Mixing
Is $T(f)=f'$ where $f$ is infinitely differentiable an additive transformation?
$begingroup$
If the map is defined as $T:V rightarrow V$ by $T(f)=f'$ where $f$ is infinitely differentiable.
I am trying to show that $T$ is a linear transformation.
I understand how to show that it is homogenous but I am struggling to show that it is additive. Please help show that it is additive.
linear-algebra linear-transformations
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
asked Jan 28 at 22:20
hoya2021hoya2021
365
$endgroup$
add a comment |
$begingroup$
If the map is defined as $T:V rightarrow V$ by $T(f)=f'$ where $f$ is infinitely differentiable.
I am trying to show that $T$ is a linear transformation.
I understand how to show that it is homogenous but I am struggling to show that it is additive. Please help show that it is additive.
linear-algebra linear-transformations
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
asked Jan 28 at 22:20
hoya2021hoya2021
365
$endgroup$
$begingroup$
Additivity means $(f+g)'=f'+g'$.
$endgroup$
– Lord Shark the Unknown
Jan 28 at 22:24
add a comment |
$begingroup$
If the map is defined as $T:V rightarrow V$ by $T(f)=f'$ where $f$ is infinitely differentiable.
I am trying to show that $T$ is a linear transformation.
I understand how to show that it is homogenous but I am struggling to show that it is additive. Please help show that it is additive.
linear-algebra linear-transformations
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
asked Jan 28 at 22:20
hoya2021hoya2021
365
$endgroup$
If the map is defined as $T:V rightarrow V$ by $T(f)=f'$ where $f$ is infinitely differentiable.
I am trying to show that $T$ is a linear transformation.
I understand how to show that it is homogenous but I am struggling to show that it is additive. Please help show that it is additive.
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
asked Jan 28 at 22:20
hoya2021hoya2021
365
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
asked Jan 28 at 22:20
hoya2021hoya2021
365
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
edited Jan 29 at 2:06
J. W. Tanner
4,0611320
4,0611320
asked Jan 28 at 22:20
hoya2021hoya2021
365
asked Jan 28 at 22:20
hoya2021hoya2021
365
asked Jan 28 at 22:20
hoya2021hoya2021
365
365
$begingroup$
Additivity means $(f+g)'=f'+g'$.
$endgroup$
– Lord Shark the Unknown
Jan 28 at 22:24
add a comment |
$begingroup$
Additivity means $(f+g)'=f'+g'$.
$endgroup$
– Lord Shark the Unknown
Jan 28 at 22:24
$begingroup$
Additivity means $(f+g)'=f'+g'$.
$endgroup$
– Lord Shark the Unknown
Jan 28 at 22:24
$begingroup$
Additivity means $(f+g)'=f'+g'$.
$endgroup$
– Lord Shark the Unknown
Jan 28 at 22:24
add a comment |
1 Answer
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$begingroup$
You should have learned, in Calculus I, that (f+ g)'= f'+ g' and that (cf)'= cf' for any constant, c.
answered Jan 28 at 22:24
user247327user247327
11.6k1516
$endgroup$
add a comment |
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$begingroup$
You should have learned, in Calculus I, that (f+ g)'= f'+ g' and that (cf)'= cf' for any constant, c.
answered Jan 28 at 22:24
user247327user247327
11.6k1516
$endgroup$
add a comment |
$begingroup$
You should have learned, in Calculus I, that (f+ g)'= f'+ g' and that (cf)'= cf' for any constant, c.
answered Jan 28 at 22:24
user247327user247327
11.6k1516
$endgroup$
add a comment |
$begingroup$
You should have learned, in Calculus I, that (f+ g)'= f'+ g' and that (cf)'= cf' for any constant, c.
answered Jan 28 at 22:24
user247327user247327
11.6k1516
$endgroup$
You should have learned, in Calculus I, that (f+ g)'= f'+ g' and that (cf)'= cf' for any constant, c.
answered Jan 28 at 22:24
user247327user247327
11.6k1516
answered Jan 28 at 22:24
user247327user247327
11.6k1516
answered Jan 28 at 22:24
user247327user247327
11.6k1516
answered Jan 28 at 22:24
user247327user247327
11.6k1516
11.6k1516
add a comment |
add a comment |
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$begingroup$
Additivity means $(f+g)'=f'+g'$.
$endgroup$
– Lord Shark the Unknown
Jan 28 at 22:24