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The equation $e^x-4x^2=0$ has a root between $4$ and $5$. Fixed point iteration with iteration function $\frac{1}{2}e^{\frac{x}{2}}$ is

$1.$ diverges .

$2.$ converges.

$3.$ oscillates.

$4.$ converges montotonically.

The same question is asked before FIxed Point Iteration (numerical analysis) but for me it seems to be convergent .

I am thinking it like if I take any initial point in between $4$ and $5$ then sequence start to decreasing and converges to the fixed point of iteration function that is before $4$. Like the picture added below enter image description hereSo according to me answer is $D$. Am I right . Please comment. Thank you .

neelkanth
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2 Answers2

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Diverges as $f(x)=\frac{e^{x/2}}{2}$ is not Lipschitz in $(4,5)$

Indeed derivative is $f'(x)=\frac{e^{x/2}}{4}$ and $|f'(x)|>1$ in the interval $(4,5)$

Out of curiosity, $f(x)=x$ has another solution in $(0,1)$ which can be found with fixed point method.

In the picture below we can see $f(x)$ and $y=x$. Should be clear the reason why $|f'(x)|<1$ is a requirement for the convergence of the fixed point method.

$$...$$

enter image description here

Raffaele
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  • But if seen by drawing graphs then it seems to be convergent out side of the interval . – neelkanth Dec 22 '20 at 18:54
  • @neelkanth I edited my answer. Indeed there is another solution in $(0,1)$. It depends on the interval: how much is "steep" the function – Raffaele Dec 22 '20 at 18:59
  • @neelkanth Added a picture, too – Raffaele Dec 22 '20 at 19:01
  • sir I added a picture in my question please have a look at this . From that picture it seems that sequence is convergent to other fixed point . Where is my mistakes . Thank you . – neelkanth Dec 23 '20 at 02:42
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Let $g(x) = \frac{1}{2}e^{x/2}$. Then $g'(x) = \frac{1}{4}e^{x/2}$ so $g'(4) = 1.85$, $g'(5) = 3.05$. Let $c$ be the solution between $4$ and $5$ Based on the equation if you define a sequence by $x_{n + 1} = g(x_{n})$ then by Taylor's theorem and using $g(c) = c$, $$x_{n + 1} - c = g'(\xi(c, x_n))(x_n - c).$$ where $\xi(c, x_n)$ is between $c$ and $x_n$. Since $|g'(c)| > 1$, the iteration moves away from the fixed point when you are close to it, so the iteration likely won't converge to $c$ unless you start exactly at $x_0 = c$. For example, suppose $x_0 > c$. Then $g'(\xi(c, x_n)) > 1$ for every $n$ so $x_n > c$ for every $n$ and $$x_{n + 1} - c > g'(c)(x_n - c) > g'(c)^{n + 1}(x_0 - c) \to \infty$$ If $x_0 < c$, then the iteration won't converge to $c$, but may converge to some other value less than $c$. Simulating it demonstrates that it converges to a solution at $0.7148059$. When in doubt, simulate it.

Mason
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  • sir I added a picture in my question please have a look at this . From that picture it seems that sequence is convergent to other fixed point . Where is my mistakes . – neelkanth Dec 23 '20 at 02:42
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    @neelkanth Yes it does converge monotonically decreasingly to the other fixed point, but not the one in between $4$ and $5$. It is easy to show that it converges to the solution at $c \approx 0.7148059$ using a similar analysis as in my post. I think in the question, "convergence" might mean convergence to the fixed point between $4$ and $5$, rather than just convergence to any value. – Mason Dec 23 '20 at 05:08