TOPICS
Search

Parseval's Theorem

If a function has a Fourier series given by

f(x)=1/2a_0+sum_(n=1)^inftya_ncos(nx)+sum_(n=1)^inftyb_nsin(nx),
(1)

then Bessel's inequality becomes an equality known as Parseval's theorem. From (1),

[f(x)]^2=1/4a_0^2+a_0sum_(n=1)^infty[a_ncos(nx)+b_nsin(nx)]+sum_(n=1)^inftysum_(m=1)^infty[a_na_mcos(nx)cos(mx)+a_nb_mcos(nx)sin(mx)+a_mb_nsin(nx)cos(mx)+b_nb_msin(nx)sin(mx)].
(2)

Integrating

int_(-pi)^pi[f(x)]^2dx=1/4a_0^2int_(-pi)^pidx+a_0int_(-pi)^pisum_(n=1)^infty[a_ncos(nx)+b_nsin(nx)]dx+int_(-pi)^pisum_(n=1)^inftysum_(m=1)^infty[a_na_mcos(nx)cos(mx)+a_nb_mcos(nx)sin(mx)+a_mb_nsin(nx)cos(mx)+b_nb_msin(nx)sin(mx)]dx =1/4a_0^2(2pi)+0+sum_(n=1)^inftysum_(m=1)^infty[a_na_mpidelta_(nm)+0+0+b_nb_mpidelta_(nm)],
(3)

so

1/piint_(-pi)^pi[f(x)]^2dx=1/2a_0^2+sum_(n=1)^infty(a_n^2+b_n^2).
(4)

For a generalized Fourier series of a complete orthogonal system {phi_i}_(i=1)^infty, an analogous relationship holds.

For a complex Fourier series,

1/(2pi)int_(-pi)^pi|f(x)|^2dx=sum_(n=-infty)^infty|a_n|^2.
(5)

See also

Bessel's Inequality, Complete Orthogonal System, Fourier Series, Generalized Fourier Series, Plancherel's Theorem, Power Spectrum

Explore with Wolfram|Alpha

References

Gradshteyn, I. S. and Ryzhik, I. M. Tables of Integrals, Series, and Products, 6th ed. San Diego, CA: Academic Press, p. 1101, 2000.Kaplan, W. Advanced Calculus, 4th ed. Reading, MA: Addison-Wesley, p. 501, 1992.

Referenced on Wolfram|Alpha

Parseval's Theorem

Cite this as:

Weisstein, Eric W. "Parseval's Theorem." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/ParsevalsTheorem.html

Subject classifications