In this problem we set out to prove the closed$-$form solution for the transformation matrix

resulting from a rotation vector of $(8\mathrm{.}17),$ i$.$e$.$ prove the following relationship:

$[$T$]$BA $=$ exp

$$

K

$$

BA B

$(8\mathrm{.}43)$

$=[$ I $]+$

$$

K

$$

BA B

$+$

$1$

$2!$

$$

K

$$

BA B

$2$

$+$

$1$

$3!$

$$

K

$$

BA B

$3$

$+$ : : :

$(8\mathrm{.}44)$

$=$ cos

BA

$[$ I $]+$

$$

$1$ cos

BA

k BA k$2$

BA B

$[$ BA $]$

B

$$

sin

BA

k BA k

K

$$

BA B

$(8\mathrm{.}45)$

To save time and space, we will use the short$-$hands $[$L$]$ :$=$

K

$$

BA

B and $[]$ :$=$

BA

B$,$ and we$$ll let

BA

B

$=$: $(1$; $2$; $3).$

We will require the infinite series expansions for sine and cosine:

sin x $=$

$1$X

n$=0$

$(1)$nx$2$n$+1$

$(2$n $+1)!$

$=$ x $$

x$3$

$3!$

$+$

x$5$

$5!$

$$

x$7$

$7!$

$+$ : : : $(8\mathrm{.}46)$

cos x $=$

$1$X

n$=0$

$(1)$nx$2$n

$(2$n$)!$

$=1$

x$2$

$2!$

$+$

x$4$

$4!$

$$

x$6$

$6!$

$+$ : : : $(8\mathrm{.}47)$

We know that $(8\mathrm{.}44)$ follows from $(8\mathrm{.}43)$ by the definition of the matrix exponential. Our

objective is to show that $(8\mathrm{.}44)$ implies $(8\mathrm{.}45).$

a$.$ Show that

$[$L$]2=[][][][][$ I $](8\mathrm{.}48)$

b$.$ Show that

$[$L$]3=([][])[$L$]=$k$[]$k$2[$L$](8\mathrm{.}49)$

c$.$ Show that $8\mathrm{.}49$ implies that, for integer n:

$($i$)$ For even powers:

$[$L$]2$n $=(1)$n k$[]$k$2$n$2[$L$]2$ for n $1(8\mathrm{.}50)$

$($ii$)$ For odd powers:

$[$L$]2$n$+1=(1)$n k$[]$k$2$n $[$L$]$ for n $0(8\mathrm{.}51)$

d$.$ By partitioning $(8\mathrm{.}44)$ into even and odd powers, show that $(8\mathrm{.}45)$ follows therefrom.