On the n-th derivative of the inverse function
$begingroup$
Given a function $f(x)$, its inverse $g(x)$ is defined as $g(f(x)) equiv x$.
In light of this, the n-th derivative of $g(x)$ can be recursively calculated as follows:
list = {};
For[n = 1, n <= 6, n++,
eqn = D[g[f[x]], {x, n}] == D[x, {x, n}];
var = Derivative[n][g][f[x]];
If[n == 1,
list = Join[list, Solve[eqn, var][[1]]],
list = Join[list, (Solve[eqn, var] /. list)[[1]]]
]
]
list // Expand // TableForm
In MMA you can get this very simply by writing something like this:
Derivative[n][InverseFunction[f]][f[x]]
The question that arises is the following: is the latter formulation using a recursive process like the one shown above or is there a non-recursive formulation?
calculus-and-analysis
asked Dec 25 '18 at 11:02
TeMTeM
2,029621
$endgroup$
add a comment |
$begingroup$
Given a function $f(x)$, its inverse $g(x)$ is defined as $g(f(x)) equiv x$.
In light of this, the n-th derivative of $g(x)$ can be recursively calculated as follows:
list = {};
For[n = 1, n <= 6, n++,
eqn = D[g[f[x]], {x, n}] == D[x, {x, n}];
var = Derivative[n][g][f[x]];
If[n == 1,
list = Join[list, Solve[eqn, var][[1]]],
list = Join[list, (Solve[eqn, var] /. list)[[1]]]
]
]
list // Expand // TableForm
In MMA you can get this very simply by writing something like this:
Derivative[n][InverseFunction[f]][f[x]]
The question that arises is the following: is the latter formulation using a recursive process like the one shown above or is there a non-recursive formulation?
calculus-and-analysis
asked Dec 25 '18 at 11:02
TeMTeM
2,029621
$endgroup$
$begingroup$
Something likeDerivative[3][InverseFunction[f]][f[x]]?
$endgroup$
– Szabolcs
Dec 25 '18 at 11:07
$begingroup$
@Szabolcs: I was just modifying writing this. Now I have again asked the question, thank you!
$endgroup$
– TeM
Dec 25 '18 at 11:16
$begingroup$
Sounds like a math question, not Mathematica ... (it's interesting though)
$endgroup$
– Szabolcs
Dec 26 '18 at 9:07
$begingroup$
@Szabolcs: In fact you are right. Sometimes, however, some topics are borderline, it is difficult to understand how and where to formulate them, at least for me.
$endgroup$
– TeM
Dec 26 '18 at 9:10
add a comment |
$begingroup$
Given a function $f(x)$, its inverse $g(x)$ is defined as $g(f(x)) equiv x$.
In light of this, the n-th derivative of $g(x)$ can be recursively calculated as follows:
list = {};
For[n = 1, n <= 6, n++,
eqn = D[g[f[x]], {x, n}] == D[x, {x, n}];
var = Derivative[n][g][f[x]];
If[n == 1,
list = Join[list, Solve[eqn, var][[1]]],
list = Join[list, (Solve[eqn, var] /. list)[[1]]]
]
]
list // Expand // TableForm
In MMA you can get this very simply by writing something like this:
Derivative[n][InverseFunction[f]][f[x]]
The question that arises is the following: is the latter formulation using a recursive process like the one shown above or is there a non-recursive formulation?
calculus-and-analysis
asked Dec 25 '18 at 11:02
TeMTeM
2,029621
$endgroup$
Given a function $f(x)$, its inverse $g(x)$ is defined as $g(f(x)) equiv x$.
In light of this, the n-th derivative of $g(x)$ can be recursively calculated as follows:
list = {};
For[n = 1, n <= 6, n++,
eqn = D[g[f[x]], {x, n}] == D[x, {x, n}];
var = Derivative[n][g][f[x]];
If[n == 1,
list = Join[list, Solve[eqn, var][[1]]],
list = Join[list, (Solve[eqn, var] /. list)[[1]]]
]
]
list // Expand // TableForm
In MMA you can get this very simply by writing something like this:
Derivative[n][InverseFunction[f]][f[x]]
The question that arises is the following: is the latter formulation using a recursive process like the one shown above or is there a non-recursive formulation?
calculus-and-analysis
calculus-and-analysis
asked Dec 25 '18 at 11:02
TeMTeM
2,029621
asked Dec 25 '18 at 11:02
TeMTeM
2,029621
edited Dec 26 '18 at 9:00
TeM
asked Dec 25 '18 at 11:02
TeMTeM
2,029621
asked Dec 25 '18 at 11:02
TeMTeM
2,029621
asked Dec 25 '18 at 11:02
TeMTeM
2,029621
2,029621
$begingroup$
Something likeDerivative[3][InverseFunction[f]][f[x]]?
$endgroup$
– Szabolcs
Dec 25 '18 at 11:07
$begingroup$
@Szabolcs: I was just modifying writing this. Now I have again asked the question, thank you!
$endgroup$
– TeM
Dec 25 '18 at 11:16
$begingroup$
Sounds like a math question, not Mathematica ... (it's interesting though)
$endgroup$
– Szabolcs
Dec 26 '18 at 9:07
$begingroup$
@Szabolcs: In fact you are right. Sometimes, however, some topics are borderline, it is difficult to understand how and where to formulate them, at least for me.
$endgroup$
– TeM
Dec 26 '18 at 9:10
add a comment |
$begingroup$
Something likeDerivative[3][InverseFunction[f]][f[x]]?
$endgroup$
– Szabolcs
Dec 25 '18 at 11:07
$begingroup$
@Szabolcs: I was just modifying writing this. Now I have again asked the question, thank you!
$endgroup$
– TeM
Dec 25 '18 at 11:16
$begingroup$
Sounds like a math question, not Mathematica ... (it's interesting though)
$endgroup$
– Szabolcs
Dec 26 '18 at 9:07
$begingroup$
@Szabolcs: In fact you are right. Sometimes, however, some topics are borderline, it is difficult to understand how and where to formulate them, at least for me.
$endgroup$
– TeM
Dec 26 '18 at 9:10
$begingroup$
Something like
Derivative[3][InverseFunction[f]][f[x]]?$endgroup$
– Szabolcs
Dec 25 '18 at 11:07
$begingroup$
Something like
Derivative[3][InverseFunction[f]][f[x]]?$endgroup$
– Szabolcs
Dec 25 '18 at 11:07
$begingroup$
@Szabolcs: I was just modifying writing this. Now I have again asked the question, thank you!
$endgroup$
– TeM
Dec 25 '18 at 11:16
$begingroup$
@Szabolcs: I was just modifying writing this. Now I have again asked the question, thank you!
$endgroup$
– TeM
Dec 25 '18 at 11:16
$begingroup$
Sounds like a math question, not Mathematica ... (it's interesting though)
$endgroup$
– Szabolcs
Dec 26 '18 at 9:07
$begingroup$
Sounds like a math question, not Mathematica ... (it's interesting though)
$endgroup$
– Szabolcs
Dec 26 '18 at 9:07
$begingroup$
@Szabolcs: In fact you are right. Sometimes, however, some topics are borderline, it is difficult to understand how and where to formulate them, at least for me.
$endgroup$
– TeM
Dec 26 '18 at 9:10
$begingroup$
@Szabolcs: In fact you are right. Sometimes, however, some topics are borderline, it is difficult to understand how and where to formulate them, at least for me.
$endgroup$
– TeM
Dec 26 '18 at 9:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
D[InverseFunction[f][x], {x, 6}]
Towards the question whether this is computed recursively: I guess so from analyzing the Trace produced by executing the code:
Table[
Length@Trace[
D[InverseFunction[f][x], {x, k}]
],
{k, 1, 12}]
{6, 7, 7, 10, 13, 18, 25, 36, 51, 73, 103, 145}
answered Dec 25 '18 at 11:06
Henrik SchumacherHenrik Schumacher
59.5k582166
$endgroup$
$begingroup$
Ok, get it, thank you!
$endgroup$
– TeM
Dec 25 '18 at 18:51
add a comment |
$begingroup$
In fact, there is a closed-form expression for the derivatives of an inverse function, if one is willing to use the (partial) Bell polynomials (implemented in Mathematica as BellY). This is related to the Lagrangian inversion formula (and see also the discussion in Charalambides).
inverseD[ff_, x_] := inverseD[ff, {x, 1}]
inverseD[ff_, {x_, k_Integer?NonNegative}] := With[{f = Function[x, ff]},
If[k == 0, Return[InverseFunction[f][x]]];
If[k == 1, Return[1/f'[InverseFunction[f][x]]]];
With[{ifun = InverseFunction[f][x]},
BellY[Table[{(k + j - 2)!, -Derivative[j][f][ifun]/f'[ifun]/j},
{j, 2, k}]]/((k - 1)! f'[ifun]^k)]]
For example:
(inverseD[g[t], {t, 4}] == D[InverseFunction[g][t], {t, 4}] /. Function[t, g[t]] -> g) // Simplify
answered Dec 31 '18 at 10:22
J. M. is away♦J. M. is away
98.9k10311467
$endgroup$
$begingroup$
(I'm still computer-less, but I had this as a draft blog post, so I thought I'd post it.)
$endgroup$
– J. M. is away♦
Dec 31 '18 at 10:22
1
$begingroup$
It is always amazing with what you come up. Good to hear from you by the way!
$endgroup$
– Henrik Schumacher
Dec 31 '18 at 10:56
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
D[InverseFunction[f][x], {x, 6}]
Towards the question whether this is computed recursively: I guess so from analyzing the Trace produced by executing the code:
Table[
Length@Trace[
D[InverseFunction[f][x], {x, k}]
],
{k, 1, 12}]
{6, 7, 7, 10, 13, 18, 25, 36, 51, 73, 103, 145}
answered Dec 25 '18 at 11:06
Henrik SchumacherHenrik Schumacher
59.5k582166
$endgroup$
$begingroup$
Ok, get it, thank you!
$endgroup$
– TeM
Dec 25 '18 at 18:51
add a comment |
$begingroup$
D[InverseFunction[f][x], {x, 6}]
Towards the question whether this is computed recursively: I guess so from analyzing the Trace produced by executing the code:
Table[
Length@Trace[
D[InverseFunction[f][x], {x, k}]
],
{k, 1, 12}]
{6, 7, 7, 10, 13, 18, 25, 36, 51, 73, 103, 145}
answered Dec 25 '18 at 11:06
Henrik SchumacherHenrik Schumacher
59.5k582166
$endgroup$
$begingroup$
Ok, get it, thank you!
$endgroup$
– TeM
Dec 25 '18 at 18:51
add a comment |
$begingroup$
D[InverseFunction[f][x], {x, 6}]
Towards the question whether this is computed recursively: I guess so from analyzing the Trace produced by executing the code:
Table[
Length@Trace[
D[InverseFunction[f][x], {x, k}]
],
{k, 1, 12}]
{6, 7, 7, 10, 13, 18, 25, 36, 51, 73, 103, 145}
answered Dec 25 '18 at 11:06
Henrik SchumacherHenrik Schumacher
59.5k582166
$endgroup$
D[InverseFunction[f][x], {x, 6}]
Towards the question whether this is computed recursively: I guess so from analyzing the Trace produced by executing the code:
Table[
Length@Trace[
D[InverseFunction[f][x], {x, k}]
],
{k, 1, 12}]
{6, 7, 7, 10, 13, 18, 25, 36, 51, 73, 103, 145}
answered Dec 25 '18 at 11:06
Henrik SchumacherHenrik Schumacher
59.5k582166
edited Dec 25 '18 at 15:56
answered Dec 25 '18 at 11:06
Henrik SchumacherHenrik Schumacher
59.5k582166
answered Dec 25 '18 at 11:06
Henrik SchumacherHenrik Schumacher
59.5k582166
answered Dec 25 '18 at 11:06
Henrik SchumacherHenrik Schumacher
59.5k582166
59.5k582166
$begingroup$
Ok, get it, thank you!
$endgroup$
– TeM
Dec 25 '18 at 18:51
add a comment |
$begingroup$
Ok, get it, thank you!
$endgroup$
– TeM
Dec 25 '18 at 18:51
$begingroup$
Ok, get it, thank you!
$endgroup$
– TeM
Dec 25 '18 at 18:51
$begingroup$
Ok, get it, thank you!
$endgroup$
– TeM
Dec 25 '18 at 18:51
add a comment |
$begingroup$
In fact, there is a closed-form expression for the derivatives of an inverse function, if one is willing to use the (partial) Bell polynomials (implemented in Mathematica as BellY). This is related to the Lagrangian inversion formula (and see also the discussion in Charalambides).
inverseD[ff_, x_] := inverseD[ff, {x, 1}]
inverseD[ff_, {x_, k_Integer?NonNegative}] := With[{f = Function[x, ff]},
If[k == 0, Return[InverseFunction[f][x]]];
If[k == 1, Return[1/f'[InverseFunction[f][x]]]];
With[{ifun = InverseFunction[f][x]},
BellY[Table[{(k + j - 2)!, -Derivative[j][f][ifun]/f'[ifun]/j},
{j, 2, k}]]/((k - 1)! f'[ifun]^k)]]
For example:
(inverseD[g[t], {t, 4}] == D[InverseFunction[g][t], {t, 4}] /. Function[t, g[t]] -> g) // Simplify
answered Dec 31 '18 at 10:22
J. M. is away♦J. M. is away
98.9k10311467
$endgroup$
$begingroup$
(I'm still computer-less, but I had this as a draft blog post, so I thought I'd post it.)
$endgroup$
– J. M. is away♦
Dec 31 '18 at 10:22
1
$begingroup$
It is always amazing with what you come up. Good to hear from you by the way!
$endgroup$
– Henrik Schumacher
Dec 31 '18 at 10:56
add a comment |
$begingroup$
In fact, there is a closed-form expression for the derivatives of an inverse function, if one is willing to use the (partial) Bell polynomials (implemented in Mathematica as BellY). This is related to the Lagrangian inversion formula (and see also the discussion in Charalambides).
inverseD[ff_, x_] := inverseD[ff, {x, 1}]
inverseD[ff_, {x_, k_Integer?NonNegative}] := With[{f = Function[x, ff]},
If[k == 0, Return[InverseFunction[f][x]]];
If[k == 1, Return[1/f'[InverseFunction[f][x]]]];
With[{ifun = InverseFunction[f][x]},
BellY[Table[{(k + j - 2)!, -Derivative[j][f][ifun]/f'[ifun]/j},
{j, 2, k}]]/((k - 1)! f'[ifun]^k)]]
For example:
(inverseD[g[t], {t, 4}] == D[InverseFunction[g][t], {t, 4}] /. Function[t, g[t]] -> g) // Simplify
answered Dec 31 '18 at 10:22
J. M. is away♦J. M. is away
98.9k10311467
$endgroup$
$begingroup$
(I'm still computer-less, but I had this as a draft blog post, so I thought I'd post it.)
$endgroup$
– J. M. is away♦
Dec 31 '18 at 10:22
1
$begingroup$
It is always amazing with what you come up. Good to hear from you by the way!
$endgroup$
– Henrik Schumacher
Dec 31 '18 at 10:56
add a comment |
$begingroup$
In fact, there is a closed-form expression for the derivatives of an inverse function, if one is willing to use the (partial) Bell polynomials (implemented in Mathematica as BellY). This is related to the Lagrangian inversion formula (and see also the discussion in Charalambides).
inverseD[ff_, x_] := inverseD[ff, {x, 1}]
inverseD[ff_, {x_, k_Integer?NonNegative}] := With[{f = Function[x, ff]},
If[k == 0, Return[InverseFunction[f][x]]];
If[k == 1, Return[1/f'[InverseFunction[f][x]]]];
With[{ifun = InverseFunction[f][x]},
BellY[Table[{(k + j - 2)!, -Derivative[j][f][ifun]/f'[ifun]/j},
{j, 2, k}]]/((k - 1)! f'[ifun]^k)]]
For example:
(inverseD[g[t], {t, 4}] == D[InverseFunction[g][t], {t, 4}] /. Function[t, g[t]] -> g) // Simplify
answered Dec 31 '18 at 10:22
J. M. is away♦J. M. is away
98.9k10311467
$endgroup$
In fact, there is a closed-form expression for the derivatives of an inverse function, if one is willing to use the (partial) Bell polynomials (implemented in Mathematica as BellY). This is related to the Lagrangian inversion formula (and see also the discussion in Charalambides).
inverseD[ff_, x_] := inverseD[ff, {x, 1}]
inverseD[ff_, {x_, k_Integer?NonNegative}] := With[{f = Function[x, ff]},
If[k == 0, Return[InverseFunction[f][x]]];
If[k == 1, Return[1/f'[InverseFunction[f][x]]]];
With[{ifun = InverseFunction[f][x]},
BellY[Table[{(k + j - 2)!, -Derivative[j][f][ifun]/f'[ifun]/j},
{j, 2, k}]]/((k - 1)! f'[ifun]^k)]]
For example:
(inverseD[g[t], {t, 4}] == D[InverseFunction[g][t], {t, 4}] /. Function[t, g[t]] -> g) // Simplify
answered Dec 31 '18 at 10:22
J. M. is away♦J. M. is away
98.9k10311467
answered Dec 31 '18 at 10:22
J. M. is away♦J. M. is away
98.9k10311467
answered Dec 31 '18 at 10:22
J. M. is away♦J. M. is away
98.9k10311467
answered Dec 31 '18 at 10:22
J. M. is away♦J. M. is away
98.9k10311467
98.9k10311467
$begingroup$
(I'm still computer-less, but I had this as a draft blog post, so I thought I'd post it.)
$endgroup$
– J. M. is away♦
Dec 31 '18 at 10:22
1
$begingroup$
It is always amazing with what you come up. Good to hear from you by the way!
$endgroup$
– Henrik Schumacher
Dec 31 '18 at 10:56
add a comment |
$begingroup$
(I'm still computer-less, but I had this as a draft blog post, so I thought I'd post it.)
$endgroup$
– J. M. is away♦
Dec 31 '18 at 10:22
1
$begingroup$
It is always amazing with what you come up. Good to hear from you by the way!
$endgroup$
– Henrik Schumacher
Dec 31 '18 at 10:56
$begingroup$
(I'm still computer-less, but I had this as a draft blog post, so I thought I'd post it.)
$endgroup$
– J. M. is away♦
Dec 31 '18 at 10:22
$begingroup$
(I'm still computer-less, but I had this as a draft blog post, so I thought I'd post it.)
$endgroup$
– J. M. is away♦
Dec 31 '18 at 10:22
1
1
$begingroup$
It is always amazing with what you come up. Good to hear from you by the way!
$endgroup$
– Henrik Schumacher
Dec 31 '18 at 10:56
$begingroup$
It is always amazing with what you come up. Good to hear from you by the way!
$endgroup$
– Henrik Schumacher
Dec 31 '18 at 10:56
add a comment |
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$begingroup$
Something like
Derivative[3][InverseFunction[f]][f[x]]?$endgroup$
– Szabolcs
Dec 25 '18 at 11:07
$begingroup$
@Szabolcs: I was just modifying writing this. Now I have again asked the question, thank you!
$endgroup$
– TeM
Dec 25 '18 at 11:16
$begingroup$
Sounds like a math question, not Mathematica ... (it's interesting though)
$endgroup$
– Szabolcs
Dec 26 '18 at 9:07
$begingroup$
@Szabolcs: In fact you are right. Sometimes, however, some topics are borderline, it is difficult to understand how and where to formulate them, at least for me.
$endgroup$
– TeM
Dec 26 '18 at 9:10