Answer
There is a decision procedure called the Risch algorithm that will either tell you that the intergral is non-elementary, or produce an elementary anti-derivative. It is not an easy algorithm to execute, or even implement in a computer algebra system (although the latter has been done), so there is no hope of finding an easy condition for the existence of an anti-derivative.
We have the sequence $d_n = \begin{cases} 1 &\text{ if } n=0 \\
\frac{n}{d_{n-1}} &\text{ if } n>0 \end{cases}$
for all natural numbers $n$.
($d_{n-1}$ is the previous number of the sequence.)
examples: $d_0 = 1$, $d_1 = 1$, $d_2 = 2$, $d_3 = \frac{3}{2}$, $d_4 = \frac{8}{3} \dots$
I have to prove using induction that $\forall n \in \mathbb N \setminus \{0\}$, $d_{2n-1}$ $\leq$ $\sqrt{2n-1}$.
so far, I've figured out the pattern that for every n greater than or equal to $2$, $d_{2n-1} = d_{2n-3} \, \frac{2n-1}{2n-2}$.
i.e. $d_5 = d_3 \, \frac{5}{4}$
In the hints section, they told me to write $d_{2k+1}$ in terms of $d_{2k-1}$ and to use the difference of squares: $(2k-1)(2k+1) = 4k^2 - 1$ for the induction step.
Any hints/tips/advice on how to solve this problem is much appreciated!
Thank you!
Answer
Your observation $$d_{2n-1} = d_{2n-3} \, \frac{2n-1}{2n-2}$$ is homologous with $$d_{2k+1} = d_{2k-1} \, \frac{2k+1}{2k}.$$
\begin{align}
d_{2k-1} \, \frac{2k+1}{2k} &\le \sqrt{2k-1} \, \frac{\bbox[yellow, 2px]{2k+1}}{2k} \tag{induction hypothesis}\\
&\le \color{blue}{\sqrt{2k-1}} \, \frac{\bbox[yellow, 2px]{\color{blue}{\sqrt{2k+1}}}}{2k} \, \bbox[yellow, 2px]{\sqrt{2k+1}} \\
&\le \frac{\color{blue}{\sqrt{4k^2-1}}}{2k} \sqrt{2k+1} \tag{hint} \\
&\le 1 \cdot \sqrt{2k+1} = \sqrt{2k+1} \tag*{$\square$}
\end{align}
Let $n > 1$ and $m$ and $r$ be positive integers. Prove that $(n^r −1)$ divides $(n^m −1)$ if and only if $r$ divides $m$.
How do I show that if $$Dm^2 - n^2D^2$$ is a perfect square for some integers $m$ and $n$ ($n \neq 0$), $D$ is the sum of two (non-zero) perfect squares? I tried solving for $D$ but that only gives me $$D = \frac{m^2}{2n^2} \pm \frac{\sqrt{m^4 - 4n^2 k^2}}{2n^2}$$ for integers $m$, $n$, and $k$, which doesn't seem easier.
EDIT: $D$ itself should not be a perfect square.
Answer
If $Dm^2-D^2n^2= a^2$, then $Dm^2$ is a sum of two squares. Now an integer is a sum of two squares if and only if all primes $\equiv 3 \mod 4$ in its factorization occur with even multiplicities. The presence of the extra square $m^2$ doesn't affect this condition, so $D$ is a sum of two squares, also.
Main purpose: For self-learning performance, neither for exam nor degree courses.
Calculus textbook using now[1]: Calculus I, Weinstein&Marsden, UTM, Springer
Question Description: I've been reading book[1] for weeks, 90% of text, 30%-40% of exercises solved. It's not bad, but for the following:
Advantages: (a)Suitably explained for concepts (b)Clear Structure
Disadvantages: (a)Not contain enough theorems (b)Too many exercises in formula-calculation/real application (c)Too little deep/proof exercises (d) Approximately 8000 exercises in total, 300-400/chapter, but 80% is simple-formula-calculation/realistic application.
My Opinion: Will it be more beneficial to start using analysis textbooks now instead of this calculus book ? For 3 reasons:
(1). Most good EU bachelor in maths, they use analysis directly in first semester instead of calculus. (e.g. Bonn University/ETHz)
(2). Since book[1] contains too many exercises of formula-using/real application ones but not deep/proof, if I continue to work with it (solve all exercises/ second time reading), book[1] will still cost several months, I doubt if it's beneficial compared with directly starting analysis.
(3). Will Analysis textbooks(e.g. book[4][5]) also contain needed calculus?(intuition/calculation skills) If it's the case, such analysis books would do both to train modern theory and calculation skills( compute derivatives/integrals which are useful later such as ODE,PDE), then there'd be no need to read calculus any more.
Future Goal: Research in Dynamic System theoretically oriented.
Note: Though [1] is UTM, but it seems engineering-oriented(not theoretical/rigorous-oriented) compared with others within series.
[2]Rose, Elementary Analysis, UTM, Springer.
[3]Serge Lang, A First Course in Calculus/Calculus of Several Variables, UTM, Springer(Even though it's still calculus, but Lang's book is more abstract-oriented)
[4]Zorich, Analysis, Universitext, Springer(As @nbubis said, analysis needs intuition behind, from the content, it seems Zorich's analysis contains many physical problems, will it works for that ?)
[5]Courant, Introduction to Calculus and Analysis I&II, Springer
Desirable answer: Advices, Discussions
Answer
I would agree. I had taken some non-proof high school Calculus, so I am not sure if my experience would be completely similar to someone who wants to go straight into analysis.
I think someone with no background in calculus could read something like Principle of Mathmatical Analysis by Walter Rudin with no great difficulty. I was able to read this book without any proof experience. In fact, the beginning of Rudin are basic metric space topology and least upper bound property results which I feel are more suitable materials for learning proofs than the more tedious proofs of theorems about derivatives and integrals found in a Calculus book. Most analysis text like Rudin will eventually cover the fundamental results of Calculus like derivatives, integrals, means values theorem, Taylor Theorem, etc. However, as you mentioned there less are emphasis on on example and calculations (which has caused me some headaches later in my studies).
So I would say if you are more interested in studying pure mathematics in the future a real analysis text like Rudin or Pugn would be a good introduction to how to do proofs. Also a Calculus book by Spivak is also a good place to learn how to do proofs and calculus as well. If you are more interested in science, applied math, you may want to take a look in a Calculus book that emphasizes Calculations.
How to simplify $\sum_{i=1}^{k}\binom{n + i - 1}{i}$? I tried reducing the sum to $\binom{n}{1}, \binom{n}{2}, \binom{n}{3}$ and so on but couldn't get a pattern.
Answer
The problem becomes trivial on using Pascal's Rule. Using it, we have,
$$\binom{n+i-1}{i}=\binom{n+i}{i}-\binom{n+i-1}{i-1}$$
Now, substituting this into our required sum (say $S$) gives us a telescoping sum (the middle terms gets cancelled out).
$$S=\sum_{i=1}^k \binom{n+i-1}{i}=\sum_{i=1}^k \left\{\binom{n+i}{i}-\binom{n+i-1}{i-1}\right\}\\ \implies S=\binom{n+1}{1}-\binom{n}{0}+\binom{n+2}{2}-\binom{n+1}{1}+\ldots +\binom{n+k}{k}-\binom{n+k-1}{k-1}\\ \implies S=\binom{n+k}{k}-\binom{n}{0}\\ \implies \boxed{S=\dbinom{n+k}{k}-1}$$
Finding sum of
$\displaystyle 1+\frac{1\cdot 3}{6}+\frac{1\cdot 3 \cdot 5}{6\cdot 8}+\frac{1\cdot 3 \cdot 5 \cdot 7}{6 \cdot 8 \cdot 10}+\cdots \cdots$
Try: We can write sum as
$$ \mathcal{S} = 4\bigg[\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3 \cdot 5}{4\cdot 6 \cdot 8}+\cdots \cdots \cdots \bigg]$$
Now Let $$a_{n} = \prod^{n}_{k=1}(2k-1)=2^n\prod^{n}_{k=1}\bigg(k-\frac{1}{2}\bigg)$$
and $$b_{n} = 2^{-1}\prod^{n}_{k=1}(2k)=2^{n-1}\prod^{n}_{k=1}k$$
So $$\frac{a_{n}}{b_{n}} = 2\cdot \Gamma\left(n+\frac{1}{2}\right)\cdot \frac{1}{\Gamma\left(\frac{1}{2}\right)\cdot \Gamma(n+1)}$$
Above I have used $$\Gamma(x+n) = (x+n-1)(x+n-2)\cdot \cdots x \cdots \Gamma(n).$$
So $$\frac{a_{n}}{b_{n}} = 2\cdot \frac{\Gamma\left(n+\frac{1}{2}\right)\cdot \Gamma\left(\frac{1}{2}\right)}{\pi \cdot \Gamma(n+1)}= \frac{2}{\pi}\int^{1}_{0}x^{n-\frac{1}{2}}(1-x)^{-\frac{1}{2}}dx$$
So our sum is $$\mathcal{S} = \frac{8}{\pi} \sum^{\infty}_{n=1}\int^{1}_{0}x^n \cdot \frac{1}{\sqrt{x-x^2}}dx$$
So $$\mathcal{S} = \frac{8}{\pi}\int^{1}_{0}\frac{x}{(1-x)\sqrt{1-x^2}}dx$$
Put $x=\sin^2 \theta$ and $dx = 2 \sin \theta \cos \theta d\theta$ and changing limits
So we have $$\mathcal{S} = \frac{16}{\pi}\int^{\frac{\pi}{2}}_{0}(\sec^2 \theta -1)d \theta = $$
I did not understand where i am wrong.
and answer is $4$
Could some help me to explain it , thanks
Answer
You almost got everything right, and the only problem you have is a minor error when you define $b_n$. You should have
$$b_n=2^n\,(n+1)!=2^n\,\Gamma(n+2)\text{ for }n=1,2,3,\ldots\,.$$
Thus, for each $n=1,2,3,\ldots$,
$$\begin{align}\frac{a_n}{b_n}&=\frac{\Gamma\left(n+\frac12\right)}{\Gamma\left(\frac12\right)\,\Gamma(n+2)}=\frac{2}{\pi}\,\left(\frac{\Gamma\left(n+\frac12\right)\,\Gamma\left(\frac32\right)}{\Gamma(n+2)}\right)\\&=\frac{2}{\pi}\,\text{B}\left(n+\frac12,\frac32\right)\,,\end{align}$$
where $\Gamma$ and $\text{B}$ are the usual gamma and beta functions, respectively. Hence,
$$\frac{a_n}{b_n}=\frac{2}{\pi}\,\int_0^1\,x^{n-\frac12}\,(1-x)^{\frac12}\,\text{d}x\,,$$
so
$$\begin{align}\sum_{n=1}^\infty\,\frac{a_n}{b_n}&=\frac{2}{\pi}\,\int_0^1\,\frac{x^{\frac12}}{1-x}\,(1-x)^{\frac12}\,\text{d}x
\\&=\frac{2}{\pi}\,\int_0^1\,x^{\frac12}\,(1-x)^{-\frac12}\,\text{d}x\,.\end{align}$$
That is, with $u:=x^{\frac12}$, we obtain
$$\begin{align}\sum_{n=1}^\infty\,\frac{a_n}{b_n}&=\frac{4}{\pi}\,\int_0^1\,\frac{u^2}{\sqrt{1-u^2}}\,\text{d}u\\&=\frac{2}{\pi}\,\left(\text{arcsin}(u)-u\,\sqrt{1-u^2}\right)\Big|_{u=0}^{u=1}\,.\end{align}$$
Ergo,
$$\sum_{n=1}^\infty\,\frac{a_n}{b_n}=1\,,$$
whence
$$1+\frac{1\cdot 3}{6}+\frac{1\cdot 3\cdot 5}{6\cdot 8}+\frac{1\cdot3\cdot 5\cdot 7}{6\cdot 8\cdot 10}+\ldots=4\,\sum_{n=1}^\infty\,\frac{a_n}{b_n}=4\,.$$
In fact, one can show that
$$f(z):=\sum_{n=0}^\infty\,\prod_{k=1}^n\,\left(\frac{k-\frac32}{k}\right)\,z^n=(1-z)^{\frac12}$$
for all $z\in\mathbb{C}$ with $|z|\leq 1$. The requested sum satisfies
$$S=8\,\Biggl(1-\frac12-f(1)\Biggr)=4\,.$$
