Anonymous at Sun, 7 Jul 2024 14:38:14 UTC No. 16271519

>>16271517
>this is reminiscent of Maclaurin series, righte?
If so, I wanna define sin and cos analogues (PS. I use Sine{...} instead of Sine(...) because I define Sine and Cose as operators over f where f := tan in the first post, hope u know what Im talking abt!)
[eqn]
e_k\{f\} = \sum_{1 \leq j_1 \leq j_2 \leq \dots \leq j_k \leq n } f(\alpha_{j_1}) \dots f(\alpha_{j_k}) \\
\mathcal{S}\mathrm{ine}\{f\} = \sum_{k=0}^n (-1)^k e_{2k + 1}\{f\} = e_1 - e_3 + e_5 - \dots + e_n \\
\mathcal{C}\mathrm{ose}\{f\} = \sum_{k=0}^n (-1)^k e_{2k}\{f\} = e_0 - e_2 + e_4 - \dots + e_{n-1}
[/eqn]
Now that we have sin and cos, we can define tan too righte?
[eqn]
\mathcal{T}\mathrm{ane}\{f\} = \frac{\mathcal{S}\mathrm{ine}\{f\}}{\mathcal{C}\mathrm{ose}\{f\}} = \left(\frac{e_1 - e_3 + e_5 - \dots + e_n }{e_0 - e_2 + e_4 - \dots + e_{n-1} }\right)\{f\} \\
[/eqn]
So the original generalization for tan(a1 + a2 + ... + an) can be expressed in terms of the operator Tane over the tan function.
[eqn]
\tan\left(\sum_{k=1}^{n}\alpha_k\right) = \left(\mathcal{T}\text{ane}\{\tan\}\right)(a_1, a_2, \dots, a_n)
[/eqn]

Soo... what kind of math did I stumble upon? O_O I have so many questions e.g. for what [math]f[/math] do Sine, Cose, Tane converge when [math]n\rightarrow\infty[/math]?

Is there some book about functional analysis(I think?) that can help me study the properties of my Sine, Cose, Tane operators?

2/2

PS. I always tought [math]\sin(x)[/math] as a girl, [math]\cos(x)[/math] as a boy, and [math]\tan(x)[/math] as non-binary. ^_^ I love other trigonometric functions too. Trigonometry is cute!

Anonymous at Sun, 7 Jul 2024 15:36:01 UTC No. 16271559

>>16271517
kill yourself the only thing you're good at being like a woman at is asking others to give you shit

Anonymous at Sun, 7 Jul 2024 15:40:14 UTC No. 16271562

>>16271559
Kys transphobe. Nobody likes you.

Anonymous at Sun, 7 Jul 2024 15:43:46 UTC No. 16271568

This seems like an interesting problem to help with. But I'm not going to because you'll probably have killed yourself by the time I finish typing out my response.

Anonymous at Sun, 7 Jul 2024 16:04:53 UTC No. 16271589

ywnbarw

Anonymous at Sun, 7 Jul 2024 16:19:41 UTC No. 16271603

I don't have much to contribute, just thought I'd add this: if F(x) is a polynomial with roots tan(alpha_j), then
Cose{f} = (F(i) - F(-i))/2
Sine{f} = (F(i) - (F-i))/(2i)

Anonymous at Sun, 7 Jul 2024 16:33:49 UTC No. 16271618

>>16271559
I don't know why I laughed so much at this

Anonymous at Sun, 7 Jul 2024 16:52:44 UTC No. 16271629

>>16271559
I like you. Total Troon Death.

Anonymous at Sun, 7 Jul 2024 17:14:16 UTC No. 16271645

Don't let these anons get to you're head. You're cute and valid.:)

Anonymous at Sun, 7 Jul 2024 18:07:02 UTC No. 16271723

>>16271517
>>16271519
These are the most mentally ill posts I've seen on 4chan in a long while, good job

Anonymous at Sun, 7 Jul 2024 18:42:30 UTC No. 16271775

>>16271723
I think they are fine. You sound mentally ill yourself.

Anonymous at Sun, 7 Jul 2024 18:52:56 UTC No. 16271787

>>16271519
I don't think these are the proper analogues of sine and cosine because you are missing the factorial.
It should actually be
[eqn]
\text{Sine}\{f\} = \left(\sum_{k=0}^\infty (-1)^k\frac{e_{2k + 1}}{(2k + 1)!}\right)\{f\} = \left(e_1 - \frac{e_3}{3!} + \frac{e^5}{5!} - \dots\right)\{f\}\\
\text{Cose}\{f\} = \left(\sum_{k=0}^\infty (-1)^k\frac{e_{2k}}{(2k)!}\right)\{f\} = \left(e_0 - \frac{e_2}{2!} + \frac{e^4}{4!} - \dots\right)\{f\}\\
\text{Expe}\{f\} = \left(\sum_{k=0}^\infty (-1)^k\frac{e_{k}}{(k)!}\right)\{f\} = \left(e_0 + e_1 + \frac{e^2}{2!} + \dots\right)\{f\}
[/eqn]

>>16271603
> if F(x) is a polynomial with roots tan(alpha_j)
In this case Vieta formulas would say
[eqn]
e_k = \frac{a_{n - k}}{a_n}
[/eqn]
Where [math]\{a_j\}_{j\leq n}[/math] are the coefficients of the polynomial.

Anonymous at Sun, 7 Jul 2024 18:59:53 UTC No. 16271796

>>16271517
I'm glad that you have a genuine curiosity for Mathematics. Such things should be celebrated.
>if you pass as a woman then you'll get less hate, that is the reality. If you are just a man in a dress you will not be respected.

🗑️ Anonymous at Sun, 7 Jul 2024 20:41:59 UTC No. 16271911

>>16271517
Let's assume that the expression [math] tan(\alpha+\beta+\gamma) = \frac{e_1+e_3}{e_0+e_2} [/math] is correct
Then [math] tan((\alpha+\beta+\gamma)+\delta+\eta) = \frac{e_{1a}+e_{3a}}{(e_0+e_{2a}} [/math]. This is a correct expression.

We set:
[math] e_{1a} = tan(\alpha+\beta+\gamma)+tan(\delta)+tan(\eta) [/math]
[math] e_{2a} = tan(\alpha+\beta+\gamma)tan(\delta)+tan(\alpha+\beta+\gamma)tan(\eta)+tan(\delta)tan(\eta)[/math]
[math] e_{3a} = tan(\alpha+\beta+\gamma)tan(\delta)tan(\eta)[/math]
Then, we have:
[math] e_{1a} = \frac{e_1+e_3}{e_0+e_2}+tan(\delta)+tan(\eta) [/math]
[math] e_{2a} = \frac{e_1+e_3}{e_0+e_2}tan(\delta)+\frac{e_1+e_3}{e_0+e_2}tan(\eta)+tan(\delta)tan(\eta)[/math]
[math] e_{3a} = \frac{e_1+e_3}{e_0+e_2}tan(\delta)tan(\eta)[/math]
According to you:
[math] \frac{e_{1a}+e_{3a}}{e_0+e_{2a}} = \frac{e_{1}+e_{3}-e_{5}}{e_0+e_{2}-e_{4}} [/math]
Or:
[math] \frac{e_{1}+e_{3}-e_{5}}{e_0+e_{2}-e_{4}} = \frac{\frac{e_1+e_3}{e_0+e_2}+tan(\delta)+tan(\eta)+\frac{e_1+e_3}{e_0+e_2}tan(\delta)tan(\eta)}{1+\frac{e_1+e_3}{e_0+e_2}tan(\delta)+\frac{e_1+e_3}{e_0+e_2}tan(\eta)+tan(\delta)tan(\eta)} [/math]
Set [math] y = \frac{e_{0}+e_{2}}{e_1+e_{3}} [/math] to get:
[math] \frac{e_{1}+e_{3}-e_{5}}{e_0+e_{2}-e_{4}} = \frac{1+ytan(\delta)+ytan(\eta)+tan(\delta)tan(\eta)}{y+tan(\delta)+tan(\eta)+ytan(\delta)tan(\eta)} [/math]
[math] \frac{e_0+e_{2}-e_{4}}{e_{1}+e_{3}-e_{5}} = \frac{y+tan(\delta)+tan(\eta)+ytan(\delta)tan(\eta)}{1+ytan(\delta)+ytan(\eta)+tan(\delta)tan(\eta)} [/math]

🗑️ Anonymous at Sun, 7 Jul 2024 20:44:07 UTC No. 16271916

>>16271911
If we set [math] \frac{1}{y} = y [/math] which must be true for some value because tan() is a continuous function that ranges from - infinity to infinity
then we get:
[math] \frac{e_{1}+e_{3}-e_{5}}{e_0+e_{2}-e_{4}} = \frac{e_0+e_{2}-e_{4}}{e_{1}+e_{3}-e_{5}} [/math]
The identity, however, only holds if y = 1,-1 or [math] e_{1}+e_{3}+e_{5} = 1+e_2+e_4 [/math]
As the values [math] tan(x), x = \alpha,\beta,\gamma [/math] etc. can be made to vary indefinitely, we can choose them such that [math] tan(\alpha) = -tan(\beta), tan(\delta) = -tan(\eta) [/math].
Then, since [math] \frac{e_{1}+e_{3}-e_{5}}{e_0+e_{2}-e_{4}} = \frac{e_{1a}+e_{3a}}{e_0+e_{2a}} [/math]
We must have [math] e_{1a}+e_{3a} = e_0+e_{2a} = \frac{e_1+e_3}{e_0+e_2}+tan(\delta)+tan(\eta) + \frac{e_1+e_3}{e_0+e_2}tan(\delta)tan(\eta) = 1+\frac{e_1+e_3}{e_0+e_2}tan(\delta)+\frac{e_1+e_3}{e_0+e_2}tan(\eta)+tan(\delta)tan(\eta) [/math]
As we have now chosen [math] tan(\beta)=-tan(\alpha), tan(\delta)=-tan(\gamma) [/math].
[math] e_{1a} = tan(\gamma) [/math]
[math] e_{2a} = tan(\delta)^2[/math]
[math] e_{3a} = tan(\gamma)tan(\delta)^2[/math]
but
[math] e_{1} = tan(\gamma) [/math]
[math] e_{2} = tan(\delta)^2[/math]
[math] e_{3} = tan(\gamma)tan(\delta)^2[/math]
[math] e_{4} = tan(\delta)^2tan(\alpha)^2 [/math]
[math] e_{5} = tan(\gamma)tan(\delta)^2tan(\alpha)^2[/math]
As [math] tan(\alpha), tan(\gamma), tan(\delta) [/math] can have any value, we see, that the rule doesn't hold true for the given choice of variables.

Anonymous at Sun, 7 Jul 2024 20:46:18 UTC No. 16271921

>>16271517
Let's assume that the expression [math] tan(\alpha+\beta+\gamma) = \frac{e_1-e_3}{e_0-e_2} [/math] is correct
Then [math] tan((\alpha+\beta+\gamma)+\delta+\eta) = e_{1a}-e_{3a}/(e_0-e_{2a}) [/math]. This is a correct expression.

We set:
[math] e_{1a} = tan(\alpha+\beta+\gamma)+tan(\delta)+tan(\eta) [/math]
[math] e_{2a} = tan(\alpha+\beta+\gamma)tan(\delta)+tan(\alpha+\beta+\gamma)tan(\eta)+tan(\delta)tan(\eta)[/math]
[math] e_{3a} = tan(\alpha+\beta+\gamma)tan(\delta)tan(\eta)[/math]
Then, we have:
[math] e_{1a} = \frac{e_1-e_3}{e_0-e_2}+tan(\delta)+tan(\eta) [/math]
[math] e_{2a} = \frac{e_1-e_3}{e_0-e_2}tan(\delta)+\frac{e_1-e_3}{e_0-e_2}tan(\eta)+tan(\delta)tan(\eta)[/math]
[math] e_{3a} = \frac{e_1-e_3}{e_0-e_2}tan(\delta)tan(\eta)[/math]
According to you:
[math] \frac{e_{1a}-e_{3a}}{e_0-e_{2a}} = \frac{e_{1}+e_{3}-e_{5}}{e_0+e_{2}-e_{4}} [/math]
Or:
[math] \frac{e_{1}+e_{3}-e_{5}}{e_0+e_{2}-e_{4}} = \frac{\frac{e_1-e_3}{e_0-e_2}+tan(\delta)+tan(\eta)+\frac{e_1-e_3}{e_0-e_2}tan(\delta)tan(\eta)}{1+\frac{e_1-e_3}{e_0-e_2}tan(\delta)+\frac{e_1-e_3}{e_0-e_2}tan(\eta)+tan(\delta)tan(\eta)} [/math]
Set [math] y = \frac{e_{0}-e_{2}}{e_1-e_{3}} [/math] to get:
[math] \frac{e_{1}+e_{3}-e_{5}}{e_0+e_{2}-e_{4}} = \frac{1+ytan(\delta)+ytan(\eta)+tan(\delta)tan(\eta)}{y+tan(\delta)+tan(\eta)+ytan(\delta)tan(\eta)} [/math]
[math] \frac{e_0+e_{2}-e_{4}}{e_{1}+e_{3}-e_{5}} = \frac{y+tan(\delta)+tan(\eta)+ytan(\delta)tan(\eta)}{1+ytan(\delta)+ytan(\eta)+tan(\delta)tan(\eta)} [/math]

Anonymous at Sun, 7 Jul 2024 20:47:47 UTC No. 16271925

>>16271921
If we set [math] \frac{1}{y} = y [/math] which must be true for some value because tan() is a continuous function that ranges from - infinity to infinity
then we get:
[math] \frac{e_{1}+e_{3}-e_{5}}{e_0+e_{2}-e_{4}} = \frac{e_0+e_{2}-e_{4}}{e_{1}+e_{3}-e_{5}} [/math]
The identity, however, only holds if y = 1,-1 or [math] e_{1}+e_{3}-e_{5} = 1+e_2-e_4 [/math]
As the values [math] tan(x), x = \alpha,\beta,\gamma [/math] etc. can be made to vary indefinitely, we can choose them such that [math] tan(\alpha) = -tan(\beta), tan(\delta) = -tan(\eta) [/math].
Then, since [math] \frac{e_{1}+e_{3}-e_{5}}{e_0+e_{2}-e_{4}} = \frac{e_{1a}-e_{3a}}{e_0-e_{2a}} [/math]
We must have [math] e_{1a}-e_{3a} = e_0-e_{2a} = \frac{e_1-e_3}{e_0-e_2}+tan(\delta)+tan(\eta) + \frac{e_1-e_3}{e_0-e_2}tan(\delta)tan(\eta) = 1+\frac{e_1-e_3}{e_0-e_2}tan(\delta)+\frac{e_1-e_3}{e_0-e_2}tan(\eta)+tan(\delta)tan(\eta) [/math]
As we have now chosen [math] tan(\beta)=-tan(\alpha), tan(\delta)=-tan(\gamma) [/math].
[math] e_{1a} = tan(\gamma) [/math]
[math] e_{2a} = tan(\delta)^2[/math]
[math] e_{3a} = tan(\gamma)tan(\delta)^2[/math]
but
[math] e_{1} = tan(\gamma) [/math]
[math] e_{2} = tan(\delta)^2[/math]
[math] e_{3} = tan(\gamma)tan(\delta)^2[/math]
[math] e_{4} = tan(\delta)^2tan(\alpha)^2 [/math]
[math] e_{5} = tan(\gamma)tan(\delta)^2tan(\alpha)^2[/math]
As [math] tan(\alpha), tan(\gamma), tan(\delta) [/math] can be made to have any value, we see, the rule doesn't hold true in the general case

Anonymous at Sun, 7 Jul 2024 20:56:00 UTC No. 16271937

>>16271925
>As tan(α),tan(γ),tan(δ) can be made to have any value, we see, the rule doesn't hold true in the general case
I just figured, this is incorrect. The value of tan(γ) is dependent on tan(α),tan(δ).
>tan(δ)=−tan(γ)
Should read [math] tan(\delta)=−tan(\eta) [/math]

Anonymous at Sun, 7 Jul 2024 20:57:34 UTC No. 16271938

>>16271921
Girl you can prove it's correct in a trivial computation.

Anonymous at Sun, 7 Jul 2024 21:10:50 UTC No. 16271952

>>16271938
>Girl you can prove it's correct in a trivial computation.
It's not correct.

Anonymous at Sun, 7 Jul 2024 21:12:25 UTC No. 16271953

>>16271921
>According to you [math]\frac{e_1 + e_3 - e_5}{e_0 + e_2 - e_4}[/math]
But OP does - first then + [math]\frac{e_1 - e_3 + e_5}{e_0 - e_2 + e_4}[/math]

Anonymous at Sun, 7 Jul 2024 21:16:57 UTC No. 16271961

>>16271953
True, I keep screwing up positive/negative signs. Still, even then. My refutal is more generic in that no series of the form [math] \frac{ae_1+be_3+ce_5...}{1+de_2+ee_4...} [/math] can be identical to [math] tan(\alpha+\beta+\gamma...) [/math]

Anonymous at Sun, 7 Jul 2024 22:03:55 UTC No. 16272034

>>16271961
>My refutal is more generic in that no series of the form ... can be identical to tan(a + b + c)
Why not?

[eqn]
\tan((\alpha + \beta + \gamma) + \delta + \eta) \\
= \frac{\left(\frac{(\tan\alpha + \tan\beta + \tan\gamma) - \tan\alpha\tan\beta\tan\gamma}{1 - \tan\alpha\tan\beta - \tan\alpha\tan\gamma - \tan\beta\tan\gamma} + \tan\delta + \tan\eta\right) - \frac{(\tan\alpha + \tan\beta + \tan\gamma) - (\tan\alpha\tan\beta\tan\gamma)}{1 - \tan\alpha\tan\beta - \tan\alpha\tan\gamma - \tan\beta\tan\gamma} \tan\delta\tan\eta}{1 - \frac{(\tan\alpha + \tan\beta + \tan\gamma) - \tan\alpha\tan\beta\tan\gamma}{1 - \tan\alpha\tan\beta - \tan\alpha\tan\gamma - \tan\beta\tan\gamma}\tan\delta - \frac{(\tan\alpha + \tan\beta + \tan\gamma) - \tan\alpha\tan\beta\tan\gamma}{1 - \tan\alpha\tan\beta - \tan\alpha\tan\gamma - \tan\beta\tan\gamma}\tan\eta - \tan\delta\tan\eta}
\\=
\frac{(a + b + c + d + e) - (abc + abd + abe + acd + ace + ade + bcd + bce + bde + cde) + abcde }{1 - (ab + ac + ad + ae + bc + bd + be + cd + ce + de) + (abcd + abce + abde + acde + bcde)}
\\=
\frac{e_1 - e_3 + e_5}{e_0 - e_2 + e_4}
[/eqn]
where
[eqn]
a = \tan\alpha\\
b = \tan\beta\\
c = \tan\gamma\\
d = \tan\delta\\
e = \tan\eta
[/eqn]
and
[eqn]
e_0 = 1\\
e_1 = a + b + c + d + e\\
e_2 = ab + ac + ad + ae + bc + bd + be + cd + ce + de\\
e_3 = abc + abd + abe + acd + ace + ade + bcd + bce + bde + cde\\
e_4 = abcd + abce + abde + acde + bcde \\
e_5 = abcde
[/eqn]

It works out when you multiply numerator and denominator with [math]1 - \tan\alpha\tan\beta - \tan\alpha\tan\gamma - \tan\beta\tan\gamma[/math]

Anonymous at Sun, 7 Jul 2024 22:08:40 UTC No. 16272043

>>16271925
>>16271921
>>16271961
You are just wrong lol

Anonymous at Sun, 7 Jul 2024 22:37:29 UTC No. 16272092

>>16272043
You need to be 18 to post here.
>>16271517
What's up with that phrase anyway
>Okey guys, so I'm a trans girl who is into math and need help from super smart math guys like u (apologies for using gendered language. I'm assuming you are all assigned male at birth, but I could be wrong, sowrry ;_;).
No, your calculations are correct. I'm pondering what is wrong with my proof by contradiction.

Anonymous at Sun, 7 Jul 2024 22:39:33 UTC No. 16272096

>>16272092
Nobody's going to go through your wall of incorrect calculations to tell you where you went wrong. You're going to have to do that yourself buddy.

Anonymous at Sun, 7 Jul 2024 23:11:29 UTC No. 16272141

>>16271519
>for what f do Sine, Cose, Tane converge when n∞?
One such example would be constant function where |f(x)|<1

Anonymous at Sun, 7 Jul 2024 23:23:00 UTC No. 16272157

>the only people legitimately posting math on /sci/ are literal fucking trannies
>these are the false flaggers constantly whining about the quality of /sci/ and want to ban dissenting opinions... Like how trannies aren't women
Holy fucking shit. You can't make this shit up

Anonymous at Mon, 8 Jul 2024 16:24:55 UTC No. 16272915

>>16272157
What in the world are you talking about?

Anonymous at Mon, 8 Jul 2024 19:14:27 UTC No. 16273099

>>16271517
Unbothered troon

Anonymous at Mon, 8 Jul 2024 19:40:44 UTC No. 16273136

Ragebaiting cuckservatives like this is actually a great way to get bumps on threads that promote real discussions in an era where every single 4chan thread naturally (or artifically) revolves around racebait/genderbait/whatever with no substantial discussion whatsoever.

Good job OP! Unfortunately, I don't see how I can help you with this. Maybe do some more research on elementary symmetric polynomials, looks like there is an even more general way of looking at these objects.

Anonymous at Tue, 9 Jul 2024 03:56:32 UTC No. 16273665

If you're trans and into math you had better start learning category theory ASAP.

Anonymous at Tue, 9 Jul 2024 04:10:21 UTC No. 16273674

>>16271559
Hahahaha! Why would anyone help a troon? At least with women there is a biological drive that makes men act stupid and waste time a resources.

Anonymous at Wed, 10 Jul 2024 00:37:05 UTC No. 16275359

>>16271517
>>16271519
>this is reminiscent of Maclaurin series of sin(x) and cos(y), right?
Congratulations you rediscovered the definition of the tangent function. tan(x) = sin(x)/cos(x). You want a cookie?
>what kind of math did I stumble upon?
you're just doing taylor series
>I always tought sin(x) as a girl, cos(x) as a boy, and tan(x) as non-binary.
ywnbarw please rope mr. he/him