#help-26
1 messages · Page 76 of 1
-35, -7, -5, -1, 1, 5, 7, 35
no
Why not?
Odd degree polys will have atleast one real root if that's what you mean
So it can be more than 3 ?
Not for a cubic no
I meant in our quation we have no x^2 term
But we can clearly observe that it cuts in three places
No it can't
If that were the case it would have to be something of degree 4 or higher
This
understand, I didn't think about it
so the question you asked:
Does a cubic function that does not contain an x² term intersect the x-axis three times?
Yes yes
I meant it is possible or not?
I am interested to know about what kind of roots will be there
so
I looked into it a bit
can cut three times again
ax³+cx = x(ax²+c)
x = 0 or x = ±√(-c/a)
I see, what you found?
There can be 3 roots if -c/a > 0
Could you make any equation with it?
So if we look at descart rule
3 real roots ☺️👌
Let's work out with our main problem with Descartes it represents (3 real roots, 1 real 2 complex roots)
This one
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I need to show $a_n = 1+ 1/a_{n-1}$ converges to the golden ratio
Junojuno
artemetra
brahim3579
iam professional
@toxic zealot Has your question been resolved?
I need to prove convergence
Formally
@toxic zealot Has your question been resolved?
@toxic zealot What exactly is the problem?
x² = x+1, then x = golden ratio
we know that already
so what is the problem?
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lol
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I crafted myself a problem to get better at integrating in the R^2 plane with double integrals, I want to calculate the Area A with double integrals
$\int_{x_1}^{x_2} dx\int_{y_1}^{y_2} dy$
tobi
where y_1 and y_2 are functions of x of x_1 and x_2 are functions of y how can would I try solving this problem?
I also calculated all the functions that represent the line segments if that helps
why? it has to work and I probably get a lot of insight in to how to solve similar problems
How do you get better at integrating without using integration?
ed was solid 5 minutes ago, but... damn, people can't read today
but wouldn’t you craft a better problem
instead of a damn triangle
which there a countless better methods to find areas of
Plus, I'm really spicy today 😉
@crisp raptor i have problem sets for finding areas in R^2
i can give you a few problems
that are integration focussed btw
what is wrong with this problem
fine
if you insist
integrate from the green line to the red line from 1 to 1.5
and then green line to blue line from 1.5 to 2.5
I could integrate von y=1 to y=2.5 (green line) and then subtract 1 to 2.5 from the red one right? and so on then I just repeat that untill im done
would that work
I dont actually wanna write all of this down I just want to understand how I would do it
y wise?
w/ respect to y you should be doing 1 to 3
intersection is at y=3
but yes that would work
ops
technically you can with a double integral if you integrate dxdy on the domain of the triangle
but pls dont
even a single integral is pushing it for calculating the area of a triangle
i already said this
$\int_{A} d\lambda^2$
tobi insisted tho
tobi
yes the first integral would just give me the function that im integrating (f,g and h right)
yeha i think thats right
ed
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
lets say I want to integrate this
$\int_{A} d\lambda^2 y^2 x^2$
tobi
then I have to do it like this
oh
thats a reason why I want to learn this
i see
I can have a whole function f(x,y) where I integrate over the domain of the triangle
because we integrate the 1 function ofc its the area
$\int_1^{\frac 32}\int_{\frac43x-\frac13}^{6x-5} x^2y^2,dy, dx$
ed
this would be one part
then you would do between the yellow and green line
similarl
y
I understand the x integral can you please explain why you choose 4/3x-1/3 on the bottom and 6x-5 on the top
thats the green line
6x-5 is the red line
would this be correct?
are we integrating the purple part
this integral is the purple part
is this true in general that if I want to integrate an area in R^2 between 2 functions, that the upper and lower bound are those functions? and then its the area between those graphs?
it should be right?
because then we only have 2 different integrals?
wdym?
wait I dont wanna integrate from x=1 to x=4 here because then I would have to solve g(x) and f(x) for x right?
that would also work
or what is the reason why you did not do that
because that would be a piecewise function
so we just take two double integrals
,w integrate from 1 to 3/2 (6x-5-(4/3x-1/3))
,w integrate from 3/2 to 4 (2/5x+17/5-(4/3x-1/3))
I dont get the correct result with this method
im confused
oh
So the area should be 3.5
thanks!!

then everything works
how did you calculate it to be 1.4
ah the norm of the vector in R^2 ?
Err no idea honestly

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We need to check the nature of the roots of the cubic equation
X^3-18x-35=0
By descartes rules of sign
that isn't a quadratic equation
@real blaze Has your question been resolved?
<@&286206848099549185>
so what is the equation then? Def not a quadratic though
or is it a typo
are you tryin to solve it or whats your question?
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how do I show that this system of equations has an infinite amount of solutions?
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How do I determine for which value of the real parameter a>0 the finite and non-zero limit exists
you find the product of the left-most terms of the series expansions of (sinx), ln(1 + x^2), and sqrt(e^4x - 1)
if a exceeds this product, the limit will go to infinity. if it falls short of this product, the limit will go to 0
use well known limits like sinx/x, ln(1+x)/x, (e^(x)-1)/x
i think
for example ln(1+x²)/x^a
a must be 2 for it to be a nonzero limit
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@obsidian otter Has your question been resolved?
<@&286206848099549185>
whats the probability of the final score being 2
(0 and 2)
*possibilities
e.g. 1 and 1
or 2 and 0
which is separate from 0 and 2
try solving the probabilities for each of these events separately and then adding them together
no
don't do that
what's the probability of 1, 1?
OH nvm I only took probability of 1 only once 🤦♂️
1/12
Thank you!
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if f(x) = x^2-2x+1 write an expression representing g(x) if (g(x) = 2f(x+1) -3
if f(x) = x^2-2x+1 write an expression representing g(x) if (g(x) = 2f(x+1) -3
Im confused how on to do it because of the x+1
what would you do if i asked for f(5)
just replace x with 5
just do the same with x+1
so I put x+1 instead of x in the f(x) one?
yup
alright thank you
eg if f(x)=5x then f(x+1)=5(x+1)
thanks
I was confused
cause I tried to do the thing +1
but i didnt work
have a good one
.close
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can someone pls help me understand what happened on the 4th step how did we get 3/2 ,3x, and 2x
They moved some things around, not much more.
what about the 3/2
Check out the 3x in the denominator in sin(3x)/(3x). That 3 wasn't there before, so they compensate with the 3 in the numerator of 3/2.
The 2 in the denominator of 3/2 cancels with the 2 in the numerator of (2x)/sin(2x).
np
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for part d why is the 4th term used and not the 5th
btw $f(x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{n+1}}{3^n\cdot n}$
Jash
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,rotate 270
i dont get diff quotient
slope
yes but how u calculate it
yeah
so like
thats exactly what the difference quotient is about
so like
do you know what a secant line is
yes
k
so we have [
m = \f{y_2 - y_1}{x_2-x_1}
] for the slope of a line
we will try and calculate the slope of the secant line
kk
thats what the difference quotient is. It is representing the slope of the secant
anyways x_1 = x like we said
so now x corresponds to f(x) on the graph yes?
yea
that's our y_1
so cool
now say the second point x_2 is like a distance h away from x
so x_2 = x+ h
clear?
here
that x_2 corresponds to f(x_2)
but we just said x_2 = x +h
so f(x+h)
putting all together this all basically means
[
m = \f{y_2 - y_1}{x_2 - x_1} = \f{\m f{x+h} - \m f x}{x+h - x} = \f{\m f{x+h} - \m f x}h
]
that's what the difference quotient is. It is a representation of the slope of the secant line
yea
if you are still confused i could link you a video maybw
check out khan academy they most definitely have something explaining it
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factor the denominator
what do u notice
yes
and if u plug in 1 u get 0
substitute
like let x=1
u get -1/0
which is undefined
so we have to do more work
there’s a vertical asymptote at x=1
so it’s going to approach either infinity or negative infinity
so we just need to know if the other factors make it negative or positive
coming from the left of 1
vertical asymptotes occur when the denominator =0
so since we’re coming from the left of one it’s really like 0.999
if we take x to be a little less than one
is x-1 positive or negative
is x-2 positive or negative
is x-3 positive or negative
(negative)(negative)(negative)=?
yes
so it should approach negative infinity
yes
the limit =negative infinity
well u can show the algebra
with the factors
and like a sign chart
showing that less than one the factors r all negative
and 3 negatives makes a negative
and a vertical asymptote at zero means itll approach either positive or negative infinity
hold i’ll show u
yea at one
sorry
because it’s in the denominator
setting the denominator =0 gives a vertical asymptote
unless u get 0/0
which dan be different
yes
hold
u make intervals
like less than one they’re all negative
so when u multiply them u get negative
in between 1 and 2 only x-1 is positive
multiplying them u get positive
and so on
and the 1,2, and 3 r just the zeros of the 3 factors
u don’t in this case im just showing u how to determine where a function is positive or negative for all intervals
for rational functions
yup
ur welcome
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I started with integration by parts
but it doesnt make any sense to me
since f(x) and g(x) arent given
Where’d you get to with integration by parts?
here
@gray harbor
wait i confused myself
i see now
the answer would only be f(1)g(1) - f(0)g(0)
You need a 2
and ignoring the second part of the formula
yes bc its -2
bc they gave me the value of the second part
i see thanks
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if i have found my nullity(T) = [-1,1,-1,1]
why does this make the dimnesion = 1
The kernel spanned by (-1, 1, -1, 1)?
yes
Then your basis of the kernel has one element, so dimension is 1
yh
find A the relative standardbasis for R^4
find ker(T) and im(T)
i dont undertand how it is one dimensional
Because the kernel is spanned by the vector $\pmqty{-1 \ 1 \ -1 \ 1}$
@vernal matrix
yes how is that a one dimensional vector
how can a vector be 1 dimesional
that isnt even a line
This is a line
how is that a line
And it's not the vector that is 1 dimensional, but the kernel, because remember that the dimension of a space is how many elements are in a basis
Saying "one dimensional vector" is a bit confusing, I think it means to state 1 dimensional vectorspace
i thought what mattered was linearly independent-ness
It's a scalar multiple of one vector, so a line
It's the line $\ds \set {\mat[b]{-1\1\-1\1}t \where t \in \R}$
yes i have wrote that on my paper
i must have a completely misunderstood idea of matrices then
Huh
i thought if you have 4 linearly independent things its would span r^4
4 linearly independent vectors
so what exactly does the coeffeicents in front it represent then
Where is 4 linearly independent things here, hmm
what is x1,x2,x3,x4
Those are the entries of your vector
Those are real numbers (considering you're in R^4)
so it isnt (x,y,z,q)
gggggg

so this is a line in a four-dimensional space
so
[-1,1,-1,1]
$\ds \set {\mat[b]{-t\t\-t\t} \where* t \in \R}$
is a point in R^4
you could also write it like this
and the span is the set of scalars you can add onto this
Yea it is also one vector
that would make a line
(-1,1,-1,1) is a point in a four dimensional space and if you times it by a scalar, lets say 2, you get (-2,2,-2,2) and if you keep this up you'll get a line
Ah, "add" part is a bit confusing there
i changed it
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yeah unfortunately we can't help on quizzes here. good luck!
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Can someone point out where these steps are wrong? The answer I get is different from the correct answer.
(2x^(ln(x)-2)(2(ln(x))^2-ln(x)+1)
what's the question
what happened here?
both exponential disappeared
oh yeah, I used lim[h->0]((e^x-1) / x) = 1
multiplying ((ln(x+2h))-(ln(x))^2)/((ln(x+2h))-(ln(x))^2)
that all sounds wrong
this is limit as h going to zero, but your function doesn't have h
oh wait i meant x -> 0 haha
even if you did fix that and apply that to the first line here, that doesn't explain why the second line still has h
once you take the limit of h, there should be no more h variables left
as i think (ln(x+2h))-(ln(x))^2 approaches 0 when h approaches 0
wait i have to write it down
i forgot to write lim[h->0] in the next few steps
explain the next line
use x^2-y^2 = (x+y)(x-y)
why is ln(x) multiplying ln(1+ 2h/x) ?
uh oh
oh sorry, the ln(x) should not be there
so that makes the answer 2x^ln(x) (ln(x))
but still different
@winter tartan Has your question been resolved?
show your answer with the mistake fixed
what did you do from the first to the second lines here
use lim[h->0] (ln(1+h)/h) = 1
and what did you do here
@winter tartan Has your question been resolved?
I have no interest in going through this kind of algebra, but the intent of my hint earlier in the day was that it would be helpful if you used the fact that, -under the limit-,
$ln(1+\frac{h}{x}) = h\ln\left[\left(1+\frac{h}{x}\right)^{\frac{1}{h}}\right] = \frac{h}{x}\ln\left[\left(1+h\right)^{\frac{1}{h}}\right]$
so that using the continuity of the natural log, the limit
$\lim_{h\to 0} \ln\left[\left(1+h\right)^{\frac{1}{h}}\right] = \ln(e).$
JessicaK
You were asked to use this fact:
That should be a sign that it should show up somewhere in your work
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this is the drivation of langrange's equation from hamilton's principle. i don't get from 2.8 to the last equation. first term vanishes but how does second term equal to the following?
from Classical Mechanics by Goldstein page 37
the second term is combined with the first integral from (2.7)
since (2.8) only deals with the second integral from (2.7)
oh shit i missed that 😭
im so dumb. thank you
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$f(x)=\begin{cases} (2x+\sqrt{x})^2 ,if \ x\geq 0 \ (2x+\sqrt{-x})^2, if \ x<0
\end{cases}$
\the slope of tangent (T) of f at A(-1,-1) is
\a)1.5
\b)1.65
\c)2.5
pirate king
isnt this question wrong or not enough information ?
or am i missing something
A(-1,-1) is not on f
so assuming that this was not chosen by mistake
then A(-1,-1) is on (T)
but thats all
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can someone give me the formula for teststastistic, if i want to do a hypothesis test about the y intercept of linear regression
I cannot find it online
@olive stream Has your question been resolved?
<@&286206848099549185>
To view a playlist and download materials shown in this eCourse, visit the course page at: http://www.jmp.com/en_us/academic/ssms.html
thank you very much
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Idk how to do a)
apply gradient (slope) formula
Line ab is perpendicular to line bc
And I calculate gradient first
And since the multiply to - 1
I just solve
Right
Wait
I got the quadratic b^2 - 7b +14 =0
I put it in desmos I got 6 and 1
And since b is greater than 3
It's 6
But
When I use quadratic equation Ion get that
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Why is the x-coordinate of the center of mass of the upper half of the circle equal to the x coordinate of the center of mass of the full circle?
because of symmetry
there is the same amount of mass on either side
which means your x coordinate would always be on the line of the center of the circle
wait could you explain that in a bit more detail
oh wait
sio like
do you intuitively get what a center of mass is
nono i get ur logic for the circle
i was just thinking about if that same logic owuld work on the caridoid too
ok so to answer "Why is the x-coordinate of the center of mass of the upper half of the circle equal to the x coordinate of the center of mass of the full circle?"
first, the line of the center of the circle is where both sides of the object will have equal amount of mass (assuming the object has constnat desnity)
so even when we're considering just the upper half of an object that is horizontally cut, because that sliced shape will have equal mass on each side from the line of the center of the circle
the x-coord of the center of mass is always on the line of the center of the circle
and this logic applies to any object of constant denisty
would this be a "correct" description?
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Good Day!
Quick question, if I have the following function, does that mean that my co-domain is between 0-1?
Or how can I read it?
Ya co domain [0,1]
Range is different
And because I cant reach all values between 0-1 my function is not surjective?
the function is not surjective and it doesn’t satisfy the intermediate value property lol
it doesn’t take any of the values from 1/4 to 3/4
for example
if you’re trying to determine whether a function is surjective like this, it’s best to draw the function, draw your codomain on the y-axis (by shading it in maybe) and seeing whether the entire thing you shaded has an x value such that f(x) = y
if all of those steps are possible
if it’s not possible to do that then you have to just reason about the function, which is a bit harder
that is right yeah
So when I sketch the function and find gaps like below, I can assume it is not surjective?
yes sort of, but like the function can go back down later (if you were given a different, slightly wackier function) if you know that over your entire domain, you’ll never reach that y value then you know that it’s not surjective, in this example i can draw the horizontal line y=1/2 and realize that my function never takes on the value 1/2, which is in our codomain, so we know it’s not surjective
And this part determines my Domain to Co-Domain relation?
So we project from (0-1) to (0-1)?
also, if we specify the codomain to be different then maybe it’ll be surjective, for example if the codomain was [0, 1/4) U [3/4, 1] then it would actually be surjective
first interval is the domain, second interval is the codomain
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what does the 15 in L_15 mean
to split it into 15 parts?
probably
.close
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Can anyone help me solve two problems about Carnot's cycle?
Oh is there one?
yeah in #old-network
Also the problems are in Spanish I'm translating them rq
Thanks I'll join that one
A reversible heat engine works between two thermal reservoirs, one of them at 443K and the other one at 783K. Find the efficiency of the engine. Find the work made by the engine if we provide 7000kcal and the heat transfered to the cold reservoir.
A reversible heat engine operates between two thermal reservoirs, one of them at temperature T and another one at 280K. If it provides 1000kJ/min of heat to the cold reservoir and develops a usable power of 40kW, find the temperature of the hot reservoir.
<@&286206848099549185>
do you know how to find the efficiency?
Norbert Baudin
@fast adder Has your question been resolved?
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hi
What is iirc/iff
okey
but its not equal its ^
wait
I misread something
Its not equal it implies
I thought that two parts should be equal, but the one part implies the second one
I understand now
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how do i know whether i can integrate a function or not. say for example
these two
left picture it should have an integral sign for 4te^2t
the left picture im supposed to do integration by parts
Second picture does not have an anti derivative in terms of elementary functions
right picture its a new method but how would i confirm that none of the normal integration methods would not work
wym by elementary functions
It means that you won't be able to integrate it using your normal methods since you can't express its primitive using only powers/trig/logs/exponentials/etc
Now the proof of that is not something of that level
how can i tell it cant integrate by just looking at it
By looking at it you just learn to recognize those integrals that won't work under any method . Like here, parts won't help whatever you do, and there's just one thing you can u-sub and is gives nothing.
Some others could be sin(sqrt(x))
ok so by looking at the left picture can i assume i can do integraion by parts because it has two different functions?
what else gives it away
i know i need two parts for it
so i have 2
u and v
Left picture gives it away because the t factor will "simplify" the integral when doing parts
Because the derivative of t is 1
wouldnt i have 4t as u
du is 4
oh wait
for picking dv, it needs to be something easy to integrate so id go with e^2t
and v would be 1/2e^2t
I'd say, usually, composition of log, sqrt and exponentials are suspicious.
suspicious as in the equation is less likely to be able to integrate?
or harder
Yeah
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$\sum_{n \geq 2} \frac{x}{n^{1 + x}}$ i need to find for which values of x this converges
uwupachkii
x>0 maybe?
but then the follow up question is the norm of it
oh yeah surely
exercise asks for largest interval
for x = 0 it works
we define $f_n : x \rightarrow \frac{x}{n^{x + 1}}$ and we need to find for $n \geq 2$, $|| f_n|^I_{\infty}$
uwupachkii
i decided i would find the derivative because i dont see obvious marjoration
and the derivative cancels when x = 1
and damn it is correct
so yeah interval is more than >1
this with x = 0 too
idk why i thought it would diverge if 1+x was not stricly superior than 2
thanks guys !
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Is 0.1 correct
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
first off be careful of your units. 0.1 feet would be a very, very tiny building
it's one mile away
try converting miles to feet first
5280
ok good. Now re-do the work with that information 🙂
yep seems right to me 🙂
plugging in 4 is already wrong
yea but your x is wrong
when the x value is left of the y axis, then that x value is negative
no idea how you got this
-b/2a
Oh
even just visually, you should know the x value of the vertex is negative and the y value of the vertex is positive
How about find the maximum or minimum value of f in that graph
Does that mean the highest/lowest x intercept
no, y = f(x), so min/max of f means min/max of y
where in the screenshot are you being asked this
I didn’t take the full picture
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Given that Megan is the second place, the formula is technically like this 5p2
Or am I wrong
second runner up means 3rd place?
I think that's 2nd place
what are champion and first runner-up then?
if megan is included in the 5 contestants (which i would assume) then it should be 4P2
uh that means she's not included in the 5 lol
but also that's information you should have included in the original question
So it's technically 5p3?
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What do I do
you use the hint
@ancient valve Has your question been resolved?
How do I know if it's right based on that tho
Because all right triangles satisfy that property, and anything that isn't a right triangle doesn't
Where tf do I get 180 from tho in the first one
I do not get it
Nvm I'm stupid
They explain in multiple steps where 180 comes from
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The tutorial said that this integral diverges as x^-1/2 is a p function where p=1/2 < 1
What does this even mean?
(Like what is a P function and why does this mean it diverges here
It refers to expressions of the form 1/x^p
The integral of 1/x diverges, so anything with a smaller power than 1 also diverges
I see I see
Why does the integral of 1/x diverge then?
Well what is the antiderivative of 1/x?
ln|x|+c
Ohhh ok I see
And if u put infinity in for x it goes to infinity so it diverges
Indeed
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How do I do this
is this a calculus problem?
the equation describes a parabola with a maximum turning point
No
Ok I know that
okay do you know how a parabola looks based on the equation?
It goes down
that is the x-coordinate of the vertex
Isn’t that part of the maximum
yes
to find the vertex from this equation, you have to use the formula u just used
x = -b/2a
to find the axis of symmetry
once u have x, you sub that value into the given equation to find y
then u have vertex (x,y)
So is y the maximum profit then?
yes
4400?
ye
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We are supposed to use limit comparison with a geometric series to find if the sum of a (n) is convergent or not. I have no clue how to even start this
for large n, which term is going to dominate in the numerator? and which term in the denominator?
@lime zenith Has your question been resolved?
does B+2=4^n?
That part is not intuitive to me, I’d say n^5 would dominate in the numerator but if it wants geometric it should be 4^n? In the denominator it would be 19^3n
I tried making b(n) = 4^n/19^3n but couldn’t figure out from there
Is that the right track?
nope, it turns out that 4^n will dominate, it grows much faster than any polynomial (n^5 is a polynomial)
😭
yes this is what you want to compare with
So then the limit would just equal 1 and b(n) would be convergent so the series converges?
I don't know what that means
why are you typing in this channel
yes to both, assuming you can justify both claims and you're not just guessing 😁
ok
Lmao
Well the limit would equal 1 just because you have terms cancelling out on top and bottom since as n approaches infinity, the other terms can be ignored
And then b(n) is convergent bc it’s a geometric series where abs value or r < 1
So for this one, would you make b(n) equal to (4^.4n/19^(n+3))^6?
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✅
@lime zenith Has your question been resolved?
@lime zenith Has your question been resolved?
You can use ratio test bro
Bruh we didn’t learn that yet
so learn it
Im saying that there’s probably a way to solve it without the ratio test that I should know
aight lemme take a gander
is this still the current quesiton
The second picture I sent is my current problem, but the instructions are the same
so using limit comparison test you need to determine what you are comparing it to
4n^2 / (19^(n+3)) because its the highest term of both
and u can use that one which happens to be geometric to know if it converges or not
the limit comparison test says that if one converges the other does too
after proving that they are both greater than 0 ofc
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in the above, if I replace n with with 1/j for some real integer j
is that allowed?
no
why btw?
n must be an integer
doing n choose 2k won't make sense then
oh right 🤦♂️
well is there like no formula for cos(x/n) then?
btw i am, to an extent, an expert on this because i wrote a 4k word essay on this stuff lol
not a general one
bro i trust u its fine
basically
i do remember bprp finding cos(theta/3) and it being multivalued or smthin like that
like there were multiple different formulas out of which only 1 was (supposed) to be correct
solving this polynomial would subdivide the angle
artemetra
why 
is discord media down again
the latex isnt loading
😐
$d$
artemetra
dead
💀
why tho? like do we replace with ((e^ix) - e^(-ix))/2?
wait
oh
yeah it's fine
that's how chebyshev polynomials were derived tho
lol
let me see if i can make sense of this
what okay this getting a lil messy
so
we have polynomials
yes
with this formula
yes
we replace x with cos(u) to get cos(nu) correct
mhm
then if replace x with cos(pi/n) why do we get cos(pi) and isnt that -1?
artemetra
Mhm WAIT OH we put pi/n in cos nx i see i see
but also here you need to define what you mean by "formula"
a generalisaiton for all integers n, so something similar to the cos(nx) formula
$\frac{e^{ix/n}-e^{-ix/n}}{2}$
artemetra
as a substitution? idrk
in place of cos(nx)? no ofc not
ye exactly
in place of cos(x/n)
i made a formula for any integer degree
like $\cos(n\textdegree)$
hmm
artemetra
wtf
i suppose i was expecting something like cos(nx) chebyshev polynomial thingy
you understood
that's correct
yes i understood
Hmm did u try doing x=1/u sub urself ever? if u wrote an essay/paper/whatever on this?
not really since that formula breaks for non-integers
you can try but you are gonna have to do (1/2)! type stuff
gamma func?
yea
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np
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