 Home

Codex Wiki

List of universities

University of Cambridge

Course of the University of Cambridge

Natural sciences course of the University of Cambridge

Year of the natural sciences course of the University of Cambridge

Year 2 of the natural sciences course of the University of Cambridge

Mathematics course of the year 2 of the natural sciences course of the University of Cambridge

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge

2015

Paper 1

Expand both sides through second order as $\epsilon\to0$, holding $X,Y$ fixed. The product is
$$
e^{\epsilon X}e^{\epsilon Y}=I+\epsilon(X+Y)+\epsilon^2\left(\frac{X^2}{2}+XY+\frac{Y^2}{2}\right)+O(\epsilon^3).
$$
Put $Z=\epsilon(X+Y)+\epsilon^2[X,Y]/2$. Its exponential is
$$
\begin{aligned}
e^Z&=I+\epsilon(X+Y)+\frac{\epsilon^2}{2}\{[X,Y]+(X+Y)^2\}+O(\epsilon^3)\\
&=I+\epsilon(X+Y)+\epsilon^2\left(\frac{X^2}{2}+XY+\frac{Y^2}{2}\right)+O(\epsilon^3).
\end{aligned}
$$
Thus
$$
\boxed{e^{\epsilon X}e^{\epsilon Y}=e^{\epsilon X+\epsilon Y+\epsilon^2[X,Y]/2}+O(\epsilon^3).}
$$
This is the quadratic part of the <Baker--Campbell--Hausdorff formula>. The remainder estimate is in any fixed finite-dimensional <matrix> <norm>, and is an asymptotic statement rather than an exact formula for arbitrary finite $\epsilon$.

Solution

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 5 / b / Solution

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 5 / b / Solution

 Ancestors (13)

 Discussion (0)

 Articles by others on the same topic (0)

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 5 / b / Solution

 Ancestors (13)

 Discussion (0)  Subscribe (1)

 Articles by others on the same topic (0)

 Discussion (0)