Determining properties of a polynomial $f$ satisfying $f(x^2)-xf(x) = x^4(x^2-1)$ for $x inBbb R^+$
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Let $f$ be a polynomial satisfying $f(x^2)-xf(x) = x^4(x^2-1), x inBbb R^+$. Then which of the following is correct?
A) $f$ is an even function
B) $f$ is an odd function
C) $displaystylelim_{xto infty} frac{f(x)}{x^3}=1$
D) $displaystylelim_{xto infty} left(frac{f(x)}{x^2}-x right)$ exist and is equal to a non-zero quantity.
I have no idea what to do here.
Looking at the options, one thing I could guess is that the question wants us to find $f(x)$. After an analysis of few minutes, I could guess $f(x) = x^3$. But that gives me the answer as B), C). But the answer given is just C).
Any help would be appreciated.
calculus limits functions even-and-odd-functions
$endgroup$
add a comment |
$begingroup$
Let $f$ be a polynomial satisfying $f(x^2)-xf(x) = x^4(x^2-1), x inBbb R^+$. Then which of the following is correct?
A) $f$ is an even function
B) $f$ is an odd function
C) $displaystylelim_{xto infty} frac{f(x)}{x^3}=1$
D) $displaystylelim_{xto infty} left(frac{f(x)}{x^2}-x right)$ exist and is equal to a non-zero quantity.
I have no idea what to do here.
Looking at the options, one thing I could guess is that the question wants us to find $f(x)$. After an analysis of few minutes, I could guess $f(x) = x^3$. But that gives me the answer as B), C). But the answer given is just C).
Any help would be appreciated.
calculus limits functions even-and-odd-functions
$endgroup$
$begingroup$
To ay that B) is true it is not enough to give one example. There may be other solutions which are not odd.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:23
1
$begingroup$
Check the condition that says this polynomial is only satisfied when $x epsilon R^+$. So you don't know if the function is odd or not.
$endgroup$
– Sauhard Sharma
Dec 20 '18 at 6:27
add a comment |
$begingroup$
Let $f$ be a polynomial satisfying $f(x^2)-xf(x) = x^4(x^2-1), x inBbb R^+$. Then which of the following is correct?
A) $f$ is an even function
B) $f$ is an odd function
C) $displaystylelim_{xto infty} frac{f(x)}{x^3}=1$
D) $displaystylelim_{xto infty} left(frac{f(x)}{x^2}-x right)$ exist and is equal to a non-zero quantity.
I have no idea what to do here.
Looking at the options, one thing I could guess is that the question wants us to find $f(x)$. After an analysis of few minutes, I could guess $f(x) = x^3$. But that gives me the answer as B), C). But the answer given is just C).
Any help would be appreciated.
calculus limits functions even-and-odd-functions
$endgroup$
Let $f$ be a polynomial satisfying $f(x^2)-xf(x) = x^4(x^2-1), x inBbb R^+$. Then which of the following is correct?
A) $f$ is an even function
B) $f$ is an odd function
C) $displaystylelim_{xto infty} frac{f(x)}{x^3}=1$
D) $displaystylelim_{xto infty} left(frac{f(x)}{x^2}-x right)$ exist and is equal to a non-zero quantity.
I have no idea what to do here.
Looking at the options, one thing I could guess is that the question wants us to find $f(x)$. After an analysis of few minutes, I could guess $f(x) = x^3$. But that gives me the answer as B), C). But the answer given is just C).
Any help would be appreciated.
calculus limits functions even-and-odd-functions
calculus limits functions even-and-odd-functions
edited Dec 20 '18 at 10:46
egreg
182k1485203
182k1485203
asked Dec 20 '18 at 6:18
TonyTony
523
523
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To ay that B) is true it is not enough to give one example. There may be other solutions which are not odd.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:23
1
$begingroup$
Check the condition that says this polynomial is only satisfied when $x epsilon R^+$. So you don't know if the function is odd or not.
$endgroup$
– Sauhard Sharma
Dec 20 '18 at 6:27
add a comment |
$begingroup$
To ay that B) is true it is not enough to give one example. There may be other solutions which are not odd.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:23
1
$begingroup$
Check the condition that says this polynomial is only satisfied when $x epsilon R^+$. So you don't know if the function is odd or not.
$endgroup$
– Sauhard Sharma
Dec 20 '18 at 6:27
$begingroup$
To ay that B) is true it is not enough to give one example. There may be other solutions which are not odd.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:23
$begingroup$
To ay that B) is true it is not enough to give one example. There may be other solutions which are not odd.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:23
1
1
$begingroup$
Check the condition that says this polynomial is only satisfied when $x epsilon R^+$. So you don't know if the function is odd or not.
$endgroup$
– Sauhard Sharma
Dec 20 '18 at 6:27
$begingroup$
Check the condition that says this polynomial is only satisfied when $x epsilon R^+$. So you don't know if the function is odd or not.
$endgroup$
– Sauhard Sharma
Dec 20 '18 at 6:27
add a comment |
3 Answers
3
active
oldest
votes
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Some questions have been raised in the comments about my answer below. A polynomial on $mathbb R$ which satisfies the given functional equation on $[0,infty)$ necessarily has the form $ax+x^{3}$ on that interval. But then it must have the same form on all of $mathbb R$. So I think the answer below is correct.
Look at the degree of the polynomial. If $f$ has degree $n$ then LHS has degree $2n$ and RHS has degree $6$. Hence $f$ is a polynomial of degree 3. The constant term is $0$ because $f(0)=0$. Show that the term in $x^{2}$ also must vanish because $xf(x)=f(x^{2})-x^{6}+x^{4}$ is a polynomial with only even powers of $x$. I guess you can take it from here. [All solutions are of the form $ax+x^{3}$]. Final anwser: A) and D) are false (the limit in D) exists but it is $0$); B) and C) are true.
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But the answer given is C).
$endgroup$
– Tony
Dec 20 '18 at 6:38
$begingroup$
I double checked my solution and it looks right. I think the answer given is wrong.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:41
$begingroup$
See Sauhard Sharma's comment. Can that be an explanation?
$endgroup$
– Tony
Dec 20 '18 at 6:50
$begingroup$
@Kavi One doubt: If a function gas the set of domain restricted to positive reals only, then how can we judge abouts odd character. I mean we can't do $f(-x) =-f(x)$ (taking x as positive) as it is not permissible for $f(x)$ to obtain negative value of x as its domain. Am I right? Or, am I missing something obvious?
$endgroup$
– jayant98
Dec 20 '18 at 6:52
1
$begingroup$
@jayant98 Please see the comments I have added to my answer. The equation is supposed to be satisfied only for $x$ positive but the polynomial is defined on the whole line and it is uniquely determined by its values on $[0,infty)$. So the question does make sense.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 7:25
|
show 1 more comment
$begingroup$
You don't need to find $f(x)$. For instance, statements A and B can be dealt with by considering
$$
f(x)=frac{f(x^2)-x^4(x^2-1)}{x} tag{*}
$$
and so
$$
f(-x)=frac{f(x^2)-x^4(x^2-1)}{-x}=-f(x)
$$
showing that $f$ is an odd function. One can object that (*) only holds for $x>0$, but the right-hand side is a polynomial, (implying $f(0)=0$) and if two polynomials agree on an infinite set they're equal.
If the degree of $f$ is $n$, then the degree of $f(x^2)$ is $2n$. By (*), the degree of $f(x^2)$ must be $6$, or the equality could not hold. Hence $n=3$.
The function $f(x)/x^3$ has finite limit $l$, owing to $deg f(x)=3$. Now
$$
frac{f(x^2)-xf(x)}{x^6}=frac{f(x^2)}{(x^2)^3}-frac{1}{x^2}frac{f(x)}{x^3}=frac{x^2-1}{x^2}=1-frac{1}{x^2}
$$
Hence $l=1$.
Similarly,
$$
frac{f(x)}{x^2}-x=frac{f(x^2)}{x^3}-x(x^2-1)-x=frac{f(x^2)}{x^3}-x^3=
xleft(frac{f(x^2)}{(x^2)^2}-x^2right)
$$
Now it's clear that statement D is false: the limit exists (finite or infinite), but if it's finite it must be $0$.
Therefore B and C are true. This allows us to find $f(x)$: it is a degree $3$ polynomial, with leading coefficient $1$ and no term of even degree. Hence we have
$$
f(x)=x^3+ax
$$
Apply the functional equation:
$$
f(x^2)-xf(x)=x^6+ax^2-x^4-ax^2=x^4(x^2-1)
$$
holds for every $x$. Therefore $a$ can be anything.
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add a comment |
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To expand on @KaviRamaMurthy's answer:
Let $f=sum^n a_kx^k$ where $a_nne 0$ for $xge 0$. Then
$$sum a_kx^{2k}-sum a_kx^{k+1}=x^6-x^4$$
$$sum a_kx^{2k}=sum a_kx^{k+1}+x^6-x^4$$
So $2nin [0,n+1]cup{6}$ which means $nle 1$ or $n=3$. So
$$f=b_0+b_1x+b_2x^2+b_3x^3$$ for $xge 0$. And then
$$b_0+b_1x^2+b_2x^4+b_3x^6=b_0x+b_1x^2+b_2x^3+b_3x^4+x^6-x^4$$
which means $b_3=1$ and then $b_2=0$: $b_1$ is unconstrained and $b_0=0$:
$$f(x)=x^3+b_1x.$$
The only way (B) can hold is if they meant
for $xin R^+$, $f$ is equal to a polynomial such that... but $f$ need not be a polynomial on all of $R$.
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Any polynomial is defined on the whole real line (in fact the whole complex plane). The functional equation is given only on the positive real axis and that is enough to determine all solutions of the equation.
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– Kavi Rama Murthy
Dec 20 '18 at 7:28
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Some questions have been raised in the comments about my answer below. A polynomial on $mathbb R$ which satisfies the given functional equation on $[0,infty)$ necessarily has the form $ax+x^{3}$ on that interval. But then it must have the same form on all of $mathbb R$. So I think the answer below is correct.
Look at the degree of the polynomial. If $f$ has degree $n$ then LHS has degree $2n$ and RHS has degree $6$. Hence $f$ is a polynomial of degree 3. The constant term is $0$ because $f(0)=0$. Show that the term in $x^{2}$ also must vanish because $xf(x)=f(x^{2})-x^{6}+x^{4}$ is a polynomial with only even powers of $x$. I guess you can take it from here. [All solutions are of the form $ax+x^{3}$]. Final anwser: A) and D) are false (the limit in D) exists but it is $0$); B) and C) are true.
$endgroup$
$begingroup$
But the answer given is C).
$endgroup$
– Tony
Dec 20 '18 at 6:38
$begingroup$
I double checked my solution and it looks right. I think the answer given is wrong.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:41
$begingroup$
See Sauhard Sharma's comment. Can that be an explanation?
$endgroup$
– Tony
Dec 20 '18 at 6:50
$begingroup$
@Kavi One doubt: If a function gas the set of domain restricted to positive reals only, then how can we judge abouts odd character. I mean we can't do $f(-x) =-f(x)$ (taking x as positive) as it is not permissible for $f(x)$ to obtain negative value of x as its domain. Am I right? Or, am I missing something obvious?
$endgroup$
– jayant98
Dec 20 '18 at 6:52
1
$begingroup$
@jayant98 Please see the comments I have added to my answer. The equation is supposed to be satisfied only for $x$ positive but the polynomial is defined on the whole line and it is uniquely determined by its values on $[0,infty)$. So the question does make sense.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 7:25
|
show 1 more comment
$begingroup$
Some questions have been raised in the comments about my answer below. A polynomial on $mathbb R$ which satisfies the given functional equation on $[0,infty)$ necessarily has the form $ax+x^{3}$ on that interval. But then it must have the same form on all of $mathbb R$. So I think the answer below is correct.
Look at the degree of the polynomial. If $f$ has degree $n$ then LHS has degree $2n$ and RHS has degree $6$. Hence $f$ is a polynomial of degree 3. The constant term is $0$ because $f(0)=0$. Show that the term in $x^{2}$ also must vanish because $xf(x)=f(x^{2})-x^{6}+x^{4}$ is a polynomial with only even powers of $x$. I guess you can take it from here. [All solutions are of the form $ax+x^{3}$]. Final anwser: A) and D) are false (the limit in D) exists but it is $0$); B) and C) are true.
$endgroup$
$begingroup$
But the answer given is C).
$endgroup$
– Tony
Dec 20 '18 at 6:38
$begingroup$
I double checked my solution and it looks right. I think the answer given is wrong.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:41
$begingroup$
See Sauhard Sharma's comment. Can that be an explanation?
$endgroup$
– Tony
Dec 20 '18 at 6:50
$begingroup$
@Kavi One doubt: If a function gas the set of domain restricted to positive reals only, then how can we judge abouts odd character. I mean we can't do $f(-x) =-f(x)$ (taking x as positive) as it is not permissible for $f(x)$ to obtain negative value of x as its domain. Am I right? Or, am I missing something obvious?
$endgroup$
– jayant98
Dec 20 '18 at 6:52
1
$begingroup$
@jayant98 Please see the comments I have added to my answer. The equation is supposed to be satisfied only for $x$ positive but the polynomial is defined on the whole line and it is uniquely determined by its values on $[0,infty)$. So the question does make sense.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 7:25
|
show 1 more comment
$begingroup$
Some questions have been raised in the comments about my answer below. A polynomial on $mathbb R$ which satisfies the given functional equation on $[0,infty)$ necessarily has the form $ax+x^{3}$ on that interval. But then it must have the same form on all of $mathbb R$. So I think the answer below is correct.
Look at the degree of the polynomial. If $f$ has degree $n$ then LHS has degree $2n$ and RHS has degree $6$. Hence $f$ is a polynomial of degree 3. The constant term is $0$ because $f(0)=0$. Show that the term in $x^{2}$ also must vanish because $xf(x)=f(x^{2})-x^{6}+x^{4}$ is a polynomial with only even powers of $x$. I guess you can take it from here. [All solutions are of the form $ax+x^{3}$]. Final anwser: A) and D) are false (the limit in D) exists but it is $0$); B) and C) are true.
$endgroup$
Some questions have been raised in the comments about my answer below. A polynomial on $mathbb R$ which satisfies the given functional equation on $[0,infty)$ necessarily has the form $ax+x^{3}$ on that interval. But then it must have the same form on all of $mathbb R$. So I think the answer below is correct.
Look at the degree of the polynomial. If $f$ has degree $n$ then LHS has degree $2n$ and RHS has degree $6$. Hence $f$ is a polynomial of degree 3. The constant term is $0$ because $f(0)=0$. Show that the term in $x^{2}$ also must vanish because $xf(x)=f(x^{2})-x^{6}+x^{4}$ is a polynomial with only even powers of $x$. I guess you can take it from here. [All solutions are of the form $ax+x^{3}$]. Final anwser: A) and D) are false (the limit in D) exists but it is $0$); B) and C) are true.
edited Dec 20 '18 at 7:23
answered Dec 20 '18 at 6:32
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
60.6k42161
$begingroup$
But the answer given is C).
$endgroup$
– Tony
Dec 20 '18 at 6:38
$begingroup$
I double checked my solution and it looks right. I think the answer given is wrong.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:41
$begingroup$
See Sauhard Sharma's comment. Can that be an explanation?
$endgroup$
– Tony
Dec 20 '18 at 6:50
$begingroup$
@Kavi One doubt: If a function gas the set of domain restricted to positive reals only, then how can we judge abouts odd character. I mean we can't do $f(-x) =-f(x)$ (taking x as positive) as it is not permissible for $f(x)$ to obtain negative value of x as its domain. Am I right? Or, am I missing something obvious?
$endgroup$
– jayant98
Dec 20 '18 at 6:52
1
$begingroup$
@jayant98 Please see the comments I have added to my answer. The equation is supposed to be satisfied only for $x$ positive but the polynomial is defined on the whole line and it is uniquely determined by its values on $[0,infty)$. So the question does make sense.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 7:25
|
show 1 more comment
$begingroup$
But the answer given is C).
$endgroup$
– Tony
Dec 20 '18 at 6:38
$begingroup$
I double checked my solution and it looks right. I think the answer given is wrong.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:41
$begingroup$
See Sauhard Sharma's comment. Can that be an explanation?
$endgroup$
– Tony
Dec 20 '18 at 6:50
$begingroup$
@Kavi One doubt: If a function gas the set of domain restricted to positive reals only, then how can we judge abouts odd character. I mean we can't do $f(-x) =-f(x)$ (taking x as positive) as it is not permissible for $f(x)$ to obtain negative value of x as its domain. Am I right? Or, am I missing something obvious?
$endgroup$
– jayant98
Dec 20 '18 at 6:52
1
$begingroup$
@jayant98 Please see the comments I have added to my answer. The equation is supposed to be satisfied only for $x$ positive but the polynomial is defined on the whole line and it is uniquely determined by its values on $[0,infty)$. So the question does make sense.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 7:25
$begingroup$
But the answer given is C).
$endgroup$
– Tony
Dec 20 '18 at 6:38
$begingroup$
But the answer given is C).
$endgroup$
– Tony
Dec 20 '18 at 6:38
$begingroup$
I double checked my solution and it looks right. I think the answer given is wrong.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:41
$begingroup$
I double checked my solution and it looks right. I think the answer given is wrong.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:41
$begingroup$
See Sauhard Sharma's comment. Can that be an explanation?
$endgroup$
– Tony
Dec 20 '18 at 6:50
$begingroup$
See Sauhard Sharma's comment. Can that be an explanation?
$endgroup$
– Tony
Dec 20 '18 at 6:50
$begingroup$
@Kavi One doubt: If a function gas the set of domain restricted to positive reals only, then how can we judge abouts odd character. I mean we can't do $f(-x) =-f(x)$ (taking x as positive) as it is not permissible for $f(x)$ to obtain negative value of x as its domain. Am I right? Or, am I missing something obvious?
$endgroup$
– jayant98
Dec 20 '18 at 6:52
$begingroup$
@Kavi One doubt: If a function gas the set of domain restricted to positive reals only, then how can we judge abouts odd character. I mean we can't do $f(-x) =-f(x)$ (taking x as positive) as it is not permissible for $f(x)$ to obtain negative value of x as its domain. Am I right? Or, am I missing something obvious?
$endgroup$
– jayant98
Dec 20 '18 at 6:52
1
1
$begingroup$
@jayant98 Please see the comments I have added to my answer. The equation is supposed to be satisfied only for $x$ positive but the polynomial is defined on the whole line and it is uniquely determined by its values on $[0,infty)$. So the question does make sense.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 7:25
$begingroup$
@jayant98 Please see the comments I have added to my answer. The equation is supposed to be satisfied only for $x$ positive but the polynomial is defined on the whole line and it is uniquely determined by its values on $[0,infty)$. So the question does make sense.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 7:25
|
show 1 more comment
$begingroup$
You don't need to find $f(x)$. For instance, statements A and B can be dealt with by considering
$$
f(x)=frac{f(x^2)-x^4(x^2-1)}{x} tag{*}
$$
and so
$$
f(-x)=frac{f(x^2)-x^4(x^2-1)}{-x}=-f(x)
$$
showing that $f$ is an odd function. One can object that (*) only holds for $x>0$, but the right-hand side is a polynomial, (implying $f(0)=0$) and if two polynomials agree on an infinite set they're equal.
If the degree of $f$ is $n$, then the degree of $f(x^2)$ is $2n$. By (*), the degree of $f(x^2)$ must be $6$, or the equality could not hold. Hence $n=3$.
The function $f(x)/x^3$ has finite limit $l$, owing to $deg f(x)=3$. Now
$$
frac{f(x^2)-xf(x)}{x^6}=frac{f(x^2)}{(x^2)^3}-frac{1}{x^2}frac{f(x)}{x^3}=frac{x^2-1}{x^2}=1-frac{1}{x^2}
$$
Hence $l=1$.
Similarly,
$$
frac{f(x)}{x^2}-x=frac{f(x^2)}{x^3}-x(x^2-1)-x=frac{f(x^2)}{x^3}-x^3=
xleft(frac{f(x^2)}{(x^2)^2}-x^2right)
$$
Now it's clear that statement D is false: the limit exists (finite or infinite), but if it's finite it must be $0$.
Therefore B and C are true. This allows us to find $f(x)$: it is a degree $3$ polynomial, with leading coefficient $1$ and no term of even degree. Hence we have
$$
f(x)=x^3+ax
$$
Apply the functional equation:
$$
f(x^2)-xf(x)=x^6+ax^2-x^4-ax^2=x^4(x^2-1)
$$
holds for every $x$. Therefore $a$ can be anything.
$endgroup$
add a comment |
$begingroup$
You don't need to find $f(x)$. For instance, statements A and B can be dealt with by considering
$$
f(x)=frac{f(x^2)-x^4(x^2-1)}{x} tag{*}
$$
and so
$$
f(-x)=frac{f(x^2)-x^4(x^2-1)}{-x}=-f(x)
$$
showing that $f$ is an odd function. One can object that (*) only holds for $x>0$, but the right-hand side is a polynomial, (implying $f(0)=0$) and if two polynomials agree on an infinite set they're equal.
If the degree of $f$ is $n$, then the degree of $f(x^2)$ is $2n$. By (*), the degree of $f(x^2)$ must be $6$, or the equality could not hold. Hence $n=3$.
The function $f(x)/x^3$ has finite limit $l$, owing to $deg f(x)=3$. Now
$$
frac{f(x^2)-xf(x)}{x^6}=frac{f(x^2)}{(x^2)^3}-frac{1}{x^2}frac{f(x)}{x^3}=frac{x^2-1}{x^2}=1-frac{1}{x^2}
$$
Hence $l=1$.
Similarly,
$$
frac{f(x)}{x^2}-x=frac{f(x^2)}{x^3}-x(x^2-1)-x=frac{f(x^2)}{x^3}-x^3=
xleft(frac{f(x^2)}{(x^2)^2}-x^2right)
$$
Now it's clear that statement D is false: the limit exists (finite or infinite), but if it's finite it must be $0$.
Therefore B and C are true. This allows us to find $f(x)$: it is a degree $3$ polynomial, with leading coefficient $1$ and no term of even degree. Hence we have
$$
f(x)=x^3+ax
$$
Apply the functional equation:
$$
f(x^2)-xf(x)=x^6+ax^2-x^4-ax^2=x^4(x^2-1)
$$
holds for every $x$. Therefore $a$ can be anything.
$endgroup$
add a comment |
$begingroup$
You don't need to find $f(x)$. For instance, statements A and B can be dealt with by considering
$$
f(x)=frac{f(x^2)-x^4(x^2-1)}{x} tag{*}
$$
and so
$$
f(-x)=frac{f(x^2)-x^4(x^2-1)}{-x}=-f(x)
$$
showing that $f$ is an odd function. One can object that (*) only holds for $x>0$, but the right-hand side is a polynomial, (implying $f(0)=0$) and if two polynomials agree on an infinite set they're equal.
If the degree of $f$ is $n$, then the degree of $f(x^2)$ is $2n$. By (*), the degree of $f(x^2)$ must be $6$, or the equality could not hold. Hence $n=3$.
The function $f(x)/x^3$ has finite limit $l$, owing to $deg f(x)=3$. Now
$$
frac{f(x^2)-xf(x)}{x^6}=frac{f(x^2)}{(x^2)^3}-frac{1}{x^2}frac{f(x)}{x^3}=frac{x^2-1}{x^2}=1-frac{1}{x^2}
$$
Hence $l=1$.
Similarly,
$$
frac{f(x)}{x^2}-x=frac{f(x^2)}{x^3}-x(x^2-1)-x=frac{f(x^2)}{x^3}-x^3=
xleft(frac{f(x^2)}{(x^2)^2}-x^2right)
$$
Now it's clear that statement D is false: the limit exists (finite or infinite), but if it's finite it must be $0$.
Therefore B and C are true. This allows us to find $f(x)$: it is a degree $3$ polynomial, with leading coefficient $1$ and no term of even degree. Hence we have
$$
f(x)=x^3+ax
$$
Apply the functional equation:
$$
f(x^2)-xf(x)=x^6+ax^2-x^4-ax^2=x^4(x^2-1)
$$
holds for every $x$. Therefore $a$ can be anything.
$endgroup$
You don't need to find $f(x)$. For instance, statements A and B can be dealt with by considering
$$
f(x)=frac{f(x^2)-x^4(x^2-1)}{x} tag{*}
$$
and so
$$
f(-x)=frac{f(x^2)-x^4(x^2-1)}{-x}=-f(x)
$$
showing that $f$ is an odd function. One can object that (*) only holds for $x>0$, but the right-hand side is a polynomial, (implying $f(0)=0$) and if two polynomials agree on an infinite set they're equal.
If the degree of $f$ is $n$, then the degree of $f(x^2)$ is $2n$. By (*), the degree of $f(x^2)$ must be $6$, or the equality could not hold. Hence $n=3$.
The function $f(x)/x^3$ has finite limit $l$, owing to $deg f(x)=3$. Now
$$
frac{f(x^2)-xf(x)}{x^6}=frac{f(x^2)}{(x^2)^3}-frac{1}{x^2}frac{f(x)}{x^3}=frac{x^2-1}{x^2}=1-frac{1}{x^2}
$$
Hence $l=1$.
Similarly,
$$
frac{f(x)}{x^2}-x=frac{f(x^2)}{x^3}-x(x^2-1)-x=frac{f(x^2)}{x^3}-x^3=
xleft(frac{f(x^2)}{(x^2)^2}-x^2right)
$$
Now it's clear that statement D is false: the limit exists (finite or infinite), but if it's finite it must be $0$.
Therefore B and C are true. This allows us to find $f(x)$: it is a degree $3$ polynomial, with leading coefficient $1$ and no term of even degree. Hence we have
$$
f(x)=x^3+ax
$$
Apply the functional equation:
$$
f(x^2)-xf(x)=x^6+ax^2-x^4-ax^2=x^4(x^2-1)
$$
holds for every $x$. Therefore $a$ can be anything.
edited Dec 20 '18 at 11:26
answered Dec 20 '18 at 9:55
egregegreg
182k1485203
182k1485203
add a comment |
add a comment |
$begingroup$
To expand on @KaviRamaMurthy's answer:
Let $f=sum^n a_kx^k$ where $a_nne 0$ for $xge 0$. Then
$$sum a_kx^{2k}-sum a_kx^{k+1}=x^6-x^4$$
$$sum a_kx^{2k}=sum a_kx^{k+1}+x^6-x^4$$
So $2nin [0,n+1]cup{6}$ which means $nle 1$ or $n=3$. So
$$f=b_0+b_1x+b_2x^2+b_3x^3$$ for $xge 0$. And then
$$b_0+b_1x^2+b_2x^4+b_3x^6=b_0x+b_1x^2+b_2x^3+b_3x^4+x^6-x^4$$
which means $b_3=1$ and then $b_2=0$: $b_1$ is unconstrained and $b_0=0$:
$$f(x)=x^3+b_1x.$$
The only way (B) can hold is if they meant
for $xin R^+$, $f$ is equal to a polynomial such that... but $f$ need not be a polynomial on all of $R$.
$endgroup$
$begingroup$
Any polynomial is defined on the whole real line (in fact the whole complex plane). The functional equation is given only on the positive real axis and that is enough to determine all solutions of the equation.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 7:28
add a comment |
$begingroup$
To expand on @KaviRamaMurthy's answer:
Let $f=sum^n a_kx^k$ where $a_nne 0$ for $xge 0$. Then
$$sum a_kx^{2k}-sum a_kx^{k+1}=x^6-x^4$$
$$sum a_kx^{2k}=sum a_kx^{k+1}+x^6-x^4$$
So $2nin [0,n+1]cup{6}$ which means $nle 1$ or $n=3$. So
$$f=b_0+b_1x+b_2x^2+b_3x^3$$ for $xge 0$. And then
$$b_0+b_1x^2+b_2x^4+b_3x^6=b_0x+b_1x^2+b_2x^3+b_3x^4+x^6-x^4$$
which means $b_3=1$ and then $b_2=0$: $b_1$ is unconstrained and $b_0=0$:
$$f(x)=x^3+b_1x.$$
The only way (B) can hold is if they meant
for $xin R^+$, $f$ is equal to a polynomial such that... but $f$ need not be a polynomial on all of $R$.
$endgroup$
$begingroup$
Any polynomial is defined on the whole real line (in fact the whole complex plane). The functional equation is given only on the positive real axis and that is enough to determine all solutions of the equation.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 7:28
add a comment |
$begingroup$
To expand on @KaviRamaMurthy's answer:
Let $f=sum^n a_kx^k$ where $a_nne 0$ for $xge 0$. Then
$$sum a_kx^{2k}-sum a_kx^{k+1}=x^6-x^4$$
$$sum a_kx^{2k}=sum a_kx^{k+1}+x^6-x^4$$
So $2nin [0,n+1]cup{6}$ which means $nle 1$ or $n=3$. So
$$f=b_0+b_1x+b_2x^2+b_3x^3$$ for $xge 0$. And then
$$b_0+b_1x^2+b_2x^4+b_3x^6=b_0x+b_1x^2+b_2x^3+b_3x^4+x^6-x^4$$
which means $b_3=1$ and then $b_2=0$: $b_1$ is unconstrained and $b_0=0$:
$$f(x)=x^3+b_1x.$$
The only way (B) can hold is if they meant
for $xin R^+$, $f$ is equal to a polynomial such that... but $f$ need not be a polynomial on all of $R$.
$endgroup$
To expand on @KaviRamaMurthy's answer:
Let $f=sum^n a_kx^k$ where $a_nne 0$ for $xge 0$. Then
$$sum a_kx^{2k}-sum a_kx^{k+1}=x^6-x^4$$
$$sum a_kx^{2k}=sum a_kx^{k+1}+x^6-x^4$$
So $2nin [0,n+1]cup{6}$ which means $nle 1$ or $n=3$. So
$$f=b_0+b_1x+b_2x^2+b_3x^3$$ for $xge 0$. And then
$$b_0+b_1x^2+b_2x^4+b_3x^6=b_0x+b_1x^2+b_2x^3+b_3x^4+x^6-x^4$$
which means $b_3=1$ and then $b_2=0$: $b_1$ is unconstrained and $b_0=0$:
$$f(x)=x^3+b_1x.$$
The only way (B) can hold is if they meant
for $xin R^+$, $f$ is equal to a polynomial such that... but $f$ need not be a polynomial on all of $R$.
answered Dec 20 '18 at 6:54
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
2,086918
$begingroup$
Any polynomial is defined on the whole real line (in fact the whole complex plane). The functional equation is given only on the positive real axis and that is enough to determine all solutions of the equation.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 7:28
add a comment |
$begingroup$
Any polynomial is defined on the whole real line (in fact the whole complex plane). The functional equation is given only on the positive real axis and that is enough to determine all solutions of the equation.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 7:28
$begingroup$
Any polynomial is defined on the whole real line (in fact the whole complex plane). The functional equation is given only on the positive real axis and that is enough to determine all solutions of the equation.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 7:28
$begingroup$
Any polynomial is defined on the whole real line (in fact the whole complex plane). The functional equation is given only on the positive real axis and that is enough to determine all solutions of the equation.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 7:28
add a comment |
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$begingroup$
To ay that B) is true it is not enough to give one example. There may be other solutions which are not odd.
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 6:23
1
$begingroup$
Check the condition that says this polynomial is only satisfied when $x epsilon R^+$. So you don't know if the function is odd or not.
$endgroup$
– Sauhard Sharma
Dec 20 '18 at 6:27