Yare yare daze

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>>7982016
Bet you think you cool.

>>7982030
No, I don't think I'm cool.

I think that I still don't fucking know the answer.

It will be years before I take a real analysis class so this will haunt me for a fucking long time.

I thought getting a PhD meant you were a genius.

>>7982035
Honestly most professors once getting their job tend to not remember anything not related to their specific field as they have no reason to. Hell most PhD students do the same shit.

>>7982016
What do you mean by combination?

>>7982044
In the moment I couldn't think of a better way of expressing myself so I thought of 'linear combinations' and said combinations.

anything like sinx + cosxtanx + sinxsecx

Just any function with trig functions. Even composite ones, I guess.

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>>7982016
>jojo poster

>>7982050
x+sinx

>>7982065
But that is trivially true.

Any non trivial ones?

To be clearer, no term can not have a trig function.

>>7982050
arcsin(x)

>>7982072
Injective for all real numbers.

arcsin is not defined for all real numbers.

>>7982077
Only sin and cos are defined for all reals and no linear combination of just those two functions is going to be injective.

>>7982087
#1: Prove it
#2: Doesn't have to be linear

>>7982092
It's easy to prove for linear combinations. "Linear combination" just means that you're considering f(x) = asin(x)+bcos(x) for some real numbers a,b. You can choose y = x+2pi, and then you get f(x) = f(y), just by periodicity of sine and cosine.

It depends on the range and domain OP.

>>7982098
Yeah, that's why I also considered terms such like sinxcosx.

Also, I don't think that it needs to be defined for all R.

1/x is not defined at x=0 but it is still injective.

But arcsin doesn't even make sense if you take it out of its restricted domain.

>>7982099
OP's asking a reasonably well defined question at this point.

They've specified that they want their function defined over all of R. This rules out anything of the form 1/sin(x) or 1/cos(x) in the function, so in particular sec(x), cot(x), tan(x), and csc(x) cannot appear in any simplified expression of the function, nor can inverse trig functions. (If OP doesn't like this, they can complain.)

This mean that the OP seems to be asking whether there is some (possibly multivariable) polynomial in terms of sin(x) and cos(x) that is injective. So for example the polynomial P(x,y) = x^2y+y^2 gives you P(sin(x),cos(x)) = sin^2(x)cos(x)+cos^2(x). These are the sorts of expressions OP seems to care about. Perhaps they would also accept things like xcos(x), and perhaps they accept infinite trigonometric series (both of which I suspect make OP's condition trivial).

>>7982105
Alright, I see, thanks.

>>7982105
sinxcosx = 0.5sin(2x)

>>7982112

No he's not faggot, he needs to specify his domain if his range is all of R.

>>7982016
>Be Me
>Be Freshman
>Go to pick classes for semester
>What Math should I take: Intro Analysis, Analysis 1, Analysis 2?
>No
>Pick Analysis on Manifolds
>Get A

What is with you brainlets taking intro classes?

>>7982016

> f(x)=sinh(x)

This board truely is a joke.

>>7982126
"want their function defined over all of R" meant that the domain is supposed to be R. Except now OP has specified that they accept countably many discontinuities (not their phrasing, but I think they'll find this ok) over R.

>>7982134
The real joke is people picking trivial counterexamples to the actually interesting question OP wants answered. Ignore hyperbolic trig functions, ignore inverse trig functions, just focus on sin(x), cos(x), 1/sin(x), 1/cos(x), and two variable polynomials in those.

>>7982071
sin - cos on the set {0}

Alternatively, if you allow for countably infinite sums of sinusoids, then we have that f:R->R,x|->x is a trigonometric function that is injective.

>>7982142
OP here.

What this guy said is what I'm interested in.

>>7982156
Then it is simply not possible.

>>7982162
Probably! But it still requires proof.

>>7982162
Prove it.

>>7982112
If this is what OP wants, then it is equivalent to asking if any injective function has a fourier transform, since it represents only functions that can be represented as a linear combination of every possible sine and cosine.

>>7982142
The answer in this case is no then. You'll have a finite amount of trigonometric terms. All trig functions have periods. The function as a whole will have a period that is the lcm of all those periods.

>>7982156
see>>7982145

>>7982175
Yes, that works!

>>7982016
Can't you just choose an interval where it is?

Sidenote: is a surjective function just a function where the image is equal to the range?

>>7982187
>>7982175
wrong, see:
>>7982172

>>7982171
I have discovered a truly marvelous proof of this which is to large to write out in this post. I leave it as an exercise for the reader.

>>7982016
>Fourier transforms

try hard

>>7982194
The third post you linked is correct (although a bit weaker since the OP hasn't even necessarily asked for infinite trig series). However, the second post you linked is also correct and gives the correct proof idea.

The point is that any term is going to look like sin^n(x), cos^n(x), 1/sin^n(x), 1/cos^n(x), sin^n(x)/cos^m(x), cos^n(x)/sin^m(x), 1/(sin^n(x)cos^m(x)), or sin^n(x)cos^m(x) for suitable positive integers n and m. Since sin^n(x) and cos^m(x) have finite periods for all positive integers n, m, you can just lcm the periods of all the functions in the expression. Then letting y = lcm(periods), you get that f(x) = f(y) but x =/= y. Since it happens for any term, it holds for the sum of all the terms.

>>7982194
So for example, if you have f(x) = sin^2(x)cos(x) + 1/(sin^3(x)cos(x)), it's easy to see that sin^2(x), cos(x), and sin^3(x) all have period 2pi, so f(x+2pi) = f(x) for all x where 1/sin(x) and 1/cos(x) are defined for both x and x+2pi. I didn't realize this earlier, but I guess all the terms will just have period 2pi.