# Is this a Taylor approximation? How is it done? [closed]

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?

$$\mathcal{L}^*_0=\mathcal{L}_0((\vec{v}+\vec{w})^2)$$

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

• 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! – knzhou Jul 4 at 4:49
• What do you mean by the square of a vector? The dot product of that vector with itself? – 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 $$w_i$$:

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

Rearrange

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

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

Taylor says:

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

substitute:

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

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.

• 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). – cmaster - reinstate monica Jul 4 at 21:43