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### Topic: inverse of a function  (Read 16435 times)

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#### xiankai

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« on: July 27, 2006, 08:47:59 AM »
as we all know, for a function to have an inverse, it must be one-one. now, i am wondering if it applies for the inverse too.

for example,

g(x) = x2 - 1

g-1(x) = ± ? x - 1

as can be seen, the inverse is not one-one. therefore the function cannot be mapped back. yet my  calculations show the inverse to have either a positive or negative sign.

how can i remedy this?

... also, what is the relationship between a function and its inverse? i am thinking they are the reflection of each other in the line y = x, but can someone confirm this?
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#### pantone159

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« Reply #1 on: July 27, 2006, 11:26:13 AM »
as can be seen, the inverse is not one-one. therefore the function cannot be mapped back. yet my  calculations show the inverse to have either a positive or negative sign.

how can i remedy this?

I think the usual 'trick' is to (arbitrarily) choose one branch of the inverse as the 'principal' one, and this branch is then one-to-one.  Without something like that, inverses of trig functions would not make much sense, for example.

... also, what is the relationship between a function and its inverse? i am thinking they are the reflection of each other in the line y = x, but can someone confirm this?

Yes, I think that is correct.  Reflecting across the line y=x is the same as swapping x and y.
Your function f defines a set of points x,f(x) and when you swap you then have the set of points f(x),x and that is the set that the inverse defines.

#### xiankai

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« Reply #2 on: July 28, 2006, 08:10:50 AM »
I think the usual 'trick' is to (arbitrarily) choose one branch of the inverse as the 'principal' one, and this branch is then one-to-one.  Without something like that, inverses of trig functions would not make much sense, for example.

hmm... it sounds like i need to define a domain/range for it, is it ok?
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#### pantone159

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« Reply #3 on: July 28, 2006, 01:57:53 PM »
That sounds right.

#### xiankai

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« Reply #4 on: July 28, 2006, 10:09:58 PM »
very much thanks, i'll serve you a dish of my speciality, scooby a la snack! one learns best by teaching

#### FeLiXe

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« Reply #5 on: August 02, 2006, 09:30:44 AM »
if you are still wondering about inverse functions, that's what I heard in maths:

a function must be bijective that means injective (one-to-one) and surjective (onto) to have an inverse function. the inverse function is then also bijective

x^2-1 is not one-to-one because y=0 is reached from x=1 and x=-1

it is not onto because y=-2 is not reached at all
Math and alcohol don't mix, so... please, don't drink and derive!

#### xiankai

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« Reply #6 on: August 03, 2006, 09:35:15 AM »
from what i can tell, surjective refers to all the elements in the range corresponding to one or more unique elements in the domain, am i wrong?

A --> 1
B --> 2
C --> 2
C --> 3
(surjective form)

because if so, i find it confusing that since injective refers to each element in the domain corresponding to one unique element in the range.

A --> 1
B --> 2
C --> 3
(injective form)

thus injectivity can be seen as a more specific subset of surjectivity, thus making surjectivity redundant one learns best by teaching

#### Yggdrasil

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« Reply #7 on: August 03, 2006, 03:47:46 PM »
Injectivity and surjectivity are only redundant if you're talking about functions whose domain has the same dimension/cardinality as its co-domain.  For example, most functions from one variable calculus map real number to real numbers.  An example of a function which maps to a set of a different cardinality is the projection function which, for example, maps elements of a two-dimensional space into a one dimensional space.  In fact, projections are good examples of functions which are surjective, but not injective.   For example, the projection onto the x-axis:

f: R2 -> R | (x,y) -> x

is surjective, but it is not injective.  An example of a function which is injective but not surjective would be a map from a one-dimensional space to a two-dimensional space:

g: R -> R2  | x -> (x,0)
« Last Edit: August 03, 2006, 04:04:00 PM by Yggdrasil »

#### xiankai

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« Reply #8 on: August 04, 2006, 07:51:17 AM »
i dont know about dimensional functions but i was asking about the one-one function in particular one learns best by teaching

#### FeLiXe

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« Reply #9 on: August 04, 2006, 07:57:08 AM »
actually injectivity and surjectivity are only the same over sets of the same finite cardinality

an exponential function R->R is injective but not surjective

a typical cubic fuction R->R is surjective but not injective

the function f:N->N f(x)=2x is injective but not surjective

...
Math and alcohol don't mix, so... please, don't drink and derive!

#### FeLiXe

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« Reply #10 on: August 04, 2006, 08:00:34 AM »
xiankai: maybe I should add that your g-1 is not a function because two values are assigned to each value of x

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#### xiankai

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« Reply #11 on: August 04, 2006, 08:30:52 PM »
i guess i may be asking the wrong questions... lets start at the basics then how do you define surjectivity and injectivity?
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#### Yggdrasil

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« Reply #12 on: August 05, 2006, 12:22:52 AM »
Well, here are the formal definitions (plus one more needed as a basis for the others):

A function, denoted f: X -> Y, maps each element of the set X to an element in the set Y.

A function, f, is injective if f(a) = f(b) implies that a = b.

A function, f: X -> Y, is surjective if for every y in Y, there exists an x in X such that f(x) = y.

#### FeLiXe

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« Reply #13 on: August 05, 2006, 09:28:54 AM »
for example
lets say we have two sets X={A,B,C} and Y={1,2,3}
and functions X->Y

A --> 1
B --> 2
C --> 2

is a function
is not injective because B and C are projected to the same value (2)
not surjective because 3 is not reached

A --> 1
B --> 2
C --> 2
C --> 3

is not a function because to one value of X two values of Y are assigned

A --> 1
B --> 3
C --> 2

is a function
is injective
is bijective
(as we said with finite sets of the same cardinality injective and surjective is the same)
Math and alcohol don't mix, so... please, don't drink and derive!

#### xiankai

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« Reply #14 on: August 05, 2006, 11:05:44 AM »
hmm i think i understand it better now; im uneducated in function language but i can infer from the examples, sweet! one learns best by teaching