I know what a Taylor series involves but you have to know the function; here $\mathcal{L}^*$ is just a function depending on $(v+w)^2$, any kind of function could be inside. How can the below approximation be done? and how this generalizes? Does this have a name?


To first order in $w$ this can be written as $$\mathcal{L}^*_0\simeq\mathcal{L}_0(v^2)+2\vec{v}\cdot\vec{w}\frac{\partial\mathcal{L}_0}{\partial v^2}$$

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    $\begingroup$ It looks like your lecture notes are just following Landau and Lifshitz, Mechanics, but in a less clear way. I recommend you just read that book instead! $\endgroup$ – knzhou Jul 4 at 4:49
  • $\begingroup$ What do you mean by the square of a vector? The dot product of that vector with itself? $\endgroup$ – Steven Gubkin Jul 4 at 22:58

Call the function variable $x$:

$$ x = v^2 $$

Expand into Cartesian components:

$$ x = v_x^2 + v_y^2 + v_z^2 $$

Now add small change ($dx$ and $\vec w$):

$$ x+dx = (v_x+w_x)^2 + (v_y+w_y)^2 + (v_z+w_z)^2 $$

To 1st order in $w_i$:

$$ x+dx = v_x^2+2v_xw_x + v_y^2+2v_yw_y + v_z^2+2v_zw_z $$


$$ x+dx = v_x^2 + v_y^2 + v_z^2+2(v_xw_x + v_yw_y + v_zw_z) $$

$$ x+dx = v^2 + 2\vec v\cdot \vec w $$

Taylor says:

$$\mathcal{L}^*(x+dx)\approx\mathcal{L(x)}+dx\frac{\partial\mathcal{L}}{\partial x}$$


$$\mathcal{L}^*((\vec v+\vec w)^2)\approx\mathcal{L(v^2)}+2\vec v\cdot \vec w\frac{\partial\mathcal{L}}{\partial v^2}$$

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It's just a normal single-variable Taylor expansion $f(x + h) = f(x) + f'(x)h + ...$ where we set $x = v^2$ and $h = 2 \mathbf{v} \cdot \mathbf{w} + \mathbf{w}^2$ in $$\mathcal{L}((\mathbf{v}+\mathbf{w})^2) = \mathcal{L}(v^2+2 \mathbf{v}\cdot \mathbf{w}+ \mathbf{w}^2) = \mathcal{L}(v^2) + \frac{\partial \mathcal{L}}{\partial v^2} (2 \mathbf{v}\cdot \mathbf{w} + \mathbf{w}^2) + ... = \mathcal{L}(\mathbf{v}^2) + \frac{\partial \mathcal{L}}{\partial v^2} 2 \mathbf{v}\cdot \mathbf{w} $$ in the last equality only keeping terms to first order in $w$ so the generalisation is just the applicability of a Taylor expansion, you don't even care that the arguments in the Lagrangian even involves vectors.

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  • $\begingroup$ Would be better if you used \approx instead of some of the = signs in your formulas. After all, a Taylor Approximation is not an equality. Not just the last = sign is erroneous, also the one before that and within the first inline formula (Taylor Approximations do not necessarily converge to the actual function they try to approximate). $\endgroup$ – cmaster - reinstate monica Jul 4 at 21:43

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