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Solving tricky functional equation resembling quadratic equation

2

$begingroup$

I have the following functional equation in hand, I can easily solve it for the case $(a, b ,c)=(1, 1,0)$ which gives $f(x)$ to be $x^2+x$.

$beginalignedg(x)=aleft[f(x)right]^2+bf(x)+c, textwhere g(x)=x^4+2x^3+2x^2+xendaligned$

I also tried setting $x=0$ that gives $af^2(0)+bf(0)+c=0$ which gives $f(0)=frac-bpmsqrtb^2-4ac2a$.

Also I've tried, using the quadratic formula to solve for $f(x)$ as follows:

$$aleft[f(x)right]^2+bf(x)+left(c-g(x)right)=0implies f(x)=dfrac-bpmsqrtb^2-4aleft(c-g(x)right)2a$$

Do both the functions solve the functional equation, if not, how can I proceed? Also the solution I found for the special case does not seem to be the same when we put in those values of $a, b, c$ in the general solution. Any hints are appreciated.

functional-equations

share|cite|improve this question

asked Apr 1 at 8:43

Paras KhoslaParas Khosla

3,285627

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This looks correct, you can see that $g(0)=0$ and so the second approach cancels to the first one.
  $endgroup$
  – George Dewhirst
  Apr 1 at 8:46
  

  
  
  
  

add a comment | 

2

$begingroup$

I have the following functional equation in hand, I can easily solve it for the case $(a, b ,c)=(1, 1,0)$ which gives $f(x)$ to be $x^2+x$.

$beginalignedg(x)=aleft[f(x)right]^2+bf(x)+c, textwhere g(x)=x^4+2x^3+2x^2+xendaligned$

I also tried setting $x=0$ that gives $af^2(0)+bf(0)+c=0$ which gives $f(0)=frac-bpmsqrtb^2-4ac2a$.

Also I've tried, using the quadratic formula to solve for $f(x)$ as follows:

$$aleft[f(x)right]^2+bf(x)+left(c-g(x)right)=0implies f(x)=dfrac-bpmsqrtb^2-4aleft(c-g(x)right)2a$$

Do both the functions solve the functional equation, if not, how can I proceed? Also the solution I found for the special case does not seem to be the same when we put in those values of $a, b, c$ in the general solution. Any hints are appreciated.

functional-equations

share|cite|improve this question

asked Apr 1 at 8:43

Paras KhoslaParas Khosla

3,285627

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This looks correct, you can see that $g(0)=0$ and so the second approach cancels to the first one.
  $endgroup$
  – George Dewhirst
  Apr 1 at 8:46
  

  
  
  
  

add a comment | 

$begingroup$

I have the following functional equation in hand, I can easily solve it for the case $(a, b ,c)=(1, 1,0)$ which gives $f(x)$ to be $x^2+x$.

$beginalignedg(x)=aleft[f(x)right]^2+bf(x)+c, textwhere g(x)=x^4+2x^3+2x^2+xendaligned$

I also tried setting $x=0$ that gives $af^2(0)+bf(0)+c=0$ which gives $f(0)=frac-bpmsqrtb^2-4ac2a$.

Also I've tried, using the quadratic formula to solve for $f(x)$ as follows:

$$aleft[f(x)right]^2+bf(x)+left(c-g(x)right)=0implies f(x)=dfrac-bpmsqrtb^2-4aleft(c-g(x)right)2a$$

Do both the functions solve the functional equation, if not, how can I proceed? Also the solution I found for the special case does not seem to be the same when we put in those values of $a, b, c$ in the general solution. Any hints are appreciated.

functional-equations

share|cite|improve this question

asked Apr 1 at 8:43

Paras KhoslaParas Khosla

3,285627

$endgroup$

share|cite|improve this question

asked Apr 1 at 8:43

Paras KhoslaParas Khosla

3,285627

share|cite|improve this question

asked Apr 1 at 8:43

Paras KhoslaParas Khosla

3,285627

asked Apr 1 at 8:43

Paras KhoslaParas Khosla

3,285627

asked Apr 1 at 8:43

Paras KhoslaParas Khosla

3,285627

1 Answer
1

active

oldest

votes

2

$begingroup$

For every $x$,

$$a[f(x)]^2+bf(x)+c-g(x)=0$$

is an ordinary quadratic equation and

$$f(x)=frac-bpmsqrtb^2-4ac+4ag(x)2a$$ indeed holds.

There are real roots when

$$b^2-4ac+4ag(x)ge0,$$ a quartic inequation.

If there are no constraints on $f$, except that it must be a function, then for any $x$ such that there are distinct real roots, you can choose either of them. So there are infinitely, non-countably many solutions.

If $f$ is restricted to be continuous, then the choice of the sign must be consistent across the intervals of the domain.

---

For the case $(1,1,0)$, the roots are

$$frac-1pmsqrt1+4(x^4+2x^3+2x^2+x)2=frac-1pm(2x^2+2x+1)2=x^2+x,-x^2-x-1.$$

A solution, among many many many others:

share|cite|improve this answer

edited Apr 1 at 9:17

answered Apr 1 at 8:56

Yves DaoustYves Daoust

133k676232

$endgroup$

add a comment | 

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2

$begingroup$

For every $x$,

$$a[f(x)]^2+bf(x)+c-g(x)=0$$

is an ordinary quadratic equation and

$$f(x)=frac-bpmsqrtb^2-4ac+4ag(x)2a$$ indeed holds.

There are real roots when

$$b^2-4ac+4ag(x)ge0,$$ a quartic inequation.

If there are no constraints on $f$, except that it must be a function, then for any $x$ such that there are distinct real roots, you can choose either of them. So there are infinitely, non-countably many solutions.

If $f$ is restricted to be continuous, then the choice of the sign must be consistent across the intervals of the domain.

---

For the case $(1,1,0)$, the roots are

$$frac-1pmsqrt1+4(x^4+2x^3+2x^2+x)2=frac-1pm(2x^2+2x+1)2=x^2+x,-x^2-x-1.$$

A solution, among many many many others:

share|cite|improve this answer

edited Apr 1 at 9:17

answered Apr 1 at 8:56

Yves DaoustYves Daoust

133k676232

$endgroup$

add a comment | 

2

$begingroup$

For every $x$,

$$a[f(x)]^2+bf(x)+c-g(x)=0$$

is an ordinary quadratic equation and

$$f(x)=frac-bpmsqrtb^2-4ac+4ag(x)2a$$ indeed holds.

There are real roots when

$$b^2-4ac+4ag(x)ge0,$$ a quartic inequation.

If there are no constraints on $f$, except that it must be a function, then for any $x$ such that there are distinct real roots, you can choose either of them. So there are infinitely, non-countably many solutions.

If $f$ is restricted to be continuous, then the choice of the sign must be consistent across the intervals of the domain.

---

For the case $(1,1,0)$, the roots are

$$frac-1pmsqrt1+4(x^4+2x^3+2x^2+x)2=frac-1pm(2x^2+2x+1)2=x^2+x,-x^2-x-1.$$

A solution, among many many many others:

share|cite|improve this answer

edited Apr 1 at 9:17

answered Apr 1 at 8:56

Yves DaoustYves Daoust

133k676232

$endgroup$

add a comment | 

$begingroup$

For every $x$,

$$a[f(x)]^2+bf(x)+c-g(x)=0$$

is an ordinary quadratic equation and

$$f(x)=frac-bpmsqrtb^2-4ac+4ag(x)2a$$ indeed holds.

There are real roots when

$$b^2-4ac+4ag(x)ge0,$$ a quartic inequation.

If there are no constraints on $f$, except that it must be a function, then for any $x$ such that there are distinct real roots, you can choose either of them. So there are infinitely, non-countably many solutions.

If $f$ is restricted to be continuous, then the choice of the sign must be consistent across the intervals of the domain.

---

For the case $(1,1,0)$, the roots are

$$frac-1pmsqrt1+4(x^4+2x^3+2x^2+x)2=frac-1pm(2x^2+2x+1)2=x^2+x,-x^2-x-1.$$

A solution, among many many many others:

share|cite|improve this answer

edited Apr 1 at 9:17

answered Apr 1 at 8:56

Yves DaoustYves Daoust

133k676232

$endgroup$

share|cite|improve this answer

edited Apr 1 at 9:17

answered Apr 1 at 8:56

Yves DaoustYves Daoust

133k676232

answered Apr 1 at 8:56

Yves DaoustYves Daoust

133k676232

answered Apr 1 at 8:56

Yves DaoustYves Daoust

133k676232