Talk:Eigenfunction

Too complicated[edit]

This explanation seems way to complicated. If someone understood this, they wouldn't need Wikipedia to explain it to them.

To understand this you have to know that functions can behave analogously to vectors, and function spaces to vector spaces, and that a linear operator converts vectors to other vectors (or functions). If you multiply a vector by a scalar, you merely change a vector's magnitude, and not it's direction. So an eigenfunction is one whose direction is unchanged by the linear operator it is an eigenfunction of. ᛭ LokiClock (talk) 08:36, 13 July 2010 (UTC)[reply]

- Right, so perhaps that should be in the article. — Preceding unsigned comment added by 71.230.108.182 (talk) 08:17, 19 February 2017 (UTC)[reply]

Slow down graphic[edit]

Please change the graphic example of the vibrating drum problem. It moves too fast and repeatedly and distracts the viewer from reading. I may even add that some pure soul with epilepsy will have a fit with it. —Preceding unsigned comment added by 87.203.201.190 (talk) 23:20, 19 August 2008 (UTC)[reply]

Hit 'Esc' and it will stop. Oleg Alexandrov (talk) 02:57, 20 August 2008 (UTC)[reply]

That's not a solution. Please change the graphic example of the vibrating drum problem.

Example in introduction[edit]

I understand why the example holds, but wouldn't f(x) = e^(kx) also be an eigenfunction of the simpler differential operation A = d/dx, with eigenvalue k? If so, why is the more complicated example offered? =/ — Preceding unsigned comment added by 72.192.213.145 (talk) 22:55, 21 January 2012 (UTC)[reply]

Special cases with finite dimension[edit]

Under section Link to eigenvalues and eigenvectors of matrices there is the text:

"In some special cases, such as the coefficients of the Fourier series of a sinusoidal function, this column vector has finite dimension."

This feels invalid to me. To my mind, the column vector is still infinite, it simply has a finite number of non-zero values. Just because the rest of the values are "unused" doesn't mean they don't exist.

Dansharkey (talk) 08:31, 9 August 2023 (UTC)[reply]