0
$\begingroup$

It's a relatively simple problem and I've been able to solve it in a different way before, but the current book I'm learning from left me with some questions:

After a few idealizations we get the following:

$F_u = \tau (\frac{\partial{u}}{\partial{x}}(x+\Delta x)-\frac{\partial{u}}{\partial{x}}(x))$

Where $F_u$ is the force in the vertical direction, $\tau$ is the tension of the string, $u$ is the displacement in the vertical direction and $x$ is the horizontal position.

Apparently, this equation should lead to:

$F_u = \tau (\frac{\partial^2{u}}{\partial{x}^2}\Delta x+\mathcal{O}(\Delta x^2))$

Now, I have no idea how to get to this and I don't have any idea what the $\mathcal{O}$ signifies (I found the Latex code thanks to detexify).

$\endgroup$
2
$\begingroup$

$F_u=\tau\left(\frac{\partial^2 u}{\partial x^2}\Delta x+\mathcal{O}(\Delta x^2)\right)$ is a Taylor expansion of $F_u$ around $x$. The $\mathcal{O}(\Delta x^2)$ means that there are further terms, which all include a factor of $\Delta x^n$ for $n>1$, i.e. for $\Delta x \ll 1$ these terms are negligable.

Generally, for a function $f(y)$ the Taylor expansion around $y_0$ is $$ f(y)\approx f(y_0)+f'(y_0)\cdot (y-y_0)+\mathcal{O}\left((y-y_0)^2\right). $$ To get to the result above we just use $f\rightarrow\frac{\partial u}{\partial x}$, $y_0\rightarrow x$ and $y-y_0\rightarrow\Delta x$. This will give you $$\frac{\partial u}{\partial x}(x+\Delta x)\approx\frac{\partial u}{\partial x}(x)+\frac{\partial^2 u}{\partial x^2}(x)\cdot\Delta x+\mathcal{O}(\Delta x^2).$$ Just plug this in your expression to get the result.

$\endgroup$
0
$\begingroup$

It comes from the definition of the (forward) derivative, it's probably only notation that is holding you back here. It might be more clear if you consider:

$\frac{\partial^2{u}}{\partial{x}^2} = \lim_{\Delta x->0} \frac{(\frac{\partial{u}}{\partial{x}}(x+\Delta x)-\frac{\partial{u}}{\partial{x}}(x))}{\Delta x}$

Just think of $\frac{\partial{u}}{\partial{x}}(x)$ as a function.

Since you are not taking the limit you introduce an error which is of the order of $\mathcal{O}(\Delta x)$, so you get.

$\frac{\partial^2{u}}{\partial{x}^2} = \frac{(\frac{\partial{u}}{\partial{x}}(x+\Delta x)-\frac{\partial{u}}{\partial{x}}(x))}{\Delta x} + \mathcal{O}(\Delta x)$

multiplying by $\Delta x$ yields

$\frac{\partial{u}}{\partial{x}}(x+\Delta x)-\frac{\partial{u}}{\partial{x}}(x)=\frac{\partial^2{u}}{\partial{x}^2}\Delta x+\mathcal{O}(\Delta x^2)$

Minus signs can be absorbed in the $\mathcal{O}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.