Skip to main content

calculus and analysis - Derivative of piecewise function


I tried to find $f'(0)$ of this function:


$$f(x) = \begin{cases} x\cdot\sin(\frac{1}{x}) & \text{if $x\ne0$} \\ 0 & \text{if $x=0$} \end{cases}$$


This is what I tried in Mathematica:


f[x_] = Piecewise[{{x*Sin[1/x], x != 0}, {0, x == 0}}]
f'[0]


Mathematica gives the answer $0$, while the answer should be $undefined$, also see this discussion: https://math.stackexchange.com/questions/1551257/derivative-of-piece-wise-function-at-x-0


Any idea why Mathematica doesn't give the correct anwer?



Answer



Mathematica is being inconsistent in how it is treating the derivative for a piecewise function (this seems like a bug to me). We can look at a simpler example to see this, which will point towards a workaround,


pwf1[x_] := Piecewise[{
{3 x, x != 0},
{5 x, x == 0}}];
pwf2[x_] := Piecewise[{
{3 x, x < 0},

{5 x, x == 0},
{3 x, x > 0}}];
pwf1'[0]
pwf2'[0]
(* 5 *)
(* 3 *)

These are both the same function, and if we take the derivative manually, then we get the same answer:


Limit[(pwf1[0 + h] - pwf1[0])/h, h -> 0]
Limit[(pwf2[0 + h] - pwf2[0])/h, h -> 0]

(* 3 *)
(* 3 *)

So apparently it is better to define the piecewise regions more explicitly,


f[x_] := Piecewise[{
{x*Sin[1/x], x > 0},
{0, x == 0},
{x*Sin[1/x], x < 0}
}]
f'[0]

(* Indeterminate *)

This is the same answer you get when you take the analytic derivative and substitute x=0,


func[x_] := x Sin[1/x];
func'[0]


During evaluation of In[149]:= Power::infy: Infinite expression 1/0 encountered. >>


During evaluation of In[149]:= Power::infy: Infinite expression 1/0 encountered. >>


During evaluation of In[149]:= Power::infy: Infinite expression 1/0 encountered. >>



During evaluation of In[149]:= General::stop: Further output of Power::infy will be suppressed during this calculation. >>



(* Indeterminate *)

Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...