How to find the derivative value at $(\pi,0)$ for this implicit function $n$ times?

Question

I am trying to take the implicit derivative at $\sin(x+y)+\sin(x)=y$ and substitute $x=\pi$ and $y=0$ at least 6-7 times since I need to find the Taylor series for this function.

Since I barely anything about coding I referred to the documentation in Mathematica and other web pages which showed to do this...

D[Sin[x]+Sin[x+y[x]] == y[x],{ x,1}]

Then I took the output of that and solved for $y[x]$

Solve[Cos[x] + Cos[x + y[x]] (1 + Derivative[1][y][x]) == 
 Derivative[1][y][x], Derivative[1][y][x]]

Then took that output and finally substituted $x=\pi$ and $y[x]=0$

    {Derivative[1][y][x] -> (-Cos[x] - Cos[x + y[x]])/(-1 + 
        Cos[x + y[x]])}/.{y[x]->0,x->Pi}
(*-1*)

Unfortunately this process was too tedious so I took some programming from here

D[Sin[x] + Sin[x+y[x]] == y, {x, 1}]
sol1 = Solve[%, y'[x]]
D[Sin[x] + Sin[x+y[x]] == y, {x, 2}]
sol2 = Solve[%, y''[x]]
sol2 /. sol1 // Simplify

And ended up with the second derivative but was too long.

Is there a more elegant way of doing this and how? (One thing you can do is solve for the first implicit derivative, substitute $x$ and $y$, then substitute the first into the second derivative and so on with the next derivatives).

Maybe Dt (total derivative) is what you need to take implicit derivatives? You could also write y as y[x], put everything in one side of the equation, and call Series. — QuantumDot, Jan 08 '16 at 19:16
Related: http://mathematica.stackexchange.com/questions/94663/implicit-function-theorem-to-higher-order (Two of the answers compute a truncated Taylor series of an implicitly defined function.) — Michael E2, Jan 08 '16 at 20:19

score 8 · Accepted Answer · edited Sep 26 '16 at 17:20

8

You can substitute a Taylor expansion in your equation

y=Sum[a[n](x-π)^n, {n,1,6}]+ O[x,π]^7

SolveAlways[ Sin[x+y]+Sin[x]-y ==0, x]

(* {{a[5]->-(1/240),a[6]->0,a[4]->0,a[3]->1/12,a[2]->0,a[1]->-1}} *)

edited Sep 26 '16 at 17:20

LCarvalho

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answered Jan 08 '16 at 19:40

Fred Simons

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score 4 · Answer 2 · edited Sep 26 '16 at 17:25

Last[Nest[Function[d, {d, Append[Last[#], Solve[0 == d /. x -> π /. Last[#]][[1, 1]]]}][
                   D[First[#], x]] &, {Sin[x] + Sin[x + y[x]] - y[x], {y[π] -> 0}}, 15]]

$$\left\{y(\pi )\to 0,y'(\pi )\to -1,y''(\pi )\to 0,y^{(3)}(\pi )\to \frac{1}{2},y^{(4)}(\pi )\to 0,y^{(5)}(\pi )\to -\frac{1}{2},y^{(6)}(\pi )\to 0,y^{(7)}(\pi )\to \frac{1}{2},y^{(8)}(\pi )\to 0,y^{(9)}(\pi )\to 17,y^{(10)}(\pi )\to 0,y^{(11)}(\pi )\to -\frac{1153}{4},y^{(12)}(\pi )\to 0,y^{(13)}(\pi )\to \frac{13297}{4},y^{(14)}(\pi )\to 0,y^{(15)}(\pi )\to \frac{108109}{2}\right\}$$

score 2 · Answer 3 · answered Jan 08 '16 at 20:09

You need to find the Taylorpolynom for your function?

The code for the Taylor formula:

taylor = (vars - point).# &;
init := D[f[vars], {vars, j}] /. Thread[vars -> point];
taylorPolynom[m_] := Sum[1/j! Nest[taylor, init, j], {j, 0, m}]

vars = {x, y};
point = {Pi, 0};
f[vars_] = Sin[x + y] + Sin[x] - y;

taylorPolynom[2]
-2 (-Pi + x) - 2 y

or

taylorPolynom[6] // FullSimplify
1/120 (2 (120 + (-20 + (Pi - x)^2) (Pi - x)^2) (Pi - x) - 
   5 (48 + (-12 + (Pi - x)^2) (Pi - x)^2) y + 
   10 (-6 + (Pi - x)^2) (Pi - x) y^2 - 
   10 (-2 + (Pi - x)^2) y^3 + 5 (Pi - x) y^4 - y^5)

score 2 · Answer 4 · answered Apr 14 '19 at 04:24

2

You can use the new in M12 function AsymptoticSolve to do this:

AsymptoticSolve[Sin[x + y] + Sin[x] == y, {y, 0}, {x, π, 10}]

{{y -> π - x + 1/12 (-π + x)^3 - 1/240 (-π + x)^5 + (-π + x)^7/10080 + (17 (-π + x)^9)/362880}}

answered Apr 14 '19 at 04:24

Carl Woll

130,679
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score 1 · Answer 5 · answered Jan 09 '16 at 13:03

The SolveAlways form is very nice, while some of the other solns didn't immediately work V7. But what is SolveAlways doing? Probably something like the following -- Compute the derivatives by brute force, and then match at the expansion center. The derivative computation becomes increasingly time consuming as the \ number of terms increases, but 9 terms is pretty fast.
The Taylor series form is used here, so the y derivatives are obtained, as shown by Coolwater. Removing the n! will match the other \ responses. (Note, subscripts are inert labels.)

nterm = 9;
Clear[yb, dyb, aList]
yb = Sum[Subscript[a, n] (x - Pi)^n/n!, {n, 1, nterm + 1}] + O[x, Pi]^(nterm + 2);
dyb = Table[   D[Sin[x + yb] + Sin[x] - yb, {x, jj}] /. x -> Pi, {jj, 1, nterm}];
aList = Table[Subscript[a, n], {n, 1, nterm}];
(* match the representation at Pi by adjusting the coefficients *)
vDerivs = Solve[dyb == 0, aList][[1]];
Print[vDerivs];
(* {Subscript[a, 8]->0,Subscript[a, 9]->17,Subscript[a, 7]->1/2,Subscript[a, 6]->0,Subscript[a, 5]->-(1/2),Subscript[a, 4]->0,Subscript[a, 3]->1/2,Subscript[a, 2]->0,Subscript[a, 1]->-1} *)

Print[Sort[vDerivs] ]; (Puts output values in the right order) — Steve C, Jan 09 '16 at 14:34

score 1 · Answer 6 · answered Jul 22 '16 at 09:36

I'm surprised that nobody followed up on QuantumDot's suggestion to use Dt[]:

With[{n = 9},
     (Table[Dt[y, {x, k}], {k, n}] /. 
      First @ Solve[Table[Dt[Sin[x] + Sin[x + y] - y, {x, k}] == 0, {k, n}], 
                    Table[Dt[y, {x, k}], {k, n}]]) /. {x -> π, y -> 0}]
   {-1, 0, 1/2, 0, -1/2, 0, 1/2, 0, 17}

How to find the derivative value at $(\pi,0)$ for this implicit function $n$ times?

6 Answers6