Solve piecewise function

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1
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I have two piecewise function

equ1 = Piecewise[0.524324 + 0.0376478x, 0.639464 <= x <= 0.839322]
equ2 = Piecewise[-0.506432 + 1.48068x, 0.658914 <= x <= 0.77085]

Now, I am trying to solve equ1 = equ2.

Firstly I tried FindRoot:

FindRoot[equ1 == equ2, x]

But the output is x = 0. I can only get the correct answer by set search starting point 0.7. How can I direct get the answer without set starting point?

Secondly, I tried code Reduce:

Reduce[equ1 == equ2, x]

However, the error appear. The good news is Reduce do provide the correct answer for my equation. The error is:

Reduce::ratnz: Reduce was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result.

Do I have other way to solve those two piecewise function?

edited 3 hours ago

Michael E2

142k11192458

asked 5 hours ago

Yubao Deng

613

2

What's wrong with Solve[0.524324 + 0.0376478 x == -0.506432 + 1.48068 x, x]? or even FindRoot[equ1 == equ2, x, .7]?
â€“Â AccidentalFourierTransform
5 hours ago

1

Try to plot then you see where is root.Plot[equ1 - equ2, x, 1/2, 1], FindRoot[equ1 == equ2, x, 7/10].
â€“Â Mariusz Iwaniuk
5 hours ago

1

Because at later stages, I need to compara 100 equations to find out solution. It is not possible to set the starting value for each one of them.
â€“Â Yubao Deng
4 hours ago

1

Your Reduce command works for me. The message is just a warning, not an error.
â€“Â Michael E2
3 hours ago

add a commentÂ |Â

up vote
1
down vote

favorite

I have two piecewise function

equ1 = Piecewise[0.524324 + 0.0376478x, 0.639464 <= x <= 0.839322]
equ2 = Piecewise[-0.506432 + 1.48068x, 0.658914 <= x <= 0.77085]

Now, I am trying to solve equ1 = equ2.

Firstly I tried FindRoot:

FindRoot[equ1 == equ2, x]

But the output is x = 0. I can only get the correct answer by set search starting point 0.7. How can I direct get the answer without set starting point?

Secondly, I tried code Reduce:

Reduce[equ1 == equ2, x]

However, the error appear. The good news is Reduce do provide the correct answer for my equation. The error is:

Reduce::ratnz: Reduce was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result.

Do I have other way to solve those two piecewise function?

edited 3 hours ago

Michael E2

142k11192458

asked 5 hours ago

Yubao Deng

613

2

What's wrong with Solve[0.524324 + 0.0376478 x == -0.506432 + 1.48068 x, x]? or even FindRoot[equ1 == equ2, x, .7]?
â€“Â AccidentalFourierTransform
5 hours ago

1

Try to plot then you see where is root.Plot[equ1 - equ2, x, 1/2, 1], FindRoot[equ1 == equ2, x, 7/10].
â€“Â Mariusz Iwaniuk
5 hours ago

1

Because at later stages, I need to compara 100 equations to find out solution. It is not possible to set the starting value for each one of them.
â€“Â Yubao Deng
4 hours ago

1

Your Reduce command works for me. The message is just a warning, not an error.
â€“Â Michael E2
3 hours ago

add a commentÂ |Â

up vote
1
down vote

favorite

I have two piecewise function

equ1 = Piecewise[0.524324 + 0.0376478x, 0.639464 <= x <= 0.839322]
equ2 = Piecewise[-0.506432 + 1.48068x, 0.658914 <= x <= 0.77085]

Now, I am trying to solve equ1 = equ2.

Firstly I tried FindRoot:

FindRoot[equ1 == equ2, x]

But the output is x = 0. I can only get the correct answer by set search starting point 0.7. How can I direct get the answer without set starting point?

Secondly, I tried code Reduce:

Reduce[equ1 == equ2, x]

However, the error appear. The good news is Reduce do provide the correct answer for my equation. The error is:

Reduce::ratnz: Reduce was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result.

Do I have other way to solve those two piecewise function?

edited 3 hours ago

Michael E2

142k11192458

asked 5 hours ago

Yubao Deng

613

I have two piecewise function

equ1 = Piecewise[0.524324 + 0.0376478x, 0.639464 <= x <= 0.839322]
equ2 = Piecewise[-0.506432 + 1.48068x, 0.658914 <= x <= 0.77085]

Now, I am trying to solve equ1 = equ2.

Firstly I tried FindRoot:

FindRoot[equ1 == equ2, x]

But the output is x = 0. I can only get the correct answer by set search starting point 0.7. How can I direct get the answer without set starting point?

Secondly, I tried code Reduce:

Reduce[equ1 == equ2, x]

However, the error appear. The good news is Reduce do provide the correct answer for my equation. The error is:

Reduce::ratnz: Reduce was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result.

Do I have other way to solve those two piecewise function?

equation-solving error piecewise

edited 3 hours ago

Michael E2

142k11192458

asked 5 hours ago

Yubao Deng

613

edited 3 hours ago

Michael E2

142k11192458

asked 5 hours ago

Yubao Deng

613

edited 3 hours ago

Michael E2

142k11192458

edited 3 hours ago

Michael E2

142k11192458

edited 3 hours ago

Michael E2

142k11192458

asked 5 hours ago

Yubao Deng

613

asked 5 hours ago

Yubao Deng

613

asked 5 hours ago

Yubao Deng

613

2

What's wrong with Solve[0.524324 + 0.0376478 x == -0.506432 + 1.48068 x, x]? or even FindRoot[equ1 == equ2, x, .7]?
â€“Â AccidentalFourierTransform
5 hours ago

1

Try to plot then you see where is root.Plot[equ1 - equ2, x, 1/2, 1], FindRoot[equ1 == equ2, x, 7/10].
â€“Â Mariusz Iwaniuk
5 hours ago

1

Because at later stages, I need to compara 100 equations to find out solution. It is not possible to set the starting value for each one of them.
â€“Â Yubao Deng
4 hours ago

1

Your Reduce command works for me. The message is just a warning, not an error.
â€“Â Michael E2
3 hours ago

add a commentÂ |Â

2

What's wrong with Solve[0.524324 + 0.0376478 x == -0.506432 + 1.48068 x, x]? or even FindRoot[equ1 == equ2, x, .7]?
â€“Â AccidentalFourierTransform
5 hours ago

1

Try to plot then you see where is root.Plot[equ1 - equ2, x, 1/2, 1], FindRoot[equ1 == equ2, x, 7/10].
â€“Â Mariusz Iwaniuk
5 hours ago

1

Because at later stages, I need to compara 100 equations to find out solution. It is not possible to set the starting value for each one of them.
â€“Â Yubao Deng
4 hours ago

1

Your Reduce command works for me. The message is just a warning, not an error.
â€“Â Michael E2
3 hours ago

What's wrong with Solve[0.524324 + 0.0376478 x == -0.506432 + 1.48068 x, x]? or even FindRoot[equ1 == equ2, x, .7]?
â€“Â AccidentalFourierTransform
5 hours ago

Try to plot then you see where is root.Plot[equ1 - equ2, x, 1/2, 1], FindRoot[equ1 == equ2, x, 7/10].
â€“Â Mariusz Iwaniuk
5 hours ago

Because at later stages, I need to compara 100 equations to find out solution. It is not possible to set the starting value for each one of them.
â€“Â Yubao Deng
4 hours ago

Your Reduce command works for me. The message is just a warning, not an error.
â€“Â Michael E2
3 hours ago

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
1
down vote

equ1 = Rationalize[
 Piecewise[0.524324 + 0.0376478 x, 0.639464 <= x <= 0.839322]]
equ2 = Rationalize[
 Piecewise[-0.506432 + 1.48068 x, 0.658914 <= x <= 0.77085]]
Reduce[equ1 == equ2, x]
x == 0.714299 || x < 79933/125000 || x > 419661/500000

answered 4 hours ago

Alex Trounev

3,3001313

add a commentÂ |Â

up vote
1
down vote

Either:

FindRoot[equ1 == equ2, x, 0.7, 0.639464, 0.839322]
(* x -> 0.714299 *)

Or:

NSolve[equ1 == equ2 && 0.639464 <= x <= 0.839322, x]

NSolve::ratnz:.... [Unimportant warning.]

(* x -> 0.714299 *)

answered 4 hours ago

Michael E2

142k11192458

add a commentÂ |Â

up vote
1
down vote

If you have 100 equations and you want to find the unique solution to the whole set and each of your Piecewise are written in exactly the same form as the two you showed and the solution must lie in the intersection of all those inequalities then perhaps

equ1 = Piecewise[0.524324 + 0.0376478 x, 0.639464 <= x <= 0.839322];
equ2 = Piecewise[-0.506432 + 1.48068 x, 0.658914 <= x <= 0.77085];
funs = equ1, equ2;
Join[Equal @@ Map[#[[1, 1, 1]] &, funs],
 Max[Map[#[[1, 1, 2, 1]] &, funs]] < x < Min[Map[#[[1, 1, 2, 3]] &, funs]]]

will give you

0.524324 + 0.0376478 x == -0.506432 + 1.48068 x, 0.658914 < x < 0.77085

and

x /. Solve[
 Join[Equal @@ Map[#[[1, 1, 1]] &, funs],
 Max[Map[#[[1, 1, 2, 1]] &, funs]] < x < Min[Map[#[[1, 1, 2, 3]] &, funs]]]
 , x][[1]]

will give you

0.714299

You can make the list of funs longer and longer and still get the solution.

The Max part of that is extracting all your lower bounds and finding the largest one. The Min part of that is extracting all your upper bounds and finding the smallest one. The Equal part of that is extracting all your equations and setting them equal to each other. We assemble those pieces and give them to Solve.

You should check carefully to make certain that there will always be one unique solution to your set of equations.

You might click on Piecewise and then Details in the help pages and read very carefully all the information there. Because you have not given any other information about the value of your functions if x happens to be outside the bounds you have given then the value of that Piecewise function will be zero for x outside those bounds. That might surprise you.

edited 3 hours ago

answered 3 hours ago

Bill

5,15058

add a commentÂ |Â

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
1
down vote

equ1 = Rationalize[
 Piecewise[0.524324 + 0.0376478 x, 0.639464 <= x <= 0.839322]]
equ2 = Rationalize[
 Piecewise[-0.506432 + 1.48068 x, 0.658914 <= x <= 0.77085]]
Reduce[equ1 == equ2, x]
x == 0.714299 || x < 79933/125000 || x > 419661/500000

answered 4 hours ago

Alex Trounev

3,3001313

add a commentÂ |Â

up vote
1
down vote

equ1 = Rationalize[
 Piecewise[0.524324 + 0.0376478 x, 0.639464 <= x <= 0.839322]]
equ2 = Rationalize[
 Piecewise[-0.506432 + 1.48068 x, 0.658914 <= x <= 0.77085]]
Reduce[equ1 == equ2, x]
x == 0.714299 || x < 79933/125000 || x > 419661/500000

answered 4 hours ago

Alex Trounev

3,3001313

add a commentÂ |Â

up vote
1
down vote

equ1 = Rationalize[
 Piecewise[0.524324 + 0.0376478 x, 0.639464 <= x <= 0.839322]]
equ2 = Rationalize[
 Piecewise[-0.506432 + 1.48068 x, 0.658914 <= x <= 0.77085]]
Reduce[equ1 == equ2, x]
x == 0.714299 || x < 79933/125000 || x > 419661/500000

answered 4 hours ago

Alex Trounev

3,3001313

equ1 = Rationalize[
 Piecewise[0.524324 + 0.0376478 x, 0.639464 <= x <= 0.839322]]
equ2 = Rationalize[
 Piecewise[-0.506432 + 1.48068 x, 0.658914 <= x <= 0.77085]]
Reduce[equ1 == equ2, x]
x == 0.714299 || x < 79933/125000 || x > 419661/500000

answered 4 hours ago

Alex Trounev

3,3001313

answered 4 hours ago

Alex Trounev

3,3001313

answered 4 hours ago

Alex Trounev

3,3001313

answered 4 hours ago

Alex Trounev

3,3001313

add a commentÂ |Â

up vote
1
down vote

Either:

FindRoot[equ1 == equ2, x, 0.7, 0.639464, 0.839322]
(* x -> 0.714299 *)

Or:

NSolve[equ1 == equ2 && 0.639464 <= x <= 0.839322, x]

NSolve::ratnz:.... [Unimportant warning.]

(* x -> 0.714299 *)

answered 4 hours ago

Michael E2

142k11192458

add a commentÂ |Â

up vote
1
down vote

Either:

FindRoot[equ1 == equ2, x, 0.7, 0.639464, 0.839322]
(* x -> 0.714299 *)

Or:

NSolve[equ1 == equ2 && 0.639464 <= x <= 0.839322, x]

NSolve::ratnz:.... [Unimportant warning.]

(* x -> 0.714299 *)

answered 4 hours ago

Michael E2

142k11192458

add a commentÂ |Â

up vote
1
down vote

Either:

FindRoot[equ1 == equ2, x, 0.7, 0.639464, 0.839322]
(* x -> 0.714299 *)

Or:

NSolve[equ1 == equ2 && 0.639464 <= x <= 0.839322, x]

NSolve::ratnz:.... [Unimportant warning.]

(* x -> 0.714299 *)

answered 4 hours ago

Michael E2

142k11192458

Either:

FindRoot[equ1 == equ2, x, 0.7, 0.639464, 0.839322]
(* x -> 0.714299 *)

Or:

NSolve[equ1 == equ2 && 0.639464 <= x <= 0.839322, x]

NSolve::ratnz:.... [Unimportant warning.]

(* x -> 0.714299 *)

answered 4 hours ago

Michael E2

142k11192458

answered 4 hours ago

Michael E2

142k11192458

answered 4 hours ago

Michael E2

142k11192458

answered 4 hours ago

Michael E2

142k11192458

add a commentÂ |Â

up vote
1
down vote

equ1 = Piecewise[0.524324 + 0.0376478 x, 0.639464 <= x <= 0.839322];
equ2 = Piecewise[-0.506432 + 1.48068 x, 0.658914 <= x <= 0.77085];
funs = equ1, equ2;
Join[Equal @@ Map[#[[1, 1, 1]] &, funs],
 Max[Map[#[[1, 1, 2, 1]] &, funs]] < x < Min[Map[#[[1, 1, 2, 3]] &, funs]]]

will give you

0.524324 + 0.0376478 x == -0.506432 + 1.48068 x, 0.658914 < x < 0.77085

and

x /. Solve[
 Join[Equal @@ Map[#[[1, 1, 1]] &, funs],
 Max[Map[#[[1, 1, 2, 1]] &, funs]] < x < Min[Map[#[[1, 1, 2, 3]] &, funs]]]
 , x][[1]]

will give you

0.714299

You can make the list of funs longer and longer and still get the solution.

You should check carefully to make certain that there will always be one unique solution to your set of equations.

edited 3 hours ago

answered 3 hours ago

Bill

5,15058

add a commentÂ |Â

up vote
1
down vote

equ1 = Piecewise[0.524324 + 0.0376478 x, 0.639464 <= x <= 0.839322];
equ2 = Piecewise[-0.506432 + 1.48068 x, 0.658914 <= x <= 0.77085];
funs = equ1, equ2;
Join[Equal @@ Map[#[[1, 1, 1]] &, funs],
 Max[Map[#[[1, 1, 2, 1]] &, funs]] < x < Min[Map[#[[1, 1, 2, 3]] &, funs]]]

will give you

0.524324 + 0.0376478 x == -0.506432 + 1.48068 x, 0.658914 < x < 0.77085

and

x /. Solve[
 Join[Equal @@ Map[#[[1, 1, 1]] &, funs],
 Max[Map[#[[1, 1, 2, 1]] &, funs]] < x < Min[Map[#[[1, 1, 2, 3]] &, funs]]]
 , x][[1]]

will give you

0.714299

You can make the list of funs longer and longer and still get the solution.

You should check carefully to make certain that there will always be one unique solution to your set of equations.

edited 3 hours ago

answered 3 hours ago

Bill

5,15058

add a commentÂ |Â

up vote
1
down vote

equ1 = Piecewise[0.524324 + 0.0376478 x, 0.639464 <= x <= 0.839322];
equ2 = Piecewise[-0.506432 + 1.48068 x, 0.658914 <= x <= 0.77085];
funs = equ1, equ2;
Join[Equal @@ Map[#[[1, 1, 1]] &, funs],
 Max[Map[#[[1, 1, 2, 1]] &, funs]] < x < Min[Map[#[[1, 1, 2, 3]] &, funs]]]

will give you

0.524324 + 0.0376478 x == -0.506432 + 1.48068 x, 0.658914 < x < 0.77085

and

x /. Solve[
 Join[Equal @@ Map[#[[1, 1, 1]] &, funs],
 Max[Map[#[[1, 1, 2, 1]] &, funs]] < x < Min[Map[#[[1, 1, 2, 3]] &, funs]]]
 , x][[1]]

will give you

0.714299

You can make the list of funs longer and longer and still get the solution.

You should check carefully to make certain that there will always be one unique solution to your set of equations.

edited 3 hours ago

answered 3 hours ago

Bill

5,15058

equ1 = Piecewise[0.524324 + 0.0376478 x, 0.639464 <= x <= 0.839322];
equ2 = Piecewise[-0.506432 + 1.48068 x, 0.658914 <= x <= 0.77085];
funs = equ1, equ2;
Join[Equal @@ Map[#[[1, 1, 1]] &, funs],
 Max[Map[#[[1, 1, 2, 1]] &, funs]] < x < Min[Map[#[[1, 1, 2, 3]] &, funs]]]

will give you

0.524324 + 0.0376478 x == -0.506432 + 1.48068 x, 0.658914 < x < 0.77085

and

x /. Solve[
 Join[Equal @@ Map[#[[1, 1, 1]] &, funs],
 Max[Map[#[[1, 1, 2, 1]] &, funs]] < x < Min[Map[#[[1, 1, 2, 3]] &, funs]]]
 , x][[1]]

will give you

0.714299

You can make the list of funs longer and longer and still get the solution.

You should check carefully to make certain that there will always be one unique solution to your set of equations.

edited 3 hours ago

answered 3 hours ago

Bill

5,15058

edited 3 hours ago

answered 3 hours ago

Bill

5,15058

answered 3 hours ago

Bill

5,15058

answered 3 hours ago

Bill

5,15058

add a commentÂ |Â

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