i see
#help-17
somber portal
have a banana :3
long flame
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
faint carbon
vocal sleetBOT
desert moth
vocal sleetBOT
desert moth
why is it not working
hybrid flicker
flat whale
,w determinant [ [-1,2,1],[-3,2,-4],[0,3,-5]]
twin meteorBOT
worthy citrus
well everything "follows the intermediate value theorem" on x^2 because its continuous, but its just not going to tell you anything in this case with a and -a
vocal sleetBOT
@vast shale Has your question been resolved?
worthy citrus
what does IVT tell you
south badge
vocal sleetBOT
south badge
Is this correct
mild flower
yes
vocal sleetBOT
@somber salmon Has your question been resolved?
vocal sleetBOT
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.
vocal sleetBOT
@somber salmon Has your question been resolved?
mild flower
creo que normal es $Ax + By = C$ para lineas
twin meteorBOT
hayley!
vocal sleetBOT
Available help channel!
thin vale
Suppose $f:(a,b)\to \mathbb{R}$ and $g:(a,b)\to \mathbb{R}$ are differentiable and $c\in (a,b)$. Suppose $f(c)=0, g(c)=0$ and $g'(x)\neq 0$ when $x\neq 0$ and that $$\lim_{x\to c} \frac{f'(x)}{g'(x)}$$ exists and is equal to
vocal sleetBOT
Suppose $f:(a,b)\to \mathbb{R}$ and $g:(a,b)\to \mathbb{R}$ are differentiable a...
twin meteorBOT
Austin
split heart
@mild flower 
livid tapir
You mean x \neq c instead of 0?
livid tapir
bet
thin vale
shut up jan
mild flower
definitely not me
livid tapir
ur mom did
thin vale
can you guys stop
livid tapir
aight lol
well for starters what have you tried
U know what I'll just go ahead and drop a hint: MVT
thin vale
honestly, it says to compare to the other exercise
but there is no work done for that one
so I thought about how to show g(x)!=0 when x!= c
but
I am completly stuck
so
a bigger hint would be apppreciated here
dull bear
What would happen if g(x) were equal to 0 for some x != c?
thin vale
im not sure, like I bet division by zero somewhere
but I dont see where or why it is important
dull bear
U know what I'll just go ahead and drop a hint: MVT
Think about this/Rolles 
you'll soon find a contradiction
thin vale
if g(c)=0
and g(x) where x not c = 0
then in between x and c
there has to be a point say h where g'(h)=0
by rolles theorem
and
h not equal c
so this would be g'(x) =0where x !=c
so g(x) != 0 when x != c
has been shown
alright
and then now we already have that lim f'(x)/g'(x) exists
but we want to show it equal to lim f(x)/g(x)
where does this previous fact help us
well
at x=c g'(x)=0=g(x)
so could we just replace this in the limit
and now have limit x->c f'(x)/g(x)
instead of g'(x) on bottom
and then perhaps we should replace the f'(x) with the difference quotient of f(x)
or am I just blabbering chartbit
thin vale
so u r the one actually blabbering
sly sierra
i never denied it
dull bear
I’m trying to figure out what the way they wanted you to do it 
thin vale
chartbit does the opposite of blabbering
dull bear
chartbit does the opposite of blabbering
Yep, just lurk and react for the most part 
sly sierra
i didn't notice, at some point did you consider writing
$$\frac{f(x)}{g(x)} = \frac{f(x) - f(c)}{g(x) - g(c)} = \frac{f(x) - f(c)}{x-c} \frac{x-c}{g(x) - g(c)}$$
twin meteorBOT
Bungo
sly sierra
where in the first equality we use that f(c) = g(c) = 0
thin vale
I considered it but I couldn't ever write it that well
sly sierra
it might be fruitful, and in that last factor you can see why it's important that $g(x) \neq 0$ when $x \neq c$
twin meteorBOT
Bungo
thin vale
why?
sly sierra
because g(x) - g(c) = g(x), since g(c) = 0
thin vale
then all we have is (x-c)/-g(c)
sly sierra
and you don't want to divide by 0
or at least i don't want to divide by 0
you might
thin vale
@ gtbot
thin vale
**Bungo**
if we take limits, this is f'(x)/g'(x)
but it isn't lim x->c f'(x)/g'(x)
or atleast we haven't shown it is
sly sierra
ok, plan b then
$$\frac{f(x)}{g(x)} = \frac{f(x) - f(c)}{g(x) - g(c)} = \frac{f(x) - f(c)}{x-c} \frac{x-c}{g(x) - g(c)}$$
twin meteorBOT
Bungo
thin vale
plan b looks a whole lot like plan a
sly sierra
what does the MVT tell you about the two factors on the RHS
no, it's the same equation but we're gonna do something different with it
thin vale
well
there is a f'(c) eequal to left factor
and there is a 1/g'(c) equal to right factor
sly sierra
maybe call it something other than c, since c is taken
thin vale
call it zeta
sly sierra
maybe f'(a) and g'(b), where a and b are numbers between x and c
thin vale
or a and b
those work too
sure
so
$$\frac{f(x)}{g(x)}=\frac{f'(a)}{g'(b)}$$ for some $a,b\in (x,c)$
twin meteorBOT
Austin
sly sierra
now we're going to look at the limit as x->c
thin vale
the interval gets squeezed
sly sierra
notice that as x->c, then so do a->c and b->c
sly sierra
not quite
thin vale
where did we even use the fact that g'(x)!=0 when x!=c
sly sierra
because the problem is asking you for $$\lim_{x \to c} \frac{f'(x)}{g'(x)}$$
twin meteorBOT
Bungo
sly sierra
note the same x in the num and denom
whereas your a and b are not necessarily the same
thin vale
but as we squeeze our interval
to c?
okay nvm I see wym
but don't we always have this inqeuality for all x in (a,b)
so we can just choose an x s.t a=b? by squeezing our intervals tightly
hmmm
sly sierra
maybe try some funny business like $f'(a) = f'(a) - f'(x) + f'(x)$ and similarly for $g'(b)$?
twin meteorBOT
Bungo
thin vale
uh
sly sierra
probably not literally that but something in that spirit
you basically want to bound $$\frac{f'(a)}{f'(b)} - \frac{f'(x)}{g'(x)}$$ when $c < a < x$ and $c < b < x$ and $x\to c$
twin meteorBOT
Bungo
thin vale
I'm confused wym by that
sly sierra
well you need to argue that it's ok that you have f'(a) / g'(b) where a and b aren't necessarily equal
and that you'll still get the same limit as if you had f'(x)/g'(x)
thin vale
that part I understand is my next goal
sly sierra
so one way to do that is to show that the difference between those two fractions is negligible as x->c
thin vale
c<a<x as x->c
so a->c
same for c<b<x
b->c
so it goes to 0
so they go to eachother
?
sly sierra
i think you need to spell out the details..
thin vale
so considering this
as x->c
c<a<x
becomes
c<a<c
and similarly for b
c<b<c
so a=b=c
so f'(a)/g'(b)=f'(c)/g'(c)
and as x->c the other term also becomes f'(c)/g'(c)
so their difference tends to 0
sly sierra
you don't necessarily have that f'(x) -> f'(c) and g'(x) -> g'(c), do you?
you just know that f'(x)/g'(x) -> f'(c)/g'(c)
thin vale
why is that?
sly sierra
isn't that what's given?
they don't say that f' and g' are assumed continuous at c
unless you can somehow deduce that, but i don't think you can in general
thin vale
they are differentiable
over (a,b)
so they are continuous over (a,b)
c in (a,b)
oh f' and g' not f and g
sly sierra
like i could have f' = g', and the quotient will be identically 1, whether or not f' is continuous
derivatives of differentiable functions have the intermediate value property, but they aren't necessarily continuous everywhere
thin vale
you don't necessarily have that f'(x) -> f'(c) and g'(x) -> g'(c), do you?
okay, so sure lets say I accept this. Where did I even use it
sly sierra
the IVP means that they can't have jump discontinuities, but they could have the bad oscillatory kind
sly sierra
okay, so sure lets say I accept this. Where did I even use it
i'm not sure, because your argument wasn't clear, but it looked like you might be assuming that
thin vale
you just know that f'(x)/g'(x) -> f'(c)/g'(c)
well maybe I was, but if I have this then I wouldn't need that assumption, then would it hold up
?
sly sierra
i think you need the more souped-up powerful version of the MVT
sly sierra
to get something of the form f'(a)/g'(a) instead of f'(a)/g'(b)
mmm, can you prove it on the fly maybe?
because i think you basically need the same argument
sly sierra
what's the context, this is an analysis class iirc?
sly sierra
here's the proof from spivak, it's all of like 3 lines long
"the simplest of tricks suffices" - maybe just replicate that trick here then
instead of citing the cauchy MVT
sly sierra
if you have that, then the argument becomes very simple
sly sierra
your f(x)/g(x) becomes f'(a)/g'(a) for some c < a < x
and then as x->c, so does a->c
boom, done
sly sierra
when you have f'(a)/g'(b) the argument becomes a lot more wriggly
thin vale
sly sierra
because you're only given that the limit of f'(x)/g'(x) exists
same x above and below
so you need to maintain quotients of that form
that's why the cauchy MVT is the key here
it gives you quotients of that form
thin vale
okay but Bungo I hate to keep asking questions but I am lost, I see your idea about the cauchy MVT but where are we applying that?
sly sierra
well let's argue in terms of epsilon/delta
you want to show that if f'(x)/g'(x) -> L, then f(x)/g(x) -> L, so given epsilon > 0, we want to show that there's a delta > 0 such that
|f(x)/g(x) - L| < epsilon for 0 < |x-c| < delta
you have shown via Cauchy MVT that f(x)/g(x) = f'(a)/g'(a) for some a between c and x
so:
thin vale
you have shown via Cauchy MVT that f(x)/g(x) = f'(a)/g'(a) for some a between c...
no
where have we shown this
sly sierra
|f(x)/g(x) - L| = |f'(a)/g'(a) - L|, and that is small because f'(x)/g'(x) -> L
sly sierra
no
you said "where are we applying that"
this is where i'm applying it
cauchy MVT gives you such an a
maybe strike the "you have shown" and reword as "Cauchy MVT tells us that there exists an a such that"
thin vale
but how are you replacing f(b)-f(a)/g(b)-g(a) with f(x)/g(x)
thin vale
yes I recall
sly sierra
look only at the second expression, ignore the last one
f(x) - f(c) / g(x) - g(c)
by the cauchy MVT, that equals f'(a)/g'(a) for some a between x and c
thin vale
true dat
sly sierra
and now we know that f'(a)/g'(a) -> L, because a is sandwiched between c and x
and hence f(x) / g(x) -> L
thin vale
and now we know that f'(a)/g'(a) -> L, because a is sandwiched between c and x
?
as x->c
thin vale
f'(a)/g'(a) goes to f'(c)/g'(c)
sly sierra
so if x->c, then a->c
thin vale
I thought
sly sierra
they just said it converges to a limit
thin vale
and now we know that f'(a)/g'(a) -> L, because a is sandwiched between c and x
okay you lose me here then
sly sierra
f'(x)/g'(x) converges to some limit L as x->c
thin vale
given to us yes
sly sierra
so given any epsilon > 0, i can make |f'(x)/g'(x) - L| < epsilon provided that x is close enough to c
and if x is close enough to c, then certainly a, which is between c and x, is close enough to c
so |f'(a)/g'(a) - L| < epsilon as well
i can do this for any epsilon
sly sierra
yep
thin vale
and it is given that f'(x)/g'(x) -> L
sly sierra
and since you had f(x)/g(x) = f'(a)/g'(a)...
you can conclude that f(x)/g(x) also -> L
thin vale
and since you had f(x)/g(x) = f'(a)/g'(a)...
we have this from cauchy mvt?
sly sierra
yep
sly sierra
no...
f(x)/g(x) = (f(x) - f(c)) / (g(x) - g(c))
because f(c) = g(c) = 0
so i'm just subtracting 0 from the top and bottom
thin vale
very confusing yes
okay ty bungo I am pretty sure there is enough here I can gather myself upon a few rereads
sly sierra
nice
sly sierra
cheers
river kettle
could someone explain what happened here
vocal sleetBOT
could someone explain what happened here
river kettle
i understand the second line
where they brought down 2 from log(x^2)
but i got totally lost after
paper depot
they subtracted 2log(x) from both sides
and also subtracted log(1/4) from both sides
river kettle
but how did they get logx alone
paper depot
$3 \log(x) - 2 \log(x) = 1 \log(x)$
twin meteorBOT
Ann
paper depot
ignorantia legis algebrae non excusat
river kettle
oh i didnt know that was legal
paper depot
you didn't know the distributive law??
or did you delude yourself into thinking that the presence of logarithms somehow overrode it??
fathom drum
hows that distributive law
subtraction is distributing over multiplication here
river kettle
wheres the multiplication here?
fathom drum
wheres the multiplication here?
3 and log x are being multiplied
fathom drum
but to get 1logx isnt it subraction
What do you mean
river kettle
to get from 3log(x) - 2log(x) = 1 log(x) you are performing a subtraction yes
because 3 - 2 = 1
fathom drum
to get from 3log(x) - 2log(x) = 1 log(x) you are performing a subtraction yes
do you understand that this becomes log x?
river kettle
what?
fathom drum
because 3 - 2 = 1
yes, this is called the distributive law
fathom drum
Yes
Subtraction distributes over multiplication, like this: $$ 3 \times \log x - 2 \times \log x = 1 \times \log x $$
twin meteorBOT
rti
paper depot
$3 \cdot \log(x) - 2 \cdot \log(x) = (3 - 2) \cdot \log(x)$
twin meteorBOT
Ann
paper depot
@river kettle do you see it now
river kettle
why is (3 - 2) taken as a factor?
fathom drum
why is (3 - 2) taken as a factor?
What do you mean
river kettle
cant i just look at is 3 - 2 and ignore the log
and it would still work
would that cause problems
fathom drum
Not here
paper depot
no you cannot just ignore the logarithm.
this is like looking at the expression 300 - 200 and saying "can't i just ignore the hundreds and say this is just 3-2?"
paper depot
no
this has nothing whatsoever to do with logs as such
this is as i have already said
the distributive law
river kettle
can you show me an example without the log
paper depot
$3 \times 42069 - 2 \times 42069 = (3 - 2) \times 42069$
twin meteorBOT
Ann
paper depot
the distributive law
river kettle
the ac + bc stuff
paper depot
you're LITERALLY quoting the distributive law
you literally quoted the distributive law just now
like verbatim
river kettle
lmao
ok
how do I simplify this $\log_{18}100x-\log_{10}xy$ when there are two different bases
twin meteorBOT
water beam
fathom drum
how do I simplify this $\log_{18}100x-\log_{10}xy$ when there are two different ...
You can use the change of base formula
outer warren
cob law
river kettle
cob law?
outer warren
change of base
fathom drum
how can you simplify ln(1/3) to ln(3)?
What do you mean
river kettle
i mean just that
fathom drum
can you give an example where you did that
fathom drum
how can you simplify ln(1/3) to ln(3)?
those two are not equal
river kettle
fathom drum
there is a minus sign
fathom drum
why does -ln(3) = ln(1/3)
because ln(1/3) + ln(3) = ln(1/3 * 3) = ln(1) = 0
so subtracting ln 3 from both sides, you get -ln 3 = ln(1/3)
river kettle
$\ln\left(\frac{1}{3}\right)=\ln\left(3^{-1}\right)=-1\ln\left(3\right)=-\ln\left(3\right)$
twin meteorBOT
water beam
paper depot
that works too
paper depot
which line
river kettle
all of them
fathom drum
all of them
what about the first?
river kettle
the first is just the equation we are trying to solve
so it starts from the second line
fathom drum
from the first line to the second line, you use the rules of exponents to get everything in terms of e^x
golden rivet
Hel please the area of triangle A and B
fathom drum
Hel please the area of triangle A and B
Please go to another help channel
outer warren
Hel please the area of triangle A and B
river kettle
from the first line to the second line, you use the rules of exponents to get ev...
i dont understand
fathom drum
i dont understand
Do you know the rules of exponents?
fathom drum
but i dont see the application here
do you see why e^(x + 1) = e^x e^1 = e^x e
river kettle
no
river kettle
everything
fathom drum
everything
even the first step?
river kettle
everything
fathom drum
everything
for the first step, I used the fact that x^(a + b) = x^a * x^b
fathom drum
everything
not even the second step?
fathom drum
for the first step, I used the fact that x^(a + b) = x^a * x^b
after this, do you understand the first step?
river kettle
e^x * e^x = e^x^2
fathom drum
e^x * e^x = e^x^2
Do you mean (e^x)^2 or e^(x^2)? They’re different things
elfin moon
In this question I got the vector
vocal sleetBOT
In this question I got the vector
elfin moon
a hat= 1/2 i + 1/√2 j + 1/2 k
waxen hawk
what is the i and j mentioned above?
elfin moon
My question is what is the component of a vector
Vector a =ax i + ay j + az k
ax,ay,az are components not (cosx,cosy, cos z)
Am i right?
paper depot
yes you are
paper depot
what is the i and j mentioned above?
i, j and k are the three standard basis vectors of R^3
elfin moon
So here their answer is (1/2,1/√2,1/2) are components
Is this not wrong? These are the values of cosx,cosy,cosz
paper depot
a is said to be a unit vector.
elfin moon
Ohh it's unit vector
Yesssss
Same line😂
Got it thanks
So when we multiply it with 1 it doesn't affect the same as cos
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
chilly oyster
e^(x + 1) = e^x e^1 = e^x e
vocal sleetBOT
e^(x + 1) = e^x e^1 = e^x e
coarse ridge
Yes?
peak matrix
e^(x + 1) = e^x e^1 = e^x e
that's true, but whats the question?
coarse ridge
Yes it's right
peak matrix
,w verify e^(x + 1) = e^x e^1 = e^x e
twin meteorBOT
vocal sleetBOT
@chilly oyster Has your question been resolved?
hollow monolith
I'm kinda stuck here 🥲
trying to prove:
\[\lim_{x\to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x} = 1\]
twin meteorBOT
ᑎᗩᑕᖇᗴᝪᑌᔑᗞᗩᗯᑎ596
split heart
,rotate
twin meteorBOT
twin meteorBOT
hollow monolith
sorry for my weird handwriting 🥲
split heart
hollow monolith
did I write anything illegal in maths there? 😭
hollow monolith
yeah
cyan falcon
did I write anything illegal in maths there? 😭
I think Ur handwriting is illegal
white acorn
I think Ur handwriting is illegal
Looks better than mine somehow
cyan falcon
I can't really shid on u cuz my handwriting is worse
white acorn
lmao
hollow monolith
I think Ur handwriting is illegal
I really apologise 🥲
hollow monolith
Let's all fix our handwriting
that's illegal tho
split heart
why didn't you take the limit yet?
livid tapir
my handwriting used to be this bad I believe in u OP
hollow monolith
why didn't you take the limit yet?
wdym?
split heart
you said you wanted to take the limit as x goes to 0
so why haven't you
that would give you 2/2
which is what you want, no?
hollow monolith
I calculated the limit, it gave me 1
now I wanna prove it using the definition
split heart
ah
hollow monolith
which is what you want, no?
that's what I found weird, I calculated it easily, but I'm stuck at proving it lol
livid tapir
do you need to use epsilon delta to prove it?
hollow monolith
do you need to use epsilon delta to prove it?
is there any other definition? (newton's one is forbidden)
split heart
eyeballing it 
livid tapir
nah I was just wondering why you couldn't just add extra words to what you already did :p
hollow monolith
I wish I could, unfortunately this is maths, not coding :c
hollow monolith
nah I was just wondering why you couldn't just add extra words to what you alrea...
I tried to prove it using the definition of a limit
white acorn
Here's what I did
split heart
That's not what he wants
white acorn
No? Alr mb
livid tapir
no I mean you could say something like "if x \neq 0 then this equals this, so this limit is equal to this limit and this is continuous so its limit as it goes to 0 equals f(0)"
hollow monolith
Here's what I did
calculating it is easy, now try to prove it
white acorn
calculating it is easy, now try to prove it
Is the proof not the calculation
Okay I'm lost there
split heart
he wants to use the formal definition of a limit to calculate it
split heart
So @hollow monolith
hollow monolith
yeah?
livid tapir
Like you don't need the epsilon delta definition to still call it math is what I mean but if the exercise is asking for it then that's different
hollow monolith
Like you don't need the epsilon delta definition to still call it math is what I...
I wish all maths teachers were like you 🥲
split heart
You want to get to a form where you have |x|
livid tapir
No im serious 🤦♂️
livid tapir
ugh w/e you seem dead set on using epsilon delta
hollow monolith
or smaller than |x|
'cause anything smaller than |x| is smaller than delta
this doesn't make sense ig 🥲
livid tapir
is this for an intro real analysis class or calc? 
Nyways what you usually do is
You assume some bound on delta to make the denominator turn into a constant
hollow monolith
a positive one to remove the abs, already done ig
hollow monolith
is this for an intro real analysis class or calc? <:hmmCat:831508801646755850>
uhm, no 🥲
just some stupid high-school stuff
hollow monolith
wdym by bound?
livid tapir
I'm saying something like assuming delta < 1
hollow monolith
we don't it like that ;-;
livid tapir
Then that gives you a max value for 1/(sqrt(x - 1) + sqrt(x + 1))
hollow monolith
we only assume delta > 0
hollow monolith
🥲
livid tapir
You always have to assume delta is less than some function of epsilon for instance
1 is technically a function of epsilon
i think you just don't know how to do these types of proofs yet 🙃but anyways
hollow monolith
here's how I usually prove limits (i chose a simple function to quickly do it)
livid tapir
That'll give you a max value for 1/(sqrt(1 +x) + sqrt(1 - x))
In the range of x such that |x| < 1
Sooo that'll be hmmm
hollow monolith
i think you just don't know how to do these types of proofs yet 🙃but anyways
I don't master them yet lol
hollow monolith
here's how I usually prove limits (i chose a simple function to quickly do it)
I also used words btw, it's much easier 😂
livid tapir
ok finding the min value of this denominator is actually low key non-trivial lol
Lemme think
Guess to find the min you could use calculus but that's kind cheating
hollow monolith
i think you just don't know how to do these types of proofs yet 🙃but anyways
I wanted to prove it this way
livid tapir
I know
hollow monolith
oh, my bad
livid tapir
Once you can like... Replace the denominator with a constant it gets a lot easier to come up with a function for delta in terms of epsilon
And then you can say delta < min(g(epsilon), 1)
And that's why it's ok to assume delta is less than say 1
livid tapir
That'll give you a max value for 1/(sqrt(1 +x) + sqrt(1 - x))
The tricky part is idk how to maximize this in the range -1< x<1 tbh lol
Which is the same as minimizing the denominator
And using AM-GM inequality we can say that's greater than or equal to 2*sqrt((1+x)(1-x))
That doesn't rly help tho hmm
See if you weren't so dead set on using epsilon-delta and allowed yourself to invoke theorems like the limit of 1/f(x) as x goes to c is 1/(lim x-> c f(x)) if limit exists and not equal to 0
Then this would be way easier and would still be provable rigorously :p
hollow monolith
that's exactly what I was thinking
hollow monolith
let's consider this last shot as a lucky shot 😂
livid tapir
uh not sureif that means we're doing the former or the latter
Oh wait hold on
We can minimize the square of that that'll be a lot easier
So (sqrt(x-1) + sqrt(x + 1))^2 = 2x - 2sqrt(x^2 - 1)
wait rip nvm
hollow monolith
lmao
hollow monolith
K brb lemme see
don't tell me this guy is still trying 😭🤣
I feel you photomath, I really feel you 🥲
vocal sleetBOT
@hollow monolith Has your question been resolved?
white acorn
And using AM-GM inequality we can say that's greater than or equal to 2*sqrt((1+...
Oh cool amgm
I'd do it if I actually listened to my teacher lmfao
livid tapir
Yeah it doesn't help tho 🙃
white acorn
Wait shit sorry for the ping
livid tapir
I know the minimum is sqrt(2) and I could prove it ez with calculus but
Idk how to prove it without calculus lol
white acorn
The what
livid tapir
Min of sqrt(1 + x) + sqrt(1 - x)
white acorn
Oh
livid tapir
I was tryna prove it so I could find the max of its reciprocal so I could replace the denominator with 1/sqrt(2)
white acorn
Ah
livid tapir
And I'm tryna do it without calculus since I figured we might as well if he was insistant on using epsilon delta lol
white acorn
I tried epsilon Delta but could only do the legitimately easy ones
This one killed my soul
white acorn
Fr
True pains is what epsilon Delta is
Also they expect a high schooler to use it??
livid tapir
He's just doing it cus apperantely
hollow monolith
Also they expect a high schooler to use it??
child abuse 🥲
livid tapir
oh or not?
hollow monolith
He's just doing it cus apperantely
no, we must use the epsilon delta
livid tapir
rip
white acorn
no, we must use the epsilon delta
Then I'm scared for what's gonna happen to me in 3 years from now
hollow monolith
actually no, it was my fault, I chose maths branch for high-school, we study only maths and physics, sometimes English and philosophy
that's why we got harder stuff than other high-schools
luckily this year is my last year lmao (I'm 16yo 🥲 )
livid tapir
wait hold on
white acorn
actually no, it was my fault, I chose maths branch for high-school, we study onl...
Wait that's so much like my school
hollow monolith
what school? lol
white acorn
Except for us it's just maths and all 3 sciences
white acorn
what school? lol
Not gonna reveal my life right here
hollow monolith
Except for us it's just maths and all 3 sciences
the only things that matter lol, like what can be done with other objects like history or music? 😂
white acorn
the only things that matter lol, like what can be done with other objects like h...
On god
hollow monolith
Not gonna reveal my life right here
oh, sorry, I didn't mean to
white acorn
Nah it's fine w me people ask when they're comfortable, that's good
Anyway imma keep trying to use epsilon delta
hollow monolith
okie
white acorn
Hope I can finish by tonight lmfao
white acorn
I mean I quite literally just learnt what epsilon Delta is, id need to try some easier examples and work my way up, no?
hollow monolith
yeah, that's exactly what I did when I was learning it
vocal sleetBOT
hollow monolith
byeee
vocal sleetBOT
Available help channel!
proud hatch
hi
proud hatch
I have to proof the following proposition : {x} a independent set -> x =! 0
to proving this on my textbook they are saying the following : " by the definition of an independent set, Lambda * x = 0 -> lambda = 0. wich means by the proposition 1.2 (2) that x =! 0 "
I don't understand how he can conclude because the proposition 1.2 (2) dosn't implies lambda = 0 -> x =! 0
also "ou" in french meang "or" in english
hard atlas
do the contrapositive
if x is zero, then you can pick lambda != 0 and have lambda *x = 0, i.e. the set {x} is lin dependent
proud hatch
if x is zero, then you can pick lambda != 0 and have lambda *x = 0, i.e. the set...
that's true but what confusing me is the implication of the text book wich seems a bit wrong no ?
hard atlas
they are using that if x was zero then lambda x = 0 wouldnt have to imply lambda=0
but well the contrapositive is really the better way to phrase this
proud hatch
they are using that if x was zero then lambda x = 0 wouldnt have to imply lambda...
there is a case where lambda and x are both equal to zero so the implication is wrong no ?
proud hatch
but well the contrapositive is really the better way to phrase this
yes it's ways more clear
hard atlas
but lambda wouldnt have to be equal to zero
so lambda x= 0 doesnt actually imply lambda=0
because lambdax=0 also works for lambda=1
or lambda=17
vast shale
i want an equation f(x) where i can put 3 cords a b and c then it draws a line from a to b to c smoothly
vocal sleetBOT
i want an equation f(x) where i can put 3 cords a b and c then it draws a line f...
soft walrus
i want an equation f(x) where i can put 3 cords a b and c then it draws a line f...
you mean interpolation? :)
also note with three points such a polynomial will always be a quadratic when using Lagrangian interpolation
the red line will be what you're looking for
move the points around to see that it always works for any point :)
and note it should be $a.x<b.x<c.x$
twin meteorBOT
MrFancy
vocal sleetBOT
@vast shale Has your question been resolved?
soft walrus
bro i love you ok
nocturne isle
vocal sleetBOT
nocturne isle
can someone please help
vocal sleetBOT
@nocturne isle Has your question been resolved?
bitter plaza
can someone please help
Did u try using similarity
AF/AD = AO/AC = FO/DC
Where O is the new point
Use the same on other triangle
Sorry i don't think that helps because it does not utilise the specificity of the problem
wary mantle
AF/AD = AO/AC = FO/DC
Does AF/FD = AO/OC also hold?
bitter plaza
Does AF/FD = AO/OC also hold?
Yes
bitter plaza
can someone please help
OD is perpendicular to AC
drowsy gulch
OD is perpendicular to AC
Can we make that guarantee?
I suppose it's plausible that we don't have enough information otherwise, but it's pretty easy to construct a representation where AC and OD are not perpendicular, while keeping the rest of the conditions
actually yeah I'm content with saying that theyre intended to be perpendicular from the diagram, otherwise we're missing information
Nevermind, there's a path through this that's independent of the angle
drowsy gulch
can someone please help
The area that we're trying to find is a triangle, yes? Do you recall the formula for the area of a triangle?
waxen portal
~~actually yeah I'm content with saying that theyre intended to be perpendicular...
I don't see it.
drowsy gulch
I don't see it.
I can DM it to you, if you like.
vocal sleetBOT
@nocturne isle Has your question been resolved?
waxen portal
I can DM it to you, if you like.
I can get every area from 0 to 20, or essentially some number as a function of the x-coordinate of the right most line, the y-coordinate of the top-most line, and the y-coordinate of the middle cutting line
There is one and only one possibility if AC is perpendicular to the unlabelled line extending from D
drowsy gulch
Might I DM you my solution then to proofread?
waxen portal
Just post it.
drowsy gulch
I would rather not give out a solution without the opportunity to help the question asker through it.
waxen portal
I am not accepting DMs for this.
drowsy gulch
That's fine. Your call.
drowsy gulch
I can get every area from 0 to 20, or essentially some number as a function of t...
I do agree that any area could be output depending on where the line FE is, but I would recommend looking at the comparison between the area of ABEF and the area of the shaded region.
waxen portal
It's 0 because I can set FE to where AB is
drowsy gulch
When FE = AB, what's the area of the shaded region?
waxen portal
0
drowsy gulch
Yep
If we pick an arbitrary placement for FE, what're the areas of ABFE and the shaded region?
vast shale
hi
vast shale
,rccw
twin meteorBOT
vast shale
can you please verify the 5th question for me
I got ans as B and D but answer key says its B and C
I had an exam today
vocal sleetBOT
@vast shale Has your question been resolved?
vocal sleetBOT
@vast shale Has your question been resolved?
vast shale
broken grail
@vast shale
- Which question do you want to ask?
- Any work?
vast shale
5th
broken grail
Hint: It is a geometric sequence
vast shale
ive solved it
but iw ant to verify the ans
because in my answer its given as BC
i got ans as BD when i solved it in exam
its a multi correct question
broken grail
give me your step and let me verify it
@vast shale
(tag me if you want me to reply. Otherwise I might miss the message)
vast shale
give me your step and let me verify it
<@456226577798135808>
(tag me if you want...
but i solved it in rough sheet
i dont have it now
ok i took log
and then differentiated
so i got D as ans
but in ans key its given as C
broken grail
How do you handle the 1 in the RHS of (b), in differentiation?
broken grail
Try to rewrite the step and see whether it is really the way to handle 1 😉
(show me the entire step (and remember to tag me after you have finished))
vast shale
i am getting
the same
thing
ok
Can you se it
See
I don't have a good phone camera
@broken grail
broken grail
I apologize that I don't quite understand the step you are writing. \
Do you mean $f'(x) = 1 - 1 - f(x) \left[ \displaystyle\sum_{r = 1}^n \dfrac{rx^{r - 1}}{1 - x^r} \right]$? \
If so, could you elaborate on how you get this from formula (write the steps)
twin meteorBOT
kelvinchan9786#0690
vast shale
see i took the product term in B as y
broken grail
Show your steps on how you obtain the formula (not conceptually, but explicitly write the mathematical formula, pls)
vast shale
so like i wrote y in terms of f(x)
broken grail
i.e. how do you take log and take derivative?
broken grail
...yes, but you did wrong.
vast shale
what exactly though
broken grail
show me how you do it and let me discover the mistake! 🙃
vast shale
seriously
okay
Lmao getting 1 + f(x) now
@broken grail @broken grail
@paper depot can you help me here
vast shale
what ?
broken grail
because f(x) = 1 - y
broken grail
so it is correct
Here is my sol. Do you get it now?
yes, f(x) = 1- y, so f'(x) = -y'(x)
broken grail
Do you get how answer C is obtained now?
broken grail
OK, great.
I hope you understand that the reason of us not directly telling the answer. We hope the askers to discover the mistakes by themselves so they can learn from it, instead of directly telling the answer, which is quite useless. In fact, this is the common practice on most of mathematics group.
We would rarely troll anyone, at least we would not intentionally tell people wrong answers even we know that they are wrong. Hope that you can trust us!
Good luck on the future on mathematics! 🦾
@vast shale
twin meteorBOT
vast shale
on simplifying I got
This would have infinite solutions right?
How do I find a solution in a specific range
vast shale
We would rarely troll anyone, at least we would not intentionally tell people wr...
I am struggling alot and can't manage alot of time to practice math
broken grail
so far you obtain $\sin \dfrac{4\pi}{n} = \sin \dfrac{3\pi}{n}$, i.e. $\sin \dfrac{4\pi}{n} - \sin \dfrac{3\pi}{n} = 0$ \
Hint: use sum-to-product formula
twin meteorBOT
kelvinchan9786#0690
broken grail
@vast shale
vast shale
wow
broken grail
Do you get it now?
vast shale
yeah i am just wondering why i could not get it myself
broken grail
so next time you will know how it works. That's rather important
broken grail
I am struggling alot and can't manage alot of time to practice math
IMO, mathematics is the easiest subject, esp. in college.
The learning progress is quite systematic:
Definition -> Theorem -> Example -> Exercises
(At most of the cases understanding geometric meaning behind definition would hasten the learning pace)
Even in university, the learning progress is similar (Pure math):
Definition -> Theorem -> Proof -> Example -> Exercises
broken grail
It would much more easier then. Don't feel bad if you can't attempt some questions. Ask and learn them, and feel satisfied that you have learnt something more.
vast shale
do you think its necessary that i practice math everyday
broken grail
...practice if you are free / in the mood or you need it
vast shale
i need it but i am also not free
well basically my questions is should i focus on learning more things in math or practicising more things that i learned earlier
cause i need both
broken grail
You should practice until you can attempt most of the questions correct (esp. in high school).
No need to be completely perfect, but most of it, like 80 - 90%.
In general College mathematics is not that hard so you don't need a long time on concentrating time.
Understand the concept first and attempt the exercises (Don't make it converse), and at most of the time you should not be reading, but attempting exercises.
I think that it will take you a very short time to finish a chapter.
broken grail
Hmm...I can't justify it because I read pure mathematics university textbook instead of watching the lecture (I self-learnt mathematics),
but the importance is to find the ways you feel comfortable, so if you think that lectures are ok, that's no problem at all!
vast shale
yes lectures are ok but like 1-2 year from now i will be in college
and there are no good lectures on college mathematics
like calculus 3 and real analysis all that
so i will have to read
broken grail
In this case you will need textbook
DM me and let me give you some recommendations.
vast shale
yeah as i am going to take on a physics course ill need alots of mathematics
broken grail
...what physics do you refer to? Something like electrodynamics, quantum mechanics, or just basic 3-credit physics subject?
vast shale
well the course will be a compilation but basically an astrophysics course
vast shale
...what physics do you refer to? Something like electrodynamics, quantum mechani...
but it'll include those
broken grail
the difference is quite significant!
broken grail
basically you would only calculus (and occasionally statistics) for college physics, so don't worry about it
vast shale
but i wanna learn math
things feel more obvious in physics after math
to me atleast
broken grail
you have already learnt single-variable calculus and vector, haven't you?
Then there is no problem on physics, in terms of mathematics skill LOL
vast shale
yes i have learned
no
alot of mathematics is required
well thats what my physics teacher told me
broken grail
...how about reading a college physics textbook so you know what exactly it will be? (esp. if you are interested in physics)
At now you might feel very confused and worried, but it is just because you didn't know what it would be. By reading the textbook (or watching the lectures), you might feel relieved about it!
vast shale
but i am still learning physics
like the classical physics
i wouldnt understand college physics now i think
broken grail
broken grail
so why are you worrying? At last you will attend to the college. You don't need to completely self-learn it.
vast shale
all those things are in my currently running syllabus
and maths used in it elementary
vast shale
so why are you worrying? At last you will attend to the college. You don't need ...
i dont know i feel insecure not knowing things
vast shale
so why are you worrying? At last you will attend to the college. You don't need ...
college proffesors dont teach that well id rather read book
broken grail
If you really want to know how it would be, try to read college physics textbook. Most of college physics textbook would be same so just find one you feel comfortable to
broken grail
Real Analysis: https://link.springer.com/book/10.1007/978-1-4939-2712-8
IDK why I sent this sticker...sorry
This is a famous textbook for real analysis. DM me if you want to read it
vast shale
yes i do obviously
broken grail
Oh. You can download for free
vast shale
i dmd you actually
broken grail
oh, sorry that I missed the question you asked in DM. Let me answer them
orchid pier
why not divide it by the pwoer of 4?
vocal sleetBOT
why not divide it by the pwoer of 4?
orchid pier
the biggest is x^4 right?
but they do 3
or do they not want to find the horizontal assympotote?
.close
fresh jacinth
I don't know how to find the domain
vocal sleetBOT
I don't know how to find the domain
worn bobcat
the cube root is defined for all of R
division by zero is not defined, but x^2 >= 0 and therefore x^2 + 9 > 0
so therefore the domain is all of R
fresh jacinth
division by zero is not defined, but x^2 >= 0 and therefore x^2 + 9 > 0
isn't x^2+9 supposed to be not equal to zero?
worn bobcat
you can solve for when it equals zero and then exclude those points from the domain
then you'll notice that you have nothing to exclude
grim juniper
x^2 is positive or zero, so the denom is never 0
fresh jacinth
I don't get it
grim juniper
whixh part
fresh jacinth
So i Have x^2+9 does not equal zero
then x^2 does not equal to -9
then square root for x
but it gives imaginary numbers, and i need domains in interval notation
worn bobcat
if you agree that x^2 >= 0
then you'll agree that x^2 + 9 >= 9
and if you agree that 9 > 0
then you'll agree that x^2 + 9 > 0
fresh jacinth
yah, then I somehow need the interval notation from there
worn bobcat
it's every real number
fresh jacinth
so negative infinity to pos infinity?
fresh jacinth
okay let me try that
ohhhh, okay I see, It says it's right now been spending some time on this question, thank you for the help
.close
flat raven
can anyone help me on part c? My tutor told me to draw a graph but Im not sure how I can derive the area from that, she also said to just use the normal formula of bh/2
vocal sleetBOT
can anyone help me on part c? My tutor told me to draw a graph but Im not sure h...
flat raven
this is my working:
one sec
there
how do i find out the area of triangle ABP?
<@&286206848099549185>
<@&286206848099549185> sorry to ping twice but please someone help me find the area of the triangle using y=3x+2 , y=4-1.5x , and y=1
<@&286206848099549185>
obtuse sierra
stop pinging again and again!
flat raven
sorry
vast shale
<@&286206848099549185> sorry to ping twice but please someone help me find the a...
Did you find P,A,B points
flat raven
Did you find P,A,B points
i found P to be (4/9, 10/3)
vast shale
A and B?
flat raven
yes
vast shale
You know y. Find x for A and B points
flat raven
right so i know the height because i can do the y of P - the 1 from the line
so the height is 7/3
vast shale
Right
flat raven
and then i can use the distance between 2 points formula
to get the base?
base length*
because A and B are the base points of the triangle
vast shale
Two points are on the y=1.
vast shale
You don't need to use that formula
flat raven
why not?
oh
right
because
right i see
i need to find the X difference of the 2 points and that is my base lenght
got it
vast shale
Right
vocal sleetBOT
vocal sleetBOT
Available help channel!
round pasture
vocal sleetBOT
round pasture
I am confused on how we got from the first step to the bottom step, where the denominator is x^2 + 9x
I have tried to multiply both sides by the conjugate of 9 + x, but I end up getting x^2 + 81 in the denominator
rancid blaze
theyre multiplying by x
round pasture
ohhh. Idk why that was so difficult for me to think of
is there like a term for that or something, or is it really just that simple
round pasture
combining fractions, making a complex fraction not complex
but when I multiplied (9 + x) with the top I just got 0
that's why I was just confused. Like multiplying by x works, but not by (9 + x)
mild flower
uh yeah because (9+x) isn't a factor of the top?
but how did you get zero from that
round pasture
uh yeah because (9+x) isn't a factor of the top?
but how did you get zero from t...
I thought if you had like a three way fraction (a/b/c) you just multiply the bottom with the top
mild flower
if you have
round pasture
so I'm not sure why 9 + x needs to be a factor at the top
mild flower
$\frac{a}{\sfrac b c}$
twin meteorBOT
hayley!
twin meteorBOT
hayley!
twin meteorBOT
hayley!
twin meteorBOT
hayley!
mild flower
make sure you understand this
mild flower
I thought if you had like a three way fraction (a/b/c) you just multiply the bot...
writing a/b/c is somewhat ambitious but would be understood as the second case
ambiguous*
round pasture
writing a/b/c is somewhat ambitious but would be understood as the second case
ohhh, I had no idea
but when I apply what you said I still run into a similar problem,
mild flower
you have
ornate ember