If f is even integrable function on [0,a] prove that f is integrable on [-a,a] and that $int_{-a}^{0} f(x) dx...
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let f be an even function and suppose that f is integrable on [0,a].prove that f is integrable on [-a,a] and that $int_{-a}^{0} f(x) dx = int_{0}^{a} f(x) dx $
The answer is given below:
But I have a difficulty understanding the third line in the solution, does it contain a typo i.e. t must be -t in the second equality??
real-analysis calculus integration analysis
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add a comment |
$begingroup$
let f be an even function and suppose that f is integrable on [0,a].prove that f is integrable on [-a,a] and that $int_{-a}^{0} f(x) dx = int_{0}^{a} f(x) dx $
The answer is given below:
But I have a difficulty understanding the third line in the solution, does it contain a typo i.e. t must be -t in the second equality??
real-analysis calculus integration analysis
$endgroup$
$begingroup$
What is confusing about the third line?
$endgroup$
– Michael Burr
Dec 3 '18 at 10:28
add a comment |
$begingroup$
let f be an even function and suppose that f is integrable on [0,a].prove that f is integrable on [-a,a] and that $int_{-a}^{0} f(x) dx = int_{0}^{a} f(x) dx $
The answer is given below:
But I have a difficulty understanding the third line in the solution, does it contain a typo i.e. t must be -t in the second equality??
real-analysis calculus integration analysis
$endgroup$
let f be an even function and suppose that f is integrable on [0,a].prove that f is integrable on [-a,a] and that $int_{-a}^{0} f(x) dx = int_{0}^{a} f(x) dx $
The answer is given below:
But I have a difficulty understanding the third line in the solution, does it contain a typo i.e. t must be -t in the second equality??
real-analysis calculus integration analysis
real-analysis calculus integration analysis
asked Dec 3 '18 at 10:24
hopefullyhopefully
189113
189113
$begingroup$
What is confusing about the third line?
$endgroup$
– Michael Burr
Dec 3 '18 at 10:28
add a comment |
$begingroup$
What is confusing about the third line?
$endgroup$
– Michael Burr
Dec 3 '18 at 10:28
$begingroup$
What is confusing about the third line?
$endgroup$
– Michael Burr
Dec 3 '18 at 10:28
$begingroup$
What is confusing about the third line?
$endgroup$
– Michael Burr
Dec 3 '18 at 10:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, the second inequality contains a $t$. This is because $f$ is even, therefore, $f(-t)=f(t)$.
This means that the values $f$ takes on the interval $[x_{k-1}, x_k]$ are the same that $f$ takes on the interval $[-x_k, -x_{k-1}]$.
A strict proof of the claim would indeed require multiplying the inequalities, in particular,a more rigorous equality would be written like so:
$$begin{align}sup{f(t)| -x_k leq t leq -x_{k-1}} &= sup{f(-t)|-x_kleq t leq -x_{k-1}}\& = sup{f(-t)|x_{k-1}leq -t leq x_k}\&=sup{f(tau)|x_{k-1}leq tauleq x_k}end{align}$$
In line one, I use only the fact that $f$ is odd. In line $2$, I just multiply the inequality by $-1$, and in line $3$, I introduce a new variable $tau=-t$.
$endgroup$
$begingroup$
but we have to multiply the inequality by -1 in the second part
$endgroup$
– hopefully
Dec 3 '18 at 10:30
1
$begingroup$
@hopefully You don't multiply the inequality. You use the fact that for every $tin[-x_k, -x_{k-1}]$, there exists some $tauin[x_{k-1}, x_k]$ such that $f(t)=f(tau)$.
$endgroup$
– 5xum
Dec 3 '18 at 10:31
1
$begingroup$
@hopefully Technically, you do multiply the inequality, but you also introduce $-t$ into the argument of $f$. I updated my answer to show exactly what happens.
$endgroup$
– 5xum
Dec 3 '18 at 10:37
1
$begingroup$
@hopefully Well you won't have $f(t)=f(-t)$ any more. To see what exactly the difference will be, I advise you try to do it, and see for yourself!
$endgroup$
– 5xum
Dec 3 '18 at 11:20
1
$begingroup$
@hopefully That sounds about right, yes
$endgroup$
– 5xum
Dec 3 '18 at 11:45
|
show 7 more comments
Your Answer
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1 Answer
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active
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1 Answer
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oldest
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active
oldest
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$begingroup$
No, the second inequality contains a $t$. This is because $f$ is even, therefore, $f(-t)=f(t)$.
This means that the values $f$ takes on the interval $[x_{k-1}, x_k]$ are the same that $f$ takes on the interval $[-x_k, -x_{k-1}]$.
A strict proof of the claim would indeed require multiplying the inequalities, in particular,a more rigorous equality would be written like so:
$$begin{align}sup{f(t)| -x_k leq t leq -x_{k-1}} &= sup{f(-t)|-x_kleq t leq -x_{k-1}}\& = sup{f(-t)|x_{k-1}leq -t leq x_k}\&=sup{f(tau)|x_{k-1}leq tauleq x_k}end{align}$$
In line one, I use only the fact that $f$ is odd. In line $2$, I just multiply the inequality by $-1$, and in line $3$, I introduce a new variable $tau=-t$.
$endgroup$
$begingroup$
but we have to multiply the inequality by -1 in the second part
$endgroup$
– hopefully
Dec 3 '18 at 10:30
1
$begingroup$
@hopefully You don't multiply the inequality. You use the fact that for every $tin[-x_k, -x_{k-1}]$, there exists some $tauin[x_{k-1}, x_k]$ such that $f(t)=f(tau)$.
$endgroup$
– 5xum
Dec 3 '18 at 10:31
1
$begingroup$
@hopefully Technically, you do multiply the inequality, but you also introduce $-t$ into the argument of $f$. I updated my answer to show exactly what happens.
$endgroup$
– 5xum
Dec 3 '18 at 10:37
1
$begingroup$
@hopefully Well you won't have $f(t)=f(-t)$ any more. To see what exactly the difference will be, I advise you try to do it, and see for yourself!
$endgroup$
– 5xum
Dec 3 '18 at 11:20
1
$begingroup$
@hopefully That sounds about right, yes
$endgroup$
– 5xum
Dec 3 '18 at 11:45
|
show 7 more comments
$begingroup$
No, the second inequality contains a $t$. This is because $f$ is even, therefore, $f(-t)=f(t)$.
This means that the values $f$ takes on the interval $[x_{k-1}, x_k]$ are the same that $f$ takes on the interval $[-x_k, -x_{k-1}]$.
A strict proof of the claim would indeed require multiplying the inequalities, in particular,a more rigorous equality would be written like so:
$$begin{align}sup{f(t)| -x_k leq t leq -x_{k-1}} &= sup{f(-t)|-x_kleq t leq -x_{k-1}}\& = sup{f(-t)|x_{k-1}leq -t leq x_k}\&=sup{f(tau)|x_{k-1}leq tauleq x_k}end{align}$$
In line one, I use only the fact that $f$ is odd. In line $2$, I just multiply the inequality by $-1$, and in line $3$, I introduce a new variable $tau=-t$.
$endgroup$
$begingroup$
but we have to multiply the inequality by -1 in the second part
$endgroup$
– hopefully
Dec 3 '18 at 10:30
1
$begingroup$
@hopefully You don't multiply the inequality. You use the fact that for every $tin[-x_k, -x_{k-1}]$, there exists some $tauin[x_{k-1}, x_k]$ such that $f(t)=f(tau)$.
$endgroup$
– 5xum
Dec 3 '18 at 10:31
1
$begingroup$
@hopefully Technically, you do multiply the inequality, but you also introduce $-t$ into the argument of $f$. I updated my answer to show exactly what happens.
$endgroup$
– 5xum
Dec 3 '18 at 10:37
1
$begingroup$
@hopefully Well you won't have $f(t)=f(-t)$ any more. To see what exactly the difference will be, I advise you try to do it, and see for yourself!
$endgroup$
– 5xum
Dec 3 '18 at 11:20
1
$begingroup$
@hopefully That sounds about right, yes
$endgroup$
– 5xum
Dec 3 '18 at 11:45
|
show 7 more comments
$begingroup$
No, the second inequality contains a $t$. This is because $f$ is even, therefore, $f(-t)=f(t)$.
This means that the values $f$ takes on the interval $[x_{k-1}, x_k]$ are the same that $f$ takes on the interval $[-x_k, -x_{k-1}]$.
A strict proof of the claim would indeed require multiplying the inequalities, in particular,a more rigorous equality would be written like so:
$$begin{align}sup{f(t)| -x_k leq t leq -x_{k-1}} &= sup{f(-t)|-x_kleq t leq -x_{k-1}}\& = sup{f(-t)|x_{k-1}leq -t leq x_k}\&=sup{f(tau)|x_{k-1}leq tauleq x_k}end{align}$$
In line one, I use only the fact that $f$ is odd. In line $2$, I just multiply the inequality by $-1$, and in line $3$, I introduce a new variable $tau=-t$.
$endgroup$
No, the second inequality contains a $t$. This is because $f$ is even, therefore, $f(-t)=f(t)$.
This means that the values $f$ takes on the interval $[x_{k-1}, x_k]$ are the same that $f$ takes on the interval $[-x_k, -x_{k-1}]$.
A strict proof of the claim would indeed require multiplying the inequalities, in particular,a more rigorous equality would be written like so:
$$begin{align}sup{f(t)| -x_k leq t leq -x_{k-1}} &= sup{f(-t)|-x_kleq t leq -x_{k-1}}\& = sup{f(-t)|x_{k-1}leq -t leq x_k}\&=sup{f(tau)|x_{k-1}leq tauleq x_k}end{align}$$
In line one, I use only the fact that $f$ is odd. In line $2$, I just multiply the inequality by $-1$, and in line $3$, I introduce a new variable $tau=-t$.
edited Dec 3 '18 at 10:32
answered Dec 3 '18 at 10:29
5xum5xum
90.1k393161
90.1k393161
$begingroup$
but we have to multiply the inequality by -1 in the second part
$endgroup$
– hopefully
Dec 3 '18 at 10:30
1
$begingroup$
@hopefully You don't multiply the inequality. You use the fact that for every $tin[-x_k, -x_{k-1}]$, there exists some $tauin[x_{k-1}, x_k]$ such that $f(t)=f(tau)$.
$endgroup$
– 5xum
Dec 3 '18 at 10:31
1
$begingroup$
@hopefully Technically, you do multiply the inequality, but you also introduce $-t$ into the argument of $f$. I updated my answer to show exactly what happens.
$endgroup$
– 5xum
Dec 3 '18 at 10:37
1
$begingroup$
@hopefully Well you won't have $f(t)=f(-t)$ any more. To see what exactly the difference will be, I advise you try to do it, and see for yourself!
$endgroup$
– 5xum
Dec 3 '18 at 11:20
1
$begingroup$
@hopefully That sounds about right, yes
$endgroup$
– 5xum
Dec 3 '18 at 11:45
|
show 7 more comments
$begingroup$
but we have to multiply the inequality by -1 in the second part
$endgroup$
– hopefully
Dec 3 '18 at 10:30
1
$begingroup$
@hopefully You don't multiply the inequality. You use the fact that for every $tin[-x_k, -x_{k-1}]$, there exists some $tauin[x_{k-1}, x_k]$ such that $f(t)=f(tau)$.
$endgroup$
– 5xum
Dec 3 '18 at 10:31
1
$begingroup$
@hopefully Technically, you do multiply the inequality, but you also introduce $-t$ into the argument of $f$. I updated my answer to show exactly what happens.
$endgroup$
– 5xum
Dec 3 '18 at 10:37
1
$begingroup$
@hopefully Well you won't have $f(t)=f(-t)$ any more. To see what exactly the difference will be, I advise you try to do it, and see for yourself!
$endgroup$
– 5xum
Dec 3 '18 at 11:20
1
$begingroup$
@hopefully That sounds about right, yes
$endgroup$
– 5xum
Dec 3 '18 at 11:45
$begingroup$
but we have to multiply the inequality by -1 in the second part
$endgroup$
– hopefully
Dec 3 '18 at 10:30
$begingroup$
but we have to multiply the inequality by -1 in the second part
$endgroup$
– hopefully
Dec 3 '18 at 10:30
1
1
$begingroup$
@hopefully You don't multiply the inequality. You use the fact that for every $tin[-x_k, -x_{k-1}]$, there exists some $tauin[x_{k-1}, x_k]$ such that $f(t)=f(tau)$.
$endgroup$
– 5xum
Dec 3 '18 at 10:31
$begingroup$
@hopefully You don't multiply the inequality. You use the fact that for every $tin[-x_k, -x_{k-1}]$, there exists some $tauin[x_{k-1}, x_k]$ such that $f(t)=f(tau)$.
$endgroup$
– 5xum
Dec 3 '18 at 10:31
1
1
$begingroup$
@hopefully Technically, you do multiply the inequality, but you also introduce $-t$ into the argument of $f$. I updated my answer to show exactly what happens.
$endgroup$
– 5xum
Dec 3 '18 at 10:37
$begingroup$
@hopefully Technically, you do multiply the inequality, but you also introduce $-t$ into the argument of $f$. I updated my answer to show exactly what happens.
$endgroup$
– 5xum
Dec 3 '18 at 10:37
1
1
$begingroup$
@hopefully Well you won't have $f(t)=f(-t)$ any more. To see what exactly the difference will be, I advise you try to do it, and see for yourself!
$endgroup$
– 5xum
Dec 3 '18 at 11:20
$begingroup$
@hopefully Well you won't have $f(t)=f(-t)$ any more. To see what exactly the difference will be, I advise you try to do it, and see for yourself!
$endgroup$
– 5xum
Dec 3 '18 at 11:20
1
1
$begingroup$
@hopefully That sounds about right, yes
$endgroup$
– 5xum
Dec 3 '18 at 11:45
$begingroup$
@hopefully That sounds about right, yes
$endgroup$
– 5xum
Dec 3 '18 at 11:45
|
show 7 more comments
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$begingroup$
What is confusing about the third line?
$endgroup$
– Michael Burr
Dec 3 '18 at 10:28