- #1

- 533

- 1

[tex]

(\mathcal O(\epsilon))^2 = \mathcal O(\epsilon^2)

[/tex]

and

[tex]

\sqrt{1 + \mathcal O(\epsilon^2)} = 1 + \mathcal O(\epsilon^2)

[/tex]

Thanks so much.

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- Thread starter AxiomOfChoice
- Start date

- #1

- 533

- 1

[tex]

(\mathcal O(\epsilon))^2 = \mathcal O(\epsilon^2)

[/tex]

and

[tex]

\sqrt{1 + \mathcal O(\epsilon^2)} = 1 + \mathcal O(\epsilon^2)

[/tex]

Thanks so much.

- #2

- 533

- 1

[tex]

\left| \frac{f(\epsilon)}{\epsilon} \right| \leq C.

[/tex]

To show that [itex](\mathcal O(\epsilon))^2 = \mathcal O(\epsilon^2)[/itex], one simply observes that

[tex]

\left| \frac{f^2(\epsilon)}{\epsilon^2} \right| \leq C^2.

[/tex]

- #3

mathman

Science Advisor

- 7,956

- 498

The second approximation can be gotten by using the binomial expansion of the left side.

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