Mixing
Applying FTC to Integral Equation (Spivak)
$begingroup$
The following is an exercise from Spivak's Calculus:
Find all continuous functions $f$ satisfying
$$int_0^xf(t)dt=(f(x))^2+C$$
for $Cneq 0$, assuming that $f$ has at most one zero.
I have several questions before bringing up the proof:
- What is significant about $Cneq 0$?
- What is significant about $f$ being continuous?
- What is significant about $f$ having at most one root?
Here is my proposed solution: (Which according to the answer key, is incorrect) By FTC, we know that $(f(x)))^2$ is a differentiable function, thus $(f(x))^2+C$ is also differentiable.
Then,
begin{align*}
frac{d}{dx}int_0^xf(t)dt & = frac{d}{dx}Big[(f(x))^2+CBig]
\ f(x) & = 2f(x)f'(x)
\ 2f(x)f'(x)-f(x) & =0
\ f(x)(2f'(x)-1) & =0
end{align*}
So either $f(x)=0$ or $f'(x)=1/2$.
The problem states that $f(x)=0$ for only one point, so $f'(x)=1/2$ everywhere else. Since $f=int_0^xf'(t)dt=int_0^x(1/2)dt=x/2+C$.
However, the answer key claims that $f$ is constant? I'm pretty confused, and I didn't use all the facts that the problem gave.
calculus integration proof-explanation integral-equations
edited Jan 22 at 19:34
José Carlos Santos
166k22132235
asked Jan 22 at 15:11
高田航高田航
1,360418
$endgroup$
add a comment |
$begingroup$
The following is an exercise from Spivak's Calculus:
Find all continuous functions $f$ satisfying
$$int_0^xf(t)dt=(f(x))^2+C$$
for $Cneq 0$, assuming that $f$ has at most one zero.
I have several questions before bringing up the proof:
- What is significant about $Cneq 0$?
- What is significant about $f$ being continuous?
- What is significant about $f$ having at most one root?
Here is my proposed solution: (Which according to the answer key, is incorrect) By FTC, we know that $(f(x)))^2$ is a differentiable function, thus $(f(x))^2+C$ is also differentiable.
Then,
begin{align*}
frac{d}{dx}int_0^xf(t)dt & = frac{d}{dx}Big[(f(x))^2+CBig]
\ f(x) & = 2f(x)f'(x)
\ 2f(x)f'(x)-f(x) & =0
\ f(x)(2f'(x)-1) & =0
end{align*}
So either $f(x)=0$ or $f'(x)=1/2$.
The problem states that $f(x)=0$ for only one point, so $f'(x)=1/2$ everywhere else. Since $f=int_0^xf'(t)dt=int_0^x(1/2)dt=x/2+C$.
However, the answer key claims that $f$ is constant? I'm pretty confused, and I didn't use all the facts that the problem gave.
calculus integration proof-explanation integral-equations
edited Jan 22 at 19:34
José Carlos Santos
166k22132235
asked Jan 22 at 15:11
高田航高田航
1,360418
$endgroup$
$begingroup$
Minor point here but I think you want $f(t) dt$ (or some other variable) inside the integral so that you don't have the same variable in your integrand and bounds
$endgroup$
– pwerth
Jan 22 at 15:15
$begingroup$
@pwerth noted, will edit.
$endgroup$
– 高田航
Jan 22 at 15:15
$begingroup$
$f(x)$ cannot be the zero function since $C ne 0$. That answers your first question. Also note that $f(0)^2 + C = 0$
$endgroup$
– Dylan
Jan 22 at 19:24
add a comment |
$begingroup$
The following is an exercise from Spivak's Calculus:
Find all continuous functions $f$ satisfying
$$int_0^xf(t)dt=(f(x))^2+C$$
for $Cneq 0$, assuming that $f$ has at most one zero.
I have several questions before bringing up the proof:
- What is significant about $Cneq 0$?
- What is significant about $f$ being continuous?
- What is significant about $f$ having at most one root?
Here is my proposed solution: (Which according to the answer key, is incorrect) By FTC, we know that $(f(x)))^2$ is a differentiable function, thus $(f(x))^2+C$ is also differentiable.
Then,
begin{align*}
frac{d}{dx}int_0^xf(t)dt & = frac{d}{dx}Big[(f(x))^2+CBig]
\ f(x) & = 2f(x)f'(x)
\ 2f(x)f'(x)-f(x) & =0
\ f(x)(2f'(x)-1) & =0
end{align*}
So either $f(x)=0$ or $f'(x)=1/2$.
The problem states that $f(x)=0$ for only one point, so $f'(x)=1/2$ everywhere else. Since $f=int_0^xf'(t)dt=int_0^x(1/2)dt=x/2+C$.
However, the answer key claims that $f$ is constant? I'm pretty confused, and I didn't use all the facts that the problem gave.
calculus integration proof-explanation integral-equations
edited Jan 22 at 19:34
José Carlos Santos
166k22132235
asked Jan 22 at 15:11
高田航高田航
1,360418
$endgroup$
The following is an exercise from Spivak's Calculus:
Find all continuous functions $f$ satisfying
$$int_0^xf(t)dt=(f(x))^2+C$$
for $Cneq 0$, assuming that $f$ has at most one zero.
I have several questions before bringing up the proof:
- What is significant about $Cneq 0$?
- What is significant about $f$ being continuous?
- What is significant about $f$ having at most one root?
Here is my proposed solution: (Which according to the answer key, is incorrect) By FTC, we know that $(f(x)))^2$ is a differentiable function, thus $(f(x))^2+C$ is also differentiable.
Then,
begin{align*}
frac{d}{dx}int_0^xf(t)dt & = frac{d}{dx}Big[(f(x))^2+CBig]
\ f(x) & = 2f(x)f'(x)
\ 2f(x)f'(x)-f(x) & =0
\ f(x)(2f'(x)-1) & =0
end{align*}
So either $f(x)=0$ or $f'(x)=1/2$.
The problem states that $f(x)=0$ for only one point, so $f'(x)=1/2$ everywhere else. Since $f=int_0^xf'(t)dt=int_0^x(1/2)dt=x/2+C$.
However, the answer key claims that $f$ is constant? I'm pretty confused, and I didn't use all the facts that the problem gave.
calculus integration proof-explanation integral-equations
calculus integration proof-explanation integral-equations
edited Jan 22 at 19:34
José Carlos Santos
166k22132235
asked Jan 22 at 15:11
高田航高田航
1,360418
edited Jan 22 at 19:34
José Carlos Santos
166k22132235
asked Jan 22 at 15:11
高田航高田航
1,360418
edited Jan 22 at 19:34
José Carlos Santos
166k22132235
edited Jan 22 at 19:34
José Carlos Santos
166k22132235
edited Jan 22 at 19:34
José Carlos Santos
166k22132235
166k22132235
asked Jan 22 at 15:11
高田航高田航
1,360418
asked Jan 22 at 15:11
高田航高田航
1,360418
asked Jan 22 at 15:11
高田航高田航
1,360418
1,360418
$begingroup$
Minor point here but I think you want $f(t) dt$ (or some other variable) inside the integral so that you don't have the same variable in your integrand and bounds
$endgroup$
– pwerth
Jan 22 at 15:15
$begingroup$
@pwerth noted, will edit.
$endgroup$
– 高田航
Jan 22 at 15:15
$begingroup$
$f(x)$ cannot be the zero function since $C ne 0$. That answers your first question. Also note that $f(0)^2 + C = 0$
$endgroup$
– Dylan
Jan 22 at 19:24
add a comment |
$begingroup$
Minor point here but I think you want $f(t) dt$ (or some other variable) inside the integral so that you don't have the same variable in your integrand and bounds
$endgroup$
– pwerth
Jan 22 at 15:15
$begingroup$
@pwerth noted, will edit.
$endgroup$
– 高田航
Jan 22 at 15:15
$begingroup$
$f(x)$ cannot be the zero function since $C ne 0$. That answers your first question. Also note that $f(0)^2 + C = 0$
$endgroup$
– Dylan
Jan 22 at 19:24
$begingroup$
Minor point here but I think you want $f(t) dt$ (or some other variable) inside the integral so that you don't have the same variable in your integrand and bounds
$endgroup$
– pwerth
Jan 22 at 15:15
$begingroup$
Minor point here but I think you want $f(t) dt$ (or some other variable) inside the integral so that you don't have the same variable in your integrand and bounds
$endgroup$
– pwerth
Jan 22 at 15:15
$begingroup$
@pwerth noted, will edit.
$endgroup$
– 高田航
Jan 22 at 15:15
$begingroup$
@pwerth noted, will edit.
$endgroup$
– 高田航
Jan 22 at 15:15
$begingroup$
$f(x)$ cannot be the zero function since $C ne 0$. That answers your first question. Also note that $f(0)^2 + C = 0$
$endgroup$
– Dylan
Jan 22 at 19:24
$begingroup$
$f(x)$ cannot be the zero function since $C ne 0$. That answers your first question. Also note that $f(0)^2 + C = 0$
$endgroup$
– Dylan
Jan 22 at 19:24
add a comment |
1 Answer
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votes
$begingroup$
First of all, note that the $C$ in your answer has nothing to do with the $C$ from the question. Furthermore, where did you read “assuming that $f$ has at most one zero”? I am looking at the third edition of Spivak's Calculus and I don't see it there.
Anyway, you were right when you got that $f(x)=frac x2$. But then$$int_0^x f(t),mathrm dt-bigl(f(x)bigr)^2=frac{x^2}4-frac{x^2}4=0,$$and $C$ is supposed to be different from $0$.
answered Jan 22 at 15:24
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
The part "assuming $f$ has at most one zero" is in the fourth edition.
$endgroup$
– 高田航
Jan 22 at 15:28
add a comment |
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1 Answer
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$begingroup$
First of all, note that the $C$ in your answer has nothing to do with the $C$ from the question. Furthermore, where did you read “assuming that $f$ has at most one zero”? I am looking at the third edition of Spivak's Calculus and I don't see it there.
Anyway, you were right when you got that $f(x)=frac x2$. But then$$int_0^x f(t),mathrm dt-bigl(f(x)bigr)^2=frac{x^2}4-frac{x^2}4=0,$$and $C$ is supposed to be different from $0$.
answered Jan 22 at 15:24
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
The part "assuming $f$ has at most one zero" is in the fourth edition.
$endgroup$
– 高田航
Jan 22 at 15:28
add a comment |
$begingroup$
First of all, note that the $C$ in your answer has nothing to do with the $C$ from the question. Furthermore, where did you read “assuming that $f$ has at most one zero”? I am looking at the third edition of Spivak's Calculus and I don't see it there.
Anyway, you were right when you got that $f(x)=frac x2$. But then$$int_0^x f(t),mathrm dt-bigl(f(x)bigr)^2=frac{x^2}4-frac{x^2}4=0,$$and $C$ is supposed to be different from $0$.
answered Jan 22 at 15:24
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
The part "assuming $f$ has at most one zero" is in the fourth edition.
$endgroup$
– 高田航
Jan 22 at 15:28
add a comment |
$begingroup$
First of all, note that the $C$ in your answer has nothing to do with the $C$ from the question. Furthermore, where did you read “assuming that $f$ has at most one zero”? I am looking at the third edition of Spivak's Calculus and I don't see it there.
Anyway, you were right when you got that $f(x)=frac x2$. But then$$int_0^x f(t),mathrm dt-bigl(f(x)bigr)^2=frac{x^2}4-frac{x^2}4=0,$$and $C$ is supposed to be different from $0$.
answered Jan 22 at 15:24
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
First of all, note that the $C$ in your answer has nothing to do with the $C$ from the question. Furthermore, where did you read “assuming that $f$ has at most one zero”? I am looking at the third edition of Spivak's Calculus and I don't see it there.
Anyway, you were right when you got that $f(x)=frac x2$. But then$$int_0^x f(t),mathrm dt-bigl(f(x)bigr)^2=frac{x^2}4-frac{x^2}4=0,$$and $C$ is supposed to be different from $0$.
answered Jan 22 at 15:24
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 15:24
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 15:24
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 15:24
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
$begingroup$
The part "assuming $f$ has at most one zero" is in the fourth edition.
$endgroup$
– 高田航
Jan 22 at 15:28
add a comment |
$begingroup$
The part "assuming $f$ has at most one zero" is in the fourth edition.
$endgroup$
– 高田航
Jan 22 at 15:28
$begingroup$
The part "assuming $f$ has at most one zero" is in the fourth edition.
$endgroup$
– 高田航
Jan 22 at 15:28
$begingroup$
The part "assuming $f$ has at most one zero" is in the fourth edition.
$endgroup$
– 高田航
Jan 22 at 15:28
add a comment |
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$begingroup$
Minor point here but I think you want $f(t) dt$ (or some other variable) inside the integral so that you don't have the same variable in your integrand and bounds
$endgroup$
– pwerth
Jan 22 at 15:15
$begingroup$
@pwerth noted, will edit.
$endgroup$
– 高田航
Jan 22 at 15:15
$begingroup$
$f(x)$ cannot be the zero function since $C ne 0$. That answers your first question. Also note that $f(0)^2 + C = 0$
$endgroup$
– Dylan
Jan 22 at 19:24