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Too complicated
editThis 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)
- 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)
Slow down graphic
editPlease 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)
- Hit 'Esc' and it will stop. Oleg Alexandrov (talk) 02:57, 20 August 2008 (UTC)
- That's not a solution. Please change the graphic example of the vibrating drum problem.
- It needs a stop/start button. Xxanthippe (talk) 03:48, 20 November 2024 (UTC).
Example in introduction
editI 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)
Special cases with finite dimension
editUnder 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.