Understanding and Evaluating the Expression ( P_n frac{sqrt{n}}{2^{n-1}} )

Introduction

In this article, we will explore the evaluation of the mathematical expression ( P_n frac{sqrt{n}}{2^{n-1}} ) and discuss its recursive formulation and derivation. We will also delve into the application of Euler's identity in evaluating the expression efficiently.

Recursive Formulation and Derivation

The expression ( P_n frac{sqrt{n}}{2^{n-1}} ) can be evaluated using a recursive approach. This means we break down the problem by using smaller subproblems, typically for values of ( n ) that are powers of two.

Initial Conditions and Base Case

To start, we need to establish the base case for ( P_n ). Here, when ( n 1 ), the product is simply the unit of multiplication, so:

( P_1 1 )

Recursive Formula for Even ( n )

For an even ( n 2m ), we can express ( P_n ) as:

( P_n prod_{r1}^{n-1} sin{left(frac{rpi}{2n}right)} )

Where ( n 2m ), we rewrite this as:

( P_n prod_{r1}^{2m-1} sin{left(frac{rpi}{4m}right)} prod_{r1}^{m-1} sin{left(frac{rpi}{4m}right)} cdot sin{left(frac{mpi}{4m}right)} cdot prod_{rm-1}^{2m-1} sin{left(frac{rpi}{4m}right)} )

This simplifies to:

( P_n sin{left(frac{pi}{4}right)} cdot prod_{r1}^{m-1} sin{left(frac{rpi}{4m}right)} cdot prod_{r1}^{m-1} cos{left(frac{rpi}{4m}right)} )

Using the double-angle identity for sine, we can rewrite the above expression as:

( P_n frac{1}{2sqrt{2}} prod_{r1}^{m-1} sin{left(frac{2rpi}{4m}right)} )

Which further simplifies to:

( P_n frac{1}{2sqrt{2}} left(frac{1}{2^{m-1}} prod_{r1}^{m-1} sin{left(frac{rpi}{2m}right)}right) )

This can be simplified to:

( P_n frac{sqrt{2}}{2^m} P_m frac{sqrt{n/m}}{2^{n-m}} P_m )

For ( m 2^k ), we can continue the iteration until ( k 1 ) by repeatedly applying this formula:

( P_n frac{sqrt{m/k}}{2^{m-k}} P_k )

Finally, substituting ( P_1 1 ) and iterating until ( k 1 ) on a power of two gives us:

( P_n frac{sqrt{n}}{2^{n-1}} )

Efficient Evaluation Using Euler's Identity

To evaluate the product more efficiently, we can use Euler's identity, which states:

( e^{ix} cos x i sin x )

and

( e^{-ix} cos x - i sin x )

From this, we can derive:

( sin x frac{e^{ix} - e^{-ix}}{2i} )

Substituting ( x frac{rpi}{2n} ), we get:

( sin{left(frac{rpi}{2n}right)} frac{e^{ifrac{rpi}{2n}} - e^{-ifrac{rpi}{2n}}}{2i} )

Factoring out ( e^{ifrac{rpi}{2n}} ) from the product, we obtain:

( prod_{r1}^{n-1} e^{ifrac{rpi}{2n}} left(e^{ifrac{pi}{2n} times frac{n(n-1)}{2}}right) left(e^{ifrac{n-1}{4}}right) )

The second factor is:

( prod_{r1}^{n-1} 2^{-frac{1}{i}} left(frac{1}{2i}right)^{n-1} )

Multiplying these two expressions together gives:

( frac{1}{2^{n-1} cdot i^{frac{n-1}{2}}} )

Conclusion

In summary, through recursive formulation and the efficient application of Euler's identity, we can evaluate the expression ( P_n frac{sqrt{n}}{2^{n-1}} ) and its underlying mathematical structures. This analysis not only showcases the elegance of mathematical derivations but also highlights the power of combining recursive techniques with identities from complex analysis.