?
i missed this step, I must assume this to do the proof, rigth; ?
weary pelican
i missed this step, I must assume this to do the proof, rigth; ?
no
you've shown it
weary pelican
and you've shown that it must also be in (A∪B)∩(A∪C), all of the time
in this step
you've shown the definition of (first set) ⊆ (second set)
so (first set) ⊆ (second set)
crude nacelle
So if an arbitrary X is in set A and set B, this must mean that A is a subset of B
?
weary pelican
not exactly it
If an arbitrary x that is in set A must also be in set B, then this must mean that A is a subset of B
odd edgeBOT
@crude nacelle Has your question been resolved?
weary pelican
So this is the second part
uh
how do you know either y in A or (y in B and y in C)
because that's what we're trying to prove
that picking an arbitrary y in the second set now
y would also have to be in the first
crude nacelle
how do you know either y in A or (y in B and y in C)
I apply the same concept that we applied in the X.
with X we said that X in set A and then prove it was also in B
Now I pick and arbitrary Y that is in set B and prove it was inA
quasi nebula
stiff kettle
odd edgeBOT
stiff kettle
my brains not braining
ik I need to find h(k(x)) and k(h(y))
but
THERES NO X OR Y
chat gpt says this but i dont get it
how can u just plug that in if it was t
does someone understand what im trying to say
hasty dome
Do you see how (h o k)(x) is h(k(x))?
static totem
you can put x into h(t) =
stiff kettle
ya i know that
so why are the variables different then
why cant it just be k(x) = 2 pix
static totem
because the letters dont matter
hasty dome
With k(p) = sin(2 pi p), you have p inside the k parentheses.
Whatever you replace p with on the left, you replace p with that on the right as well.
So,
k(p) = sin(2 pi p)
k(x) = sin(2 pi x)
You replaced the p with an x on the left, so you replace the p with an x on the right.
stiff kettle
mhm
stiff kettle
mhm
hasty dome
k(x) = sin(2 pi x)
So, you can replace k(x) with sin(2 pi x).
That's why h(k(x)) can become h(2 pi x).
Does that make sense?
hasty dome
OK, now we have:
h(t) = 3t^2 + 1
h(2 pi x) = 3**(2 pi x)**^2 + 1
stiff kettle
so after i do that do i just leave it
stiff kettle
OK, now we have:
h(**t**) = 3**t**^2 + 1
h(**2 pi x**) = 3**(2 pi x)**^2 + 1
yep i got the same thing
wait what about the sin
3(sin(2 pi x)^2) + 1
hasty dome
Ahh, sorry.
stiff kettle
its ok i also forgot it LOLLL
hasty dome
So, you get 3 sin^2(2 pi x) + 1.
stiff kettle
yep
hasty dome
OK, so what would k(h(y)) become?
stiff kettle
i got
sin(2pi(3t^2 + 1))
i did it pretty fast idk if its ight
oops
instead of t its y
thorny herald
<@&286206848099549185> Prove that a circle is not a polyhedral convex set
fathom mulch
lol probs a troll
More like someone who's desparate
stiff kettle
LOL
hasty dome
sin(2pi(3t^2 + 1))
Almost.
k(h(y))
k(3y^2 + 1)
sin(2 pi (3y^2 + 1))
stiff kettle
yep
hasty dome
Oh, you found it.
hasty dome
Yeah.
stiff kettle
thank you so much though you actually explained it very well
hasty dome
You're welcome.
odd edgeBOT
odd edgeBOT
Available help channel!
wary forge
odd edgeBOT
wary forge
i tried using by parts
but it got more confusing
solutions online are confusing too
ping me please
wooden gorge
!show
odd edgeBOT
Show your work, and if possible, explain where you are stuck.
wooden gorge
Tried using IBP on I_{n+1} ?
wary forge
wait
idk what to do from here
i can maybe apply ibp again but it wouldnt make sense
wary forge
Tried using IBP on I_{n+1} ?
no
wary forge
just replace n by n+1 in this right?
wooden gorge
It's not x^{k-1} but x^k
wary forge
why
wary forge
doesnt k come down cuz of power rul
wooden gorge
Because you forgot to multiply the x from \int 1
wooden gorge
And then you rewrite x^k = 1 - (1 - x^k)
wooden gorge
Ofc
wary forge
wait
wooden gorge
What'd you get
wooden gorge
Something feels wrong though but if it could just be me
wary forge
?
wooden gorge
Should've been but the question disagrees
wary forge
whatever it is now i know how to solve it
next time ill do the same
thank you
.close
odd edgeBOT
odd edgeBOT
Available help channel!
true musk
Trying to evaluate $$ \int_1^{e^{\pi/4}} (1+{\ln^2{x}})^{-1} dx $$
I was able to convert it to
$$ \int_0^{\pi/4} \frac{e^u}{1+u^2} du $$
clever fjordBOT
MetuMortis
low locust
WA ends up just giving the result in terms of the exponential integral Ei(x)
low locust
if you want that I think you should be able to do partial fractions for 1/(1+u^2)
wolframalpha
true musk
if you want that I think you should be able to do partial fractions for 1/(1+u^2...
It is actually $$ e^u \cdot (arctan(u))' $$
clever fjordBOT
MetuMortis
low locust
yes. and?
true musk
IBP?
low locust
I have a feeling it wont help
you'd be left with integrating e^u arctan(u) which also seems horrid
(and WA doesnt find a closed form without Ei(x) so thats a bad sign)
odd edgeBOT
@true musk Has your question been resolved?
true musk
<@&286206848099549185>
odd edgeBOT
@true musk Has your question been resolved?
odd edgeBOT
odd edgeBOT
Available help channel!
lofty sigil
odd edgeBOT
lofty sigil
Hi I'm here cus I'm confused af on something basically I have the anwser sheet and it says the anwser is 56
Can someone help me
sand horizon
Which ?
viscid flint
to test if something is divisible by 12, you can test if it's divisible by 3 and 4
trim trellis
to test if something is divisible by 12, you can test if it's divisible by 3 and...
no
the question is
20224ab8'
idk what it is
its like
quesiton is made wrong
lofty sigil
to test if something is divisible by 12, you can test if it's divisible by 3 and...
Yea but idk
lofty sigil
quesiton is made wrong
The thing is I don't think so cus the question were made by SASMO
trim trellis
the number with those as digits
wdym
trim trellis
The thing is I don't think so cus the question were made by SASMO
ohh
oh the bar means those are digits?
viscid flint
like it might be 20224698
lofty sigil
like it might be 20224**69**8
Oh wait ●_● I think ik
Cus as you said we have to check if the number can be divided by 4 and 3
And cus there's 8 we don't need to worry about the 4
And we need a + b to be divisible by 3 but the largest number so it's 7 and 8
□_□ I just sloved my own question hm
.close
odd edgeBOT