Mixing
If $f'(x)$ is odd then is $f(x)$ even?
$begingroup$
I am trying to prove but every proof I encounter can also prove that if $f'(x)$ is even then $f(x)$ is odd and this is not correct ($x^3 + 1$ for example).
Thanks!
calculus
edited Jan 6 at 10:26
Saad
19.7k92352
asked Jan 2 '15 at 18:27
Lior CohenLior Cohen
52
$endgroup$
add a comment |
$begingroup$
I am trying to prove but every proof I encounter can also prove that if $f'(x)$ is even then $f(x)$ is odd and this is not correct ($x^3 + 1$ for example).
Thanks!
calculus
edited Jan 6 at 10:26
Saad
19.7k92352
asked Jan 2 '15 at 18:27
Lior CohenLior Cohen
52
$endgroup$
1
$begingroup$
$x^3+1$ is neither odd nor even. What proofs have you encountered?
$endgroup$
– abiessu
Jan 2 '15 at 18:30
1
$begingroup$
The obvious proof is by integrating $f'$ from $0$ to $x$. On the other hand, when $f'$ is even, you should get $f(x)-f(0)=-big(f(-x)-f(0)big)$, which does not imply that $f$ is odd.
$endgroup$
– Rahul
Jan 2 '15 at 18:35
2
$begingroup$
@Henrik $cos x$ is even.
$endgroup$
– Mark Bennet
Jan 2 '15 at 18:46
$begingroup$
math.stackexchange.com/a/357232, when I tried this solution for both cases it succeeded. Im curios what am I doing wrong.. thanks for the comments and solution though! It's really helpful!
$endgroup$
– Lior Cohen
Jan 2 '15 at 19:21
add a comment |
$begingroup$
I am trying to prove but every proof I encounter can also prove that if $f'(x)$ is even then $f(x)$ is odd and this is not correct ($x^3 + 1$ for example).
Thanks!
calculus
edited Jan 6 at 10:26
Saad
19.7k92352
asked Jan 2 '15 at 18:27
Lior CohenLior Cohen
52
$endgroup$
I am trying to prove but every proof I encounter can also prove that if $f'(x)$ is even then $f(x)$ is odd and this is not correct ($x^3 + 1$ for example).
Thanks!
calculus
calculus
edited Jan 6 at 10:26
Saad
19.7k92352
asked Jan 2 '15 at 18:27
Lior CohenLior Cohen
52
edited Jan 6 at 10:26
Saad
19.7k92352
asked Jan 2 '15 at 18:27
Lior CohenLior Cohen
52
edited Jan 6 at 10:26
Saad
19.7k92352
edited Jan 6 at 10:26
Saad
19.7k92352
edited Jan 6 at 10:26
Saad
19.7k92352
19.7k92352
asked Jan 2 '15 at 18:27
Lior CohenLior Cohen
52
asked Jan 2 '15 at 18:27
Lior CohenLior Cohen
52
asked Jan 2 '15 at 18:27
Lior CohenLior Cohen
52
52
1
$begingroup$
$x^3+1$ is neither odd nor even. What proofs have you encountered?
$endgroup$
– abiessu
Jan 2 '15 at 18:30
1
$begingroup$
The obvious proof is by integrating $f'$ from $0$ to $x$. On the other hand, when $f'$ is even, you should get $f(x)-f(0)=-big(f(-x)-f(0)big)$, which does not imply that $f$ is odd.
$endgroup$
– Rahul
Jan 2 '15 at 18:35
2
$begingroup$
@Henrik $cos x$ is even.
$endgroup$
– Mark Bennet
Jan 2 '15 at 18:46
$begingroup$
math.stackexchange.com/a/357232, when I tried this solution for both cases it succeeded. Im curios what am I doing wrong.. thanks for the comments and solution though! It's really helpful!
$endgroup$
– Lior Cohen
Jan 2 '15 at 19:21
add a comment |
1
$begingroup$
$x^3+1$ is neither odd nor even. What proofs have you encountered?
$endgroup$
– abiessu
Jan 2 '15 at 18:30
1
$begingroup$
The obvious proof is by integrating $f'$ from $0$ to $x$. On the other hand, when $f'$ is even, you should get $f(x)-f(0)=-big(f(-x)-f(0)big)$, which does not imply that $f$ is odd.
$endgroup$
– Rahul
Jan 2 '15 at 18:35
2
$begingroup$
@Henrik $cos x$ is even.
$endgroup$
– Mark Bennet
Jan 2 '15 at 18:46
$begingroup$
math.stackexchange.com/a/357232, when I tried this solution for both cases it succeeded. Im curios what am I doing wrong.. thanks for the comments and solution though! It's really helpful!
$endgroup$
– Lior Cohen
Jan 2 '15 at 19:21
1
1
$begingroup$
$x^3+1$ is neither odd nor even. What proofs have you encountered?
$endgroup$
– abiessu
Jan 2 '15 at 18:30
$begingroup$
$x^3+1$ is neither odd nor even. What proofs have you encountered?
$endgroup$
– abiessu
Jan 2 '15 at 18:30
1
1
$begingroup$
The obvious proof is by integrating $f'$ from $0$ to $x$. On the other hand, when $f'$ is even, you should get $f(x)-f(0)=-big(f(-x)-f(0)big)$, which does not imply that $f$ is odd.
$endgroup$
– Rahul
Jan 2 '15 at 18:35
$begingroup$
The obvious proof is by integrating $f'$ from $0$ to $x$. On the other hand, when $f'$ is even, you should get $f(x)-f(0)=-big(f(-x)-f(0)big)$, which does not imply that $f$ is odd.
$endgroup$
– Rahul
Jan 2 '15 at 18:35
2
2
$begingroup$
@Henrik $cos x$ is even.
$endgroup$
– Mark Bennet
Jan 2 '15 at 18:46
$begingroup$
@Henrik $cos x$ is even.
$endgroup$
– Mark Bennet
Jan 2 '15 at 18:46
$begingroup$
math.stackexchange.com/a/357232, when I tried this solution for both cases it succeeded. Im curios what am I doing wrong.. thanks for the comments and solution though! It's really helpful!
$endgroup$
– Lior Cohen
Jan 2 '15 at 19:21
$begingroup$
math.stackexchange.com/a/357232, when I tried this solution for both cases it succeeded. Im curios what am I doing wrong.. thanks for the comments and solution though! It's really helpful!
$endgroup$
– Lior Cohen
Jan 2 '15 at 19:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose $f'(x)$ is odd. Then $f'(-x) = -f'(x)$.
Consider that by the fundamental theorem of calculus, $f(t) - f(0)= int_{0}^t f'(x), dx$. Now let $t mapsto -t$, and we have
$$f(-t) - f(0) = int_{0}^{-t} f'(x), dx = -int_{-t}^0 f'(x), dx = int_{-t}^0 f'(-x), dx.$$
Substitute $u = -x$, and we get
$$f(-t) - f(0) = -int_{t}^0 f'(u), du = int_0^t f'(u), du.$$
So $f(-t) = f(t)$.
answered Jan 2 '15 at 19:10
EmilyEmily
29.4k468111
$endgroup$
add a comment |
$begingroup$
Let $g(x)=f(-x)$, then $g'(x)=[f(-x)]'=-f'(-x)$. Since $f'(x)$ is odd, $f'(x)=-f'(-x)=g'(x)$ and $f'(0)=g'(0)=0$. Thus $f(x)=g(x)=f(-x)$. Thus $f(x)$ is even.
answered Jan 2 '15 at 19:14
DashermanDasherman
1,020817
$endgroup$
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',
autoActivateHeartbeat: false,
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%2f1088772%2fif-fx-is-odd-then-is-fx-even%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
$begingroup$
Suppose $f'(x)$ is odd. Then $f'(-x) = -f'(x)$.
Consider that by the fundamental theorem of calculus, $f(t) - f(0)= int_{0}^t f'(x), dx$. Now let $t mapsto -t$, and we have
$$f(-t) - f(0) = int_{0}^{-t} f'(x), dx = -int_{-t}^0 f'(x), dx = int_{-t}^0 f'(-x), dx.$$
Substitute $u = -x$, and we get
$$f(-t) - f(0) = -int_{t}^0 f'(u), du = int_0^t f'(u), du.$$
So $f(-t) = f(t)$.
answered Jan 2 '15 at 19:10
EmilyEmily
29.4k468111
$endgroup$
add a comment |
$begingroup$
Suppose $f'(x)$ is odd. Then $f'(-x) = -f'(x)$.
Consider that by the fundamental theorem of calculus, $f(t) - f(0)= int_{0}^t f'(x), dx$. Now let $t mapsto -t$, and we have
$$f(-t) - f(0) = int_{0}^{-t} f'(x), dx = -int_{-t}^0 f'(x), dx = int_{-t}^0 f'(-x), dx.$$
Substitute $u = -x$, and we get
$$f(-t) - f(0) = -int_{t}^0 f'(u), du = int_0^t f'(u), du.$$
So $f(-t) = f(t)$.
answered Jan 2 '15 at 19:10
EmilyEmily
29.4k468111
$endgroup$
add a comment |
$begingroup$
Suppose $f'(x)$ is odd. Then $f'(-x) = -f'(x)$.
Consider that by the fundamental theorem of calculus, $f(t) - f(0)= int_{0}^t f'(x), dx$. Now let $t mapsto -t$, and we have
$$f(-t) - f(0) = int_{0}^{-t} f'(x), dx = -int_{-t}^0 f'(x), dx = int_{-t}^0 f'(-x), dx.$$
Substitute $u = -x$, and we get
$$f(-t) - f(0) = -int_{t}^0 f'(u), du = int_0^t f'(u), du.$$
So $f(-t) = f(t)$.
answered Jan 2 '15 at 19:10
EmilyEmily
29.4k468111
$endgroup$
Suppose $f'(x)$ is odd. Then $f'(-x) = -f'(x)$.
Consider that by the fundamental theorem of calculus, $f(t) - f(0)= int_{0}^t f'(x), dx$. Now let $t mapsto -t$, and we have
$$f(-t) - f(0) = int_{0}^{-t} f'(x), dx = -int_{-t}^0 f'(x), dx = int_{-t}^0 f'(-x), dx.$$
Substitute $u = -x$, and we get
$$f(-t) - f(0) = -int_{t}^0 f'(u), du = int_0^t f'(u), du.$$
So $f(-t) = f(t)$.
answered Jan 2 '15 at 19:10
EmilyEmily
29.4k468111
answered Jan 2 '15 at 19:10
EmilyEmily
29.4k468111
answered Jan 2 '15 at 19:10
EmilyEmily
29.4k468111
answered Jan 2 '15 at 19:10
EmilyEmily
29.4k468111
29.4k468111
add a comment |
add a comment |
$begingroup$
Let $g(x)=f(-x)$, then $g'(x)=[f(-x)]'=-f'(-x)$. Since $f'(x)$ is odd, $f'(x)=-f'(-x)=g'(x)$ and $f'(0)=g'(0)=0$. Thus $f(x)=g(x)=f(-x)$. Thus $f(x)$ is even.
answered Jan 2 '15 at 19:14
DashermanDasherman
1,020817
$endgroup$
add a comment |
$begingroup$
Let $g(x)=f(-x)$, then $g'(x)=[f(-x)]'=-f'(-x)$. Since $f'(x)$ is odd, $f'(x)=-f'(-x)=g'(x)$ and $f'(0)=g'(0)=0$. Thus $f(x)=g(x)=f(-x)$. Thus $f(x)$ is even.
answered Jan 2 '15 at 19:14
DashermanDasherman
1,020817
$endgroup$
add a comment |
$begingroup$
Let $g(x)=f(-x)$, then $g'(x)=[f(-x)]'=-f'(-x)$. Since $f'(x)$ is odd, $f'(x)=-f'(-x)=g'(x)$ and $f'(0)=g'(0)=0$. Thus $f(x)=g(x)=f(-x)$. Thus $f(x)$ is even.
answered Jan 2 '15 at 19:14
DashermanDasherman
1,020817
$endgroup$
Let $g(x)=f(-x)$, then $g'(x)=[f(-x)]'=-f'(-x)$. Since $f'(x)$ is odd, $f'(x)=-f'(-x)=g'(x)$ and $f'(0)=g'(0)=0$. Thus $f(x)=g(x)=f(-x)$. Thus $f(x)$ is even.
answered Jan 2 '15 at 19:14
DashermanDasherman
1,020817
edited Jan 2 '15 at 19:21
answered Jan 2 '15 at 19:14
DashermanDasherman
1,020817
answered Jan 2 '15 at 19:14
DashermanDasherman
1,020817
answered Jan 2 '15 at 19:14
DashermanDasherman
1,020817
1,020817
add a comment |
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.
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%2f1088772%2fif-fx-is-odd-then-is-fx-even%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
1
$begingroup$
$x^3+1$ is neither odd nor even. What proofs have you encountered?
$endgroup$
– abiessu
Jan 2 '15 at 18:30
1
$begingroup$
The obvious proof is by integrating $f'$ from $0$ to $x$. On the other hand, when $f'$ is even, you should get $f(x)-f(0)=-big(f(-x)-f(0)big)$, which does not imply that $f$ is odd.
$endgroup$
– Rahul
Jan 2 '15 at 18:35
2
$begingroup$
@Henrik $cos x$ is even.
$endgroup$
– Mark Bennet
Jan 2 '15 at 18:46
$begingroup$
math.stackexchange.com/a/357232, when I tried this solution for both cases it succeeded. Im curios what am I doing wrong.. thanks for the comments and solution though! It's really helpful!
$endgroup$
– Lior Cohen
Jan 2 '15 at 19:21