If-f-x-x-x-3-x-5-x-n-and-lim-x-1-f-2-x-f-2-1-x-1-2-10-then-n-

Question Number 166627 by cortano1 last updated on 23/Feb/22

If f (x) = x + x^{3} + x^{5} + \dots + x^{n} and

\lim_{x \to 1} \frac{f^{2} (x) - f^{2} (1)}{x - 1} = 2^{10} then n = ?

Commented by MJS_new last updated on 23/Feb/22

15

Commented by cortano1 last updated on 23/Feb/22

why ?

Answered by MJS_new last updated on 23/Feb/22

I took f^{2} (x) = {(f (x))}^{2}

f (x) = x + x^{3} + x^{5} + \dots + x^{2 k - 1} \Rightarrow f (1) = k

f (x) = \underset{j = 1}{\sum^{k}} x^{2 j - 1} = \frac{x (x^{2 k} - 1)}{x^{2} - 1}; \lim_{x \to 1} \frac{x (x^{2 k} - 1)}{x^{2} - 1} = k \Rightarrow f (1) = k

\frac{f^{2} (x) - f^{2} (1)}{x - 1} = \frac{x^{4 k + 2} - 2 x^{2 k + 2} - k^{2} x^{2} + x^{2} - k^{2}}{x^{5} - x^{4} - 2 x^{3} + 2 x^{2} + x - 1}

\lim_{x \to 1} \frac{f^{2} (x) - f^{2} (1)}{x - 1} =

= \lim_{x \to 1} \frac{\frac{d^{3}}{{d x}^{3}} [x^{4 k + 2} - 2 x^{2 k + 2} - k^{2} x^{2} + x^{2} - k^{2}]}{\frac{d^{3}}{{d x}^{3}} [x^{5} - x^{4} - 2 x^{3} + 2 x^{2} + x - 1]} =

= \lim_{x \to 1} \frac{2 k x ((2 k + 1) (4 k + 1) x^{4 k - 2} - (k + 1) (2 k + 1) x^{2 k - 2} - 3 k x)}{3 (5 x^{2} - 2 x - 1)} = 2 k^{3}

2 k^{3} = 2^{10}

k^{3} = 2^{9}

k = 2^{3} = 8

n = 2 k - 1 = 15

Commented by cortano1 last updated on 24/Feb/22

oo i think f^{2} (x) = (fof) (x)

what standart form (fof) (x) ?

Commented by MJS_new last updated on 24/Feb/22

I^{'} m not sure {what}^{'} s standard nowadays .

also f \circ g for some means f (g (x)) and for

others g (f (x))

too much confusion

if \sin^{2} x means {(\sin x)}^{2} then why \sin^{- 1} x

{doesn}^{'} t mean {(\sin x)}^{- 1} ?

what about \ln^{- 1} x ? is it \frac{1}{\ln x} or e^{x} ?

if we use f^{n} (x) for the n^{th} derivate then why

f^{- 1} (x) \neq \int f (x) d x ?

better ask in each case what is meant \dots

Commented by cortano1 last updated on 24/Feb/22

thanks you sir

Answered by mahdipoor last updated on 23/Feb/22

\lim_{x \to 1} \frac{f^{2} (x) - f^{2} (1)}{x - 1} = \lim_{x \to 1} (\frac{f (x) - f (1)}{x - 1}) (f (x) + f (1)) =

f^{^{'}} (1) \times 2 f (1) = (1 + 3 + 5 + \dots + n) \times 2 (1 + \dots + 1) =

{(\frac{n + 1}{2})}^{2} \times 2 \times (\frac{n + 1}{2}) = 2^{10} \Rightarrow \frac{n + 1}{2} = 2^{3} \Rightarrow

n = 15