Prove that $x = x^{3}+0.001$ has a root near 0

Question

How can i prove that $x = x^{3}+0.001$ has a root near $0$? And how does this root look like? I think that i need to use contraction mapping here, but i don't know how exactly. Any hints?

Answer 1

Hint:

It looks like posted solutions, while correct, did not address the role of theory of contraction mappings.

There are iterative processes for finding real roots of an algebraic equations which can be justified by the the theory of contraction mappings. One such method is Newton-Raphson. Also check this: Polynomial Roots using Contractual Mapping

Source: Applications of Contraction Mapping

Answer 2

Note that we can factor as follows: $$x^3-x = x(x+1)(x-1) = -0.001.$$ Let $x = 1/\epsilon,$ where $|\epsilon| << 1$. $$\implies x^3-x \approx \frac{1}{\epsilon^3},$$ which is not consistent. However, if $x = \epsilon$, then $$x^3-x \approx -\epsilon,$$ which is consistent with the small constant $-0.001$ on the right hand side.

So, the equation must have a root near zero.

Answer 3

Let $f(x)=x^3-x+0.001$.

Observing that close to $0$, $x^3$ is negligible, we try

$$f(0.001)=0.000000001>0$$ and which is very close to a root.

From this, the first Newton's iterate gives

$$0.001-\frac{f(0.001)}{f'(0.001)}=0.001000001$$

with $$f(0.001000001)=3\cdot10^{-15}>0,$$ even closer.

For safety we double the step,

$$0.001-2\frac{f(0.001)}{f'(0.001)}=0.001000002$$

and

$$f(0.001000002)=-0.000000001<0.$$

This gives us full guarantee of a root.

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The next Newton's iterate is

$$0.001000001000003\cdots,$$ with $f=1.2\cdot10^{-20}$.

Answer 4

Let $f(x)=x^3+c-x$ where $0<|c|<1/\sqrt 8$ (e.g., $c=0.001$). Then $f(c)=c^3$ and $f(2c)=8c^3-c=c(8c^2-1)$, so $f(c)f(2c)=c^4(8c^2-1)<0$. From the Intermediate Value Theorem, $f$ has a root between $c$ and $2c$.

Answer 5

Let $f(x)=x^3-x+.001.$ Then $f(0)=.001>0$ and $f(.1)=.002-.1<0$ so $f$ has a root between $0$ and $.1$.

Answer 6

Or take $20$ seconds and plot the functions:

Answer 7

There is, BTW, a nice series expansion for the root of equation $x - x^3 = \epsilon$ near $x=0$:

$$ x = \sum_{k=0}^\infty \frac{(3k)!}{k! (2k+1)!} \epsilon^{2k+1} $$

which can be derived using the Lagrange inversion theorem.