$f(z)=\frac{1}{z+1}+\frac{1}{z-1}$
$\gamma_j:[0,2\pi]\rightarrow \mathbb{C} \ (j=1,2,3)$
$\gamma_j(t)=(-1)^j+\frac{1}{2}\exp(it)$ for $j=1,2$
$\gamma_3(t)=4\exp(it)$
I need to compute
$\int_{\gamma_1}f(z) \mathrm{d}z + \int_{\gamma_2}f(z) \mathrm{d}z$ and $\int_{\gamma_3}f(z) \mathrm{d}z$.
Are there any better ways to compute these than using
$$\int_{0}^{2\pi}f(\gamma(t))\gamma'(t) \mathrm{d}t$$?
I recently ran into the following exercise:
Evaluate
$$\oint_\Gamma\frac{\cos z}{(z-\pi)^2}dz,$$where $\Gamma$ is a complete circuit of the circle $|z|=1$.
Clearly, the singularity lies outside the contour:
However, recall Cauchy's Residue Theorem:
If $\Gamma$ is a simple closed positively oriented contour and $f$ is analytic inside and on $\Gamma$ except at the points $z_1$, $z_2$, ..., $z_n$, then
$$\int_\Gamma f(z)dz=2\pi i\sum_{j=1}^{n}\text{Res}(z_j),$$
where
$$\text{Res}(f;z_0)=\lim_{z\to z_0}\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mf(z)].$$
If I use this to evaluate the integral, I will have that
$$\text{Res}(f;\pi)=\lim_{z\to\pi}\frac{d}{dz}[\cos z]=0,$$and hence the value of the integral will evaluate to $0$, which is the correct answer.
Is this purely coincidental? Because, as I understand it, for Cauchy's theorem to hold, the singularities must lie within the contour (or perhaps not?). Thanks in advance!
Answer
You don't have to do anything to evaluate your integral, because $\cos(z)/(z-\pi)^2$ is holomorphic on the unit disc. So by Cauchy's theorem for a disk (or whichever more general variant you'd prefer), the integral is zero. You can also evaluate this using the residue theorem, but none of the residues are inside the disk so once again you get zero.
The function does indeed have a pole of order $2$ at $\pi$ and the residue is $0$. This does not mean that the function has a removable singularity at the pole, it just means that the function has an analytic antiderivative in some punctured disk about $\pi$.
Problem:
How to find the following limit :
$$\lim_{n \to \infty}[(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})]^{\frac{1}{n}}$$ is equal to
(a) $\frac{4}{e}$
(b) $\frac{3}{e}$
(c) $\frac{1}{e}$
(d) $e$
Please suggest how to proceed in this problem thanks...
Answer
$$\log\left(\lim_{n \to \infty}[(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})]^{\frac{1}{n}}\right) =\lim_{n \to \infty}\frac{\log(1+\frac{1}{n})+\log(1+\frac{2}{n})+\cdots+\log(1+\frac{n}{n})}{n} =\int_{1}^2 \log(1+x)dx= [x\log(x)-x]_{x=1}^{x=2}=2\log(2)-1$$
This yields the solution $e^{2\log(2)-1}=4/e$.
The series is
$$1 + \frac{1}{2}.\frac{x^2}{4} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6}.\frac{x^4}{8} + \frac{1\cdot3\cdot5\cdot7\cdot9}{2\cdot4\cdot6\cdot8\cdot10}.\frac{x^6}{12}+... , x\gt0$$
I just stuck over the nth term finding and once I get nth term than I can do different series test but here I am unable to find the nth term of the given series please help me out of this.
The question is different as it contains x terms and its nth term will be totally different from marked as duplicate question
Answer
Hint 1: Since $(4n-2)^2\gt(4n-1)(4n-3)$, we have
$$
\begin{align}
\frac{1\cdot3\cdots(4n-3)}{2\cdot4\cdots(4n-2)}
&\le\sqrt{\frac{1\cdot3\cdots(4n-3)}{3\cdot5\cdots(4n-1)}\cdot\frac{1\cdot3\cdots(4n-3)}{1\cdot3\cdots(4n-3)}}\\
&=\sqrt{\frac1{4n-1}}
\end{align}
$$
Hint 2: Since $(4n-3)^2\gt(4n-2)(4n-4)$, we have
$$
\begin{align}
\frac12\cdot\frac{3\cdot5\cdots(4n-3)}{4\cdot6\cdots(4n-2)}
&\ge\frac12\sqrt{\frac{4\cdot6\cdots(4n-2)}{4\cdot6\cdots(4n-2)}\cdot\frac{2\cdot4\cdots(4n-4)}{4\cdot6\cdots(4n-2)}}\\
&=\frac12\sqrt{\frac1{2n-1}}
\end{align}
$$
Thus,
$$
\frac12\sqrt{\frac1{2n-1}}\,\frac{x^{2n}}{4n}\le\frac{1\cdot3\cdots(4n-3)}{2\cdot4\cdots(4n-2)}\frac{x^{2n}}{4n}\le\sqrt{\frac1{4n-1}}\,\frac{x^{2n}}{4n}
$$
and therefore, since $4n-1\ge3n$ for $n\ge1$, we have
$$
\bbox[5px,border:2px solid #C0A000]{\frac{x^{2n}}{8\sqrt2\,n^{3/2}}\le\frac{1\cdot3\cdots(4n-3)}{2\cdot4\cdots(4n-2)}\frac{x^{2n}}{4n}\le\frac{x^{2n}}{4\sqrt3\,n^{3/2}}}
$$
Prove using induction on k that for any natural number 'n'
$$\sum_{i=1}^n i^k \le \frac{n^k(n+1)}{2}$$
I know I first need to start off with a base case, which I think will be $k=0$
$$\sum_{i=1}^n i^0 \le \frac{n^0(n+1)}{2}$$
But I'm not quite sure how to prove this base case, honestly. And beyond that, I know I need an induction hypothesis and induction step where I increase k by 1. I'm a bit lost on how to solve this problem. If anyone could point me in the right direction, I'd be grateful.
Answer
Hint (without induction): for $k=1$ the equality holds:
$$\sum_{i=1}^n \;i \;=\; \frac{n(n+1)}{2} \quad \iff \quad \sum_{i=1}^n \;\frac{i}{n} \;=\; \frac{n+1}{2}$$
Note that $\;\cfrac{i}{n} \le 1\,$ for $1 \le i \le n\,$, which implies $\;\left(\cfrac{i}{n}\right)^k \le \cfrac{i}{n}\;$ for all $\;k \ge 1\,$. Then:
$$\frac{1}{n^k}\;\sum_{i=1}^n \;i^k \;=\; \sum_{i=1}^n \left(\frac{i}{n}\right)^k \;\le\; \sum_{i=1}^n \frac{i}{n} \;=\; \frac{n+1}{2}$$
Every proof of this fact that I've seen relies on guessing a "formula" for the GCD first, such as "the smallest positive integer of the form $ax+by$" or $\frac{ab}{\text{lcm}(a,b)}$. Then one shows that the guess was indeed correct and proves the result. I don't find these proofs very intuitive and I would like to know if there's a simpler proof that doesn't involve guessing what the GCD looks like (this includes the fundamental theorem of arithmetic, which seems like overkill).
The proof should go like this:
The statement is trivially true for $1$ and $(a,b)$ itself. Let $(a,b)=d$. Suppose $\exists c$ such that $1
Now suppose $1<(c,d)
EDIT:
So my original proof only used multiplicative properties of $\Bbb Z$, but I have learned that the very existence of the GCD requires additive properties as well. However, I've found a new proof that doesn't seem to use any additive properties (not even duality with LCM). I believe it is closer to what I was looking for. The reasoning behind this proof relies on additive properties of $\Bbb Z$, but they seem to disappear in my formal proof. What's going on here? How is this proof equivalent to other proofs?
Proof. Let $c$ be a common divisor of $a$ and $b$ ($a
Since $c \not \mid d$, we can't have $a=d$, so $a=kd$ for some $k>1$. Also $a=tc$, for some $t>k$. We have $kd=tc \implies c=\frac kt d$. Observe that $k \not \mid t$, otherwise $d=\frac tk c \implies c \mid d$. Let $v=(k,t)$; then $1 \le v < k$. Of course $(\frac kv, \frac tv)=1$.
Now, $$b=k'd=t'c=t' \frac kt d \implies k'=t' \frac kt= t' \frac{k/v}{t/v} \implies t/v \mid t'$$ But then $b= t' \frac kt d = t' \frac {k/v}{t/v} d=\frac {t'}{t/v} \left(\frac kv d \right)$. Also $a=kd=v \left( \frac kv d \right)$. This shows that $\frac kv d > d$ is a common divisor and completes the proof. $\square$
Note that $c \mid \frac kv d$ as well.
This proof is a formalization of the following hand-waving:
suppose that $a=4d=6c$. Then the respective times $d$ and $c$ are contained in any common multiple of $d$ and $c$ must always have a ratio of $2:3$. This means that there must be a factor of $2d$ (and therefore $3c$) in any common multiple. If, for example, $b=5d$, then $b=6c+d$. But $c \mid b$ and $c \mid 6c$ imply $c \mid d$. This is impossible, because $3c=2d \implies 3=2 \frac dc$, a contradiction. This situation arises every time there are two common divisors and neither divides the other.
Answer
This is not so much a direct answer to your question as an indication of how one of the standard approaches might naturally be motivated
Suppose $c|a$ and $c|b$ then $c|ha+kb$ for any integer choice of $h$ and $k$.
It is natural to constrain $c$ as much as possible, and we do this by taking the least positive value of $ha+kb$. Let's call this $f$, so we have $c|f$.
Now let's think about how this relates to $a$. We have $f\le a$ since $1a+0b=a$ and so we can divide $a$ by $f$ to get $a=mf+n$ with $0\le n\lt f\le a$. But $n=a-mf=(1-mh)a-mkb$ can't be a positive value, so must be zero. We therefore have $f|a$. Likewise $f|b$.
We now know that any common factor of $a$ and $b$ divides $f$, and also that $f$ is a common factor.
The tricky part of the proof, which you can do by uniqueness of prime factorisation as well, is to show that any common factor divides the highest common factor. Note that proving uniqueness of prime factorisation uses the additive properties of the integers and doesn't just depend on multiplicative properties.
So you will find that, at least implicit within your argument is an appeal to the additive properties of the integers.
This is quite a subtle point, and is the reason why the most efficient proofs are written the way they are. I agree they can seem a bit like magic, but they can also be motivated, as I have tried to illustrate.
[disclaimer: I've studied a lot of logic but never been good at analysis, so that's the angle I'm coming from below]
in my attempt to find a precise version of the 'definitions' usually given when first introducing leibniz notation in single or multivariable calculus or analysis, wherein there is no appeal to differentials or infinitesimals, I've discovered that I'm confused about a few interellated low-level issues around formalization and notation, etc. I don't know which questions are the more basic ones here, so I'll just ask them as go along explaining what I think I do understand about proposed formal defintions of the sort in question.
I'm concerned with both the dy/dx notation and the 'del y/del x' notation for partial derivatives, but just the real-valued case.
Since I suspect that my confusions stem from use-mention confusions, and the confusion of functions, variables, and their values, I will use, and assume familiarity with lambda notation, metavariables, and quasi-quotation throughout. Where not stated, ⌜λx.φ⌝ refers to the function on the largest real domain on which φ is a real number. Assume only definitions for real variables are sought after below.
Anyway, on with the definitions:
These short articles by the author Thurston (the first five results) all give roughly the same formal definition of leibniz notation:
http://scholar.google.ca/scholar?hl=en&as_sdt=0,5&q=thurston+leibniz
whereas the top results for this search are by the author Harrison, and each give one of a few slight variants on a different definition:
here is my understanding of their definitions:
HARRISON:
⌜dφ/dψ⌝ is a shorthand for ⌜D(λψ.φ)(ψ)⌝ i.e.
"dy/dx" stands for "D(λx.y)(x)"
This means that:
ψ must be a variable of the underlying logic (and it has a free and a bound occurence here) and
φ must be a string such that ⌜∀x, φ = f(x)⌝ is true for some function f:S→R where S⊆R.
Q1. does anyone have a simpler way to state the restriction on φ?
Q1.1 what is the proper name for the sort of string φ must be?
THURSTON:
⌜dφ/dψ⌝ is a shorthand for ⌜ψ'/φ'⌝ i.e.
"dy/dx" is short for "y'/x'"
This means that φ and ψ must be names of functions from the reals to the reals.
Q1.2 Are there other formal definitions in the literature I should compare to these? I haven't found any yet...
Harrison's defintion does not return a value because a free variable is uninstntiated, in the same sense in which wffs which are not sentences do not return a truth value. Thurston's version, however returns a function.
For instance,
df/dx = f'(x) for harrison,
df/dx = f' for thurston
More concretely
d(x²+x)/dx =
2x+1 for harrison
λx.2x+1 for thurston
Q2 is the value '2x+1' which contains a free variable being returned, or are we implicitly quantifying over x, or, let's call it 'ξ' for the purposes of quasi quotation, and saying: ∀ξ, ⌜2ξ+1⌝ refers to (not 'is') the value returned?
However, consider the following 'typical' calculus problem:
"
y = f(x)
x = g(u)
g(x) = x³ - 7
find df/dx
"
Thurston gets us df/dx = λx.1/(3x²)
This is undefined on harison's approach, since x is not a variable of the logic, but the name of a function.
Q3 should I be considering the possibility that the logic allows for variables ranging over functions?
And if we pretended that it was we'd still get df/dx = D(λx.f)(x) but this is mal-formed since λx needs something like 'f(x)' rather than 'f' as input.
And if we added a case to harrison's definition to append '(x)' or the like when it's missing, we'd still get df/dx = 1, which is not equal to the result we got with thurston's definition--but more strikingly, the function we evaluated to get 1 was, unike the f' vs f'(x) case above, not even the same function as was returned by thurston's definition.
Q4 What should I conclude from the fact that these definitions diverge in this manner?
Now consider:
"
y = f(x)
f(x) = x⁹
find dy/dx
"
On harrison's account, we could view y as a metavariable, so that f(x) is placed substitutionally into the defining string, but I have a feeling that is not the right way to understand it. However, if y is merely a variable of the logic, it is a free variable in the result, and we end up with one free variable too many..
On thurston's account, y must be the name of a function, but "y = f(x)" sets y equal to an expression with a free variable, not equal to the name of a function
Q5 should i view statements like "y=f(x)" as involving a supressed "(x)" and "∀" so that we get "∀x,y(x)=f(x)" ? Or should I see y as a metavariable? Or, should I imagine the logic extended to allow some new syntactic category of 'dependent' variables, while thinking of the usual variables in the logic as 'independent' variables--i.e. those whose value does not depend on others? I think I am very confused about what happends when one variable depends on another.
Q6 On a related note, I saw a passage recently that spoke in terms like "x(u) is the inverse function of u(x)"--how should this be understood more precisely? I've come to discover that I don't understand expressions of this sort at all!
Q7 Does either of these defintions clearly capture 'practice' better than another?
Q8 How should similar attempts be made for the 'del' notation for patial derivatives?
Q9 Can someone give me an example of where d/dx and del/delx return different values on the same input? If I'm not mistaken, in some formalizations this never happens, and in other formalizations it does--I think harrison's would not allow for this since it just returns an 'expression' rather than one of the various functions that can be formed by an expression when you apply a lambda operator to it.
I started also trying to read this article on revising patial derivative notation:
[I've hit my link limit as I'm new here, but google "revised notation for partial derivatives" (with quotes). It's by WC Hassenpflug]
but I got stuck on the sentence:
"If we have a function u = f(x,y) and the transformation y=g(x∩) is made, then it is not clear whether del u / dx means del u / dx |y or del y / dx | n"
Can someone explain that one to me?
Q10 This all bears some superficial similarity to the relationship between so-called 'random variables', which are actually functions, and what are called 'variables' in the underlying logic--this has also confused me, and I see many operations done in text on random variables where the operators have only been defined for values in the random variable's domain, and not on functions. Can anyone comment on this? I would be nice if I could dismiss two long-standing confusions with one stone :p
