Linear transformation with polynomial
$begingroup$
Determine if its linear the transformation
$f:Re_{n}[x] rightarrow Re$
such as
$f(p(x))=p(x)+1$
for any $p(x) in Re_{n}$[x]
The solution says it's not linear but I worked it out like this:
Condition 1. $T(u+v)=T(u)+T(v)$
$f(p(x))+p'(x))=((p(x)+1)+(p'(x)+1))=f(p(x))+f(p'(x))$
Condition 2. $alpha T(u)=T(alpha u)$
$alpha f(p(x))= alpha (p(x)+1) = alpha p(x)+alpha = f(alpha p(x))$
What am I doing wrong?
linear-transformations
asked Jan 5 at 21:40
JakcjonesJakcjones
828
$endgroup$
add a comment |
$begingroup$
Determine if its linear the transformation
$f:Re_{n}[x] rightarrow Re$
such as
$f(p(x))=p(x)+1$
for any $p(x) in Re_{n}$[x]
The solution says it's not linear but I worked it out like this:
Condition 1. $T(u+v)=T(u)+T(v)$
$f(p(x))+p'(x))=((p(x)+1)+(p'(x)+1))=f(p(x))+f(p'(x))$
Condition 2. $alpha T(u)=T(alpha u)$
$alpha f(p(x))= alpha (p(x)+1) = alpha p(x)+alpha = f(alpha p(x))$
What am I doing wrong?
linear-transformations
asked Jan 5 at 21:40
JakcjonesJakcjones
828
$endgroup$
1
$begingroup$
Take a very careful look at the first equality of condition 1
$endgroup$
– user328442
Jan 5 at 21:44
$begingroup$
Your fourth line contradicts your second one: the $;f;$ you're defining there is not a function to $;Bbb R;$ (as written in second line) as it doesn't give a number (=a scalar), but a new polynomial. It also confusing what you wrote in the line below "condition 1", with derivatives and etc. Where does this all come from? So what exactly is the counter domain of $;f;$ and what exactly is the definition of $;f;$ ?
$endgroup$
– DonAntonio
Jan 5 at 22:11
$begingroup$
My notation was unfortunate, the p' is not the derivative
$endgroup$
– Jakcjones
Jan 5 at 22:40
$begingroup$
As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
$endgroup$
– Jakcjones
Jan 5 at 22:42
add a comment |
$begingroup$
Determine if its linear the transformation
$f:Re_{n}[x] rightarrow Re$
such as
$f(p(x))=p(x)+1$
for any $p(x) in Re_{n}$[x]
The solution says it's not linear but I worked it out like this:
Condition 1. $T(u+v)=T(u)+T(v)$
$f(p(x))+p'(x))=((p(x)+1)+(p'(x)+1))=f(p(x))+f(p'(x))$
Condition 2. $alpha T(u)=T(alpha u)$
$alpha f(p(x))= alpha (p(x)+1) = alpha p(x)+alpha = f(alpha p(x))$
What am I doing wrong?
linear-transformations
asked Jan 5 at 21:40
JakcjonesJakcjones
828
$endgroup$
Determine if its linear the transformation
$f:Re_{n}[x] rightarrow Re$
such as
$f(p(x))=p(x)+1$
for any $p(x) in Re_{n}$[x]
The solution says it's not linear but I worked it out like this:
Condition 1. $T(u+v)=T(u)+T(v)$
$f(p(x))+p'(x))=((p(x)+1)+(p'(x)+1))=f(p(x))+f(p'(x))$
Condition 2. $alpha T(u)=T(alpha u)$
$alpha f(p(x))= alpha (p(x)+1) = alpha p(x)+alpha = f(alpha p(x))$
What am I doing wrong?
linear-transformations
linear-transformations
asked Jan 5 at 21:40
JakcjonesJakcjones
828
asked Jan 5 at 21:40
JakcjonesJakcjones
828
edited Jan 5 at 22:37
Jakcjones
asked Jan 5 at 21:40
JakcjonesJakcjones
828
asked Jan 5 at 21:40
JakcjonesJakcjones
828
asked Jan 5 at 21:40
JakcjonesJakcjones
828
828
1
$begingroup$
Take a very careful look at the first equality of condition 1
$endgroup$
– user328442
Jan 5 at 21:44
$begingroup$
Your fourth line contradicts your second one: the $;f;$ you're defining there is not a function to $;Bbb R;$ (as written in second line) as it doesn't give a number (=a scalar), but a new polynomial. It also confusing what you wrote in the line below "condition 1", with derivatives and etc. Where does this all come from? So what exactly is the counter domain of $;f;$ and what exactly is the definition of $;f;$ ?
$endgroup$
– DonAntonio
Jan 5 at 22:11
$begingroup$
My notation was unfortunate, the p' is not the derivative
$endgroup$
– Jakcjones
Jan 5 at 22:40
$begingroup$
As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
$endgroup$
– Jakcjones
Jan 5 at 22:42
add a comment |
1
$begingroup$
Take a very careful look at the first equality of condition 1
$endgroup$
– user328442
Jan 5 at 21:44
$begingroup$
Your fourth line contradicts your second one: the $;f;$ you're defining there is not a function to $;Bbb R;$ (as written in second line) as it doesn't give a number (=a scalar), but a new polynomial. It also confusing what you wrote in the line below "condition 1", with derivatives and etc. Where does this all come from? So what exactly is the counter domain of $;f;$ and what exactly is the definition of $;f;$ ?
$endgroup$
– DonAntonio
Jan 5 at 22:11
$begingroup$
My notation was unfortunate, the p' is not the derivative
$endgroup$
– Jakcjones
Jan 5 at 22:40
$begingroup$
As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
$endgroup$
– Jakcjones
Jan 5 at 22:42
1
1
$begingroup$
Take a very careful look at the first equality of condition 1
$endgroup$
– user328442
Jan 5 at 21:44
$begingroup$
Take a very careful look at the first equality of condition 1
$endgroup$
– user328442
Jan 5 at 21:44
$begingroup$
Your fourth line contradicts your second one: the $;f;$ you're defining there is not a function to $;Bbb R;$ (as written in second line) as it doesn't give a number (=a scalar), but a new polynomial. It also confusing what you wrote in the line below "condition 1", with derivatives and etc. Where does this all come from? So what exactly is the counter domain of $;f;$ and what exactly is the definition of $;f;$ ?
$endgroup$
– DonAntonio
Jan 5 at 22:11
$begingroup$
Your fourth line contradicts your second one: the $;f;$ you're defining there is not a function to $;Bbb R;$ (as written in second line) as it doesn't give a number (=a scalar), but a new polynomial. It also confusing what you wrote in the line below "condition 1", with derivatives and etc. Where does this all come from? So what exactly is the counter domain of $;f;$ and what exactly is the definition of $;f;$ ?
$endgroup$
– DonAntonio
Jan 5 at 22:11
$begingroup$
My notation was unfortunate, the p' is not the derivative
$endgroup$
– Jakcjones
Jan 5 at 22:40
$begingroup$
My notation was unfortunate, the p' is not the derivative
$endgroup$
– Jakcjones
Jan 5 at 22:40
$begingroup$
As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
$endgroup$
– Jakcjones
Jan 5 at 22:42
$begingroup$
As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
$endgroup$
– Jakcjones
Jan 5 at 22:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It should be
$$f(p(x)+p'(x)) = p(x)+p'(x)+1 ne p(x)+p'(x)+2 = f(p(x)) + f(p'(x))$$
Anyway, it is faster to note that $f(0) = 1 ne 0$, so $f$ cannot be linear.
answered Jan 5 at 21:44
mechanodroidmechanodroid
28.9k62548
$endgroup$
add a comment |
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$begingroup$
It should be
$$f(p(x)+p'(x)) = p(x)+p'(x)+1 ne p(x)+p'(x)+2 = f(p(x)) + f(p'(x))$$
Anyway, it is faster to note that $f(0) = 1 ne 0$, so $f$ cannot be linear.
answered Jan 5 at 21:44
mechanodroidmechanodroid
28.9k62548
$endgroup$
add a comment |
$begingroup$
It should be
$$f(p(x)+p'(x)) = p(x)+p'(x)+1 ne p(x)+p'(x)+2 = f(p(x)) + f(p'(x))$$
Anyway, it is faster to note that $f(0) = 1 ne 0$, so $f$ cannot be linear.
answered Jan 5 at 21:44
mechanodroidmechanodroid
28.9k62548
$endgroup$
add a comment |
$begingroup$
It should be
$$f(p(x)+p'(x)) = p(x)+p'(x)+1 ne p(x)+p'(x)+2 = f(p(x)) + f(p'(x))$$
Anyway, it is faster to note that $f(0) = 1 ne 0$, so $f$ cannot be linear.
answered Jan 5 at 21:44
mechanodroidmechanodroid
28.9k62548
$endgroup$
It should be
$$f(p(x)+p'(x)) = p(x)+p'(x)+1 ne p(x)+p'(x)+2 = f(p(x)) + f(p'(x))$$
Anyway, it is faster to note that $f(0) = 1 ne 0$, so $f$ cannot be linear.
answered Jan 5 at 21:44
mechanodroidmechanodroid
28.9k62548
answered Jan 5 at 21:44
mechanodroidmechanodroid
28.9k62548
answered Jan 5 at 21:44
mechanodroidmechanodroid
28.9k62548
answered Jan 5 at 21:44
mechanodroidmechanodroid
28.9k62548
28.9k62548
add a comment |
add a comment |
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1
$begingroup$
Take a very careful look at the first equality of condition 1
$endgroup$
– user328442
Jan 5 at 21:44
$begingroup$
Your fourth line contradicts your second one: the $;f;$ you're defining there is not a function to $;Bbb R;$ (as written in second line) as it doesn't give a number (=a scalar), but a new polynomial. It also confusing what you wrote in the line below "condition 1", with derivatives and etc. Where does this all come from? So what exactly is the counter domain of $;f;$ and what exactly is the definition of $;f;$ ?
$endgroup$
– DonAntonio
Jan 5 at 22:11
$begingroup$
My notation was unfortunate, the p' is not the derivative
$endgroup$
– Jakcjones
Jan 5 at 22:40
$begingroup$
As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
$endgroup$
– Jakcjones
Jan 5 at 22:42