calculate-Q-If-f-x-2-e-x-1-e-x-3-2-e-x-f-1-pi-4-

Question Number 199405 by mnjuly1970 last updated on 03/Nov/23

calculate \dots

Q : I f, f (x) = 2 e^{x} - 1 + ⌊ e^{x} + \frac{3}{2} + ⌊ e^{x} ⌋ ⌋

\Rightarrow f^{- 1} (\frac{π}{4}) = ?

Answered by Frix last updated on 03/Nov/23

e^{x} = i + f; i \in Z \land 0 ⩽ f < 1

e^{π} = 23 + ζ

2 (i + f) - 1 + ⌊ i + f + 1 + \frac{1}{2} + ⌊ i + f ⌋ ⌋ = 23 + ζ

2 i - 1 + 2 f + ⌊ 2 i + 1 + f + \frac{1}{2} ⌋ = 23 + ζ

4 i + 2 f + ⌊ f + \frac{1}{2} ⌋ = 23 + ζ

0 ⩽ f < \frac{1}{2}

4 i + 2 f = 23 + ζ \Rightarrow i = \frac{23}{4} \notin Z

\frac{1}{2} ⩽ f < 1 \Rightarrow 0 ⩽ 2 f - 1 < 1

4 i + 2 + (2 f - 1) = 23 + ζ \Rightarrow i = \frac{21}{4} \notin Z

\Rightarrow f (x) \neq e^{π}

\Rightarrow f^{- 1} (e^{π}) does not exist

Commented by Frix last updated on 03/Nov/23

Please correct me if I^{'} m wrong .

Commented by mnjuly1970 last updated on 03/Nov/23

t h a n k y o u s o m u c h s i r

f o r y o u r e f f o r t .

Answered by mnjuly1970 last updated on 03/Nov/23

m y s o l u t i o n

f (x) = 2 e^{x} - 1 + ⌊ e^{x} ⌋ + ⌊ e^{x} + \frac{1}{2} ⌋ + 1

= 2 e^{x} + ⌊ 2 e^{x} ⌋

- - - - -

y = 2 e^{x} + ⌊ 2 e^{x} ⌋ \Rightarrow x = 2 e^{y} + ⌊ 2 e^{y} ⌋

⌊ x ⌋ = 2 ⌊ 2 e^{y} ⌋ \Rightarrow ⌊ 2 e^{y} ⌋ = \frac{⌊ x ⌋}{2}

- - - -

x = 2 e^{y} + \frac{⌊ x ⌋}{2} \Rightarrow y = l n (\frac{x}{2} - \frac{⌊ x ⌋}{4})

f^{- 1} (x) = l n (\frac{x}{2} - \frac{⌊ x ⌋}{4})

f^{- 1} (\frac{π}{4}) = l n (\frac{π}{8}) \Leftrightarrow f (l n (\frac{π}{8})) = \frac{π}{4}

Commented by Frix last updated on 03/Nov/23

f (x) = {2 e}^{x} + ⌊ {2 e}^{x} ⌋

\lim_{x \to {(\ln 6)}^{-}} f (x) = 23

\lim_{x \to {(\ln 6)}^{+}} f (x) = 24

e^{π} \approx 23.1407

\Rightarrow \forall x \in R : f (x) \neq e^{π}

======================

f (x) = {2 e}^{x} + ⌊ {2 e}^{x} ⌋

f^{- 1} (x) = \ln \frac{2 x - ⌊ x ⌋}{4}

f^{- 1} (f (x)) = x but f (f^{- 1} (x)) \neq x

This is strange \dots

f^{- 1} (e^{π}) = \ln \frac{{2 e}^{π} - 23}{4}

But

f (\ln \frac{{2 e}^{π} - 23}{4}) = e^{π} - \frac{1}{2}

So {there}^{'} s something wrong .

Commented by Frix last updated on 03/Nov/23

v = u + ⌊ u ⌋

The range of this function is

R ∖ {v \in R \land n \in Z ∣ 2 n - 1 ⩽ v < 2 n}

\Rightarrow v^{- 1} is not defined in these intervals

\Rightarrow

f^{- 1} (x) = {\begin{cases} \ln \frac{2 x - ⌊ x ⌋}{4}, 2 n ⩽ x < 2 n + 1 \\ undefined, 2 n - 1 ⩽ x < 2 n \end{cases} \forall n \in Z

23 < e^{π} < 24

2 \times 12 - 1 < e^{π} < 2 \times 12

\Rightarrow

f^{- 1} (e^{π}) {doesn}^{'} t exist

Commented by mnjuly1970 last updated on 03/Nov/23

t h a n k s a l o t s i r . y o u a r e r i g h t \underset{―}{\underset{⏟}{⋚}}

Commented by Frix last updated on 03/Nov/23

{You}^{'} re welcome, working on these things

keeps my brain young .

Facebook Tweet Pin