Let $f$ be continuous on $[0,1]$ such that $int_0^x f = int_x^1 f$, what is $f$?
$begingroup$
Let $f$ be continuous on $[0,1]$, and suppose that for all $x$, $0<x<1$,
$int_0^x f = int_x^1 f$
Can you determine $f$?
I've argued the following:
Since $f$ is continuous on $[0,1]$, it follows that $f$ has an antiderivative $F$ on $I=[0,1]$.
Evaluating both sides of the equation we get
$F(x)-F(0) = F(1)-F(x)$=
$2F(x) = F(1) + F(0)$
$F(x) = frac{F(1) + F(0)}{2}$
Taking the derivative of both sides we get
$F'(x) = f(x) = frac{F'(1) + F'(0)}{2} = 0$ since $F(1) + F(0)$ is a constant.
I'm not sure if I'm going about this problem correctly.
calculus analysis
$endgroup$
add a comment |
$begingroup$
Let $f$ be continuous on $[0,1]$, and suppose that for all $x$, $0<x<1$,
$int_0^x f = int_x^1 f$
Can you determine $f$?
I've argued the following:
Since $f$ is continuous on $[0,1]$, it follows that $f$ has an antiderivative $F$ on $I=[0,1]$.
Evaluating both sides of the equation we get
$F(x)-F(0) = F(1)-F(x)$=
$2F(x) = F(1) + F(0)$
$F(x) = frac{F(1) + F(0)}{2}$
Taking the derivative of both sides we get
$F'(x) = f(x) = frac{F'(1) + F'(0)}{2} = 0$ since $F(1) + F(0)$ is a constant.
I'm not sure if I'm going about this problem correctly.
calculus analysis
$endgroup$
3
$begingroup$
Aside from the "since $F(1) + F(0)$ missing a conclusion, yes this is correct. This is a ridiculously strong assumption: It says that no matter how you split the interval $[0, 1]$ into two subintervals, the mass that $f$ carries on each one is the same. That's too strong for any non-zero function.
$endgroup$
– T. Bongers
Dec 1 '18 at 1:19
add a comment |
$begingroup$
Let $f$ be continuous on $[0,1]$, and suppose that for all $x$, $0<x<1$,
$int_0^x f = int_x^1 f$
Can you determine $f$?
I've argued the following:
Since $f$ is continuous on $[0,1]$, it follows that $f$ has an antiderivative $F$ on $I=[0,1]$.
Evaluating both sides of the equation we get
$F(x)-F(0) = F(1)-F(x)$=
$2F(x) = F(1) + F(0)$
$F(x) = frac{F(1) + F(0)}{2}$
Taking the derivative of both sides we get
$F'(x) = f(x) = frac{F'(1) + F'(0)}{2} = 0$ since $F(1) + F(0)$ is a constant.
I'm not sure if I'm going about this problem correctly.
calculus analysis
$endgroup$
Let $f$ be continuous on $[0,1]$, and suppose that for all $x$, $0<x<1$,
$int_0^x f = int_x^1 f$
Can you determine $f$?
I've argued the following:
Since $f$ is continuous on $[0,1]$, it follows that $f$ has an antiderivative $F$ on $I=[0,1]$.
Evaluating both sides of the equation we get
$F(x)-F(0) = F(1)-F(x)$=
$2F(x) = F(1) + F(0)$
$F(x) = frac{F(1) + F(0)}{2}$
Taking the derivative of both sides we get
$F'(x) = f(x) = frac{F'(1) + F'(0)}{2} = 0$ since $F(1) + F(0)$ is a constant.
I'm not sure if I'm going about this problem correctly.
calculus analysis
calculus analysis
edited Dec 1 '18 at 1:21
K.M
asked Dec 1 '18 at 1:17
K.MK.M
686412
686412
3
$begingroup$
Aside from the "since $F(1) + F(0)$ missing a conclusion, yes this is correct. This is a ridiculously strong assumption: It says that no matter how you split the interval $[0, 1]$ into two subintervals, the mass that $f$ carries on each one is the same. That's too strong for any non-zero function.
$endgroup$
– T. Bongers
Dec 1 '18 at 1:19
add a comment |
3
$begingroup$
Aside from the "since $F(1) + F(0)$ missing a conclusion, yes this is correct. This is a ridiculously strong assumption: It says that no matter how you split the interval $[0, 1]$ into two subintervals, the mass that $f$ carries on each one is the same. That's too strong for any non-zero function.
$endgroup$
– T. Bongers
Dec 1 '18 at 1:19
3
3
$begingroup$
Aside from the "since $F(1) + F(0)$ missing a conclusion, yes this is correct. This is a ridiculously strong assumption: It says that no matter how you split the interval $[0, 1]$ into two subintervals, the mass that $f$ carries on each one is the same. That's too strong for any non-zero function.
$endgroup$
– T. Bongers
Dec 1 '18 at 1:19
$begingroup$
Aside from the "since $F(1) + F(0)$ missing a conclusion, yes this is correct. This is a ridiculously strong assumption: It says that no matter how you split the interval $[0, 1]$ into two subintervals, the mass that $f$ carries on each one is the same. That's too strong for any non-zero function.
$endgroup$
– T. Bongers
Dec 1 '18 at 1:19
add a comment |
2 Answers
2
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Your method is generally correct, but I think this problem is easier than you think. We have:
$$int_0^x f(t)dt=int_x^1 f(t)dt=-int_1^x f(t)dt$$
Use the fundamental theorem of calculus to take the derivative of both sides:
$$f(x)=-f(x)rightarrow 2f(x)=0rightarrow f(x)=0$$
Thus, $f(x)=0$ for all $xin [0, 1]$.
$endgroup$
1
$begingroup$
Beat me to it by 2 minutes. +1
$endgroup$
– triple_sec
Dec 1 '18 at 1:26
add a comment |
$begingroup$
The use of the fundamental theorem of calculus makes the solution simpler. Since the two functions $xmapstoint_0^x f$ and $xmapstoint_x^1 f$ are identical on the interval $(0,1)$, it follows that their derivatives are also equal: $$f(x)=-f(x)quadtext{for every $xin(0,1)$.}$$ Hence, $f(x)=0$ for every $xin(0,1)$ and, by continuity, also for the endpoints $x=0$ and $x=1$.
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add a comment |
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2 Answers
2
active
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2 Answers
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active
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$begingroup$
Your method is generally correct, but I think this problem is easier than you think. We have:
$$int_0^x f(t)dt=int_x^1 f(t)dt=-int_1^x f(t)dt$$
Use the fundamental theorem of calculus to take the derivative of both sides:
$$f(x)=-f(x)rightarrow 2f(x)=0rightarrow f(x)=0$$
Thus, $f(x)=0$ for all $xin [0, 1]$.
$endgroup$
1
$begingroup$
Beat me to it by 2 minutes. +1
$endgroup$
– triple_sec
Dec 1 '18 at 1:26
add a comment |
$begingroup$
Your method is generally correct, but I think this problem is easier than you think. We have:
$$int_0^x f(t)dt=int_x^1 f(t)dt=-int_1^x f(t)dt$$
Use the fundamental theorem of calculus to take the derivative of both sides:
$$f(x)=-f(x)rightarrow 2f(x)=0rightarrow f(x)=0$$
Thus, $f(x)=0$ for all $xin [0, 1]$.
$endgroup$
1
$begingroup$
Beat me to it by 2 minutes. +1
$endgroup$
– triple_sec
Dec 1 '18 at 1:26
add a comment |
$begingroup$
Your method is generally correct, but I think this problem is easier than you think. We have:
$$int_0^x f(t)dt=int_x^1 f(t)dt=-int_1^x f(t)dt$$
Use the fundamental theorem of calculus to take the derivative of both sides:
$$f(x)=-f(x)rightarrow 2f(x)=0rightarrow f(x)=0$$
Thus, $f(x)=0$ for all $xin [0, 1]$.
$endgroup$
Your method is generally correct, but I think this problem is easier than you think. We have:
$$int_0^x f(t)dt=int_x^1 f(t)dt=-int_1^x f(t)dt$$
Use the fundamental theorem of calculus to take the derivative of both sides:
$$f(x)=-f(x)rightarrow 2f(x)=0rightarrow f(x)=0$$
Thus, $f(x)=0$ for all $xin [0, 1]$.
answered Dec 1 '18 at 1:23
Noble MushtakNoble Mushtak
15.2k1735
15.2k1735
1
$begingroup$
Beat me to it by 2 minutes. +1
$endgroup$
– triple_sec
Dec 1 '18 at 1:26
add a comment |
1
$begingroup$
Beat me to it by 2 minutes. +1
$endgroup$
– triple_sec
Dec 1 '18 at 1:26
1
1
$begingroup$
Beat me to it by 2 minutes. +1
$endgroup$
– triple_sec
Dec 1 '18 at 1:26
$begingroup$
Beat me to it by 2 minutes. +1
$endgroup$
– triple_sec
Dec 1 '18 at 1:26
add a comment |
$begingroup$
The use of the fundamental theorem of calculus makes the solution simpler. Since the two functions $xmapstoint_0^x f$ and $xmapstoint_x^1 f$ are identical on the interval $(0,1)$, it follows that their derivatives are also equal: $$f(x)=-f(x)quadtext{for every $xin(0,1)$.}$$ Hence, $f(x)=0$ for every $xin(0,1)$ and, by continuity, also for the endpoints $x=0$ and $x=1$.
$endgroup$
add a comment |
$begingroup$
The use of the fundamental theorem of calculus makes the solution simpler. Since the two functions $xmapstoint_0^x f$ and $xmapstoint_x^1 f$ are identical on the interval $(0,1)$, it follows that their derivatives are also equal: $$f(x)=-f(x)quadtext{for every $xin(0,1)$.}$$ Hence, $f(x)=0$ for every $xin(0,1)$ and, by continuity, also for the endpoints $x=0$ and $x=1$.
$endgroup$
add a comment |
$begingroup$
The use of the fundamental theorem of calculus makes the solution simpler. Since the two functions $xmapstoint_0^x f$ and $xmapstoint_x^1 f$ are identical on the interval $(0,1)$, it follows that their derivatives are also equal: $$f(x)=-f(x)quadtext{for every $xin(0,1)$.}$$ Hence, $f(x)=0$ for every $xin(0,1)$ and, by continuity, also for the endpoints $x=0$ and $x=1$.
$endgroup$
The use of the fundamental theorem of calculus makes the solution simpler. Since the two functions $xmapstoint_0^x f$ and $xmapstoint_x^1 f$ are identical on the interval $(0,1)$, it follows that their derivatives are also equal: $$f(x)=-f(x)quadtext{for every $xin(0,1)$.}$$ Hence, $f(x)=0$ for every $xin(0,1)$ and, by continuity, also for the endpoints $x=0$ and $x=1$.
answered Dec 1 '18 at 1:26
triple_sectriple_sec
15.8k21851
15.8k21851
add a comment |
add a comment |
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$begingroup$
Aside from the "since $F(1) + F(0)$ missing a conclusion, yes this is correct. This is a ridiculously strong assumption: It says that no matter how you split the interval $[0, 1]$ into two subintervals, the mass that $f$ carries on each one is the same. That's too strong for any non-zero function.
$endgroup$
– T. Bongers
Dec 1 '18 at 1:19