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HI, i need someone who can write down each question by very detail steps. When you down just send me by picture. Thank you
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I wrote the solutions. I described in detail but it is always possible to write more:) Please read and ask if something is unclear.docx and pdf files are identical (docx is better if you'll need to edit it while pdf is more universal). I made pictures, too:)
1 𝑛
1. The mentioned known sequence is {(1 + 𝑛) } and its known limit is 𝑒.
The given sequence is {(1 +
1
1 𝑛
) }
4𝑛
1
= {((1 +
1 4𝑛 4
) ) }.
4𝑛
4𝑛
1 𝑛
The sequence {((1 + 4𝑛) )} is a subsequence of {(1 + 𝑛) } and hence has the same limit 𝑒.
1
The function 𝑓(𝑥) = 𝑥 4 is continuous and therefore
1
1
𝟏
1 4𝑛 4
1 4𝑛 4
lim ((1 + ) ) = ( lim (1 + ) ) = 𝒆𝟒 .
𝑛→∞
𝑛→∞
4𝑛
4𝑛
2. 𝑠1 = 1, 𝑠𝑛+1 = √2𝑠𝑛 + 3, 𝑛 ≥ 1.
We will start proof from (2) because it is used in proof of (1).
(2) Prove that 𝑠𝑛 is bounded from the above by the number 3.
Use induction: the statement 𝑠𝑛 < 3 is true for 𝑛 = 1 (the base).
Induction step: let 𝑠𝑛 < 3, consider 𝑛 + 1: 𝑠𝑛+1 = √2𝑠𝑛 + 3 < √2 ∙ 3 + 3 = √9 = 3, Q.E.D.
(1) Prove that 𝑠𝑛 is monotone increasing, 𝑠𝑛+1 > 𝑠𝑛 .
Consider the (quadratic) function 𝑓(𝑥) = 𝑥 2 − 2𝑥 − 3. Its roots are 𝑥 = −1 and 𝑥 = 3 and it is negative
between them. We know that 0 < 𝑠𝑛 < 3 for all 𝑛, therefore 𝑓(𝑠𝑛 ) = 𝑠𝑛2 − 2𝑠𝑛 − 3 < 0 which is the
same as 𝑠𝑛2 < 2𝑠𝑛 + 3, or 𝑠𝑛 < √2𝑠𝑛 + 3 = 𝑠𝑛+1 , Q.E.D.
Bounded from above and increasing sequence has a limit.
(3) Because 𝑠𝑛 has a limit, 𝑠𝑛+1 has the same limit 𝑠. Apply limit to both sides of the equality
𝑠𝑛+1 = √2𝑠𝑛 + 3 and get the equation 𝑠 = √2𝑠 + 3, or 𝑠 2 = 2𝑠 + 3. The root 𝑠 = −1 is not a possible
limit of 𝑠𝑛 while 𝑠 = 3 is. Thus lim 𝑠𝑛 = 𝟑.
𝑛→∞
3.
(1) It is known that the set ℚ is everywhere dense in ℝ (I can prove this but believe you can use this fact).
This means every interval contains at least one rational number. Therefore 𝐸 = ℝ\ℚ has no whole interval
inside it, which proves that its interior is empty.
In turn, ℝ\ℚ is dense in ℝ, too, because any interval is uncountable but ℚ is countable. This shows that no
interval can consist of rational numbers only and thus each interval contains at least one irrational number,
i.e. a member of 𝐸. Therefore the set of accumulation points of 𝑬 is entire ℝ.
(2) If 𝐴 is open and 𝐵 is closed we can say (and prove) that 𝑩\𝑨 is closed.
It is because 𝐵\𝐴 = 𝐵 ∩ 𝐴𝐶 where 𝐴𝐶 is 𝐴’s complement. The...
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