Mixing
Composition and Addition on a Set of Function and Consequences for Distributitivity
Clash Royale CLAN TAG#URR8PPP
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I'm working on the following question:
Let $S$ be the set of all functions $f: mathbbR rightarrow mathbbR$. Consider the following pointwise operations $*$ and $+$:
$$(f+g)(x) = f(x) + g(x) \ (f*g))(x) = f(g(x))$$
Which of the following is true:
- $f * (g + h) = f*g + f*h$
- $(g + h)*f = g*f + h*f$
By using the example of constant functions one can eliminate 1. leaving 2. But, I only know this because I peeked at the answer- I didn't think of this beforehand.
Is there a way of seeing this directly without examples? For instance $S$ appears to have some algebraic structure.
abstract-algebra
asked 27 mins ago
yoshi
1,008816
add a comment |Â
up vote
1
down vote
favorite
I'm working on the following question:
Let $S$ be the set of all functions $f: mathbbR rightarrow mathbbR$. Consider the following pointwise operations $*$ and $+$:
$$(f+g)(x) = f(x) + g(x) \ (f*g))(x) = f(g(x))$$
Which of the following is true:
- $f * (g + h) = f*g + f*h$
- $(g + h)*f = g*f + h*f$
By using the example of constant functions one can eliminate 1. leaving 2. But, I only know this because I peeked at the answer- I didn't think of this beforehand.
Is there a way of seeing this directly without examples? For instance $S$ appears to have some algebraic structure.
abstract-algebra
asked 27 mins ago
yoshi
1,008816
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm working on the following question:
Let $S$ be the set of all functions $f: mathbbR rightarrow mathbbR$. Consider the following pointwise operations $*$ and $+$:
$$(f+g)(x) = f(x) + g(x) \ (f*g))(x) = f(g(x))$$
Which of the following is true:
- $f * (g + h) = f*g + f*h$
- $(g + h)*f = g*f + h*f$
By using the example of constant functions one can eliminate 1. leaving 2. But, I only know this because I peeked at the answer- I didn't think of this beforehand.
Is there a way of seeing this directly without examples? For instance $S$ appears to have some algebraic structure.
abstract-algebra
asked 27 mins ago
yoshi
1,008816
I'm working on the following question:
Let $S$ be the set of all functions $f: mathbbR rightarrow mathbbR$. Consider the following pointwise operations $*$ and $+$:
$$(f+g)(x) = f(x) + g(x) \ (f*g))(x) = f(g(x))$$
Which of the following is true:
- $f * (g + h) = f*g + f*h$
- $(g + h)*f = g*f + h*f$
By using the example of constant functions one can eliminate 1. leaving 2. But, I only know this because I peeked at the answer- I didn't think of this beforehand.
Is there a way of seeing this directly without examples? For instance $S$ appears to have some algebraic structure.
abstract-algebra
abstract-algebra
asked 27 mins ago
yoshi
1,008816
asked 27 mins ago
yoshi
1,008816
asked 27 mins ago
yoshi
1,008816
asked 27 mins ago
yoshi
1,008816
asked 27 mins ago
yoshi
1,008816
1,008816
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
For the second one apply the definitions: Let's take some $xinmathbbR$ and look at $((g+h)*f)(x)$. By the definition of $*$, this is the same as $(g+h)(f(x))$, which we can further simplify using the definition of $+$ to $(g(f(x))+h(f(x))$. But this is nothing else than $(g*f)(x)+(h*f)(x)$, so $(g+h)*f=g*f+h*f$.
answered 21 mins ago
Jannik Pitt
239313
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up vote
2
down vote
At least the first one is false. Take $f(x) = x^2 $, $g=x$ and $h=1$, then
$$ f * (g+h) = f(x+1) = (x+1)^2 $$
but
$$ f * g = x^2 $$
$$ f * h = 1 $$
answered 24 mins ago
Jimmy Sabater
1,19013
add a comment |Â
up vote
1
down vote
For (1): It is saying given $f,g,h$ can we have
$$f(g+h)=f(g)+f(h).$$
It almost reminds you of linearity. So any function $f$ which is not behaving "linearly" (so to speak) can provide a counter example. For example, take $f(x)=sin x$ and take $g(x)=h(x)=x$. Then if this was true, we would have $sin(2x)=2sin x$, which as you may know is not true for all $x$.
For (2). This one is true. Here we need to show that $(g+h)f=g(f)+h(f)$. Let $a in mathbbR$, then
beginalign*
(g+h)f(a) & = g(f(a))+h(f(a)) && (textby definition)\
& = g(f)(a)+h(f)(a).
endalign*
So the two functions agree at all real numbers, hence they are equal functions.
answered 19 mins ago
Anurag A
23.2k12245
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For the second one apply the definitions: Let's take some $xinmathbbR$ and look at $((g+h)*f)(x)$. By the definition of $*$, this is the same as $(g+h)(f(x))$, which we can further simplify using the definition of $+$ to $(g(f(x))+h(f(x))$. But this is nothing else than $(g*f)(x)+(h*f)(x)$, so $(g+h)*f=g*f+h*f$.
answered 21 mins ago
Jannik Pitt
239313
add a comment |Â
up vote
2
down vote
accepted
For the second one apply the definitions: Let's take some $xinmathbbR$ and look at $((g+h)*f)(x)$. By the definition of $*$, this is the same as $(g+h)(f(x))$, which we can further simplify using the definition of $+$ to $(g(f(x))+h(f(x))$. But this is nothing else than $(g*f)(x)+(h*f)(x)$, so $(g+h)*f=g*f+h*f$.
answered 21 mins ago
Jannik Pitt
239313
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For the second one apply the definitions: Let's take some $xinmathbbR$ and look at $((g+h)*f)(x)$. By the definition of $*$, this is the same as $(g+h)(f(x))$, which we can further simplify using the definition of $+$ to $(g(f(x))+h(f(x))$. But this is nothing else than $(g*f)(x)+(h*f)(x)$, so $(g+h)*f=g*f+h*f$.
answered 21 mins ago
Jannik Pitt
239313
For the second one apply the definitions: Let's take some $xinmathbbR$ and look at $((g+h)*f)(x)$. By the definition of $*$, this is the same as $(g+h)(f(x))$, which we can further simplify using the definition of $+$ to $(g(f(x))+h(f(x))$. But this is nothing else than $(g*f)(x)+(h*f)(x)$, so $(g+h)*f=g*f+h*f$.
answered 21 mins ago
Jannik Pitt
239313
answered 21 mins ago
Jannik Pitt
239313
answered 21 mins ago
Jannik Pitt
239313
answered 21 mins ago
Jannik Pitt
239313
239313
add a comment |Â
add a comment |Â
up vote
2
down vote
At least the first one is false. Take $f(x) = x^2 $, $g=x$ and $h=1$, then
$$ f * (g+h) = f(x+1) = (x+1)^2 $$
but
$$ f * g = x^2 $$
$$ f * h = 1 $$
answered 24 mins ago
Jimmy Sabater
1,19013
add a comment |Â
up vote
2
down vote
At least the first one is false. Take $f(x) = x^2 $, $g=x$ and $h=1$, then
$$ f * (g+h) = f(x+1) = (x+1)^2 $$
but
$$ f * g = x^2 $$
$$ f * h = 1 $$
answered 24 mins ago
Jimmy Sabater
1,19013
add a comment |Â
up vote
2
down vote
up vote
2
down vote
At least the first one is false. Take $f(x) = x^2 $, $g=x$ and $h=1$, then
$$ f * (g+h) = f(x+1) = (x+1)^2 $$
but
$$ f * g = x^2 $$
$$ f * h = 1 $$
answered 24 mins ago
Jimmy Sabater
1,19013
At least the first one is false. Take $f(x) = x^2 $, $g=x$ and $h=1$, then
$$ f * (g+h) = f(x+1) = (x+1)^2 $$
but
$$ f * g = x^2 $$
$$ f * h = 1 $$
answered 24 mins ago
Jimmy Sabater
1,19013
answered 24 mins ago
Jimmy Sabater
1,19013
answered 24 mins ago
Jimmy Sabater
1,19013
answered 24 mins ago
Jimmy Sabater
1,19013
1,19013
add a comment |Â
add a comment |Â
up vote
1
down vote
For (1): It is saying given $f,g,h$ can we have
$$f(g+h)=f(g)+f(h).$$
It almost reminds you of linearity. So any function $f$ which is not behaving "linearly" (so to speak) can provide a counter example. For example, take $f(x)=sin x$ and take $g(x)=h(x)=x$. Then if this was true, we would have $sin(2x)=2sin x$, which as you may know is not true for all $x$.
For (2). This one is true. Here we need to show that $(g+h)f=g(f)+h(f)$. Let $a in mathbbR$, then
beginalign*
(g+h)f(a) & = g(f(a))+h(f(a)) && (textby definition)\
& = g(f)(a)+h(f)(a).
endalign*
So the two functions agree at all real numbers, hence they are equal functions.
answered 19 mins ago
Anurag A
23.2k12245
add a comment |Â
up vote
1
down vote
For (1): It is saying given $f,g,h$ can we have
$$f(g+h)=f(g)+f(h).$$
It almost reminds you of linearity. So any function $f$ which is not behaving "linearly" (so to speak) can provide a counter example. For example, take $f(x)=sin x$ and take $g(x)=h(x)=x$. Then if this was true, we would have $sin(2x)=2sin x$, which as you may know is not true for all $x$.
For (2). This one is true. Here we need to show that $(g+h)f=g(f)+h(f)$. Let $a in mathbbR$, then
beginalign*
(g+h)f(a) & = g(f(a))+h(f(a)) && (textby definition)\
& = g(f)(a)+h(f)(a).
endalign*
So the two functions agree at all real numbers, hence they are equal functions.
answered 19 mins ago
Anurag A
23.2k12245
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For (1): It is saying given $f,g,h$ can we have
$$f(g+h)=f(g)+f(h).$$
It almost reminds you of linearity. So any function $f$ which is not behaving "linearly" (so to speak) can provide a counter example. For example, take $f(x)=sin x$ and take $g(x)=h(x)=x$. Then if this was true, we would have $sin(2x)=2sin x$, which as you may know is not true for all $x$.
For (2). This one is true. Here we need to show that $(g+h)f=g(f)+h(f)$. Let $a in mathbbR$, then
beginalign*
(g+h)f(a) & = g(f(a))+h(f(a)) && (textby definition)\
& = g(f)(a)+h(f)(a).
endalign*
So the two functions agree at all real numbers, hence they are equal functions.
answered 19 mins ago
Anurag A
23.2k12245
For (1): It is saying given $f,g,h$ can we have
$$f(g+h)=f(g)+f(h).$$
It almost reminds you of linearity. So any function $f$ which is not behaving "linearly" (so to speak) can provide a counter example. For example, take $f(x)=sin x$ and take $g(x)=h(x)=x$. Then if this was true, we would have $sin(2x)=2sin x$, which as you may know is not true for all $x$.
For (2). This one is true. Here we need to show that $(g+h)f=g(f)+h(f)$. Let $a in mathbbR$, then
beginalign*
(g+h)f(a) & = g(f(a))+h(f(a)) && (textby definition)\
& = g(f)(a)+h(f)(a).
endalign*
So the two functions agree at all real numbers, hence they are equal functions.
answered 19 mins ago
Anurag A
23.2k12245
answered 19 mins ago
Anurag A
23.2k12245
answered 19 mins ago
Anurag A
23.2k12245
answered 19 mins ago
Anurag A
23.2k12245
23.2k12245
add a comment |Â
add a comment |Â
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