Linear transformation with polynomial
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
588
add a comment |
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
588
1
Take a very careful look at the first equality of condition 1
– user328442
Jan 5 at 21:44
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 $;f;$ ?
– DonAntonio
Jan 5 at 22:11
My notation was unfortunate, the p' is not the derivative
– Jakcjones
Jan 5 at 22:40
As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
– Jakcjones
Jan 5 at 22:42
add a comment |
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
588
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
588
asked Jan 5 at 21:40
JakcjonesJakcjones
588
edited Jan 5 at 22:37
Jakcjones
asked Jan 5 at 21:40
JakcjonesJakcjones
588
asked Jan 5 at 21:40
JakcjonesJakcjones
588
asked Jan 5 at 21:40
JakcjonesJakcjones
588
588
1
Take a very careful look at the first equality of condition 1
– user328442
Jan 5 at 21:44
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 $;f;$ ?
– DonAntonio
Jan 5 at 22:11
My notation was unfortunate, the p' is not the derivative
– Jakcjones
Jan 5 at 22:40
As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
– Jakcjones
Jan 5 at 22:42
add a comment |
1
Take a very careful look at the first equality of condition 1
– user328442
Jan 5 at 21:44
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 $;f;$ ?
– DonAntonio
Jan 5 at 22:11
My notation was unfortunate, the p' is not the derivative
– Jakcjones
Jan 5 at 22:40
As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
– Jakcjones
Jan 5 at 22:42
1
1
Take a very careful look at the first equality of condition 1
– user328442
Jan 5 at 21:44
Take a very careful look at the first equality of condition 1
– user328442
Jan 5 at 21:44
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 $;f;$ ?
– DonAntonio
Jan 5 at 22:11
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 $;f;$ ?
– DonAntonio
Jan 5 at 22:11
My notation was unfortunate, the p' is not the derivative
– Jakcjones
Jan 5 at 22:40
My notation was unfortunate, the p' is not the derivative
– Jakcjones
Jan 5 at 22:40
As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
– Jakcjones
Jan 5 at 22:42
As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
– Jakcjones
Jan 5 at 22:42
add a comment |
1 Answer
1
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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
27k62446
add a comment |
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1 Answer
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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
27k62446
add a comment |
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
27k62446
add a comment |
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
27k62446
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
27k62446
answered Jan 5 at 21:44
mechanodroidmechanodroid
27k62446
answered Jan 5 at 21:44
mechanodroidmechanodroid
27k62446
answered Jan 5 at 21:44
mechanodroidmechanodroid
27k62446
27k62446
add a comment |
add a comment |
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1
Take a very careful look at the first equality of condition 1
– user328442
Jan 5 at 21:44
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 $;f;$ ?
– DonAntonio
Jan 5 at 22:11
My notation was unfortunate, the p' is not the derivative
– Jakcjones
Jan 5 at 22:40
As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
– Jakcjones
Jan 5 at 22:42