Composite functions help

[hotlip1]

Given these functions
f(x)=x^2
g(x,y)=x+y^2
h(t)=(2t-1,2-t)
i(x,y)=(x-2y,x+y)

I'm asked to calculate
ioi
hoi
ioh
fogoi
goh
hog
iohof


I know how to do composite functions for R->R but these compositions are too complicated for me D:
Explanation please?

Thanks in advance.

 

[Ishuda]

Given these functions
f(x)=x^2
g(x,y)=x+y^2
h(t)=(2t-1,2-t)
i(x,y)=(x-2y,x+y)

I'm asked to calculate
ioi
hoi
ioh
fogoi
goh
hog
iohof


I know how to do composite functions for R->R but these compositions are too complicated for me D:
Explanation please?

Thanks in advance.

For functions of more than one dimension, i.e. i(x,y) above, one needs to know what is meant by the composition of that function with another. For example by ioi do we mean
ioi = i(i(x,y),y) = (i(x,y)-y, i(x,y) + y)
and, if so, what does (x-2y,x+y)-y mean? Or maybe it means
ioi = i(i(x,y),y) = (i(x,y), x+y) = (((x-2y),x+y), x+y)
or maybe ...

So, what does it mean? Not only for ioi, but for the multiply dimensional composition in general?

 

[hotlip1]

For functions of more than one dimension, i.e. i(x,y) above, one needs to know what is meant by the composition of that function with another. For example by ioi do we mean
ioi = i(i(x,y),y) = (i(x,y)-y, i(x,y) + y)
and, if so, what does (x-2y,x+y)-y mean? Or maybe it means
ioi = i(i(x,y),y) = (i(x,y), x+y) = (((x-2y),x+y), x+y)
or maybe ...

So, what does it mean? Not only for ioi, but for the multiply dimensional composition in general?

I assume ioi=i(i(x,y),i(x,y)), though this wasn't specified.
As for the rest of what you said, I don't really understand you :\



One thing I didn't say is that the exercise says "Calculate IF POSSIBLE the following compositions.."
In which cases would that not be possible?

 

[Ishuda]

I assume ioi=i(i(x,y),i(x,y)), though this wasn't specified.
As for the rest of what you said, I don't really understand you :\



One thing I didn't say is that the exercise says "Calculate IF POSSIBLE the following compositions.."
In which cases would that not be possible?

Normally, when one has a composite function, the place where the second (and subsequent) function(s) goes is designated. If there is only one argument, that is not necessary. For example suppose we have two functions f(x) and g(x), then fog = f(g(x)) since there is only one place g(x) can go. However, if there are more than two independent variables, then one must be specific. That is, if we have two functions f(x,y) and g(x,y) then fog is ambiguous, it could mean f(g(x,y), y) or f(x, g(x,y)) or f(g(x,y), g(x,y)). So instead, it is explicitly written out which is meant [or defined before hand for the discussion at hand].

In addition, the range of the second function has the same dimensions [same number of dependent variables] as the original function. That is, if we have a function f(x, y, z) where x, y, and z are scalar variables (one dimensional variables), then the g which replaces the x (or y or z) in the composite function is also a scalar variable. If x were two dimensional, i.e. x = (x₁, x₂) where x₁ and x₂ were scalars, then (the output of) g should be two dimensional.

So, under normal circumstances, ioi is meaningless since, first of all, you don't know which argument, the x or y, the i is supposed to replace in i(x,y) and, secondly i has a two dimensional range [is an ordered pair] while both x and y are scalars [one dimensional].

 

[pka]

Given these functions
f(x)=x^2
g(x,y)=x+y^2
h(t)=(2t-1,2-t)
i(x,y)=(x-2y,x+y)

I'm asked to calculate
ioi
hoi
ioh
fogoi
goh
hog
iohof

First, one must know the initial set and the final set for each function.
In any composition the correct sets must 'fit' together.

\(\displaystyle f:{\mathbb{R}^1} \to {\mathbb{R}^1},\quad g:{\mathbb{R}^2} \to {\mathbb{R}^1},\quad h:{\mathbb{R}^1} \to {\mathbb{R}^2},\quad i:{\mathbb{R}^2} \to {\mathbb{R}^2}\)

\(\displaystyle f \circ g\) is defined because \(\displaystyle f \circ g: \mathbb{R}^2\to\mathbb{R}^1\)

These are also defined:
\(\displaystyle g \circ h\quad \& \quad h \circ g\\g \circ i\\i \circ h\\h \circ f\\i \circ h \circ f\\\)

 

[hotlip1]

First, one must know the initial set and the final set for each function.
In any composition the correct sets must 'fit' together.

\(\displaystyle f:{\mathbb{R}^1} \to {\mathbb{R}^1},\quad g:{\mathbb{R}^2} \to {\mathbb{R}^1},\quad h:{\mathbb{R}^1} \to {\mathbb{R}^2},\quad i:{\mathbb{R}^2} \to {\mathbb{R}^2}\)

\(\displaystyle f \circ g\) is defined because \(\displaystyle f \circ g: \mathbb{R}^2\to\mathbb{R}^1\)

These are also defined:
\(\displaystyle g \circ h\quad \& \quad h \circ g\\g \circ i\\i \circ h\\h \circ f\\i \circ h \circ f\\\)

Ok, I think I get it.
Say we have f:Rⁿ->R^m and g:R^s->R^t
fog is defined only if n=t. Right?

If that's correct, then given i(x,y)=(x-2y,x+y), ioi is defined and is equal to ((x-2y)-2(x+y),(x-2y)+(x+y)) right?

 

[Ishuda]

Ok, I think I get it.
Say we have f:Rⁿ->R^m and g:R^s->R^t
fog is defined only if n=t. Right?

If that's correct, then given i(x,y)=(x-2y,x+y), ioi is defined and is equal to ((x-2y)-2(x+y),(x-2y)+(x+y)) right?

The way I learned it, you almost have it: Let f be a function f(R^j1,R^j2,R^j3,...,R^jn), then where g goes must be specified, i.e. a
j_k, k=1, 2, 3, ...,n
must be specified and t must be equal to that specified j_k.

As an example for ioi: for the initial i; both R^j1 and R^j2 are (assumed to be) scalars, i.e. belong to R¹, k equals either 1 or 2 and both j₁ and j₂ are 1. However the output of the second i is R² so you can replace neither of the R^j1 argument (x) nor the R^j2 argument (y) by i(x,y). So, either ioi is not defined or you will have to make a special definition just for that composition.

As an example for iog: for the initial i; both R^j1 and R^j2 are (assumed to be) scalars, i.e. belong to R¹, k equals either 1 or 2 and both j₁ and j₂ are 1. The output of g is R¹, a scalar, and you can replace either the R^j1 argument (x) or the R^j2 argument (y) by g(x,y). Now, you have to choose which (x or y) you want to replace to get
iog = i(g(x,y),y) = (g(x,y) - 2 y, g(x,y) + y)
or
iog = i(x, g(x,y)) = (x - 2 g(x,y), x + g(x,y))

 

[pka]

Say we have f:Rⁿ->R^m and g:R^s->R^t
fog is defined only if n=t. Right? YES
If that's correct, then given i(x,y)=(x-2y,x+y), ioi is defined and is equal to ((x-2y)-2(x+y),(x-2y)+(x+y)) right?

That is absolutely correct. But lets look at an example.
\(\displaystyle f(x)=x^2,\quad g(x,y)=x+y^2~\quad\&\quad h(t)=(2t-1,1-2t)\)

\(\displaystyle f\circ g(u,v)=\left(u+v^2\right)^2\) and \(\displaystyle h\circ f\circ g(u,v)=\left(2(u+v^2)^2-1,\left(1-2(u+v^2)\right)^2\right)\)

The way I learned it,
1. The output of g is R¹, a scalar, and you can replace either the R^j1 argument (x) or the R^j2 argument (y) by g(x,y). Now, you have to choose which (x or y) you want to replace to get
iog = i(g(x,y),y) = (g(x,y) - 2 y, g(x,y) + y)
or
iog = i(x, g(x,y)) = (x - 2 g(x,y), x + g(x,y))

I do not understand any of that. It is not constant with any thing in axiomatic set theory. It greatly distresses me to think that is taught anywhere.

 

[hotlip1]

The way I learned it, you almost have it: Let f be a function f(R^j1,R^j2,R^j3,...,R^jn), then where g goes must be specified, i.e. a
j_k, k=1, 2, 3, ...,n
must be specified and t must be equal to that specified j_k.

As an example for ioi: for the initial i; both R^j1 and R^j2 are (assumed to be) scalars, i.e. belong to R¹, k equals either 1 or 2 and both j₁ and j₂ are 1. However the output of the second i is R² so you can replace neither of the R^j1 argument (x) nor the R^j2 argument (y) by i(x,y). So, either ioi is not defined or you will have to make a special definition just for that composition.

As an example for iog: for the initial i; both R^j1 and R^j2 are (assumed to be) scalars, i.e. belong to R¹, k equals either 1 or 2 and both j₁ and j₂ are 1. The output of g is R¹, a scalar, and you can replace either the R^j1 argument (x) or the R^j2 argument (y) by g(x,y). Now, you have to choose which (x or y) you want to replace to get
iog = i(g(x,y),y) = (g(x,y) - 2 y, g(x,y) + y)
or
iog = i(x, g(x,y)) = (x - 2 g(x,y), x + g(x,y))

That is absolutely correct. But lets look at an example.
\(\displaystyle f(x)=x^2,\quad g(x,y)=x+y^2~\quad\&\quad h(t)=(2t-1,1-2t)\)

\(\displaystyle f\circ g(u,v)=\left(u+v^2\right)^2\) and \(\displaystyle h\circ f\circ g(u,v)=\left(2(u+v^2)^2-1,\left(1-2(u+v^2)\right)^2\right)\)


I do not understand any of that. It is not constant with any thing in axiomatic set theory. It greatly distresses me to think that is taught anywhere.

Thank you both, I now understand everything you said. Don't miss me though, I'll be around the forums asking many more things

 

[Ishuda]

That is absolutely correct. But lets look at an example.
\(\displaystyle f(x)=x^2,\quad g(x,y)=x+y^2~\quad\&\quad h(t)=(2t-1,1-2t)\)

\(\displaystyle f\circ g(u,v)=\left(u+v^2\right)^2\) and \(\displaystyle h\circ f\circ g(u,v)=\left(2(u+v^2)^2-1,\left(1-2(u+v^2)\right)^2\right)\)


I do not understand any of that. It is not constant with any thing in axiomatic set theory. It greatly distresses me to think that is taught anywhere.

pka,

Let's start over, please. I may have put it badly, but what I meant to mean is what I think is correct. What is meant by gof given the g and f of
g(x,y)=x+y²
f(t)= t²

 

[pka]

pka,

Let's start over, please. I may have put it badly, but what I meant to mean is what I think is correct. What is meant by gof given the g and f of
g(x,y)=x+y²
f(t)= t²

In that case \(\displaystyle g\circ f\) cannot exist.
Because \(\displaystyle f:\Re^1\to\Re^1~\&~g:\Re^2\to\Re^1\), so \(\displaystyle \large g;\Re^1\not\to\Re^1\) does not fit.

However \(\displaystyle f\circ g\) does exist and \(\displaystyle \Re^2\to\Re^1\).

Again: If \(\displaystyle j:\Re^n\to\Re^m~\&~k:\Re^s\to\Re^t\) then \(\displaystyle k\circ j\) exist \(\displaystyle \iff~s=m\)
Hence we have \(\displaystyle k\circ j:\Re^n\to\Re^t\).

 

[Ishuda]

In that case \(\displaystyle g\circ f\) cannot exist.
Because \(\displaystyle f:\Re^1\to\Re^1~\&~g:\Re^2\to\Re^1\), so \(\displaystyle \large g;\Re^1\not\to\Re^1\) does not fit.

However \(\displaystyle f\circ g\) does exist and \(\displaystyle \Re^2\to\Re^1\).

Again: If \(\displaystyle j:\Re^n\to\Re^m~\&~k:\Re^s\to\Re^t\) then \(\displaystyle k\circ j\) exist \(\displaystyle \iff~s=m\)
Hence we have \(\displaystyle k\circ j:\Re^n\to\Re^t\).

If that is the case, what do you call
g(f(t), y) = f(t) + y²
and what do you call
g(x, f(t)) = x + f²(t)

 

[pka]

If that is the case, what do you call
g(f(t), y) = f(t) + y²
and what do you call
g(x, f(t)) = x + f²(t)

Those are both correct. But neither is an example of function composition.

I don't know what one should call them other than both are examples of how the function \(\displaystyle g\) operates.

The function \(\displaystyle g\) maps pairs to numbers: \(\displaystyle (x,y)\mapsto x+y^2\). In other words, \(\displaystyle g\) adds the first term of the pair to the square of the second term.