# Is there a way to Connect a function ''normally'' through its Pole?

Basicaly I have this function:

Vff[x] = (1.33872 (-x^2 +
x^2 (0.0337677 + 3.53187 x - 8.1484 x^2 + 12.6818 x^3 -
12.3699 x^4 + 7.65446 x^5 - 3.04756 x^6 + 0.777227 x^7 -
0.122575 x^8 + 0.010878 x^9 - 0.00041522 x^10)))/(x (-1 +
x^2 (0.0337677 + 3.53187 x - 8.1484 x^2 + 12.6818 x^3 -
12.3699 x^4 + 7.65446 x^5 - 3.04756 x^6 + 0.777227 x^7 -
0.122575 x^8 + 0.010878 x^9 - 0.00041522 x^10)))


One can do the plot:

Plot[Vff[x], {x, 0, 5}, PlotStyle -> {{Thick, Blue}},
PlotRange -> All]


To get this image:

Is there any way to find the function of this and somehow make it pas through (1,0.2) normally?

Any help is appreciated :)

EDIT: (Progress?)

I have made two lists, one for x<0.85 and one for x>1.15 and after merging them:

Leftt = Table[{x, N[Vff[x]]}, {x, 0.01, 0.85, .2}];
Rightt = Table[{x, N[Vff[x]]}, {x, 1.15, 5, .2}];
VFF = Riffle[Leftt, Rightt];


I tried to FindFormula, resulating in a constant 0.154???

VFFF[x_] =
FindFormula[VFF, x, PerformanceGoal -> "Quality",
SpecificityGoal -> "Low"]
VFFF[x]
0.154565


Edit2: Fixed it!! Just made the analysis better!

• Depends on what you mean by Normal. You can try the option Exclusions -> None which will remove the disconnect. However the pole will still be there. If you want to remove the pole completely then create a Table of values for x and Vff[x] with x=1 removed. Then ListPlot this data. – Lotus Jul 11 '18 at 11:07
• @Lotus Hmm will try to find this way with table. don't know how to do it(new to mathematica) Thanks for your Help! – billy Jul 11 '18 at 11:15
• @billy isn't that more of a math problem? We can help implement any solution/algorithm you find but it is rather broad now, e.g. you can draw a straight line from {.5 Vff[.5]} to {1.5, Vff[1.5]} and it fits description. – Kuba Jul 11 '18 at 11:29
• @kuba I just didn't know if there was anycommand like (Flatten) or stuf like this that would bypass this pole. – billy Jul 11 '18 at 12:05
• Looks like Vff could be a Padé approximation and you're trying to use it outside its range of validity. It may be better to go back to the origin of the approximation from which Vff is derived, and choose different parameters there. – Jens Jul 11 '18 at 16:37