Changed layout for more manageable volumes

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-#LyX 2.4 created this file. For more info see https://www.lyx.org/
-\lyxformat 620
-\begin_document
-\begin_header
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-\index Index
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-\end_header
-
-\begin_body
-
-\begin_layout Standard
-\begin_inset Note Note
-status open
-
-\begin_layout Plain Layout
-TODO 1,
- 2,
- 4,
- 6,
- 11,
- 13 (1 p.) (est.
- 0:21:19)
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc1[HM01]
-\end_layout
-
-\end_inset
-
-What is 
-\begin_inset Formula $\lim_{n\to\infty}O(n^{-1/3})$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-0.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc2[M10]
-\end_layout
-
-\end_inset
-
-Mr.
- B.
- C.
- Dull obtained astonishing results by using the 
-\begin_inset Quotes eld
-\end_inset
-
-self-evident
-\begin_inset Quotes erd
-\end_inset
-
- formula 
-\begin_inset Formula $O(f(n))-O(f(n))=0$
-\end_inset
-
-.
- What was his mistake,
- and what should the right-hand side of his formula has been?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $O(f(n))$
-\end_inset
-
- is a set of functions,
- not a single function,
- so the two 
-\begin_inset Formula $O(f(n))$
-\end_inset
-
- do not cancel out (for example,
- the first 
-\begin_inset Formula $O(f(n))$
-\end_inset
-
- could represent 
-\begin_inset Formula $f(n)$
-\end_inset
-
- and the second one could be 0).
- The right hand side should be another 
-\begin_inset Formula $O(f(n))$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc4[M15]
-\end_layout
-
-\end_inset
-
-Give an asymptotic expansion of 
-\begin_inset Formula $n(\sqrt[n]{a}-1)$
-\end_inset
-
-,
- if 
-\begin_inset Formula $a>0$
-\end_inset
-
-,
- to terms 
-\begin_inset Formula $O(1/n^{3})$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We can do this by considering the function in terms of 
-\begin_inset Formula $x\coloneqq\frac{1}{n}$
-\end_inset
-
-,
- which would be 
-\begin_inset Formula $\frac{1}{x}(a^{x}-1)$
-\end_inset
-
-.
- Then 
-\begin_inset Formula $a^{x}=\text{e}^{x\ln a}=1+x\ln a+\frac{1}{2}x^{2}\ln^{2}a+\frac{1}{6}x^{3}\ln^{3}a+O(x^{4}\ln^{4}a)$
-\end_inset
-
-,
- so
-\begin_inset Formula 
-\[
-a^{x}-1=x\ln a+\frac{x^{2}\ln^{2}a}{2}+\frac{x^{3}\ln^{3}a}{6}+O(x^{4})=\frac{\ln a}{n}+\frac{\ln^{2}a}{2n^{2}}+\frac{\ln^{3}a}{6n^{3}}+O(n^{-4})
-\]
-
-\end_inset
-
-and finally
-\begin_inset Formula 
-\[
-n(\sqrt[n]{a}-1)=\ln a+\frac{\ln^{2}a}{2n}+\frac{\ln^{3}a}{6n^{2}}+O(n^{-3}).
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc6[M20]
-\end_layout
-
-\end_inset
-
-What is wrong with the following arguments?
- 
-\begin_inset Quotes eld
-\end_inset
-
-Since 
-\begin_inset Formula $n=O(n)$
-\end_inset
-
-,
- and 
-\begin_inset Formula $2n=O(n)$
-\end_inset
-
-,
- ...,
- we have
-\begin_inset Formula 
-\[
-\sum_{k=1}^{n}kn=\sum_{k=1}^{n}O(n)=O(n^{2}).\text{''}
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-The second expression does not make sense.
- It is a sum of 
-\begin_inset Formula $n$
-\end_inset
-
- functions each of which has 
-\begin_inset Formula $n$
-\end_inset
-
- as a parameter,
- but if 
-\begin_inset Formula $n$
-\end_inset
-
- is a parameter,
- it cannot be an 
-\begin_inset Quotes eld
-\end_inset
-
-external
-\begin_inset Quotes erd
-\end_inset
-
- parameter of the function represented by 
-\begin_inset Formula $O(n)$
-\end_inset
-
- as well.
- In this case,
- it is invalid to move the big O inside the sum as the range of the sum depends on 
-\begin_inset Formula $n$
-\end_inset
-
- and so does the domain of values 
-\begin_inset Formula $k$
-\end_inset
-
- can take,
- so 
-\begin_inset Formula $k$
-\end_inset
-
-,
- which depends on 
-\begin_inset Formula $n$
-\end_inset
-
-,
- cannot be treated like a constant.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc11[M11]
-\end_layout
-
-\end_inset
-
-Explain why Eq.
- (18) is true.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-The first identity is because 
-\begin_inset Formula $\sqrt[n]{n}=\text{e}^{\ln\sqrt[n]{n}}=\text{e}^{\ln n^{1/n}}=\text{e}^{\ln n/n}$
-\end_inset
-
-,
- taking the power 
-\begin_inset Formula $1/n$
-\end_inset
-
- out of the logarithm.
- The second identity is a direct application of Eq.
- (12),
- using that 
-\begin_inset Formula $\ln n/n\to0$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc13[M10]
-\end_layout
-
-\end_inset
-
-Prove or disprove:
- 
-\begin_inset Formula $g(n)=\Omega(f(n))$
-\end_inset
-
- if and only if 
-\begin_inset Formula $f(n)=O(g(n))$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $g(n)=\Omega(f(n))$
-\end_inset
-
- if and only if there exist 
-\begin_inset Formula $n_{0}$
-\end_inset
-
- and 
-\begin_inset Formula $L$
-\end_inset
-
- such that 
-\begin_inset Formula $|g(n)|\geq L|f(n)|$
-\end_inset
-
- for each 
-\begin_inset Formula $n\geq n_{0}$
-\end_inset
-
-,
- but this is the same as saying that,
- for all such 
-\begin_inset Formula $n$
-\end_inset
-
-,
- 
-\begin_inset Formula $|f(n)|\leq\frac{1}{L}|g(n)|$
-\end_inset
-
-,
- which is to say that 
-\begin_inset Formula $f(n)=O(g(n))$
-\end_inset
-
-.
- Since arguably this 
-\begin_inset Formula $L$
-\end_inset
-
- must be positive (otherwise the statement is trivial),
- taking 
-\begin_inset Formula $\frac{1}{L}$
-\end_inset
-
- is valid,
- and likewise,
- if 
-\begin_inset Formula $|f(n)|\leq M|g(n)|$
-\end_inset
-
- for each 
-\begin_inset Formula $n\geq n_{0}$
-\end_inset
-
- and some 
-\begin_inset Formula $M$
-\end_inset
-
-,
- we can always make this 
-\begin_inset Formula $M$
-\end_inset
-
- positive to make the reverse argument.
-\end_layout
-
-\end_body
-\end_document