best way of solving a system of equations
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My question is what is the most straight forward of solving a system equations such as these. I usually end up isolating a variable and substituting but then there are so many steps I tend to make mistakes.
How would you approach this problem?
$$x-2lambda x^3=0$$
$$y-2lambda y^3=0$$
$$x^4+y^4=16$$
linear-algebra
asked Aug 10 at 16:51
ojd
424
add a comment |Â
up vote
1
down vote
favorite
My question is what is the most straight forward of solving a system equations such as these. I usually end up isolating a variable and substituting but then there are so many steps I tend to make mistakes.
How would you approach this problem?
$$x-2lambda x^3=0$$
$$y-2lambda y^3=0$$
$$x^4+y^4=16$$
linear-algebra
asked Aug 10 at 16:51
ojd
424
Is this a lagrangian optimisation ;)? If so you know $x=0$ and $y=0$ are (separately) possible solutions, you can plug these into the constraint to get one solution set, then you can eliminate $lambda$ in the top 2 and plug into the last one possibly to get another set
– Mehness
Aug 10 at 16:54
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My question is what is the most straight forward of solving a system equations such as these. I usually end up isolating a variable and substituting but then there are so many steps I tend to make mistakes.
How would you approach this problem?
$$x-2lambda x^3=0$$
$$y-2lambda y^3=0$$
$$x^4+y^4=16$$
linear-algebra
asked Aug 10 at 16:51
ojd
424
My question is what is the most straight forward of solving a system equations such as these. I usually end up isolating a variable and substituting but then there are so many steps I tend to make mistakes.
How would you approach this problem?
$$x-2lambda x^3=0$$
$$y-2lambda y^3=0$$
$$x^4+y^4=16$$
linear-algebra
asked Aug 10 at 16:51
ojd
424
edited Aug 10 at 17:05
user582949
asked Aug 10 at 16:51
ojd
424
asked Aug 10 at 16:51
ojd
424
asked Aug 10 at 16:51
ojd
424
424
Is this a lagrangian optimisation ;)? If so you know $x=0$ and $y=0$ are (separately) possible solutions, you can plug these into the constraint to get one solution set, then you can eliminate $lambda$ in the top 2 and plug into the last one possibly to get another set
– Mehness
Aug 10 at 16:54
add a comment |Â
Is this a lagrangian optimisation ;)? If so you know $x=0$ and $y=0$ are (separately) possible solutions, you can plug these into the constraint to get one solution set, then you can eliminate $lambda$ in the top 2 and plug into the last one possibly to get another set
– Mehness
Aug 10 at 16:54
Is this a lagrangian optimisation ;)? If so you know $x=0$ and $y=0$ are (separately) possible solutions, you can plug these into the constraint to get one solution set, then you can eliminate $lambda$ in the top 2 and plug into the last one possibly to get another set
– Mehness
Aug 10 at 16:54
Is this a lagrangian optimisation ;)? If so you know $x=0$ and $y=0$ are (separately) possible solutions, you can plug these into the constraint to get one solution set, then you can eliminate $lambda$ in the top 2 and plug into the last one possibly to get another set
– Mehness
Aug 10 at 16:54
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
1.) If $x=0$ then $y = pm 2$ so $$lambda =1over 2y^2 = 1over 8$$
2.) If $xne 0$ then $x^2 =1over 2lambda$
$...$ a) If $yne 0$ then $y^2=1over 2lambda$ so $$21over 4lambda ^2 =16implies lambda = pm1over 4sqrt2 implies x=y=...$$
$...$ b) If $y=0$ then $x = pm 2$ so $$lambda =1over 2y^2 = 1over 8$$
Thank you ojd!!
– user582949
Aug 10 at 17:30
add a comment |Â
up vote
1
down vote
Hint: Write your system in thew form
$$x(1-2lambda x^2)=0$$
$$y(1-2lambda y^2)=0$$
From here we get $x=0$ or $y=0$ or
$$x^2=frac12lambda$$ or $$y^2=frac12lambda$$ so we get
$$frac14lambda^2+frac14lambda^2=16$$ and you will get $lambda$.
answered Aug 10 at 17:01
Dr. Sonnhard Graubner
67.1k32660
add a comment |Â
up vote
0
down vote
$xyne0$ because if not we would have $0=1$(Why?).
Then $x^2=y^2=dfrac12lambda$ from which we have
$$(x,y,lambda)=left(2sqrt2,space2sqrt2,spacedfracsqrt28right)$$
answered Aug 10 at 17:18
Piquito
17.4k31234
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
1.) If $x=0$ then $y = pm 2$ so $$lambda =1over 2y^2 = 1over 8$$
2.) If $xne 0$ then $x^2 =1over 2lambda$
$...$ a) If $yne 0$ then $y^2=1over 2lambda$ so $$21over 4lambda ^2 =16implies lambda = pm1over 4sqrt2 implies x=y=...$$
$...$ b) If $y=0$ then $x = pm 2$ so $$lambda =1over 2y^2 = 1over 8$$
Thank you ojd!!
– user582949
Aug 10 at 17:30
add a comment |Â
up vote
1
down vote
accepted
1.) If $x=0$ then $y = pm 2$ so $$lambda =1over 2y^2 = 1over 8$$
2.) If $xne 0$ then $x^2 =1over 2lambda$
$...$ a) If $yne 0$ then $y^2=1over 2lambda$ so $$21over 4lambda ^2 =16implies lambda = pm1over 4sqrt2 implies x=y=...$$
$...$ b) If $y=0$ then $x = pm 2$ so $$lambda =1over 2y^2 = 1over 8$$
Thank you ojd!!
– user582949
Aug 10 at 17:30
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
1.) If $x=0$ then $y = pm 2$ so $$lambda =1over 2y^2 = 1over 8$$
2.) If $xne 0$ then $x^2 =1over 2lambda$
$...$ a) If $yne 0$ then $y^2=1over 2lambda$ so $$21over 4lambda ^2 =16implies lambda = pm1over 4sqrt2 implies x=y=...$$
$...$ b) If $y=0$ then $x = pm 2$ so $$lambda =1over 2y^2 = 1over 8$$
1.) If $x=0$ then $y = pm 2$ so $$lambda =1over 2y^2 = 1over 8$$
2.) If $xne 0$ then $x^2 =1over 2lambda$
$...$ a) If $yne 0$ then $y^2=1over 2lambda$ so $$21over 4lambda ^2 =16implies lambda = pm1over 4sqrt2 implies x=y=...$$
$...$ b) If $y=0$ then $x = pm 2$ so $$lambda =1over 2y^2 = 1over 8$$
answered Aug 10 at 17:02
user582949
Thank you ojd!!
– user582949
Aug 10 at 17:30
add a comment |Â
Thank you ojd!!
– user582949
Aug 10 at 17:30
Thank you ojd!!
– user582949
Aug 10 at 17:30
Thank you ojd!!
– user582949
Aug 10 at 17:30
add a comment |Â
up vote
1
down vote
Hint: Write your system in thew form
$$x(1-2lambda x^2)=0$$
$$y(1-2lambda y^2)=0$$
From here we get $x=0$ or $y=0$ or
$$x^2=frac12lambda$$ or $$y^2=frac12lambda$$ so we get
$$frac14lambda^2+frac14lambda^2=16$$ and you will get $lambda$.
answered Aug 10 at 17:01
Dr. Sonnhard Graubner
67.1k32660
add a comment |Â
up vote
1
down vote
Hint: Write your system in thew form
$$x(1-2lambda x^2)=0$$
$$y(1-2lambda y^2)=0$$
From here we get $x=0$ or $y=0$ or
$$x^2=frac12lambda$$ or $$y^2=frac12lambda$$ so we get
$$frac14lambda^2+frac14lambda^2=16$$ and you will get $lambda$.
answered Aug 10 at 17:01
Dr. Sonnhard Graubner
67.1k32660
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: Write your system in thew form
$$x(1-2lambda x^2)=0$$
$$y(1-2lambda y^2)=0$$
From here we get $x=0$ or $y=0$ or
$$x^2=frac12lambda$$ or $$y^2=frac12lambda$$ so we get
$$frac14lambda^2+frac14lambda^2=16$$ and you will get $lambda$.
answered Aug 10 at 17:01
Dr. Sonnhard Graubner
67.1k32660
Hint: Write your system in thew form
$$x(1-2lambda x^2)=0$$
$$y(1-2lambda y^2)=0$$
From here we get $x=0$ or $y=0$ or
$$x^2=frac12lambda$$ or $$y^2=frac12lambda$$ so we get
$$frac14lambda^2+frac14lambda^2=16$$ and you will get $lambda$.
answered Aug 10 at 17:01
Dr. Sonnhard Graubner
67.1k32660
answered Aug 10 at 17:01
Dr. Sonnhard Graubner
67.1k32660
answered Aug 10 at 17:01
Dr. Sonnhard Graubner
67.1k32660
answered Aug 10 at 17:01
Dr. Sonnhard Graubner
67.1k32660
67.1k32660
add a comment |Â
add a comment |Â
up vote
0
down vote
$xyne0$ because if not we would have $0=1$(Why?).
Then $x^2=y^2=dfrac12lambda$ from which we have
$$(x,y,lambda)=left(2sqrt2,space2sqrt2,spacedfracsqrt28right)$$
answered Aug 10 at 17:18
Piquito
17.4k31234
add a comment |Â
up vote
0
down vote
$xyne0$ because if not we would have $0=1$(Why?).
Then $x^2=y^2=dfrac12lambda$ from which we have
$$(x,y,lambda)=left(2sqrt2,space2sqrt2,spacedfracsqrt28right)$$
answered Aug 10 at 17:18
Piquito
17.4k31234
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$xyne0$ because if not we would have $0=1$(Why?).
Then $x^2=y^2=dfrac12lambda$ from which we have
$$(x,y,lambda)=left(2sqrt2,space2sqrt2,spacedfracsqrt28right)$$
answered Aug 10 at 17:18
Piquito
17.4k31234
$xyne0$ because if not we would have $0=1$(Why?).
Then $x^2=y^2=dfrac12lambda$ from which we have
$$(x,y,lambda)=left(2sqrt2,space2sqrt2,spacedfracsqrt28right)$$
answered Aug 10 at 17:18
Piquito
17.4k31234
answered Aug 10 at 17:18
Piquito
17.4k31234
answered Aug 10 at 17:18
Piquito
17.4k31234
answered Aug 10 at 17:18
Piquito
17.4k31234
17.4k31234
add a comment |Â
add a comment |Â
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Is this a lagrangian optimisation ;)? If so you know $x=0$ and $y=0$ are (separately) possible solutions, you can plug these into the constraint to get one solution set, then you can eliminate $lambda$ in the top 2 and plug into the last one possibly to get another set
– Mehness
Aug 10 at 16:54