Is the following solution correct? Show that $f(x)=0$ for all real numbers $x$.
$begingroup$
I have a calculus 1 exam soon, and I just read a solution to a problem from last year's exam. I'm pretty sure the solution is almost correct, but not correct. Here's the problem:
$f:mathbb{R} rightarrow mathbb{R}$ is a function. There is a polynomial $P(x)=a_n cdot x^n+...+a_0$, $n$ is odd, $a_n neq 0$, and $P$ satisfies that for any natural number $n$ and a real number $x$, $|f^{(n)}(x)| leq |P(x)|$.
Show that for all real numbers $x$, $f(x)=0$. Hint: Show that there exists $x_0$ such that $f^{(n)}(x_0)=0$ for all values of $n$.
Here's the (wrong) solution: First we'll prove the hint: WLOG $a_n>0$, and then $lim_{x to infty} P(x) = infty$, and $lim_{x to -infty} P(x) = -infty$, so since $P(x)$ is continuous there exists $x_0$ such that $P(x_0)=0$. Now it is easy to show that $x_0$ is the needed point in the hint. Now, look at Taylor Polynomial of order $n$ of $f$ around $x_0$, call it $P_n(x)$. It holds that $P_n(x)=0$, So the remainder $R_n(x)$ is just $f(x)$. Take a real number $x$. We'll show that $f(x)=0$ According to lagrange remainder theorem, there exists $c_x$ between $x$ and $x_0$ such that $f(x)=R_n(x)=frac{f^{(n+1)}(c_x)}{(n+1)!} cdot(x-x_0)^{n+1}$. Now, $|f^{n+1}(c_x)| leq |P(c_x)|$, and that means that $|R_n(x)| leq |P(c_x)| cdot frac{(x-x_0)^{n+1}}{(n+1)!}$. Since $P(c_x)$ is constant, $lim_{n to infty}|R_n(x)|=0$, but $lim_{n to infty}|R_n(x)|=f(x)$, so $f(x)=0$.
I think that the wrong part here is the use of the wrong fact that $P(c_x)$ is a constant, because $c_x$ also depends on $n$, so it needs to be $c_{n,x}$. Of course this can be easily fixed because since $c_{n,x}$ is between $x$ and $x_0$ for all $n$, it's easy to find a constant $M$ such that $P(c_{n,x})<M$ for all $n$. I just want to make sure that I understand it. Am I right, and the solution is not precisely accurate?Thank you.
calculus limits taylor-expansion
asked Jan 20 at 18:36
OmerOmer
3619
$endgroup$
add a comment |
$begingroup$
I have a calculus 1 exam soon, and I just read a solution to a problem from last year's exam. I'm pretty sure the solution is almost correct, but not correct. Here's the problem:
$f:mathbb{R} rightarrow mathbb{R}$ is a function. There is a polynomial $P(x)=a_n cdot x^n+...+a_0$, $n$ is odd, $a_n neq 0$, and $P$ satisfies that for any natural number $n$ and a real number $x$, $|f^{(n)}(x)| leq |P(x)|$.
Show that for all real numbers $x$, $f(x)=0$. Hint: Show that there exists $x_0$ such that $f^{(n)}(x_0)=0$ for all values of $n$.
Here's the (wrong) solution: First we'll prove the hint: WLOG $a_n>0$, and then $lim_{x to infty} P(x) = infty$, and $lim_{x to -infty} P(x) = -infty$, so since $P(x)$ is continuous there exists $x_0$ such that $P(x_0)=0$. Now it is easy to show that $x_0$ is the needed point in the hint. Now, look at Taylor Polynomial of order $n$ of $f$ around $x_0$, call it $P_n(x)$. It holds that $P_n(x)=0$, So the remainder $R_n(x)$ is just $f(x)$. Take a real number $x$. We'll show that $f(x)=0$ According to lagrange remainder theorem, there exists $c_x$ between $x$ and $x_0$ such that $f(x)=R_n(x)=frac{f^{(n+1)}(c_x)}{(n+1)!} cdot(x-x_0)^{n+1}$. Now, $|f^{n+1}(c_x)| leq |P(c_x)|$, and that means that $|R_n(x)| leq |P(c_x)| cdot frac{(x-x_0)^{n+1}}{(n+1)!}$. Since $P(c_x)$ is constant, $lim_{n to infty}|R_n(x)|=0$, but $lim_{n to infty}|R_n(x)|=f(x)$, so $f(x)=0$.
I think that the wrong part here is the use of the wrong fact that $P(c_x)$ is a constant, because $c_x$ also depends on $n$, so it needs to be $c_{n,x}$. Of course this can be easily fixed because since $c_{n,x}$ is between $x$ and $x_0$ for all $n$, it's easy to find a constant $M$ such that $P(c_{n,x})<M$ for all $n$. I just want to make sure that I understand it. Am I right, and the solution is not precisely accurate?Thank you.
calculus limits taylor-expansion
asked Jan 20 at 18:36
OmerOmer
3619
$endgroup$
add a comment |
$begingroup$
I have a calculus 1 exam soon, and I just read a solution to a problem from last year's exam. I'm pretty sure the solution is almost correct, but not correct. Here's the problem:
$f:mathbb{R} rightarrow mathbb{R}$ is a function. There is a polynomial $P(x)=a_n cdot x^n+...+a_0$, $n$ is odd, $a_n neq 0$, and $P$ satisfies that for any natural number $n$ and a real number $x$, $|f^{(n)}(x)| leq |P(x)|$.
Show that for all real numbers $x$, $f(x)=0$. Hint: Show that there exists $x_0$ such that $f^{(n)}(x_0)=0$ for all values of $n$.
Here's the (wrong) solution: First we'll prove the hint: WLOG $a_n>0$, and then $lim_{x to infty} P(x) = infty$, and $lim_{x to -infty} P(x) = -infty$, so since $P(x)$ is continuous there exists $x_0$ such that $P(x_0)=0$. Now it is easy to show that $x_0$ is the needed point in the hint. Now, look at Taylor Polynomial of order $n$ of $f$ around $x_0$, call it $P_n(x)$. It holds that $P_n(x)=0$, So the remainder $R_n(x)$ is just $f(x)$. Take a real number $x$. We'll show that $f(x)=0$ According to lagrange remainder theorem, there exists $c_x$ between $x$ and $x_0$ such that $f(x)=R_n(x)=frac{f^{(n+1)}(c_x)}{(n+1)!} cdot(x-x_0)^{n+1}$. Now, $|f^{n+1}(c_x)| leq |P(c_x)|$, and that means that $|R_n(x)| leq |P(c_x)| cdot frac{(x-x_0)^{n+1}}{(n+1)!}$. Since $P(c_x)$ is constant, $lim_{n to infty}|R_n(x)|=0$, but $lim_{n to infty}|R_n(x)|=f(x)$, so $f(x)=0$.
I think that the wrong part here is the use of the wrong fact that $P(c_x)$ is a constant, because $c_x$ also depends on $n$, so it needs to be $c_{n,x}$. Of course this can be easily fixed because since $c_{n,x}$ is between $x$ and $x_0$ for all $n$, it's easy to find a constant $M$ such that $P(c_{n,x})<M$ for all $n$. I just want to make sure that I understand it. Am I right, and the solution is not precisely accurate?Thank you.
calculus limits taylor-expansion
asked Jan 20 at 18:36
OmerOmer
3619
$endgroup$
I have a calculus 1 exam soon, and I just read a solution to a problem from last year's exam. I'm pretty sure the solution is almost correct, but not correct. Here's the problem:
$f:mathbb{R} rightarrow mathbb{R}$ is a function. There is a polynomial $P(x)=a_n cdot x^n+...+a_0$, $n$ is odd, $a_n neq 0$, and $P$ satisfies that for any natural number $n$ and a real number $x$, $|f^{(n)}(x)| leq |P(x)|$.
Show that for all real numbers $x$, $f(x)=0$. Hint: Show that there exists $x_0$ such that $f^{(n)}(x_0)=0$ for all values of $n$.
Here's the (wrong) solution: First we'll prove the hint: WLOG $a_n>0$, and then $lim_{x to infty} P(x) = infty$, and $lim_{x to -infty} P(x) = -infty$, so since $P(x)$ is continuous there exists $x_0$ such that $P(x_0)=0$. Now it is easy to show that $x_0$ is the needed point in the hint. Now, look at Taylor Polynomial of order $n$ of $f$ around $x_0$, call it $P_n(x)$. It holds that $P_n(x)=0$, So the remainder $R_n(x)$ is just $f(x)$. Take a real number $x$. We'll show that $f(x)=0$ According to lagrange remainder theorem, there exists $c_x$ between $x$ and $x_0$ such that $f(x)=R_n(x)=frac{f^{(n+1)}(c_x)}{(n+1)!} cdot(x-x_0)^{n+1}$. Now, $|f^{n+1}(c_x)| leq |P(c_x)|$, and that means that $|R_n(x)| leq |P(c_x)| cdot frac{(x-x_0)^{n+1}}{(n+1)!}$. Since $P(c_x)$ is constant, $lim_{n to infty}|R_n(x)|=0$, but $lim_{n to infty}|R_n(x)|=f(x)$, so $f(x)=0$.
I think that the wrong part here is the use of the wrong fact that $P(c_x)$ is a constant, because $c_x$ also depends on $n$, so it needs to be $c_{n,x}$. Of course this can be easily fixed because since $c_{n,x}$ is between $x$ and $x_0$ for all $n$, it's easy to find a constant $M$ such that $P(c_{n,x})<M$ for all $n$. I just want to make sure that I understand it. Am I right, and the solution is not precisely accurate?Thank you.
calculus limits taylor-expansion
calculus limits taylor-expansion
asked Jan 20 at 18:36
OmerOmer
3619
asked Jan 20 at 18:36
OmerOmer
3619
asked Jan 20 at 18:36
OmerOmer
3619
asked Jan 20 at 18:36
OmerOmer
3619
asked Jan 20 at 18:36
OmerOmer
3619
3619
add a comment |
add a comment |
1 Answer
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I don't think the claim is correct unless you know more about $f$, like it being analytic. Here is a counterexample. Let $P(x)=x$. Then let $$f(x)=begin {cases} 0& |x=0 \ e^{-1/x^2} &xneq 0end {cases}$$
A plot is below. Blue is $|P(x)|=|x|$ and orange is $f(x)$. $f(x)$ is a standard example of a function that has all derivatives at every point, all derivatives are $0$ at $x=0$, but is not constant. I have not checked that all derivatives are less than $|x|$ at all points, but I believe it to be true.
answered Jan 20 at 19:05
Ross MillikanRoss Millikan
297k23198371
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$begingroup$
I don't think the claim is correct unless you know more about $f$, like it being analytic. Here is a counterexample. Let $P(x)=x$. Then let $$f(x)=begin {cases} 0& |x=0 \ e^{-1/x^2} &xneq 0end {cases}$$
A plot is below. Blue is $|P(x)|=|x|$ and orange is $f(x)$. $f(x)$ is a standard example of a function that has all derivatives at every point, all derivatives are $0$ at $x=0$, but is not constant. I have not checked that all derivatives are less than $|x|$ at all points, but I believe it to be true.
answered Jan 20 at 19:05
Ross MillikanRoss Millikan
297k23198371
$endgroup$
add a comment |
$begingroup$
I don't think the claim is correct unless you know more about $f$, like it being analytic. Here is a counterexample. Let $P(x)=x$. Then let $$f(x)=begin {cases} 0& |x=0 \ e^{-1/x^2} &xneq 0end {cases}$$
A plot is below. Blue is $|P(x)|=|x|$ and orange is $f(x)$. $f(x)$ is a standard example of a function that has all derivatives at every point, all derivatives are $0$ at $x=0$, but is not constant. I have not checked that all derivatives are less than $|x|$ at all points, but I believe it to be true.
answered Jan 20 at 19:05
Ross MillikanRoss Millikan
297k23198371
$endgroup$
add a comment |
$begingroup$
I don't think the claim is correct unless you know more about $f$, like it being analytic. Here is a counterexample. Let $P(x)=x$. Then let $$f(x)=begin {cases} 0& |x=0 \ e^{-1/x^2} &xneq 0end {cases}$$
A plot is below. Blue is $|P(x)|=|x|$ and orange is $f(x)$. $f(x)$ is a standard example of a function that has all derivatives at every point, all derivatives are $0$ at $x=0$, but is not constant. I have not checked that all derivatives are less than $|x|$ at all points, but I believe it to be true.
answered Jan 20 at 19:05
Ross MillikanRoss Millikan
297k23198371
$endgroup$
I don't think the claim is correct unless you know more about $f$, like it being analytic. Here is a counterexample. Let $P(x)=x$. Then let $$f(x)=begin {cases} 0& |x=0 \ e^{-1/x^2} &xneq 0end {cases}$$
A plot is below. Blue is $|P(x)|=|x|$ and orange is $f(x)$. $f(x)$ is a standard example of a function that has all derivatives at every point, all derivatives are $0$ at $x=0$, but is not constant. I have not checked that all derivatives are less than $|x|$ at all points, but I believe it to be true.
answered Jan 20 at 19:05
Ross MillikanRoss Millikan
297k23198371
answered Jan 20 at 19:05
Ross MillikanRoss Millikan
297k23198371
answered Jan 20 at 19:05
Ross MillikanRoss Millikan
297k23198371
answered Jan 20 at 19:05
Ross MillikanRoss Millikan
297k23198371
297k23198371
add a comment |
add a comment |
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