Solve from interpolation function
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1
$begingroup$
I have the following list and its interpolating finction x1:
point=0., 0., 0.2, 1., 0.25, 1.5, 0.3, 1., 0.5, 0.;x1 =
Interpolation[point]
I would like to obtain values of x for which y=1 and I try:
Solve[x1[x] == 1, x]
but the output is:
NSolve::ifun: Inverse functions are being used by NSolve, so some solutions
may not be found; use Reduce for complete solution information.
InterpolatingFunction[0., 0.5, <>] -> 0.3
with a unique solution x=0.3. From the plot of x1, I expect that will be two different solutions. How can I solve this?Thanks
interpolation
asked Feb 13 at 9:10
Gae PGae P
1589
$endgroup$
add a comment |
1
$begingroup$
I have the following list and its interpolating finction x1:
point=0., 0., 0.2, 1., 0.25, 1.5, 0.3, 1., 0.5, 0.;x1 =
Interpolation[point]
I would like to obtain values of x for which y=1 and I try:
Solve[x1[x] == 1, x]
but the output is:
NSolve::ifun: Inverse functions are being used by NSolve, so some solutions
may not be found; use Reduce for complete solution information.
InterpolatingFunction[0., 0.5, <>] -> 0.3
with a unique solution x=0.3. From the plot of x1, I expect that will be two different solutions. How can I solve this?Thanks
interpolation
asked Feb 13 at 9:10
Gae PGae P
1589
$endgroup$
add a comment |
1
1
1
$begingroup$
I have the following list and its interpolating finction x1:
point=0., 0., 0.2, 1., 0.25, 1.5, 0.3, 1., 0.5, 0.;x1 =
Interpolation[point]
I would like to obtain values of x for which y=1 and I try:
Solve[x1[x] == 1, x]
but the output is:
NSolve::ifun: Inverse functions are being used by NSolve, so some solutions
may not be found; use Reduce for complete solution information.
InterpolatingFunction[0., 0.5, <>] -> 0.3
with a unique solution x=0.3. From the plot of x1, I expect that will be two different solutions. How can I solve this?Thanks
interpolation
asked Feb 13 at 9:10
Gae PGae P
1589
$endgroup$
I have the following list and its interpolating finction x1:
point=0., 0., 0.2, 1., 0.25, 1.5, 0.3, 1., 0.5, 0.;x1 =
Interpolation[point]
I would like to obtain values of x for which y=1 and I try:
Solve[x1[x] == 1, x]
but the output is:
NSolve::ifun: Inverse functions are being used by NSolve, so some solutions
may not be found; use Reduce for complete solution information.
InterpolatingFunction[0., 0.5, <>] -> 0.3
with a unique solution x=0.3. From the plot of x1, I expect that will be two different solutions. How can I solve this?Thanks
interpolation
interpolation
asked Feb 13 at 9:10
Gae PGae P
1589
asked Feb 13 at 9:10
Gae PGae P
1589
asked Feb 13 at 9:10
Gae PGae P
1589
asked Feb 13 at 9:10
Gae PGae P
1589
asked Feb 13 at 9:10
Gae PGae P
1589
1589
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
1
$begingroup$
Try
FindRoot[x1[x] == 1, x, .25, 0., 0.25, Method -> "Secant"]
(*x -> 0.2*)
FindRoot[x1[x] == 1, x, .25, 0.25, 0.5, Method -> "Secant"]
(*x -> 0.3*)
answered Feb 13 at 9:40
Ulrich NeumannUlrich Neumann
9,503616
$endgroup$
add a comment |
1
$begingroup$
Use numerical solver,
NSolve[x1[x] == 1, x]
x -> 0.3
answered Feb 13 at 9:41
zhkzhk
9,70911433
$endgroup$
$begingroup$
UsuallyNDSolvewould be my first choice, but in this example I'm wondering whyNSolve[x1[x] == 1, 0 <= x <= 0.5, x]doesn't evaluate the two(!) solutions.
$endgroup$
– Ulrich Neumann
Feb 13 at 9:54
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Try
FindRoot[x1[x] == 1, x, .25, 0., 0.25, Method -> "Secant"]
(*x -> 0.2*)
FindRoot[x1[x] == 1, x, .25, 0.25, 0.5, Method -> "Secant"]
(*x -> 0.3*)
answered Feb 13 at 9:40
Ulrich NeumannUlrich Neumann
9,503616
$endgroup$
add a comment |
1
$begingroup$
Try
FindRoot[x1[x] == 1, x, .25, 0., 0.25, Method -> "Secant"]
(*x -> 0.2*)
FindRoot[x1[x] == 1, x, .25, 0.25, 0.5, Method -> "Secant"]
(*x -> 0.3*)
answered Feb 13 at 9:40
Ulrich NeumannUlrich Neumann
9,503616
$endgroup$
add a comment |
1
1
1
$begingroup$
Try
FindRoot[x1[x] == 1, x, .25, 0., 0.25, Method -> "Secant"]
(*x -> 0.2*)
FindRoot[x1[x] == 1, x, .25, 0.25, 0.5, Method -> "Secant"]
(*x -> 0.3*)
answered Feb 13 at 9:40
Ulrich NeumannUlrich Neumann
9,503616
$endgroup$
Try
FindRoot[x1[x] == 1, x, .25, 0., 0.25, Method -> "Secant"]
(*x -> 0.2*)
FindRoot[x1[x] == 1, x, .25, 0.25, 0.5, Method -> "Secant"]
(*x -> 0.3*)
answered Feb 13 at 9:40
Ulrich NeumannUlrich Neumann
9,503616
answered Feb 13 at 9:40
Ulrich NeumannUlrich Neumann
9,503616
answered Feb 13 at 9:40
Ulrich NeumannUlrich Neumann
9,503616
answered Feb 13 at 9:40
Ulrich NeumannUlrich Neumann
9,503616
9,503616
add a comment |
add a comment |
1
$begingroup$
Use numerical solver,
NSolve[x1[x] == 1, x]
x -> 0.3
answered Feb 13 at 9:41
zhkzhk
9,70911433
$endgroup$
$begingroup$
UsuallyNDSolvewould be my first choice, but in this example I'm wondering whyNSolve[x1[x] == 1, 0 <= x <= 0.5, x]doesn't evaluate the two(!) solutions.
$endgroup$
– Ulrich Neumann
Feb 13 at 9:54
add a comment |
1
$begingroup$
Use numerical solver,
NSolve[x1[x] == 1, x]
x -> 0.3
answered Feb 13 at 9:41
zhkzhk
9,70911433
$endgroup$
$begingroup$
UsuallyNDSolvewould be my first choice, but in this example I'm wondering whyNSolve[x1[x] == 1, 0 <= x <= 0.5, x]doesn't evaluate the two(!) solutions.
$endgroup$
– Ulrich Neumann
Feb 13 at 9:54
add a comment |
1
1
1
$begingroup$
Use numerical solver,
NSolve[x1[x] == 1, x]
x -> 0.3
answered Feb 13 at 9:41
zhkzhk
9,70911433
$endgroup$
Use numerical solver,
NSolve[x1[x] == 1, x]
x -> 0.3
answered Feb 13 at 9:41
zhkzhk
9,70911433
answered Feb 13 at 9:41
zhkzhk
9,70911433
answered Feb 13 at 9:41
zhkzhk
9,70911433
answered Feb 13 at 9:41
zhkzhk
9,70911433
9,70911433
$begingroup$
UsuallyNDSolvewould be my first choice, but in this example I'm wondering whyNSolve[x1[x] == 1, 0 <= x <= 0.5, x]doesn't evaluate the two(!) solutions.
$endgroup$
– Ulrich Neumann
Feb 13 at 9:54
add a comment |
$begingroup$
UsuallyNDSolvewould be my first choice, but in this example I'm wondering whyNSolve[x1[x] == 1, 0 <= x <= 0.5, x]doesn't evaluate the two(!) solutions.
$endgroup$
– Ulrich Neumann
Feb 13 at 9:54
$begingroup$
Usually
NDSolvewould be my first choice, but in this example I'm wondering why NSolve[x1[x] == 1, 0 <= x <= 0.5, x] doesn't evaluate the two(!) solutions.$endgroup$
– Ulrich Neumann
Feb 13 at 9:54
$begingroup$
Usually
NDSolvewould be my first choice, but in this example I'm wondering why NSolve[x1[x] == 1, 0 <= x <= 0.5, x] doesn't evaluate the two(!) solutions.$endgroup$
– Ulrich Neumann
Feb 13 at 9:54
add a comment |
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