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pacman2003

A tuff question...

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pacman2003    122
Hi, I m reading a text-book on differential calculus and i have a doubt regarding a solved example given in the book. The question given is: A function R->R (from R to R) where R is the set of all real numbers is defined by: f(x) = {A(x^2) + 6x - 8} / {A + 6x - 8(x^2)} Find integral values of ''A'' for which the function ''f'' is onto. The answer for A turns out to be: A >= 2 and A <= 14 My doubt: If A is substituted in f(x) as 2, then f(x) is not defined for x = 1 and x = -1/4, in which case f(x) ceases to be a function (because f(x) is not defined for all the elements of the domain viz. set of all real numbers). Can anybody explain this? Would typing in the solution given in the book help? Thank you!

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Guest Anonymous Poster   
Guest Anonymous Poster
Obviously f(x) isn''t defined for all x in R, but it is still ''onto'' since for any y in R there exist some x such that y=f(x).

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pacman2003    122
Well, the book defines a function as "a relation from one set (called domain) to another (called co-domain) such that EACH element of the domain is related to exactly one element of the co-domain".

Any ideas?

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VolkerG    151
The only case I can see where f(x) is not a function from R to R is when the denominator has one or more roots. Thus find the A''s so that no real root exists (the discriminant is less than 0). But I can''t get the values you gave us.

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quote:
Original post by pacman2003
Well, the book defines a function as "a relation from one set (called domain) to another (called co-domain) such that EACH element of the domain is related to exactly one element of the co-domain".

True, but the domain need not be the entire set of real numbers. It could be any subset, like in this case. The same is true for the co-domain (or "range" as I thought it was called in english).

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grhodes_at_work    1385
This is dangerously close to a homework question, may actually be a homework question. The discussion so far is okay, but please keep this at a general discussion level and do not provide a derived answer.

Review the Forum FAQ for our policy on homework!

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

[edited by - grhodes_at_work on January 25, 2004 11:18:38 AM]

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Guest Anonymous Poster   
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pacmanerhaps ud be better off asking ur question on a forum specially dedicated to math. this one is for math and physics related only to game development

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furby100    102
quote:
Original post by Anonymous Poster
True, but the domain need not be the entire set of real numbers. It could be any subset, like in this case. The same is true for the co-domain (or "range" as I thought it was called in english).



The co-domain and range are not the same thing. The codomain is the set onto which you are mapping, and the range is the actual set of points which are mapped onto, e.g. for f(x) = x*x, the domain could be all the real numbers and the codomain all the real numbers, but the range would only be all the positive real numbers and zero.

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pacman2003    122
(No grhodes, this is not a homework question. I do not need the derived answer because its already given in the book - this is a SOLVED example.)

If A = 2 is substituted in F[x], we have

f[x] = (2x^2 + 6x - 8)/(-8x^2 + 6x + 2)
= (x+4)(x-1)/((4x+1)(x-1))

and it is actually improper to cancel the factor of (x-1) from the numerator and denominator, because for x = 1, we have F[1] = 0/0 which is undefined. Thus there is a gap in F at x = 1, and F is not onto, since there is no value of x such that F[x] = -1.
There is a similar gap at x = -1 when A = 14.

So I think that the given answer IS wrong in this respect: it should be 2 < A < 14 and NOT 2 <= A <= 14.

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alvaro    21263
I have to agree with the original poster that a function that is not defined for some real number is not a function R->R. You can call it a function from a subset of R to R. It is a common abuse of language to call those things "functions R->R", when technically they are not.



[edited by - Alvaro on January 27, 2004 3:15:07 PM]

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f(x)=0/0 is not undefined per se... see Hospital rule

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Timkin    864
The AP way up above gave the answer you''re looking for, although they didn''t explain it particularly clearly, which may be why it was overlooked.

A function f is onto if and only if, for all y in the range, there exists an x in the domain that maps to y through f .

Thus, to prove that f is onto , you must show that for all y satisfying x =f -1(y ), the resultant x satisfies y =f (x ). Given that there is some tedious rearrangement required for this proof, given this example, I''ll leave that up to you!

Cheers,

Timkin

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Guest Anonymous Poster   
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However, as pacman2003 says, for A=2 and y=-1, there is no x such that f(x)=y (unless you cancel out the factor (x-1)).

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Timkin    864
When A=2, x=1 cannot be considered to be part of the domain, because f(x=1) is not defined. Thus, it''s irrelevant unless you find a value for y such that f-1(y) = 1... then you have proved that f is NOT ''onto'', because that point y maps onto a point that is not a part of the domain.

It seems to me that there is a fundamental misunderstanding of what ''onto'' means in this thread and how you must deal with points that are and are not in the domain. Think of it this way...

D & C are manifolds. f maps points on D to points on C. f is 1-1 if, obviously, each point in D is mapped to exactly 1 exclusive point in C. f is ''onto'' IF all points in C map to points in D via the mapping f-1 (if its 1-1 and onto then each point in D has exactly 1 point in C that it maps to and all points on both manifolds are connected via f... a straight line in any number of dimensions is a perfect example of this). It is completely irrelevant that D may be a subset of R (the reals) in that some points in R may not exist in D (which is the case in this problem for A=2,x=1).

I hope this helps clear things up...

Timkin

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davepermen    1047
quote:
Original post by Anonymous Poster
However, as pacman2003 says, for A=2 and y=-1, there is no x such that f(x)=y (unless you cancel out the factor (x-1)).



in programming terms:

it''s not required, that, for every input, you get an output. it''s only required that every possible output has to be reached with some input.

y = sin(x) for example doesn''t. there isn''t any x that returns y = 2..

y = tan(x) for example does. it returns values from -inf to +inf. it _does_not_ return a value as .. pi/2 ? (or pi:D hey, it''s a long time), etc..


hope that helps.

if f is onto, it only has to be able to return all values in the domain. the input domain doesn''t mather.




If that''s not the help you''re after then you''re going to have to explain the problem better than what you have. - joanusdmentia

davepermen.net

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quote:
Original post by Timkin
When A=2, x=1 cannot be considered to be part of the domain, because f(x=1) is not defined. Thus, it''s irrelevant unless you find a value for y such that f<sup>-1</sup>(y) = 1... then you have proved that f is NOT ''onto'', because that point y maps onto a point that is not a part of the domain.[/qoute]

That sounds weird. Can you give an example of a function that is not ''onto'' in that sense? As I understand it being ''onto'' is the same as being a surjection. Would you agree with that?


quote:
Original post by davepermen
it''s not required, that, for every input, you get an output. it''s only required that every possible output has to be reached with some input.


What do you mean by "every possible output"? If you mean the output you can actually get from the function (its range), then what you say is tautologically true for every function. If you by "every possible output" mean all of the real numbers then it is not true for y=-1 above.

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alvaro    21263
I''m sorry to contradict most people in this thread, but an object that is not a function cannot be an onto function. A function f from X to Y is, by definition, a set of pairs (x,y), x in X, y in Y such that every element in X appears in one and only one pair. Then we introduce convinient notation, like f(x) to denote the element of Y that appears in the pair in which x appears.

If f(x) is not defined for some real value x, that means that f is not a function from R to R, so it can''t possibly be an onto function from R to R.

We can change the problem a little bit and say that a function f:A -> R, where A is a subset of R, is given by some formula for all real numbers in which the formula makes sense.

The problem in the book was just poorly worded.

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Timkin    864
quote:
Original post by alvaro
I'm sorry to contradict most people in this thread, but an object that is not a function cannot be an onto function.


I don't believe anyone is talking about non-functions being onto... we're discussing functions being onto... if you want to raise the point that the original question, as posed, does not present a function, that's fine... but don't accuse people of being wrong without citing specifics please.

quote:
Original post by Anonymous Poster
As I understand it being 'onto' is the same as being a surjection. Would you agree with that?



Yes, the words are used interchangeably.

quote:
Original post by Anonymous Poster
That sounds weird. Can you give an example of a function that is not 'onto' in that sense?



It sounds weird probably because it's not the typical way in which something is described as being 'onto' (or not 'onto'). I was trying to point out that if such an unusual situation arose, you would know that the function was not onto. Typically, one would say that a function is not 'onto' if there exists values 'a' in the domain and 'b' in the range for which f(a)=b and for which f-1(b)!=a (I'm using != to mean not equal to). A function would also not be onto if there were values a and b such that f-1(b)=a and 'b' was an element of the range but 'a' was not an element of the domain. Of course, f wouldn't be a function in this case either, so I probably chose a bad way to explain what I meant. My apologies if it caused confusion. What I was trying to get at was that we can ignore points that are not in the domain when we are considering whether a mapping (function) is 'onto'.

Cheers,

Timkin
Sure can...



[edited by - Timkin on January 28, 2004 7:55:54 PM]

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quote:
Original post by Timkin
Typically, one would say that a function is not ''onto'' if there exists values ''a'' in the domain and ''b'' in the range for which f(a)=b and for which f<sup>-1</sup>(b)!=a (I''m using != to mean not equal to).

That still sounds weird to me. If there exists an inverse for f(x), I can''t see how that condition could ever be met, as it seems to contradict the very definition of an inverse function. And if an inverse doesn''t exist (as for most functions), the condition would not even be applicable. Could you give an example of a function that is not ''onto'' according to that condition?

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pacman2003    122
Well, briefly heres what I have been able to gather about relations and functions. Please do correct me if I go wrong somewhere.

A mathematical rule associating i.e. relating the elements of one set (lets call this set A) to the elements of another set (lets call this set B) is called a relation from set A to set B.

The relation may or may not relate all the elements of set A to elements of set B. Also every element in B may or may not be such that an element of A is related to it.

many-many relation:
If there is at least one element in A which is related to more than one elements in set B AND there is at least one element in B such that more than one elements in set A relate to it, then such a relation is called a ''many-many'' relation.

one-many relation:
If there is at least one element in A which is related to more than one elements in set B AND there is no element in B such that more than one elements in set A relate to it, then such a relation is called a ''one-many'' relation.

many-one relation:
If there is at least one element in B such that more than one elements in set A relate to it AND there no element in A such that more than one elements in set B are related to it, then such a relation is called a ''many-one'' relation.

one-one relation:
If there is no element in A which is related to more than one elements in set B AND there is no element in B such that more than one elements in set A relate to it, then such a relation is called a ''one-one'' relation.


A function (generally denoted by f(x)) is a special case of a relation which relates EVERY element of set A to not more than one element of set B. However every element of B need not be such that some element of A is related to it. Set A is called the domain of the function and set B is called the codomain.
Thus many-one and one-one relations MAY be (not necessarily are) functions but many-many and one-many relations can never be functions.
Those elements of the codomain, to which some element of the domain relate to, form a set called the range of the function.
A function for which range = codomain is called an onto function.
An inverse function exists only for a one-one onto function.

(I hope somebody does read all this )

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grhodes_at_work    1385
quote:
Original post by pacman2003
(No grhodes, this is not a homework question. I do not need the derived answer because its already given in the book - this is a SOLVED example.)


It wasn''t the fact that the final answer existed or not that caused me to make that comment. What I meant by my comment was that I didn''t want to see anyone show their derivation of the answer. Many times teachers will assign home work students asking them to find out how to reach the answer in the back---and my concern was that you were asking folks here to figure that derivation out for you.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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Timkin    864
quote:
Original post by Anonymous Poster
That still sounds weird to me. If there exists an inverse for f(x), I can't see how that condition could ever be met, as it seems to contradict the very definition of an inverse function.


Okay, in Euclidean space it would be hard to find such a function if we just thought about typical geometry. One obvious function that comes to mind though: y=Ez(p(z|x)) for stochastic variables x and z. y is not stochastic. Ez is the expectation taken w.r.t z. For nonlinear functions p(), the inverse of this function does not map y back to the original value(s) of x.

I hope this helps to highlight what I was saying above.

Cheers,

Timkin

[edited by - Timkin on January 29, 2004 7:18:43 PM]

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quote:
Original post by Timkin
I hope this helps to highlight what I was saying above.

Not really. In fact, it just sounds even weirder! I can''t see how bringing stochastic variables and nonlinear probability functions into the picture would make an inverse function not being an inverse function. I would be thrilled though if you would care to explain things a bit further, especially your definition of an inverse function.

Anyway, you say above that this is the ''typical'' way to describe a function not being ''onto'', yet I have not seen anything quite like it in any of my books or searching the internet. Rather, the typical way to put it seems to be that a function is not ''onto'' if the range differs from the codomain. To clarify your position, would you say that your definition is compatible with this view, and would you agree that in pacman2003''s example, f(x) is not ''onto'', for A=2?

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