Applying FTC to Integral Equation (Spivak)












1












$begingroup$


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:




  1. What is significant about $Cneq 0$?

  2. What is significant about $f$ being continuous?

  3. 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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















1












$begingroup$


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:




  1. What is significant about $Cneq 0$?

  2. What is significant about $f$ being continuous?

  3. 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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$


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:




  1. What is significant about $Cneq 0$?

  2. What is significant about $f$ being continuous?

  3. 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







share|cite|improve this question











$endgroup$




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:




  1. What is significant about $Cneq 0$?

  2. What is significant about $f$ being continuous?

  3. 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 19:34









José Carlos Santos

164k22131234




164k22131234










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




















  • $begingroup$
    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












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 $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The part "assuming $f$ has at most one zero" is in the fourth edition.
    $endgroup$
    – 高田航
    Jan 22 at 15:28













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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 $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The part "assuming $f$ has at most one zero" is in the fourth edition.
    $endgroup$
    – 高田航
    Jan 22 at 15:28


















1












$begingroup$

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 $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The part "assuming $f$ has at most one zero" is in the fourth edition.
    $endgroup$
    – 高田航
    Jan 22 at 15:28
















1












1








1





$begingroup$

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 $0$.






share|cite|improve this answer









$endgroup$



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 $0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 22 at 15:24









José Carlos SantosJosé Carlos Santos

164k22131234




164k22131234












  • $begingroup$
    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






$begingroup$
The part "assuming $f$ has at most one zero" is in the fourth edition.
$endgroup$
– 高田航
Jan 22 at 15:28




















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