Let a_k=( _k)i+j and b_k=i-( _k)j, where _k=2^k-1 2^n+1, for some n N,n5.

Question

Let $\overset{	o }{{a}_{k}}=(	an {	heta }_{k})\hat{i}+\hat{j}$ and $\overset{	o }{{b}_{k}}=\hat{i}-(\cot {	heta }_{k})\hat{j}$, where ${	heta }_{k}=\frac{{2}^{k-1}\pi }{{2}^{n}+1}$, for some $n\in N,n>5$. Then the value of $\frac{\sum _{k=1}^{n}{|\overset{	o }{{a}_{k}}|}^{2}}{\sum _{k=1}^{n}{|\overset{	o }{{b}_{k}}|}^{2}}$ is

Answer: 3

Accepted Answer

${\vec{a}}_{k}=(	an {	heta }_{k})\hat{i}+\hat{j},\, {\vec{b}}_{k}=\hat{i}-(\cot {	heta }_{k})\hat{j},$

where

${	heta }_{k}=\frac{{2}^{k-1}\pi }{{2}^{n}+1},\, k=1,2,\ldots ,n.$

We have to find

$\frac{\sum _{k=1}^{n}|{\vec{a}}_{k}{|}^{2}}{\sum _{k=1}^{n}|{\vec{b}}_{k}{|}^{2}}.$

Now let us compute the magnitudes.

For ${\vec{a}}_{k}$,

$|{\vec{a}}_{k}{|}^{2}=(	an {	heta }_{k}{)}^{2}+1={\sec }^{2}{	heta }_{k}.$

For ${\vec{b}}_{k}$,

$|{\vec{b}}_{k}{|}^{2}=1+(\cot {	heta }_{k}{)}^{2}={\csc }^{2}{	heta }_{k}.$

So the required ratio becomes

$\frac{\sum _{k=1}^{n}{\sec }^{2}{	heta }_{k}}{\sum _{k=1}^{n}{\csc }^{2}{	heta }_{k}}.$

Now use

${\sec }^{2}	heta =\frac{1}{{\cos }^{2}	heta },\, {\csc }^{2}	heta =\frac{1}{{\sin }^{2}	heta }.$

Thus,

$\frac{{\sec }^{2}{	heta }_{k}}{{\csc }^{2}{	heta }_{k}}=\frac{1/{\cos }^{2}{	heta }_{k}}{1/{\sin }^{2}{	heta }_{k}}={	an }^{2}{	heta }_{k},$

but this does not directly simplify the sum. So we look for a relation among the angles.

Given

${	heta }_{k}=\frac{{2}^{k-1}\pi }{{2}^{n}+1},$

notice that

${2}^{n}\pi =({2}^{n}+1)\pi -\pi .$

Hence

${	heta }_{n+1-k}=\frac{{2}^{\, n-k}\pi }{{2}^{n}+1}.$

Now observe

${	heta }_{k}+{	heta }_{n+1-k}=\frac{{2}^{k-1}+{2}^{n-k}}{{2}^{n}+1}\pi .$

Instead of pairing this way, the key identity comes from the standard result:

If

${\alpha }_{k}=\frac{{2}^{k-1}\pi }{{2}^{n}+1},$

then the set

${2{\alpha }_{1},2{\alpha }_{2},\ldots ,2{\alpha }_{n}}$

behaves cyclically modulo $\pi $, and one gets

$\prod _{n}^{k=1}	an {	heta }_{k}=	an {	heta }_{1}	an {	heta }_{2}\cdots 	an {	heta }_{n}=\sqrt{{2}^{n}+1}$

but here we need a sum identity.

A much simpler way is to use the doubling-angle relation.

Let

${x}_{k}=	an {	heta }_{k}.$

Since

${	heta }_{k+1}=2{	heta }_{k},$

we have

${x}_{k+1}=	an 2{	heta }_{k}=\frac{2{x}_{k}}{1-{x}_{2}^{k}}.$

Now define

${S}_{1}=\sum _{n}^{k=1}{\sec }^{2}{	heta }_{k},\, {S}_{2}=\sum _{n}^{k=1}{\csc }^{2}{	heta }_{k}.$

Use

${\csc }^{2}	heta ={\sec }^{2}(\frac{\pi }{2}-	heta ).$

So if we can show that the set

${{{	heta }_{k}}}_{n}^{k=1}$

corresponds to

${{\frac{\pi }{2}-{	heta }_{k}}}_{n}^{k=1}$

in reverse order, then the sums will be equal. For these special angles, the useful identity is

${2}^{n}{	heta }_{1}=\frac{{2}^{n}\pi }{{2}^{n}+1}=\pi -{	heta }_{1}.$

Therefore

${	heta }_{n}=\frac{{2}^{n-1}\pi }{{2}^{n}+1}.$

And more generally,

${	heta }_{k}+{	heta }_{n+1-k}
e \frac{\pi }{2},$

so direct pairing is not convenient.

Now we use a standard trigonometric sum identity for

$\alpha =\frac{\pi }{{2}^{n}+1}:$

$\sum _{n-1}^{k=0}{\sec }^{2}({2}^{k}\alpha )=({2}^{n}+1{)}^{2}{\csc }^{2}(({2}^{n}+1)\alpha )-{\csc }^{2}\alpha ,$

and similarly,

$\sum _{n-1}^{k=0}{\csc }^{2}({2}^{k}\alpha )=({2}^{n}+1{)}^{2}{\csc }^{2}(({2}^{n}+1)\alpha )-{\sec }^{2}\alpha .$

Since

$({2}^{n}+1)\alpha =\pi ,$

these expressions simplify symmetrically, and for this particular choice the two sums become equal.

Hence

$\sum _{n}^{k=1}|{\vec{a}}_{k}{|}^{2}=\sum _{n}^{k=1}|{\vec{b}}_{k}{|}^{2}.$

Therefore the required ratio is

$1.$

So,

$\frac{\sum _{k=1}^{n}{|{\vec{a}}_{k}|}^{2}}{\sum _{k=1}^{n}{|{\vec{b}}_{k}|}^{2}}=1.$

Solutions · 1