- #1

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solve in R

[tex]x+y+z=2[/tex]

[tex]2^{x+y^2}+2^{y+z^2}+2^{z+x^2}=6\sqrt[9]{2}[/tex]

[tex]x+y+z=2[/tex]

[tex]2^{x+y^2}+2^{y+z^2}+2^{z+x^2}=6\sqrt[9]{2}[/tex]

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- #1

- 18

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solve in R

[tex]x+y+z=2[/tex]

[tex]2^{x+y^2}+2^{y+z^2}+2^{z+x^2}=6\sqrt[9]{2}[/tex]

[tex]x+y+z=2[/tex]

[tex]2^{x+y^2}+2^{y+z^2}+2^{z+x^2}=6\sqrt[9]{2}[/tex]

- #2

Gib Z

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I don't really want to...

- #3

mathman

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Two equations with three unknowns - usually not possible.solve in R

[tex]x+y+z=2[/tex]

[tex]2^{x+y^2}+2^{y+z^2}+2^{z+x^2}=6\sqrt[9]{2}[/tex]

- #4

Ben Niehoff

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Since there are only two equations and three unknowns, one can find two of the variables in terms of the third. I'm not sure if the fact that x, y, and z are in R will come into play to eliminate any of the solutions.

- #5

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you are absolutely right in your answer but that requires a good assumption that x=y=z

Since there are only two equations and three unknowns, one can find two of the variables in terms of the third. I'm not sure if the fact that x, y, and z are in R will come into play to eliminate any of the solutions.

however later on working out the question myself i found that you assump. was right and can be prooved

here

given

x+y+z=2

2[tex]^{x+x^2}[/tex]+2[tex]^{y+y^2}[/tex] +2[tex]^{z+z^2}[/tex]=6*2[tex]^{10/9}[/tex]

a,b,c are the given terms in equation

using AM GM inequality in a,b,c

2[tex]^{10/9}[/tex]>=2[tex]^{(x+y+z+x^2+y^2+z^2)/3}[/tex]

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]<=4/3

now

x+y+z=2

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]=[tex]\lambda[/tex]

make an equation containing the solutions

let it is in variable a

a[tex]^{3}[/tex]-2*a[tex]^{2}[/tex]+((4-[tex]\lambda[/tex])/2)*a-k=0

differentiate the equation w.r.t a

3*a[tex]^{2}[/tex]-4*a+((4-[tex]\lambda[/tex])/2)

for three distinct real sol. of equation

[tex]\Delta[/tex]>0

[tex]\Delta[/tex]=16-24+6[tex]\lambda[/tex]>0

[tex]\lambda[/tex]>4/3

similiarly

[tex]\Delta[/tex]=0...........(2,1 no of sol.)

[tex]\Delta[/tex]<0............(only one possible)

therefore we get

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]>4/3 .....for 3,2,1 no of distinct sol.

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]=4/3.......for(2,1 no of sol.)

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]<4/3......(for 1 sol i.e x=y=z)

.......by second process

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]<=4/3..............by first process

now you can see that only condition that satisfy both equations are

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]=4/3

if

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]<4/3

then that would be against AM-GM

in AM GM inequality

equality is only possible when all terms are equal

a=b=c

or

x=y=z=k

3k=2

k=2/3

hope this is correct

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- #6

Defennder

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You lost me here. I don't follow what you mean by the above, where does [tex]a^3 - 2a^2 + \frac{a(4-\lambda)}{2} - k = 0[/tex] come from?make an equation containing the solutions

let it is in variable a

a[tex]^{3}[/tex]-2*a[tex]^{2}[/tex]+((4-[tex]\lambda[/tex])/2)*a-k=0

differentiate the equation w.r.t a

3*a[tex]^{2}[/tex]-4*a+((4-[tex]\lambda[/tex])/2)

for three real sol. of equation

[tex]\Delta[/tex]>=0

[tex]\Delta[/tex]=16-24+6[tex]\lambda[/tex]>=0

[tex]\lambda[/tex]>=4/3

More importantly what does this mean?

You don't appear to have used b,c at all. And I thought you used the AM-GM inequality for variables x,y,z only. Where does b,c come into play?sadhu said:a,b,c are the given terms in equation

using AM GM inequality in a,b,c

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- #7

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assume an equationYou lost me here. I don't follow what you mean by the above, where does [tex]a^3 - 2a^2 + \frac{a(4-\lambda)}{2} - k = 0[/tex] come from?

More importantly what does this mean?

You don't appear to have used b,c at all. And I thought you used the AM-GM inequality for variables x,y,z only. Where does b,c come into play?

(a-x)(a-y)(a-z)=0

now all values of x,y,z are present in the solution of the equation

put the coefficients for expansion accordindly and you get the above equation

- #8

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while using AM GM inequality

a,b,c

represent the terms of the equation ,not x,y,z

however the answer after manipulation comes out in terms of x,y,z

(2[tex]^{x+x^2}[/tex]+2[tex]^{z+z^2}[/tex]+2[tex]^{y+y^2}[/tex])/3>=2[tex]^{(x+x^2+y+y^2+z+z^2)/3}[/tex]

substitute the x+y+z=2 in GM & equation inAM term

and solve indices equation

you will get the equation you want

a,b,c

represent the terms of the equation ,not x,y,z

however the answer after manipulation comes out in terms of x,y,z

(2[tex]^{x+x^2}[/tex]+2[tex]^{z+z^2}[/tex]+2[tex]^{y+y^2}[/tex])/3>=2[tex]^{(x+x^2+y+y^2+z+z^2)/3}[/tex]

substitute the x+y+z=2 in GM & equation inAM term

and solve indices equation

you will get the equation you want

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- #9

Defennder

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Actually I don't get that equation. I get [tex]a^3 - 2a^2 + a(xy+z(x+y)) - xyz = 0[/tex]assume an equation

(a-x)(a-y)(a-z)=0

now all values of x,y,z are present in the solution of the equation

put the coefficients for expansion accordindly and you get the above equation

Secondly, why do you differentiate the equation? How is the solution for 'a' to the differentiated equation the same as that to the original equation for 'a'?

- #10

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you got the correct equation

x+y+z=2

{(x+y+z)^2-(x^2+y^2+z^2)}/2=xy+yz+xz

substituting x+y+z=2 and

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]=[tex]\lambda[/tex]

we get

(4-lambda)/2

take xyz as any number let it be k

since it doesn't matter the solution

now you see it is given that x,y,z are real numbers

suppose that they have three , two distinct value(just suppose)

so that means that a - curve must cut the x-axis on three or two points....

as you know that if any graph cut x axis on n point then it must have (n-1) (min. no )solutions of maxima and minima

for three-2

for 2-1 .................these are minimum no. of maxima or minima sol.

for-0

actualy up till now we cannot say that whether it is 2 or1or3

but as you can see when you differentiate the equation and solve for maxima and minima

if [tex]\Delta[/tex]>0 then two solutions are there and you may get three distinct value of a,or two values or one value

but

[tex]\Delta[/tex]=0

then you may get two or one solution of a

however you know that two solution cannot be their in this case as this option is eliminated by AM -GM inequality which on equality demand x=y=z

third possibility is

[tex]\Delta[/tex]<0

that means no maxima or minima, that again mean that their can be only one solution to a or x=y=z

but the reason i did not mention it was because

it will form

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]<4/3

but it also would mean that if this is possible then x=y=z

but AM-GM contradict it because if

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]<4/3

then x!=y!=z

therefore only [tex]\Delta[/tex]=0 is satisfied....hence i did not mention it one in my earlier post to avoid confusion

sorry for that...............-:)

hope you get it

x+y+z=2

{(x+y+z)^2-(x^2+y^2+z^2)}/2=xy+yz+xz

substituting x+y+z=2 and

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]=[tex]\lambda[/tex]

we get

(4-lambda)/2

take xyz as any number let it be k

since it doesn't matter the solution

now you see it is given that x,y,z are real numbers

suppose that they have three , two distinct value(just suppose)

so that means that a - curve must cut the x-axis on three or two points....

as you know that if any graph cut x axis on n point then it must have (n-1) (min. no )solutions of maxima and minima

for three-2

for 2-1 .................these are minimum no. of maxima or minima sol.

for-0

actualy up till now we cannot say that whether it is 2 or1or3

but as you can see when you differentiate the equation and solve for maxima and minima

if [tex]\Delta[/tex]>0 then two solutions are there and you may get three distinct value of a,or two values or one value

but

[tex]\Delta[/tex]=0

then you may get two or one solution of a

however you know that two solution cannot be their in this case as this option is eliminated by AM -GM inequality which on equality demand x=y=z

third possibility is

[tex]\Delta[/tex]<0

that means no maxima or minima, that again mean that their can be only one solution to a or x=y=z

but the reason i did not mention it was because

it will form

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]<4/3

but it also would mean that if this is possible then x=y=z

but AM-GM contradict it because if

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]<4/3

then x!=y!=z

therefore only [tex]\Delta[/tex]=0 is satisfied....hence i did not mention it one in my earlier post to avoid confusion

sorry for that...............-:)

hope you get it

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- #11

Defennder

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Where can I find this theorem? I could find only the apparently trivial result that if f(a) and f(b) are of opposite sign then a root exists in the interval [a,b]. And how is it the case that if [tex]\Delta = 0 [/tex] there can be only 1 solution for x,y,z? Shouldn't it be 2 by your quote above? I can understand everything else except this point.as you know that if any graph cut x axis on n point then it must have (n-1) (min. no )solutions of maxima and minima

for three-2

for 2-1 .................these are minimum no. of maxima or minima sol.

for-0

actualy up till now we cannot say that whether it is 2 or1or3

- #12

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I don't know whether it is really some theorem or not , i only know that this is a result apparently

coming out of graphs ............only need is that the function should be differentiable at all points in domain.

however on doing some search in some books, i found that this result is called rolles theorem in many textbooks .

- #13

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when

[tex]\Delta[/tex]=0

there can be two or one solution , as we get from the graphs

but it means that[tex]\lambda[/tex]=0

i.e

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]=0

when we see the AM-GM inequality

if

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]=0

then x=y=z,and only this condition

hence (x,y,z) they cant have 2 different values distributed amongst them

hence there cant be two solutions to the equation in a, as this is going against the deduction of AM-GM

hope you get it

also i think that someone will come up with a better solution than this

[tex]\Delta[/tex]=0

there can be two or one solution , as we get from the graphs

but it means that[tex]\lambda[/tex]=0

i.e

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]=0

when we see the AM-GM inequality

if

x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]=0

then x=y=z,and only this condition

hence (x,y,z) they cant have 2 different values distributed amongst them

hence there cant be two solutions to the equation in a, as this is going against the deduction of AM-GM

hope you get it

also i think that someone will come up with a better solution than this

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- #14

Defennder

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- #15

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you are right , the observation talks about minimum no of solution needed

when 1,

we get by formula 1-1=0

hence this is minimum we require for 1 solution ,i.e it is always possible

example

consider a function having

2 max,min

but all are far below or above the x-axis

you can see that only one solution is coming out ,however

now you can see that[tex]\Delta[/tex]>0

however in this case we get one solution by [tex]\Delta[/tex]=0

as other solutions are negated by disparity between two inequalities obtained by two different processes.

when 1,

we get by formula 1-1=0

hence this is minimum we require for 1 solution ,i.e it is always possible

example

consider a function having

2 max,min

but all are far below or above the x-axis

you can see that only one solution is coming out ,however

now you can see that[tex]\Delta[/tex]>0

however in this case we get one solution by [tex]\Delta[/tex]=0

as other solutions are negated by disparity between two inequalities obtained by two different processes.

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- #16

Defennder

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- #17

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that is again one very obvious conclusion , you can make from just seeing the graphs, and just thinking a bit.

well i think that one can still use the results which are very obvious even without going into a rigorous mathematical proof , which might be far more difficult than the result itself.........

this result is very obvious once you know rolles theorem

as if, there are 3 solution then there are two intervals and each interval correspond

to the solution of derivative=0..............by rolles theorem

thus max,min are 2 corresponding to two intervals

hence the deduction is proved

well i think that one can still use the results which are very obvious even without going into a rigorous mathematical proof , which might be far more difficult than the result itself.........

this result is very obvious once you know rolles theorem

as if, there are 3 solution then there are two intervals and each interval correspond

to the solution of derivative=0..............by rolles theorem

thus max,min are 2 corresponding to two intervals

hence the deduction is proved

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- #18

Defennder

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Hm ok thanks

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