Antiderivative of an odd function
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
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
46 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
23 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
20 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
46 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
23 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
20 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
46 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
23 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
20 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
46 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
46 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
23 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
23 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
20 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
20 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 38 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
25 mins ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043594%2fantiderivative-of-an-odd-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 38 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 38 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 38 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 38 mins ago
Fred
43.7k1644
answered 38 mins ago
Fred
43.7k1644
answered 38 mins ago
Fred
43.7k1644
answered 38 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
25 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
25 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
25 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
25 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
25 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
25 mins ago
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043594%2fantiderivative-of-an-odd-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
46 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
23 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
20 mins ago