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Talk:Formal power series: Difference between revisions
→About solving equations: The problem is with the task
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assuming that d/d''x'' denotes the differential operator. The task uses the integral operator instead and sin''x'' as a temporal. Shouldn't it work with '''any''' operator defined defined in the task, in any combination of? --[[User:Dmitry-kazakov|Dmitry-kazakov]] 10:32, 17 February 2009 (UTC)
:But also <math>\sin x = \frac{-d^2}{dx^2} \sin x</math>; and in fact any linear combination of sin and cos will satisfy that differential equation you gave. The basic reason why we don't use differentiation is that it loses information, so equations based on differentiation will not have unique solutions. --[[Special:Contributions/76.167.241.45|76.167.241.45]] 06:00, 18 February 2009 (UTC)
::No, the actual problem is that the solutions proposed in the task (and lazy expressions in particular) do not solve what the task is supposed to require. It only happens so that formal integration of 1 plus an infinite self recursion occasionally gives Taylor series of cos:
:::<math>1</math><br>
:::<math>1 - \int\int 1dx = 1 - x/2</math><br>
:::<math>1 - \int\int (1 - \int\int 1dy) dx = 1 - x^2/2 + x^4/2*3*4</math><br>
:::<math>1 - \int\int (1 - \int\int (1 - \int\int 1dz)dy)dx = 1 - x^2/2 + x^4/2*3*4 - x^6/2*3*4*5*6</math><br>
:::. . .<br>
::But it is a wrong solution in other cases, which are mathematically equivalent. Strange if that would be otherwise. Because mathematically it is about solving equations (integral, differential, basically any).
::I would suggest to remove laziness as irrelevant to Taylor series, as well as self-recursive and also incomputable task. Instead of this I would take some finite series and add, integrate them etc. --[[User:Dmitry-kazakov|Dmitry-kazakov]] 10:38, 18 February 2009 (UTC)
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