but i will do the other ones. Let me see why this is fucking up 
#help-13
crimson sedge
\env{alignat*}{{2}
x + x & = (x+x) \cd 1 \q &\c b{x \cd 1 &= x} \\
&= (x+x)\cd (x+x') &\c b{x +x' &= 1} \\
&= \bs{x\cd(x+x')} + \bs{x \cd(x+x')} &\c b{x \cd (y + z) &= (x\cd y) + (x \cd z)} \\
&= \bs{(x\cd x) + (x\cd x')} + \bs{(x\cd x) + (x\cd x')} &\\
&=\bs[\big]{(x\cd x) + \underbrace{(x\cd x')}_{=0}} + \bs[\big]{\underbrace{(x'\cd x)}_{=0} + (x\cd x)} &\c b{x \cd x' &= 0} \\
&=(x\cd x) + (x\cd x)
}
yeah so this is what im getting
which is like really not helping
wraith daggerBOT
opal basin
Maybe you can expand (1 * x) as (x + x') * x
crimson sedge
Maybe you can expand (1 * x) as (x + x') * x
oh so u want me to do it in reverse?
crimson sedge
god bless you equality for being symmetric
ok but anyways
lets do this
[
x = 1\cd x = (x+x')\cd x = x\cd x + x\cd x' = x \cd x
]
wraith daggerBOT
opal basin
I found it while trying this $$x+x=x+\left(1\cdot x\right)$$
wraith daggerBOT
Jelle
crimson sedge
yeah its much easier this way
okay thats great
now proving x + 1 = x and x*0 = 0
lemme try
hm
so i cant say 1 = (x + x') because i dont have associativity
so that sucks
opal basin
can you use all of these btw?
crimson sedge
,tex i did like
\env{align*}{
x+1 &= (x+1)\cd 1 \ &= (x+1)\cd (x+x') \ &= \bs{x\cd (x+x')} + \bs{1\cd (x+x')}
}
wraith daggerBOT
crimson sedge
x*1 = x, 1\*1 = 1
yeah ig but my statement is still not really going muchh
opal basin
oh, yeah it simplifies to x + 1
crimson sedge
so question
when proving like stuff
can your premise contain whatever you want?
like we are only using the postulates here
but what if we take the other theorems as fact in our premise
opal basin
Btw, I'm not really qualified to do anything here, I just happened to solve the first problem
crimson sedge
no worries
so like
i tried doing it the other way
you get like [
1 = xx + x'x'
]
wraith daggerBOT
crimson sedge
but idk how u r meant to simplify that
Well we already know x*x = x so i will use that
so 1 = x + x'x'
but like
how do i recover the fact that x'x' = 1
that doesnt seem right...
opal basin
wait, does this work $$x+1=x\cdot1+1=1\cdot\left(x+1\right)=\left(x+x'\right)\left(x+1\right)$$
wraith daggerBOT
Jelle
opal basin
$$x+1=1\cdot x+1=1\cdot\left(x+1\right)=\left(x+x'\right)\left(x+1\right)$$$$=\left(x\cdot x+x\cdot1\right)+\left(x'\cdot x+x'\cdot1\right)=x+\left(0+x'\right)=x+x'=1$$
wraith daggerBOT
Jelle
opal basin
yeah, I'm not sure why my proof does work
maybe the order of expanding
you had (x + 1)(x + x')
opal basin
yeah, so both are valid, but they might give a different looking result
crimson sedge
mathematical trolling
ok wait
wait but
jelle
where did this term go
oh i see
you absorbed it using the previous theorem
opal basin
yeah, you should probably use multiple steps
is it possible for elements to not be 0 or 1 actually? It seems there is no use for other ones
crimson sedge
is it possible for elements to not be 0 or 1 actually? It seems there is no use ...
yeah they could
i mean im only dealing with two-valued boolean algebras
so only containing {0,1}
oh but ig (x')' = x
cant we just say like
x + x' = 1
opal basin
Maybe you need associativity first, or maybe it isn't even true
crimson sedge
wdym
opal basin
or is (x')' = x a postulate
crimson sedge
nah its not a postulate
it should be proveable
anyways i think its fine
the rest i dont really care about
thanks for the help ill close this now
.close
crimson sedge
hey
crimson sedge
well we have the following problem that i got wrong apparently
the first matrix is
$A = \begin{bmatrix}
2 & 0 \
0 & 3
\end{bmatrix}$
wraith daggerBOT
snow
crimson sedge
second matrix is $B = \begin{bmatrix}
1 & 1 \
1 & 0
\end{bmatrix}$
wraith daggerBOT
snow
crimson sedge
the task is to calculate C = BA(B)^-1 and to determine C^20
the way i did it is i realized that B * B^-1 is just matrix I which is
$I = \begin{bmatrix}
1 & 0 \
0 & 1
\end{bmatrix}$
wraith daggerBOT
snow
crystal raptor
BAB^-1 is not equal to ABB^-1 in general
crystal raptor
Matrix multiplication does not commute
south tundra
B * A and B^-1 and multiply
crimson sedge
yes
there is a problem tho
i couldn't get B ^-1
i know that B * B^-1 has to equal I , but when i calculate it, it doesnt give me the right result
i will do my calculations again i guess
for matrix A * B i got
$AB = \begin{bmatrix}
2 & 2 \
3 & 0
\end{bmatrix}$
wraith daggerBOT
snow
south tundra
Again, A * B is irrelevant
crimson sedge
it's not
south tundra
It's B * A * B^-1
crimson sedge
i could multiply them 2 by 1
south tundra
You are multiplying B * A and B^-1
south tundra
Not A * B and B^-1
crimson sedge
i am doing it by steps
wait..
$BA = \begin{bmatrix}
2 & 3 \
2 & 0
\end{bmatrix}$
true
wraith daggerBOT
snow
crimson sedge
its not the same
now i have to find out B^-1
$B^{-1} = \begin{bmatrix}
0 & 1 \
1 & -1
\end{bmatrix}$
wraith daggerBOT
snow
crimson sedge
$C = \begin{bmatrix}
3 & -1 \
0 & 2
\end{bmatrix}$
wraith daggerBOT
snow
south tundra
So C = BAB^-1
For demonstration, expand (BAB^-1)^3
And see what happens
You should notice a pattern
crimson sedge
you want me to calculate C^3?
south tundra
So that you can predict what C^20 is going to look like
crimson sedge
shouldnt i calculate like C^2?
south tundra
Just do it and you will see what I am talking about
crimson sedge
because 20 does not divide to 3
south tundra
I am not telling you to calculate C^3 for the sake of calculating C^20
crimson sedge
yes i know
south tundra
I want you to see the general rule for calculating C^n by first trying out some small values of n
crimson sedge
$C^3 = \begin{bmatrix}
27 & -19 \
0 & 8
\end{bmatrix}$
wraith daggerBOT
snow
south tundra
So what I meant is noticing the following
\begin{align*} C^3 = \left(BAB^{-1}\right)^3 = BAB^{-1}BAB^{-1}BAB^{-1} = BAAAB^{-1} = BA^3B^{-1} \end{align*}
Can you see guess what $C^{20}$ is based on this?
wraith daggerBOT
A Lonely Bean
south tundra
Yeah
And you can easily calculate A^20
Because the only nonzero entries are on the diagonal
Making $A^{20} = \begin{bmatrix} 2^{20} & 0 \ 0 & 3^{20} \end{bmatrix}$
wraith daggerBOT
A Lonely Bean
crimson sedge
yes i just calculated it
so now i have to multiply B*A^20
then multiply the matrix with B^-1
and that is the result
correct?
$BA^{20} = \begin{bmatrix}
2^{20} & 3^{20} \
2^{20} & 0
\end{bmatrix}$
wraith daggerBOT
snow
wraith daggerBOT
snow
crimson sedge
and this indeed checks out to be the right result.
well, thank you so much for helping me out!!
south tundra
You are welcome
cedar kilnBOT
south tundra
You too
cedar kilnBOT
Available help channel!
cloud inlet
whats the domain of fractional part function and -1
y = 1- {x}
tropic oxide
the domain of a function is the set of all numbers that the function accepts as input.
tropic oxide
so you are asking what the range of y = 1 - {x} is?
cloud inlet
like previously a good samaritan told me that it lies between -1 and 1 but when i do it in paper i dont get it
tropic oxide
you were told that the range is [-1, 1]?
this is incorrect even if you fiddle with the endpoint inclusions.
cloud inlet
yeah wait i will send a pic
crimson sedge
like previously a good samaritan told me that it lies between -1 and 1 but when ...
it was about domain of arccos
cloud inlet
tropic oxide
uncrop please
crimson sedge
not about the range of function inside
cloud inlet
let me get this straight obviously fractional part lies in 0 to 1
so 1-fractional part lies in -1 to 0 right?
lavish bobcat
Find all x ∈ R that solve the inequality depending on the parameter a ∈ R. Sketch the set of all pairs (a, x) where x solves the inequality for a given a.
lavish bobcat
<@&286206848099549185>
cosmic torrent
first of all , , transposes the elements into the same side and reduces to the same denominator .
After that you just use this : study the sign of $(x-a)/(x-b)$
wraith daggerBOT
Natural7
cedar kilnBOT
@lavish bobcat Has your question been resolved?
lavish bobcat
.close
twin panther
AB ∥ CD and AD + BC = 4√ 10 are given in trapezium ABCD. The area of this trapezoid with height 6 is 72 and it is possible to draw a circle outside. Find the radius of the inscribed circle of this trapezoid.
simple egret
!status
cedar kilnBOT
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
@twin panther Has your question been resolved?
proud isle
How do i solve this trigonometry question, how do i start?
cedar kilnBOT
How do i solve this trigonometry question, how do i start?
subtle harbor
the diagram involves using SOHCAHTOA. have you heard of that mnemonic?
proud isle
Yeah i haved heard abt it
subtle harbor
proud isle
Alright thanks
cedar kilnBOT
@proud isle Has your question been resolved?
lyric narwhal
I know that for a cubic $$ax^3+bx^2+cx+d$$ the cubic determinant $\Delta_3$ is given by $$\Delta_3=18abcd+b^2c^2-4ac^3-4b^3d-27a^2d^2$$
wraith daggerBOT
kheerii
lyric narwhal
if $\Delta_3>0$, the cubic has 3 unique roots
wraith daggerBOT
kheerii
lyric narwhal
if $\Delta_3<0$ the cubic has 1 real root and 2 complex conjugate roots
wraith daggerBOT
kheerii
lyric narwhal
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
raw sluice
could someone help me understand this question?
cedar kilnBOT
could someone help me understand this question?
floral salmon
I think they ask you to write $\frac{t^4+t^2+1}{(t^2+1)(t^2+4)^2}$ as some sum $\frac{p(t)}{t^2+1}+\frac{q(t)}{(t^2+4)^2}$ for some polynomials $p(t)$ and $q(t)$.
wraith daggerBOT
Crystopher
raw sluice
oh
floral salmon
At least it is something like that, you can find more info here: https://en.wikipedia.org/wiki/Partial_fraction_decomposition in case I'm wrong.
raw sluice
Is this correct?
FUCK MY WIFI
FFS
also how the hell do i solve this using partial fraction
floral salmon
It may be correct but you used x instead of t as a variable. I'm guessing that you need some substitution in the integral and then somehow use partial decomposition on the substitution.
raw sluice
It may be correct but you used x instead of t as a variable. I'm guessing that y...
alrrr thanks
raw sluice
also how the hell do i solve this using partial fraction
got any idea for this?
stoic gale
u = sqrt(x) maybe
floral salmon
right now I'm trying substituting $t=x^{\frac{1}{6}}$, so that $\sqrt{x}=t^3$ and $\sqrt[3]{x}=t^2$. Let's see how it goes.
wraith daggerBOT
Crystopher
raw sluice
right now I'm trying substituting $t=x^{\frac{1}{6}}$, so that $\sqrt{x}=t^3$ an...
Yh thats a smart one
got any hints for this one too?
tried substituting u = cosx
but i just ended up with smth harder
floral salmon
try u=sin x instead
raw sluice
**Crystopher**
dx will be 1/6 x^-5/6 ?
floral salmon
I noticed that, this is how I did:
$dt=\frac{1}{6}x^{\frac{-5}{6}}dt\iff\frac{6dt}{x^{\frac{-5}{6}}}=dx\iff \frac{6dt}{t^{-5}}=dx$
wraith daggerBOT
Crystopher
floral salmon
Although that made using partial decomposition a bit wacky.
raw sluice
Although that made using partial decomposition a bit wacky.
it says evaluate soo
floral salmon
It gets wacky because the numerator gets to have greater degree than the denominator, basically rendering decomposition useless.
raw sluice
yes
floral salmon
It is easier to perform polynom division instead.
cedar kilnBOT
@raw sluice Has your question been resolved?
lyric narwhal
it says evaluate soo
which one are you not able to do?
3?
for that one, substitute $u=\sqrt[6]{x}$ so that $\sqrt{x}=u^3$ and $\sqrt[3]{x}=u^2$and then use partial fractions
wraith daggerBOT
kheerii
raw sluice
bro i think my brain froze
do we need to do long divsion for t^3/(t-1)
wtf
why the hell is this q messing with my basic maths knowledge
floral salmon
What's wrong, care to show how is the division going?
raw sluice
i got it finally
floral salmon
In that one you will actually need decomposition.
raw sluice
yes
btw
q
how do u konw to sub u = sinx
instead of u = cosx
is it all try and error?
or u look at the denominator
floral salmon
I just looked at denominator and numerator and realized that if u = sin x then the numerator would conveniently disappear, leaving also a polynomial in the denominator.
Which would be fit to the theme of decomposition. If I wasn't 'alert' about decompositions it probably would fly over my head to do that substitution.
raw sluice
I just looked at denominator and numerator and realized that if u = sin x then t...
these are some impressive skills tbh, thanks a lot for helping me btw
floral salmon
yw
raw sluice
any ideas for this one?
i tried subs t= x^2, t^2 = x^4 and so on
Got this far
and i thought by partial fractions was gonna be the easiest under integration💀
floral salmon
Try factorizing the denominator in terms of x as is, then decompose the fraction in x.
raw sluice
Like that?
floral salmon
Ye, but you dont need Ax nor Cx, else the degree of the numerator would be 3.
But the problem then becomes that x^1 cannot be directly expressed in the decomposition...
raw sluice
Ye, but you dont need Ax nor Cx, else the degree of the numerator would be 3.
shouldnt the degree of the dominator be less by 1 from the nominator
floral salmon
True, then lets do as you stated.
floral salmon
What do you mean?
floral salmon
$\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+2}=\frac{x^3(C+A)+x^2(D+B)+x(2A+C)+(2B+D)}{(x^2+1)(x^2+2)}$
wraith daggerBOT
Crystopher
raw sluice
oh so i expand
floral salmon
Ye, that should make an equation system you can solve for A, B, C and D.
raw sluice
damnn alr thanks lemme see what i can do now
okay
im finally done with maths for today
thats enough
time for electricity
thanks a lot for helping me, idk what was i gonna do without u @floral salmon
floral salmon
Alright, good luck.
covert nimbus
is there a nice little notation to use when showing that you're finding the perpendicular gradient of a line
covert nimbus
for example, i have y=2x+2 and i want to show that i am doing the perpendicular line (y=-1/2x+c)
instead of just clumsily writing "perpendicular"
open canyon
Well, you use the formula m1*m2=-1 (m=gradient) when two lines are perpendicular to each other
So there's nothing clumsy or wrong in writing this
covert nimbus
oh ok i didnt know about that formula i was just taught it was the negative reciprocal of the first gradient
thanks
.close
remote willow
i dont know how to continue from here, how do i calculate the e^2/2+2 - e^0/2
dire geode
i dont know how to continue from here, how do i calculate the e^2/2+2 - e^0/2
Those are just numbers
,calc e^2
wraith daggerBOT
Result:
7.3890560989306
You should know what e^0 is
remote willow
i might have missed a chapter, the e^x is still confusing
dire geode
i might have missed a chapter, the e^x is still confusing
That's something you should have learned before calculus
crimson sedge
Is $\int_{a}^{b} -f(x) , dx = \int_{b}^{a} f(x) , dx$?
cedar kilnBOT
Is $\int_{a}^{b} -f(x) \, dx = \int_{b}^{a} f(x) \, dx$?
wraith daggerBOT
Bla
floral salmon
I think the rule is $-\int_{a}^{b} f(x) , dx = \int_{b}^{a} f(x) , dx$, but $\int_{a}^{b} -f(x) , dx = -\int_{a}^{b} f(x) , dx$, so ultimately yes, $\int_{a}^{b} -f(x) , dx = \int_{b}^{a} f(x) , dx$
wraith daggerBOT
Crystopher
crimson sedge
**Crystopher**
Okay and is there a short proof or intuition for this?
Ah wait you can prove it by the fundamental theorem of calculus right
wicked brook
Yes, you can also just think about how you partition [a,b]:
when considering the Riemann sum you multiply by $\Delta x = x_{i+1} - x_i.$ If you wanna go from b to a, you instead have to multiply by $x_i-x_{i+1} = -\Delta x.$
wraith daggerBOT
Zander
cedar kilnBOT
@crimson sedge Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
cedar kilnBOT
green jungle
which side is correct?
u do know that u could just take minus one out of the integral sign right?
that makes life easier
u should and like for the partial fractions i recommend just equating the thing as in you could common out x and write x(A+B)and then a plus b would equal to -1
then b-2a would be equal to -7
yes
to find out a and b basically thats an easier way
chill listen
the second step on the left
kindred bridge
hi
green jungle
x+7/(x+2)(x-2)
kindred bridge
<@&286206848099549185>
green jungle
yeah take out the minus sign
tropic oxide
!redir
cedar kilnBOT
This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.
green jungle
<@&268886789983436800>
then it would just be x plus 7 on the numerator
out of the integral sign i mean
you can take constants in and out of the integral sign btw if u didnt know
yes
wraith daggerBOT
bob420
green jungle
now x plus 7 / (x+1)(x-2) = A/x+1 + B/x+2
now do the lcm thing and cancel out the denominators on each side
yeah so now cancel out the denominators
now you can common out x and write x(a+b)-2a+b = x+7
yeah its easier right?
now integrate
dont forget the mins sign in the integral
yes sir
whats wrong
no
i think the sites bugged prolly the final answer should be 2ln(x+1)-3ln(x-2)
or did u write x^2-x-2
in the denominator ?
oh u did
nvm'
its ln x-2
its not wrong chill
try another integral simulator and see if the answers match up
cedar kilnBOT
@atomic pecan Has your question been resolved?
cedar kilnBOT
@atomic pecan Has your question been resolved?
winter tulip
How do I solve this?
green notch
So i'm reading this paper that uses the MAP Elites method for the evolution of flying machines in minecraft.
As far as I understand, it uses a multidimensional space where each dimension is a characterization of the solution, it then spreads out bins in that space, and those bins are occupied with the best performer that happens to have these characterizations.
What I don't understand is how we can make graphs such as the one attached (Where ME.C, ME.CN and ME.PO are different dimensions), where different dimensions are compared to one another. Since any bin will have the characterization of all dimensions, how can a dimension have a higher rate of succesful runs ?So i'm reading this paper that uses the MAP Elites method for the evolution of flying machines in minecraft.
As far as I understand, it uses a multidimensional space where each dimension is a characterization of the solution, it then spreads out bins in that space, and those bins are occupied with the best performer that happens to have these characterizations.
What I don't understand is how we can make graphs such as the one attached (Where ME.C, ME.CN and ME.PO are different dimensions), where different dimensions are compared to one another. Since any bin will have the characterization of all dimensions, how can a dimension have a higher rate of succesful runs ?
green notch
Mayb I wasn't clear, but i'm rather asking how they made them in the first place ? Any solution in the archive would be on each dimension, so I don't get how they say "x solution is from y dimension"
cedar kilnBOT
@green notch Has your question been resolved?
cedar kilnBOT
@green notch Has your question been resolved?
zinc pagoda
Can someone help me with this question please?
cedar kilnBOT
Can someone help me with this question please?
dull oxide
<@&286206848099549185>
!15m
cedar kilnBOT
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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.
cosmic torrent
If the X is center of C and r it radius , its equation is $(x-3)^2+(y-5)^2=r^2$ Then if C and I intercept , this means that the difference of its equtions is 0 . Expand the equation of C , transposes the $r^2$ and make the difference .
zinc pagoda
@cosmic torrent Do I need to use b^2-4ac
wraith daggerBOT
Natural7
cosmic torrent
<@1141379801656086598> Do I need to use b^2-4ac
For the question a no , just expand the equation .
cosmic torrent
<@1141379801656086598> Can you help me with part b please?
There , remark that , we know that I is tangent to C . This means the solution of our equation is one and its belongs both C and I .
cosmic torrent
$B \in C$ and $\in I$ , then try to find the value of x using $\delta = -(4k-26)^2-20(k^2-10k+34-r^2)$ use the positive solution .
wraith daggerBOT
Natural7
cedar kilnBOT
@zinc pagoda Has your question been resolved?
void sand
i'm finding this question a bit difficult
cedar kilnBOT
i'm finding this question a bit difficult
void sand
more precisely, I can't figure out how to use induction here
any suggestions?
<@&286206848099549185>
cedar kilnBOT
@void sand Has your question been resolved?
bold hinge
hint: does row reduction change the determinant?
how?
hint 2: how do you find the det of an upper triangular matrix?
void sand
hint: does row reduction change the determinant?
Swapping two rows reverses the sign of detA, scaling a row by c scales the determinant by c, adding a multiple of one row to another does nothing
void sand
hint 2: how do you find the det of an upper triangular matrix?
Take the product of the diagonal entries
bold hinge
oh nvm you already did 20
void sand
I guess start by finding det(A-tI) and equate coefficients to f(t)
what do you mean?
dire geode
what do you mean?
What didn't you understand
void sand
i don't see why that helps
dire geode
Which part
void sand
equating coefficients to f(t)
void sand
Doing this will help you see why deg(q) < n
i've done both 2 and 3 already
i can see why it's true
but not how to prove it lol idk
void sand
more precisely, I can't figure out how to use induction here
well yes, but i cant figure out how to do induction here, as stated before
it's not clear to me how the induction hypothesis is to be used
dire geode
When you proved n=3 case, that depends on the n=2 case
Then it's the same pattern for n to n+1
void sand
When you proved n=3 case, that depends on the n=2 case
did it?
dire geode
It can if you do it that way
void sand
i just laplace expanded the 3x3 case
i didnt see the 3x3 case being dependent on the 2x2 case...
dire geode
Well find a way to squeeze the n=2 case in there
cedar kilnBOT
@void sand Has your question been resolved?
void sand
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
wicked sphinx
cedar kilnBOT
undone epoch
(cos x)' = - sin (x)
undone epoch
Sure is.
undone epoch
Welcome.
wicked sphinx
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
umbral tapir
hii im a bit confused of how the first expression transformed into the next, can someone explain the gist to me? thank you sm! 🥲
tropic oxide
adding and subtracting -x+1
umbral tapir
is this the concept of complete squares?
or is it different
oh i dont think so cuz the numberator is not in the form ax^2+bx+c but i could be wrong
but the denominator is
just a bit confused why it was applied on the numberator and when do i know i should do it
humble karma
No it's not square completion.
Just adding and subtracting something, as Ann pointed out.
umbral tapir
is it an option when the degree of the numerator and the denominator is the same?
humble karma
I feel like you'd do synthetic division in those cases for a general solution, but it's usually equivalent to adding and subtracting something either way.
umbral tapir
thank you so much!!
humble karma
I don't think completing the square would get you anywhere in a situation like this, but methods for solving an integral may vary drastically even when changing something small, so it's possible that some integrals out there do benefit from completing the square. In this case, it serves no purpose.
cedar kilnBOT
@umbral tapir Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
limpid plume
I'll ask about something very simple but this confuses me. So, I've been tasked with the proof of R's completeness by our measure theory professor.
limpid plume
What I did is - I showed every Cauchy Sequence is bounded, then I used Bolzano-Weierstrass Theorem to get a convergent subsequence.
That subsequence converges to some x. I claim that the original sequence also converges at the same point x.
Now I continue like this -
The prof now says this is incorrect, and instead I need to use the triangular inequality such that we get 4 different terms instead of just two. But I don't see the error. When asked, he said to find it myself.
I've been trying to spot the error ever since, but am unsuccessful.
mental trail
N2 is false
You've shown convergence of the subsequence x_n_k, not x_n itself
So as you wrote it, such an N2 possibly doesn't exist
limpid plume
Wait, I don't get it. The subsequence also goes upto countably many terms right?
limpid plume
You mean I'll have an N_2 such that for all k >= N_2 i'll have |x_n_k - x| < eps, not n
mental trail
You mean I'll have an N_2 such that for all k >= N_2 i'll have |x_n_k - x| < eps...
This is correct yes
mental trail
So first, use the fact that your original sequence is cauchy
cedar kilnBOT
@limpid plume Has your question been resolved?
lament kelp
2x/x-2 > 1
cedar kilnBOT
2x/x-2 > 1
royal lark
2x/x-2 > 1
What is your question?
lament kelp
solve x
lament kelp
No I think we meant to do something else
lament kelp
Right
autumn fox
If x - 2 < 0 then the sign will change. So you need to split your work into cases
lament kelp
2x/x-2 > 1
How do I make this in big white text
royal lark
x>-2 means (-2, infinity)
autumn fox
How do I make this in big white text
$\frac{2x}{x - 2} > 1$
wraith daggerBOT
ℝ ⊇ Mikkel ∉ 𝐵(ℝ)
royal lark
What?
lament kelp
0 then?
royal lark
x+1>-1
lament kelp
no bro wyd
royal lark
2x/x-2 -1 >0
lament kelp
Yes bro
royal lark
It is incorrect
royal lark
The subtract is wrong
lament kelp
The subtract is wrong
I don’t know
royal lark
x+2>0
lament kelp
like can’t we just say
it cannot be 2
wait
Hmm
Fuck
the x can be -
<@&286206848099549185>
Cus if x was -
It could be
0
Which would mean
Lower than -2
x-2<0
No
No that should be correct
x+2<0
crimson sedge
2x/x-2 > 1
$$ \frac{2x}{x-2} >1$$
Check at what point funtion is undefined(denominator is 0 ) then subtract it from R
$$ Range = R-{..}$$
wraith daggerBOT
Snow
crimson sedge
ok so at what value of x the denominator is 0?
lament kelp
2
crimson sedge
yes
crimson sedge
2
now add it on a number line .
Add numbers that are not defined in the function like this:
crimson sedge
x+2/x-2 > 0 , x=/2
2x or x+2?
lament kelp
2x-x+2= x+2
lament kelp
now add it on a number line .
Add numbers that are not defined in the function l...
Hmm I suppose -2 and 2 > 0
I don’t understand
It’s meant to be -2 and 2 < 0
Also
so
hmm
crimson sedge
Hmm I suppose -2 and 2 > 0
at negative value function is defined so its not -2 🥴
At x=2, denominator would be 0 that is undefined so the range of this function is
R-{2} (all real number except 2 )
lament kelp
at negative value function is defined so its not -2 🥴
At x=2, denominator woul...
What do you mean it’s not -2
x-2>0
x>-2
x<2
x<-2
crimson sedge
What do you mean it’s not -2
is this IxI ? or x
function belongs (infinity , 1) U (3, infinity )
crimson sedge
x>-2
ahh i see
You forgot the sign rule
When we shift a constant from LHS to RHS then their sign is also changed
$$ x-2 >0$$
$$ x > 2$$
wraith daggerBOT
Snow
crimson sedge
It’s x cus we didn’t have power up or anything
No i mean modulus of x?
crimson sedge
**Snow**
then this should be right answer to your question
if its not modulus of x
crimson sedge
X<-2, x>2
i see i made a mistake
I accidently took it as ** > 0 not >1**
lemme me re-do it again
crimson sedge
**Snow**
$$ 2x/x-2 > 1$$
$$ \frac{2x}{x-2} >1$$
If the value of x is less than 2 then it would be an undefined function and if the value of x is more than -2
it would be bigger than 1
$$ R \in (-2 , 2)$$
$$ 2x/x-2 > 1$$
$$ \frac{2x}{x-2} >1$$
If the value of x is less than 2(max value) then it would be an undefined function and if the value of x is less than -2 then its a smaller value then 1
$$ R \in (-2 , 2)$$
wraith daggerBOT
Snow
crimson sedge
(2, ♾️), (-2, -infinity)?
If you put $x = \infty$ here then it would be a value less than 1 bc of denominator
wraith daggerBOT
Snow
lament kelp
right
crimson sedge
yea ._.
opal basin
<@&268886789983436800> ugh
cedar kilnBOT
@stoic arrow Has your question been resolved?
hasty pelican
quick question, would this be c_1 and c_2 or would they both be the same? It doesn't matter that much because one of them is eliminated but I would still like to be correct here
cedar kilnBOT
quick question, would this be c_1 and c_2 or would they both be the same? It doe...
@hasty pelican Has your question been resolved?
cedar kilnBOT
@hasty pelican Has your question been resolved?
hasty pelican
<@&286206848099549185>
cedar kilnBOT
@hasty pelican Has your question been resolved?
cedar kilnBOT
@hasty pelican Has your question been resolved?
bitter finch
quick question, would this be c_1 and c_2 or would they both be the same? It doe...
It doesn’t matter. Your solution is saying that T can be equal to sqrt(2s+c_1) for any real c_1, or it can be equal to -sqrt(2s+c_2) for any real c_2. That’s equivalent to saying T is equal to plus or minus sqrt(2s+c) for any real c
neat pike
if H is the orthrocenter of a triangle ABC then does BHA=AHC=BHC=120o imply that ABC must be equilateral?
peak salmon
a, b are real numbers. Knowing that a - b = pi, show that cosa * cosb <= 0
crimson sedge
a=π+β
b=α-π
mild arrow
How are you typing pi
crimson sedge
im greek
sharp kernel
Greek
mild arrow
No way
mild arrow
That’s such an advantage lmao
crimson sedge
Moes ligo dm
yee sure sned me whatever ya need
sharp kernel
αβγδεζ
peak salmon
a=π+β
b=α-π
and what do you do after?
mild arrow
Wait so on your keyboard you have English and Greek characters?
sharp kernel
Yes
crimson sedge
a, b are real numbers. Knowing that a - b = pi, show that cosa * cosb <= 0
wait lemme write it on a piece of paper so icam do
sharp kernel
Im greek too
peak salmon
wait lemme write it on a piece of paper so icam do
alright
dire thorn
Wait so on your keyboard you have English and Greek characters?
if u have a keyboard u can hold alt and press 227 on keypad
to get pi
sharp kernel
αβγδεζηθικλμνξοπρστυφχψω
dire thorn
other number combinations give other characters
idle tusk
a, b are real numbers. Knowing that a - b = pi, show that cosa * cosb <= 0
if it's <= 0, then either one is 0 or exclusively one is negative
sharp kernel
idle tusk
if it's <= 0, then either one is 0 or exclusively one is negative
a=b+pi
peak salmon
if it's <= 0, then either one is 0 or exclusively one is negative
cant they just both be 0?
i mean negative
idle tusk
so now we have $\cos(b)\cdot\cos(b+\pi)$
wraith daggerBOT
artemetra
peak salmon
**artemetra**
and now i apply the product to sum formula?
or?..
but that doesnt really help no?
idk
crimson sedge
**artemetra**
yeeee is $cos(\pi+b)$ and its negative cuz it is in the 3rd quadrant
wraith daggerBOT
Prime Minister
idle tusk
yeah
crimson sedge
so its negativeand at most 0
idle tusk
idk what formula is that but $\cos(x+\pi) = -\cos(x)$
wraith daggerBOT
artemetra
crimson sedge
idk what formula is that but $\cos(x+\pi) = -\cos(x)$
its not a formula its just that the cosine is negative in the 2nd and 3rd quadrant and positive in the 1st and 4th
crimson sedge
yems
restive willow
how do i solve this
cedar kilnBOT
how do i solve this
crimson estuary
Two people start walking from P at the same time. One person walks due west at 2.5 km/h and the other walks on a bearing of 250 degrees at 3km/h How far apart are the walkers after 4 hours?
crimson estuary
Is there something wrong with my diagram?
dire thorn
Is there something wrong with my diagram?
What did you do after this? I think u might be assuming the person walking west is directly above the person walking at 250 degrees
crimson estuary
I just used cosine rule for it
dire thorn
what did u get
crimson estuary
i got some negative number
Oh wait i put it into the calculator wrong sorry
Ok i got the correct answer now thanks for the help
.close
crimson sedge
cedar kilnBOT
warped drum
crimson sedge
i guess 90+196 so 286?
oh so to find the compliment you just subtract theta from 90
ok ty
.close
nimble mountain
cedar kilnBOT
nimble mountain
i dont get this... arent matrix entries numbers, why do they say each matrix entry is a differentiable function..
warped drum
matrices are defined on vector spaces
$\mathbb{R}^n$ is the usual vector space (the number entries), but differentiable functions form a vector space too (its pretty easier easy to check actually)
wraith daggerBOT
LF
cedar kilnBOT
@nimble mountain Has your question been resolved?
royal lark
cedar kilnBOT
royal lark
How to solve it?
bleak harbor
what are a,b,c. Postive? Real?
dire geode
Did you find the discriminant
royal lark
what are a,b,c. Postive? Real?
Not given
I thought that if the right side is 0
Then (x-a)(x-b) will be 0
Here x should be positive a,b
Same as c and d
so all a,b,c,d will be positive
And equal to a,b,c
bleak harbor
Okay I think maybe just expand the entire expression out and use the quadratic formula/discriminant
dire geode
so all a,b,c,d will be positive
What is d
bleak harbor
I'm not sure what d is but I would just expand it out and look at the discriminant
royal lark
🫣
royal lark
Ohh yes
Discriminant will be
√{4(a+b+c)^2 - 4(ab+bc+ca)}
I got this
Are you there guys?
I am stucked
meager jungle
thats just 1/2((a+b)²+(b+c)²+(a+c)²)
cedar kilnBOT
@royal lark Has your question been resolved?
royal lark
thats just 1/2((a+b)²+(b+c)²+(a+c)²)
How did you get this?
meager jungle
a²+b²+c²+ab+bc+ac = 1/2((a+b)²+(b+c)²+(a+c)²)
cedar kilnBOT
@royal lark Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
weary mesa
cedar kilnBOT
cedar kilnBOT
@weary mesa Has your question been resolved?
cerulean harness
first of all according to the reminder theroam i.e (ax - b) being a factor for f(x), f(b/a) = 0
2z-3 is factor for the function where a is a constant [We have to figure out the value of 'a' first to express the function]
so f(3/2) = 0
Hence get the value of 'a' (It'd be -1)
Now having one factor (2z-3) , we have to express the function in such order that we can get extra factors
like this->
f(z) = 2z^3 - z^2 - z - 3
= 2z^3 - 3z^2 + 2z^2 -3z + 2z - 3
= z^2(2z -3) + z (2z -3) + 1 (2z-3)
= (2z-3) ( z^2 + z + 1)
Hence the other factor for the function is (z^2 + z +1)
cedar kilnBOT
hollow trail
do you know how to find maxima/minima of functions in general?
cedar kilnBOT
@crimson sedge Has your question been resolved?
lyric narwhal
what did you get for the slope of that curve?
basically the slope of the curve is the derivative
they are asking you to find the maximum value of the derivative of y in the domain (-2,2)
cedar kilnBOT
@crimson sedge Has your question been resolved?
opal basin
What's the derivative of the function that's given?
you have to minimize that on the interval (-2, 2)
more negative is indeed better
spring shard
last saffron
In implicit differentiation, can we just separately calculate the derivatives of x and y (eg. by power rule?)
sinful geyser
Show an example
last saffron
So like if I want to differentiate dx/dy, and x = t^2 and y = t^3, then will dx/dy be 2t/3t^2
sinful geyser
Indian?
last saffron
Yeah
sinful geyser
Abhi me rd Sharma krra hu
last saffron
Same
sinful geyser
Join karega?
last saffron
Class?
sinful geyser
Meet ya kuch
last saffron
Nah, can’t talk
last saffron
Sure
sinful geyser
You got headphones?
runic ingot
cedar kilnBOT
runic ingot
we are talking in the world of Naturals(with 0), for what values of i this holds true?
i was thinking about it this way, if i =1, well of course it holds true, since i divides any number j.
But if we take for example any other i, for example i =0 or i = 5, we get that F iff F, isn't it true?
cedar kilnBOT
@runic ingot Has your question been resolved?
dull oxide
Are you saying that $p(i)$ is a set defined as $p(i)={i \in\bN:\forall (j\in \bN)(i|j)}$?
wraith daggerBOT
SWR
cedar kilnBOT
@runic ingot Has your question been resolved?
reef whale
What is the p(i) supposed to stand for
cedar kilnBOT
cedar kilnBOT
Available help channel!
cloud inlet
guys a elementary doubt
what does 3 number a,b,c in ratio represent?
a : b : c
cloud inlet
can we form individual relations based on them?
a : b , b : c, a : c
can we break them like this
nimble mountain
its one number in relation to another, yes you can split them
tropic oxide
yes, you can drop one or more numbers from a three-way or higher ratio
cloud inlet
its one number in relation to another, yes you can split them
thank u
so in the end we get 3 seperate equations
right?
nimble mountain
i mean you can get 3 separate equations from it yes. the 3 you mentioned above. but of those 3, any 2 is able to represent all the information
cedar kilnBOT
cedar kilnBOT
Available help channel!
rain nacelle
Damn
olive crescent
well because the noncommmon factors are not overlapping, we can split ab into two parts
let f = a without noncommon factors, g = b without... and x and y be the oncommon factors we have 500=fgxy, and xy=20
olive crescent
which means the common factors multiply to 500/20 = 25
and because they are common factors, we have 2 of them; that is f = g so f^2 = 25 and therefore f = 5
rain nacelle
let f = a without noncommon factors, g = b without... and x and y be the oncommo...
I quite did not get it a litte, missed in this part ngl
olive crescent
we can factor a (and b) into the bits which are unique to a and the bits which aren't
olive crescent
yeah pretty much
a = fx
for f being all the noncommon factors and x being the common ones
then for b = gx, g being noncommon and we have x again because they are the same as for a
then fgx^2 = 500 and fg = 20
rain nacelle
now i get it a lil
gud
okay now i get it
@olive crescent, thanks
merru crismas
.close
cedar kilnBOT
rain nacelle
yuppie
cedar kilnBOT
Available help channel!
vital pumice
cedar kilnBOT
vital pumice
sqrt?