- #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?

- 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

- 105

[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

- 21

- 0

Thanks!

- Replies
- 3

- Views
- 2K

- Replies
- 9

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 1K

- Replies
- 3

- Views
- 1K

- Replies
- 2

- Views
- 2K

- Replies
- 7

- Views
- 4K

- Last Post

- Replies
- 47

- Views
- 12K

- Last Post

- Replies
- 5

- Views
- 3K

- Last Post

- Replies
- 14

- Views
- 3K

- Last Post

- Replies
- 6

- Views
- 3K