x∈ ( 0 , 1 ) , k ∈ N prove that : kx^( k)


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Question Number 174439 by mnjuly1970 last updated on 01/Aug/22

$x \in (0, 1), k \in N$ $p r o v e t h a t :$ ${k x}^{k} < \frac{x}{1 - x} (m a t h a n a l y s i s)$
	Answered by behi834171 last updated on 01/Aug/22

	$f o r : k = 1 : 1 \times x^{1} \overset{?}{<} \frac{x}{1 - x} \Rightarrow x \overset{?}{<} \frac{x}{1 - x}$ $0 < x < 1 \Rightarrow - 1 < - x < 0 \Rightarrow 0 < 1 - x < 1$ $s o : 1 - x < 1 \Rightarrow \frac{x}{1 - x} > x,$ $i . e : i t i s t r u e f o r : k = 1$ $l e t a s s u m e t h a t i s t r u e f o r k = n$ $w e s h o u l d p r o v e t h a t i s t r u e f o r k = n + 1$ $(n + 1) x^{n + 1} \overset{?}{<} \frac{x}{1 - x}$ $(n + 1) x^{n + 1} (1 - x) = (n + 1) x^{n + 1} - (n + 1) x^{n + 2} <$ $< {n x}^{n} . x + x^{n + 1} - {n x}^{n} . x^{2} - x^{n + 2}$ $< \frac{x}{1 - x} . x - \frac{x}{1 - x} . x^{2} + x^{n + 1} - x^{n + 2} =$ $= \frac{x^{2} - x^{3}}{1 - x} + x^{n + 1} - x^{n + 2} = x^{2} + x^{n + 1} - x^{n + 2} <$ $< x + x - x = x . h e n c e p r o v e d . ◼$ $[0 < x < 1 \Rightarrow x^{p \in N} < x]$
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