let-P-x-0-i-lt-j-n-x-i-j-1-calculate-P-x-2-find-0-1-P-x-dx-

Question Number 74013 by mathmax by abdo last updated on 17/Nov/19

l e t P (x) = \sum_{0 ⩽ i < j ⩽ n} x^{i + j}

1) c a l c u l a t e P^{^{'}} (x)

2) f i n d \int_{0}^{1} P (x) d x

Commented by abdomathmax last updated on 18/Nov/19

1) w e h a v e {(\sum_{i = 0}^{n} x^{i})}^{2} = \sum_{i = 0}^{n} x^{2 i} + \sum_{0 ⩽ i < j ⩽ n} x^{i + j}

\Rightarrow P (x) = {(\sum_{i = 0}^{n} x^{i})}^{2} - \sum_{i = 0}^{n} x^{2 i}

c a s e 1 x \neq 1 \Rightarrow P (x) = {(\frac{1 - x^{n + 1}}{1 - x})}^{2} - \frac{x^{2 n + 2} - 1}{x^{2} - 1}

\Rightarrow P^{^{'}} (x) = 2 {(\frac{x^{n + 1} - 1}{x - 1})}^{^{'}} (\frac{x^{n + 1} - 1}{x - 1})

- \frac{(2 n + 2) x^{2 n + 1} (x^{2} - 1) - 2 x (x^{2 n + 2} - 1)}{{(x^{2} - 1)}^{2}}

= 2 \frac{x^{n + 1} - 1}{x - 1} \times \frac{(n + 1) x^{n} (x - 1) - (x^{n + 1} - 1)}{{(x - 1)}^{2}}

- \frac{(2 n + 2) x^{2 n + 3} - (2 n + 2) x^{2 n + 2} - 2 x^{2 n + 3} + 1}{{(x^{2} - 1)}^{2}}

= 2 \frac{x^{n + 1} - 1}{{(x - 1)}^{3}} ({n x}^{n + 1} - (n + 1) x^{n} + 1)

- \frac{2 n x^{3 n + 3} - (2 n + 2) x^{2 n + 2} + 1}{{(x^{2} - 1)}^{2}}

c a s e 2 x = 1 \Rightarrow P (x) = P (1) = {(n + 1)}^{2} - (n + 1)

= n^{2} + 2 n + 1 - n - 1 = n^{2} + n

Commented by abdomathmax last updated on 18/Nov/19

e r r o r o f t y p o f o r x \neq 1

P^{^{'}} (x) = \frac{(2 x^{n + 1} - 2) ({n x}^{n + 1} - (n + 1) x^{n} + 1)}{{(x - 1)}^{3}}

- \frac{2 {n x}^{2 n + 3} - (2 n + 2) x^{2 n + 2} + 1}{{(x^{2} - 1)}^{2}}

Commented by abdomathmax last updated on 18/Nov/19

2) \int_{0}^{1} P (x) d x = \sum_{0 ⩽ i < j ⩽ n} \int_{0}^{1} x^{i + j} d x

= \sum_{0 ⩽ i < j ⩽ n} \frac{1}{i + j + 1}

= \sum_{j = 1}^{n} (\sum_{i = 0}^{j - 1} \frac{1}{i + j + 1}) c h a n g e m e n t o f i n d i c e

i + j + 1 = k g i v e

\int_{0}^{1} P (x) d x = \sum_{j = 1}^{n} (\sum_{k = j + 1}^{2 j} \frac{1}{k})

= \sum_{j = 1}^{n} (\sum_{k = 1}^{j} \frac{1}{k} + \sum_{k = j + 1}^{2 j} \frac{1}{k} - \sum_{k = 1}^{j} \frac{1}{k})

= \sum_{j = 1}^{n} (H_{2 j} - H_{j}) = \sum_{j = 1}^{n} H_{2 j} - \sum_{n = 1}^{n} H_{j}

Answered by mind is power last updated on 18/Nov/19

let Q (x) = (\underset{i = 0}{\sum^{n}} x^{i})

Q {(x)}^{2} = 2 p (x) + \underset{i = 1}{\sum^{n}} x^{2 i} = 2 p (x)

Q (x) = {(\frac{1 - x^{n + 1}}{1 - x})}^{2}, \forall x \in C - {1}

\Rightarrow p (x) = \frac{1}{2} [{(\frac{1 - x^{n + 1}}{1 - x})}^{2} - \frac{1 - x^{2 n + 2}}{1 - x^{2}}], \forall x \in C - {1, - 1}

= p (x) = \frac{1}{2} [\frac{(x^{2 n + 2} + 1 - {2 x}^{n + 1}) (1 + x) - (1 - x^{2 n + 2}) (1 - x)}{(1 - x^{2}) (1 - x)}]

\Rightarrow p (x) = \frac{1}{2} [\frac{x^{2 n + 3} + x - {2 x}^{n + 2} + x^{2 n + 2} + 1 - {2 x}^{n + 1} - 1 + x + x^{2 n + 2} - x^{2 n + 3}}{(1 - x^{2}) (1 - x)}]

\Rightarrow p (x) = \frac{1}{2} [\frac{{2 x}^{2 n + 2} - {2 x}^{n + 2} - {2 x}^{n + 1} + 2 x}{(1 - x^{2}) (1 - x)}]

p (x) = \frac{x^{2 n + 2} - x^{n + 2} - x^{n + 1} + x}{(1 - x^{2}) (1 - x)}

p^{'} (x) = \sum_{1 ⩽ i < j ⩽ n} (i + j) x^{i + j - 1} = p^{'} (x), x \in C - {1, - 1}

if x = 1 \Rightarrow p^{'} (x) = \sum_{1 ⩽ i < j ⩽ n} (i + j)

= \sum_{2 ⩽ j ⩽ n} \underset{i = 1}{\sum^{j - 1}} (i + j)

= \sum_{2 ⩽ j ⩽ n} . {\frac{(1 + j + 2 j - 1)}{2} . (j - 1)

= \sum_{2 ⩽ j ⩽ n} \frac{3}{2} (j^{2} - j) = \frac{3}{2} . \sum_{2 ⩽ j ⩽ n} j^{2} - \frac{3}{2} . \frac{n - 1}{2} . \frac{(n + 2)}{}

= \frac{3}{2} [{. \frac{n (n + 1) (2 n + 1)}{6} - \frac{1}{6}} - \frac{(n - 1) (n + 2)}{2}

2) \int_{0}^{1} P (x) = \sum_{0 ⩽ i < j ⩽ n} \frac{1}{i + j + 1}

= \underset{j = 1}{\sum^{n}} . \underset{i = 0}{\sum^{j - 1}} . \frac{1}{i + j + 1}

= \underset{j = 1}{\sum^{n}} . \underset{i = 0}{\sum^{j - 1}} . (\frac{1}{i + j + 1} + \underset{k = 1}{\sum^{j + i}} \frac{1}{k} - \underset{k = 1}{\sum^{j + i}} \frac{1}{k})

= \underset{j = 1}{\sum^{n}} . \underset{i = 0}{\sum^{j - 1}} {H_{2 j} - H_{j + i}}

= \underset{j = 1}{\sum^{n}} {jH}_{2 j} - \underset{j = 1}{\sum^{n}} \underset{i = 0}{\sum^{j - 1}} H_{i + j}

Commented by abdomathmax last updated on 18/Nov/19

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