Determine whether f is a function, an injection, a surjection Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)How to determine whether a sequence of functions converges uniformly or pointwise to a function?Is $z=x^2+y^2$ a bijection?Identify as an injection, surjection, bijection or non-functionDetermine whether each of these functions is a bijection from R to RWhat is the mistake in this proof?Determine Total/Partial Functions and Injection/Surjection/BijectionExplaining whether a function is injective/surjection ($fcolonBbb Nto P(Bbb N)$)Is this a surjection? (Elementary real analysis)Help on Surjection, Injection, and BijectionFunction One to One with coordinates
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Determine whether f is a function, an injection, a surjection
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)How to determine whether a sequence of functions converges uniformly or pointwise to a function?Is $z=x^2+y^2$ a bijection?Identify as an injection, surjection, bijection or non-functionDetermine whether each of these functions is a bijection from R to RWhat is the mistake in this proof?Determine Total/Partial Functions and Injection/Surjection/BijectionExplaining whether a function is injective/surjection ($fcolonBbb Nto P(Bbb N)$)Is this a surjection? (Elementary real analysis)Help on Surjection, Injection, and BijectionFunction One to One with coordinates
$begingroup$
Let $P= $p(x)$ is a polynomial of degree $n$, $n in Bbb Z^+cup0 $ with coefficients in $Bbb R $. Define $f : Prightarrow P$ where $f(p(x)) =p'(x)$, the derivative of $p(x)$.
Determine whether $f$ is a function, an injection, a surjection, a bijection.
Now I have the solutions, and I understand that it is a function because each polynomial has a unique derivative. And it is not an injection as the antiderivative of a given polynomial is not unique.
However, I do not understand the book's solution for determining whether it is a surjection, nor am I able to come up with one myself. In all honesty, I think I am having trouble understanding the mapping from P to P. The solution states that it is a surjection. Why is this so?
calculus functions derivatives elementary-set-theory
asked Apr 14 at 20:07
John ArgJohn Arg
546
$endgroup$
|
show 8 more comments
$begingroup$
Let $P= $p(x)$ is a polynomial of degree $n$, $n in Bbb Z^+cup0 $ with coefficients in $Bbb R $. Define $f : Prightarrow P$ where $f(p(x)) =p'(x)$, the derivative of $p(x)$.
Determine whether $f$ is a function, an injection, a surjection, a bijection.
Now I have the solutions, and I understand that it is a function because each polynomial has a unique derivative. And it is not an injection as the antiderivative of a given polynomial is not unique.
However, I do not understand the book's solution for determining whether it is a surjection, nor am I able to come up with one myself. In all honesty, I think I am having trouble understanding the mapping from P to P. The solution states that it is a surjection. Why is this so?
calculus functions derivatives elementary-set-theory
asked Apr 14 at 20:07
John ArgJohn Arg
546
$endgroup$
2
$begingroup$
I interpreted it to mean that it is the set of all polynomials.
$endgroup$
– Tony S.F.
Apr 14 at 20:13
1
$begingroup$
I read it more as the set of "all polynomials with degree that is a nonnegative integer." Though I do wonder of the need to specify that.
$endgroup$
– Eevee Trainer
Apr 14 at 20:15
1
$begingroup$
I think the implication is that $n$ is not fixed, that it is simply a nonnegative integer.
$endgroup$
– Eevee Trainer
Apr 14 at 20:23
2
$begingroup$
The derivative of a constant, nonzero polynomial (of degree $0$) is the zero polynomial (of degree $-infty$). Surely then $f : P to P$ is not well defined at all, because some of its images are not in $P$?
$endgroup$
– wchargin
Apr 15 at 4:57
1
$begingroup$
You want to write $(f(p))(x)=p‘(x)$ or just $f(p)=p‘$.
$endgroup$
– Carsten S
Apr 15 at 6:31
|
show 8 more comments
$begingroup$
Let $P= $p(x)$ is a polynomial of degree $n$, $n in Bbb Z^+cup0 $ with coefficients in $Bbb R $. Define $f : Prightarrow P$ where $f(p(x)) =p'(x)$, the derivative of $p(x)$.
Determine whether $f$ is a function, an injection, a surjection, a bijection.
Now I have the solutions, and I understand that it is a function because each polynomial has a unique derivative. And it is not an injection as the antiderivative of a given polynomial is not unique.
However, I do not understand the book's solution for determining whether it is a surjection, nor am I able to come up with one myself. In all honesty, I think I am having trouble understanding the mapping from P to P. The solution states that it is a surjection. Why is this so?
calculus functions derivatives elementary-set-theory
asked Apr 14 at 20:07
John ArgJohn Arg
546
$endgroup$
Let $P= $p(x)$ is a polynomial of degree $n$, $n in Bbb Z^+cup0 $ with coefficients in $Bbb R $. Define $f : Prightarrow P$ where $f(p(x)) =p'(x)$, the derivative of $p(x)$.
Determine whether $f$ is a function, an injection, a surjection, a bijection.
Now I have the solutions, and I understand that it is a function because each polynomial has a unique derivative. And it is not an injection as the antiderivative of a given polynomial is not unique.
However, I do not understand the book's solution for determining whether it is a surjection, nor am I able to come up with one myself. In all honesty, I think I am having trouble understanding the mapping from P to P. The solution states that it is a surjection. Why is this so?
calculus functions derivatives elementary-set-theory
calculus functions derivatives elementary-set-theory
asked Apr 14 at 20:07
John ArgJohn Arg
546
asked Apr 14 at 20:07
John ArgJohn Arg
546
asked Apr 14 at 20:07
John ArgJohn Arg
546
asked Apr 14 at 20:07
John ArgJohn Arg
546
asked Apr 14 at 20:07
John ArgJohn Arg
546
546
2
$begingroup$
I interpreted it to mean that it is the set of all polynomials.
$endgroup$
– Tony S.F.
Apr 14 at 20:13
1
$begingroup$
I read it more as the set of "all polynomials with degree that is a nonnegative integer." Though I do wonder of the need to specify that.
$endgroup$
– Eevee Trainer
Apr 14 at 20:15
1
$begingroup$
I think the implication is that $n$ is not fixed, that it is simply a nonnegative integer.
$endgroup$
– Eevee Trainer
Apr 14 at 20:23
2
$begingroup$
The derivative of a constant, nonzero polynomial (of degree $0$) is the zero polynomial (of degree $-infty$). Surely then $f : P to P$ is not well defined at all, because some of its images are not in $P$?
$endgroup$
– wchargin
Apr 15 at 4:57
1
$begingroup$
You want to write $(f(p))(x)=p‘(x)$ or just $f(p)=p‘$.
$endgroup$
– Carsten S
Apr 15 at 6:31
|
show 8 more comments
2
$begingroup$
I interpreted it to mean that it is the set of all polynomials.
$endgroup$
– Tony S.F.
Apr 14 at 20:13
1
$begingroup$
I read it more as the set of "all polynomials with degree that is a nonnegative integer." Though I do wonder of the need to specify that.
$endgroup$
– Eevee Trainer
Apr 14 at 20:15
1
$begingroup$
I think the implication is that $n$ is not fixed, that it is simply a nonnegative integer.
$endgroup$
– Eevee Trainer
Apr 14 at 20:23
2
$begingroup$
The derivative of a constant, nonzero polynomial (of degree $0$) is the zero polynomial (of degree $-infty$). Surely then $f : P to P$ is not well defined at all, because some of its images are not in $P$?
$endgroup$
– wchargin
Apr 15 at 4:57
1
$begingroup$
You want to write $(f(p))(x)=p‘(x)$ or just $f(p)=p‘$.
$endgroup$
– Carsten S
Apr 15 at 6:31
2
2
$begingroup$
I interpreted it to mean that it is the set of all polynomials.
$endgroup$
– Tony S.F.
Apr 14 at 20:13
$begingroup$
I interpreted it to mean that it is the set of all polynomials.
$endgroup$
– Tony S.F.
Apr 14 at 20:13
1
1
$begingroup$
I read it more as the set of "all polynomials with degree that is a nonnegative integer." Though I do wonder of the need to specify that.
$endgroup$
– Eevee Trainer
Apr 14 at 20:15
$begingroup$
I read it more as the set of "all polynomials with degree that is a nonnegative integer." Though I do wonder of the need to specify that.
$endgroup$
– Eevee Trainer
Apr 14 at 20:15
1
1
$begingroup$
I think the implication is that $n$ is not fixed, that it is simply a nonnegative integer.
$endgroup$
– Eevee Trainer
Apr 14 at 20:23
$begingroup$
I think the implication is that $n$ is not fixed, that it is simply a nonnegative integer.
$endgroup$
– Eevee Trainer
Apr 14 at 20:23
2
2
$begingroup$
The derivative of a constant, nonzero polynomial (of degree $0$) is the zero polynomial (of degree $-infty$). Surely then $f : P to P$ is not well defined at all, because some of its images are not in $P$?
$endgroup$
– wchargin
Apr 15 at 4:57
$begingroup$
The derivative of a constant, nonzero polynomial (of degree $0$) is the zero polynomial (of degree $-infty$). Surely then $f : P to P$ is not well defined at all, because some of its images are not in $P$?
$endgroup$
– wchargin
Apr 15 at 4:57
1
1
$begingroup$
You want to write $(f(p))(x)=p‘(x)$ or just $f(p)=p‘$.
$endgroup$
– Carsten S
Apr 15 at 6:31
$begingroup$
You want to write $(f(p))(x)=p‘(x)$ or just $f(p)=p‘$.
$endgroup$
– Carsten S
Apr 15 at 6:31
|
show 8 more comments
4 Answers
4
active
oldest
votes
$begingroup$
To see that $f$ is a surjection we take an arbitrary element $y$ in $P$ and show that $exists xin P$ such that $f(x)=y$. This is what it means to be surjective; we cover the entire space with the image of $f$ on $P$.
Let $pin P$, i.e. $p$ is some polynomial $p(x)$. Then $p(x)$ has an antiderivative, $q(x) = int p(x)dx$. This antiderivative is a polynomial (easy to check), so it is in $P$. Then $f(q) = p$, and since $p$ was arbitrary, $f$ is therefore surjective.
Here is a more concrete analogy to help you understand what a surjection is.
Imagine you have two lists of names, one for adults and one for children. Imagine that someone says they have a way to assign names from the adult list to names in children list. Their way of assigning the names is surjective if every name is the children list is assigned at least one name in the adult list. This would mean that if you pick a name from the children list as random, there is at least one adult name corresponding to it.
answered Apr 14 at 20:13
Tony S.F.Tony S.F.
3,85121031
$endgroup$
$begingroup$
This is exactly the solution presented in my book (albeit reworded) yet for some reason I am having trouble understanding it. It seems to be that I am having a problem with the first two lines. Perhaps I do not properly understand what surjection is?
$endgroup$
– John Arg
Apr 14 at 20:23
$begingroup$
@JohnArg A surjection is simply a function onto, which means its image covers the whole codomain. In other words, it takes each value in the codomain. You may want to see also the comparision of Bijection, injection and surjection
$endgroup$
– CiaPan
Apr 15 at 10:37
add a comment |
$begingroup$
For any polynomial $p (x) $ there exists its integral $P (x) $, for which $p $ is a derivative: $$p (x)=P'(x)$$ and which itself is a real-coefficients polynomial, too $$P(x)in P$$ Hence each $p(x) in P$ is in the image of your function $f$. Then $f$ is surjective.
answered Apr 14 at 20:24
CiaPanCiaPan
10.4k11248
$endgroup$
add a comment |
$begingroup$
Think about what the inverse mapping for $f$ would be if indeed it would be a surjection: it would take the derivatives to the original function. Phrased another way, it would take a function to its antiderivative, right?
Well, we know that all polynomials have an antiderivative (which is itself a polynomial), and since the codomain of $f$ is basically the set of real polynomials, we have a pre-image for each polynomial in $P$ - its antiderivative. Thus, $f$ is surjective.
answered Apr 14 at 20:13
Eevee TrainerEevee Trainer
10.6k31842
$endgroup$
$begingroup$
What do you mean by pre-image?
$endgroup$
– John Arg
Apr 14 at 20:20
$begingroup$
You know how, say, $f(x)$ might be called the "image" for $x$? Same sort of deal: we would also say $x$ is the pre-image for $f(x)$. To use, arguably, more familiar terminology: inputs are to outputs, as pre-images are to images.
$endgroup$
– Eevee Trainer
Apr 14 at 20:21
add a comment |
$begingroup$
Your main problem seems to be with what surjection actually means. To add to the other, excellent answers, I will add short "translations" for the words function, surjection and injection in (hopefully) very clear language.
- "$f$ is a function": Every polynomial has a derivative (which is unique, and itself a polynomial as well).
- "$f$ is a surjection": Every polynomial is a derivative (of at least one other polynomial).
- "$f$ is an injection": The anti-derivative of a polynomial is unique (this one is not true).
You should maybe go back and try to understand these definitions, and how to prove that they are satisfied, in some easier examples -- I am sure there are some in your book.
answered Apr 15 at 8:57
NoiralefNoiralef
357112
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To see that $f$ is a surjection we take an arbitrary element $y$ in $P$ and show that $exists xin P$ such that $f(x)=y$. This is what it means to be surjective; we cover the entire space with the image of $f$ on $P$.
Let $pin P$, i.e. $p$ is some polynomial $p(x)$. Then $p(x)$ has an antiderivative, $q(x) = int p(x)dx$. This antiderivative is a polynomial (easy to check), so it is in $P$. Then $f(q) = p$, and since $p$ was arbitrary, $f$ is therefore surjective.
Here is a more concrete analogy to help you understand what a surjection is.
Imagine you have two lists of names, one for adults and one for children. Imagine that someone says they have a way to assign names from the adult list to names in children list. Their way of assigning the names is surjective if every name is the children list is assigned at least one name in the adult list. This would mean that if you pick a name from the children list as random, there is at least one adult name corresponding to it.
answered Apr 14 at 20:13
Tony S.F.Tony S.F.
3,85121031
$endgroup$
$begingroup$
This is exactly the solution presented in my book (albeit reworded) yet for some reason I am having trouble understanding it. It seems to be that I am having a problem with the first two lines. Perhaps I do not properly understand what surjection is?
$endgroup$
– John Arg
Apr 14 at 20:23
$begingroup$
@JohnArg A surjection is simply a function onto, which means its image covers the whole codomain. In other words, it takes each value in the codomain. You may want to see also the comparision of Bijection, injection and surjection
$endgroup$
– CiaPan
Apr 15 at 10:37
add a comment |
$begingroup$
To see that $f$ is a surjection we take an arbitrary element $y$ in $P$ and show that $exists xin P$ such that $f(x)=y$. This is what it means to be surjective; we cover the entire space with the image of $f$ on $P$.
Let $pin P$, i.e. $p$ is some polynomial $p(x)$. Then $p(x)$ has an antiderivative, $q(x) = int p(x)dx$. This antiderivative is a polynomial (easy to check), so it is in $P$. Then $f(q) = p$, and since $p$ was arbitrary, $f$ is therefore surjective.
Here is a more concrete analogy to help you understand what a surjection is.
Imagine you have two lists of names, one for adults and one for children. Imagine that someone says they have a way to assign names from the adult list to names in children list. Their way of assigning the names is surjective if every name is the children list is assigned at least one name in the adult list. This would mean that if you pick a name from the children list as random, there is at least one adult name corresponding to it.
answered Apr 14 at 20:13
Tony S.F.Tony S.F.
3,85121031
$endgroup$
$begingroup$
This is exactly the solution presented in my book (albeit reworded) yet for some reason I am having trouble understanding it. It seems to be that I am having a problem with the first two lines. Perhaps I do not properly understand what surjection is?
$endgroup$
– John Arg
Apr 14 at 20:23
$begingroup$
@JohnArg A surjection is simply a function onto, which means its image covers the whole codomain. In other words, it takes each value in the codomain. You may want to see also the comparision of Bijection, injection and surjection
$endgroup$
– CiaPan
Apr 15 at 10:37
add a comment |
$begingroup$
To see that $f$ is a surjection we take an arbitrary element $y$ in $P$ and show that $exists xin P$ such that $f(x)=y$. This is what it means to be surjective; we cover the entire space with the image of $f$ on $P$.
Let $pin P$, i.e. $p$ is some polynomial $p(x)$. Then $p(x)$ has an antiderivative, $q(x) = int p(x)dx$. This antiderivative is a polynomial (easy to check), so it is in $P$. Then $f(q) = p$, and since $p$ was arbitrary, $f$ is therefore surjective.
Here is a more concrete analogy to help you understand what a surjection is.
Imagine you have two lists of names, one for adults and one for children. Imagine that someone says they have a way to assign names from the adult list to names in children list. Their way of assigning the names is surjective if every name is the children list is assigned at least one name in the adult list. This would mean that if you pick a name from the children list as random, there is at least one adult name corresponding to it.
answered Apr 14 at 20:13
Tony S.F.Tony S.F.
3,85121031
$endgroup$
To see that $f$ is a surjection we take an arbitrary element $y$ in $P$ and show that $exists xin P$ such that $f(x)=y$. This is what it means to be surjective; we cover the entire space with the image of $f$ on $P$.
Let $pin P$, i.e. $p$ is some polynomial $p(x)$. Then $p(x)$ has an antiderivative, $q(x) = int p(x)dx$. This antiderivative is a polynomial (easy to check), so it is in $P$. Then $f(q) = p$, and since $p$ was arbitrary, $f$ is therefore surjective.
Here is a more concrete analogy to help you understand what a surjection is.
Imagine you have two lists of names, one for adults and one for children. Imagine that someone says they have a way to assign names from the adult list to names in children list. Their way of assigning the names is surjective if every name is the children list is assigned at least one name in the adult list. This would mean that if you pick a name from the children list as random, there is at least one adult name corresponding to it.
answered Apr 14 at 20:13
Tony S.F.Tony S.F.
3,85121031
edited Apr 14 at 20:26
answered Apr 14 at 20:13
Tony S.F.Tony S.F.
3,85121031
answered Apr 14 at 20:13
Tony S.F.Tony S.F.
3,85121031
answered Apr 14 at 20:13
Tony S.F.Tony S.F.
3,85121031
3,85121031
$begingroup$
This is exactly the solution presented in my book (albeit reworded) yet for some reason I am having trouble understanding it. It seems to be that I am having a problem with the first two lines. Perhaps I do not properly understand what surjection is?
$endgroup$
– John Arg
Apr 14 at 20:23
$begingroup$
@JohnArg A surjection is simply a function onto, which means its image covers the whole codomain. In other words, it takes each value in the codomain. You may want to see also the comparision of Bijection, injection and surjection
$endgroup$
– CiaPan
Apr 15 at 10:37
add a comment |
$begingroup$
This is exactly the solution presented in my book (albeit reworded) yet for some reason I am having trouble understanding it. It seems to be that I am having a problem with the first two lines. Perhaps I do not properly understand what surjection is?
$endgroup$
– John Arg
Apr 14 at 20:23
$begingroup$
@JohnArg A surjection is simply a function onto, which means its image covers the whole codomain. In other words, it takes each value in the codomain. You may want to see also the comparision of Bijection, injection and surjection
$endgroup$
– CiaPan
Apr 15 at 10:37
$begingroup$
This is exactly the solution presented in my book (albeit reworded) yet for some reason I am having trouble understanding it. It seems to be that I am having a problem with the first two lines. Perhaps I do not properly understand what surjection is?
$endgroup$
– John Arg
Apr 14 at 20:23
$begingroup$
This is exactly the solution presented in my book (albeit reworded) yet for some reason I am having trouble understanding it. It seems to be that I am having a problem with the first two lines. Perhaps I do not properly understand what surjection is?
$endgroup$
– John Arg
Apr 14 at 20:23
$begingroup$
@JohnArg A surjection is simply a function onto, which means its image covers the whole codomain. In other words, it takes each value in the codomain. You may want to see also the comparision of Bijection, injection and surjection
$endgroup$
– CiaPan
Apr 15 at 10:37
$begingroup$
@JohnArg A surjection is simply a function onto, which means its image covers the whole codomain. In other words, it takes each value in the codomain. You may want to see also the comparision of Bijection, injection and surjection
$endgroup$
– CiaPan
Apr 15 at 10:37
add a comment |
$begingroup$
For any polynomial $p (x) $ there exists its integral $P (x) $, for which $p $ is a derivative: $$p (x)=P'(x)$$ and which itself is a real-coefficients polynomial, too $$P(x)in P$$ Hence each $p(x) in P$ is in the image of your function $f$. Then $f$ is surjective.
answered Apr 14 at 20:24
CiaPanCiaPan
10.4k11248
$endgroup$
add a comment |
$begingroup$
For any polynomial $p (x) $ there exists its integral $P (x) $, for which $p $ is a derivative: $$p (x)=P'(x)$$ and which itself is a real-coefficients polynomial, too $$P(x)in P$$ Hence each $p(x) in P$ is in the image of your function $f$. Then $f$ is surjective.
answered Apr 14 at 20:24
CiaPanCiaPan
10.4k11248
$endgroup$
add a comment |
$begingroup$
For any polynomial $p (x) $ there exists its integral $P (x) $, for which $p $ is a derivative: $$p (x)=P'(x)$$ and which itself is a real-coefficients polynomial, too $$P(x)in P$$ Hence each $p(x) in P$ is in the image of your function $f$. Then $f$ is surjective.
answered Apr 14 at 20:24
CiaPanCiaPan
10.4k11248
$endgroup$
For any polynomial $p (x) $ there exists its integral $P (x) $, for which $p $ is a derivative: $$p (x)=P'(x)$$ and which itself is a real-coefficients polynomial, too $$P(x)in P$$ Hence each $p(x) in P$ is in the image of your function $f$. Then $f$ is surjective.
answered Apr 14 at 20:24
CiaPanCiaPan
10.4k11248
edited Apr 14 at 21:24
answered Apr 14 at 20:24
CiaPanCiaPan
10.4k11248
answered Apr 14 at 20:24
CiaPanCiaPan
10.4k11248
answered Apr 14 at 20:24
CiaPanCiaPan
10.4k11248
10.4k11248
add a comment |
add a comment |
$begingroup$
Think about what the inverse mapping for $f$ would be if indeed it would be a surjection: it would take the derivatives to the original function. Phrased another way, it would take a function to its antiderivative, right?
Well, we know that all polynomials have an antiderivative (which is itself a polynomial), and since the codomain of $f$ is basically the set of real polynomials, we have a pre-image for each polynomial in $P$ - its antiderivative. Thus, $f$ is surjective.
answered Apr 14 at 20:13
Eevee TrainerEevee Trainer
10.6k31842
$endgroup$
$begingroup$
What do you mean by pre-image?
$endgroup$
– John Arg
Apr 14 at 20:20
$begingroup$
You know how, say, $f(x)$ might be called the "image" for $x$? Same sort of deal: we would also say $x$ is the pre-image for $f(x)$. To use, arguably, more familiar terminology: inputs are to outputs, as pre-images are to images.
$endgroup$
– Eevee Trainer
Apr 14 at 20:21
add a comment |
$begingroup$
Think about what the inverse mapping for $f$ would be if indeed it would be a surjection: it would take the derivatives to the original function. Phrased another way, it would take a function to its antiderivative, right?
Well, we know that all polynomials have an antiderivative (which is itself a polynomial), and since the codomain of $f$ is basically the set of real polynomials, we have a pre-image for each polynomial in $P$ - its antiderivative. Thus, $f$ is surjective.
answered Apr 14 at 20:13
Eevee TrainerEevee Trainer
10.6k31842
$endgroup$
$begingroup$
What do you mean by pre-image?
$endgroup$
– John Arg
Apr 14 at 20:20
$begingroup$
You know how, say, $f(x)$ might be called the "image" for $x$? Same sort of deal: we would also say $x$ is the pre-image for $f(x)$. To use, arguably, more familiar terminology: inputs are to outputs, as pre-images are to images.
$endgroup$
– Eevee Trainer
Apr 14 at 20:21
add a comment |
$begingroup$
Think about what the inverse mapping for $f$ would be if indeed it would be a surjection: it would take the derivatives to the original function. Phrased another way, it would take a function to its antiderivative, right?
Well, we know that all polynomials have an antiderivative (which is itself a polynomial), and since the codomain of $f$ is basically the set of real polynomials, we have a pre-image for each polynomial in $P$ - its antiderivative. Thus, $f$ is surjective.
answered Apr 14 at 20:13
Eevee TrainerEevee Trainer
10.6k31842
$endgroup$
Think about what the inverse mapping for $f$ would be if indeed it would be a surjection: it would take the derivatives to the original function. Phrased another way, it would take a function to its antiderivative, right?
Well, we know that all polynomials have an antiderivative (which is itself a polynomial), and since the codomain of $f$ is basically the set of real polynomials, we have a pre-image for each polynomial in $P$ - its antiderivative. Thus, $f$ is surjective.
answered Apr 14 at 20:13
Eevee TrainerEevee Trainer
10.6k31842
answered Apr 14 at 20:13
Eevee TrainerEevee Trainer
10.6k31842
answered Apr 14 at 20:13
Eevee TrainerEevee Trainer
10.6k31842
answered Apr 14 at 20:13
Eevee TrainerEevee Trainer
10.6k31842
10.6k31842
$begingroup$
What do you mean by pre-image?
$endgroup$
– John Arg
Apr 14 at 20:20
$begingroup$
You know how, say, $f(x)$ might be called the "image" for $x$? Same sort of deal: we would also say $x$ is the pre-image for $f(x)$. To use, arguably, more familiar terminology: inputs are to outputs, as pre-images are to images.
$endgroup$
– Eevee Trainer
Apr 14 at 20:21
add a comment |
$begingroup$
What do you mean by pre-image?
$endgroup$
– John Arg
Apr 14 at 20:20
$begingroup$
You know how, say, $f(x)$ might be called the "image" for $x$? Same sort of deal: we would also say $x$ is the pre-image for $f(x)$. To use, arguably, more familiar terminology: inputs are to outputs, as pre-images are to images.
$endgroup$
– Eevee Trainer
Apr 14 at 20:21
$begingroup$
What do you mean by pre-image?
$endgroup$
– John Arg
Apr 14 at 20:20
$begingroup$
What do you mean by pre-image?
$endgroup$
– John Arg
Apr 14 at 20:20
$begingroup$
You know how, say, $f(x)$ might be called the "image" for $x$? Same sort of deal: we would also say $x$ is the pre-image for $f(x)$. To use, arguably, more familiar terminology: inputs are to outputs, as pre-images are to images.
$endgroup$
– Eevee Trainer
Apr 14 at 20:21
$begingroup$
You know how, say, $f(x)$ might be called the "image" for $x$? Same sort of deal: we would also say $x$ is the pre-image for $f(x)$. To use, arguably, more familiar terminology: inputs are to outputs, as pre-images are to images.
$endgroup$
– Eevee Trainer
Apr 14 at 20:21
add a comment |
$begingroup$
Your main problem seems to be with what surjection actually means. To add to the other, excellent answers, I will add short "translations" for the words function, surjection and injection in (hopefully) very clear language.
- "$f$ is a function": Every polynomial has a derivative (which is unique, and itself a polynomial as well).
- "$f$ is a surjection": Every polynomial is a derivative (of at least one other polynomial).
- "$f$ is an injection": The anti-derivative of a polynomial is unique (this one is not true).
You should maybe go back and try to understand these definitions, and how to prove that they are satisfied, in some easier examples -- I am sure there are some in your book.
answered Apr 15 at 8:57
NoiralefNoiralef
357112
$endgroup$
add a comment |
$begingroup$
Your main problem seems to be with what surjection actually means. To add to the other, excellent answers, I will add short "translations" for the words function, surjection and injection in (hopefully) very clear language.
- "$f$ is a function": Every polynomial has a derivative (which is unique, and itself a polynomial as well).
- "$f$ is a surjection": Every polynomial is a derivative (of at least one other polynomial).
- "$f$ is an injection": The anti-derivative of a polynomial is unique (this one is not true).
You should maybe go back and try to understand these definitions, and how to prove that they are satisfied, in some easier examples -- I am sure there are some in your book.
answered Apr 15 at 8:57
NoiralefNoiralef
357112
$endgroup$
add a comment |
$begingroup$
Your main problem seems to be with what surjection actually means. To add to the other, excellent answers, I will add short "translations" for the words function, surjection and injection in (hopefully) very clear language.
- "$f$ is a function": Every polynomial has a derivative (which is unique, and itself a polynomial as well).
- "$f$ is a surjection": Every polynomial is a derivative (of at least one other polynomial).
- "$f$ is an injection": The anti-derivative of a polynomial is unique (this one is not true).
You should maybe go back and try to understand these definitions, and how to prove that they are satisfied, in some easier examples -- I am sure there are some in your book.
answered Apr 15 at 8:57
NoiralefNoiralef
357112
$endgroup$
Your main problem seems to be with what surjection actually means. To add to the other, excellent answers, I will add short "translations" for the words function, surjection and injection in (hopefully) very clear language.
- "$f$ is a function": Every polynomial has a derivative (which is unique, and itself a polynomial as well).
- "$f$ is a surjection": Every polynomial is a derivative (of at least one other polynomial).
- "$f$ is an injection": The anti-derivative of a polynomial is unique (this one is not true).
You should maybe go back and try to understand these definitions, and how to prove that they are satisfied, in some easier examples -- I am sure there are some in your book.
answered Apr 15 at 8:57
NoiralefNoiralef
357112
answered Apr 15 at 8:57
NoiralefNoiralef
357112
answered Apr 15 at 8:57
NoiralefNoiralef
357112
answered Apr 15 at 8:57
NoiralefNoiralef
357112
357112
add a comment |
add a comment |
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2
$begingroup$
I interpreted it to mean that it is the set of all polynomials.
$endgroup$
– Tony S.F.
Apr 14 at 20:13
1
$begingroup$
I read it more as the set of "all polynomials with degree that is a nonnegative integer." Though I do wonder of the need to specify that.
$endgroup$
– Eevee Trainer
Apr 14 at 20:15
1
$begingroup$
I think the implication is that $n$ is not fixed, that it is simply a nonnegative integer.
$endgroup$
– Eevee Trainer
Apr 14 at 20:23
2
$begingroup$
The derivative of a constant, nonzero polynomial (of degree $0$) is the zero polynomial (of degree $-infty$). Surely then $f : P to P$ is not well defined at all, because some of its images are not in $P$?
$endgroup$
– wchargin
Apr 15 at 4:57
1
$begingroup$
You want to write $(f(p))(x)=p‘(x)$ or just $f(p)=p‘$.
$endgroup$
– Carsten S
Apr 15 at 6:31