What is the solution of the difference equation for this equation?

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Discussion Overview

The discussion revolves around determining the range of the function $\displaystyle f(x) = x^2+1+\frac{1}{x^2+x+1}$ for all real numbers $x$. Participants explore the function's behavior, particularly its minimum value and the implications for its range. The conversation also touches on the methods used to find real roots of a related polynomial equation and the connection to difference equations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates that the limits of $f(x)$ as $x$ approaches positive and negative infinity are both $+\infty$, suggesting the need to find the absolute minimum to define the range.
  • Another participant states that the polynomial equation $2x^5 + 4x^4 + 6x^3 + 4x^2 - 1 = 0$ has a single real root, approximately $x_0 \sim 0.379093$, leading to a minimum value of $y_0 \sim 1.800394$ for $f(x)$.
  • Some participants express concern that the problem may not belong in the "Pre-Calculus" category if calculus is required to solve it.
  • There is a request for methods to calculate the real roots of the polynomial equation mentioned, indicating a need for further exploration of root-finding techniques.
  • A later reply connects the solutions of the equation $\varphi(x)=0$ to the limit of the solution of a difference equation, introducing a more complex mathematical framework.

Areas of Agreement / Disagreement

Participants generally agree on the need to find the minimum of the function to determine its range, but there is disagreement regarding the classification of the problem and the methods required to solve it. The discussion remains unresolved regarding the best approach to finding the real roots of the polynomial.

Contextual Notes

Participants mention the dependence on calculus for finding the minimum and roots, which may limit the discussion's accessibility to those without calculus knowledge. The connection between the polynomial equation and difference equations introduces additional complexity that is not fully resolved.

juantheron
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Range of $\displaystyle f(x) = x^2+1+\frac{1}{x^2+x+1}\forall x\in \mathbb{R}$
 
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jacks said:
Range of $\displaystyle f(x) = x^2+1+\frac{1}{x^2+x+1}\forall x\in \mathbb{R}$

It is clear that $\displaystyle \lim_{x \rightarrow + \infty} f(x) = \lim_{x \rightarrow - \infty} f(x)= + \infty$ so that, the range is defined if we find the absolute minimum of f(x). Proceeding with the standard method the values of x where $f^{\ '}(x)$ vanishes are the root of the equation...

$\displaystyle 2\ x^{5} +4\ x^{4} +6\ x^{3} +4\ x^{2} -1=0$ (1)

The (1) has a single real root in $x_{0} \sim .379093$, so that the minimum is $y_{0}=f(x_{0}) \sim 1.800394$ and the range of f(*) is $[y_{0}, + \infty)$...

Kind regards

$\chi$ $\sigma$
 
jacks said:
Range of $\displaystyle f(x) = x^2+1+\frac{1}{x^2+x+1}\forall x\in \mathbb{R}$
jacks,

if a solver can't do this without using calculus, then this problem should not
be posted under "Pre-Calculus." You would post it under "Calculus."

---------- Post added at 11:07 ---------- Previous post was at 10:57 ----------

chisigma said:
The (1) has a single real root in $x_{0} \sim > > .379093$, < < so that the minimum is $y_{0}=f(x_{0}) \sim > > 1.800394$ < <
and the range of f(*) is $[y_{0}, + \infty)$...

My TI-83 gives rounded values of (which when rounded to 6 decimal places) as

0.379095 and 1.800395, respectively.
 
Yes http://www.mathhelpboards.com/member.php?219-checkittwice next time i will consider it

Thanks chisigma

but how can i calculate the real roots of the equation $2x^5+4x^4+6x^3+4x^2-1=0$
 
jacks said:
Yes http://www.mathhelpboards.com/member.php?219-checkittwice next time i will consider it

Thanks chisigma

but how can i calculate the real roots of the equation $2x^5+4x^4+6x^3+4x^2-1=0$

As explained in...

http://www.mathhelpboards.com/showthread.php?426-Difference-equation-tutorial-draft-of-part-I

... the solutions of the equation $\displaystyle \varphi (x)=0$ is the limit [if limit exists...] of the solution of the difference equation...

$\displaystyle \Delta_{n}= a_{n+1}-a_{n}= f(a_{n})\ ;\ f(x)=- \frac {\varphi(x)}{\varphi^{\ '}(x)}$ (1)

... with a proper 'initial value' $a_{0}$ and the conditions about $a_{0}$ for convergence to the solution are also explained...

Kind regards

$\chi$ $\sigma$
 

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