According to equation 2.4.26 of Polchinski's String Theory Volume 1, the finite transformation of the energy momentum tensor is
$$ (\partial z')^2 T'(z') = T(z) - \frac{c}{12} \{z',z\}\tag{2.4.26}$$
Where $$\{z',z\} = \frac{2 \partial^3 z' \partial z' - 3 \partial^2 z' \partial^2 z' }{2 \partial z' \partial z'}. \tag{2.4.27}$$
From here I need to show that the infinitesimal form of the transformation equation 2.4.24
$$\epsilon^{-1} \delta T(z) = -\frac{c}{12 } \partial^3 v(z) - 2 \partial v(z) T(z) - v(z) \partial T(z).\tag{2.4.24} $$
How do I go about this? I tried to start by assuming that $$ z' = z + \epsilon v(z)$$
Then I ignored terms in $\epsilon$ and and higher powers of $\epsilon$. This is my first encounter with CFT and so am unsure if this is the right way to do it.
EDIT: Here is my work.
$$ \Big(\frac{\partial z'}{\partial z}\Big)^2 = (1 + \epsilon v')^2$$
The Schwarzian derivative is:
$$ \{z',z\} = \frac{2 \epsilon v'''(z) (1 + \epsilon v'(z)) - 3 \epsilon^2 v'(z)^2}{2(1 + \epsilon v'(z))^2}.$$
I multiplied both sides of the equation by $2 (1 + \epsilon v'(z))^2$ and ended up with, up to terms of leading order in $\epsilon$
$$ 2(1 + 4 \epsilon v'(z)) T'(z') = 2(1+ 2 \epsilon v'(z)) T(z) - \frac{c}{12} (2 \epsilon v'''(z)).$$
Divide both sides by $2 \epsilon$:
$$ \epsilon^{-1} (T'(z') - T(z)) = -\frac{c}{12} \partial^3 v(z) + 2 \partial v(z) T(z) - 4 \partial v(z) T'(z').$$
Thus, we have
$$ \epsilon^{-1} \delta T (z) = -\frac{c}{12} \partial^3 v(z) - 2 \partial v(z) T(z) + 4 \partial v(z) T(z) - 4 \partial v(z) T'(z').$$
In particular, I am unsure if the following is true:
$$ \delta T(z) = T'(z') - T(z) .$$
I am also unsure about how to bring about the $2 \partial v(z) T(z)$ term.
How do I proceed?
