If (1+z)^n=p0+p1 z+p2 z^2+…+pn z^n, show that (i) P0-P2+P4-+…=2^n / 2 cos(n π)/(4) (ii) p1-p3+P5

If (1+z)^n=p0+p1 z+p2 z^2+…+pn z^n, show that (i)...

Key Concepts

Binomial Theorem

The Binomial Theorem states that (a + b)^n can be expanded into a sum involving binomial coefficients, namely, (a + b)^n = ? (n choose k) a^(n-k) b^k. This principle underlies the expansion given in the problem and is essential for extracting and analyzing the coefficients of the power series, especially when summing over particular indices or alternating them.

Complex Numbers and Euler's Formula

Complex numbers extend the real numbers by including an imaginary unit i, and Euler's formula, e^(i?) = cos ? + i sin ?, provides a powerful connection between exponential and trigonometric functions. This connection is crucial in the problem when transformations are applied to the series expansion by substituting z with complex numbers to turn alternating sums of coefficients into expressions involving cosine and sine.

Roots of Unity Filter

The Roots of Unity Filter is a technique used to isolate particular groups of coefficients from a generating function. By substituting roots of unity (such as i, which is a primitive 4th root of unity) into the expansion and using their orthogonality properties, one can selectively sum over terms that meet certain congruence conditions (like even or odd indices). This strategy provides the bridge from the combinatorial coefficients to trigonometric identities in the problem.

De Moivre's Theorem

De Moivre's Theorem asserts that (cos ? + i sin ?)^n = cos(n?) + i sin(n?) for any integer n, and it facilitates raising complex numbers (especially those in polar form) to a power. It is a key tool in converting the expressions obtained via the substitution of complex numbers back into trigonometric form. In the problem, it enables the simplification of sums involving alternating signs to elegant forms with cosine and sine functions.

Transcript

00:01 In this question it is given that 1 plus z to the power n is equal to p0 plus p1 z plus p2 z 2 plus p3 z plus so on plus p n and we need to prove p0 minus p2 plus p 4 minus so on is equal to 2 to the power n.

00:35 N by 2, cos n pi by 4.

00:42 Second part, p1 minus p3 plus p 5 and so on equals to 2 to 2 to 5, sine, n pi by 4.

01:01 And third part, p 0 plus p 4 plus p 8 plus so on is equal to 2 to 2 to 2 to the power, n by 2 minus 1 cos n pi by 4 plus 2 to the power n minus 2 to solve this substitute z is equal to iota in the given equation we get 1 plus iota to the power n is equal to p0 plus p1 iota plus p2 iota square plus p3 iota cube plus so on p n iota to the power n we know that iota equals to minus 1 therefore iota to the power 4 is equal to plus 1 on the basis of these values we can write 1 plus iota to the power n is equal to p0 plus p b0 plus p 1 i minus p2 minus p3 i plus p4 and so on from removers theorem theorem we can say that 1 plus iota to the power n is equals to root 2 cos pi by 4 plus iota sine pi by 4 to the power n.

03:25 Substitute this value in the above equation we get 2 to the power n by 2 cos n pi by 4 plus iota sign n pi by 4 to the power n and this will be equals to p 0 plus p 1 iota minus p 3 iota and so on.

04:19 Now equate the real parts.

04:27 Equate the real parts.

04:34 We get p0 minus p2 plus p4 and so on is equal to 2 to 2 to the power n by 2 cos n pi by 4.

05:02 And this is the required expression of the part a.

05:10 Hence part a is proved.

05:19 Now part second.

05:30 To get the part second equate the imaginary parts we get p1 minus p3 plus p5 and so on is equal to 2 to 2 xin pi by 4...