On the n-th derivative of the inverse function
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 2 days ago
TeM
1,886620
add a comment |
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 2 days ago
TeM
1,886620
Something likeDerivative[3][InverseFunction[f]][f[x]]?
– Szabolcs
2 days ago
@Szabolcs: I was just modifying writing this. Now I have again asked the question, thank you!
– TeM
2 days ago
Sounds like a math question, not Mathematica ... (it's interesting though)
– Szabolcs
yesterday
@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.
– TeM
yesterday
add a comment |
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 2 days ago
TeM
1,886620
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 2 days ago
TeM
1,886620
asked 2 days ago
TeM
1,886620
edited yesterday
asked 2 days ago
TeM
1,886620
asked 2 days ago
TeM
1,886620
asked 2 days ago
TeM
1,886620
1,886620
Something likeDerivative[3][InverseFunction[f]][f[x]]?
– Szabolcs
2 days ago
@Szabolcs: I was just modifying writing this. Now I have again asked the question, thank you!
– TeM
2 days ago
Sounds like a math question, not Mathematica ... (it's interesting though)
– Szabolcs
yesterday
@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.
– TeM
yesterday
add a comment |
Something likeDerivative[3][InverseFunction[f]][f[x]]?
– Szabolcs
2 days ago
@Szabolcs: I was just modifying writing this. Now I have again asked the question, thank you!
– TeM
2 days ago
Sounds like a math question, not Mathematica ... (it's interesting though)
– Szabolcs
yesterday
@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.
– TeM
yesterday
Something like
Derivative[3][InverseFunction[f]][f[x]]?– Szabolcs
2 days ago
Something like
Derivative[3][InverseFunction[f]][f[x]]?– Szabolcs
2 days ago
@Szabolcs: I was just modifying writing this. Now I have again asked the question, thank you!
– TeM
2 days ago
@Szabolcs: I was just modifying writing this. Now I have again asked the question, thank you!
– TeM
2 days ago
Sounds like a math question, not Mathematica ... (it's interesting though)
– Szabolcs
yesterday
Sounds like a math question, not Mathematica ... (it's interesting though)
– Szabolcs
yesterday
@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.
– TeM
yesterday
@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.
– TeM
yesterday
add a comment |
1 Answer
1
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oldest
votes
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 2 days ago
Henrik Schumacher
48.6k467137
Ok, get it, thank you!
– TeM
2 days ago
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 2 days ago
Henrik Schumacher
48.6k467137
Ok, get it, thank you!
– TeM
2 days ago
add a comment |
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 2 days ago
Henrik Schumacher
48.6k467137
Ok, get it, thank you!
– TeM
2 days ago
add a comment |
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 2 days ago
Henrik Schumacher
48.6k467137
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 2 days ago
Henrik Schumacher
48.6k467137
edited 2 days ago
answered 2 days ago
Henrik Schumacher
48.6k467137
answered 2 days ago
Henrik Schumacher
48.6k467137
answered 2 days ago
Henrik Schumacher
48.6k467137
48.6k467137
Ok, get it, thank you!
– TeM
2 days ago
add a comment |
Ok, get it, thank you!
– TeM
2 days ago
Ok, get it, thank you!
– TeM
2 days ago
Ok, get it, thank you!
– TeM
2 days ago
add a comment |
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Something like
Derivative[3][InverseFunction[f]][f[x]]?– Szabolcs
2 days ago
@Szabolcs: I was just modifying writing this. Now I have again asked the question, thank you!
– TeM
2 days ago
Sounds like a math question, not Mathematica ... (it's interesting though)
– Szabolcs
yesterday
@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.
– TeM
yesterday