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Let $f,g \subset C[a,b] $ where $f$ satisfies $f(a)f(b)<0$. Consider the bisection method for the numerical approximation of the root in $fg^2$ has atleast one root in $[a,b]$, and the sequence $y_1, y_2$ the method produces. Explain why $fg^2$ has atleast one root and also prove analyticaly that $y_n \to y*$ as $n \to \infty$

I'm trying to think of a way to prove this, say for example if we take $f(g) = 2g+3g+1g$ in $[a,b] = [-2,1]$, then $f(a)f(b) < 0$ is satisfied. Then $g^2$ would result in a positive and negative number for both $a$ and $b$. So it is obvious that $fg^2$ has atleast one root, because we have an $[a,b]$ that has both a positive and a negative number... But what about a general case?

If we apply the bisection method on $(2g+3g+1g)^2$ it will converge... again, how to prove for the general case?

$y* = root$

  • Multiplying by $g^2$ does not change the sign and only adds more zeros. – nicomezi Apr 26 '21 at 10:05
  • $2g+3g+1g=6g$, what does this all mean ? –  Apr 26 '21 at 10:05
  • The claim is wrong. The iterations can very well get stuck on a root of $g$. –  Apr 26 '21 at 10:06
  • What does "f(g) = 2g+3g+1g" mean when $g$ has been introduced as an element of $C[a,b]$? – Christian Blatter Apr 26 '21 at 10:06
  • it hasn't this is just my attempt at solving this @ChristianBlatter –  Apr 26 '21 at 10:08
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    "Consider the bisection method for the numerical approximation of the root in $fg^2$" has atleast one root in $[a,b]$": please make this a meaningful sentence. –  Apr 26 '21 at 10:08
  • @nicomezi: if the method hits a root of $g$, the next iterates are undefined. –  Apr 26 '21 at 10:14
  • @nicomezi: Well, the details of the method implementation are not disclosed, so you don't know if the next $y_n$ are defined or not. And having a definition of $y^*$ would indeed not be luxury. –  Apr 26 '21 at 10:17
  • @nicomezi: I know, but the OP has to say that. –  Apr 26 '21 at 10:20
  • Then address this comment to him, not to me. :p Can we clean this comment section ? it seems we agree on all the topics in the end. – nicomezi Apr 26 '21 at 10:21
  • $y*$ is a root.. –  Apr 26 '21 at 10:21
  • "If we apply the bisection method on $(2g+3g+1g)^2$ it will converge": most probably not. There is a lot of confusion in your post. –  Apr 26 '21 at 10:22
  • @mathmindius: a root of what ?? –  Apr 26 '21 at 10:22
  • @YvesDaoust A root of $g^2$ –  Apr 26 '21 at 10:33
  • $2g+3g+1g)^2$ is just my attempt at solving, the original question is the first paragraph @YvesDaoust –  Apr 26 '21 at 10:34

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