Determine-interval-A-of-real-numbers-for-which-a-a-1-a-1-a-whenever-a-A-

Question Number 1716 by Rasheed Soomro last updated on 02/Sep/15

D e t e r m i n e i n t e r v a l A o f r e a l n u m b e r s f o r w h i c h

a^{a + 1} ⩾ {(a + 1)}^{a} w h e n e v e r a \in A

Commented by Rasheed Ahmad last updated on 03/Sep/15

a^{a + 1} ⩾ {(a + 1)}^{a}

a^{a} . a ⩾ a^{a} {(1 + \frac{1}{a})}^{a}

W e a s s u m e h e r e a > 0 b e c a u s e

f o r a < 0 a^{a + 1}, {(a + 1)}^{a} m a y b e

i m a g i n a r y . (A t a = 0 t h e s t a t e m e n t

i s f a l s e)

S o a > 0 \Rightarrow a^{a} > 0

a^{a} . a ⩾ a^{a} {(1 + \frac{1}{a})}^{a} \Rightarrow {(1 + \frac{1}{a})}^{a} ⩽ a

{(1 + \frac{1}{a})}^{a} ⩽ {(a^{\frac{1}{a}})}^{a}

\Rightarrow 1 + \frac{1}{a} ⩽ a^{\frac{1}{a}}

a^{\frac{1}{a}} - \frac{1}{a} - 1 ⩾ 0

a^{\frac{1}{a}} - \frac{1}{a} - 1 = 0 \lor a^{\frac{1}{a}} - \frac{1}{a} - 1 > 0

[[W i t h t h e h e l p o f g r a p h i n g c a l c u l a t o r

a ⩾ 2.293 \dots .]]

B y u s i n g n u m e i c a l m e t h o d s \dots .

Answered by 123456 last updated on 03/Sep/15

x^{x + 1} ⩾ {(x + 1)}^{x}

h (x) = x^{x + 1} - {(x + 1)}^{x}

h (x) < 0 \Leftrightarrow x^{x + 1} < {(x + 1)}^{x}

h (x) = 0 \Leftrightarrow x^{x + 1} = {(x + 1)}^{x}

h (x) > 0 \Leftrightarrow x^{x + 1} = {(x + 1)}^{x}

h (0) = h (1) = h (2) = - 1

h (3) = 17

for x \in [0, + \infty) its continuous and since

h (2) < 0 < h (3) by inthermediare value theorem (rolle theorem)

\exists ξ \in [2, 3], h (ξ) = 0

comtinue

Commented by 123456 last updated on 03/Sep/15

h (x) = 0, bissection method (\approx values)

a b x h (a) h (b) h (x)

2, 00 ∣ 3, 00 ∣ 2, 50 ∣ - 1, 00 ∣ + 17, 00 ∣ + 1, 79

2, 00 ∣ 2, 50 ∣ 2, 25 ∣ - 1, 00 ∣ + 01, 79 ∣ - 0, 23

2, 25 ∣ 2, 50 ∣ 2, 38 ∣ - 0, 23 ∣ + 01, 79 ∣ + 0, 59

2, 25 ∣ 2, 38 ∣ 2, 32 ∣ - 0, 23 ∣ + 00, 59 ∣ + 0, 16

2, 25 ∣ 2, 32 ∣ 2, 29 ∣ - 0, 23 ∣ + 00, 16 ∣ - 0, 02

x \approx 2, 29, ∣ h (x) ∣\approx 0, 02

Commented by Rasheed Ahmad last updated on 03/Sep/15

Approach is really appreciable!

Commented by Rasheed Ahmad last updated on 14/Sep/15

h (x) > 0 \Leftrightarrow x^{x + 1} > {(x + 1)}^{x}