Fourier Series Convergence at the Origin

In summary, The theorem being discussed is Theorem 1 in Kreyszig's Advanced Engineering Mathematics, 3rd Ed., which states that if a periodic function with period ##2\pi## is piecewise continuous and has left-and right-hand derivatives at each point in the interval ##-\pi \leq x \leq \pi##, then its Fourier series is convergent and equals the function except at points of discontinuity where it is the average of the left- and right-hand limits. The numbers in parentheses represent the usual definition of a Fourier series and the formulas for its coefficients. The theorem does not state that the value of the series at x = -π/2 is 1/2, but rather that the
  • #1
robertjford80
388
0

Homework Statement



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The Attempt at a Solution



Obviously brackets mean something other than parentheses because .5[0 + 0] ≠ .5
 
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  • #2
The brackets mean the same thing as parentheses...

I was unable to "Dirichlet's theorem" on the web, except as it applies to number theory and prime numbers. In Kreyszig's Advanced Engineering Mathematics, 3rd Ed., there is a theorem (identified only as Theorem 1) in the section titled "Fourier Series. Euler Formulas," and I believe that's the one you're mentioning.

If a periodic function f(x) with period ##2\pi## is piecewise continuous in the interval ##-\pi \leq x \leq \pi## and has a left-and right-hand derivative at each point of that interval, then the corresponding Fourier series (7) [with coefficients (6)] is convergent. Its sum is f(x), except at a point x0 at which f(x) is discontinuous and the sum of the series is the average of the left- and right-hand limits of f(x) at x0.
The numbers in parentheses are, respectively, the usual definition of a Fourier series, and the formulas for the coefficients of the series.

The theorem is talking more about the function represented by the Fourier series, and not so much about the function the series represents. At most points, the two are identical, but at discontinuities, they are different.

The theorem doesn't say that f(##-\pi/2##) = 1/2. It says that the value of the series at x = ##-\pi/2## is 1/2.
 
  • #3
robertjford80 said:


Screenshot2012-06-17at15615AM.png




The Attempt at a Solution



Obviously brackets mean something other than parentheses because .5[0 + 0] ≠ .5


That's just a typo. It should read$$\frac 1 2[f(0^+)+f(0^-) =\frac 1 2 [0+1]=\frac 1 2$$and it isn't f(0); it is the value of the FS at 0.
 

What is a Fourier series convergence?

A Fourier series convergence is a mathematical concept that describes how a periodic function can be represented as a sum of simple sine and cosine functions. It is used to approximate and analyze functions in fields such as engineering, physics, and signal processing.

How is the convergence of a Fourier series determined?

The convergence of a Fourier series is determined by evaluating the partial sums of the series and checking if they approach the original function as the number of terms increases. If the partial sums converge to the function, the Fourier series is said to converge.

What is the importance of Fourier series convergence?

Fourier series convergence is important because it allows us to approximate complex functions with simpler trigonometric functions, making it easier to analyze and manipulate them. It also has many practical applications in fields such as electrical engineering, acoustics, and image processing.

What are some conditions for a Fourier series to converge?

Some conditions for a Fourier series to converge include the function being represented being periodic, continuous, and piecewise smooth. The function must also have a finite number of discontinuities and the integral of the function over one period must exist.

Can a Fourier series converge to a non-periodic function?

No, a Fourier series can only converge to a periodic function. If a function is not periodic, its Fourier series will not be able to accurately represent it. However, by extending the function periodically, its Fourier series can still be used to approximate it.

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