this means that you can have everything relating to x on one side and everything relating to t on the other side
#help-19
clever fjordBOT
LF
wet lichen
And thats what i had here right
wet lichen
So what IS a differential equation attempting to find? just wanna really zero it in
dim thicket
you are solving for a function given a relation between its derivatives
dim thicket
wet lichen
that looks scary
dim thicket
for example those differential equations (maxwell) are very useful in electromagnetism
but if its for a basic calculus course, you wont go much further than the example you posted
only separable equations
wet lichen
i think i got the general intuition for now
the hard part for me was just knowing what a differential equation was because i didnt really know what i was doing
sick plaza
i used to hate that version of maxwells equations but when you work with them you get to understand them more fundamentally and truly learn the concepts embedded in them
gray locust
quick question
odd edgeBOT
quick question
gray locust
its more so physics than mathematics
I'm drafting a box on an incline with an atwood machine
I'm forgetting the name of force that would pull the box downwards
its not perpencidular force as that would be straight down from the center of the box, when I say downwards I mean pulling it down the incline not downwards in a sense of gravity
red thorn
I'm forgetting the name of force that would pull the box downwards
Are you referring to the component of gravity along the direction of the incline?
gray locust
Are you referring to the component of gravity along the direction of the incline...
yes
along the incline in the direction of the drawn arrow
red thorn
As far as I know, there isn't a special name for that force
red thorn
its the parallel force
Oh never mind
near marten
I'm not sure how to start 85
odd edgeBOT
I'm not sure how to start 85
forest sky
do you know a taylor series that could approximate ln(1.75)?
near marten
do you know a taylor series that could approximate ln(1.75)?
no
Yeah I honestly have no clue at all how to start
forest sky
one of the series you may already know is $$\frac{1}{1-x} = \sum_{n=0}^\infty x^n = 1+x+x^2+x^3 + \cdots$$ What happens if we integrate both sides?
clever fjordBOT
cloud
near marten
one of the series you may already know is $$\frac{1}{1-x} = \sum_{n=0}^\infty x^...
Wait so you do this then use the intergral remainder therom?
forest sky
once you have that, you have a maclaurin series for ln(|1-x|) (and for ln(|1+x|) if you substitute in -x), which is then part of your standard toolkit alongside e^x, sin(x), etc.
near marten
once you have that, you have a maclaurin series for ln(|1-x|) (and for ln(|1+x|...
Wait to find a similar already known series?
forest sky
**cloud**
the series for ln(1+x) is considered a "known series", but this is how it can be derived
near marten
the series for ln(1+x) is considered a "known series", but this is how it can be...
😦 i'm sorry I still don't understand.
forest sky
,tex .maclaurin
clever fjordBOT
cloud
forest sky
the series for ln(1+x) is common enough that we include it in lists of common maclaurin series. but if you didn't already know it, you can find it by doing term-by-term integration of the series for $\frac{1}{1+x} = \frac{1}{1-(-x)}$
clever fjordBOT
cloud
near marten
the series for ln(1+x) is common enough that we include it in lists of common ma...
So we find a similar series to the original question and then find the terms for that?
forest sky
what do you mean by "similar series"?
near marten
what do you mean by "similar series"?
I'm not sure how to do the concept at all:(. I don't remeber my teacher teaching this at all but he still gave it as homework
forest sky
ok, so we know that given infinitely many terms, $$\sin(x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} + \cdots$$ But we can't really calculate an infinite sum. so instead, we take the first few terms, since that will be very close: $$\sin(x) \approx x - \frac{x^3}{3!}$$ If we want to find an approximate value for $\sin(0.1)$, then all we'd need to do is plug in $x=0.1$ to the polynomial $x - \frac{x^3}{3!}$. \ If we want to know the maximum error we'd get out of this, then we could apply any of our error bound tests to find an upper limit on how inaccurate our approximation is (the easiest one would be alternating series error bound)
clever fjordBOT
cloud
forest sky
**cloud**
we can apply a similar logic to approximating $\ln(1.75)$ given that we know the series for $\ln(1+x)$
clever fjordBOT
cloud
near marten
we can apply a similar logic to approximating $\ln(1.75)$ given that we know the...
Thank you! i'll try it out
near marten
we can apply a similar logic to approximating $\ln(1.75)$ given that we know the...
Actually nvm I got the answer. Thank You so much
.close
wheat apex
How do you evaluate these?
odd edgeBOT
How do you evaluate these?
orchid torrent
How do you evaluate these?
You will need to ||expand out the parentheses before using these||
wheat apex
oh that makes sense
wheat apex
is the 2nd and 3rd power something i should just memorize? or is the proof/example of it simple?
because i understand where the first power came from
didnt look at any proof for it however
orchid torrent
is the 2nd and 3rd power something i should just memorize? or is the proof/examp...
The proof is actually quite nice
I’d encourage you to look it up
“Sum of squares formula proof” or “sum of cubes formula proof” should suffice
odd edgeBOT
@wheat apex Has your question been resolved?
odd edgeBOT
Available help channel!
coarse shuttle
Why is that
odd edgeBOT
Why is that
signal crown
sinus hyperbolic
signal crown
ln = natural logarithm
cold urchin
[\ln\p[\sinh(x)]]
weary pelican
$sh(x) = \frac{e^x-e^{-x}}{2}$
mystic saffron
[
\m\sinh x = x + \f{x^3}{3!} + \dots = x + \m o{x}
]
clever fjordBOT
mystic saffron
sh(x) = x + o(x)
yeah
coarse shuttle
oooh
I see
So to calculate the limit i can just taylor series the things ?
at 0 obv
mystic saffron
yeah obviously
cold urchin
cold
odd edgeBOT
cold urchin
odd edgeBOT
Available help channel!
left seal
odd edgeBOT
sacred whale
Need help to find it’s minimum point
So easiest way is probably to recognize it’s a quadratic equation in vertex form
left seal
I understand
sacred whale
So you can basically just look at the function given
left seal
I am but when I try put x value it gives a wrong y
mystic saffron
You can just find the vertex by looking at the equation
iron grotto
<@&286206848099549185>
question
vestal sand
What's your x
iron grotto
equation?
vestal sand
Do I not put x value in the expression?
No
left seal
This
vestal sand
y=k(x-a)^2+b formula
The b here is your y value
left seal
Not the expression of x?
vestal sand
The x value is negative -a
vestal sand
Not the expression of x?
No
left seal
And y value was -49/12
What am I doing wrong bro
<@&286206848099549185>
.ckose
.close
odd edgeBOT
odd edgeBOT
Available help channel!
errant heart
hello so i dont understand why we choose the z value here, apparently we have to choose a value for z, i dont get it why?
low locust
well, for one of the variables we can choose a value
and once we do that, we can calculate the correct value for the other
its just pretty standard that we start with the last variables when we choose values
but we could also for example choose y to have some value and then express everything in terms of y
odd edgeBOT
@errant heart Has your question been resolved?
errant heart
well, for one of the variables we can choose a value
so i can choose any numbers i like? or is there a specific pattern?
low locust
I dont know why exactly they chose z=1/3. what is the context here?
the point is, y and z are connected
they have to satisfy y=z+5/3
if you choose y then you get a value for z and if you choose z then you get a value for y
edgy bay
Let $\varphi$ be defined by $\varphi(x)=\frac{15}{16}(x^2-1)^2$ for $|x|<1$ and $\varphi(x)=0$ otherwise. Let $f$ be a function with a continuous derivative. Find the limit $$\lim_{n\to\infty}\int_{-1}^1n^2\varphi '(nx)f(x)dx.$$
edgy bay
However, I'm stuck here. This is a problem from a section on positive summability kernels, but I have been unable to verify what the kernel is in this exercise, if there is any. Appreciate any help.
odd edgeBOT
@edgy bay Has your question been resolved?
odd edgeBOT
@edgy bay Has your question been resolved?
odd edgeBOT
@edgy bay Has your question been resolved?
quasi sparrow
**Philip**
Did you try dominating the last integral with something
hoary steeple
Hello,
I'm doing an exercise on the analaytical way of solving the quantum harmonic oscillator.
I first start with schrodinger's equation
$-\frac{\hbar^2}{2m} \frac{d^2\phi(x)}{dx^2} + \frac{1}{2}m\omega^2x^2\phi(x) = E\phi(x)$
Then I do a substitution to nondimensionalize x $x =\tilde x / \beta$ with $\beta = \sqrt{\frac{m\omega}{\hbar}}$
By doing so I find :
$\frac{d^2\psi(\tilde{x})}{d\tilde{x}^2} - (\tilde{x}^2 - 2\epsilon)\psi(\tilde{x}) = 0$ with $\epsilon=\frac{2E}{\hbar\omega}$
Then it is suggested to use a solution of the form :
$\psi(\tilde{x}) = e^{-\frac{\tilde{x}^2}{2}}h(\tilde{x})$
So I plug it in and derive the following differential equation for $h(\tilde{x})$:
(From now on i'll use Lagrange's prime notation for derivative with respect to $\tilde{x}$)
$h''(\tilde{x}) - 2\tilde{x}h'(\tilde{x}) + h(\tilde{x})(2\epsilon - 1) = 0$
It is then said to use a solution of the form :
$h(\tilde{x}) = \sum_{m=0}^{\infty} a_{2m}\tilde{x}^{2m + p}, \quad a_0 \neq 0$
And now I have to show that there exist two independent solutions that corresponds to p=0 and p=1.
So by plugging this solution into the above differential equation I find two solutions and a recurrence relation.
$\frac{a_{2m+2}}{a_{2m}}=\frac{4m+2p+1-2\epsilon}{(2m+p+2)(2m+p+1)}$ and $ p=0 ; p=1$
hoary steeple
This is the exercise subject :
clever fjordBOT
Lucas
hoary steeple
I'm basically stuck at question 6
odd edgeBOT
@hoary steeple Has your question been resolved?
odd edgeBOT
@hoary steeple Has your question been resolved?
modern sundial
I don't really want to look at this problem in detail, but the part where you take a limit as m tends to infinity and... get a function of m is nonsense.
hoary steeple
It is a physics problem it doesn't mean "infinity" just really big
and by looking at the equation if you make m really big you can neglect the p, epsilon etc and you get 1/m or lambda/m
modern sundial
I'm of the opinion that if you come to a mathematics forum and ask for help, you should write it in notation understood by the audience. Don't write that you are taking a limit but mean it is big O of a function instead.
hoary steeple
Yes, you're right, I apologize
odd edgeBOT
@hoary steeple Has your question been resolved?
hardy fossil
Yes, you're right, I apologize
do better.
humble stone
can someone explain why its D
odd edgeBOT
can someone explain why its D
latent scaffold
You can think of it as the constant term 6 getting streched by -3 as well.
steady tide
can someone explain why its D
stretching a function vertically by a factor of k is essentially multiplying that function by a factor of k
so 3 * (1/2sqrt(x+5) + 6)
you forgot to also multiply 6 by 3
humble stone
You can think of it as the constant term 6 getting streched by -3 as well.
ohh wait, i was thinking about function notation.
if it was $\frac{-1}{2}f\left(x\right)+6$ would i be doing it right the way i was doing it?
clever fjordBOT
hypox
latent scaffold
Yeah if f was sqrt(x+5) then that would be the case.
But since f(x) = -1/2*sqrt(x+5) + 6, the transformed function is -3f(x) .
mystic saffron
odd edgeBOT
mystic saffron
Again
quasi sparrow
Where did you get from other people's help?
odd edgeBOT
@mystic saffron Has your question been resolved?
mystic saffron
I didn't get why we did this
odd edgeBOT
@mystic saffron Has your question been resolved?
near gulch
math is for nerds
regal chasm
,rotate
clever fjordBOT
mystic saffron
what do you think
odd edgeBOT
@mystic saffron Has your question been resolved?
fathom falcon
mystic saffron
send it in another channel
the ones available
<@&268886789983436800> some guidance she needs
lavish jackal
odd edgeBOT
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).
near wedge
odd edgeBOT
near wedge
algebric are numbers that have real roots. i would assume things like e, and pi are non algebraic terms
wooden python
algebric are numbers that have real roots.
incorrect.
near wedge
okay
wooden python
look up the actual definition of "algebraic number".
also this is going to have to be a cardinality argument.
wooden python
wooden python
overuse of the word "it"
"it's finite so it's countable" is poorly communicated at best and just says nothing at worst
also what is 2.12? you haven't shown us, nor do we have a copy of your textbook on hand.
near wedge
youre right
wooden python
(rather, we don't know what textbook you're using.)
near wedge
wooden python
did you wish to actually say that some set is countable?
near wedge
well thats what i was implying with 2.2
wooden python
yes or no
wooden python
what's finite?
near wedge
the an
wooden python
...
near wedge
the intergers¨'
wooden python
you are once again not communicating properly
wooden python
and i am also trying my best to understand what you are saying
as you can see i am not very successful at it, so sue me
wooden python
you are saying things that could go in a fully written out proof, but unfortunately you're saying them in a manner that is way way too fragmented.
wooden python
what are you going to type up
a proof of 2.2? or a proof of 2.3? or something else?
near wedge
2.2 i guess since were talking about it. i dont know what to do for 2.3 but i would rather see if you think im right or not on 2.2 since were on it i thought i had it but now you got me dobuting myself
wooden python
ok
near wedge
wooden python
im gonna ask you for one thing tho
in the stuff you're about to type out
don't use the word "it". at all.
near wedge
got it 😛
Let An be the set that represents the solutions that satisies the condition given in the hint. Since the number of solutions to this type of polynomial is finite then the set is finite. using theorm 2.12 this is countable. Then the set of alebgraic numbers for this polynomial are countable
wooden python
Let An be the set that represents the solutions that satisies the condition given in the hint.
"represents"?
and you say A_n, did you mean A_N?
near wedge
and you say A_n, did you mean A_N?
teach me the difference?
wooden python
n is lowercase, N is uppercase...
near wedge
okay
wooden python
i have a feeling you don't have a good idea of what set(s) you actually want to talk about
near wedge
nah i dont your right
wooden python
would you like me to try and explain how the proof of 2.2 actually goes?
near wedge
i of course would love to learn ann, its why im here 🙂
wooden python
ok
i'll format my explanation as a series of claims. once i am done, i will ask you to name the earliest claim that you don't understand. for example, if you say "I am stuck on claim 10", I will assume that you understand all claims from 1 to 9 inclusive.
do you understand this instruction? y/n
near wedge
y
wooden python
ok
for the purposes of making my own life easier, i will introduce some ad-hoc terminology along with necessary notations.
Definition 1: The rank of a polynomial with integer coefficients is defined as the sum of its degree and the absolute values of its coefficients. In other words, the rank of a_n x^n + ... + a_1 x + a_0 is defined as n + |a_0| + |a_1| + ... + |a_n|. [For example, the rank of the polynomial x^4 + x^3 - 7x^2 + 2x - 3 is 4+1+1+7+2+3 = 19.]
Definition 2: For a positive integer N, i will denote by E_N the set of all polynomials of rank N.
Definition 3: For a positive integer N, I will denote by R_N the set of all roots of polynomials from E_N.
Definition 4: I will denote by A the set of all algebraic numbers.
near wedge
i understand
odd edgeBOT
@near wedge Has your question been resolved?
wooden python
Claim 1: For every integer N, E_N is finite.
Proof: Given directly in the hint.
Claim 2: Every polynomial in E_N has no more than N roots.
Proof: Let f be a polynomial in E_N. From the definition of rank, we have that deg(f) <= N. From the fundamental theory of algebra, we have the number of roots of f is at most its degree.
Claim 3: For every positive integer N, R_N is finite.
Proof: By claim 2, every polynomial in E_N has at most N roots. By claim 1, E_N is finite. From this it follows that |R_N| <= N * |E_N|, thus R_N is also finite.
Claim 4: The union of R_N for N ranging from 1 to infinity equals A.
Proof:
[If this feels like stating the obvious, or bureaucratic, that's because it is both of those things.] Every number in R_N is by definition the root of a polynomial with integer coefficients, and thus belongs to A. Conversely, for every x in A, there is a polynomial f with integer coefficients such that f(x)=0. Letting N = rank(f), we have that f is in E_N, thus x is in R_N, thus x belongs to the union of the R_N.
Claim 5: A is countable.
Proof: By claims 4 and 3, A is the union of a countable collection of finite sets (R_N). By theorem 2.12, it follows that A is countable.
near wedge
i understand this. i need to start writing like this
so all roots of polynomials are algebraic numbers
this makes a lot of sense
and for every x in the algebraic numbers theres a polynomial that has zero root
tall veldt
Again you need to be specific in what you say
Not all roots of polynomials are algebraic numbers
near wedge
the roots of polynomals with integer coefs
are rational coefs still algebraic?
i understand if you put pi out in front it wont be
near wedge
**Claim 1:** For every integer N, E_N is finite.
*Proof:* Given directly in the...
thank you very much ann. i really do appreciate the time you took. thank you
.close
mystic saffron
im meant to express the function $F = x'y+xz$ as a product of maxterms, but im not sure how to do that
clever fjordBOT
odd edgeBOT
mystic saffron
woops forgot to close that
,align
F = x'y + xz &= (x'y +x)(x'y +z) \ &=(x+x')(x+y)(x'+z)(y+z)
clever fjordBOT
blissful depot
Find the elements of A
odd edgeBOT
Find the elements of A
thin garnet
well
the nested roots are ugly right
what if i make it look better
5 - 2sqrt(6) looks inda like
1 - 2sqrt6 + 6 to me
that looks neat right
try using that
elfin solar
use this formula
wise idol
5-2root(6) = (root(3) - root(2))^2
odd edgeBOT
@blissful depot Has your question been resolved?
blissful depot
use this formula
Okay.
I'm not sure about that 2sqrt(6)
Gotten to this
<@&286206848099549185>
mystic saffron
Write 4+2sqrt(3) as 3+2sqrt(3)+1
blissful depot
okay
mystic saffron
And it should be -sqrt(2)+1 instead of -sqrt(2)-1 on the LHS
blissful depot
And it should be -sqrt(2)+1 instead of -sqrt(2)-1 on the LHS
Why?
I rationalized the fraction
by amplifying with sqrt(2) - 1
mystic saffron
There are brackets around the sqrt(2)-1 when you rationalizw the fraction
blissful depot
I see.
mystic saffron
And -(a-b)=-a+b
mystic saffron
👍👍
blissful depot
xsqrt(3) - xsqrt(2) - sqrt(2) + 1 = xsqrt(3)+x
mystic saffron
RHS should be xsqrt(3)+x
blissful depot
Yeah, I typed it wrong
blissful depot
RHS should be xsqrt(3)+x
should I isolate x to LHS
mystic saffron
Yes
Get it in the form ax=b
Wait I made a typo
It was 4-2sqrt(3)
So the RHS is xsqrt(3)-x
mystic saffron
✅
mystic saffron
-x(sqrt(2)-1) = sqrt(2) - 1
No this is correct
blissful depot
-x(sqrt(2)-1) = sqrt(2) - 1
and here x is -1 or 1
blissful depot
No this is correct
Great.
mystic saffron
How do you get 2 solutions from a lineir equation?
mystic saffron
Yeah thats right
blissful depot
.close
dire pelican
How do you factorise with ‘X3-7X+6’ I only know how to do it with the start of the equation being X2
tacit parcel
How do you factorise with ‘X3-7X+6’ I only know how to do it with the start of t...
u mean a cubic?
dire pelican
Yeah like X cubed do we need to do something to make it an X squared?
crisp wadi
Find one root of the cubic; Via root-factor theorem and polynomial division, you will be able to reduce that to a quadratic
dire pelican
So do we do the sum and product equation?
tacit parcel
in this case you can actually find all 3 roots with rrt
so u dont neet to do division
dire pelican
Because the question im struggling with, show (x-2) is a factor so would i use that first?
Sorry if im wording this wierdly
tacit parcel
Because the question im struggling with, show (x-2) is a factor so would i use ...
if (x-2) is a factor, that means x=? is a solution?
dire pelican
So would I replace the ‘X’ in the equation with 2?
tacit parcel
So would I replace the ‘X’ in the equation with 2?
yeah that means x=2 should be a solution, and if you put that into the function and it works, that means (x-2) is a factor
dire pelican
Okk so for example i should replace ‘x3-7x +6’ as 2(3)-7(2)+6?
tacit parcel
Okk so for example i should replace ‘x3-7x +6’ as 2(3)-7(2)+6?
yeah, and if that equals 0, that means that 2 is a solution and thus, (x-2) is a factor
dire pelican
Ok! Thank you so much!
odd edgeBOT
@dire pelican Has your question been resolved?
odd edgeBOT
odd edgeBOT
Available help channel!
solid rain
What is the value of the term that does not contain x
solid rain
I wrote this (n!/i!j!k!)* 1^i * x^j * x^-k
j-k=0
and since i does not matter the options are j=k=1 and i=3 , j=k=2 and i=1
the answer I'm supposed to get is 51
but I did not get that
surreal pulsar
Let's go back one step. You have (x^0 + x^1 + x^-1)^5 and want to know the x^0 term in the expansion?
solid rain
why x^0?
surreal pulsar
That's another way of saying 1
solid rain
then yes
surreal pulsar
OK, so when you multiply out these terms, what sorts of things do you get? Just as an example
surreal pulsar
Yes 🙂 see what happens there and then work up to 3, 4, and 5
Just to find the general pattern or idea of how these exponents multiply out
surreal pulsar
OK. Actually, let me go back a step.. you were using a factorial formula. Do you know how you get that formula?
surreal pulsar
OK, great. Sorry, I probably should have gone down that path first. OK, explain in a little more detail how you were using the multinomial theorem?
solid rain
the sum of coefficients must be 5
and i want x^0 so j-k=0
so I found 2 solutions
but i did not get the correct answer
surreal pulsar
OK, so you understand your terms will look like (x^0)^i * (x^1)^j * (x^-1)^k where i+j+k=5?
surreal pulsar
Sorry, wait
Corrected, thank you
So you're looking for cases where 0*i + 1*j + -1*k = 0 while satisfying i+j+k = 5?
solid rain
yes
surreal pulsar
OK, keep in mind that i can be something other than 0
For example, if i=1 then j=k=2 works
young plover
help
leaden karma
maybe try writing (x+y+z)^3 in terms of the expressions you are given
young plover
maybe try writing (x+y+z)^3 in terms of the expressions you are given
wait damn yeah
wait how would that help without know
x^3 + y^3 + z^3
leaden karma
it does
leaden karma
yeah, it's a pretty common factorisation in these olympiad problems
young plover
yeah, it's a pretty common factorisation in these olympiad problems
this isnt even oly 💀
😭
leaden karma
as 3(x-y)(y-z)(z-x)=0
sorry this was wrong
you have $(x+y+z)^3=x^3+y^3+z^3\implies 3(x+y)(y+z)(z+x)=0$
clever fjordBOT
kheerii
leaden karma
WLOG let $x=-y$, then $y^2z=2\implies z=\frac{2}{y^2}$
clever fjordBOT
kheerii
leaden karma
replace all the x and z in one of the first two equations and find out what y is
from there, you can find out x+y+z
young plover
yeah
young plover
replace all the x and z in one of the first two equations and find out what y is
wait
i did it with x = -y
and when i added the first two subsituted eq i just got 2y^2 z = 4
which always makes sense
leaden karma
why are you adding?
young plover
oh shit i cant add them
young plover
why are you adding?
just to cancel out terms
leaden karma
you don't need to
young plover
but nvm i dont think i can dot hat
leaden karma
but nvm i dont think i can dot hat
you can of course, but there's no reason to
young plover
you don't need to
y^3 + 2 - z^2 y = - 1
leaden karma
replace the z as well
young plover
and then i sub in z and get
like a
^6
polynomial
oh wait
its a quadratic
but u take the ^ 1/3 of it
that seems messy in order to find the sum
leaden karma
but u take the ^ 1/3 of it
yep
leaden karma
that seems messy in order to find the sum
shouldn't be too much
young plover
alr..
leaden karma
$y^3+2-\frac{2}{y^3}=-1$
clever fjordBOT
kheerii
leaden karma
$y^6+3y^3-2=0$
clever fjordBOT
kheerii
young plover
**kheerii**
isnt it - 4
not - 2
@leaden karma
?
i got 2 and 2^-1/3 for x = -y
case
leaden karma
isnt it - 4
you're right
leaden karma
i got 2 and 2^-1/3 for x = -y
you mean the value of x+y+z?
young plover
ye
leaden karma
that seems correct
young plover
im wondering if there is a better sol
wait what i get the same values
for x = -z
oh nvm
nvm i do lol
but wtvr
ye they are all the same
idk that seems weird but wtbvr
leaden karma
well yeah
you can choose any of x+y,y+z or z+x to be 0
it will give you the same resuly
since all the given expressions are cyclic in x,y, and z
young plover
since all the given expressions are cyclic in x,y, and z
wait how is this symmetric
oh
cyclic is diff from symmetric
oh but
that makes sense
leaden karma
cyclic is diff from symmetric
indeed
young plover
ayy
leaden karma
equation 1 and 2 together make a symmetric condition I believe
kind of
but it's a bit iffy
young plover
equation 1 and 2 together make a symmetric condition I believe
yeah thats why
i couldnt add them
when plugging the stuff in cause
we treated them as one eq
rather than 2
odd edgeBOT
yeah that does make sense
leaden karma
.close
odd edgeBOT
young plover
yeah that does make sense
wtf
young plover
it's a common substitution when you have cyclic expressions like these
oh prolly oly
stuff
ig
leaden karma
yeah basically
young plover
r u in us?
leaden karma
nope
young plover
o alr
leaden karma
i'm from India
odd edgeBOT
Available help channel!
lost pivot
Help
lost pivot
I am not sure where to start with the complex problems
nocturne belfry
well for 1 i think you can just use definitions?
and some algebra
do you know what all those symbols and words mean?
lost pivot
only some
nocturne belfry
which dont you know?
lost pivot
still not sure about these symbols
nocturne belfry
its just saying that neither a or b are 0
lost pivot
what about the R
lost pivot
right right
nocturne belfry
this is set notation
lost pivot
how about writing A
in the standard form
of course the standard form is given with Z= a+bi?
nocturne belfry
for z
lost pivot
"bi" is the imaginary number
Whilst a is the real number
lost pivot
ok ok
nocturne belfry
and a+bi is just some complex number
nocturne belfry
i know its Z to the power of 2 but what are those two lines
modulus
,w modulus of a complex number
can you do it wa
clever fjordBOT
Requested by janniku
clever fjordBOT
jan Niku
nocturne belfry
presumably you have these definitions somewhere?
lost pivot
nope
lost pivot
our teacher get straight into the problem without much definitions
and we don't even study from the book
he has his own lessons in a powerpoint
odd edgeBOT
Available help channel!
prime scroll
odd edgeBOT
prime scroll
using reverse chain rule
what would i let y equal (i know i could use u sub but i want to use reverse chain rule)
zinc glacier
Whats the difference between usub and reverse chain rule
prime scroll
you let y= ln of the denominator
tough mesa
What is reverse chain rule
vivid girder
I guess chain rule is more 'formal' than substitution
prime scroll
but will reverse chain rule work
vivid girder
reverse chain rule should work since substitution is built upon it. But I don't know how to use reverse chain rule per se here.
quasi sparrow
but will reverse chain rule work
Just try it and see if it does
odd edgeBOT
@prime scroll Has your question been resolved?
prime scroll
.close
brazen sleet
hey guys i have a question rq, how can i get rid of the squared for this triple integral problem?
brazen sleet
that's the question if u want it written
vivid girder
Wait, why is the square problematic? I took multivariable calculus two semesters ago, so I don't quite remember all.
brazen sleet
Wait, why is the square problematic? I took multivariable calculus two semesters...
i don't know if i did a mistake by the order of the integrals but i think if i can get rid of that square i can keep on going by integrating of the next limit of integration
is there a way to get rid of that square ?
vivid girder
Can't you just expand it and the integrate as usual?
brazen sleet
yes but i have to expand the square first
otherswise i can't keep up the integration
vivid girder
so you want to avoid the square just to be sure that you don't commit mistakes/ not make it harder.
brazen sleet
hey guys i have a question rq, how can i get rid of the squared for this triple ...
and ngl it's shameful, but i forgot if there's a way to expand what is inside the brackets (if there's 3 values just like it's shown here)
what should i do to expand it ?
i learned only (a+b)^2 = a^2 + 2ab + b^2
brazen sleet
hey guys i have a question rq, how can i get rid of the squared for this triple ...
but there're multiple things to look around in this case
vivid girder
use distribution, so $(2-x-y)^2=(2-x-y)(2-x-y)=(2-(x+y))(2-(x+y))=4-4(x+y)+(x+y)^2=4-4x-4y+x^2+2xy+y^2$.
clever fjordBOT
Crystopher
vivid girder
Larger terms can be divided in smaller ones this even if you only know the good old $(x\pm y)^2$
clever fjordBOT
Crystopher
brazen sleet
woah that's a new method
vivid girder
But I don't think there is some shortcut in this case since the bounds are not independent. The last resort may be a substitution but I think that may be worse than expanding.
brazen sleet
But I don't think there is some shortcut in this case since the bounds are not i...
yeah probably
what if i change the order?
will the final answer be wrong by any chance?
vivid girder
Well, we can assume that we integrate first by x, for instance, you would then need to integrate $2yz$ on bounds involving $x$, so ultimately you wouldn't be able to get a concrete answer really.
clever fjordBOT
Crystopher
brazen sleet
Well, we can assume that we integrate first by x, for instance, you would then n...
so im not mistaken by the order
it's just about how i expand the brackets and i can go on
vivid girder
Your order is alright since it 'eliminates' varibles one by one.
vivid girder
you're welcome
brazen sleet
.close
odd edgeBOT
odd edgeBOT
Available help channel!
green cradle
odd edgeBOT
green cradle
can anyone explain why the integral was broken down to two integrals
brittle beacon
So you can remove the absolute value bars and write it more explicitly
green cradle
okay but why was the 1-e^x put in the integral with upper limit 0
brittle beacon
Because when x<0, you have |e^x - 1| = -(e^x - 1) = 1 - e^x
green cradle
i might be stupid but i still dont understand
brittle beacon
You're familiar that
[
\abs{x} =
\begin{cases}
x & x \geq 0\
-x & x < 0
\end{cases}
]
right?
clever fjordBOT
@brittle beacon
green cradle
yes indeed
oh okay i get that oh okay but still why was -x placed in the first integral
and not the second
brittle beacon
For the first one, between -1 and 0, you have that $e^x - 1<0$ and so $\abs{e^x - 1} = -(e^x - 1)$, but for the second one, between 0 and 1, you have $e^x - 1 \geq 0$ and so $\abs{e^x - 1} = e^x - 1$
clever fjordBOT
@brittle beacon
green cradle
what if the upper was 1 and lower was 0 would that mean that I e^x -1I = e^x -1
odd edgeBOT
@green cradle Has your question been resolved?
brittle beacon
what if the upper was 1 and lower was 0 would that mean that I e^x -1I = e^x -1
Yea if upper was 1 and lower was 0, you have that being true because over that interval your function you're taking the absolute value of is positive
odd edgeBOT
odd edgeBOT
Available help channel!
ebon jay
odd edgeBOT
ebon jay
Hi, I just wanted to check that this was a valid proof for "if xn -> 1 as n -> inf, xn^1/n -> 1"
@spiral basalt sorry for the ping this was from your suggestion
spiral basalt
yeah if you know x -> x^1/n brings things closer to 1 it's easy
but how would you prove that ?
ebon jay
yeah good point i did just kinda assume that
is it binomial expansion/Bernoulli's inequality? (1 + ε)^n for ε > 0 is greater than 1 + nε > 1 + ε so 1 + ε > (1 + ε)^1/n
but we can't necessarily use that with (1 - ε)^n since we could have ε > 1 and then bernoulli's inequality wouldn't hold, let me think
ebon jay
is it binomial expansion/Bernoulli's inequality? (1 + ε)^n for ε > 0 is greater ...
I should also mention that we've shown pretty early on x^n > y ^n <=> x > y for all x, y > 0 so i know this is by induction and contrapositive
spiral basalt
but we can't necessarily use that with (1 - ε)^n since we could have ε > 1 and t...
that's why I originally said "for small x at least"
I let you figure out the meaning of small x here
oh nvm that was a previous version of my message
ebon jay
ah okay so do we assume ε < 1 here? but then how do we prove it for ε > 1
spiral basalt
who cares
ebon jay
or is it not true
spiral basalt
are you gonna take the nth root of a negative number ?
ebon jay
ahh very true okay
so if ε > 1, 1 - ε < 0 < xn^1/n by default gotcha
so that combined with dividing everything by 3^n works, thanks :)
spiral basalt
but we can't necessarily use that with (1 - ε)^n since we could have ε > 1 and t...
you only care about small epsilons anyways
though arguably it's not a very short argument, but it's a simple one
ebon jay
yeah i mean are there any counter examples in analysis where a convergence works for small epsilons but not large ones?
can't think of any off the top of my head
seems illogical
spiral basalt
seems illogical
prove it
you may restrict yourself from
forall e > 0
to
forall 0 < e < a
for a fixed constant a, and it's an equivalent definition of the limit
ebon jay
since if you have say ε1 < ε2, if |an - l| < ε1 we have l - ε2 < l - ε1 < an < l + ε1 < l + ε2 so |an - l| < ε2 automatically
wait sorry that way round
spiral basalt
so what delta do you choose ?
spiral basalt
you may restrict yourself from
forall e > 0
to
forall 0 < e < a
for a fixed con...
one implication is immediate
What about the other one
ebon jay
so what delta do you choose ?
where does delta come in sorry?
spiral basalt
N
ebon jay
oh right, this was just a generic example
spiral basalt
we're dealing with sequences, but it's the same for functions
ebon jay
so pick N such that for all n >= N, |an - l| < ε1
and then for all n >= N, |an - l| < ε2 by the above
since ε2 > ε1
spiral basalt
you may restrict yourself from
forall e > 0
to
forall 0 < e < a
for a fixed con...
yeah, if e >= a
just pick N that works for a/2, it works for e
spiral basalt
3rd year of undergrad
ebon jay
cool cool, im 1st year
got my first exam in january so i've gone pretty hard on analysis
probably my favourite module so far i'll likely try and specialise in it
spiral basalt
Don't worry about specializing just yet
You're barely getting started with real math
ebon jay
true yeah
spiral basalt
Odds are, you won't even stay a math major
spiral basalt
odds are doesn't mean it's more than 50%
At least not with me
Just
uni math is different
Some people end up preferring physics or CS
Sometimes even chemistry
I myself am doing my first year as a CS major
And probably my last year as a math major
spiral basalt
cool cool, im 1st year
What courses do you have ?
ebon jay
What courses do you have ?
In our first term we did:
- Analysis I
- Algebra I
- Methods of Mathematical Modelling I
- Foundations (everything they didn't fit in the other modules)
- Introduction to Probability
- Mathematics by Computer
spiral basalt
What country is that?
spiral basalt
Supposedly your historical enemy
spiral basalt
We've had a few
How about the one just down south
spiral basalt
In our first term we did:
- Analysis I
- Algebra I
- Methods of Mathematical Mod...
What's in the last one ?
ebon jay
ah just some basic python stuff, numpy, sympy, things like that
did very elementary machine learning at the end of our final week
I'm interested in it but i prefer practising maths more than coding so i'm quite out of practice in coding
spiral basalt
So scientific computing
ebon jay
Yeah
they try to tie it into our other modules so we had some analysis and algebra to do with computing
and obviously some stats with the machine learning
odd edgeBOT
@ebon jay Has your question been resolved?
spiral basalt
.close
odd edgeBOT
odd edgeBOT
Available help channel!
pale roost
i have trouble with fractions of amounts.
odd edgeBOT
i have trouble with fractions of amounts.
wooden python
do you have a problem/question we can look at?
odd edgeBOT
odd edgeBOT
Available help channel!
elfin halo
oooooooooooooooooo nice
odd edgeBOT
oooooooooooooooooo nice
royal herald
@elfin halo
odd edgeBOT
@elfin halo Has your question been resolved?
mystic saffron
oooooooooooooooooo nice
whats the question?
mystic saffron
hey someone help me w chinese reminder theorem pls?
odd edgeBOT
hey someone help me w chinese reminder theorem pls?
mystic saffron
the proof and the theorem
latent oxide
Just put any example
mystic saffron
its actually a theorem
state nd prove chinese reminder theorem
<@&286206848099549185>
tough mesa
What is the Chinese remainder theorem
mystic saffron
state nd prove chinese reminder theorem
any example ? or smt
quartz slate
Don’t you mean remainder?
tough mesa
Yes yes 😂
mystic saffron
this is it but i dont understand shit
mystic saffron
Don’t you mean remainder?
yes lol sorry
mystic saffron
lol help me out its an important question for 15 marks
quartz slate
Sure give me like a minute please
mystic saffron
yessir
quartz slate
Do you not understand it at all?
mystic saffron
yea i dont
mystic saffron
yes thanku so much
mystic saffron
sorryy wrong msg
quartz slate
Oh okie
mystic saffron
thanks :))
quartz slate
All goods :3
odd edgeBOT
Available help channel!
odd edgeBOT
deep arrow
This free
odd edgeBOT
odd edgeBOT
Available help channel!
charred yoke
Is this correct?
odd edgeBOT
Is this correct?
frigid canopy
I don;t think so, you haven't performed the substitution properly
jolly halo
yeah if you do that you will still miss a variable x there
charred yoke
what
frigid canopy
doubt it has a nice closed form, is this high school calc?
if so, it most likely should be x^2
not x^4
but again, I could be wrong
charred yoke
doubt it has a nice closed form, is this high school calc?
nope
university
oka im getting the news that the question was written wrong
there shouldve been x above
jolly halo
oh thats nice
charred yoke
fuck the doc
frigid canopy
that helps a lot
jolly halo
you can use your method now
jolly halo
yeah
charred yoke
any idea how to do this one?
alpine garnet
any idea how to do this one?
theres a nice u sub in this one
alpine garnet
me too, but thatll also work i think
alpine garnet
du will be 2lnx * 1/x right?
alpine garnet
so itll be 1/2 integ(du/rootu)
alpine garnet
@charred yoke u as lnx is another way to do it
charred yoke
ah
alpine garnet
its a direct formula then
charred yoke
here i will split it from -2 to 0 then from 0 to 2 right
jolly halo
<@579275211652726797> u as lnx is another way to do it
yeee
soft ravine
For X_7 x⊕y is the rest of x+y devided by 7 how do i prove it is a group?
odd edgeBOT
For X_7 x⊕y is the rest of x+y devided by 7 how do i prove it is a group?
@soft ravine Has your question been resolved?
crisp wadi
!status
odd edgeBOT
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
spiral tangle
any method to calculate the function of a function 100 times quickly?
eg/ f(x) = (3x+1)/(2x-3)
(fofofofofo...100 times..fofofof)(x + 1)
spiral tangle
the answer is x + 1 by the way. I can not get my head around to proove this without actually doing the thing
frigid canopy
maybe there;s a pattern? Have you tried upto say 4 terms?
odd edgeBOT
@spiral tangle Has your question been resolved?
worthy dawn
odd edgeBOT
worthy dawn
In question 11d, how do we know that it's meant to be in the form: Ax + b?
In all my previous questions, the partial fractions were also in fraction form. So how do I know when do do it in what form?
worthy dawn
(This is the worked solution, not my working)
odd edgeBOT
@worthy dawn Has your question been resolved?
odd edgeBOT
@worthy dawn Has your question been resolved?
odd edgeBOT
@worthy dawn Has your question been resolved?
worthy dawn
Why I'm trying to say is how do I make partial fractions!
meager juniper
@worthy dawn if you perform polynomial long division of the numerator and denominator, you'll get Ax + B with some remainder.
most of your previous problems, I am assuming, the degree of the numerator was less than or equal to the denominator.
this is one way of handling it when it's greater than
You can also just perform polynomial long division to start with, and do partial fractions on the remainder
worthy dawn
Thanks, I think that makes sense.
Is that what the worked solution did?
meager juniper
well, the worked solution just assumed the polynomial long division could be done, and used the coefficients of the quotient as unknowns and then solved them within the system of equations
both ways are completely valid.
fossil vault
i managed to clamp the e function between (0, 1) for domain (0, 1). but im wondering how i can get a more exaggerated form (green line in the second image) by manipulating the exp function
fossil vault
here is the desmos
you dont have to use my already existing expression. if you have an alternative solution using exp than feel free to tell me
quasi sparrow
i managed to clamp the e function between (0, 1) for domain (0, 1). but im wonde...
Define "clamp"
fossil vault
sorry for fancy language. between 0 and 1
the domain is (0, 1) and the range is (0, 1) using exp
thats what i mean
clever fjordBOT
quasi sparrow
,w plot log(x+1)
clever fjordBOT
fossil vault
does x = 1 map to 1?
quasi sparrow
,calc sqrt(1)
clever fjordBOT
Result:
1
Square root does
You can multiply the log by a number so that it does too
,calc log(1+1) / log(2)
clever fjordBOT
Result:
1
tiny heart
You have a grand total of 5 marbles (lmao pathetic), to pull out one that has a siiick pattern on it, you got a 40% chance. How many marbles don't got a super sick pattern ?
How do I figure this out?
trim owl
You have a grand total of 5 marbles (lmao pathetic), to pull out one that has a ...
Suppose 1/5 have a "sickkkk" pattern. What's the chance you pull one with a "sickkkk" pattern? Try that out and you should be able to solve it
tiny heart
Well the chance is 20%
And
Like I’m trying to figure out the mathematical calculation behind it
marbles = 5
those with sick pattern = x
Percentage = 0.4
trim owl
Like I’m trying to figure out the mathematical calculation behind it
20% = .2...you want to double that chance
trim owl
Mhm, so if becomes 40%
Try 2 marbles with a sick pattern
odd edgeBOT
@tiny heart Has your question been resolved?
mortal hinge
odd edgeBOT
mortal hinge
hi can someone help pls ?
