Question-93225

Question Number 93225 by Ajao yinka last updated on 11/May/20

Commented by prakash jain last updated on 12/May/20

1 + x + \dots + x^{n} has no root in R for n even

\int_{- \infty}^{\infty} x^{n} δ (1 + x + \dots + x^{n}) d x = 0

for even n . So inequality does nold

for even n .

Commented by prakash jain last updated on 12/May/20

for n odd

δ (1 + x + \dots + x^{n}) = \frac{δ (x + 1)}{∣ 1 + 2 x + 3 x^{2} + . . + {n x}^{n - 1} ∣_{x = - 1}}

S = 1 + 2 x + 3 x^{2} + \dots + {n x}^{n - 1}

x S = x + 2 x^{2} + \dots . + (n - 1) x^{n - 1} + {n x}^{n}

(1 - x) S = (1 + x + x^{2} + . . + x^{n - 1}) - {n x}^{n}

x = - 1

2 S = n + 1 (n o d d)

S = \frac{n + 1}{2}

δ (1 + x + x^{2} + . . + x^{n}) = \frac{2 δ (x + 1)}{n + 1} (n o d d)

So for odd n

\int_{- \infty}^{+ \infty} x^{n + 1} δ (1 + x + x^{2} + \dots + x^{n}) d x = \frac{2}{n + 1}