What's the only way that "5+something" and "5-something" are actually the same?
#help-49
urban sentinel
0
urban sentinel
Yes
visual tiger
The only problem is the square root
If we want the solutions to exist in reals, the square root must exist
urban sentinel
Isn’t it a=0 a real solution
visual tiger
What's a
visual tiger
How does it solve the equation
urban sentinel
It doesn’t …
grand pondBOT
Rafilouyear2026
urban sentinel
Yes
visual tiger
So, if those solutions are to be real
We need the square root to be real
So we need 27+q to be a valid thing to put inside the square root
urban sentinel
Yes
grand pondBOT
Rafilouyear2026
visual tiger
And I'm assuming the question asks for two distinct real solutions
urban sentinel
Yes it does
visual tiger
Because we already know what happens in the case of double root
urban sentinel
Yes
visual tiger
So combine $27+q \geq 0$ and $q \neq -27$
grand pondBOT
Rafilouyear2026
visual tiger
We get ...
grand pondBOT
Rafilouyear2026
visual tiger
Can you rewrite this with only q on the left side?
grand pondBOT
Rafilouyear2026
grand pondBOT
Rafilouyear2026
visual tiger
And that's it
urban sentinel
Ooooh
Hold on
That’s it?
I don’t really get the steps we took that well
But kinda
visual tiger
We have a formula that gives us two solutions for our equation
So we can ask ourselves two questions:
- first question: when is the formula valid (when does it give us real solutions here)
- second question: when does it give us two different values, and when does it give us the same value twice
visual tiger
So, if those solutions are to be real
This is the answer to the first question
That's the answer to the second question
midnight plankBOT
I got it
If you are done with this channel, please mark your problem as solved by typing .close
urban sentinel
.close
midnight plankBOT
midnight plankBOT
Available help channel!
hazy fable
midnight plankBOT
hazy fable
Can someone help me solve this
midnight plankBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
small warren
Can I please get help understanding this problem? Here's my work, I don't understand how to algebraically get this.
midnight plankBOT
Can I please get help understanding this problem? Here's my work, I don't unders...
small warren
My work on the right is incorrect, x isn't equal to 6.
But my left-side work is.
Idk how, I mean I know how and I can explain it but I'm frustrated.
I know that the side opposite to the 30 degree angle which is the missing angle is x, and the side opposite to the 60 degree angle or angle V is x times sqrt 3 because of my lesson earlier and it corresponds to UW which is what we're trying to find in the question, and that the 90 degree angle's opposite side is VU which is just 2x.
Is this how I'm supposed to get the answer?
small warren
My work on the right is incorrect, x isn't equal to 6.
In-fact it's not even an option.
And $(6)^2+(6/2)^2 doesn't equal (18)^2$
grand pondBOT
Aurora
small warren
And $(6)^2+(6/2)^2 doesn't equal (18)^2$
I didn't multiply both sides by 2 🤦♀️
I was about to but forgot to, it wouldn't have made the statement correspond to x in the diagram anyway though. Right?
small warren
Can I please get help understanding this problem? Here's my work, I don't unders...
2x=18, x=9
2(9)=18
small warren
Is this how I'm supposed to get the answer?
I suppose I was supposed to make 2x=18, I didn't know that though so I just intuitively thought what value of x had to split 18 into 2 and came up with the answer 9 which is why i was so frustrated 
I've realized just now what just happened though, I figured out all 3 lengths of this special right triangle which has only one given length (the hypotenuse) based on the ratios I've learned about this type of triangle in general. Tysm anyway to anyone here trying to help. I think I've found my answer?
midnight plankBOT
@small warren Has your question been resolved?
sturdy summit
(You can ping helpers once as it has been over 15 minutes)
small warren
(You can ping helpers once as it has been over 15 minutes)
i'm gonna close, ty. I forgot i could ping them
.close
midnight plankBOT
midnight plankBOT
Available help channel!
azure bolt
guys i made a proof to explain to someone a while back and remembered but now i overthought it and i dont understand my working
midnight plankBOT
guys i made a proof to explain to someone a while back and remembered but now i ...
grand pondBOT
calvin
azure bolt
it was “show that the sum of exterior angles of a polygon is 360 degrees”
nova yoke
do you still have the diagram?
violet storm
it was “show that the sum of exterior angles of a polygon is 360 degrees”
that doesn't seem right
azure bolt
violet storm
maybe an exterior angle isnt what i think it is
fossil knot
do you know what the sum of the interior angles is
fossil knot
yes
azure bolt
no but can someone explain why i subbed back in
idk i made this proof myself
i cant believe i forgot it
fossil knot
what's the total sum of the angles of all the interior plus exterior angles
azure bolt
its n sides
violet storm
n sides -> n vertices
azure bolt
oh
violet storm
so 180 * n (sum of interior + exterior) - 180(n - 2) (sum of interior angles) = 360, isnt that all you need for the proof?
violet storm
if you are allowed the 180(n-2) knowledge
azure bolt
ohh i get it now
maybe
wait no
i dont
why is theta 360 and why am i subbing back into
sum of interior
fossil knot
notice how the sum of the interior angle of C and the exterior angle of C is 180
because they just make a line
grand pondBOT
calvin
fossil knot
so if you have n vertices, the sum of all interior + exterior angles is 180n
Now, you know the sum of the interior angles is 180(n-2)
violet storm
so 180 * n (sum of interior + exterior) - 180(n - 2) (sum of interior angles) = ...
i just said all this ^
azure bolt
notice how the sum of the interior angle of C and the exterior angle of C is 180
yes
ohhh
fossil knot
=> sum of exterior angles is 180n - 180(n-2) = 360
shell wigeon
In your proof you're using interior and exterior wrong
azure bolt
=> sum of exterior angles is 180n - 180(n-2) = 360
ohh that makes sense
shell wigeon
You first call "one exterior angle" theta, and then a few lines later "... means that one [interior] angle, theta, has ..."
shell wigeon
(plus at the start you can't just say "let theta be any angle between 0º and 180º" and then fix its value again)
shell wigeon
doubt
shell wigeon
If x is the interior, yes
azure bolt
yeah
wait ill fix it and redi it
redo
so how should i do this proof
like
im overthinkng
i wrote that x is interior
and 180-x is exterior
then the sum of interior is nx
shell wigeon
Tbh it seems easier to first find the value of exterior angles (from the "central" angles) and then deduce the value of interior angles
fossil knot
I think the way I presented it earlier is pretty straightforward logically
the proof that interior angles sum to 180(n-2) isn't hard
actually idk the direct way of proving exterior angles sum to 360 without this
wanton rover
and x = 180(n-2)/n
Wrong.
small jasper
Tbh it seems easier to first find the value of exterior angles (from the "centra...
hi 
azure bolt
br
wanton rover
cinnamon roles have more brain than you.
shell wigeon
all math is bullsh*t
<@&268886789983436800> troll
azure bolt
ari i think i understand how you did it now
wanton rover
and @wanton rover
azure bolt
cinnamon roles have more brain than you.
whatever you say
wanton rover
Give me food and feed a pig.
azure bolt
alpha_1 is a central angle, since n of them make up a full 360º, alpha_1 = 360º/...
yeah
wanton rover
9+10=???
shell wigeon
Then alpha_2 is 180º - beta_1 - beta_2, so 180º - (180º-alpha_1) = alpha_1
tribal temple
azure bolt
Then alpha_2 is 180º - beta_1 - beta_2, so 180º - (180º-alpha_1) = alpha_1
i never knew this proof was so complicated
shell wigeon
i never knew this proof was so complicated
It's really not
It doesn't use the fact that the sum of interior angles = 180º(n-2)
shell wigeon
Then alpha_2 is 180º - beta_1 - beta_2, so 180º - (180º-alpha_1) = alpha_1
So you know alpha_2 is the same as alpha_1, meaning it's also 360º/n
The sum of all is then, of course 360º
grand pondBOT
calvin
Compile Error! Click the 
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(You may edit your message to recompile.)
shell wigeon
Sure, it's valid
azure bolt
okay thank you
shell wigeon
My point was that knowing the formula for the sum of interior angles beforehand is a bit odd
midnight plankBOT
midnight plankBOT
Available help channel!
scenic wharf
Can ratio test for convergence of a Sequence, be used to prove divergence of a sequence?
midnight plankBOT
Can ratio test for convergence of a Sequence, be used to prove divergence of a s...
junior flower
it could help, yes
scenic wharf
help...how
junior flower
it can tell you a sequence diverges to infinity for example
scenic wharf
I am given some sequences and I have to tell whether it's convergent of divergent
junior flower
ohhh sorry i misread your question
scenic wharf
Sequence is b^n/n²
Where b>1
junior flower
it can tell you a sequence diverges to infinity for example
i edited this
scenic wharf
it can tell you a sequence diverges to infinity for example
Sequence diverges to infinity => limit doesn't exist ...
junior flower
if lim |a_{n+1}/a_n| > 1 then then (a_n) diverges
scenic wharf
Is it fully Rigorous?
junior flower
for any sequence that doesn’t have 0 terms, it is true
junior flower
if lim |a_{n+1}/a\_n| > 1 then then (a\_n) diverges
this basically says the terms grow exponentially or faster in absolute value. so the sequence diverges
midnight plankBOT
midnight plankBOT
Available help channel!
near geyser
midnight plankBOT
near geyser
I want to understand this solution
In mathematical induction we first see....f(1)....f(n)....then if it is true for f(n+1)
midnight plankBOT
Available help channel!
twilit field
question 1
midnight plankBOT
question 1
twilit field
$S_n^2= \frac{1}{n-1} \sum_{i=1}^{n} (X_i- \overline{X})^2$
\
We then have $V(S_n^2)= \frac{1}{(n-1)^2} V \left( \sum_{i=1}^{n} \left( X_i - \frac{\sum_{i=1}^{n} X_i}{n} \right)^2\right)$
My question is as it's iid X_i=X , no?
Where X~Ber(P)
grand pondBOT
wai
midnight plankBOT
@twilit field Has your question been resolved?
twilit field
<@&286206848099549185>
zealous schooner
What’s 5.2.23
runic hamlet
(5.2.23) is clearly an equation
untold sentinel
$\int_{0}^{1}\frac{\ln^{2}\left(x\right)\ln\left(1+x\right)}{1+x}dx$
midnight plankBOT
$\int_{0}^{1}\frac{\ln^{2}\left(x\right)\ln\left(1+x\right)}{1+x}dx$
grand pondBOT
zeta theta beta eta
untold sentinel
reduced it to $$\int_{0}^{1}\left(\frac{\frac{3}{2}\zeta\left(3\right)+2\text{Li}{3}\left(-t\right)}{t-1}-\frac{2\text{Li}{3}\left(-t\right)}{t}\right)dt,$$ no idea how to continue
grand pondBOT
zeta theta beta eta
untold sentinel
or wondering if there are alternate quicker methods
midnight plankBOT
@untold sentinel Has your question been resolved?
midnight plankBOT
@untold sentinel Has your question been resolved?
midnight plankBOT
@untold sentinel Has your question been resolved?
ember sinew
**zeta theta beta eta**
This looked okay,
ember sinew
**zeta theta beta eta**
but where the hell did Li and -t and the zeta come from
untold sentinel
but where the hell did Li and -t and the zeta come from
ill send my working
\begin{align*}
I &= \int_{0}^{1}\frac{\ln^{2}\left(x\right)\ln\left(1+x\right)}{1+x}dx \
&= \int_{0}^{1}\frac{\ln^{2}\left(x\right)}{1+x}\int_{0}^{1}\frac{x}{1+tx}dtdx \
&= \int_{0}^{1}\int_{0}^{1}\frac{x\ln^{2}\left(x\right)}{\left(1+x\right)\left(1+tx\right)}dxdt \
&= \int_{0}^{1}\frac{1}{t-1}\left(\int_{0}^{1}\frac{\ln^{2}\left(x\right)}{1+x}dx-\int_{0}^{1}\frac{\ln^{2}\left(x\right)}{1+tx}dx\right)dt \
&= \int_{0}^{1}\frac{1}{t-1}\left(\sum_{n=0}^{\infty}\left(-1\right)^{n}\int_{0}^{1}x^{n}\ln^{2}\left(x\right)dx-\sum_{n=0}^{\infty}\left(-t\right)^{n}\int_{0}^{1}x^{n}\ln^{2}\left(x\right)dx\right)dt \
&= \int_{0}^{1}\frac{1}{t-1}\left(\sum_{n=0}^{\infty}\left(-1\right)^{n}\int_{0}^{\infty}e^{-x\left(n+1\right)}x^{2}dx-\sum_{n=0}^{\infty}\left(-t\right)^{n}\int_{0}^{\infty}e^{-x\left(n+1\right)}x^{2}dx\right)dt \
&= \int_{0}^{1}\frac{1}{t-1}\left(2\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(n+1\right)^{3}}-2\sum_{n=0}^{\infty}\frac{\left(-t\right)^{n}}{\left(n+1\right)^{3}}\right)dt \
&= \int_{0}^{1}\frac{1}{t-1}\left(\frac{3}{2}\zeta\left(3\right)+\frac{2}{t}\operatorname{Li}{3}\left(-t\right)\right)dt \
&= \int{0}^{1}\left(\frac{\frac{3}{2}\zeta\left(3\right)+2\operatorname{Li}{3}\left(-t\right)}{t-1}-\frac{2\operatorname{Li}{3}\left(-t\right)}{t}\right)dt
\end{align*}
grand pondBOT
zeta theta beta eta
untold sentinel
the target is $4\operatorname{Li}_{4}\left(\frac{1}{2}\right)-\frac{\pi^{4}}{24}+\frac{7}{2}\ln\left(2\right)\zeta\left(3\right)-\frac{\pi^{2}}{6}\ln^{2}\left(2\right)+\frac{1}{6}\ln^{4}\left(2\right)$
grand pondBOT
zeta theta beta eta
midnight plankBOT
@untold sentinel Has your question been resolved?
small jasper
midnight plankBOT
@untold sentinel Has your question been resolved?
untold sentinel
$\int_{0}^{1}\frac{\ln^{2}\left(x\right)\ln\left(1+x\right)}{1+x}dx$
midnight plankBOT
$\int_{0}^{1}\frac{\ln^{2}\left(x\right)\ln\left(1+x\right)}{1+x}dx$
grand pondBOT
zeta theta beta eta
untold sentinel
https://math.stackexchange.com/questions/457371/alternating-harmonic-sum-sum-k-g...
i want to uncover the answer myself
without following a worked solution or i wont improve
midnight plankBOT
@untold sentinel Has your question been resolved?
fallow scarab
Wants to do it yourself. But asks for help from other people. Pick both
untold sentinel
?
midnight plankBOT
@untold sentinel Has your question been resolved?
candid gull
Hint: What would you do without the $\ln^2(x)$ term? Is $\frac{\ln(1+x)}{1+x}$ easy to integrate?
grand pondBOT
Mirror
midnight plankBOT
@untold sentinel Has your question been resolved?
untold sentinel
**Mirror**
what does that exactly help with here
marble hedge
what does that exactly help with here
Can you try putting 1/(1+x) = t
untold sentinel
$$\int_{\frac{1}{2}}^{1}\frac{2\ln\left(1-t\right)\ln^{2}\left(t\right)-\ln^{2}\left(1-t\right)\ln\left(t\right)-\ln^{2}\left(t\right)\ln\left(t\right)}{t}dt$$
$$=2\int_{\frac{1}{2}}^{1}\frac{\ln\left(1-t\right)\ln^{2}\left(t\right)}{t}dt-\int_{\frac{1}{2}}^{1}\frac{\ln^{2}\left(1-t\right)\ln\left(t\right)}{t}dt+\frac{\ln\left(2\right)^{4}}{4}$$
grand pondBOT
zeta theta beta eta
untold sentinel
$$=2\int_{0}^{\frac{1}{2}}\frac{\ln\left(t\right)\ln^{2}\left(1-t\right)}{1-t}dt-\int_{0}^{\frac{1}{2}}\frac{\ln\left(t\right)\ln^{2}\left(1-t\right)}{t}dt+\frac{3\ln\left(2\right)^{4}}{4}$$
grand pondBOT
zeta theta beta eta
untold sentinel
now what
dim shoal
Im trying a alternative approach
using this conceptually so I can wrap my head around and catch up to the progress. (let me know if I made an error on this function)
dim shoal
i want to uncover the answer myself
how did this function begin and take off?
this seems to be the part it gets interesting...give me some time? might can help.
maybe using this? but Im not sure if that is the answer unless the implication is the function can be continuous.
dim shoal
Think the issue is it stops being a arbitrary integral and it becomes the difference of two polylog kernels that cancel out.
midnight plankBOT
@untold sentinel Has your question been resolved?
dim shoal
pi im rough with, thought this was geometry as my first approach it was making shapes XDD
let me know if this helps
wanton spade
!occupied
midnight plankBOT
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
midnight plankBOT
@untold sentinel Has your question been resolved?
wind oxide
INTEGRATION HEHE
untold sentinel
<@522997196560531476> u still stuck?
from $\int_{0}^{1}\left(\frac{\frac{3}{2}\zeta\left(3\right)+2\text{Li}{3}\left(-t\right)}{t-1}-\frac{2\text{Li}{3}\left(-t\right)}{t}\right)dt,$ yes
grand pondBOT
zeta theta beta eta
visual pier
**zeta theta beta eta**
ah.. another one of these
zenith ibex
someone should make a bot trap
cedar berry
add an honeypot channel
midnight plankBOT
@untold sentinel Has your question been resolved?
dim shoal
Thought the aim was to reduce and conclude this mystery function
bold peak
.close
midnight plankBOT
bold peak
.reopen
midnight plankBOT
✅ Original question: #help-49 message
untold sentinel
**zeta theta beta eta**
i'm still stuck on this
also i got a new question that's unrelated but idk if i can send it
solid iris
@pliant peak we dont allow pirating here
civic lynx
!occupied
midnight plankBOT
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grand canopy
Wants to do it yourself. But asks for help from other people. Pick both
what an incredibly naive thing to say. do you know what giving hints are/nudging in the right direction?
fallow scarab
blocked
untold sentinel
i agree with them
but keep drama out of this channel or unnecessary comments if its not helpful
untold sentinel
https://math.stackexchange.com/questions/457371/alternating-harmonic-sum-sum-k-g...
if anyone is up to explaining whatever this harmonic stuff means and how it can be motivated im up for that as well
wind oxide
if anyone is up to explaining whatever this harmonic stuff means and how it can ...
well i skimmed through the MSE link and similar integrals appear when computing that harmonic sum
and ur integral also appears
not sure why civil linked it tho
@untold sentinel are you familiar with feynmanns technique?
untold sentinel
yes
untold sentinel
**zeta theta beta eta**
that landed me here
wind oxide
dont you feel your integrals looks feynmannable?
untold sentinel
\begin{align*}
I &= \int_{0}^{1}\frac{\ln^{2}\left(x\right)\ln\left(1+x\right)}{1+x}dx \
&= \int_{0}^{1}\frac{\ln^{2}\left(x\right)}{1+x}\int_{0}^{1}\frac{x}{1+tx}dtdx \
&= \int_{0}^{1}\int_{0}^{1}\frac{x\ln^{2}\left(x\right)}{\left(1+x\right)\left(1+tx\right)}dxdt \
&= \int_{0}^{1}\frac{1}{t-1}\left(\int_{0}^{1}\frac{\ln^{2}\left(x\right)}{1+x}dx-\int_{0}^{1}\frac{\ln^{2}\left(x\right)}{1+tx}dx\right)dt \
&= \int_{0}^{1}\frac{1}{t-1}\left(\sum_{n=0}^{\infty}\left(-1\right)^{n}\int_{0}^{1}x^{n}\ln^{2}\left(x\right)dx-\sum_{n=0}^{\infty}\left(-t\right)^{n}\int_{0}^{1}x^{n}\ln^{2}\left(x\right)dx\right)dt \
&= \int_{0}^{1}\frac{1}{t-1}\left(\sum_{n=0}^{\infty}\left(-1\right)^{n}\int_{0}^{\infty}e^{-x\left(n+1\right)}x^{2}dx-\sum_{n=0}^{\infty}\left(-t\right)^{n}\int_{0}^{\infty}e^{-x\left(n+1\right)}x^{2}dx\right)dt \
&= \int_{0}^{1}\frac{1}{t-1}\left(2\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(n+1\right)^{3}}-2\sum_{n=0}^{\infty}\frac{\left(-t\right)^{n}}{\left(n+1\right)^{3}}\right)dt \
&= \int_{0}^{1}\frac{1}{t-1}\left(\frac{3}{2}\zeta\left(3\right)+\frac{2}{t}\operatorname{Li}{3}\left(-t\right)\right)dt \
&= \int{0}^{1}\left(\frac{\frac{3}{2}\zeta\left(3\right)+2\operatorname{Li}{3}\left(-t\right)}{t-1}-\frac{2\operatorname{Li}{3}\left(-t\right)}{t}\right)dt
\end{align*}
grand pondBOT
zeta theta beta eta
wind oxide
err thats not feynman..
untold sentinel
same thing as feynman i expressed ln(1 + x) as an integral
wind oxide
**zeta theta beta eta**
and also the second and first integral (if u split everything) is quite easy
untold sentinel
its more compact
wind oxide
same thing as feynman i expressed ln(1 + x) as an integral
uhm okay what about double feynmann tho
untold sentinel
???
wind oxide
so the issue is $\int_0^1 \frac{ \text{Li} ( -t)}{t-1} \mathrm{d}t$
grand pondBOT
rak³en
untold sentinel
ok
wind oxide
**zeta theta beta eta**
yes so u shouldnt have a problem with zeta(3) and li_3(-t) integrals i suppose?
untold sentinel
how do i continue
wind oxide
well imo the most straightforward thing would to be split everything and deal with them separately
untold sentinel
you cant split the first one because they diverge
wind oxide
(i havent solved it every using this way, so i'll need some time to check if this is right)
untold sentinel
ok
wind oxide
you cant split the first one because they diverge
its log(t- oh right right
wind oxide
okay i cant do it ur way
@untold sentinel write an integral $I(a,b) = \int_0^1 x^a (1+x)^{b-1} \mathrm{d}x$
grand pondBOT
rak³en
midnight plankBOT
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
wind oxide
hey guys
if u dont have a question and just want to chat: #discussion
balmy sage
if u dont have a question and just want to chat: <#268882317391429632>
hey bro is it possible to connect with u?
im struggling and juggling my way up in maths
untold sentinel
<@522997196560531476> write an integral $I(a,b) = \int_0^1 x^a (1+x)^{b-1} \math...
that's gonna be tedious
since the only closed form? is hypergeometrics
and then differentiating a hypergeometric thrice 😵💫
wind oxide
since the only closed form? is hypergeometrics
u dont need the full closed form
just expand (1+x)^(b-1) for |x| < 1
wind oxide
yes
grand pondBOT
zeta theta beta eta
wind oxide
now diff wr.r.t b
wind oxide
and use gamma function properties to prove that around b=0 this is (-1)^k-1/k
atleast thats what i remember doing
untold sentinel
u said differentiate with respect to b
we have factorials
differentiation of generalized factorial (gamma function) involves digamma
,w D[Sum[Binomial[b - 1, k]/(k + a + 1), {k, 0, Infinity}], b]
grand pondBOT
Requested by opusclavicembalisticum
untold sentinel
bruh
wind oxide
we have factorials
no we have a product..?
untold sentinel
$\left(b-1\right)!=\Gamma\left(b\right)$
grand pondBOT
zeta theta beta eta
wind oxide
**zeta theta beta eta**
yes but we dont need to write it like that
untold sentinel
$\frac{d}{db}\left(\left(b-1\right)!\right)=\frac{d}{db}\left(\Gamma\left(b\right)\right)=\psi\left(b\right)\Gamma\left(b\right)$
grand pondBOT
zeta theta beta eta
untold sentinel
do you just want to differentiate it three times
im not up to this level of tedium
and it being a series as well
grand pondBOT
rak³en
wind oxide
do you just want to differentiate it three times
its not that tedious gng
its just looks like it
untold sentinel
did you get the answer from that?
wind oxide
also tf r u expecting
majority of these integrals are like this
alot of IBP, differentiation, known results
wind oxide
did you get the answer from that?
yes i have it
wind oxide
could you show me your working
will type the latex up gimme 10 mins
untold sentinel
where did you originally work it out?
you can show that
i dont mind if its messy
grand pondBOT
zeta theta beta eta
untold sentinel
this is what mathematica returned
untold sentinel
where did you originally work it out?
@wind oxide
untold sentinel
majority of these integrals are like this
show me an integral that's been done by differentiating a generalized series thrice
the only thing ive seen being differentiated thrice like that is the beta function and that's nice at least because of all the polygammas it ends up spewing out
wind oxide
where did you originally work it out?
i dont have the nb
untold sentinel
wdym
untold sentinel
how
why would you trash your working
after getting the solution
that doesn't make sense
very sus 🤨
wind oxide
indeed very sus
wind oxide
why would you trash your working
because i dont solve these for a living
me is a jee kid i do these for fun
untold sentinel
that doesn't add up
wind oxide
that doesn't add up
why do u even care gng 😭
untold sentinel
because I have a speculation you asked AI and since you have no working
for the idea of differentiating a generalized series three times... ive never seen anyone do that
untold sentinel
**zeta theta beta eta**
anyways after differentiating i believe you get this?
horrendous series when you sub a = 0 and b = 1
wind oxide
anyways after differentiating i believe you get this?
no because i never involved gammas but this should still give the same thing
untold sentinel
how did you simplify the series?
hello?
doesnt even numerically match up
unless i typed it wrong
wind oxide
doesnt even numerically match up
u typed it wrong :\ also just 5 more minutes trust
untold sentinel
**zeta theta beta eta**
@wind oxide could you explain the process of simplifying this series?
wind oxide
no because thats now how i did it
Define
$$
I(a,b)=\int_0^1 x^a(1+x)^{b-1} \mathrm{d}x
$$
Then
$$
\int_0^1 \frac{\ln^2 x,\ln(1+x)}{1+x},dx
=\left.\frac{\partial^3 I}{\partial a^2,\partial b}\right|_{(a,b)=(0,0)}
$$
For $|x| \leq 1$ , the binomial series gives
$$
(1+x)^{b-1}=\sum_{k=0}^\infty \binom{b-1}{k}x^k,
$$
where, for fixed integer (k),
$
\binom{b-1}{k}=\frac{(b-1)(b-2)\cdots(b-k)}{k!}
$
is a polynomial in (b) and is therefore differentiable for all real (b).
Interchanging summation and integration:
$$
I(a,b)=\sum_{k=0}^\infty \binom{b-1}{k}\int_0^1 x^{a+k},dx
=\sum_{k=0}^\infty \binom{b-1}{k}\frac{1}{a+k+1}.
$$
Now, differentiating the series termwise and using the identity
$
\left.\frac{d}{db}\binom{b-1}{k}\right|{b=0}=\frac{(-1)^{k-1}}{k}
$
we get
$$
\left.\frac{\partial I}{\partial b}\right|*{b=0}
=\sum{k=1}^\infty \frac{(-1)^{k-1}}{k}\frac{1}{a+k+1}.
$$
Now partially differentiate w.r.t $a$ twice and evaluate the series using results.
Proof of identity:
$(1+x)^(b-1) = \sum_{k \geq 0} \binom{b-1}{k} x^k$
Differentiate w.r.t b and set b=0:
$\frac{\ln (1+x)}{(1+x)} = \sum_{k \geq 0} \frac{\mathrm{d}}{\mathrm{d}b} \binom{b-1}{k} x^k$
but the LHS has gen func $\frac{ \ln(1+x)}{1+x} = \sum_{k \geq 0} (-1)^{k-1} \frac{x^k}{k}$
after which we do comparision
untold sentinel
💀 ???????
grand pondBOT
rak³en
wind oxide
guh whatever i aint fixing this
wind oxide
values of the eta and zeta functions my guy
untold sentinel
show me
you didn't show an evaluation of the series either? where's the solution you said you had
it also reads like AI
untold sentinel
im not sure how you get the polylog of order 4 from that unless im dumb
what final result did the AI return
wind oxide
in fact why tf r u even asking here if u think an ai can do it
wind oxide
i do not like resorting to ai
u think i do?
untold sentinel
calm down
i just want to see how you evaluated that series
because it's tedious as hell 😭
wind oxide
because it's tedious as hell 😭
all euler sums and related integrals are tedious
just take a look at this
untold sentinel
i just want to see how you evaluated that series
?
where did you throw your notebook
untold sentinel
just take a look at this
what of it? i dont see any answer triple differentiating a series and each step is very clear and well motivated enough
the answer by robjohn
wind oxide
what of it? i dont see any answer triple differentiating a series and each step ...
its still tedious
thats what i meant ...
untold sentinel
and yet, you still managed to omit that tedium in your "working" and call it "obvious from [there]"
dim shoal
we still reducing it down sizes yea?
best I got for it (but I got lost in the sauce of whatever the current conversation is about, hope it helps)
untold sentinel
also my progress now has reduced the integral to evaluating the remaining $$\int_{0}^{\frac{1}{2}}\frac{\ln^{3}\left(x\right)}{1-x}dx$$
grand pondBOT
zeta theta beta eta
untold sentinel
\begin{align*}
I &= \int_{0}^{1}\frac{\ln^{2}\left(x\right)\ln\left(1+x\right)}{1+x}dx \
&= \int_{0}^{\frac{1}{2}}\frac{\ln^{2}\left(\frac{x}{1-x}\right)\ln\left(\frac{1}{1-x}\right)}{\frac{1}{1-x}}\frac{1}{\left(1-x\right)^{2}}dx \
&= \int_{0}^{\frac{1}{2}}\frac{\left(\ln^{2}\left(x\right)-2\ln\left(x\right)\ln\left(1-x\right)+\ln^{2}\left(1-x\right)\right)\ln\left(1-x\right)}{x-1}dx \
&= \underbrace{\int_{0}^{\frac{1}{2}}\frac{\ln^{2}\left(x\right)\ln\left(1-x\right)}{x-1}dx}{=K,;\text{Domain split with }x \to 1-x}+2\underbrace{\int{0}^{\frac{1}{2}}\frac{\ln\left(x\right)\ln^{2}\left(1-x\right)}{1-x}dx}{\text{I.B.P.}}-\int{0}^{\frac{1}{2}}\frac{\ln^{3}\left(1-x\right)}{1-x}dx \
&= -\left(\int_{0}^{1}\frac{\ln^{2}\left(x\right)\ln\left(1-x\right)}{1-x}dx-\underbrace{\int_{0}^{\frac{1}{2}}\frac{\ln^{2}\left(1-x\right)\ln\left(x\right)}{x}dx}{\text{I.B.P.}}\right)-\frac{2}{3}\left(\frac{\pi^{4}}{15}+\underbrace{\int{0}^{\frac{1}{2}}\frac{\ln^{3}\left(x\right)}{1-x}dx}{x \to 1-x}\right)-\frac{5}{12}\ln^{4}\left(2\right) \
&= -\left(\int{0}^{1}\frac{\ln^{2}\left(x\right)\ln\left(1-x\right)}{1-x}dx-\frac{1}{2}\ln^{4}\left(2\right)-\underbrace{\int_{0}^{\frac{1}{2}}\frac{\ln^{2}\left(x\right)\ln\left(1-x\right)}{1-x}dx}{K}\right)-\frac{2\pi^{4}}{45}-\frac{2}{3}\int{0}^{\frac{1}{2}}\frac{\ln^{3}\left(x\right)}{1-x}dx-\frac{5}{12}\ln\left(2\right)^{4} \
&= -\left(\frac{1}{2}\underbrace{\int_{0}^{1}\frac{\ln^{2}\left(x\right)\ln\left(1-x\right)}{1-x}dx}{I_1}-\frac{1}{4}\ln^{4}\left(2\right)\right)-\frac{2}{3}\int{0}^{\frac{1}{2}}\frac{\ln\left(x\right)^{3}}{1-x}dx-\frac{2\pi^{4}}{45}-\frac{5}{12}\ln^{4}\left(2\right) \
&= \frac{\pi^{4}}{360}-\frac{2}{3}\int_{0}^{\frac{1}{2}}\frac{\ln^{3}\left(x\right)}{1-x}dx-\frac{2\pi^{4}}{45}-\frac{1}{6}\ln^{4}\left(2\right)\
\end{align*}
grand pondBOT
zeta theta beta eta
untold sentinel
so yeah that's the remaining integral
untold sentinel
best I got for it (but I got lost in the sauce of whatever the current conversat...
i misread this mb
but I don't believe that's right
trim quest
\begin{align*}
I &= \int_{0}^{1}\frac{\ln^{2}\left(x\right)\ln\left(1+x\right)}{...
If this is correct then there is no good repersentation by usual constants. It can be expressed as Li_4(1/2) which is expressed as multiple zeta, but no simple expression through pi, zeta(n) is known for that.
Of course pi^4/120 is not correct
grand pondBOT
zeta theta beta eta
untold sentinel
i finally got it
untold sentinel
thanks to whoever mentioned that substitution (1 - x)/x
i cant find it
i still wonder if there can be a shorter method to this...
untold sentinel
!occupied
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worldly ingot
mb
pure verge
i still wonder if there can be a shorter method to this...
is your doubt cleared?
untold sentinel
looking at that mse post I'm seeing people solve it with harmonic sums
wondering how that works
junior flower
mse meme integrals? in my help channels?
foggy obsidian
**zeta theta beta eta**
buoyant linden
blocked
💀
@grand canopy did he deadass block u?
he blocked me
i can tell cuz i couldnt put reactions on his msg
dull pendant
💀
i also got into an argument with him
once is normal. twice is concidential. thrice is a pattern.
buoyant linden
it seems like they blocked me as well for no reason
THEY BLOCKED YOU? 😭
ts (this) guy's such a clown
they lack any actual helping capacity
void kernel
move on, this is neither relevant nor appropriate for a help channel
and as a matter of fact, dogpiling a user for blocking you is not appropriate anywhere in the server
surreal viper
So this is where u guys been lurking
buoyant linden
So this is where u guys been lurking
hi
surreal viper
Hewo
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@untold sentinel Has your question been resolved?
celest turret
I've found the rotation matrix on the standard basis in R2
Now, I'm trying to find the rotation matrix for the basis {(sqrt2 / 2, sqrt2 / 2), (sqrt2 / 2, -sqrt2 / 2)}
I've got that the the representation of this new basis under the rotation matrix is
(sqrt2 / 2)(cos(theta)-sin(theta), cos(theta)+sin(theta) and
(sqrt2 / 2)(cos(theta)+sin(theta), -cos(theta)+sin(theta)
can i just use these two vectors as the columns of a matrix and that is the rotation matrix for this new basis?
midnight plankBOT
I've found the rotation matrix on the standard basis in R2
Now, I'm trying to fi...
celest turret
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hasty flume
i dont really get
midnight plankBOT
i dont really get
hasty flume
why rhs > lhs
i dont get how they get the rhs either, but i know how they did it
i integrated it and got 2e^sqrtx with limits n and 1
and then they did that integral + y when x=n
heres the mark scheme
hybrid goblet
why rhs > lhs
So you don't understand how rhs>lhs? Is that all?
hasty flume
yep
hybrid goblet
You see RHS is the the integration of the function which gives us the area under the given curve within x=1 and n
While LHS we are considering rectangles and adding their areas for example the first rectangle comes between x=1 and x=2
So the area would be the value of the function at this point (length) times the breadth(unit)
Now while adding the rectangles notice that few strips of areas are omitted out but while doing the integration it is considered, so RHS>LHS
hybrid goblet
excluded b asically
Yup excluded
These highlighted areas are excluded whole calculating the area of the rectangles
hasty flume
is it because when i integrate for n limit i still have a triangle left
ik its smth to do w induction
slow thorn
one more rectangle is not covered in the integral
x = 1 to x = n is spaned by n-1 rectangles
n th rectangle goes from x = n to n + 1
midnight plankBOT
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cursive granite
hey could someone tell me the area for this question
midnight plankBOT
hey could someone tell me the area for this question
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hybrid goblet
is the perimeter okay?
Yes it is correct
midnight plankBOT
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steel cedar
How does that integral not depend on x?
midnight plankBOT
How does that integral not depend on x?
fresh sparrow
How does that integral not depend on x?
the theorem statement is "if the integral is independent of path in D, then F is a conservative vector field on D"
so we assume the integral is independent of the path, and since its value doesn't change regardless of the path taken, the value is constant wrt x
steel cedar
then why don't the C2 integrals doesn't goes to zero when differentiated wrt x?
midnight plankBOT
@steel cedar Has your question been resolved?
steel cedar
I also still don't understand what it means to differentiate it wrt x
steel cedar
then why don't the C2 integrals doesn't goes to zero when differentiated wrt x?
Is it because we specifically ask it to be a horizontal line?
While C1 is arbitrary?
fresh sparrow
I also still don't understand what it means to differentiate it wrt x
(x,y) are the coordinates of a point
the path (a,b) to (x_1, y) doesn't change if you change the variable x
but the path C2 does since it's from (x_1, y) to (x, y)
steel cedar
the path (a,b) to (x_1, y) doesn't change if you change the variable x
Like slide it along x?
fresh sparrow
no
we define a function f(x,y) = integral bla bla
and we differentiate it with respect to x
here (x,y) is a point that is the end point of C2
but there is no "x" in (a,b) or (x_1, y)
x is the name of the variable
let me rewrite it more clearly because x and y are ambiguous
steel cedar
Sure"
fresh sparrow
let
$$
f(p,q) = \int_{(a,b)}^{(p,q)} \bf{F}\cdot d \bf{r}
$$
then we differentiate $f$ wrt $p$
grand pondBOT
artemetra
fresh sparrow
yes
fresh sparrow
They should've written (x', y') or sumn
does it make more sense now?
steel cedar
partially
fresh sparrow
ok
then i'll continue
the paths become $C_1: (a,b) \to (p_1, q)$ and $C_2: (p_1, q)\to(p,q)$
grand pondBOT
artemetra
fresh sparrow
and as you can see the first one does not depend on p
so the integral is constant over C_1 and doesn't change if p changes
fresh sparrow
and as you can see the first one does not depend on p
like it's literally not part of the endpoints
thus the derivative wrt p is 0
steel cedar
okay, that part is clear now, but what does differentiate the C2 integral wrt x really represent?
I mean wrt p
fresh sparrow
that means you move point (p,q) to another point (p+small distance, q)
wait no let me reread
yes it's how the integral changes when you move (p,q) a bit along the x-axis (or the p-axis i guess)
steel cedar
Like slide it along x?
So umm, slide it along x (infitely small)
fresh sparrow
yup
steel cedar
Alright! That's it i guess
fresh sparrow
awesome
steel cedar
Tysm!
fresh sparrow
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leaden seal
what does w.r.t. mean?
midnight plankBOT
what does w.r.t. mean?
low arrow
with respect to.
in the case of derivatives, the chosen variable (like dx, or dt, or whatever) is the variable being differentiated with respect to.
e.g.: dy/dx is the derivative of y wrt x.
leaden seal
thanks
low arrow
anything else you would like to ask?
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obsidian glen
Show that Cantor set C ⊂ [0, 1] contains no closed interval [a, b] with a < b
obsidian glen
i have no clue how I'd start
I know an open interval is one, for which we have an open ball of positive radius centred at each element in the intetrval and closed is not open
so I am guessing we proceed by contradiction
nova yoke
oddly stated question, it suffices to show that C contains no open interval (a,b)
which is how it's usually worded
obsidian glen
oddly stated question, it suffices to show that C contains no open interval (a,b...
closed no?
dreamy lichen
closed no?
[a,b] itself contains (a,b)
nova yoke
well if it contained [a,b] it would contain (a,b)
obsidian glen
ah yeah
nova yoke
show that if a,b are two points in the cantor set with a < b, then one of the open intervals excluded when constructing the cantor set lies in [a,b]
obsidian glen
so if I have the excluded intervals as $\cup_{n \in \mathbb{N}} J_n$ then I just show some $J_k \in [a,b]$ and use the general expression for these $J_n$s?
grand pondBOT
ginny
nova yoke
$\subset$, not $\in$, but otherwise yes
grand pondBOT
Bungo
obsidian glen
yeah mb
nova yoke
there's also a simpler argument involving basic lebesgue measure theory but i'm guessing you aren't doing measure theory here?
nova yoke
the essence of the argument is that the cantor set has length zero and any nonempty open interval has nonzero length
so the cantor set can't contain such an interval
obsidian glen
so if I have the excluded intervals as $\cup_{n \in \mathbb{N}} J_n$ then I just...
having a hard time with this, I have to show that an arbitrary element of this supposed interval J_k is in [a,b] as well, how would I do that though/
radiant roost
I think (2n+1)/(2*3^k) is never in the cantor set
Its not in J_(k+1)
Maybe this can be proven with induction
obsidian glen
hold up, J_k for me is the excluded portion
radiant roost
Yeah
obsidian glen
if that number isn't in the cantor set nor in J_{k+1} then where is it omg😭
radiant roost
Oh sorry, it IS in J_(k+1)
radiant roost
Maybe this can be proven with induction
Then it suffices to find (2n+1)/(2*3^k) in (a, b)
obsidian glen
so we essentially find k and n in terms of a and b ?
radiant roost
We probably have to use denseness or something
Make k large enough that the distance between (2n+1)/(2*3^k) and (2(n+1)+1)/(2*3^k) is less than b-a
I wonder if theres a more elegant proof
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uneven sandal
Let f be a real valued continues, monotone function.
Then is the following true?
The number of points where f is not differentiable is a Null Set (with respect to the Lebesgue measure)
midnight plankBOT
Let f be a real valued continues, monotone function.
Then is the following true...
junior flower
what do you think?
uneven sandal
what do you think?
I think it should be true
fallow scarab
I'm not even sure continuous is necessary
midnight plankBOT
@uneven sandal Has your question been resolved?
junior flower
yea it’s not necessary
well
if the domain is connected it’s not necessary
maybe still not necessary for disconnected domains but i don’t have a proof of that in mind
crisp granite
I am working on quadratic equations and I came across this equation that I dont know how to solve: x^2 = 1/2(5x+12)
Theres also a second one: x+(2/x)=-9/2
midnight plankBOT
I am working on quadratic equations and I came across this equation that I dont ...
crisp granite
Im stuck on how to start
I dont know where to start working on these
ping me if someone sees this and is here to help (im doing other questions)
crisp granite
oh hi
pastel garden
What do you know about solving a quadratic equation
crisp granite
I have learnt the cross method
where you use the x with the square and the c value to make up bx
kinda like x^2 - 3x -18 = 0
and then u do the thing
pastel garden
Alright let's solve the first equation. Rearrange it for me first so it's a simplified trinomial equal to zero
crisp granite
x^2 - 5x/2 - 6 = 0 (this is the point where i got stuck on)
pastel garden
don't forget to change the sign of the constant
pastel garden
yes so we have a rational term here (I.E. not an integer) do you know how to complete the square or use the quadratic formula?
crisp granite
I have learnt kinda how to use completing the square but its really annoying and a lot of the time I cant get it to work
i havent learnt the quad formula just yet :0
pastel garden
alright well can you try completing the square for me I'll help
crisp granite
k
pastel garden
The quadratic formula isn't necessary here.
crisp granite
so is the first step halving the coefficient of x and then squaring it?
crisp granite
(5x/4)^2
so it becomes that
and then -6
= that again?
(5x/4)^2 -6 = (5x/4)^2
pastel garden
yes and make sure to move the constant to the other side
crisp granite
But doesnt this become strange though?
because we're saying that a = a-6
which technically shouldnt be right
uhh
pastel garden
we're just moving the constant term to the right side
and then adding (b/2)^2 to both sides
crisp granite
(5x/4)^2 = (5x/4)^2 - 6
pastel garden
okay ill take you back to the previous steps
crisp granite
thx
pastel garden
we rearrange the equation to x^2 - 5x/2 = 6
(I've moved the six back because we need it that way if we are completing the square)
crisp granite
ye
pastel garden
then as you figured out we add 25/16
(or (5x/4)^2 )
to both sides
pastel garden
and sorry i made a mistake, and this may be what's causing the issue
pastel garden
we are taking the coefficient of b not the whole term
pastel garden
so that's adding (5/4)^2
crisp granite
so we add that to both sides
does the 5x/2 stay on the LHS?
pastel garden
yes it does we were just creating a value to add to the equation using b
crisp granite
oo k
pastel garden
I'm sure you know how to factor out x^2 - 5x/2 + 25/16?
crisp granite
I have been using the cross method but i feel like this would be hard
because fraction
pastel garden
what's easy about this part is we have already figured out what the binomial term would be
when we calculated b/2 to be 5/4
can you factor this for me now?
pastel garden
(x + k)^2 = x^2 + 2kx + k^2
crisp granite
i think i have seen this somehwere...
im not familiar with it
but
i think i can se it
so it becomes (x+25/16)^2?
pastel garden
close we don't square it in the factored form because it has yet to be squared in the binomial
pastel garden
(x + k)^2 = (x + k)(x + k) = (x * x) + (x * k) + (k * x) + (k * k)
pastel garden
since it's a perfect square we can also use the (x + k)^2 form
crisp granite
and to clarify k is 25/16?
pastel garden
k is the square root of 25/16
pastel garden
remember the middle term is negative so your sign here is wrong
pastel garden
I know what you're trying to write and the answer is yes
so can you write out the factored form
#help-49 message
you were close here
crisp granite
uhh
Im not sure what else to do tho
becuase if its that then what else would there be?
pastel garden
to make sure we're on the same page can you write out the equation that we have at this point
crisp granite
I am kind of confused where we are
to clarify we're not even looking at the RHS right now?
pastel garden
we'll bring that back in when you get this right
pastel garden
so it becomes (x+sqrt(25/16) ^2
you got to this point and i told you the sign of sqrt(25/16) was wrong
pastel garden
yes the bracket looks weird but yes
crisp granite
oh lol
pastel garden
so now you have (x - 5/4)^2
pastel garden
yes now that we've completed the square try solving the whole equation for x.
(x - 5/4)^2 = 6 + 25/16
crisp granite
(x−5/4)^2=121/16
thats the first step
and then...
take the root of both sides?
pastel garden
yes
pastel garden
yes
crisp granite
oh and then i just solve it by moving -5/4 over
ok
that makes sense now
Thank you for your help :)\
pastel garden
yeah sure
crisp granite
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gilded jolt
In a three player game of nim, assume players a and b will try to make themselves win, and if not, try to make the other win, while player c will only try to make themselves win, (we count a win for players a and b as the same win) is there any game in which the outcome would be different from a game of two player nim where one player goes twice in a row
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In a three player game of nim, assume players a and b will try to make themselve...
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gilded jolt
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amber thorn
So probability?
frozen talon
In a three player game of nim, assume players a and b will try to make themselve...
yeah, in an N game with the solo player going first; they reach 0, teammate 1 disturbs it, teammate 2 returns to 0
gilded jolt
What about games with nim stacks, like the 3 2 2 game
Wait
That just the sum of 2 player games, which is 2 players
Right?
frozen talon
(1 2) works; solo reaches (1 1), teammates take both stacks
gilded jolt
Alright
frozen talon
I guess it's not nim anymore because one side has access to moves the other doesn't
e.g. the solo player can't take from two different stacks on the same turn
hence not impartial
gilded jolt
I guess it's not nim anymore because one side has access to moves the other does...
hmm, i supose
so its a partial two player game practicaly
I think that’s all I need
Thanks!
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compact copper
Are u tamil
.I don't speak Tamil but that looks like gibberish to me
compact copper
So thats why I asked
.alright np man
pearl hull
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✅ Original question: #help-49 message
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upbeat timber
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upbeat timber
I'm doing the first question
Which asks is the function injective?
Its a real function
So I first chosed x,y \in R
And assumed f(x)=f(y)
And I found that there is another case which is the invert of x
Is my work correct?
upbeat timber
Yep
clever sedge
Do you understand injection from this, by the way?
subtle zinc
idk
upbeat timber
Do you understand injection from this, by the way?
The definition?
upbeat timber
I guess ,it's just that 2 functions having the same image should have the same x
image as in f(x)
So 2 distincts x , can't give the same value
clever sedge
Yes exactly
Different inputs must always give different outputs
And you got the opposite
You found that $f\left(\frac{1}{x}\right) = f(x)$
grand pondBOT
USS-Enterprise
clever sedge
But we know $x \neq \frac{1}{x}$
grand pondBOT
USS-Enterprise
upbeat timber
Yep
clever sedge
👍
upbeat timber
So for surjectivity , it means theres always a f(x)=y
upbeat timber
Meaning for every y , there's an antecedent x
clever sedge
So how would we find if it is surjective
upbeat timber
Assume f(x)=y and solve ?
subtle zinc
yup
clever sedge
Sure
upbeat timber
But what am I solving for ?
lyric charm
x
upbeat timber
,, 2x = y + yx^2
grand pondBOT
criRata
clever sedge
the lack of blackboard bold font in R is bothering me
clever sedge
You can't find x from this
Why not?
Get everything to one side and you've got a quadratic in x
lyric charm
You can't find x from this
you can, it's a quadratic in x
it'll be in terms of y
you can find its discriminant assuming that's legal for you to do
upbeat timber
Just solved the quadratic
clever sedge
Right we've got $yx^2 - 2x + y = 0$
grand pondBOT
USS-Enterprise
clever sedge
?
upbeat timber
I meant that I've did a mistake
clever sedge
Ah
upbeat timber
So we'd have delta= 4(1-y^2)
clever sedge
Yes
lyric charm
is delta always positive
upbeat timber
Nope
lyric charm
then what can we conclude
upbeat timber
Only if y in [-1,1]
upbeat timber
then what can we conclude
Nice
subtle zinc
wait f is obvi not surjective bc it goes to 0 on both sides
lyric charm
wait f is obvi not surjective bc it goes to 0 on both sides
correct conclusion wrong reasoning
a function can have limits at ±infty equal to 0 and still be surjective onto R
midnight plankBOT
subtle zinc
a function can have limits at ±infty equal to 0 and still be surjective onto R
ok but not if it's continuous and defined on all of R
clever sedge
you can't just imply it silently
Ann just hasn't mastered telepathy like me yet 😒
lyric charm
ok teacher
dunno if that was sarcastic or genuine but ok
upbeat timber
saluting should be genuine XD
midnight plankBOT
Available help channel!
thick mist
is the answer pie/3? since i wrote sin 2pie/3 as pie/pie/3 which becomes pie/3 so is the answer none of these?
midnight plankBOT
is the answer pie/3? since i wrote sin 2pie/3 as pie/pie/3 which becomes pie/3 s...
zealous schooner
yes the value is pi/3
dreamy lichen
btw it's pi, not 🥧
thick mist
btw it's pi, not 🥧
mbmb i always type pi as pie
uneven sandal
btw it's pi, not 🥧
now im hungry 😭
lyric charm
mbmb i always type pi as pie
habit worth breaking
lyric charm
is the answer pie/3? since i wrote sin 2pie/3 as pie/pie/3 which becomes pie/3 s...
also i think you meant pi minus pi/3
midnight plankBOT