Pointwise convergence for all real x

  • #1
lys04
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4
Homework Statement
pointwise convergence of a series
Relevant Equations
$$ \sum_{n=1}^\infty sin(\frac{n\pi}{2})sin(nx) $$
How do I know whether or not the series
$$ \sum_{n=1}^\infty sin(\frac{n\pi}{2})sin(nx)$$

converges pointwise for all real x or not?

By the way am I right in thinking that converging pointwise for all real x means whatever x i plug into the series it converges to some finite value?

I was thinking if i plug in x=pi/2 then I'd get

$$ \sum_{n=1}^\infty sin^2(\frac{n\pi}{2}) $$

Which diverges, does that prove that the series doesn't converge pointwise for all x?
 
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  • #2
lys04 said:
How do I know whether or not the series
$$ \sum_{n=1}^\infty sin(\frac{n\pi}{2})sin(nx)$$

converges pointwise for all real x or not?
You either show that it does or find a way to show that it does not.


lys04 said:
By the way am I right in thinking that converging pointwise for all real x means whatever x i plug into the series it converges to some finite value?
Yes.

lys04 said:
I was thinking if i plug in x=pi/2 then I'd get

$$ \sum_{n=1}^\infty sin^2(\frac{n\pi}{2}) $$

Which diverges, does that prove that the series doesn't converge pointwise for all x?
Yes.
 
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  • #3
lys04 said:
Homework Statement: pointwise convergence of a series
Relevant Equations: $$ \sum_{n=1}^\infty sin(\frac{n\pi}{2})sin(nx) $$

How do I know whether or not the series
$$ \sum_{n=1}^\infty sin(\frac{n\pi}{2})sin(nx)$$

converges pointwise for all real x or not?
Note that ##\sin(\frac{n\pi}{2}) = 0## if ##n## is even and alternates between ##\pm1## if ##n## is odd. Using the alternating series test, ##\sum_{n=1}^\infty sin(\frac{n\pi}{2})a_n## converges iff ##a_n \to 0## for odd ##n##.

In this case, the series converges iff ##\sin((2n+1)x) \to 0## as ##n \to \infty##. That's only the case when ##x = k\pi## for some integer ##k##. Although, it's not that easy from first principles to prove that it does not converge for any other ##x##. You found a good example,
 
  • #4
PeroK said:
In this case, the series converges iff ##\sin((2n+1)x) \to 0## as ##n \to \infty##. That's only the case when ##x = k\pi## for some integer ##k##. Although, it's not that easy from first principles to prove that it does not converge for any other ##x##. You found a good example,
The series expansion of the sine function can help. Not really first principles but at least in the realm of the question.

Another idea is to substitute ##y=x-(\pi/2)## which turns the series members into ##\sin^2(n \pi /2)\cos(ny)## which reduces the question to: Under what circumstances does ##\sum \sin^2(n \pi /2)\cos(ny)## converge, if ##\sum \sin^2(n \pi /2)## diverges?

One should also keep the Weierstraß- or half-angle substitution in mind whenever it comes to trig functions and integrals or sums.
 
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  • #5
PeroK said:
Note that ##\sin(\frac{n\pi}{2}) = 0## if ##n## is even and alternates between ##\pm1## if ##n## is odd. Using the alternating series test, ##\sum_{n=1}^\infty sin(\frac{n\pi}{2})a_n## converges iff ##a_n \to 0## for odd ##n##.

One small point, we don't actually know the ##a_n## are positive in this question so the alternating test doesn't work.
 
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  • #6
Office_Shredder said:
One small point, we don't actually know the ##a_n## are positive in this question so the alternating test doesn't work.
Yes, well spotted.
 

FAQ: Pointwise convergence for all real x

What is pointwise convergence?

Pointwise convergence refers to a type of convergence of a sequence of functions. A sequence of functions {f_n} defined on a set D converges pointwise to a function f if, for every point x in D, the sequence of real numbers {f_n(x)} converges to f(x) as n approaches infinity.

How do you determine if a sequence of functions converges pointwise?

To determine if a sequence of functions converges pointwise, you need to check for each point x in the domain D whether the limit of f_n(x) exists as n approaches infinity. If the limit exists for all x in D, then the sequence converges pointwise to the function f defined by f(x) = lim (n→∞) f_n(x).

What is the difference between pointwise convergence and uniform convergence?

The key difference between pointwise convergence and uniform convergence lies in the manner of convergence. Pointwise convergence requires that for each individual point x in the domain, the function values converge. In contrast, uniform convergence requires that the functions converge to the limit function uniformly across the entire domain, meaning that the speed of convergence does not depend on the choice of x and is uniform for all x in the domain.

Can pointwise convergence fail to preserve continuity?

Yes, pointwise convergence can fail to preserve continuity. A sequence of continuous functions can converge pointwise to a function that is not continuous. A classic example is the sequence of functions f_n(x) = x^n defined on the interval [0, 1]. As n approaches infinity, f_n(x) converges pointwise to a function that is continuous at x = 1 and discontinuous at every point in [0, 1) as it converges to 0.

Is pointwise convergence sufficient for integration and differentiation?

No, pointwise convergence is not sufficient for interchanging limits with integration or differentiation. For example, the limit of the integrals of a sequence of functions may not equal the integral of the limit function, and similar issues arise with differentiation. Conditions such as uniform convergence or the Dominated Convergence Theorem are often required to ensure that such interchange is valid.

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