How to evaluate this function?
$begingroup$
I’m trying to evaluate
y[t] = InverseFunction[ NIntegrate[Sqrt[2]/( -8+Exp[y]+2Exp[-y]), {y,0,t}]]
Some of the problems I’m facing, is that i’d like to get the integral’s result in terms of t as a definite integral.
Any help to get y[0], y[1] and y’[0] ?
calculus-and-analysis numerical-integration numerics
asked Apr 21 at 6:54
S.S.S.S.
1266
$endgroup$
add a comment |
$begingroup$
I’m trying to evaluate
y[t] = InverseFunction[ NIntegrate[Sqrt[2]/( -8+Exp[y]+2Exp[-y]), {y,0,t}]]
Some of the problems I’m facing, is that i’d like to get the integral’s result in terms of t as a definite integral.
Any help to get y[0], y[1] and y’[0] ?
calculus-and-analysis numerical-integration numerics
asked Apr 21 at 6:54
S.S.S.S.
1266
$endgroup$
add a comment |
$begingroup$
I’m trying to evaluate
y[t] = InverseFunction[ NIntegrate[Sqrt[2]/( -8+Exp[y]+2Exp[-y]), {y,0,t}]]
Some of the problems I’m facing, is that i’d like to get the integral’s result in terms of t as a definite integral.
Any help to get y[0], y[1] and y’[0] ?
calculus-and-analysis numerical-integration numerics
asked Apr 21 at 6:54
S.S.S.S.
1266
$endgroup$
I’m trying to evaluate
y[t] = InverseFunction[ NIntegrate[Sqrt[2]/( -8+Exp[y]+2Exp[-y]), {y,0,t}]]
Some of the problems I’m facing, is that i’d like to get the integral’s result in terms of t as a definite integral.
Any help to get y[0], y[1] and y’[0] ?
calculus-and-analysis numerical-integration numerics
calculus-and-analysis numerical-integration numerics
asked Apr 21 at 6:54
S.S.S.S.
1266
asked Apr 21 at 6:54
S.S.S.S.
1266
edited Apr 21 at 7:10
S.S.
asked Apr 21 at 6:54
S.S.S.S.
1266
asked Apr 21 at 6:54
S.S.S.S.
1266
asked Apr 21 at 6:54
S.S.S.S.
1266
1266
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assuming the integral is not possible to evaluate symbolically, so that Roman's approach was not available, it is still possible to construct an interpolating function solution of the inverse. First, rewrite your relationship as follows:
eqn = y == Inactive[Integrate][Sqrt[2]/(-8+Exp[s]+2Exp[-s]),{s,0,t[y]}];
eqn //TeXForm
$y=int _0^{t(y)}frac{sqrt{2}}{2 e^{-s}+e^s-8}ds$
where I renamed the integration variable to avoid confusion. Then, differentiating with respect to y produces an ODE:
ode = D[eqn, y];
ode //TeXForm
$1=frac{sqrt{2} t'(y)}{2 e^{-t(y)}+e^{t(y)}-8}$
For initial conditions, clearly t[0] == 0. So, we need to solve:
sol = NDSolveValue[{ode, t[0]==0}, t, {y, -1, 1}];
Visualization:
Plot[sol[y], {y, -1, 1}]
Comparison with the exact answer (as provided by Roman):
sol[0]
sol[1]
sol'[0]
0.
-1.33756
-3.53553
vs:
roman[y_] := Log[4-Sqrt[14] Tanh[Sqrt[7] y+ArcTanh[3/Sqrt[14]]]]
roman[0]
roman[1] //N
roman'[0] //N
0
-1.33756
-3.53553
answered Apr 21 at 15:17
Carl WollCarl Woll
75.8k3100198
$endgroup$
add a comment |
$begingroup$
No need to do numerical integration, you can do this one analytically.
Start with the integral:
Integrate[Sqrt[2]/(-8 + Exp[y] + 2 Exp[-y]), {y, 0, t}]
(* ConditionalExpression[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]),
E^t == 4 || (-4 + E^t)^2 <= 0] *)
It turns out that this result is valid for $ln(4-sqrt{14})<t<ln(4+sqrt{14})$:
Plot[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]),
{t, Log[4 - Sqrt[14]], Log[4 + Sqrt[14]]}]
Invert it:
Solve[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]) == y, t]
(expression that I simplify by hand to get rid of branches:)
t[y_] = Log[4 - Sqrt[14] Tanh[Sqrt[7] y + ArcTanh[3/Sqrt[14]]]];
Plot[t[y], {y, -1, 1}]
From this you get your points of interest:
t[0]
(* 0 *)
t[1]
(* Log[4 - Sqrt[14] Tanh[Sqrt[7] + ArcTanh[3/Sqrt[14]]]] *)
t'[0]
(* -5/Sqrt[2] *)
answered Apr 21 at 8:11
RomanRoman
6,07611132
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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active
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oldest
votes
$begingroup$
Assuming the integral is not possible to evaluate symbolically, so that Roman's approach was not available, it is still possible to construct an interpolating function solution of the inverse. First, rewrite your relationship as follows:
eqn = y == Inactive[Integrate][Sqrt[2]/(-8+Exp[s]+2Exp[-s]),{s,0,t[y]}];
eqn //TeXForm
$y=int _0^{t(y)}frac{sqrt{2}}{2 e^{-s}+e^s-8}ds$
where I renamed the integration variable to avoid confusion. Then, differentiating with respect to y produces an ODE:
ode = D[eqn, y];
ode //TeXForm
$1=frac{sqrt{2} t'(y)}{2 e^{-t(y)}+e^{t(y)}-8}$
For initial conditions, clearly t[0] == 0. So, we need to solve:
sol = NDSolveValue[{ode, t[0]==0}, t, {y, -1, 1}];
Visualization:
Plot[sol[y], {y, -1, 1}]
Comparison with the exact answer (as provided by Roman):
sol[0]
sol[1]
sol'[0]
0.
-1.33756
-3.53553
vs:
roman[y_] := Log[4-Sqrt[14] Tanh[Sqrt[7] y+ArcTanh[3/Sqrt[14]]]]
roman[0]
roman[1] //N
roman'[0] //N
0
-1.33756
-3.53553
answered Apr 21 at 15:17
Carl WollCarl Woll
75.8k3100198
$endgroup$
add a comment |
$begingroup$
Assuming the integral is not possible to evaluate symbolically, so that Roman's approach was not available, it is still possible to construct an interpolating function solution of the inverse. First, rewrite your relationship as follows:
eqn = y == Inactive[Integrate][Sqrt[2]/(-8+Exp[s]+2Exp[-s]),{s,0,t[y]}];
eqn //TeXForm
$y=int _0^{t(y)}frac{sqrt{2}}{2 e^{-s}+e^s-8}ds$
where I renamed the integration variable to avoid confusion. Then, differentiating with respect to y produces an ODE:
ode = D[eqn, y];
ode //TeXForm
$1=frac{sqrt{2} t'(y)}{2 e^{-t(y)}+e^{t(y)}-8}$
For initial conditions, clearly t[0] == 0. So, we need to solve:
sol = NDSolveValue[{ode, t[0]==0}, t, {y, -1, 1}];
Visualization:
Plot[sol[y], {y, -1, 1}]
Comparison with the exact answer (as provided by Roman):
sol[0]
sol[1]
sol'[0]
0.
-1.33756
-3.53553
vs:
roman[y_] := Log[4-Sqrt[14] Tanh[Sqrt[7] y+ArcTanh[3/Sqrt[14]]]]
roman[0]
roman[1] //N
roman'[0] //N
0
-1.33756
-3.53553
answered Apr 21 at 15:17
Carl WollCarl Woll
75.8k3100198
$endgroup$
add a comment |
$begingroup$
Assuming the integral is not possible to evaluate symbolically, so that Roman's approach was not available, it is still possible to construct an interpolating function solution of the inverse. First, rewrite your relationship as follows:
eqn = y == Inactive[Integrate][Sqrt[2]/(-8+Exp[s]+2Exp[-s]),{s,0,t[y]}];
eqn //TeXForm
$y=int _0^{t(y)}frac{sqrt{2}}{2 e^{-s}+e^s-8}ds$
where I renamed the integration variable to avoid confusion. Then, differentiating with respect to y produces an ODE:
ode = D[eqn, y];
ode //TeXForm
$1=frac{sqrt{2} t'(y)}{2 e^{-t(y)}+e^{t(y)}-8}$
For initial conditions, clearly t[0] == 0. So, we need to solve:
sol = NDSolveValue[{ode, t[0]==0}, t, {y, -1, 1}];
Visualization:
Plot[sol[y], {y, -1, 1}]
Comparison with the exact answer (as provided by Roman):
sol[0]
sol[1]
sol'[0]
0.
-1.33756
-3.53553
vs:
roman[y_] := Log[4-Sqrt[14] Tanh[Sqrt[7] y+ArcTanh[3/Sqrt[14]]]]
roman[0]
roman[1] //N
roman'[0] //N
0
-1.33756
-3.53553
answered Apr 21 at 15:17
Carl WollCarl Woll
75.8k3100198
$endgroup$
Assuming the integral is not possible to evaluate symbolically, so that Roman's approach was not available, it is still possible to construct an interpolating function solution of the inverse. First, rewrite your relationship as follows:
eqn = y == Inactive[Integrate][Sqrt[2]/(-8+Exp[s]+2Exp[-s]),{s,0,t[y]}];
eqn //TeXForm
$y=int _0^{t(y)}frac{sqrt{2}}{2 e^{-s}+e^s-8}ds$
where I renamed the integration variable to avoid confusion. Then, differentiating with respect to y produces an ODE:
ode = D[eqn, y];
ode //TeXForm
$1=frac{sqrt{2} t'(y)}{2 e^{-t(y)}+e^{t(y)}-8}$
For initial conditions, clearly t[0] == 0. So, we need to solve:
sol = NDSolveValue[{ode, t[0]==0}, t, {y, -1, 1}];
Visualization:
Plot[sol[y], {y, -1, 1}]
Comparison with the exact answer (as provided by Roman):
sol[0]
sol[1]
sol'[0]
0.
-1.33756
-3.53553
vs:
roman[y_] := Log[4-Sqrt[14] Tanh[Sqrt[7] y+ArcTanh[3/Sqrt[14]]]]
roman[0]
roman[1] //N
roman'[0] //N
0
-1.33756
-3.53553
answered Apr 21 at 15:17
Carl WollCarl Woll
75.8k3100198
answered Apr 21 at 15:17
Carl WollCarl Woll
75.8k3100198
answered Apr 21 at 15:17
Carl WollCarl Woll
75.8k3100198
answered Apr 21 at 15:17
Carl WollCarl Woll
75.8k3100198
75.8k3100198
add a comment |
add a comment |
$begingroup$
No need to do numerical integration, you can do this one analytically.
Start with the integral:
Integrate[Sqrt[2]/(-8 + Exp[y] + 2 Exp[-y]), {y, 0, t}]
(* ConditionalExpression[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]),
E^t == 4 || (-4 + E^t)^2 <= 0] *)
It turns out that this result is valid for $ln(4-sqrt{14})<t<ln(4+sqrt{14})$:
Plot[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]),
{t, Log[4 - Sqrt[14]], Log[4 + Sqrt[14]]}]
Invert it:
Solve[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]) == y, t]
(expression that I simplify by hand to get rid of branches:)
t[y_] = Log[4 - Sqrt[14] Tanh[Sqrt[7] y + ArcTanh[3/Sqrt[14]]]];
Plot[t[y], {y, -1, 1}]
From this you get your points of interest:
t[0]
(* 0 *)
t[1]
(* Log[4 - Sqrt[14] Tanh[Sqrt[7] + ArcTanh[3/Sqrt[14]]]] *)
t'[0]
(* -5/Sqrt[2] *)
answered Apr 21 at 8:11
RomanRoman
6,07611132
$endgroup$
add a comment |
$begingroup$
No need to do numerical integration, you can do this one analytically.
Start with the integral:
Integrate[Sqrt[2]/(-8 + Exp[y] + 2 Exp[-y]), {y, 0, t}]
(* ConditionalExpression[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]),
E^t == 4 || (-4 + E^t)^2 <= 0] *)
It turns out that this result is valid for $ln(4-sqrt{14})<t<ln(4+sqrt{14})$:
Plot[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]),
{t, Log[4 - Sqrt[14]], Log[4 + Sqrt[14]]}]
Invert it:
Solve[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]) == y, t]
(expression that I simplify by hand to get rid of branches:)
t[y_] = Log[4 - Sqrt[14] Tanh[Sqrt[7] y + ArcTanh[3/Sqrt[14]]]];
Plot[t[y], {y, -1, 1}]
From this you get your points of interest:
t[0]
(* 0 *)
t[1]
(* Log[4 - Sqrt[14] Tanh[Sqrt[7] + ArcTanh[3/Sqrt[14]]]] *)
t'[0]
(* -5/Sqrt[2] *)
answered Apr 21 at 8:11
RomanRoman
6,07611132
$endgroup$
add a comment |
$begingroup$
No need to do numerical integration, you can do this one analytically.
Start with the integral:
Integrate[Sqrt[2]/(-8 + Exp[y] + 2 Exp[-y]), {y, 0, t}]
(* ConditionalExpression[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]),
E^t == 4 || (-4 + E^t)^2 <= 0] *)
It turns out that this result is valid for $ln(4-sqrt{14})<t<ln(4+sqrt{14})$:
Plot[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]),
{t, Log[4 - Sqrt[14]], Log[4 + Sqrt[14]]}]
Invert it:
Solve[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]) == y, t]
(expression that I simplify by hand to get rid of branches:)
t[y_] = Log[4 - Sqrt[14] Tanh[Sqrt[7] y + ArcTanh[3/Sqrt[14]]]];
Plot[t[y], {y, -1, 1}]
From this you get your points of interest:
t[0]
(* 0 *)
t[1]
(* Log[4 - Sqrt[14] Tanh[Sqrt[7] + ArcTanh[3/Sqrt[14]]]] *)
t'[0]
(* -5/Sqrt[2] *)
answered Apr 21 at 8:11
RomanRoman
6,07611132
$endgroup$
No need to do numerical integration, you can do this one analytically.
Start with the integral:
Integrate[Sqrt[2]/(-8 + Exp[y] + 2 Exp[-y]), {y, 0, t}]
(* ConditionalExpression[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]),
E^t == 4 || (-4 + E^t)^2 <= 0] *)
It turns out that this result is valid for $ln(4-sqrt{14})<t<ln(4+sqrt{14})$:
Plot[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]),
{t, Log[4 - Sqrt[14]], Log[4 + Sqrt[14]]}]
Invert it:
Solve[-((ArcTanh[3/Sqrt[14]] + ArcTanh[(-4 + E^t)/Sqrt[14]])/Sqrt[7]) == y, t]
(expression that I simplify by hand to get rid of branches:)
t[y_] = Log[4 - Sqrt[14] Tanh[Sqrt[7] y + ArcTanh[3/Sqrt[14]]]];
Plot[t[y], {y, -1, 1}]
From this you get your points of interest:
t[0]
(* 0 *)
t[1]
(* Log[4 - Sqrt[14] Tanh[Sqrt[7] + ArcTanh[3/Sqrt[14]]]] *)
t'[0]
(* -5/Sqrt[2] *)
answered Apr 21 at 8:11
RomanRoman
6,07611132
edited Apr 21 at 8:29
answered Apr 21 at 8:11
RomanRoman
6,07611132
answered Apr 21 at 8:11
RomanRoman
6,07611132
answered Apr 21 at 8:11
RomanRoman
6,07611132
6,07611132
add a comment |
add a comment |
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