Is sin(x+y) Analytic in ℝ with y Fixed?

[Shoelace Thm.]

Homework Statement

or y fixed in ℝ, show f(x)=sin(x+y) is analytic in ℝ.

Homework Equations

The Attempt at a Solution

Clearly $f(0 + x) = \sin(x+y) = \sum_{n=0}^{\infty} a_n y^n$ by the analyticity of $\sin x$. Now how should I proceed?

 

[Shoelace Thm.]

Can someone provide some guidance?

 

[Dick]

Can someone provide some guidance?

I don't know what you are looking for as a proof. In general if f(x) is analytic then f(x+c) is analytic for c a constant. Tell me why. If you don't know try expanding sin(x+y) using trig and the fact that sin(x) is analytic and cos(x) is also analytic.

 

[Shoelace Thm.]

Ok here is why: using the same notation as above, f is analytic in a neighborhood of 0. Now $f(x) = \sin(x+y) = \sum_{n=0}^{\infty} (-1)^n\frac{(x+y)^{2n+1}}{(2n+1)!}$ by the definition of sin(x) as a power series, which converges on R because the domain of sin(x+y) is R. Thus because f can be represented as a formal power series about 0, and that formal power series converges on all of R, then f is analytic in R (this is a theorem).

Will this do?

 

[Shoelace Thm.]

Can you check this Dick?

 

[Dick]

Can you check this Dick?

Can you say what your exact definition of 'analytic' is? What you have is a not a power series expansion in x.

 

[Shoelace Thm.]

Yes what I wrote won't suffice for that very reason. Alternatively, computing the nth degree Taylor polynomial for f about 0, the coefficients are all bounded by 1 and so the remainder term goes to 0 as $n$ goes to infinity. Thus f is analytic in a neighborhood of 0. Because its Taylor series converges in R ($\frac{1}{\limsup_{n \to \infty} \frac{c}{n!}} = \infty, |c| \le 1$ ), f is then analytic in R.

Does this work?

 

[Dick]

Yes what I wrote won't suffice for that very reason. Alternatively, computing the nth degree Taylor polynomial for f about 0, the coefficients are all bounded by 1 and so the remainder term goes to 0 as $n$ goes to infinity. Thus f is analytic in a neighborhood of 0. Because its Taylor series converges in R ($\frac{1}{\limsup_{n \to \infty} \frac{c}{n!}} = \infty, |c| \le 1$ ), f is then analytic in R.

Does this work?

sin(x+y) is not bounded by 1 when (x+y) is a complex variable, if that's what you are thinking in your argument. Expanding in powers of (x+y) is a power series expansion around x=(-y). To get it's radius of convergence why don't you just try something simple, like a ratio test?

 

[Shoelace Thm.]

No; y is fixed in R. By analytic, I mean 'can be represented locally as a power series.'

The nth degree polynomial for f about 0 is:

$\sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!} x^k = \sin y + \cos y \cdot x - \frac{ \sin y }{2} x^2 + \cdots$

Looking at the remainder term,

$R_n(x) = \frac{ f^{(n+1)}(\theta) }{(n+1)!} x^{n+1} \le \frac{1}{(n+1)!} x^{n+1} \to 0$ as $n \to \infty, 0 < \theta < x$

Thus in a neighborhood of 0, f is given by its taylor series about 0 and so f is analytic in neighborhood of 0. This taylor series converges in R, and so f is analytic in R.

 

[Dick]

No; y is fixed in R. By analytic, I mean 'can be represented locally as a power series.'

The nth degree polynomial for f about 0 is:

$\sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!} x^k = \sin y + \cos y \cdot x - \frac{ \sin y }{2} x^2 + \cdots$

Looking at the remainder term,

$R_n(x) = \frac{ f^{(n+1)}(\theta) }{(n+1)!} x^{n+1} \le \frac{1}{(n+1)!} x^{n+1} \to 0$ as $n \to \infty, 0 < \theta < x$

Thus in a neighborhood of 0, f is given by its taylor series about 0 and so f is analytic in neighborhood of 0. This taylor series converges in R, and so f is analytic in R.

Ok. Yes, you are just talking about real analytic.

 

[Shoelace Thm.]

How do I prove this generally though, like you we're saying in your first post?

 

[Bacle2]

Why don't you expand sin(x+y) in terms of sinx, cosx, siny, cosy , both of which are analytic? Then you can show that the product of analytic functions is analytic; don't think should be hard; maybe a bit annoying, keeping track of a lot of things, but don't think too hard.

You can use the proof that e^z as e^(x+iy) is (complex) analytic, and e^(x+iy)=(e^x)(e^iy)

to aid in your argument.

 

[Shoelace Thm.]

I'm looking for something more general, i.e. if f(x) is analytic, then f(x+c) is analytic.

 

[Bacle2]

If I understood you well, the same idea would work. Expand , then you have Sin(x)Cos(c)+Cos(x)Sin(c)

Then you just have the scaled sum (by Cos(c), Sin(c) respectively) of two analytic functions, which is analytic.

 

[Shoelace Thm.]

No I mean for a general function f (not necessarily the one I've specified).

 

[Shoelace Thm.]

Ok here is an argument: Because f is analytic, $f(c+x) = \sum_{n=0}^{\infty} a_n x^n$. Then $g(0+x) = f(c+x) = f(x+c) = \sum_{n=0}^{\infty} a_n x^n$, so g is analytic in a neighborhood of 0. Then g is analytic in (-R,R), R being the radius of convergence of the series.

Can someone confirm the validity of this argument?

 

[Bacle2]

Well, if f is analytic in (-oo,oo) and x+c is in (-oo,oo) , then f is analytic in (-oo,oo) , i.e., x+c lies in the domain of analyticity of f.

 

[Shoelace Thm.]

I don't quite understand what you've written; could you explain? Is what I've written correct?

 

[Bacle2]

Yes, what you said in #4 is correct; sorry,I started reading at post #10-or-so. What I meant is something similar to what you said: f(x) is analytic for all x in R , by assumption. Then x+c is a number in R, so that f should be analytic at x+c . Imagine if the problem were different and you had g analytic in some (-a,a). Then
g would be analytic at x+c if -a<x+c<a .

 

[Shoelace Thm.]

I meant what I wrote in #16.

Also, I think what I wrote in #4 is incorrect because of what Dick said, i.e. the power series i had written is not a power series in x.

 

[Bacle2]

I don't see why the series has to be centered on any specific point; is that part of your conditions?

Also, I can't tell what your function g stands for.

Sorry, I will be gone for a while.

Edit:I'm back.

 

[Bacle2]

i'm back for a while.