- #1

- 21

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But after taking a log and dividing by 2 he arrives at

$$ln[m_{ph}] = ln[m] \left [ \left ( \frac{5}{12}\alpha(ln \frac{\mu}{m}) +\frac{1}{2} c' \right ) + O(\alpha^2)\right]$$

Why is there no ln on the $$\frac{5}{12} \alpha$$ term?

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- Thread starter Higgsy
- Start date

- #1

- 21

- 0

But after taking a log and dividing by 2 he arrives at

$$ln[m_{ph}] = ln[m] \left [ \left ( \frac{5}{12}\alpha(ln \frac{\mu}{m}) +\frac{1}{2} c' \right ) + O(\alpha^2)\right]$$

Why is there no ln on the $$\frac{5}{12} \alpha$$ term?

- #2

- 1,006

- 108

[tex]\ln(1 + x) = x + O(x^2)[/tex]

Here ##x## has an ##\alpha## in it, so the ##O(x^2)## term is absorbed into the ##O(\alpha^2)##.

- #3

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Thanks!

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