An antiderivative of an odd function is even. Proof in general.most general antiderivative involving sec xHow to solve problem like this - even & odd functionIs my proof correct? Proof: $n^2$ is odd then $n$ is odd.Show that integral of even $times$ odd = 0, even $times$ even is not necessarily zeroAre there any exceptions to the rule where a product of two odd functions is even?if a function is bounded then both its even part and odd part are bounded?Prove that if $3n^2 + 2n$ is even, then $n$ is evenConsidering that $a$ and $b$ are odd functions, prove that the solution of the differential equation $y' + ay = b$ is evenHow to determine whether a function is even or odd in case the function has discontinuity at the origin?Determining if a function is even or odd using a system of equations and solving for unknown constants
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An antiderivative of an odd function is even. Proof in general.
most general antiderivative involving sec xHow to solve problem like this - even & odd functionIs my proof correct? Proof: $n^2$ is odd then $n$ is odd.Show that integral of even $times$ odd = 0, even $times$ even is not necessarily zeroAre there any exceptions to the rule where a product of two odd functions is even?if a function is bounded then both its even part and odd part are bounded?Prove that if $3n^2 + 2n$ is even, then $n$ is evenConsidering that $a$ and $b$ are odd functions, prove that the solution of the differential equation $y' + ay = b$ is evenHow to determine whether a function is even or odd in case the function has discontinuity at the origin?Determining if a function is even or odd using a system of equations and solving for unknown constants
$begingroup$
We have:
$int f(x)dx=F(x)+C$,
$f(x) = -f(-x)$ - odd function.
Proof that $F(x)$ - even function.
My suggetion (please, check it):
Since
$int f(x)dx = F(x) + C$,$int f(-x)dx = -F(-x) + C$.
Since
$f(x)=-f(-x)$,$int f(x)dx=-int f(-x)dx$
Hence,
$F(x) + C = F(-x) + C Rightarrow F(x)=F(-x)$
The nuance is that those constants are not necessarily the same. I mean, I know that this statement is correct, but I cant find the justification of this C-issue.
Even more. If eberything above is true, can we use the same technique to prove the reverse statement, i.e. an antiderivative of an even function is odd?
calculus integration proof-verification even-and-odd-functions
asked Mar 20 at 18:05
GeorgiiGeorgii
336
$endgroup$
add a comment |
$begingroup$
We have:
$int f(x)dx=F(x)+C$,
$f(x) = -f(-x)$ - odd function.
Proof that $F(x)$ - even function.
My suggetion (please, check it):
Since
$int f(x)dx = F(x) + C$,$int f(-x)dx = -F(-x) + C$.
Since
$f(x)=-f(-x)$,$int f(x)dx=-int f(-x)dx$
Hence,
$F(x) + C = F(-x) + C Rightarrow F(x)=F(-x)$
The nuance is that those constants are not necessarily the same. I mean, I know that this statement is correct, but I cant find the justification of this C-issue.
Even more. If eberything above is true, can we use the same technique to prove the reverse statement, i.e. an antiderivative of an even function is odd?
calculus integration proof-verification even-and-odd-functions
asked Mar 20 at 18:05
GeorgiiGeorgii
336
$endgroup$
$begingroup$
Looks correct. It might be clearer to write the integral as $int_0^x f(t) dt$ and then do your argument, since $int_0^-x f(t) dt = -int_-x^0 f(t) dt = C -F(-x)$.
$endgroup$
– Kyle C
Mar 20 at 18:08
$begingroup$
No, the problem appears to be asking about indefinite integrals (hence the constant $C$). I think step 1 of his proof is in need of justification, since no elementary property of indefinite integrals says that is true.
$endgroup$
– Mark Fischler
Mar 20 at 18:10
$begingroup$
@KyleC I'm not sure that it will still be the same problem. I try to provide a proof for an indefinite integral.
$endgroup$
– Georgii
Mar 20 at 18:19
$begingroup$
$int_a^x f(t) dt$ is by definition the indefinite integral. @MarkFischler, he has used change of variables with $u = -x$ to get step 1 (u-substitution)
$endgroup$
– Kyle C
Mar 20 at 18:38
add a comment |
$begingroup$
We have:
$int f(x)dx=F(x)+C$,
$f(x) = -f(-x)$ - odd function.
Proof that $F(x)$ - even function.
My suggetion (please, check it):
Since
$int f(x)dx = F(x) + C$,$int f(-x)dx = -F(-x) + C$.
Since
$f(x)=-f(-x)$,$int f(x)dx=-int f(-x)dx$
Hence,
$F(x) + C = F(-x) + C Rightarrow F(x)=F(-x)$
The nuance is that those constants are not necessarily the same. I mean, I know that this statement is correct, but I cant find the justification of this C-issue.
Even more. If eberything above is true, can we use the same technique to prove the reverse statement, i.e. an antiderivative of an even function is odd?
calculus integration proof-verification even-and-odd-functions
asked Mar 20 at 18:05
GeorgiiGeorgii
336
$endgroup$
We have:
$int f(x)dx=F(x)+C$,
$f(x) = -f(-x)$ - odd function.
Proof that $F(x)$ - even function.
My suggetion (please, check it):
Since
$int f(x)dx = F(x) + C$,$int f(-x)dx = -F(-x) + C$.
Since
$f(x)=-f(-x)$,$int f(x)dx=-int f(-x)dx$
Hence,
$F(x) + C = F(-x) + C Rightarrow F(x)=F(-x)$
The nuance is that those constants are not necessarily the same. I mean, I know that this statement is correct, but I cant find the justification of this C-issue.
Even more. If eberything above is true, can we use the same technique to prove the reverse statement, i.e. an antiderivative of an even function is odd?
calculus integration proof-verification even-and-odd-functions
calculus integration proof-verification even-and-odd-functions
asked Mar 20 at 18:05
GeorgiiGeorgii
336
asked Mar 20 at 18:05
GeorgiiGeorgii
336
edited Mar 20 at 18:33
Georgii
asked Mar 20 at 18:05
GeorgiiGeorgii
336
asked Mar 20 at 18:05
GeorgiiGeorgii
336
asked Mar 20 at 18:05
GeorgiiGeorgii
336
336
$begingroup$
Looks correct. It might be clearer to write the integral as $int_0^x f(t) dt$ and then do your argument, since $int_0^-x f(t) dt = -int_-x^0 f(t) dt = C -F(-x)$.
$endgroup$
– Kyle C
Mar 20 at 18:08
$begingroup$
No, the problem appears to be asking about indefinite integrals (hence the constant $C$). I think step 1 of his proof is in need of justification, since no elementary property of indefinite integrals says that is true.
$endgroup$
– Mark Fischler
Mar 20 at 18:10
$begingroup$
@KyleC I'm not sure that it will still be the same problem. I try to provide a proof for an indefinite integral.
$endgroup$
– Georgii
Mar 20 at 18:19
$begingroup$
$int_a^x f(t) dt$ is by definition the indefinite integral. @MarkFischler, he has used change of variables with $u = -x$ to get step 1 (u-substitution)
$endgroup$
– Kyle C
Mar 20 at 18:38
add a comment |
$begingroup$
Looks correct. It might be clearer to write the integral as $int_0^x f(t) dt$ and then do your argument, since $int_0^-x f(t) dt = -int_-x^0 f(t) dt = C -F(-x)$.
$endgroup$
– Kyle C
Mar 20 at 18:08
$begingroup$
No, the problem appears to be asking about indefinite integrals (hence the constant $C$). I think step 1 of his proof is in need of justification, since no elementary property of indefinite integrals says that is true.
$endgroup$
– Mark Fischler
Mar 20 at 18:10
$begingroup$
@KyleC I'm not sure that it will still be the same problem. I try to provide a proof for an indefinite integral.
$endgroup$
– Georgii
Mar 20 at 18:19
$begingroup$
$int_a^x f(t) dt$ is by definition the indefinite integral. @MarkFischler, he has used change of variables with $u = -x$ to get step 1 (u-substitution)
$endgroup$
– Kyle C
Mar 20 at 18:38
$begingroup$
Looks correct. It might be clearer to write the integral as $int_0^x f(t) dt$ and then do your argument, since $int_0^-x f(t) dt = -int_-x^0 f(t) dt = C -F(-x)$.
$endgroup$
– Kyle C
Mar 20 at 18:08
$begingroup$
Looks correct. It might be clearer to write the integral as $int_0^x f(t) dt$ and then do your argument, since $int_0^-x f(t) dt = -int_-x^0 f(t) dt = C -F(-x)$.
$endgroup$
– Kyle C
Mar 20 at 18:08
$begingroup$
No, the problem appears to be asking about indefinite integrals (hence the constant $C$). I think step 1 of his proof is in need of justification, since no elementary property of indefinite integrals says that is true.
$endgroup$
– Mark Fischler
Mar 20 at 18:10
$begingroup$
No, the problem appears to be asking about indefinite integrals (hence the constant $C$). I think step 1 of his proof is in need of justification, since no elementary property of indefinite integrals says that is true.
$endgroup$
– Mark Fischler
Mar 20 at 18:10
$begingroup$
@KyleC I'm not sure that it will still be the same problem. I try to provide a proof for an indefinite integral.
$endgroup$
– Georgii
Mar 20 at 18:19
$begingroup$
@KyleC I'm not sure that it will still be the same problem. I try to provide a proof for an indefinite integral.
$endgroup$
– Georgii
Mar 20 at 18:19
$begingroup$
$int_a^x f(t) dt$ is by definition the indefinite integral. @MarkFischler, he has used change of variables with $u = -x$ to get step 1 (u-substitution)
$endgroup$
– Kyle C
Mar 20 at 18:38
$begingroup$
$int_a^x f(t) dt$ is by definition the indefinite integral. @MarkFischler, he has used change of variables with $u = -x$ to get step 1 (u-substitution)
$endgroup$
– Kyle C
Mar 20 at 18:38
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your idea is correct, but can be streamlined.
You can prove that two differentiable functions are equal by proving that they have the same derivative and the same value at some point. Equivalently, that their difference is constant (derivative zero) and has value zero at some point.
Let $f$ be continuous over $mathbbR$ and odd. Let $F$ be an antiderivative of $f$. Then we can consider $G(x)=F(x)-F(-x)$. Then, using the chain rule,
$$
G'(x)=F'(x)+F'(-x)=f(x)+f(-x)=f(x)-f(x)=0
$$
and $G(0)=F(0)-F(0)=0$.
Can you prove that an antiderivative of an even function $f$ is odd? Not in general. There is only one antiderivative of an even function that's odd, precisely the only antiderivative $F$ such that $F(0)=0$.
answered Mar 20 at 18:38
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
Basically, in step $3$ you use the initial value for the odd function $f(-0)=-f(0)=0$:
$$f(-x)=-f(x) iff \
int f(-x)dx = int -f(x)dx iff \
-F(-x)+A=-F(x)+B iff \
requirecancelcancel-F(-0)+A=cancel-F(0)+B iff \
A=B.$$
answered Mar 22 at 10:26
farruhotafarruhota
21.7k2842
$endgroup$
add a comment |
$begingroup$
If $f$ is odd,
$$F(x):=int_0^x f(x),dx+Cimplies
\F(-x)=int_0^-xf(x),dx+C=-int_0^xf(-x),dx+C=F(x).$$
If $f$ is even,
$$F(x):=int_0^x f(x),dx+Cimplies F(0)=C$$ and in general $F$ is not odd.
answered Mar 22 at 10:34
Yves DaoustYves Daoust
132k676229
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your idea is correct, but can be streamlined.
You can prove that two differentiable functions are equal by proving that they have the same derivative and the same value at some point. Equivalently, that their difference is constant (derivative zero) and has value zero at some point.
Let $f$ be continuous over $mathbbR$ and odd. Let $F$ be an antiderivative of $f$. Then we can consider $G(x)=F(x)-F(-x)$. Then, using the chain rule,
$$
G'(x)=F'(x)+F'(-x)=f(x)+f(-x)=f(x)-f(x)=0
$$
and $G(0)=F(0)-F(0)=0$.
Can you prove that an antiderivative of an even function $f$ is odd? Not in general. There is only one antiderivative of an even function that's odd, precisely the only antiderivative $F$ such that $F(0)=0$.
answered Mar 20 at 18:38
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
Your idea is correct, but can be streamlined.
You can prove that two differentiable functions are equal by proving that they have the same derivative and the same value at some point. Equivalently, that their difference is constant (derivative zero) and has value zero at some point.
Let $f$ be continuous over $mathbbR$ and odd. Let $F$ be an antiderivative of $f$. Then we can consider $G(x)=F(x)-F(-x)$. Then, using the chain rule,
$$
G'(x)=F'(x)+F'(-x)=f(x)+f(-x)=f(x)-f(x)=0
$$
and $G(0)=F(0)-F(0)=0$.
Can you prove that an antiderivative of an even function $f$ is odd? Not in general. There is only one antiderivative of an even function that's odd, precisely the only antiderivative $F$ such that $F(0)=0$.
answered Mar 20 at 18:38
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
Your idea is correct, but can be streamlined.
You can prove that two differentiable functions are equal by proving that they have the same derivative and the same value at some point. Equivalently, that their difference is constant (derivative zero) and has value zero at some point.
Let $f$ be continuous over $mathbbR$ and odd. Let $F$ be an antiderivative of $f$. Then we can consider $G(x)=F(x)-F(-x)$. Then, using the chain rule,
$$
G'(x)=F'(x)+F'(-x)=f(x)+f(-x)=f(x)-f(x)=0
$$
and $G(0)=F(0)-F(0)=0$.
Can you prove that an antiderivative of an even function $f$ is odd? Not in general. There is only one antiderivative of an even function that's odd, precisely the only antiderivative $F$ such that $F(0)=0$.
answered Mar 20 at 18:38
egregegreg
185k1486206
$endgroup$
Your idea is correct, but can be streamlined.
You can prove that two differentiable functions are equal by proving that they have the same derivative and the same value at some point. Equivalently, that their difference is constant (derivative zero) and has value zero at some point.
Let $f$ be continuous over $mathbbR$ and odd. Let $F$ be an antiderivative of $f$. Then we can consider $G(x)=F(x)-F(-x)$. Then, using the chain rule,
$$
G'(x)=F'(x)+F'(-x)=f(x)+f(-x)=f(x)-f(x)=0
$$
and $G(0)=F(0)-F(0)=0$.
Can you prove that an antiderivative of an even function $f$ is odd? Not in general. There is only one antiderivative of an even function that's odd, precisely the only antiderivative $F$ such that $F(0)=0$.
answered Mar 20 at 18:38
egregegreg
185k1486206
answered Mar 20 at 18:38
egregegreg
185k1486206
answered Mar 20 at 18:38
egregegreg
185k1486206
answered Mar 20 at 18:38
egregegreg
185k1486206
185k1486206
add a comment |
add a comment |
$begingroup$
Basically, in step $3$ you use the initial value for the odd function $f(-0)=-f(0)=0$:
$$f(-x)=-f(x) iff \
int f(-x)dx = int -f(x)dx iff \
-F(-x)+A=-F(x)+B iff \
requirecancelcancel-F(-0)+A=cancel-F(0)+B iff \
A=B.$$
answered Mar 22 at 10:26
farruhotafarruhota
21.7k2842
$endgroup$
add a comment |
$begingroup$
Basically, in step $3$ you use the initial value for the odd function $f(-0)=-f(0)=0$:
$$f(-x)=-f(x) iff \
int f(-x)dx = int -f(x)dx iff \
-F(-x)+A=-F(x)+B iff \
requirecancelcancel-F(-0)+A=cancel-F(0)+B iff \
A=B.$$
answered Mar 22 at 10:26
farruhotafarruhota
21.7k2842
$endgroup$
add a comment |
$begingroup$
Basically, in step $3$ you use the initial value for the odd function $f(-0)=-f(0)=0$:
$$f(-x)=-f(x) iff \
int f(-x)dx = int -f(x)dx iff \
-F(-x)+A=-F(x)+B iff \
requirecancelcancel-F(-0)+A=cancel-F(0)+B iff \
A=B.$$
answered Mar 22 at 10:26
farruhotafarruhota
21.7k2842
$endgroup$
Basically, in step $3$ you use the initial value for the odd function $f(-0)=-f(0)=0$:
$$f(-x)=-f(x) iff \
int f(-x)dx = int -f(x)dx iff \
-F(-x)+A=-F(x)+B iff \
requirecancelcancel-F(-0)+A=cancel-F(0)+B iff \
A=B.$$
answered Mar 22 at 10:26
farruhotafarruhota
21.7k2842
answered Mar 22 at 10:26
farruhotafarruhota
21.7k2842
answered Mar 22 at 10:26
farruhotafarruhota
21.7k2842
answered Mar 22 at 10:26
farruhotafarruhota
21.7k2842
21.7k2842
add a comment |
add a comment |
$begingroup$
If $f$ is odd,
$$F(x):=int_0^x f(x),dx+Cimplies
\F(-x)=int_0^-xf(x),dx+C=-int_0^xf(-x),dx+C=F(x).$$
If $f$ is even,
$$F(x):=int_0^x f(x),dx+Cimplies F(0)=C$$ and in general $F$ is not odd.
answered Mar 22 at 10:34
Yves DaoustYves Daoust
132k676229
$endgroup$
add a comment |
$begingroup$
If $f$ is odd,
$$F(x):=int_0^x f(x),dx+Cimplies
\F(-x)=int_0^-xf(x),dx+C=-int_0^xf(-x),dx+C=F(x).$$
If $f$ is even,
$$F(x):=int_0^x f(x),dx+Cimplies F(0)=C$$ and in general $F$ is not odd.
answered Mar 22 at 10:34
Yves DaoustYves Daoust
132k676229
$endgroup$
add a comment |
$begingroup$
If $f$ is odd,
$$F(x):=int_0^x f(x),dx+Cimplies
\F(-x)=int_0^-xf(x),dx+C=-int_0^xf(-x),dx+C=F(x).$$
If $f$ is even,
$$F(x):=int_0^x f(x),dx+Cimplies F(0)=C$$ and in general $F$ is not odd.
answered Mar 22 at 10:34
Yves DaoustYves Daoust
132k676229
$endgroup$
If $f$ is odd,
$$F(x):=int_0^x f(x),dx+Cimplies
\F(-x)=int_0^-xf(x),dx+C=-int_0^xf(-x),dx+C=F(x).$$
If $f$ is even,
$$F(x):=int_0^x f(x),dx+Cimplies F(0)=C$$ and in general $F$ is not odd.
answered Mar 22 at 10:34
Yves DaoustYves Daoust
132k676229
answered Mar 22 at 10:34
Yves DaoustYves Daoust
132k676229
answered Mar 22 at 10:34
Yves DaoustYves Daoust
132k676229
answered Mar 22 at 10:34
Yves DaoustYves Daoust
132k676229
132k676229
add a comment |
add a comment |
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$begingroup$
Looks correct. It might be clearer to write the integral as $int_0^x f(t) dt$ and then do your argument, since $int_0^-x f(t) dt = -int_-x^0 f(t) dt = C -F(-x)$.
$endgroup$
– Kyle C
Mar 20 at 18:08
$begingroup$
No, the problem appears to be asking about indefinite integrals (hence the constant $C$). I think step 1 of his proof is in need of justification, since no elementary property of indefinite integrals says that is true.
$endgroup$
– Mark Fischler
Mar 20 at 18:10
$begingroup$
@KyleC I'm not sure that it will still be the same problem. I try to provide a proof for an indefinite integral.
$endgroup$
– Georgii
Mar 20 at 18:19
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$int_a^x f(t) dt$ is by definition the indefinite integral. @MarkFischler, he has used change of variables with $u = -x$ to get step 1 (u-substitution)
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– Kyle C
Mar 20 at 18:38