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
164k22131234
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
164k22131234
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
164k22131234
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
164k22131234
asked Jan 22 at 15:11
高田航高田航
1,360418
edited Jan 22 at 19:34
José Carlos Santos
164k22131234
asked Jan 22 at 15:11
高田航高田航
1,360418
edited Jan 22 at 19:34
José Carlos Santos
164k22131234
edited Jan 22 at 19:34
José Carlos Santos
164k22131234
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
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
1
active
oldest
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
164k22131234
$endgroup$
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3083283%2fapplying-ftc-to-integral-equation-spivak%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
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
164k22131234
$endgroup$
$begingroup$
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
164k22131234
$endgroup$
$begingroup$
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
164k22131234
$endgroup$
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
164k22131234
answered Jan 22 at 15:24
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 22 at 15:24
José Carlos SantosJosé Carlos Santos
164k22131234
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
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3083283%2fapplying-ftc-to-integral-equation-spivak%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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