bruh alr
#help-19
solar marsh
how abt this
i have no idea how to start
i tried to put the Cs into the different groups and add them up
wooden python
you need to break down into cases based on where the O's are
COO C__ ___
CO_ CO_ ___
CO_ C__ O__
C__ C__ OO_
work out the number of possibilities in each case separately
note that some of the trios may be interchangeable, which will reduce the number of possibilities
solar marsh
o yea thats just me i usually put it to see the working better
for the second one i have no idea
so like 5C1*4C1?
dim sentinel
Im back and please save me
odd edgeBOT
Im back and please save me
dim sentinel
I don't wanna do this for this long
I might actually surrender
Same figure as before
Angle ABC=38 is the given
Using the quadrilateral idk makes minor arc AC=142 then that means angle AXC=142 TOO
Now how do i get the measurement of major arc ADC
Uumt
odd adder
lol
odd adder
Now how do i get the measurement of major arc ADC
oh wait btw i didn't see that you already got minor arc AC
odd edgeBOT
odd edgeBOT
Available help channel!
novel venture
Hello, im preparing for a math olympiad exam type exam... and honestly i have no experience with olympiad math, i have done highschool math in a good depth where it was just calculation and slight manipulation with mcq questions. In olympiad i am supposed to know every theory in great detail. My mind blanks out whenever i see a question where i have to prove something. I dont even know where to start, i feel very helpless, the thought process seems so ambiguous. And throughout my highschool math, ive learnt through videos, its how i learnt best, by knowing their thought process in each step which also looks logical, but in olympiad math i feel helpless, is there any way to undo this(especially in number theory and geometry) any resources which will be helpful or any general tips on how to not ragequit? (Im not really that much of a book guy, i already have a book dedicated to it but i just cant do the problems no matter how much i try it feels like it wont click)
novel venture
Also the fact that i cant dedicate much time, i dont have a lot of time, around 3 months or so
odd edgeBOT
@novel venture Has your question been resolved?
fading bloom
this is for homework
taking a linear algebra class and it's going ok but i don't understand what i'm supposed to be doing for the questions in the form of ```
Find the equation of the plane through the point P that is perpendicular to the line r=(ai + bj + ck)+t(di + ej + fk). Write the equation in the form indicated: qx+ry+sz=v.
i've removed the exact numbers from here for obv reasons but i don't understand how to go about doing this
i would think my first steps would be to get the equation for r into general equation form but i don't know how to do thati would think for a question like this the point doesn't matter
eager wraith
do you know about the vector point-normal form of a plane?
wooden python
in your notation, the point thru which P must pass affects the value of v
fossil sail
not if it is homework i dont think
wooden python
is it against the rules for me 2 post the question verbatim ?
no it's not... in fact we prefer you do this
eager wraith
perpendicular to the line means the direction vector of the line is the normal vector of the plane
warped glacier
and if you multiply any vector by 2, it still points in the same direction
warped glacier
the answer i got was like 2x+20y-6z=16
wrong normal vector
fading bloom
whag
wooden python
the answer i got was like 2x+20y-6z=16
how did those 20 and -6 happen
cold swift
what’s the direction vector for the line
wooden python
yeah so you should scale this up to make the x coordinate 2
(because the first term is set in stone as 2x)
fading bloom
i don't really understand why
lone elbow
perpendicular to the line means the direction vector of the line is the normal v...
do you understand this robin?
fading bloom
yeah that makes sense
eager wraith
scaling a vector doesn't change the direction it points in right
fading bloom
yeah but why am i scaling it
fading bloom
the input being what here
wooden python
answer input
eager wraith
my memory is a bit shaky here but iirc, for an equation of the form ax + by + cx = d, the vector (a, b, c) is the vector normal to the plane (correct me if i'm wrong)
lone elbow
after you found out the normal (A,B,C)
you plug the point the plane passes through in
Ax + By + Cz = D
and you find D
eager wraith
you could also expand out the vector point normal form but that's just extra effort
fading bloom
the thing i was confused about was i did <1,5,-2>*<2,4,3> to get those values which results in <2,20,-6> which already started with a 2 which was why i didnt think about that
oh i am reading the wrong thing
i need more sleep
it makes sense now
.close
odd edgeBOT
cold zinc
I have a question about limits
odd edgeBOT
Available help channel!
cold zinc
odd edgeBOT
cold zinc
So i have problem m here
wooden python
uh huh
cold zinc
The solution is -1
But i dont get how thats possible because of the rule of sums or however uts called
This rule
Tge limit of each fraction individually is undefined
woeful briar
This rule is only when both limits exist and are finite
wooden python
sum rule is inapplicable if a or b or a+b don't exist
cold zinc
So how can they have a limit together
cold zinc
This rule is only when both limits exist and are finite
Ohh
odd edgeBOT
odd edgeBOT
Available help channel!
mystic saffron
odd edgeBOT
woeful briar
Is the modulus term negative or positive?
if x < 4 then x - 2 < 2
so x-2 can be both positive and negative, depending on x
mystic saffron
Oh
On solving i get a and b = 1, but book says a = 1 and b = -1
Which means that term is negative, only then we get b=-1
mystic saffron
so x-2 can be both positive and negative, depending on x
Isnt it +ve since we consider left neighborhood point,which approaches from left side, which is very very close to 4, which means its greater than 2 right?
woeful briar
Isnt it +ve since we consider left neighborhood point,which approaches from left...
if x is approaching 4 from the left side then x < 4
if its approaching from the right its x > 4
mystic saffron
Ye, but we consider some number really close to 4 from left side, right?
So LT x->4^- such that 4-h, where h is very small positive number right?
Am i wrong?
mystic saffron
The ans depends on the sign of mod of (x-2)
I took +(x-2), but im wrong according to book
Book took -(x-2)
Which is correct?
U here?
odd edgeBOT
@mystic saffron Has your question been resolved?
mystic saffron
.close
odd edgeBOT
odd edgeBOT
Available help channel!
lusty quiver
odd edgeBOT
late dust
+21 or -21 ?
lusty quiver
Yea
brazen hare
-3 * -7 = ?
late dust
Not sure how to help you with that
errant dew
$…-3(x^{2}-7)=…+(-3)\times(x^{2}-7)$
clever fjordBOT
BBMaths
tender cloak
quasi sparrow
.close
mystic saffron
I solved it like this, I moved F2 in the place of CD while its perpendicular on F1 cause i think its fine to move vectors as long as its parallel to the original vector position, i moved it to make a right angled triangle and then i just solved for F1 (adj) and T (hyp) and the play around with it a little to get the value of T which should be D but somehow its not. Can someone check?
mystic saffron
the guide answer i checked from could potentially be incorrect so until i ask my teacher can someone solve it?
gloomy cipher
letme try
errant dew
Multiply both sides by cos(20)
mystic saffron
Multiply both sides by cos(20)
which step?
errant dew
Oh sorry I was starting at the answer
tacit wasp
errant dew
Yeah it’s correct
tacit wasp
I solved it like this, I moved F2 in the place of CD while its perpendicular on ...
You already have the answer
errant dew
$T\cos(20^\circ)=F_{1}$
mystic saffron
ah so it should be D?
clever fjordBOT
BBMaths
mystic saffron
the guide answer had something else so i got confused
errant dew
None of the other answers are correct
odd edgeBOT
gloomy cipher
None of the other answers are correct
d is
gloomy cipher
mb baby
odd edgeBOT
Available help channel!
vivid orchid
how did the partial red gp series formed by the initial yellow series in the question what property was used to bifurcate it, and write different smaller series im kinda confused as to how to take common term out of the progression.
errant dew
I’m not sure what you’re asking as the question
odd edgeBOT
@vivid orchid Has your question been resolved?
vivid orchid
I’m not sure what you’re asking as the question
like howd the partial red series came from the yellow one
ignore the green part
errant dew
Imagine a times table grid where the input row is 1, 1/4, 1/16, … and then the column is 1, 1/3, 1/9, … (but only fill in the ones you want)
You will get all the fractions
The red sum is summing the fractions reading them row by row and the other is reading them column by column
vivid orchid
oh okay got it, but when do we apply this logic , only in combination series like these?
@errant dew
errant dew
$$\begin{bmatrix}
1 & & & \
1/4 & 1/12 & & \
1/16 & 1/48 & 1/144 & \
\vdots & \vdots & \vdots & \ddots
\end{bmatrix}$$ The red and yellow sums comes from summing either horizontally or vertically first and then doing the other one
odd edgeBOT
errant dew
.reopen
odd edgeBOT
✅ Original question: #help-19 message
errant dew
In general, swapping the orders of summing terms can actually change the answer, but since all the terms are positive we don’t have any issue
vivid orchid
In general, swapping the orders of summing terms can actually change the answer,...
oh okay
got it
clever fjordBOT
BBMaths
static locust
hey if i have 555= + 2 |x-4| + 5
odd edgeBOT
hey if i have 555= + 2 |x-4| + 5
modest nacelle
Yes
modest nacelle
Don't remove the absolute function first.
static locust
like that?
chilly ridge
you can’t + 4 to both sides
at this point you should consider both the positive and negative case
static locust
at this point you should consider both the positive and negative case
how
modest nacelle
a = |b| means a = b or a = -b
steep mantle
static locust
so how do i know which one it is
chilly ridge
lol huh
modest nacelle
Well, depending on context.
static locust
u can have negative meters?
chilly ridge
maybe?
modest nacelle
Yes
static locust
i see
chilly ridge
what’s the question
static locust
i forgot now but i did it like thismorning
it asked me for the amount of meters when the altitude was 555
something like that
chilly ridge
😭
shrewd trellis
odd edgeBOT
modest nacelle
What are you doing with f(x)
$f(x) = \frac{e^x-e^{-x}}{2}$, not $f(x) = e^x-\frac{e^{-x}}{2}$
clever fjordBOT
Xwtek
shrewd trellis
**Xwtek**
oh i wrote it wrong
I think I needed to put parantehsis so the 1/2 is distrubuted evenly
amber schooner
Both functions are always positive
no
you don't just have to show that
false
oh is this a different question
😭
i thought this was the one from before
shrewd trellis
this was my answer
amber schooner
did you figure out the one from before?
i don't know that i like your argument at the end
not very rigorous
probably better to say e^x > 0 so 1/e^x > 0 and from that the sum is > 0 and multiplying by 1/2 is also > 0
also you wrote $\frac{1}{2e^{-x}}$
clever fjordBOT
knief
amber schooner
🤔
shrewd trellis
Which line
Are you referring
I don’t understand up to here " and from that the sum is > 0 and multiplying by 1/2 is also > 0"
odd edgeBOT
@shrewd trellis Has your question been resolved?
shrewd trellis
this was my answer
<@&286206848099549185>
brazen hare
this was my answer
Your last statement "Plug in..." is unacceptable
shrewd trellis
got it but
shrewd trellis
I don’t understand up to here " and from that the sum is > 0 and multiplying by ...
^
brazen hare
lmao
brazen hare
I don’t understand up to here " and from that the sum is > 0 and multiplying by ...
e^x > 0, you know that right?
brazen hare
so 1/(e^x) > 0
brazen hare
do u mean the range is greater than 0?
i mean that the the quantity e^x + e^(-x) is greater than 0 for all x
clever fjordBOT
brazen hare
notice that this is the same expression as your f'(x)
brazen hare
you can get directly from line 1 to line 3
shrewd trellis
i mean that the the quantity e^x + e^(-x) is greater than 0 for all x
for all x? I dont see it
brazen hare
for all x? I dont see it
wdym
is there a value of x for which that quantity would not be greater than 0?
shrewd trellis
you can get directly from line 1 to line 3
first to the third line?
brazen hare
first to the third line?
yeah
brazen hare
put your therefore symbol ($\therefore$) on the last line not before e^x > 0
clever fjordBOT
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
shrewd trellis
that is in the last line
shrewd trellis
make a new line for what?
brazen hare
f'(x) > 0
cause that is what we're ultimately trying to show
and then move your therefore symbol to that last line
brazen hare
correct?
yeah
i would say so
but of course
if your teacher takes issue with it, adapt to what he or she wants
odd edgeBOT
wait what about for this
brazen hare
wait what about for this
i think having the $\implies$ is important here, so that one can follow the logic
clever fjordBOT
brazen hare
oh is this a different problem?
shrewd trellis
same question but different part
brazen hare
and i assume f''(x) = -1/(2e^x)?
shrewd trellis
Im trying to do e rn
to sketch it, I would need the nature of the graph right?
or maybe not, since there is no stationary points
because f'(x)>0
I havnt learned how to sketch hyperbolic exponential functions but my prof said we can sketch this since we have enough info
brazen hare
what did you get for your x and y intercepts?
shrewd trellis
0
well I got y intercept as 0 but I stated that there is no x intercept
since i coudnt solve for x
brazen hare
are you sure?
brazen hare
x-intercept occurs when y = 0
shrewd trellis
yes
shrewd trellis
but why cant i find the x intercept from here
brazen hare
just solve for x
yes, it's true that e^x and e^(-x) cannot be 0
but their difference can
shrewd trellis
is it because their bases are the same? even if they are one is a -, while the other is +
short terrace
brazen hare
@shrewd trellis uhh do you really need help solving for x in the equation
brazen hare
nope I got it, but the graph yes
so what's the x-intercept?
shrewd trellis
0,0
brazen hare
alright
shrewd trellis
and its an odd function so it reflects on the origin
jaunty minnow
Could we do 25?
odd edgeBOT
Could we do 25?
jaunty minnow
I’m so confused
I got to a point where y’ is on both sides
But idk where to go from there
quasi sparrow
set z = y' and solve for z
quasi sparrow
set z = y' and solve for z
yea now just do this
jaunty minnow
One sec
jaunty minnow
yea now just do this
But how???
It doesn’t make sense
Either way it would lead me to the samething
quasi sparrow
It doesn’t make sense
did you try it
jaunty minnow
Ye
jaunty minnow
Still doesn’t make sense
jaunty minnow
quasi sparrow
where did you try to solve for z
quasi sparrow
set z = y' and solve for z
the second half of this
jaunty minnow
Uhhh
How
do
i
solve
for
z
subtract z?
hold up
let me try something
is the answer
nvm
I'm stuck on
y^-1/2 times z + z = 1
odd edgeBOT
@jaunty minnow Has your question been resolved?
quasi sparrow
y^-1/2 times z + z = 1
$az + z = z(?)$
clever fjordBOT
riemann
tribal gull
need help with this integral
odd edgeBOT
need help with this integral
tribal gull
well i did the sub. found x= arctan(t). and dt = 1/(1+t^2) dt
so integral of 1/(1+cos^2(arctan(t))) * 1/(1+t^2) dt
found cos(arctan(t)) = 1/sqrt(t^2+1)
got stuck there
rotund steeple
need help with this integral
lmfao this is funny
they intended a weierstrass substitution but you can just divide both sides by cos squared and manipulate the denom into tan
clever fjordBOT
k
tribal gull
yeah thats is what i have rn well except the (1/1+t^2) is squared
i am unsure if i should multiply out the (1/1+t^2)^2 and somehow figure it out from there but i feel like it would get way too complicated and im unsure if thats really the right way
narrow crypt
yeah thats is what i have rn well except the (1/1+t^2) is squared
[ x = \arctan(t) = \arccos\left(\frac{1}{\sqrt{1+t^2}}\right) \implies \cos^2(x) = \frac{1}{1+t^2}? ]
tribal gull
ohh
clever fjordBOT
k
clever fjordBOT
k
narrow crypt
Hence the hint
tribal gull
ohhhh yeahhhhhh
would i end up using the integral of 1/a^2+x^2 = 1/a *arctan(x/a) + C formula
narrow crypt
👍
tribal gull
so would the answer be 1/sqrt(2) * arctan (tan(x)/sqrt(2)) + C?
narrow crypt
[ \frac{1}{\sqrt{2}}\arctan(\frac{\tan(x)}{\sqrt{2}}) + C]
Let’s see
,w int 1/(1+cos^2 x) dx
narrow crypt
Wonderfuul
odd edgeBOT
odd edgeBOT
Available help channel!
raven finch
odd edgeBOT
raven finch
am i tripping how the h**k is num 3 not linear
y' = yt^2
or known as y' - yt^2 = 0
should that not be linear?
wooden python
it isn't linear until manipulated ig
raven finch
ok then follow up question
cuz the answer to this one was C
but the answer to
from the same exam maker
wooden python
raven finch
same for this one , A is correct
but u have to manipulate it
for refernce
i think ima just cry
cuz thats so inconsistent
warped glacier
it's not much of a rearrangement
a first-order linear equation is of the form $y' + Py = Q$ for functions $P, Q$
clever fjordBOT
south
warped glacier
you have $y' + (-1/t) y = 0$, so it matches with $P = -1/t, Q = 0$
clever fjordBOT
south
raven finch
shouldnt 3 be linear here then ?
warped glacier
similarly, these can be rearranged into $y''' - y' + y = 0$, $y'' - y \sin t = 0$ and $d$ is $y''' - ty = 0$
clever fjordBOT
south
raven finch
mb
warped glacier
but if you have y'' + |y| = 0, that's automatically non-linear
you can't have |y|
you can have sin(t) * y + (1 - e^t + 2^(pi t) * y) * y' = cos(t)
in short, any function of t multiplied by any derivative of y
and the right hand side can be any function of t
warped glacier
or known as y' - yt^2 = 0
this is linear
raven finch
ok so typo in question
warped glacier
not a typo, the correct option is d
(2), (3), (4) are all linear
raven finch
not a typo, the correct option is d
(2), (3), (4) are all linear
or yea , typo in answer key
warped glacier
or yea , typo in answer key
oh right I didn't know what the answer key said for that
raven finch
why does gpt say its 2 & 4 only tho ?
warped glacier
why does gpt say its 2 & 4 only tho ?
that's a reason to not trust GPT
raven finch
oh right I didn't know what the answer key said for that
yea said it was only 2 &4
warped glacier
GPT doesn't know anything
raven finch
kk i doubted myself cuz gpt and answer key both aligned
warped glacier
there's been a recent news announcement by Claude AI that if 0.002% of the data has wrong information, the entire language model will give you wrong information if you prompt it a certain way
in short, these models assume you can trust everything and everyone on the internet
and that's of course not true
warped glacier
yeah.....
warped glacier
was losing my mind for half and hour
no worries!!!
graceful parrot
Can someone expkain to me why dr isnt -3 to 3 and theta isnt 0 to 2pi
mental lotus
the domain over which you integrate is all in positive y half space
mental lotus
you dont integrate over any negative y
mental lotus
because of the sqrt
yea
mental lotus
so your theta is only 0 to pi, and corresponding radius is also positive
odd edgeBOT
warped glacier
the domain over which you integrate is all in positive y half space
wait, you mean positive z right
mental lotus
wait, you mean positive z right
no, I mean yes, but OPs question is related to y
z gets copied as is
warped glacier
oh right okay I didn't know what they were asking
mental lotus
its just cartesian to cylinderical coordinate system
graceful parrot
,w (9(3)^3)/3 - (((3)^5)/5)
clever fjordBOT
warped glacier
because of the sqrt
yes, the bounds for dy tell you 0 <= y <= sqrt(9 - x^2)
odd edgeBOT
Available help channel!
keen mural
what does "projection onto the section component" mean?
green elm
odd edgeBOT
@keen mural Has your question been resolved?
merry finch
keen mural
i think it means projection onto the second component
i don't quite remember what a projection is defined as though
solar marsh
no idea how to start, tried to just do
odd edgeBOT
no idea how to start, tried to just do
solar marsh
do 7C2 5C2
pastel steeple
do 7C2 5C2
i believe they mean the two c's, if used, must be in different groups, not that the two c's must be in the groups separately
wait i am dumb
three groups lol
i read two idk why
odd edgeBOT
vernal yacht
nvm i got it now im just stupid i just needed to lay out the things
You mean list all the possible conditions?
solar marsh
You mean list all the possible conditions?
yea
vernal yacht
Sure, but you might want to post it here as well for double or triple checks by helpers
Lest you miss anything
odd edgeBOT
Available help channel!
novel venture
The product of the last two digits of
(1919)^1919 is?
bitter folio
The product of the last two digits of
(1919)^1919 is?
Gist of it: reduce it to 19^1919 mod 100, then note that 19^phi(100) = 1 mod 100 by euler totient theorem/fermat little theorem
Then reduce it further, can you try that?
novel venture
Gist of it: reduce it to 19^1919 mod 100, then note that 19^phi(100) = 1 mod 100...
I just keep getting a loop trying to reduce
bitter folio
I just keep getting a loop trying to reduce
Show your work
Its supposed to be a loop
bitter folio
Gist of it: reduce it to 19^1919 mod 100, then note that 19^phi(100) = 1 mod 100...
.
low locust
Gist of it: reduce it to 19^1919 mod 100, then note that 19^phi(100) = 1 mod 100...
phi(100) is not 99
bitter folio
phi(100) is not 99
oh mb
odd edgeBOT
@novel venture Has your question been resolved?
gaunt loom
.
eager tinsel
I don't know how to solve this problem on combinatorics (don't understand the logic behind it) can anyone help me out? Here is it is: There are 7 chairs with 7 different colors, red, yellow, orange, green, blue, purple and pink. How many sequences is it possible to form in a way that the red, yellow, and green stay together, while yellow between the 2 other chairs.
Notice I said the 2 other. The exercise does not specify if it should be a sequence of 6 or 7.
My first reasoning was 7!/(7-6)!, then I'd devide by 4 to get the position of those chairs, can anybody help me out?
nocturne brook
I don't know how to solve this problem on combinatorics (don't understand the lo...
so yellow is between red and green, and the remaining colors are present, for a total of 7 colors? that's how I read it
eager tinsel
so yellow is between red and green, and the remaining colors are present, for a ...
yes... that basically it, the question is kind of confusing on pourpose. You know how exam questions are...
nocturne brook
yes... that basically it, the question is kind of confusing on pourpose. You kno...
is that the original wording of the question, or is this a translation?
eager tinsel
is that the original wording of the question, or is this a translation?
no, I'm portuguese, this is a schoolbook translated question
as the teacher doesn't help me a lot, I have to learn from khan academy
the doubts were accumulating
and chat gpt wasn't cutting it
so I decided to take the largest doubt here
nocturne brook
let's just assume there are 7 chairs, and yellow is meant to be between red and green
eager tinsel
I have the solution
nocturne brook
makes sense yeah
nocturne brook
I don't know how to solve this problem on combinatorics (don't understand the lo...
so explain your logic here
how do you count the permutations?
eager tinsel
so explain your logic here
there are 7 chairs, I thought 7!
After that
I needed to account for the chairs
If the 3 are together in a specific order
it's
4!
unless
it's on other postions (AAABBBB, or BAAABBB - be A the chairs green, yellow or red),
so it's actually 4!*5
right?
nocturne brook
4!*5 is better known as 5!
but what thing are you counting when you say "it's 5!"?
eager tinsel
well, the total ammount of possibilities right?
oh wait
you can swap them green with red
so.. 5! * 2 = 240, is that it?
nocturne brook
it sounded like you wanted to start from all 7! permutations at first, so the logic didn't flow. you probably have the logic right, but you didn't really say the words in a convincing way
eager tinsel
wdym?
nocturne brook
but yeah that's it. you can assume "ryg" is all one chair, and permute the 5 chairs. for every permutation with "ryg", you can swap r and g to get another unique permutation, doubling your list
so you double the list of 5! permutations featuring "ryg" to include those featuring "gyr", and you have 2*5! = 240 total
eager tinsel
Well yeah... it wasn't obvious when I looked at the schoolbook
but after I realized something in the question, (so an interpretation issue) it was easy
two more questions: to improve this thinking logic ability, it's just practice right? How much pratice?
and lastly: how do I close this?
nocturne brook
practice is good yeah. I always thought combinatorics logic was kinda hard to enact confidently (like, you can make an argument that sounds good, but it has a flaw that's hard to find). so you can make practice more meaningful by trying to explain your solutions, and why they're correct
how much practice is hard to say, because it differs from person to person
and, you type .close to close the channel
odd edgeBOT
nocturne brook
best of luck in your counting adventures
odd edgeBOT
Available help channel!
misty marsh
Are my answers right? im not rlly sure how to do inequalties on a graph
misty marsh
,rotate 270
clever fjordBOT
viscid flint
for each line think about whether we're shading under or over that line
if we're shading under it, then the ineq will be y ≤ stuff
odd edgeBOT
odd edgeBOT
Available help channel!
pure tulip
I have to verify the existence of the limit shown in the excercise 141
this is done by putting f(x)> M
thats what i’ve done so far, but the answer would be a parabole , not a line function
pure tulip
there is definitely something about 2nd degree inequalities that i’m missing
the result is shown on the page and it should, in fact, be a line equation
warped glacier
I have to verify the existence of the limit shown in the excercise 141
this is ...
the key is that M is arbitrary
given any (real) value of M, you can always find an x such that f(x) > M
so that means that f(x) has no upper bound
pure tulip
so that means that f(x) has no upper bound
then why here there is no lower bound?
pure tulip
given any (real) value of M, you can always find an x such that f(x) > M
and since f(x) is a 2nd degree equation, shouldn’t there be 2 solutions anyway?
warped glacier
I need time to explain this properly
pure tulip
I need time to explain this properly
yes sure take your time
warped glacier
given any (real) value of M, you can always find an x such that f(x) > M
one way to make more sense of the result is that for $M > 0$, $x > \frac{M + \sqrt{M^2 + 4}}{2} > \frac{M + M}{2} = M$
so assume by contradiction that $f(x) \le M$ for all real $x$
we can just choose $x = M + 1$ and then $f(x) > M$
clever fjordBOT
south
warped glacier
I know that if you solve the inequality, the full solution for x is this
but remember to demonstrate a contradiction, we just need one value of x such that f(x) > M
warped glacier
I know that if you solve the inequality, the full solution for x is this
we already can find one such x in the first inequality
pure tulip
**south**
hold on
what do you get that inequality from?
or is that just a thought process
warped glacier
what do you get that inequality from?
$M > 0 \implies \sqrt{M^2 + 4} > M$
clever fjordBOT
south
pure tulip
is that so? thank you I was never told that afaik
warped glacier
yes that's what I used
pure tulip
and why do you assume that x <= M?
warped glacier
note the condition that M is arbitrary
warped glacier
and why do you assume that x <= M?
then f(x) <= M is the definition of being bounded above
there exists an M such that f(x) <= M is true for all real x
warped glacier
I don’t understand why it’s bounded above
assume it's bounded above
pure tulip
the graph should be one of a positive 2nd grade
warped glacier
then we aim to disprove this by contradiction
pure tulip
ok
so by saying that x <= M and then plugging x = M + 1 in the equation you disprove that
warped glacier
so by saying that x <= M and then plugging x = M + 1 in the equation you disprov...
yes
given that M > 0, then x = M + 1 satisfies x > sqrt(M^2 + 4)/2 + M/2 for all M
pure tulip
I know that if you solve the inequality, the full solution for x is this
but why does my book give as a solution only the first one then?
warped glacier
but remember to demonstrate a contradiction, we just need one value of x such th...
.
pure tulip
for limits where the x or the y tends to a fixed value it gives 2 results
warped glacier
for limits where the x or the y tends to a fixed value it gives 2 results
this isn't about solving quadratics at all
it's about understanding proofwriting and logical argumentation techniques
pure tulip
but like in 147 both solutions are given
warped glacier
but like in 147 both solutions are given
right, cause clearly the limit in 147 converges
pure tulip
although they become one
warped glacier
a nice strategy for proving a limit diverges to +infinity or -infinity is by contradiction
assume there is some maximum value M for all x
pure tulip
it's about understanding proofwriting and logical argumentation techniques
so in order to demonstrate that a limit tends to infinity I only need one of the two solutions?
warped glacier
then show that the maximum M is always exceeded for some choice of x
warped glacier
so in order to demonstrate that a limit tends to infinity I only need one of the...
yes
pure tulip
sorry can I ask you one more thing
warped glacier
if the limit approaches infinity for $x \to \infty$, then $x \to -\infty$ is irrelevant
warped glacier
sorry can I ask you one more thing
sure
clever fjordBOT
south
pure tulip
what is the difference between these limits and those limits in finding their existence?
I know for these you have to find both solutions, so maybe you don’t need to find both if the x tends to infinity as well as the y?
oh wait
now I get it
so
warped glacier
what is the difference between these limits and those limits in finding their ex...
ah, so in this situation, the limit can fail to exist
pure tulip
if the x tends to + infinity you want the upper solution
if the x tends to - infinity you want the lower solution?
warped glacier
$\lim_{x \to 0^-} \frac{1}{x} = - \infty, \lim_{x \to 0^+} \frac{1}{x} = + \infty$
clever fjordBOT
south
warped glacier
you don't want this to happen
so you must check that the limits as x approaches 0 from the left and the right are the same
warped glacier
question 101 is a bit different: the function isn't even defined for negative x
so you automatically have 0 < x by definition
warped glacier
I solved these kind of things by doing this for instance
I feel the way they are teaching you limit proofs in liceo is very wishy-washy
they're not teaching you the 'why' of things
pure tulip
I feel the way they are teaching you limit proofs in liceo is very wishy-washy
how do you know I got a liceo lol
im tryna translate hold on
the thing that the limit is defined by
we call it “around”
what’s that in english lol
they’re teaching that in order to prove that a limit exists we need to prove that there is an “around”
warped glacier
they’re teaching that in order to prove that a limit exists we need to prove tha...
what's the term in Italian?
pure tulip
what's the term in Italian?
intorno
sand oak
the blue brackets in your textbook would be the intorno
pure tulip
since a limit is defined as
for each neighborhood delta of x0, there exists a neighborhood of y called epsilon, so that if |x - x0| < delta then |f(x) - l | < epsilon
hold on I got a pic
sand oak
exactly
warped glacier
yes, if you can find an open neighbourhood (a neighbourhood which is an open set, so a < x < b is an open set of the real numbers)
I see what they are doing now
pure tulip
even though this definition always seemed off for me
warped glacier
yes, those theorems are correct
this is the epsilon-delta definition of a limit in English
pure tulip
yes, those theorems are correct
ok so that’s what I’ve been doing in the easier excercises
but I can’t do that in the ones I sent here?
this is the one if both y and x tend to infinite
pure tulip
this is the epsilon-delta definition of a limit in English
I’m sorry if I’m making this harder for you
but idk why our teacher would tell us to demonstrate by contradiction without even explaining us that we had to demonstrate by contradicting
sand oak
so they told us that if we can find such neighborhood we know the limit is real?
epsilon is an "error" of some kind. error here refers to how close the function value is to the limit. and it can be anything. if you can find a neighbourhood around x_0 such that the function value is only epsilon amount of error from the limit, you can say the limit exists.
warped glacier
sta bene, il mio italiano non é lo migliore
ma io comprendo tutto nella pagina
pure tulip
sta bene, il mio italiano non é lo migliore
ma io comprendo tutto nella pagina
yoo you speak italian that’s crazy
pure tulip
epsilon is an "error" of some kind. error here refers to how close the function ...
ok and I get that, so if that’s the definition of a limit and I can demonstrate its existence by doing that, why can’t I do the same if both x and y tend to infinite and I have to resort to contradiction?
pure tulip
I don't claim to exactly 'know' Italian
you know more italian than my russian per say
you gotta know nothing before knowing something
sand oak
ok and I get that, so if that’s the definition of a limit and I can demonstrate ...
so infinity isn't a real number. you can't surround it with two real numbers, i.e., there's no neighbourhood around infinity
warped glacier
again all of these drawings indicate that you can show that a function converges to some limit, if you can always draw a 'window' around the centre of the limit
if the function is not continuous
then you can always draw a window where the box does not intersect the function
pure tulip
so infinity isn't a real number. you can't surround it with two real numbers, i....
I thought neighborhoods were used exactly to elude the issue where in a function you have a x and you need a y
like the neighborhood of infinity would be infinity +- 1 and that would enable you to use it in a function
this wasn’t explained to me, I made it up to understand why we would switch from functions to limits and introduce neighborhoods so it would make sense if it’s wrong
warped glacier
I thought neighborhoods were used exactly to elude the issue where in a function...
oh yeah, the cases for when x approaches infinity are messy, I agree
pure tulip
if the x tends to + infinity you want the upper solution
if the x tends to - inf...
is this correct tho?
or it’s more about the y
no I think it’s the x
warped glacier
is this correct tho?
this doesn't work for infinity
pure tulip
this doesn't work for infinity
istg all the excercises where x tends to -infinity want the lower solution while all the ones there x tends to + infinity want the upper solution
warped glacier
the way we reason about infinity is through some arbitrary finite number M
if we can find a neighbourhood $x \in (x_0 - \delta, x_0 + \delta)$ such that $f(x) > M$ for any $M$
we can keep increasing $M$ higher and higher, say $M = 999, 9999, 999999 \cdots$
and there will always be at least one real $x$ in our neighbourhood
clever fjordBOT
south
warped glacier
thus showing that the limit is +infinity
pure tulip
so we demonstrate this by demonstrating that we can raise M as much as we want?
warped glacier
this is the one if both y and x tend to infinite
and then when we have $f(x) \to \infty$ only when $x \to \infty$ (the definition on the very bottom here), it's even further from the standard definition
clever fjordBOT
south
warped glacier
so we demonstrate this by demonstrating that we can raise M as much as we want?
yes!
sand oak
let's first distinguish limits with infinity. first, there are functions that approach (+/-) infinity as x goes to (+/-) infinity. second, functions that approach (+/-) infinity when x goes to a real number
warped glacier
we found a neighbourhood that works for all M
that's the point
the neighbourhood is expressed using the variable M
warped glacier
let's first distinguish limits with infinity. first, there are functions that a...
yep, that's what I wanted to say too
pure tulip
I think I get it now
warped glacier
so far we're on functions that approach (+/-) infinity when x goes to a real number
pure tulip
so M is a number that I can choose, and if I can demonstrate that the M can be anything on the upper bound it tends to infinity, to do so, I make it so that I can’t actually raise it to infinity and demonstrate that that’s not true, by using
x>M and giving x = M +1
x<M*
warped glacier
so M is a number that I can choose, and if I can demonstrate that the M can be a...
well x = M + 1 was a specific choice that worked for that one particular question
for other questions, you might have x = 2M or x = e^M
x = some function of M
pure tulip
but hold on
why do we make it so that the x < M and not the Y?
isn’t M a Y number and N a X number?
warped glacier
why do we make it so that the x < M and not the Y?
maybe it's too confusing to explain it using contradiction
from this picture
does the definition make sense to you?
pure tulip
what I don’t get is why N and M have to be necessarily positive
I asked it to my teacher and she said “per convenzione”
so for convention
I hate doing stuff not understanding what I do it for tho
warped glacier
I asked it to my teacher and she said “per convenzione”
yes that's correct
we could have for all M > 999999 there exists an N > 0
sand oak
what I don’t get is why N and M have to be necessarily positive
good point. they don't have to be. but if you want to show that a function goes to positive infinity, then a big enough M will be positive eventually. same with the N
warped glacier
or for all M > 999999 there exists an N > 99999
basically any finite number works
2, 3, whatever
pure tulip
good point. they don't have to be. but if you want to show that a function goes ...
so the reason is that having it low doesn’t just help you?
but it CAN be low
it’s just useless
sand oak
it could be low indeed
pure tulip
or for all M > 999999 there exists an N > 99999
gotcha
so yea that formula makes sense to me
warped glacier
maybe it's too confusing to explain it using contradiction
the lower bounds of M and N are arbitrary
you just want to show that the function is always inside the green window
sand oak
it’s just useless
in calc class maybe, but for some problems and applications you might be interested in the first N or M that works
having a tight boundary could be useful
pure tulip
maybe it's too confusing to explain it using contradiction
f(x) > M as infinity needs to be bigger than any given number that’s not infinity itself
sand oak
if you study computer science, this becomes relevant when you analyse algorithms
warped glacier
maybe the function could do this
but if the limit as x goes to infinity really is f(x) = infinity
eventually the function will have its final minimum value
then it will be always increasing after that
warped glacier
maybe it's too confusing to explain it using contradiction
so yes, a shortcoming of this diagram is that it assumes f(x) is an increasing function
f(x) may not be increasing for all x
but there will always be an N such that f(x) is always increasing for x > N
warped glacier
so why is it to complex to explain it using contradiction?
it's not actually that complex but you are learning a lot of new material
warped glacier
you already have these useful definitions
I just didn't want to overconfuse things for you
pure tulip
it's not actually that complex but you are learning a lot of new material
well I’m always complaining how school feels too easy and I’m learning stuff way too slowly
so this is fun
warped glacier
well I’m always complaining how school feels too easy and I’m learning stuff way...
nice
time to learn the structure of proof by contradiction I guess
pure tulip
time to learn the structure of proof by contradiction I guess
well I do know how proving by contradiction works
but I’ve never used it outside of geometry
so I haven’t used it in 4 years lol
the thing I don’t get is that tho
why do we say that x < M?
and not Y?
if we say that Y < M we assume that the Y is less than the arbitrary boundary we set
hence the function doesn’t get to infinity ever
but for X I don’t see the argument
sand oak
you could theoretically pick an M such that x<M and y<M at the same time
warped glacier
one way to make more sense of the result is that for $M > 0$, $x > \frac{M + \sq...
oh wait I did type it wrong
should be assume by contradiction that f(x) <= M for all real x
yeah
pure tulip
oh ok
warped glacier
that explains why
sand oak
you could theoretically pick an M such that x<M and y<M at the same time
wait this doesn't make sense. disregard that
warped glacier
and then we actually need to show that the function is not bounded above by f(x) = 0, or f(x) = (some negative number) for all real x
just sub in x = 2, so f(x) = (2^2 - 1)/2 > 0
cause we assumed that M > 0
so for no positive number M, is f(x) <= M for all M: this is what we have shown
pure tulip
and then we actually need to show that the function is not bounded above by f(x)...
makes sense
warped glacier
there's a small detail here to check
pure tulip
so for no positive number M, is f(x) <= M for all M: this is what we have shown
and then we can choose only the upper solution cause if x approaches infinity there is no way it could go toward negative infinity?
warped glacier
or actually it doesn't matter, cause if f(x) <= -2 for all x
then f(x) <= 2 for all x
I guess this is evidence of why choosing M > 0 doesn't even matter
warped glacier
and then we can choose only the upper solution cause if x approaches infinity th...
yes
so you always want to have an clear, intuitive idea of what you want to prove
pure tulip
ok this makes so much sense thank you guys
warped glacier
you want to prove something about $x \to +\infty$
clever fjordBOT
south
warped glacier
so you need to have a condition that respects that, which we can do by creating some fixed real number N, and then only examining x > N
as we take x being larger and larger, such that x approaches +infinity, we want to of course prove that f(x) <= M is impossible for any M, for the x that satisfy x > N
but you don't need any condition that doesn't obey x > N
warped glacier
ok this makes so much sense thank you guys
alright, no worries!
pure tulip
so you need to have a condition that respects that, which we can do by creating ...
f(x)?
isn’t it f(x) >=M?
warped glacier
isn’t it f(x) >=M?
yes, okay yes I used M twice
pure tulip
cause you’d forgotten to put f(x) instead of x
sand oak
you could let K be the max between N and M. and use it for both x and f(x)
warped glacier
I need to think about how to explain to you what I meant there
pure tulip
yes, okay yes I used M twice
so aren’t we stuck by this at the same point since putting f(x) > M is what I tried to do before?
wdym by “used M twice”?
both for x and y?
sand oak
so aren’t we stuck by this at the same point since putting f(x) > M is what I tr...
if you're looking for x>N such that f(x)>M, you can let K=max(N,M) and then use K for both x and f(x). it'll still be true
warped glacier
if you're looking for x>N such that f(x)>M, you can let K=max(N,M) and then use ...
yeah that's a great point
pure tulip
so if N = 7 and M = 9 K =9?
warped glacier
if we have x > max(M, N), then x > N
so if we can prove the stronger claim that f(x) > max(M, N) for all these x, then f(x) > M
pure tulip
and since both N and M are arbitrary numbers I can just pick the bigger between the two and that could be used for both?
sand oak
yes that would work
warped glacier
and since both N and M are arbitrary numbers I can just pick the bigger between ...
yes
warped glacier
if we have x > max(M, N), then x > N
so if we can prove the stronger claim tha...
you can make the problem a bit harder by showing that f(x) > max(M, N) instead of f(x) > M
pure tulip
so I can just put x > M cause
if M > N => f ( x ) > M > N => X>M > N
if N > M => x > N > M
if N = M => x > M
the first if condition is weird tho
say N = 9 and M = 10
x>9 but is not necessarily bigger than 10
the other two make sense
warped glacier
I have to go but it's been nice helping you!
sand oak
say N = 9 and M = 10
x>9 but is not necessarily bigger than 10
if you're showing that f(x) goes to positive infinity as x goes to positive infinity, you have to show that for all M, there is some N such that for all x bigger than N, you get f(x)>M
pure tulip
if you're showing that f(x) goes to positive infinity as x goes to positive infi...
yes
but if N is the bigger between the two how can I prove that x is bigger than M?
sand oak
it's more that you need to show that f(x)>M on the interval x>N
pure tulip
it's more that you need to show that f(x)>M on the interval x>N
no ok and that’s easy to do
sand oak
and if M is bigger than N, your interval (N, inf) can be replaced with (M, inf)
bc it'll still be true by you already proving for the wider interval
pure tulip
so I can just put x > M cause
if M > N => f ( x ) > M > N => X>M > N
if N > M =>...
so I can’t prove it by doing this, it’s more like, since it’s an arbitrary number, infinite will be bigger so it doesn’t matter
sand oak
so I can’t prove it by doing this, it’s more like, since it’s an arbitrary numbe...
yeah, it's not so much you can't do that. it's what you just said. it's the arbitrary part
pure tulip
aight
pure tulip
if x => infinity, x will approach infinity and infinity is bigger than any given number, hence x > M > N
sand oak
you got it
sand oak
you too 🙂
pure tulip
hold on how do I close the page
sand oak
.close i think
pure tulip
.close
odd edgeBOT
odd edgeBOT
Available help channel!
lapis magnet
Asking help for this one
odd edgeBOT
Asking help for this one
stoic cloud
What have you tried and what are you stuck on?
lapis magnet
This one, I'm having a hard time processing this. I don't understand this one.
hard drum
convert the statements into mathematical relations friend
lapis magnet
.close
odd edgeBOT
odd edgeBOT
Available help channel!
real fjord
Okay. I'm not sure if this belongs here or elsewhere. I'm not sure if this is considered Calculus or Linear Algebra and I'm not sure if this is considered Pre-University or Early University.
It's an optimization problem regarding 5 variables that I set up.
I need to minimize the equation 0.99x + 4.99y + 19.99z + 49.99a + 99.99b
I need to accomplish the equation x + 3y + 12z + 25a + 50b ≥ 25
And I have the limitations of
x ≤ 5
y ≤ 5
z ≤ 5
a ≤ 1
b ≤ 1
Oh and all values need to be positive. So x, y, z, a, b ≥ 0
crude hollow
wouldnt AM GM work?
real fjord
wouldnt AM GM work?
I don't know what AM GM is
crude hollow
real fjord
I'm not sure how to apply that...
crude hollow
u have the terms
x + 3y + 12z + 25a + 50b
put that in formula
least value = 25
which is the under root part
mystic nova
but it like, 5 variables 
crude hollow
this is a linear programming problem
whats different here tho
cause sum is constant
wooden python
AM-GM will have you working w the product of these things
real fjord
this is a linear programming problem
I thought it was optimization? Also I'm not sure if this makes it harder or easier but I forgot to mention that the variables need to be natural numbers
mystic nova
0.99x + 4.99y + 19.99z + 49.99a + 99.99b can we round this to x+5b+20z+50a+100b
wooden python
it just looks unhelpful here @crude hollow
wooden python
I thought it was optimization? Also I'm not sure if this makes it harder or easi...
it is a type of optimization
crude hollow
it just looks unhelpful here <@1061639142208978975>
alr
mystic nova
I think we can cancel out a and b term using this x + 3y + 12z + 25a + 50b ≥ 25
so it likes 2x + 6y + 24z + 50a + 100b ≥ 50 and maybe...
we can graph now
ehh we got... that thing >= -x-y-4z+25
this's 3d.....
odd edgeBOT
@real fjord Has your question been resolved?
real fjord
No
Glad to know I apparently made a really complicated real world problem
Like a = 1, and the rest is 0 is a solution, z = 2, x = 1, and the rest is 0 is a better solution, y = 4, x = 1, and the rest is 0, is a better solution, I know I can brute force finding solutions that work, I just was trying to figure out how to find the optimal solution
I wonder if it would be easier to make a spreadsheet testing all 864 combinations instead?
Yes. Apparently it was faster to make a spreadsheet to test all possible combinations. The optimal solution is x = 4, y = 3, z = 1, a = 0, b = 0.
.close
wooden oar
Can i write $\int{(4x^3+9x^2-8x+1)}dx$ like this $$\int{(4x^3)dx}+\int{(9x^2)dx}-\int{(8x+1)dx}$$
clever fjordBOT
dream
wooden oar
thanks in advance!
wooden python
Can i write $\int{(4x^3+9x^2-8x+1)}dx$ like this $$\int{(4x^3)dx}+\int{(9x^2)dx}...
no, sign error in the last integral
wooden oar
no, sign error in the last integral
really?
wooden python
you can have ...+int (-8x+1) dx
or ...-int (8x-1) dx
wooden oar
but its negative
wooden python
only the -8x term has a negative coeff
wooden oar
you can have ...+int (-8x+1) dx
or ...-int (8x-1) dx
oh
wooden python
the 1 at the end isn't
wooden oar
only the -8x term has a negative coeff
ohh you are right
thanks
$\int{(4x^3+9x^2-8x+1)}dx = \int{(4x^3)dx}+\int{(9x^2)dx}-\int{(8x-1)dx}$
clever fjordBOT
dream
odd edgeBOT
odd edgeBOT
Available help channel!
wide drift
Wtf is the 'intuition' whether its true or false here? 😭
