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#help-39
fringe fjord
pearl pondBOT
pearl pondBOT
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daring bay
can anyone help here, so
pearl pondBOT
can anyone help here, so
daring bay
for i) its just use inverse normal distribute function
with calculator
but wtihout calculator id just find phi (z) = 0,24 right
toxic lichen
but wtihout calculator id just find phi (z) = 0,24 right
you would need tables for that
idk if that counts as non-technology
daring bay
you would need tables for that
oh
yeah lowkey it does
uhh
or actually i dont think so, because how tf are uu supposed to solve it then
so
do you have an idea how to do the iii)
actuallyw e dont have to do this
nvm
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pearl pondBOT
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vestal thistle
Hello, could someone check if this proof looks good please?
\newtheorem{case}{Case}
\begin{Definition}[Subset of a set]
Suppose $A$ and $B$ are sets. If every element in $A$ is also an element $B$,
then $A$ is a \emph{subset} of $B$, which is denoted $A \subseteq B$.
\end{Definition}
\begin{Definition}[Cartisian product]
Suppose $A$ and $B$ are sets. The \emph{Cartisian product} of $A$ and $B$
is the set $A \times B = { (x,y) : x \in A \text{ and } y \in B}$.
\end{Definition}
% --------------------------------------------------------------------------------
\begin{Theorem}
Assume $A$, $B$ and $C$ are sets, where $C$ is non-empty.
If $A \times C = B \times C$, then $A = B$.
\end{Theorem}
\begin{proof}
Let $A$, $B$ and $C$ be sets, where $C$ is non-empty, and $A \times C = B \times C$.
Without loss of generality, either both $A$ or $B$ is empty or non-empty.
\begin{case}
Suppose $A = \emptyset$.
Then, by the definition of Cartisian product, $$A \times C = \emptyset.$$
So $B \times C = \emptyset$, then $B = \emptyset$, and thus $$A = B = \emptyset.$$
The rest follow from a similar case with $B = \emptyset$.
\end{case}
\begin{case}
Now, suppose $A$ is non-empty.
Let $x \in A$ and $y \in C$.
Then, by the definition of the Cartisian product, $$(x,y) \in A \times C.$$
Since $A \times C = B \times C$, we have that $$(x,y) \in B \times C.$$
This means, by the definition of the Cartisian product, that $$x \in B.$$
Since $x \in A$ implies that $x \in B$, then, by the definition of subset,
$$A \subseteq B.$$
The rest follow from a similar case with $B$ non-empty, thus $B \subseteq A$.
Therefore, since $A \subseteq B$ and $B \subseteq A$, this means that $$A = B.$$
\end{case}
\end{proof}
jolly parrotBOT
Mor Bras
thorny radish
this is fine
modern talon
**Mor Bras**
looks good, other than some minor stylistic tweaks, but if this is for hw then its chill
actually now that i think about it, i think you shouldnt say wlog both A and B are both empty or nonempty; you’re trying to show that they ARE both empty or nonempty (in which case they are equal)
i.e assume A empty, then (proof) hence B also empty
then assume A nonempty (proof) hence B also nonempty, and A = B
waxen agate
it's worth stating that B x C = Ø implies B = Ø because C ≠ Ø
vestal thistle
actually now that i think about it, i think you shouldnt say wlog both A and B a...
Is the wording better now?
\newtheorem{case}{Case}
\begin{proof}
Let $A$, $B$ and $C$ be sets, where $C$ is non-empty, and $A \times C = B \times C$.
Let break this proof in cases: either $A$ (or $B$) is empty or non-empty.
\begin{case}
Suppose $A = \emptyset$.
Then, by the definition of Cartisian product, $$A \times C = \emptyset.$$
So $B \times C = \emptyset$, then $B = \emptyset$, and thus $$A = B = \emptyset.$$
The rest follow from a similar case with $B = \emptyset$.
\end{case}
\begin{case}
Now, suppose $A$ is non-empty.
Let $x \in A$ and $y \in C$.
Then, by the definition of the Cartisian product, $$(x,y) \in A \times C.$$
Since $A \times C = B \times C$, we have that $$(x,y) \in B \times C.$$
This means, by the definition of the Cartisian product, that $$x \in B.$$
Since $x \in A$ implies that $x \in B$, then, by the definition of subset,
$$A \subseteq B.$$
The rest follow from a similar case with $B$ non-empty, thus $B \subseteq A$.
Therefore, since $A \subseteq B$ and $B \subseteq A$, this means that $$A = B.$$
\end{case}
\end{proof}
jolly parrotBOT
Mor Bras
pearl pondBOT
@vestal thistle Has your question been resolved?
modern talon
**Mor Bras**
looks better, yeah
pearl pondBOT
pearl pondBOT
Available help channel!
iron basin
what can this mystery sequence be
pearl pondBOT
what can this mystery sequence be
iron basin
i tried taking LCM assuming numerator to be of the form 9n^2 or denominator to be of the form 13 multiple something
neither seemed to work
limber nimbus
The denominators are definitely 13(n + 4) as far as I can tell
That would make the numerators
9, 39, 81, 117
hmm
+30, +42, +36
36 is the average of 30 and 42 
eager jewel
That would make the numerators
9, 39, 81, 117
3^2..3x13..3^4..3^2x13..
limber nimbus
I could make up a rule that works here LOL but idk it doesn’t seem super compelling
hmm that also seems plausible
At the same time if there are interlacing sequences i feel like you need more terms to justify it
eager jewel
what can this mystery sequence be
49/47=1.04255.
36/35=1.02
25/21=1.19
21/19=1.10
9/8 = 1.125 :)
earnest warren
49/47=1.04255.
36/35=1.02
25/21=1.19
21/19=1.10
9/8 = 1.125 :)
Oh that's funny
iron basin
49/47=1.04255.
36/35=1.02
25/21=1.19
21/19=1.10
9/8 = 1.125 :)
well you can eliminate one option then
earnest warren
Doesn't that imply 25/21 is the answer
eager jewel
well you can eliminate one option then
wdym
they are asking greatest term
of sequence
only 1 term is greater
eager jewel
Doesn't that imply 25/21 is the answer
25/21..
earnest warren
Ah, mb
eager jewel
now we can close this ❤️
limber nimbus
that’s so lame 😭 there has to be an actual solution
eager jewel
that’s so lame 😭 there has to be an actual solution
hey ur lame 😡
earnest warren
that’s so lame 😭 there has to be an actual solution
Can't agree more
limber nimbus
If only there was oeis for rational sequences
pearl pondBOT
@iron basin Has your question been resolved?
pearl pondBOT
thorn agate
Where did you pull this from
iron basin
wdym
thorn agate
Source
thorn agate
Oh,
pearl pondBOT
Available help channel!
cyan jungle
3 of the heaviest cats take up 60% of the weight of the litter and 2 of the lightest cats take up 25% of the litter
how many cats are there?
cyan jungle
i was already told that the answer is 6 but i dont understand it that well
proper nova
hm
cyan jungle
the teacher said
3 heaviest cats = 0.6x
so 0.2x = 1 heaviest cat
and 2 lightest cats = 0.25x
so 0.125x=1 lightest cat
but the part where i get confused is when they wrote
0.125<0.15<0.2
(since theres 15% left)
and that somehow means theres 1 cat so 3+2+1=6
this question had me tweaking for the entire lesson 😭
dark plume
the teacher said
3 heaviest cats = 0.6x
so 0.2x = 1 heaviest cat
and 2 lightest ...
Because you could not have 2 cats
They would be too light
dark plume
but how can you assume that 3 heaviest= 0.6x means that 1 heaviest=0.2x
This is a question phrasing matter. I would typically assume unless otherwise indicated that the cats are tied for weight, otherwise the question is impossible.
cyan jungle
ah ok
cyan jungle
Because you could not have 2 cats
is it because 0.15 goes into 0.2 once?
stuck hatch
3 of the heaviest cats take up 60% of the weight of the litter and 2 of the ligh...
we can start by assuming about the lightest cats: if one is 25%, the other one is 0%. the smallest that both of the smallest cats can be is 12.5%. therefore, no cat that is NOT one of the two smaller can be lighter than 12.5%.
to complete 100%, we are missing 15% between the heaviest and the lightest. Since no middle cat can be lighter than 12.5%, the only way to distribute 15% is only one cat
thorn agate
The lightest cat weigh 0.125x, heaviest 0.2 now 15 % of litter is left.
If you divide 0.15 in two or more parts each part will be either greater than 0.2 or less than 0.125 contradicting the initial statement because weight of litter by heaviest and lightest cats is bounded
stuck hatch
so there are 6 total: 3 heaviest, 2 lightest, and one middle
thorn agate
Oh , i couldn't see ypu already typed im on phone
stuck hatch
np
pearl pondBOT
@cyan jungle Has your question been resolved?
pearl pondBOT
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ebon aurora
Yesterday I argued with the teacher about this problem:
"There are 8 boxes. Each contains 10 white and 5 black balls. One ball is drawn at random from each box. Which event is more likely: drawing 6 white and 2 black balls or drawing 5 white and 3 black balls?"
My friend and I think the event with 6 white and 2 black balls is more likely.
Using the formula Pn(k) = Cn(k) × p^k × q^(n-k), the teacher calculated that the probabilities are equal.
What's the truth?
-# English isn't my native language
west sapphire
i computed them to be equal (0.273129)
west sapphire
did you try using the formula?
ebon aurora
like...
we have 7 equals events and 1 is different
ebon aurora
did you try using the formula?
teacher tryed
west sapphire
like...
we have 7 equals events and 1 is different
wdym by this?
ebon aurora
wdym by this?
wdym - why do you mind?
west sapphire
no "what do you mean"
ebon aurora
oh, ok
ebon aurora
wdym by this?
we have 8 events by drawing 8 balls from boxes
in this 2 events (when 6w2b and 5w3b) we must to draw 7 equals balls (5w and 2b)
different in just 1 ball, when we have 2/3 chance to draw white and 1/3 to draw black
west sapphire
oh i see what you're saying
notice also though, there are more ways to choose 5 white balls than there are to choose 6 white balls
8 choose 5 = 56
8 choose 6 = 28
west sapphire
they differ by a factor of 2
which balances out the factor of 2 that you get when you change 2/3 chance to 1/3 chance for one ball
west sapphire
yw
ebon aurora
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pearl pondBOT
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boreal laurel
can i have this
pearl pondBOT
can i have this
toxic lichen
well you just claimed it
do you have a math question to ask?
also tbh that nickname is kind of sus
boreal laurel
boreal laurel
also tbh that nickname is kind of sus
sorry my apologies fellow discord user
plush bramble
,w define swastika
jolly parrotBOT
plush bramble
oof uh <@&268886789983436800> maybe idk
boreal laurel
,w define swastika
its a indian religious name s
can somebody actual help me solve the question
dim linden
swastik is a common indian name
boreal laurel
plush bramble
!show
pearl pondBOT
Show your work, and if possible, explain where you are stuck.
boreal laurel
oh ok
toxic lichen
apologies for the misunderstanding earlier
boreal laurel
its fine as long as u help me solve the question 🤡🤡
boreal laurel
!show
i know the integeral concentrates near pi/2
i change the variable near the point pi/2
i just cant get the monstrosity ahead of that
plush bramble
do you think the limit diverges then?
boreal laurel
dominant convergence ?
plush bramble
that could work if it were convergent
boreal laurel
thats what i want to say
plush bramble
i know the integeral concentrates near pi/2
but this seems to imply you think it diverges
boreal laurel
but this seems to imply you think it diverges
i dont really know for sure wbu
toxic fractal
pearl pondBOT
plush bramble
tf i think i got it anyways thanks for doing nothing
amazing attitude
plush bramble
tf i think i got it anyways thanks for doing nothing
!vol
pearl pondBOT
tf i think i got it anyways thanks for doing nothing
Helpers are just people volunteering their time to help you. Be polite and patient.
buoyant panther
tf i think i got it anyways thanks for doing nothing
Glad to help
nocturne grail
tf i think i got it anyways thanks for doing nothing
perioddd
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Available help channel!
stone imp
pearl pondBOT
stone imp
It’s telling me to solve the equation
I simplified it and then applied the pq formula
It looked like i was doing right but the answer is wrong
feral olive
pq formula?
stone imp
This
feral olive
you mean quadratic?
stone imp
stone imp
you mean quadratic?
Probably called that
feral olive
this is what you were taught?
stone imp
Yep
feral olive
normally we have the same formula but where theres a coefficient on x^2 as well: $x=\frac{-b\pm{}\sqrt{b^2-4ac}}{2a}$
but this is fine
jolly parrotBOT
ImOakley
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
feral olive
what is happening bruh
stone imp
lol
feral olive
$x=\frac{-b\pm{}\sqrt{b^2-4ac}}{2a}$
jolly parrotBOT
ImOakley
feral olive
the hell it works there ok
stone imp
I’ll show u what I did and where I stopped cause I realised it ain’t gonna be right
feral olive
yeah you dont have to use that one im just saying
but also theres an even faster way to solve it
how can you make x(x-5)=0?
NOT AGAIN
stone imp
Omg
feral olive
<@&268886789983436800>
stone imp
It’s always me I swear
stone imp
how can you make x(x-5)=0?
I made it x^2-5x=0
feral olive
yeah
stone imp
Then I made it this
Bit chaotic
I stopped at the side over there cuz I realised 2,5^2 was not making sense
feral olive
oh yeah you dont
mb
yeah -(-2.5) just reverses the minus again to make it positive
stone imp
I peeked the answer and it was 0 and 5 so when I got the second part to be 2,5^2 I realised that ain’t gonna work
feral olive
why is that wrong
stone imp
… omg it’s not wrong I was just doubting myself
feral olive
see what happens when you follow through with it
lol
ik why
you thought because it was a decimal and squaring it gives more decimals
stone imp
Yeah
feral olive
but theres a much faster way as well
if you think about what you multiply something by to get 0
what could you multiply it by
stone imp
0
feral olive
yes
so if you go back to x(x-5)=0
to make that 0 you can just make x 0 and it would work
same with x-5
stone imp
Ooooh
feral olive
if you can factorise the equation you dont even need to touch the pq formula
you just make everything equal 0
stone imp
Isn’t that a bit complicated
feral olive
no its less complicated
its just knowing that the only way to make it equal 0 is by making the terms 0
i remember i learned the factorising method before learning the formula
stone imp
It’ll be 0(0-5)=0
feral olive
yeah and its 0
stone imp
Wait can u type the two different equations for the two different x
jolly parrotBOT
ImOakley
stone imp
Huuuh
feral olive
heres an example: $(x-6)(x+2)=0$
jolly parrotBOT
ImOakley
eternal plover
u can also think of it as like
if u multiply sometihng by 0, it becomes zero
so since both the brackets are being multiplied
for it to equal to zero, one of them has to be zero
since we have a variable
either bracket can be zero
thus u can get two values from it
so first the first bracket can be zero
and so u equate it to zero to find x
then do the same for the second bracket
stone imp
I’m so confused if we take the equation I have for example what’s the two different ways to include 0 in it
eternal plover
okay so what do u understand when an equation is equation to 0?
stone imp
That the answer is gonna be 0
eternal plover
hmm almost
but if we have a variable
lets say
x-5=0
what r we trying to say here?
what do u think
feral olive
I’m so confused if we take the equation I have for example what’s the two differ...
its either x=0 or (x-5)=0
eternal plover
yes
so here the answer isnt zero, but 5
so when i say
(x-5)(x-5)= 0
im saying that when i multiply (x-5) with (x-5), i get 0
stone imp
Ohhh okay yeah
eternal plover
the only way u can get 0 through multiplication is by multiplying with zero itself
so that means that either of the brackets, (x-5) or (x-5) (second one) have to equal to zero
do u understand that much?
stone imp
Yeah
eternal plover
okay so now since x is a variable, meaning it can take any value, we have to solve to find every possible value of x
so that we can get the complete result
so u have to equate both brackets with 0
since both brackets are same (x-5) and (x-5)
ull get the same value of x in both cases
therefore in this case
there exists only 1 value of x
stone imp
Oooh
eternal plover
yes
which is 5 as u said earlier
now if we have two different brackets
(x-3)(x-7)=0
we have to find two values of x
stone imp
Ooooooooohhhhh
stone imp
So if they’re the same it’s always gonna be one value of x
stone imp
Can I use this same technique in the equation 2x^2-50=0 cause I tried using the pq formula again and it just came out wrong at the second part
eternal plover
u can but it wouldnt be needed
in this case
since u dont have just multiplication
but addition too
try to rearrange the equation for x
stone imp
I divided everything by 2
eternal plover
so we can also write it as 2x^2=50
yes then u can divide by 2
x^2=25
now u can take square root on both sides
stone imp
Oh ur right I don’t even need to do anything unnecessary and complicated here
stone imp
It’s just a normal equation right
stone imp
Yeah what’s this thing called in English
eternal plover
such as for x^2+2x+1=0
stone imp
Wait
eternal plover
factorization
eternal plover
quadratic?
eternal plover
square root
stone imp
Thank uuuu
stone imp
Yeah now it’s easier thanks
eternal plover
np!
stone imp
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pearl pondBOT
stone imp
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pearl pondBOT
✅ Original question: #help-39 message
stone imp
Here we go again
I had the equation (x+5)(3x-2)=0
I simplified it
And it became
3x^2+15x-10
And I thought I could use the pq formula here since I have all the parts I need
But when I divide everything by 3 to make 3x^2 into x^2
10 becomes decimal
buoyant panther
Here we go again
You're overcomplicating it once again, just apply what you did in the previous questions
stone imp
To make one bracket 0 and then the other 0?
buoyant panther
Yes
buoyant panther
Because multiplication of two numbers is zero if and only if at least one of those numbers is zero
stone imp
How do u know which
ruby cargo
$a \cdot b = 0 \iff a = 0 \vee b = 0$
jolly parrotBOT
@ruby cargo
buoyant panther
It doesn't matter
stone imp
Do I just make all the numbers in one of the brackets 0
buoyant panther
If let's say the first one is zero, then the second can even be the speed of light, the result is still 0
buoyant panther
It is
brisk current
How do I get the math symbol?
buoyant panther
It’s -5 and 2/3
Those are x values, not values of the function
brisk current
next fo my name
buoyant panther
In order to obtain 0, plug them in
ruby cargo
Given $(x + 5)(3x - 2) = 0$
$\implies$ either x + 5 = 0 or 3x - 2 = 0
Yea if we make one of the brackets 0 it will make the answer 0 how will that get me to the answer
brisk current
I’m not laughing this is no joke
brisk current
no mwrh
buoyant panther
Yea if we make one of the brackets 0 it will make the answer 0 how will that get...
You've just solved it, don't you see the answer?
jolly parrotBOT
@ruby cargo
stone imp
Made one of the brackets 0
brisk current
Go to a different channel
you can’t solve a simple quadratic equation are you in elementary??
or sum?
buoyant panther
Made one of the brackets 0
Yes, and then?
brisk current
Yes, and then?
can you help me
buoyant panther
(x+5)(3x-2) = 0
let's say you did
x + 5 = 0
What was you next step?
stone imp
you can’t solve a simple quadratic equation are you in elementary??
<@&268886789983436800>
ruby cargo
can you help me
This channel is occupied if you need help, please open your own channel #❓how-to-get-help
stone imp
(x+5)(3x-2) = 0
let's say you did
x + 5 = 0
What was you next step?
Ohhh so I completely remove one bracket
slow oak
you can’t solve a simple quadratic equation are you in elementary??
- This channel is occupied
- You don't have to be an elitist snob to someone because they don't want to help you, you can try to get help again later.
simple ruin
stone imp
It’s okay
buoyant panther
What was you next step?
stone imp
Let me try with the second one since I think the first 5 is obvious and easy rn
(3x-2)=0 right
buoyant panther
There you go
stone imp
Ohhh okay when do i know when I can just use this step instead of all that extra stuff with the pq formula cause i always automatically use it when I see the opportunity to
buoyant panther
When you have the factored form
i.e. if you see they gave you BRACKET * BRACKET = 0
In other words, if all terms are multiplied, no addition or subtraction
stone imp
Cause I was able to do the Same with x(x-5)=0 and that only had one bracket
solemn kindle
you have to think of it as multiples
stone imp
Wdym
buoyant panther
Cause I was able to do the Same with x(x-5)=0 and that only had one bracket
It still has factored form, you can even think about it as (x-0) * (x-5) = 0 if it helps
rugged rover
buoyant panther
Your method is correct as well, but you see it's like using a sledgehammer to crack a nut
Not worth it in such cases
pearl pondBOT
pearl pondBOT
Available help channel!
fiery lance
pearl pondBOT
fiery lance
for 2a can someone tell me real quick
where these values come from
i understand everything else
buoyant panther
If you don't understand it, you can firstly find this angle:
pearl pondBOT
buoyant panther
,w sin(arctan(1/2)-pi/9)*(sqrt(5)/2)
jolly parrotBOT
pearl pondBOT
Available help channel!
swift spindle
ik i havent gave optimal context, but in general what is the point of 0 elsewhere if all cases of t is already laid out?
pearl pondBOT
flat badge
pearl pondBOT
Available help channel!
fading nexus
pearl pondBOT
fading nexus
So the second pic is the answer key
Why are we reflecting over the midline?
And not the x axis
How would I know which goes first
pearl pondBOT
@fading nexus Has your question been resolved?
hasty raptor
Why are we reflecting over the midline?
because the negative sign in front of the trig function reflects the function across its midline NOT THE X AXIS.
maybe try playing around with trig functions on desmos
fading nexus
because the negative sign in front of the trig function reflects the function ac...
Oh so the midline’s are not affected by reflections?
Or can it be? If so what would it look like
wraith stag
Why are we reflecting over the midline?
-3 is outside of cosine
thats why
oh thats not the question
anyways
it's because of order of operations
the +2 at the very end is applied last during transformations
pearl pondBOT
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Available help channel!
tawdry solar
Find the angle $\th$, between $0^{\circ}$ and $360^{\circ}$, inclusive such that [\tan(\th) = -\frac 1{\sqrt3}]
jolly parrotBOT
calvin
fierce aspen
Find the angle $\th$, between $0^{\circ}$ and $360^{\circ}$, inclusive such that...
yk tan^-1?
tawdry solar
no calculator
fierce aspen
that doesnt matter
tawdry solar
yes
fierce aspen
do yk inverse is what im asking
tawdry solar
yes
fierce aspen
simple qs, where all is tan negative
fierce aspen
now i have an value alpha (assume positive)
tan^-1(alpha)= theta
tawdry solar
oh its just 30
fierce aspen
oh its just 30
nope.
fierce aspen
?
tawdry solar
degrees
i think its just trial and error
because [ \tan(\th) = \frac{\sin(\th)}{\cos(\th)}]
jolly parrotBOT
calvin
tawdry solar
what why
tawdry solar
sin(-b)/cos(-b)??
fierce aspen
yk property of inverses right?
tawdry solar
idk their properties
fierce aspen
sigh k, know this
tan^-1(-b) = -tan^-1(b)
tawdry solar
yh
fierce aspen
tan^-1(1/root3) = 30
-is there so -30
or 360-30.
tawdry solar
well
fierce aspen
k fine
fierce aspen
30 degrees
tawdry solar
as an acute angle
fierce aspen
or -30
pearl pondBOT
pearl pondBOT
Available help channel!
formal radish
how to find this
pearl pondBOT
how to find this
turbid summit
Try using L'Hôpital's rule
turbid summit
how to find this
This is a 0/0 case, so L'Hôpital's should work
If you still get another 0/0 case, apply the rule again
formal radish
uh
I dont know the rule
oh
yeah i cant do it
i tried dividing and multiplying both num and denominator by x
brave sluice
this is l'hopital's rule
brave sluice
this is l'hopital's rule
it's only valid if the numerator and denominator are both approaching 0, or both approaching infinity
brave sluice
you can use a common trig limit
i was able to factor out || x/sin(x) but i was stuck with the e^x - 1 ||
formal radish
lim (e^x - 1)/ x = 1
x->0
mystic geode
i was able to factor out || x/sin(x) but i was stuck with the e^x - 1 ||
write it as (e^x -1 / x) ( x^2 / (1-cos(x))
mystic geode
true
versed mica
write it as (e^x -1 / x) ( x^2 / (1-cos(x))
+?
formal radish
so is it possible to do it using normal limit identities
jolly parrotBOT
InterGalactic
formal radish
or do i have to differentiate both num and denom
formal radish
we havent been taught l'hopital rule yet
versed mica
we havent been taught l'hopital rule yet
what have you been taught
formal radish
the limit identities for 1-cos/x and e^x - 1/x
brave sluice
oh you have the e^x-1 limit, i see
versed mica
the limit identities for 1-cos/x and e^x - 1/x
this makes it simpler
formal radish
yes
turbid summit
If you treat x as 2×x/2, what would 1-cosx be?
versed mica
i was thinking just x^2(1 + cosx)/sin^2x —> 2
brave sluice
versed mica
yep
brave sluice
i didn't use any double angle formulas, but i used sin^2 + cos^2 = 1
mystic geode
is that a known limit?
brave sluice
it's not a limit, it's a trig identity
versed mica
x/sinx and e^x - 1 / x are pretty common, the rest is just algebra
brave sluice
a trick i used is multiplying by (1+cos)/(1+cos)
formal radish
a trick i used is multiplying by (1+cos)/(1+cos)
how am i supposed to know i have to do that 😔
versed mica
this is a common trick
when you see something like 1 - x
to exploit difference of two squares
especially here because of the pythagorean identity
brave sluice
there's a similar trick with square roots and with complex numbers
formal radish
i got the answer
versed mica
a denominator that doesn’t involve a sum/difference is nicer than one that does
formal radish
its 2 right
brave sluice
ye
mystic geode
there's a similar trick with square roots and with complex numbers
what is that?
brave sluice
$\frac{2}{1+\sqrt{3}}=\frac{2(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}$
jolly parrotBOT
Axe
brave sluice
like this
mystic geode
oh right
formal radish
also lim x->0 for sinx/x and x/sinx = 1?
versed mica
yes
versed mica
no
how could they both be zero
they are both 1 in the previous example since the reciprocal of 1 is just 1
formal radish
oh yeah
true
sorry
so same goes for e^x-1/x and its reciprocal and log e (1+x) / x and its reciprocal
brave sluice
,align
\lim_{x\to 0}\frac{x}{\sin(x)}&=\lim_{x\to 0}\frac{1}{\frac{\sin(x)}{x}}\
&=\frac{\lim_{x\to 0}1}{\lim_{x\to 0}\frac{\sin(x)}{x}}\
&=\frac{1}{1}
jolly parrotBOT
Axe
formal radish
.close
pearl pondBOT
brave sluice
sorry, yeah, the same would go for (e^x - 1)/x
pearl pondBOT
Available help channel!
queen knot
I’m confused on this question. Im using properties of exponents to simplify it
vernal jasper
$x^{-1} = \frac{1}{x}$
jolly parrotBOT
Maddie
vernal jasper
I dunno.
I would simplify 6/2 first to 3/...
Then i see the inverses, I want to remove the negatives.
Once you do that, you should get a clean expression with x on top, y below.
toxic lichen
i would go a bit more radical: $$\frac62\cdot\frac{x^2}{x^{-3}}\cdot\frac{y^{-2}}{y}$$
and simplify each bit separately
jolly parrotBOT
Ann
queen knot
Yea I definitely understand how to get it started now
queen knot
Thank you 🙂
Would that equal 3xy?
vernal jasper
Review this ^
toxic lichen
!nosols btw
pearl pondBOT
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.
mystic geode
!nosols btw
can't you remove that, or you are not mod
mystic geode
👍
vernal jasper
Aw... Okay anyways
queen knot
I just realized that random dude just showed the answer to it 😭
vernal jasper
From what I said just now.. If
$x^{-1} = \frac{1}{x}$
jolly parrotBOT
Maddie
jolly parrotBOT
Maddie
mystic geode
$$a^{-2} = \frac{1}{a^2}$$
mystic geode
then x^2
queen knot
It’s a little confusing still 😅
vernal jasper
**Maddie**
Do you recognize that this can also be written this way?
$\frac{1}{\frac{1}{x^2}}$
jolly parrotBOT
Maddie
queen knot
Oh wait with that yea
vernal jasper
then,
$\frac{1}{\frac{1}{x^2}} = 1 \div \frac{1}{x^2}$
jolly parrotBOT
Maddie
vernal jasper
do you know this property?
$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$
jolly parrotBOT
Maddie
queen knot
I don’t think I learned that one, I can send you a picture of the ones I learned from my notes
vernal jasper
**Maddie**
Oh, this is just a simple property of fraction division
vernal jasper
**Maddie**
Not anything about indices, I was just showing u how we get x^2 from 1/1/x^2
queen knot
I’m still confused a little on what I’m suppose to do. I understand how you said about the 6/2 to just three but I can’t really understand on what I do with the rest of the problem
vernal jasper
ok
Ill use what Ann posted
i would go a bit more radical: $$\frac62\cdot\frac{x^2}{x^{-3}}\cdot\frac{y^{-2}}{y}$$
and simplify each bit separately
jolly parrotBOT
Maddie
vernal jasper
for $\frac{x^2}{x^{-3}}$
jolly parrotBOT
Maddie
queen knot
That would be x^-1?
vernal jasper
Can we remove the "-3" and make it positive?
vernal jasper
You can
$\frac{1}{\frac{1}{x^2}} = 1 \div \frac{1}{x^2}$
particularly this is what im saying
jolly parrotBOT
Maddie
eternal plover
yes negative powers denote reciprocals so if the take the reciprocal it will just go on top
vernal jasper
**Maddie**
a more "down to the foundation" way of saying
vernal jasper
yes negative powers denote reciprocals so if the take the reciprocal it will jus...
but simply, this is what will happen
eternal plover
yep
vernal jasper
I think it's more important for you to understand why it just "goes on top" instead of memorizing this way
$\frac{1}{x^{-2}} = \frac{1}{\frac{1}{x^2}} = 1 \div \frac{1}{x^2} = 1 \times x^2 = x^2$
eternal plover
i mean its simpler to understand that its reciprocal
jolly parrotBOT
Maddie
eternal plover
but yes maddie is teaching u the theory behind it
vernal jasper
Or simply yes, $\frac{1}{x^{-a}} = x^{a}$
jolly parrotBOT
Maddie
queen knot
So with everything you said wouldn’t that mean that the x^-3 would just change into x^3 then you would add the 2 and 3?
vernal jasper
Yes.....
queen knot
With Y would you change it into a positive number or would it stay a negative?
vernal jasper
Positive
We want to make it a clean expression
Do you remember for y^-2, how would you make it positive?
queen knot
It wouldn’t right?
queen knot
It’d be 3x^5/y^3?
queen knot
Thank you so much for helping me! I understand it better now :).
Have a good one!
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golden herald
Calc 2: If I am integrating a rational function, say (x3 + x2)/(x+1) can I immediately simply the integrand down to (x2(x+1))/(x+1) and then to just x2, or does that created like missed solutions/ have a different problem? I'm pretty sure that it doesn't because everything I've integrated so far only has 1 antiderivative (ignoring the constant) but I wouldn't be surprised if there are some functions with multiple distinct antiderivatives
tropic saddle
all antiderivatives only differ by a constant
it can happen that the formulas look very different
(most often in the case of trig functions due to the various identities)
but there is no problem like that here
cancelling out factors like this is fine
it only changes the function at a single point and integrals dont care about single points
pearl pondBOT
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✅ Original question: #help-39 message
golden herald
it can happen that the formulas *look* very different
I don't understand what you are trying to convey by this
Do you mean that if there is trig involved there can be multiple distinct antiderivatives?
I know there can be equivalent trigonometric functions that can be transformed into each other with identities
But that wouldn't be distinct
Because I could take any arbitrary antiderivative and write it in terms of different trigonometric functions without changing the meaning
golden herald
it only changes the function at a single point and integrals dont care about sin...
This part does make sense to me though
Definite intergals effectively give you area between the limits of integration
And antiderivatives are just a more generalized form of that
stuck hatch
Do you mean that if there is trig involved there can be multiple distinct antide...
the opposite. like you said, two functions can look very different from each other, but if they are equivalent expressions, then they will have the same antiderivative
pearl pondBOT
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royal galleon
Can someone help clarify the contradiction. He says that if we have q(x) is positive in so small interval meaning f(x) > f(c) when x>c. Wouldn't this only contradict the fact that f(c) was a max. Because if f(c) was a min this is what we would want. But we never assume it is a max or min so how can we contradict anything?
brazen hare
Can someone help clarify the contradiction. He says that if we have q(x) is posi...
relative maximum or minimum
we are assuming that c is the extremum in the function
still hamlet
Can someone help clarify the contradiction. He says that if we have q(x) is posi...
yes but in that interval when c>x we have f(c)>f(x) so this accounts for the min too
royal galleon
we are assuming that c is the extremum in the function
How can we contradict two things?
royal galleon
yes but in that interval when c>x we have f(c)>f(x) so this accounts for the min...
But does not happen if q(x) is positive only when it is negative
The only way I see this working is if he says the extremum is a min when q(x) is positive
And show it is a max but that just shows that it could be a min
still hamlet
if Q(x) is positive
say we have c>x,
then we have 0>x-c
that means the only way to have (f(x)-f(c))/(x-c) > 0 is if we have f(x)-f(c)<0
which implies f(x)<f(c)
royal galleon
if Q(x) is positive
say we have c>x,
then we have 0>x-c
that means the only way...
Did you not already contradict yourself if q(x) is positive then c>x cannot be true
still hamlet
im considering all points in some open interval around c
that includes points that are smaller than c
royal galleon
But how can we do this. We said that we are going to assume that q(c)>0 and then we can find a small interval where this is true and that means all the x and c in that interval must be x>c
still hamlet
we are assuming q(c)>0
since q is continuous, I can find some interval centered around c, where q(c)>0
royal galleon
Yes and then x>c must always be true
still hamlet
that interval will be of the form $(c-\delta, c+\delta)$
jolly parrotBOT
Green
royal galleon
But I still don't see what I'm not seeing
still hamlet
you are assuming we can only take all points x>c
there is no reason for that
could you tell me why you think x>c
royal galleon
Ok so we want to first assume that q(c) is positive
Oh wait I'm was closing my mind off. I kept saying that if q(c) is positive then they both must be positive. Which is silly. They can both be negative
So then our contradiction has how can you have an extrumum that is both a max or a min
Max and a min
still hamlet
the contradiction is that its not even an extremum if Q(c) =/= 0
royal galleon
Is that because what I said because it is showing it is both a max and min which makes no sense?
still hamlet
hmm
the proof is showing if Q(c)>0, then for some point near it, i have f(x)>f(c) and for some other point near it i have f(y)<f(c)
royal galleon
Ok so I'm assuming that they are happening at the same points which isn't necessarily what they are saying
But then we have many extrumums????
still hamlet
But then we have many extrumums????
why do u think that
royal galleon
Because we define a max by f(c)≥f(x) and min by f(c)≤f(x) for min
We are getting both possibilities
sonic patrol
Can someone help clarify the contradiction. He says that if we have q(x) is posi...
suppose $Q(c) > 0$, meaning it is positive in a small interval around $c$. This implies that, for $x > c$ in this small interval, $f(x) > f(c)$, so $c$ cannot be a local maximum. but also, for $x < c$, $f(x) < f(c)$, so $c$ cannot be a local minimum. thus, $c$ cannot be a local extremum
jolly parrotBOT
Pseudo (Cat theory #1 Fan)
still hamlet
yea but we're not getting both possibilities for all points around c
sonic patrol
if $Q(c) < 0$, meaning it is negative in a small interval around $c$, then for $x > c, f(x) < f(c)$, so $c$ cannot be a local minimum. but also, for $x < c, f(x) > f(c)$, so $c$ cannot be a local maximum. thus, $c$ cannot be a local extremum
jolly parrotBOT
Pseudo (Cat theory #1 Fan)
royal galleon
if $Q(c) < 0$, meaning it is negative in a small interval around $c$, then for $...
Why did you say that min is f(x) < f(c) isn't that a max and likewise for what you wrote for min
sonic patrol
Why did you say that min is f(x) < f(c) isn't that a max and likewise for what y...
If f(x) < f(c) near c, then c cannot be a local minimum
I wasn’t saying “if c is a local minimum, then f(x) < f(c) near c”
What we’re doing here is using “if c is a local minimum, then f(x) >= f(c) near c”, and then taking the contrapositive
It seems like you might have mixed up the order of implication?
royal galleon
It seems like you might have mixed up the order of implication?
Wdym?
sonic patrol
Wdym?
You said “why did you say min is f(x) < f(c)”
I did not say that
I said “f(x) < f(c) near c, thus c cannot be a min”
royal galleon
I don't see why we can't just say it is a max and for the other one saying it is a min
sonic patrol
I said “f(x) < f(c) near c, thus c cannot be a min”
Do you agree with this statement
royal galleon
Yes
sonic patrol
Then we’ve shown that c cannot be a local minimum, right?
royal galleon
Yes
sonic patrol
The other statement is “f(x) > f(c) for x < c, meaning c cannot be a local maximum”
Do you agree with that statement?
royal galleon
Yes this is all see. You are saying that wr can't haven't it be a min when x is near c and then likewise for max. But what I don't see is why you can't say it is a max and a min
sonic patrol
If it was both a maximum and a minimum, then c would be a local minimum
But we’ve shown that it can’t be
We’ve shown that c cannot be A, and it cannot be B
That also excludes the possibility of it being “A and B”
If you a know a person is not white, and that they’re not a woman, then you definitely know they’re not a white woman
royal galleon
If it was both a maximum and a minimum, then c would be a local minimum
Wait how can it be one?
sonic patrol
If c is both a local maximum and a local minimum, then in particular it is a local minimum
This is because “A and B” implies “A”
If a person is both white and a woman, then they’re definitely white
sonic patrol
It can help to draw out Venn diagrams for these
royal galleon
What I don't understand is there an issue approaching it the other way. I see what your doing. You are showing it cannot be a min and cannot be a max
royal galleon
I didn't see that when I was reading and read it as both a max and a min which seems like a contradiction too
sonic patrol
Which have some intersection
royal galleon
They can't?
sonic patrol
I didn't see that when I was reading and read it as both a max and a min which s...
It’s possible for c to be both a local maximum and a local minimum
They aren’t mutually exclusive properties
royal galleon
How can it be a max and a min?
royal galleon
But that doesn't fit our definition of max or min
Oh wait nvm it is great or equal
Ok so what is the flaw in logic I am making when I say that instead of saying it is not a max I will call it a min. Is it just that in the end we wont obtain the contradiction we are looking for?
sonic patrol
Ok so what is the flaw in logic I am making when I say that instead of saying it...
By looking on one side of x, we conclude that c cannot be a minimum. You could say “ah but maybe c is a maximum”, but then looking on the other side of x we conclude that c cannot be a maximum
So we’ve eliminated both possibilities
If c is a local extremum, it has to be a minimum or a maximum
But we’ve shown it cannot be either
royal galleon
So for the max part we say oh maybe it is a min but we eliminated that earlier so it is nothing right?
sonic patrol
Yes
Once we’ve shown that c cannot be a local minimum, we’re free to use that fact in future arguments
royal galleon
I see. Ok this all makes sense then. Thx for the help
pearl pondBOT
@royal galleon Has your question been resolved?
kindred forge
4
kindred forge
My idea is that if you’re revisiting a point on a walk to get from odd vertex to odd vertex, it’s redundant to even revisit said point
sullen cedar
in a graph, the sum of the degrees over all vertices must be an even number. So it cant be that the only two vertices of odd degree are in different connected components
kindred forge
in a graph, the sum of the degrees over all vertices must be an even number. So ...
Yeah I’m gonna say that too let me post my proof in a second
kindred forge
in a graph, the sum of the degrees over all vertices must be an even number. So ...
Does this make sense?
sullen cedar
it does, but i dont see the need for the first part. you just say that they cant be in two connected components and you are done.
it is true in any simple graph that if a walk between two vertices exists, then there is a path.
kindred forge
it does, but i dont see the need for the first part. you just say that they cant...
I wanted to be very thorough about when they’re in connected components just because
Going as overkill as possible since it’s my first proof in this class
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viscid moon
<@&286206848099549185>
pearl pondBOT
<@&286206848099549185>
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
Asking the actual question right away is more likely to get responses.
Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.
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robust lynx
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✅ Original question: #help-39 message
pearl pondBOT
robust lynx
ummm does the channel like explain lesson and topics?
No. It is to recieve help from other people.
Did you have a question?
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sharp smelt
pearl pondBOT
sharp smelt
kind of confused as to why the beta function is needed here
isn't it just $c \cdot \int_{0}^{1} x-2x^2+x^3dx =1$
jolly parrotBOT
wai
sharp smelt
isn't it just $c \cdot \int_{0}^{1} x-2x^2+x^3dx =1$
as the function is positive on the domain
inland ivy
It’s not needed but this is an example of a beta distribution
pearl pondBOT
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still hamlet
does part a amount to showing that we cannot have a non zero x_k because of linear independence?
still hamlet
ye
still hamlet
oh wait 💀
worldly jewel
still hamlet
does part a amount to showing that we cannot have a non zero x_k because of line...
well, if i were to show that a is even true, would this be the right way to go about it
worldly jewel
Do you know about the matrix of a linear map?
still hamlet
yea
worldly jewel
You can view the Aij as entries of a matrix A, which corresponds to a map F^n -> F^n
Now (a) says ker A = (fill in the blank), (b) says im A = (fill in the blank)
still hamlet
Now (a) says ker A = (fill in the blank), (b) says im A = (fill in the blank)
hmm {0} and F^n?
worldly jewel
yep
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wary vault
,tex
Hi there. I wanna write the negation of the first statement, which states that f diverges as x approaches a certain point. Is this correct?
$\lnot \left( \left(\forall M\in\mathbb{R} \right) \left( \exists \delta >0 \right) \left( \forall x\in\mathbb{R} \right) \left(\lvert x-x_0 \rvert <\delta\implies f(x) >M \right)\right) $
\vspace{0.23cm}
$ = \left( \exists M\in\mathbb{R} \right) \left( \forall \delta >0 \right) \left( \exists x\in\mathbb{R} \right) \left( \lvert x-x_0 \rvert <\delta \implies f(x) \leq M \right) $
wary vault
oops
jolly parrotBOT
fijokazż
worldly jewel
change the \implies to \land
unborn abyss
ummmm ok well you didn't negate it properly:
A -> B
is negated by
A & -B
unborn abyss
(remember that A -> B is equivalent to -A || B)
wary vault
does || mean or?
unborn abyss
yes sorry computer science brain
pearl pondBOT
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prime stream
Problem. Let $f$ be twice differentiable on an open interval $A$ containing $a$, and suppose $f''$ is continuous at $a$. Show that
$$\lim_{h \to 0} \dfrac{f(a+h) - 2f(a) + f(a-h)}{h^2}$$.
Remark. It was previously proved that if a function $g$ is differentiable at a point $c$ of an open interval, then
$$g'(c) = \lim_{h \to 0} \dfrac{g(c+h) - g(c-h)}{2h}$$.
I decided to apply L’Hôpital’s rule, but I have doubts, since it seems that the continuity of the second derivative is not used anywhere.
The function $f$ is continuous and differentiable on $A$, and also at $h = 0$ the numerator and denominator are equal to zero. Also (took derivatives with respect to $h$ of the numerator and denominator):
$$\lim_{h \to 0} \dfrac{f'(a+h) - f'(a-h)}{2h} = f''(a)$$
The last equality is by the remark above.
Therefore, by L’Hôpital’s rule,
$$\lim_{h \to 0} \dfrac{f(a+h) - 2f(a) + f(a-h)}{h^2} = \lim_{h \to 0} \dfrac{f'(a+h) - f'(a-h)}{2h}
= f''(a)$$.
Where is the mistake?
jolly parrotBOT
Dedekind
still hamlet
<@&268886789983436800>
still hamlet
**Dedekind**
hmm i dont think there's a mistake
prime stream
so the continuity of f'' is indeed not required?
still hamlet
so the continuity of f'' is indeed not required?
yea you don't necessarily need it
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cerulean smelt
can someone explain to me why limiting the digits in a number to 0 <= x <= n - 1 (where n is the base) ensures there is only one unique representation of any number
proper nova
probably remainder properties or smth
sharp vigil
that's not fully enough, you also have to ensure there are no 0.9999.... situations
proper nova
well assume for integers ig
cerulean smelt
yh
cerulean smelt
probably remainder properties or smth
what properties exactly
toxic lichen
can someone explain to me why limiting the digits in a number to 0 <= x <= n - ...
imagine that in decimal we used a digit A with value equal to ten, in addition to the standard digits 0 through 9.
is the number ten now 10 or A? is twenty 20 or 1A? is one hundred 100, A0 or 9A?
cerulean smelt
is the number _ten_ now `10` or `A`? is _twenty_ `20` or `1A`? is _one hundred_ ...
alright this makes sense, but why does restricting the digits as such prevents this problem?
it's not clear to me
toxic lichen
have you seen the euclidean division lemma?
its statement boils down to "the quot and rem of a euclidean division are unique"
cerulean smelt
no 🥲
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graceful garden
pearl pondBOT
graceful garden
How do I show this in brackets rather than < or >
Ik the sign with the equal is a closed bracket and the one without is an open bracket
graceful garden
[-2,2]
Yeah ik but do I just write this next to the equation or is there something I need to write
X=[-2,2]
Something like this?
versed mica
no
versed mica
$x \in [-2, 2]$
jolly parrotBOT
knief
ebon skiff
yeah
ebon skiff
yes
ebon skiff
yeah
graceful garden
How do I write a piecewise function in one
glass meadow
\left\{x<2:x^{2},x>2:x^{3}-22\right\}