How to evaluate A Taylor series at a specific value

08242018, 02:08 PM
Post: #30




RE: How to evaluate A Taylor series at a specific value
(08242018 06:21 AM)parisse Wrote: This is wrong: x:=1 followed by sin(x)x=2 returns sin(2), and that's not easy to evaluate, you must have a "local" value of x (2) that takes precedence over the global value of x (1). It would be nice if you could use the "where" operator "" to temporarily undefine variables. For example: x:=1 followed by sin(x)x=? (I'm not sure what the syntax ought to be here, but basically mark x as undefined) would return sin(x), ignoring the value assigned to x. — Ian Abbott 

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