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Abstract algebra functions.

by mtayab1994
Tags: abstract, algebra, functions
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mtayab1994
#1
Apr30-12, 03:20 PM
P: 584
1. The problem statement, all variables and given/known data

f is a quadratic function from the second degree and [tex]f(a)=bc;f(b)=ac;f(c)=ab[/tex]

2. Relevant equations

Calculate : [tex]f(a+b+c)[/tex]


3. The attempt at a solution

Can we say that [tex]f(a+b+c)=f(a)+f(b)+f(c)[/tex] and the go on from there plugging in the values of each one are do i have to do something else?
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Mark44
#2
Apr30-12, 03:44 PM
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Quote Quote by mtayab1994 View Post
1. The problem statement, all variables and given/known data

f is a quadratic function from the second degree and [tex]f(a)=bc;f(b)=ac;f(c)=ab[/tex]

2. Relevant equations

Calculate : [tex]f(a+b+c)[/tex]


3. The attempt at a solution

Can we say that [tex]f(a+b+c)=f(a)+f(b)+f(c)[/tex] and the go on from there plugging in the values of each one are do i have to do something else?
No, you can't do that. You haven't used the fact that f is a quadratic function. You know what that means, right?
mtayab1994
#3
Apr30-12, 07:01 PM
P: 584
Quote Quote by Mark44 View Post
No, you can't do that. You haven't used the fact that f is a quadratic function. You know what that means, right?
What do you mean a quadratic function to the second degree is written in the form ax^2+bx+c that's correct right?

Mark44
#4
Apr30-12, 07:11 PM
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Abstract algebra functions.

Quote Quote by mtayab1994 View Post
What do you mean a quadratic function to the second degree is written in the form ax^2+bx+c that's correct right?
Yes. Note that "quadratic function to the second degree" is redundant. Quadratic means "squared".

So f(x) = ax2 + bx + c.
SammyS
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Apr30-12, 07:43 PM
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Quote Quote by Mark44 View Post
Yes. Note that "quadratic function to the second degree" is redundant. Quadratic means "squared".

So f(x) = ax2 + bx + c.
Of course, for your function, you will want to different variable names for your coefficients, perhaps:
[itex]\displaystyle f(x)=Dx^2+Ex+G[/itex]
Are you sure you aren't working with a more limited form of a quadratic? ... It's possible that is the case due to the relationships that were stated.
Mark44
#6
Apr30-12, 08:05 PM
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Quote Quote by SammyS View Post
Of course, for your function, you will want to different variable names for your coefficients, perhaps:
[itex]\displaystyle f(x)=Dx^2+Ex+G[/itex]
That ambiguity occurred to me, too, but I'm leaning toward the view that a, b, and c in the problem description are the same as the coefficients of the terms in the quadratic. I could be wrong, though.
Quote Quote by SammyS View Post

Are you sure you aren't working with a more limited form of a quadratic? ... It's possible that is the case due to the relationships that were stated.
mtayab1994
#7
May1-12, 04:14 AM
P: 584
Quote Quote by SammyS View Post
Of course, for your function, you will want to different variable names for your coefficients, perhaps:
[itex]\displaystyle f(x)=Dx^2+Ex+G[/itex]
Are you sure you aren't working with a more limited form of a quadratic? ... It's possible that is the case due to the relationships that were stated.
Yes i could work with other variables too say alpha β or gamma. And the a,b,and c have nothing to do with the form of the quadratic equation itself.
SammyS
#8
May1-12, 09:39 AM
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Quote Quote by mtayab1994 View Post
Yes i could work with other variables too say alpha β or gamma. And the a,b,and c have nothing to do with the form of the quadratic equation itself.
Letting [itex]\displaystyle f(x)=Dx^2+Ex+G[/itex], you might look at [itex]\displaystyle f(a)-f(b)[/itex] for instance.
[itex]\displaystyle f(a)-f(b)=bc-ac[/itex]
which becomes
[itex]\displaystyle Da^2+Ea+G-Db^2-Eb-G=c(b-a)[/itex]
A little algebra will give a relationship between the constants D & E and the quantities a+b and c.

That will eliminate the constant, G.

BTW: What course is this for ?
mtayab1994
#9
May1-12, 02:29 PM
P: 584
Quote Quote by SammyS View Post
Letting [itex]\displaystyle f(x)=Dx^2+Ex+G[/itex], you might look at [itex]\displaystyle f(a)-f(b)[/itex] for instance.
[itex]\displaystyle f(a)-f(b)=bc-ac[/itex]
which becomes
[itex]\displaystyle Da^2+Ea+G-Db^2-Eb-G=c(b-a)[/itex]
A little algebra will give a relationship between the constants D & E and the quantities a+b and c.

That will eliminate the constant, G.

BTW: What course is this for ?
I'm a junior in high school in Morocco, and this is a problem our teacher gave us and told us think about it.
mtayab1994
#10
May1-12, 02:41 PM
P: 584
Ok so this is what i got:

[tex]f(a+b+c)=Da^{2}+Ea+G+Db^{2}+Db+G+Dc^{2}+Dc+G[/tex] then you factor it out and you get:

[tex]D(a^{2}+b^{2}+c^{2})+E(a+b+c)+3G=c(a+b)+ab[/tex]

Now can we use substitution for a+b+c or what from here on?
SammyS
#11
May1-12, 11:38 PM
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Quote Quote by mtayab1994 View Post
Ok so this is what i got:

[tex]f(a+b+c)=Da^{2}+Ea+G+Db^{2}+Db+G+Dc^{2}+Dc+G[/tex] then you factor it out and you get:

[tex]D(a^{2}+b^{2}+c^{2})+E(a+b+c)+3G=c(a+b)+ab[/tex]

Now can we use substitution for a+b+c or what from here on?
No. You're still treating this as if [itex]\displaystyle f(a+b+c)=f(a)+f(b)+f(c)\,,[/itex] which is definitely not the case.

[itex]\displaystyle f(a+b+c)=D(a+b+c)^2+E(a+b+c)+G[/itex]

and of course,
[itex]\displaystyle (a+b+c)^2=a^2+2ab+b^2+2bc+c^2+2ac\ .[/itex]

Etc.
mtayab1994
#12
May2-12, 05:47 AM
P: 584
Quote Quote by SammyS View Post
No. You're still treating this as if [itex]\displaystyle f(a+b+c)=f(a)+f(b)+f(c)\,,[/itex] which is definitely not the case.

[itex]\displaystyle f(a+b+c)=D(a+b+c)^2+E(a+b+c)+G[/itex]

and of course,
[itex]\displaystyle (a+b+c)^2=a^2+2ab+b^2+2bc+c^2+2ac\ .[/itex]

Etc.
Yea i solved it . Thank you for your help.
SammyS
#13
May2-12, 09:54 AM
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Quote Quote by mtayab1994 View Post
Yea i solved it . Thank you for your help.
What was your result ?


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