if-y-1-x-2-x-3-x-n-x-y-

Question Number 203910 by Davidtim last updated on 01/Feb/24

i f y = (1 - x) (2 - x) (3 - x) \cdot \cdot \cdot (n - x)

y^{^{'}} = ?

Answered by mr W last updated on 01/Feb/24

y = \underset{k = 1}{\prod^{n}} (k - x)

\ln y = \underset{k = 1}{\sum^{n}} \ln (k - x)

\frac{y^{'}}{y} = - \underset{k = 1}{\sum^{n}} \frac{1}{k - x}

y^{'} = - \underset{k = 1}{\prod^{n}} (k - x) \underset{k = 1}{\sum^{n}} \frac{1}{k - x}

Answered by witcher3 last updated on 02/Feb/24

f (x) = \underset{i = 1}{\prod^{n}} p_{i}

You can't use 'macro parameter character #' in math mode

proof n = 2

f (x) = p_{1} . p_{2} \Rightarrow f^{'} (x) = p_{1}^{'} . p_{2} + p_{2}^{'} . p_{1} = \underset{i = 1}{\sum^{n}} (p_{i}) \underset{l = 1 \neq i}{\prod^{2}} p_{l}

suppose \forall n \in N f^{'} (x) = Σ p_{i}^{'} Π p_{l} true

shows for f (x) = p_{1} \dots . p_{n + 1}

You can't use 'macro parameter character #' in math mode

f (x) = (p_{n + 1}) . (p_{1} \dots . p_{n}) \Rightarrow f^{'} (x) = p_{n + 1}^{'} . (p_{1} \dots p_{n})

You can't use 'macro parameter character #' in math mode

You can't use 'macro parameter character #' in math mode

= \underset{k = 1}{\sum^{n + 1}} p_{k}^{'} \underset{l = 1 \neq k}{\prod^{n + 1}} p_{l}

prooved

in our exempl p_{i} = (i - x) \Rightarrow p_{i}^{'} = - 1

y^{'} = \underset{k = 1}{\sum^{n}} (- 1) \underset{l = 1 \neq k}{\prod^{n}} (l - x) = - \underset{k = 1}{\sum^{n}} \frac{1}{k - x} \underset{l = 1}{\prod^{n}} (l - x)

= - y \underset{k = 1}{\sum^{n}} \frac{1}{k - x}

Commented by York12 last updated on 02/Feb/24

GJ

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