## #1033 taylor at infinity for non algebraic

closed
nobody
3
2007-10-19
2006-11-29
No

Consider:

(%i1) log(x) + exp(x^2)\$
(%i2) e : diff(%,x)/%;
(%o2) (2*x*%e^x^2+1/x)/(log(x)+%e^x^2)

Towards infinity, a good approximation to e is 2 x. But

(%i3) taylor(e,x,inf,2);
(%o3) 4/x+...

And

(%i4) taylor(e,x,inf,5);
(%o4) 240/x^5+...

And

(%i5) taylor(ratsimp(e),x,inf,5);
1/(x*log(x)+x*%e^x^2)
Assumed to be zero in `taylor'
(%o5) 0+...

It would be better if taylor just gave up. All
the results are not right.

Barton

## Discussion

• Barton Willis - 2006-12-01

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Two related problems:

(%i1) log(log(x)) + exp(-x)\$
(%i2) diff(%,x)/%\$
(%i3) taylor(%,x,inf,2);
Invalid call to var-expand

(%i4) log(log(x)) + x\$
(%i5) diff(%,x)/%\$
(%i6) taylor(%,x,inf,2);
(%o6) 1/x+(1/log(x)+(-log(log(x))+zeroa+...)+...)/x^2+...

Isn't this unsimplified? I don't know what taylor
is trying to do with expressions similar to %o5.
Again, maybe it would be better if taylor gave up.

• Dan Gildea - 2007-10-04

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These seem ok in current cvs.

(%i7) log(x) + exp(x^2)\$

(%i8) e : diff(%,x)/%;
(%o8) (2*x*%e^x^2+1/x)/(log(x)+%e^x^2)

(%i9) taylor(e,x,inf,2);
(%o9) +2*x+((+(-2*log(x)))*x+1/x)*%e^-x^2
+((+2*log(x)^2)*x+(+(-log(x)))/x)*(%e^-x^2)^2

(%i10) taylor(e,x,inf,5);
(%o10) +2*x+((+(-2*log(x)))*x+1/x)*%e^-x^2
+((+2*log(x)^2)*x+(+(-log(x)))/x)*(%e^-x^2)^2
+((+(-2*log(x)^3))*x+(+log(x)^2)/x)*(%e^-x^2)^3
+((+2*log(x)^4)*x+(+(-log(x)^3))/x)*(%e^-x^2)^4
+((+(-2*log(x)^5))*x+(+log(x)^4)/x)*(%e^-x^2)^5

(%i11) taylor(ratsimp(e),x,inf,5);
(%o11) +2*x+((+(-2*log(x)))*x+1/x)*%e^-x^2
+((+2*log(x)^2)*x+(+(-log(x)))/x)*(%e^-x^2)^2
+((+(-2*log(x)^3))*x+(+log(x)^2)/x)*(%e^-x^2)^3
+((+2*log(x)^4)*x+(+(-log(x)^3))/x)*(%e^-x^2)^4
+((+(-2*log(x)^5))*x+(+log(x)^4)/x)*(%e^-x^2)^5

(%i12) log(log(x)) + exp(-x)\$

(%i13) diff(%,x)/%;
(%o13) (1/(x*log(x))-%e^-x)/(log(log(x))+%e^-x)

(%i14) taylor(%,x,inf,2);
(%o14) +(+(+1/log(log(x)))/log(x))/x+(+(-1/log(log(x)))
+(+(+(-1/log(log(x))^2))/log(x))/x)
*%e^-x+(+1/log(log(x))^2)*(%e^-x)^2

(%i15) log(log(x)) + x\$

(%i16) diff(%,x)/%;
(%o16) (1/(x*log(x))+1)/(log(log(x))+x)

(%i17) taylor(%,x,inf,2);
(%o17) 1/x+(+(-log(log(x)))+1/log(x))/x^2

• Dan Gildea - 2007-10-04
• status: open --> pending

• SourceForge Robot - 2007-10-19

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user_id=1312539
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This Tracker item was closed automatically by the system. It was
previously set to a Pending status, and the original submitter
did not respond within 14 days (the time period specified by

• SourceForge Robot - 2007-10-19
• status: pending --> closed