Antiderivative of an odd function
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Is the antiderivative of an odd function even?
The answer given by the book is yes.
However, I found a counterexample defined in $mathbb{R}setminus {0}$:
$$f(x)=begin{cases}ln |x|+1& x<0\ln |x|&x>0end{cases}$$
Its derivative is $frac 1x$, which is an odd function.
Question: is my counterexample right?
real-analysis calculus integration
asked 1 hour ago
Kemono Chen
2,137435
add a comment |
up vote
2
down vote
favorite
Is the antiderivative of an odd function even?
The answer given by the book is yes.
However, I found a counterexample defined in $mathbb{R}setminus {0}$:
$$f(x)=begin{cases}ln |x|+1& x<0\ln |x|&x>0end{cases}$$
Its derivative is $frac 1x$, which is an odd function.
Question: is my counterexample right?
real-analysis calculus integration
asked 1 hour ago
Kemono Chen
2,137435
I think the question implies that the odd function in question must contain $0$ in its domain. Otherwise, you can't integrate it across a symmetric interval
– Dylan
48 mins ago
You can actually do this for any odd function if you allow a piecewise function definition, as there are infinitely many anti-derivatives for a given function --- simply pick two that differ by a constant, then piece them together. E.g. consider $F(x) = x^4 + [x>0]$, where $[cdot]$ is the Iverson bracket (equal to $1$ when the condition is true, otherwise $0$), which is an antiderivative of $f(x) = x^3$.
– apnorton
24 mins ago
@apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
– Kemono Chen
21 mins ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is the antiderivative of an odd function even?
The answer given by the book is yes.
However, I found a counterexample defined in $mathbb{R}setminus {0}$:
$$f(x)=begin{cases}ln |x|+1& x<0\ln |x|&x>0end{cases}$$
Its derivative is $frac 1x$, which is an odd function.
Question: is my counterexample right?
real-analysis calculus integration
asked 1 hour ago
Kemono Chen
2,137435
Is the antiderivative of an odd function even?
The answer given by the book is yes.
However, I found a counterexample defined in $mathbb{R}setminus {0}$:
$$f(x)=begin{cases}ln |x|+1& x<0\ln |x|&x>0end{cases}$$
Its derivative is $frac 1x$, which is an odd function.
Question: is my counterexample right?
real-analysis calculus integration
real-analysis calculus integration
asked 1 hour ago
Kemono Chen
2,137435
asked 1 hour ago
Kemono Chen
2,137435
asked 1 hour ago
Kemono Chen
2,137435
asked 1 hour ago
Kemono Chen
2,137435
asked 1 hour ago
Kemono Chen
2,137435
2,137435
I think the question implies that the odd function in question must contain $0$ in its domain. Otherwise, you can't integrate it across a symmetric interval
– Dylan
48 mins ago
You can actually do this for any odd function if you allow a piecewise function definition, as there are infinitely many anti-derivatives for a given function --- simply pick two that differ by a constant, then piece them together. E.g. consider $F(x) = x^4 + [x>0]$, where $[cdot]$ is the Iverson bracket (equal to $1$ when the condition is true, otherwise $0$), which is an antiderivative of $f(x) = x^3$.
– apnorton
24 mins ago
@apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
– Kemono Chen
21 mins ago
add a comment |
I think the question implies that the odd function in question must contain $0$ in its domain. Otherwise, you can't integrate it across a symmetric interval
– Dylan
48 mins ago
You can actually do this for any odd function if you allow a piecewise function definition, as there are infinitely many anti-derivatives for a given function --- simply pick two that differ by a constant, then piece them together. E.g. consider $F(x) = x^4 + [x>0]$, where $[cdot]$ is the Iverson bracket (equal to $1$ when the condition is true, otherwise $0$), which is an antiderivative of $f(x) = x^3$.
– apnorton
24 mins ago
@apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
– Kemono Chen
21 mins ago
I think the question implies that the odd function in question must contain $0$ in its domain. Otherwise, you can't integrate it across a symmetric interval
– Dylan
48 mins ago
I think the question implies that the odd function in question must contain $0$ in its domain. Otherwise, you can't integrate it across a symmetric interval
– Dylan
48 mins ago
You can actually do this for any odd function if you allow a piecewise function definition, as there are infinitely many anti-derivatives for a given function --- simply pick two that differ by a constant, then piece them together. E.g. consider $F(x) = x^4 + [x>0]$, where $[cdot]$ is the Iverson bracket (equal to $1$ when the condition is true, otherwise $0$), which is an antiderivative of $f(x) = x^3$.
– apnorton
24 mins ago
You can actually do this for any odd function if you allow a piecewise function definition, as there are infinitely many anti-derivatives for a given function --- simply pick two that differ by a constant, then piece them together. E.g. consider $F(x) = x^4 + [x>0]$, where $[cdot]$ is the Iverson bracket (equal to $1$ when the condition is true, otherwise $0$), which is an antiderivative of $f(x) = x^3$.
– apnorton
24 mins ago
@apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
– Kemono Chen
21 mins ago
@apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
– Kemono Chen
21 mins ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
I think that $f$ should be defined on an interval $I$ which contains $0$ and is symmetric to $0$. If $F$ is an antiderivative of $f$ on $I$, then there is a constant $c$ such that
$$F(x)=int_0^x f(t) dt+c.$$
If you now calculate $F(-x)$ with the substitution $s=-t$ you will get $F(-x)=-F(x).$
Try it !
answered 39 mins ago
Fred
43.7k1644
add a comment |
up vote
2
down vote
For a function $f$ to be even, we must have $f(-x) = f(x)$.
In your function $f$, consider $f(1)$ and $f(-1)$:
$$f(1) = ln|1| = ln(1) = 0$$
$$f(-1) = ln|-1|+1 = ln(1)+1 = 0+1 = 1$$
$f$ is not even, since $f(1) neq f(-1)$.
Or odd, for that matter, since $f(1) neq -f(-1)$.
answered 1 hour ago
Eevee Trainer
3,340225
What I unsure is the answer given by the book. $int_0^x f(x)dx=int_0^{-x}f(x)dx$. But the integral does not converge for this case. So is this "proof" given by the book wrong?
– Kemono Chen
1 hour ago
This answer isn't relevant to the question, as the question is claiming to show an odd function ($f(x) = 1/x$) with an anti-derivative that isn't even ($F(x) = text{piecewise function}$).
– apnorton
26 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I think that $f$ should be defined on an interval $I$ which contains $0$ and is symmetric to $0$. If $F$ is an antiderivative of $f$ on $I$, then there is a constant $c$ such that
$$F(x)=int_0^x f(t) dt+c.$$
If you now calculate $F(-x)$ with the substitution $s=-t$ you will get $F(-x)=-F(x).$
Try it !
answered 39 mins ago
Fred
43.7k1644
add a comment |
up vote
2
down vote
accepted
I think that $f$ should be defined on an interval $I$ which contains $0$ and is symmetric to $0$. If $F$ is an antiderivative of $f$ on $I$, then there is a constant $c$ such that
$$F(x)=int_0^x f(t) dt+c.$$
If you now calculate $F(-x)$ with the substitution $s=-t$ you will get $F(-x)=-F(x).$
Try it !
answered 39 mins ago
Fred
43.7k1644
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I think that $f$ should be defined on an interval $I$ which contains $0$ and is symmetric to $0$. If $F$ is an antiderivative of $f$ on $I$, then there is a constant $c$ such that
$$F(x)=int_0^x f(t) dt+c.$$
If you now calculate $F(-x)$ with the substitution $s=-t$ you will get $F(-x)=-F(x).$
Try it !
answered 39 mins ago
Fred
43.7k1644
I think that $f$ should be defined on an interval $I$ which contains $0$ and is symmetric to $0$. If $F$ is an antiderivative of $f$ on $I$, then there is a constant $c$ such that
$$F(x)=int_0^x f(t) dt+c.$$
If you now calculate $F(-x)$ with the substitution $s=-t$ you will get $F(-x)=-F(x).$
Try it !
answered 39 mins ago
Fred
43.7k1644
answered 39 mins ago
Fred
43.7k1644
answered 39 mins ago
Fred
43.7k1644
answered 39 mins ago
Fred
43.7k1644
43.7k1644
add a comment |
add a comment |
up vote
2
down vote
For a function $f$ to be even, we must have $f(-x) = f(x)$.
In your function $f$, consider $f(1)$ and $f(-1)$:
$$f(1) = ln|1| = ln(1) = 0$$
$$f(-1) = ln|-1|+1 = ln(1)+1 = 0+1 = 1$$
$f$ is not even, since $f(1) neq f(-1)$.
Or odd, for that matter, since $f(1) neq -f(-1)$.
answered 1 hour ago
Eevee Trainer
3,340225
What I unsure is the answer given by the book. $int_0^x f(x)dx=int_0^{-x}f(x)dx$. But the integral does not converge for this case. So is this "proof" given by the book wrong?
– Kemono Chen
1 hour ago
This answer isn't relevant to the question, as the question is claiming to show an odd function ($f(x) = 1/x$) with an anti-derivative that isn't even ($F(x) = text{piecewise function}$).
– apnorton
26 mins ago
add a comment |
up vote
2
down vote
For a function $f$ to be even, we must have $f(-x) = f(x)$.
In your function $f$, consider $f(1)$ and $f(-1)$:
$$f(1) = ln|1| = ln(1) = 0$$
$$f(-1) = ln|-1|+1 = ln(1)+1 = 0+1 = 1$$
$f$ is not even, since $f(1) neq f(-1)$.
Or odd, for that matter, since $f(1) neq -f(-1)$.
answered 1 hour ago
Eevee Trainer
3,340225
What I unsure is the answer given by the book. $int_0^x f(x)dx=int_0^{-x}f(x)dx$. But the integral does not converge for this case. So is this "proof" given by the book wrong?
– Kemono Chen
1 hour ago
This answer isn't relevant to the question, as the question is claiming to show an odd function ($f(x) = 1/x$) with an anti-derivative that isn't even ($F(x) = text{piecewise function}$).
– apnorton
26 mins ago
add a comment |
up vote
2
down vote
up vote
2
down vote
For a function $f$ to be even, we must have $f(-x) = f(x)$.
In your function $f$, consider $f(1)$ and $f(-1)$:
$$f(1) = ln|1| = ln(1) = 0$$
$$f(-1) = ln|-1|+1 = ln(1)+1 = 0+1 = 1$$
$f$ is not even, since $f(1) neq f(-1)$.
Or odd, for that matter, since $f(1) neq -f(-1)$.
answered 1 hour ago
Eevee Trainer
3,340225
For a function $f$ to be even, we must have $f(-x) = f(x)$.
In your function $f$, consider $f(1)$ and $f(-1)$:
$$f(1) = ln|1| = ln(1) = 0$$
$$f(-1) = ln|-1|+1 = ln(1)+1 = 0+1 = 1$$
$f$ is not even, since $f(1) neq f(-1)$.
Or odd, for that matter, since $f(1) neq -f(-1)$.
answered 1 hour ago
Eevee Trainer
3,340225
answered 1 hour ago
Eevee Trainer
3,340225
answered 1 hour ago
Eevee Trainer
3,340225
answered 1 hour ago
Eevee Trainer
3,340225
3,340225
What I unsure is the answer given by the book. $int_0^x f(x)dx=int_0^{-x}f(x)dx$. But the integral does not converge for this case. So is this "proof" given by the book wrong?
– Kemono Chen
1 hour ago
This answer isn't relevant to the question, as the question is claiming to show an odd function ($f(x) = 1/x$) with an anti-derivative that isn't even ($F(x) = text{piecewise function}$).
– apnorton
26 mins ago
add a comment |
What I unsure is the answer given by the book. $int_0^x f(x)dx=int_0^{-x}f(x)dx$. But the integral does not converge for this case. So is this "proof" given by the book wrong?
– Kemono Chen
1 hour ago
This answer isn't relevant to the question, as the question is claiming to show an odd function ($f(x) = 1/x$) with an anti-derivative that isn't even ($F(x) = text{piecewise function}$).
– apnorton
26 mins ago
What I unsure is the answer given by the book. $int_0^x f(x)dx=int_0^{-x}f(x)dx$. But the integral does not converge for this case. So is this "proof" given by the book wrong?
– Kemono Chen
1 hour ago
What I unsure is the answer given by the book. $int_0^x f(x)dx=int_0^{-x}f(x)dx$. But the integral does not converge for this case. So is this "proof" given by the book wrong?
– Kemono Chen
1 hour ago
This answer isn't relevant to the question, as the question is claiming to show an odd function ($f(x) = 1/x$) with an anti-derivative that isn't even ($F(x) = text{piecewise function}$).
– apnorton
26 mins ago
This answer isn't relevant to the question, as the question is claiming to show an odd function ($f(x) = 1/x$) with an anti-derivative that isn't even ($F(x) = text{piecewise function}$).
– apnorton
26 mins ago
add a comment |
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I think the question implies that the odd function in question must contain $0$ in its domain. Otherwise, you can't integrate it across a symmetric interval
– Dylan
48 mins ago
You can actually do this for any odd function if you allow a piecewise function definition, as there are infinitely many anti-derivatives for a given function --- simply pick two that differ by a constant, then piece them together. E.g. consider $F(x) = x^4 + [x>0]$, where $[cdot]$ is the Iverson bracket (equal to $1$ when the condition is true, otherwise $0$), which is an antiderivative of $f(x) = x^3$.
– apnorton
24 mins ago
@apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
– Kemono Chen
21 mins ago