text:
# In dual slope type of ADCs, an input hold time is _____

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1. Almost zero
2. Higher than that of flash type ADCs
3. Longest
4. All of the above

Option 2 : Higher than that of flash type ADCs
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## Detailed Solution

Explanation:

In dual slope type ADC, the integrator generates two different ramps, one with the known analog input voltage VA and another with a known reference voltage –Vref. Hence it is called a dual-slope A to D converter. The logic diagram for the same is shown below.

The dual ramp output waveform is shown below:

The Output Voltage of Integrator is given by:

• $${V_0} = - \frac{{{V_a}RC}}{{{T_1}}} + \frac{{{V_{ref}}\left( {t - {T_1}} \right)}}{{RC}}$$
• It is mostly used in digital voltmeters.
• In the dual-slope A/D converter, the leakage current of the capacitor can cause errors in the integration and consequentially, an error.
• These effects, in the dual-slope AID converter, will manifest themselves as a reading of the DVM when no input voltage is present.

Its main disadvantage is of slow conversion rate and also due to the integration process during input hold time.

Hence flash type ADC needs only one clock with minimum conversion time.

So input hold time of dual slope type ADC is higher than that of flash type ADCs.

Hence option (2) is the correct answer.

Important Points

The conversion time of different types of ADC is shown :

 Type of ADC No. of Clocks Time Counter Type 2n-1 2n-1 Tclk SAR n nTclk Flash 1 Tclk Dual Slope 2n+1 2n+1 Tclk

• From the above table, it is clear that the dual-slope is the slowest ADC, and Flash Type is the fastest ADC.
• The conversion type of Flash type is independent of the number of bits.
• As we require (2n – 1) comparators for n-bit conversion, the circuit is complex and expensive.
• These are employed on very high-speed digital acquisition systems.


text:
# A body is in equilibrium when three forces act on it. Two of the forces are F1 of magnitude 8N...

## Question:

A body is in equilibrium when three forces act on it. Two of the forces are F1 of magnitude 8N acting due east and F2 which is 5N acting N 60degree E. Find the magnitude and direction of the third force.

## Equilibrium of Forces:

When to or more forces are acting on a body the body remains in equilibrium if vector sum of all the forces acting on the body is zero. Here we shall find the resultant of the 8 N and 5 N forces. The magnitude of the required force will be equal to the magnitude of resultant and just opposite to the resultant.

The magnitude of third force 'F' is given by

{eq}F^2=F_1^2+F_2^2+2F_1F_2\cos60 \\ F^2=8^2+5^2+2 \times 8 \times 5\cos60 \\ F=11.35\ N {/eq}

The direction of this force with east is given by

{eq}tan \alpha= \dfrac{F_2 \sin 60}{F_1+F_2 cos60} \\ tan \alpha= \dfrac{5 \sin 60}{8+5 cos60} \\ \Rightarrow \alpha =22.4^\circ {/eq}

Hence the third force should make an angle of 180+22.4=200.4 degree with east

or 22.4 degree south of west

format: plain


text:
Kronecker Product?
02-17-2019, 02:39 PM
Post: #1
 DrD Senior Member Posts: 1,128 Joined: Feb 2014
Kronecker Product?
How could a Kronecker Product be found with the hp prime?
(This would be like Matlab's kron() command, or Wolfram's KroneckerProduct() command).

https://en.wikipedia.org/wiki/Kronecker_product

I'm no expert on this subject matter, but have recently encountered a need for this in connection with an energy flow problem. Here is an example of a 4x4 identity matrix, with a 2x2 matrix, resulting in an 8x8 Kronecker product:

Example (using wxmaxima here):

(%i3) a:ident(4);
b:matrix ([1,-1],[-1,1]);
kronecker_product(a,b);

(a) matrix(
[1, 0, 0, 0],
[0, 1, 0, 0],
[0, 0, 1, 0],
[0, 0, 0, 1]
)
(b) matrix(
[1, -1],
[-1, 1]
)
0 errors, 0 warnings

(%o3) matrix(
[1, -1, 0, 0, 0, 0, 0, 0],
[-1, 1, 0, 0, 0, 0, 0, 0],
[0, 0, 1, -1, 0, 0, 0, 0],
[0, 0, -1, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 1, -1, 0, 0],
[0, 0, 0, 0, -1, 1, 0, 0],
[0, 0, 0, 0, 0, 0, 1, -1],
[0, 0, 0, 0, 0, 0, -1, 1]
)
02-17-2019, 03:47 PM (This post was last modified: 02-17-2019 04:25 PM by informach.)
Post: #2
 informach Member Posts: 109 Joined: Jul 2015
RE: Kronecker Product?
Hi!, DrD :
You can use, for calculate matrices, with Hadamard, as …
Syntax :
From … User_Guide_EN_2018_01_12_1.pdf (page 475).
02-17-2019, 05:48 PM
Post: #3
 DrD Senior Member Posts: 1,128 Joined: Feb 2014
RE: Kronecker Product?
(02-17-2019 03:47 PM)informach Wrote:  Hi!, DrD :
You can use, for calculate matrices, with Hadamard, as …
Syntax :
From … User_Guide_EN_2018_01_12_1.pdf (page 475).

The hadamard() command fails. Please note that the matricies in the example I provided are of differing sizes: a(4x4) and b(2x2), and produce an 8x8 Kronecker product.

I'm not seeing any in-built commands that produce the Kronecker product. It may have to be created as an external command, or function.

[attachment=6953]
02-17-2019, 07:52 PM (This post was last modified: 02-17-2019 08:01 PM by Didier Lachieze.)
Post: #4
 Didier Lachieze Senior Member Posts: 1,181 Joined: Dec 2013
RE: Kronecker Product?
Here is a CAS program to calculate the Kronecker product of two matrices:

Code:
#cas kronecker(a,b):= BEGIN  local m,n,p,q;  m:=rowDim(a); n:=colDim(a);  p:=rowDim(b); q:=colDim(b);  makemat((j,k)→a(iquo(j-1,p)+1,iquo(k-1,q)+1)*b(((j-1) MOD p)+1,((k-1) MOD q)+1),m*p,n*q); END; #end
02-17-2019, 08:06 PM
Post: #5
 DrD Senior Member Posts: 1,128 Joined: Feb 2014
RE: Kronecker Product?
(02-17-2019 07:52 PM)Didier Lachieze Wrote:  Here is a CAS program to calculate the Kronecker product of two matrices:

Code:
#cas kronecker(a,b):= BEGIN  local m,n,p,q;  m:=rowDim(a); n:=colDim(a);  p:=rowDim(b); q:=colDim(b);  makemat((j,k)→a(iquo(j-1,p)+1,iquo(k-1,q)+1)*b(((j-1) MOD p)+1,((k-1) MOD q)+1),m*p,n*q); END; #end

Thank you, kind sir! Excellent, as always. I was a bit surprised not to find this in the suite of similar commands, but I'll add this into my collection!
02-18-2019, 08:58 AM
Post: #6
 Didier Lachieze Senior Member Posts: 1,181 Joined: Dec 2013
RE: Kronecker Product?
Here is a slightly updated version, replacing the Home MOD function by the CAS irem function to be more consistent:

Code:
#cas kronecker(a,b):= BEGIN  local m,n,p,q;  m:=rowDim(a); n:=colDim(a);  p:=rowDim(b); q:=colDim(b);  makemat((j,k)→a(iquo(j-1,p)+1,iquo(k-1,q)+1)*b(irem(j-1,p)+1,irem(k-1,q)+1),m*p,n*q); END; #end
02-18-2019, 10:26 AM
Post: #7
 DrD Senior Member Posts: 1,128 Joined: Feb 2014
RE: Kronecker Product?
Thanks for the update. I was studying your makelist construction, and thinking to myself how nicely crafted, made even better, now, by using only CAS commands.

There is a lot, in just your short program, to be learned about the hp prime, CAS/HOME, and programming; embedded in it. This would make a great example for the user guide, to introduce CAS programming structures, and techniques.

Short, efficient, and to the point! Very well done!

-Dale-
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text:
# [Tugindia] Changing font size of footnote label

Anoop Sinha dr.anoopsinha at gmail.com
Mon Mar 5 05:54:13 CET 2012

Hi everyone,

I'm preparing a presentation in Beamer, and I need to add footnotes
(mainly references to the journals I used in making the presentation).
I managed to reduce the font size of the footnotes by doing the
following:

\def\Tiny{\fontsize{5pt}{5pt}\selectfont}
\makeatletter
\patchcmd{\@makefntext}{\insertfootnotetext{#1}}{\insertfootnotetext{\Tiny#1}}{}{}
\makeatother

While this worked with the text in the footnotes, the footnote labels


text:
1

NEET 2016 Phase 2

Suppose the elements X and Y combine to form two compounds XY2 and X3Y2. When 0.1 mole of XY2 weights 10 g and 0.05 mole of X3Y2 weighs 9 g, the atomic weights of X and Y are
A
40, 30
B
60, 40
C
20, 30
D
30, 20

Explanation

Assume the atomic weight of element X is x and element Y is y.

For XY2, $n = {w \over {Mol.\,\,wt.}}$

$0.1 = {{10} \over {x + 2y}}$

$\Rightarrow x + 2y = {{10} \over {0.1}} = 100$    ...(i)

For X3Y2, $n = {w \over {Mol.\,\,wt.}}$

$0.05 = {9 \over {3x + 2y}}$

$\Rightarrow 3x + 2y = {9 \over {0.05}} = 180$    ...(ii)

On solving equations (i) and (ii),

we get y = 30 g mol-1

Put the value of y in equation. (i)

x + 2(30) = 100

$\Rightarrow$ x = 100 - 60 = 40 g mol-1
2

AIPMT 2015 Cancelled Paper

A mixture of gases contains H2 and O2 gases in the ratio of 1 : 4 (w/w). What is the molar ratio of the two gases in the mixture ?
A
16 : 1
B
2 : 1
C
1 : 4
D
4 : 1

Explanation

Ratio of weight of gases = wH2 : wO2 = 1 : 4

$\therefore$ Number of moles of H2 = $1 \over 2$

$\therefore$ Number of moles of O2 = $4 \over 32$

Ratio of moles of gases

= nH2 : nO2

= $1 \over 2$ : $4 \over 32$

= $1 \over 2$ $\times$ $32 \over 4$ = 4 : 1
3

AIPMT 2015

The number of water molecules is maximum in
A
1.8 gram of water
B
18 gram of water
C
18 moles of water
D
18 molecules of water

Explanation

1.8 g of water = ${{6.023 \times {{10}^{23}}} \over {18}} \times 1.8$

= 6.023 $\times$ 1022 molecules

$\therefore$ 18 g of water = 6.023 $\times$ 1023 molecules = 1 mole of water

$\therefore$ 18 moles of water = 18 $\times$ 6.023 $\times$ 1023 molecules

4

AIPMT 2015

If Avogadro number NA, is changed from 6.022 $\times$ 1023 mol$-$1 to 6.022 $\times$ 1020 mol$-$1, this would change
A
the mass of one mole of carbon
B
the ratio of chemical species to each other in a balanced equation
C
the ratio of elements to each other in a compound
D
the definition of mass in units of grams.

Explanation

We know mass of 1 mol (6.022 $\times$ 1023) atoms of carbon = 12 g

If Avogadro number is changed to 6.022 $\times$ 1020 then mass of 1 mol of cabon is

= ${{12 \times 6.022 \times {{10}^{20}}} \over {6.022 \times {{10}^{23}}}}$ = $12 \times {10^{ - 3}}\,g$

$\therefore$ It would change the mass of one mole of carbon.

NEET


text:
### May 26, 2011

I redesigned the underlying code of my Books page, making it cleaner and more easily maintainable. As an unforeseen by-product of this, the two bugs which have been bugging me for some time (the Multiple Role Bug and the String Length Restriction Bug, as I call them) magically vanished! Also, browsing by category or author (or more generally, by any method which requires many-to-many tables to be joined) used to mean that under each book, only the category/author under consideration would show (even though a book might be written by many authors or be in many categories); no more! And there's a new feature: For all non-book views on the main page, you get the number of books in a category in parenthesis.

The Multiple Role Bug meant that for books where some authors had roles (editor, illustrator, translator, etc.), I'd get two sets of roles. (Example: The Solaris Book of New Science Fiction.) I had this fixed with a kludge, deleting every other role. However, this worked most places, but not all. It turned out that I'd get n sets of roles, where n is the number of categories in which a book is (most books are in two categories).

The String Length Restriction Bug would cut off the comma-separated strings containing the authors' names and ids long before its actual limit was reached, so that anthology books with lots of contributing authors were bungled. (Example: Dangerous Visions.) I never could figure out how to fix it.

However, I looked at my database queries, and designed them a little more logically. Before, I had one master query per view (by category, by series, single books, special criteria, etc.) and sent along GROUP_CONCAT()s for author names, ids, and roles, and for category names and ids, and then had my home-grown print_book() function work on the strings. Now, I do two separate queries inside the function to fish out author and category information, so that instead of one massive and unwieldy master query, I have three smaller ones. The lesson? Don't be afraid to use several queries.

I'm so happy with the way it works now that in a spirit of why-not-ism I added all the rest of my books to the database. So in the future when I've read a book, making it appear in the various views is as easy as flipping a boolean value. This should serve as an incentive to read more books (something I've really been neglecting lately).


text:
## Monday, December 22, 2014

### Compiling regular expressions (II)

Automata are modeled as 'state' records with two fields. The pos field contains the set of positions that are valid for recognition in the given state. Transitions are modeled as lists of pairs of symbols and states. In this way a state may contain transitions that reference itself.

type state = {
pos : Int_set.t;
mutable trans : (char * state) list ;
}

We will require a function that for each input symbol $a$ and a given set of positions $s$, computes the list of pairs $(a, s')$ where $s'$ is the subset of $s$ that correspond to $a$.

let (partition : char option array -> Int_set.t
-> (char option * Int_set.t) list) =
fun chars s ->
let f acc c =
match c with
| Some _ ->
if List.mem_assoc c acc then acc
else
let f i acc =
if chars.(i) <> c then acc else Int_set.add i acc in
(c, Int_set.fold f s (Int_set.empty)) :: acc
| None ->
if List.mem_assoc c acc then acc else (c, Int_set.empty) :: acc in
List.rev (Array.fold_left f [] chars)

This function makes a list from a set of ints.
let list_of_int_set : Int_set.t -> Int_set.elt list =
fun s -> List.rev (Int_set.fold (fun e acc -> e :: acc) s [])

This function, accessible given a state, computes the list of sets that accessible from that state.
let (accessible : state -> Int_set.t array -> char option array
-> (char * Int_set.t) list) =
let part = partition chars s.pos in
let f p rest =
match p with
| (Some c, l) ->
(c,
List.fold_left
(Int_set.union)
(Int_set.empty)
(List.map (Array.get follow) (list_of_int_set l))
) :: rest
| _ -> rest  in
List.fold_right f part []

find_state takes a set $s$ and two lists of states (marked and unmarked). It searches for a state which has a pos field equal to $s$ and returns this state or it fails.
let (find_state : Int_set.t -> state list -> state list -> state) =
fun s l m ->
let test e = e.pos = s in
try
List.find test l
with
| Not_found -> List.find test m

The algorithm to compute the automata works like this. Two lists are maintained, marked and unmarked states. The algorithm is initialized such that the only state is unmarked with a pos field containing first_pos $n_{0}$ where $n_{0}$ is the root of the syntax tree; the list of transitions is empty.

For an unmarked state $st$, the algorithm does these things:

• Calculate a set of numbers accessible from $st$. That is, a set of pairs $(c, s)$, where $c$ is a character and $s$ a set of positions. A position $j$ is accessible from $st$ by $c$ if there is an $i$ in st.pos such that $j$ is in follow $i$ and $i$ numbers the character $c$.
• For each of the pairs $(c, s)$
• If there exists a state st' (whether marked or unmarked) such that $s =$st'.pos, it adds $(c, st')$ to the transitions of $st$;
• Otherwise, a new state $st'$ without transitions is created, added to the transitions of $st$, and $st'$ is added to the list of unmarked states.
• It marks $st$.
The algorithm terminates only when there are no remaining unmarked states. The result is an array of states obtained from the list of marked states. The terminal states are all those containing the number associated with Accept. Here then is the algorithm in code.
let rec (compute_states : state list -> state list -> Int_set.t array
-> char option array -> state array) =
fun marked unmarked follow chars ->
match unmarked with
| [] -> Array.of_list marked
| st :: umsts ->
let access = accessible st follow chars in
let marked1 = st :: marked in
let f (c, s) umsts =
if Int_set.is_empty s then
umsts (*Suppress empty sets*)
else
try
st.trans <- (c, find_state s marked1 umsts) ::st.trans ;
umsts
with
| Not_found ->
let state1 = {pos = s; trans = []} in
st.trans <- (c, state1) :: st.trans;
state1 :: umsts in
let unmarked1 = List.fold_right f access umsts in

We are just about ready to write the function to compute the automaton. It is fundamentally a call to compute_states but does one more thing. That is, it searches the resulting array for the index of the initial state and puts the index in the first slot of the array. To do this it uses the utility function array_indexq which performs the search for the index using physical equality. This is because the usual test using structural equality will not terminate on structures that loop.

let (array_indexq : 'a array -> 'a -> int) =
fun arr e ->
let rec loop i =
if i = Array.length arr then
raise (Not_found)
else if Array.get arr i == e then i
else loop (i + 1) in
loop 0

So, here it is, dfa_of, the function to compute the automaton.
let (dfa_of : augmented_regexp * Int_set.t array * char option array
-> state array) =
fun (e, follow, chars) ->
let init_state = {pos = first_pos e; trans = []} in
let dfa = compute_states [] [init_state] follow chars in
(*Installing initial state at index 0*)
let idx_start = array_indexq dfa init_state in
dfa.(idx_start) <- dfa.(0);
dfa.(0) <- init_state;
dfa

We are now on the home stretch. All that remains is to write a function to interpret the automaton. To do this, we'll make use of a mini-combinator library of recognizers. I'll not provide the OCaml code for that today - you could reverse engineer from my earlier 'Recognizers' blog-post or, consult [1].

let (interpret_dfa : state array -> int -> char Recognizer.recognizer) =
fun dfa accept ->
let num_states = Array.length dfa in
let fvect = Array.make (num_states) (fun _ -> failwith "no value") in
for i = 0 to num_states - 1 do
let trans = dfa.(i).trans in
let f (c, st) =
let pc = Recognizer.recognizer_of_char c in
let j = array_indexq dfa st in
Recognizer.compose_and pc (fun l -> fvect.(j) l) in
let parsers = List.map f trans in
if Int_set.mem accept (dfa.(i).pos) then
fvect.(i) <- compose_or_list
(Recognizer.end_of_input) parsers
else match parsers with
| [] -> failwith "Impossible"
| p :: ps -> fvect.(i) <- Recognizer.compose_or_list p ps
done;
fvect.(0)

We wrap up with a couple of high level convenience functions : compile produces a recognizer from a string representation of a regular expression and match takes a recognizer (that is, a compiled regular expression) and a string and uses the recognizer to categorize the given string as admissible or not (where explode is a simple function that transforms a string into a char list - recognizers operate on lists).
let compile xpr =
let ((e, follow, chars) as ast) = regexp_follow xpr in
let dfa = dfa_of ast in
let parser = interpret_dfa dfa (Array.length chars - 1) in
fun s -> parser (explode s)

let re_match xpr s =
let result = xpr s in
match result with
| Recognizer.Remains [] -> true
| _ -> false

Here's a simple test driver that shows how these functions can be used.

let test xpr s =
match re_match xpr s with
| true -> Printf.printf "\"%s\" : success\n" s
| false -> Printf.printf "\"%s\" : fail\n" s

let _ =
try
let xpr = compile "(a|b)*abb" in
Printf.printf "Pattern: \"%s\"\n" "(a|b)*abb" ;
test xpr "abb" ;
test xpr "aabb" ;
test xpr "baabb" ;
test xpr "bbbbbbbbbbbbbaabb" ;
test xpr "aaaaaaabbbaabbbaabbabaabb" ;
test xpr "baab" ;
test xpr "aa" ;
test xpr "ab" ;
test xpr "bb" ;
test xpr "" ;
test xpr "ccabb" ;
with
| Failure msg -> print_endline msg

So that's it for this series of posts on building recognizers for regular expressions. Hope you enjoyed it!

References
[1] "The Functional Approach to Programming" - Cousineau & Mauny
[2] "Compilers Principles, Techniques & Tools" - Aho et. al.


text:
# Prove that $v_1,v_2,…,v_k$ are eigenvectors if $A=\lambda_1v_1v_1^{T}+\lambda_2v_2v_2^{T}+…+\lambda_kv_kv_k^{T}$

Given are vectors $v_1,v_2,...,v_k \in \mathbb{R}^{n}$ with $k<n$, all orthogonal to the standard inner product on $\mathbb{R}^{n}$. Now take $\lambda_1,\lambda_2,...,\lambda_k \in \mathbb{R}$ and suppose

$A=\lambda_1v_1v_1^{T}+\lambda_2v_2v_2^{T}+...+\lambda_kv_kv_k^{T} \in \mathbb{R}^{n\times n}$

Now I'm supposed to prove that $v_1,v_2,...,v_k$ are eigenvectors of $A$, but I don't know how to go about it. I already tried proving it based on the properties of symmetric matrices, but that didn't seem to get me anywhere.

• Isn't $x x^T$ just the definition of $<x, x> = \| x \|^2$? Should it instead say $A = \sum_{j = 1}^{k} \lambda_j v_{j}^{T} v_j$? – AJY Jan 18 '17 at 12:33
• @AJY No, $x^Tx$ is equal to $\langle x,x\rangle$, while $xx^T$ is a $n\times n$ matrix. – 5xum Jan 18 '17 at 12:34
• Okay, so I had it mixed up. In that case, we should still have $\sum_{j = 1}^k \lambda_j v_j (v_{j}^{T} x) = \sum_{j = 1}^k \lambda_j v_j \left< v_j, x \right>$. – AJY Jan 18 '17 at 12:37

For all $i=1,2,\ldots k$: $$Av_i=\left(\sum_{j=1}^k\lambda_jv_jv_j^T\right)v_i=\sum_{j=1}^k\lambda_jv_j(\underbrace{v_j^Tv_i}_{=\langle v_j,v_i \rangle})\underbrace{=}_{\{v_j\}\text{ orthog. system}}\lambda_iv_i|| v_i||^2=\left(\lambda_i|| v_i||^2\right)v_i.$$

• Is $||v_i||^{2}$ always $1$? I don't exactly understand why. – Simon Jan 18 '17 at 12:55
• Sorry, I read orthonormal instead of orthogonal. I've edited the answer. – Fernando Revilla Jan 18 '17 at 13:00

Hint

Notice that if $(v_i)$ is orthonormal then

$$v_iv_i^Tv_j=v_i(v_i^Tv_j)=(v_i^Tv_j)v_i=\delta_{i,j} v_i$$

Hint:

What is $Av_1$?

Remember, $(A+B)v = Av + Bv$...


text:
how to find the Allele Number of specific variant from gnomAD
1
0
Entering edit mode
6 months ago
Sara ▴ 220

I am working with a dataset from gnomAD for which i need to find the "Allele Number" for the variant we are working on. I tried to find such information from this link:

https://gnomad.broadinstitute.org/downloads#v2-coverage

but it is not in those files. do you know from which file we can find Allele Number? here is a screenshot of the page showing some information (however I am looking for a file to extract required information including Allele Number):

0
Entering edit mode
6 months ago

The VCF files. column info: AN

\$ wget -q -O - "https://storage.googleapis.com/gcp-public-data--gnomad/release/2.1.1/vcf/exomes/gnomad.exomes.r2.1.1.sites.22.vcf.bgz" | gunzip -c | grep ID=AN,
##INFO=<ID=AN,Number=1,Type=Integer,Description="Total number of alleles in samples">

0
Entering edit mode

@PierreLindenbaum : Thanks. using this command I only get info about ancestries. but what I am looking for is the "Allele Number" for each variant. like the screenshot I attached in the question. do you know how I can get this number?

1
Entering edit mode


text:
# Role of Statistics in Data Science

6K
• by Amit Diwan
• 18th Sep, 2020
• Last updated on 17th Mar, 2021

• In this article, we understand why data is important, and talk about the importance of statistics in data analysis and data science.
• We also understand some basic statistics concepts and terminologies.
• We see how statistics and machine learning work in sync to give deep insights into data
• We understand the fundamentals behind Bayesian thinking and how Bayesian theorem works.

Introduction

Data plays a huge role in today’s tech world. All technologies are data-driven, and humongous amounts of data are produced on a daily basis. data scientist is a professional who is able to analyse data sources, clean and process the data, understand why and how such data has been generated, take insights from it, and make changes such that they profit the organization. These days, everything revolves around data.

• Data Cleaning: It deals with gathering the data and structuring it so that it becomes easy to pass this data as input to any machine learning algorithm. This way, redundant, irrelevant data and noise can also be eliminated.
• Data Analysis: This deals with understanding more about the data, why the data has yielded certain results, and what can be done to improve it. It also helps calculate certain numerical values like mean, variance, the distributions, and the probability of a certain prediction.

## How the basics of statistics will serve as a foundation to manipulate data in data science

The basics of statistics include terminologies, and methods of applying statistics in data science. In order to analyze the data, the important tool is statistics. The concepts involved in statistics help provide insights into the data to perform quantitative analysis on it. In addition to this, as a foundation, the basics and working of linear regression and classification algorithms must also be known to a data science aspirant.

### Terminologies associated with statistics

• Population: It is an entire pool of data from where a statistical sample is extracted. It can be visualized as a complete data set of items that are similar in nature
• Sample: It is a subset of the population, i.e. it is an integral part of the population that has been collected for analysis.
• Variable: A value whose characteristics such as quantity can be measured, it can also be addressed as a data point, or a data item.
• Distribution: The sample data that is spread over a specific range of values.
• Parameter: It is a value that is used to describe the attributes of a complete data set (also known as ‘population’). Example: Average, Percentage
• Quantitative analysis: It deals with specific characteristics of data- summarizing some part of data, such as its mean, variance, and so on.
• Qualitative analysis: This deals with generic information about the type of data, and how clean or structured it is.

## How does analyzing data using statistics help gain deep insights into data?

Statistics serve as a foundation while dealing with data and its analysis in data science. There are certain core concepts and basics which need to be thoroughly understood before jumping into advanced algorithms.

Not everyone understand the performance metrics of machine learning algorithms like f-score, recall, precision, accuracy, root mean squared error, and so on. Instead, visual representation of the data and the performance of the algorithm on the data serves as a good metric for the layperson to understand the same.

Also, visual representation helps identify outliers, specific trivial patterns, and certain metric summary such as mean, median, variance, that helps in understanding the middlemost value, and how the outlier affects the rest of the data.

## Statistical Data Analysis

Statistical data analysis deals with the usage of certain statistical tools that need knowledge of statistics. Software can also help with this, but without understanding why something is happening, it is impossible to get considerable work done in statistics and data science.

Statistics deals with data variables that are either univariate or multivariate. Univariate, as the name suggests deals with single data values, whereas multivariate data deals with the multiple number of values. Discriminant data analysis, factor data analysis can be performed on multivariate data. On the other hand, univariate data analysis, Z-test, F-test can be performed if we are dealing with univariate data.

Data associated with statistics is of many types. Some of them have been discussed below.

Categorical data represents characteristics of people, such as marital status, gender, food they like, and so on. It is also known as ‘qualitative data’ or ‘yes/no data’. It takes numerical values like ‘1’, ‘2’, where these numbers indicate one or other type of characteristics. These numbers are not mathematically significant, which means it can’t be associated with each other.

Continuous data deals with data that is immeasurable, and can’t be counted, which basically continual forms of values are. Predictions from a linear regression are continuous in nature. It is a continuous distribution that is also known as probability density function.

On the other hand, discrete values can be measured, counted, and are discontinuous. Predictions from logistic regression are considered to be discrete in nature. Discrete data is non-continuous, and density concept doesn’t come into the picture here. The distribution is known as probability mass function.

## The Best way to Learn Statistics for Data Science

The best way to learn anything is by implementing it, by working on it, by making mistakes and again learning from it.  It is important to understand the concepts, either by going through standard books or well-known websites, before implementing them.

Before jumping into data science, the core statistics concepts like such as regression, maximum likelihood, distributions, priors, posteriors, conditional probabilityBayesian theorem and basics of machine learning have to be understood clearly.

### Core statistics concepts

Descriptive statistics: As the name suggests, it uses the data to give out more information about every aspect of the data with the help of graphs, plots, or numbers. It organizes the data into a structure, and helps think about the attributes that highlight the important parts of the data.

• Inferential statistics: It deals with drawing inferences/conclusions on the sample data set which is obtained from the population (entire data set) based on the relationship identified between data points in the data set. It helps in generalizing the relationship to the entire dataset. It is important to remember that the dataset drawn from the population is relevant and represents the population accurately.
• Regression: The term ‘regression’ which is a part of statistics and machine learning, talks about how data can be fit to a line, and how every point from the straight line gives some insights. In terms of machine learning, it can be understood as tasks that can be solved without explicitly being programmed. They discuss how a line can be fit to a given set of data points, and how it can be further extrapolated for the predictions to be done.
• Maximum likelihood: It is a method that helps in finding values of parameters for a specific model. The values of the parameters have to be such that the likelihood of the predictions that occur have to be maximum in comparison to the data values that were actually observed. This means the difference between the actual and predicted value has to be less, thereby reducing the error and increasing the accuracy of the predictions.

Note: This concept is generally used with Logistic regression when we are trying to find the output as 0 or 1, yes or no, wherein the maximum likelihood tells about how likely a data point is near to 0 or 1.

### Bayesian thinking

Bayesian thinking deals with using probability to model the process of sampling, and being able to quantify the uncertainty associated with the data that would be collected.

This is known as prior probabilitywhich means the level of uncertainty that is associated with the data before it is collected to be analysed.

Posterior probability deals with the uncertainty that occurs after the data has been collected.

Machine learning algorithms are usually focussed on giving the best predictions as output with minimal errors, exact probabilities of specific events occurring and so on. Bayes theorem is way of calculating the probability of a hypothesis (situation, which might not have occurred in reality) based on our previous experiences and the knowledge we have gained by it. This is considered as a basic concept that needs to be known.

Bayes theorem can be stated as follows:

P(hypo | data) = (P(data | hypo) * P(hypo)) / P(data)

In the above equation,

P(hypo | data) is the probability of a hypothesis ‘hypo’ when data ‘data’ is given, which is also known as posterior probability.

P(data | hypo) is the probability of data ‘data’ when the specific hypothesis ‘hypo’ is known to be true.

P(hypo) is the probability of a hypothesis ‘hypo’ being true (irrespective of the data in hand), which is also known as prior probability of ‘hypo’.

P(data) is the probability of the data (irrespective of the hypothesis).

The idea here is to get the value of the posterior probability, given other data. The posterior probability for a variety of different hypotheses has to be found out, and the probability that has the highest value is selected. This is known as the maximum probable hypothesis, and is also known as the maximum a posteriori (MAP) hypothesis.

MAP(hypo) = max(P(hypo | data))

If the value of P(hypo | data) is replaced with the value we saw before, the equation would become:

MAP(hypo) = max((P(data | hypo) * P(hypo)) / P(data))

P(data) is considered as a normalizing term that helps in determining the probability. This value can be safely ignored when required, since it is a constant value.

### Naïve Bayes classifier

It is an algorithm that can be used with binary or multi-class classification problems. It is a simple algorithm wherein the probability for every hypothesis is simplified.

This is done in order to make the data more traceable. Instead of calculating value of every attribute like P(data1, data2,..,datan|hypo), we assume that every data point is independent of every other data point in the data set when the respective output is given.

This way, the equation becomes:

P(data1 | hypo) * P(data2 |hypo) * … * P(data-n| hypo).

This way, the attributes would be independent of each other. This classifier performs quite well even in the real world with real data when the assumption of data points being independent of each other doesn’t hold good.

Once a Naïve Bayes classifier has learnt from the data, it stores a list of probabilities in a data structure. Probabilities such as ‘class probability’ and ‘condition probability’ are stored. Training such a model is quick since the probability of every class and its associated value needs to be determined, and this doesn’t involve any optimization processes or changing of coefficient to give better predictions.

• Class probability: It tells about the probability of every class that is present in the training dataset. It can be calculated by finding the frequency of values that belongs to each class divided by the total number of values.
• Class probability = (number of classes/(number of classes of group 0 + number of classes of group 1))
• Conditional probability: It talks about the conditional probability of every input that is associated with a class value. It can be calculated by finding the frequency of every data attribute in the data for a given class, and this can be determined by the number of data values that have that data label/class value.
• Conditional probability P(condition | result ) = number of ((values with that condition and values with that result)/ (number of values with that result))

Not just the concept, once the user understands the way in which data scientist needs to think, they will be able to focus on getting cleaner data, with better insights that would lead to performing better analysis, which in turn would give great results.

## Introduction to Statistical Machine Learning

The methods used in statistics are important to train and test the data that is used as input to the machine learning model. Some of these include outlier/anomaly detection, sampling of data, data scaling, variable encoding, dealing with missing values, and so on.

Statistics is also essential to evaluate the model that has been used, i.e. see how well the machine learning model performs on test dataset, or on data that it has never seen before.

Statistics is essential in selecting the final and appropriate model to deal with that specific data in a predictive modelling situation.

It is also needed to show how well the model has performed, by taking various metrics and showing how the model has fared.

## Metrics used in Statistics

Most of the data can be fit to a common pattern that is known as Gaussian distribution or normal distribution. It is a bell-shaped curve that can be used to summarize the data with the below mentioned two parameters:

• Mean: It is understood as the central most value when the data points are arranged in a descending or ascending order, or the most likely value.Mode: It can be understood as the data point that occurs the greatest number of times, i.e. The frequency of the value in the dataset would be very high.
• Median: It is a measure of central tendency of the data set. It is the middle number, that can be found by sorting all the data points in a dataset and picking the middle-most element. If the number of data points in a dataset is odd, one single middle value is picked up, whereas two middle values are picked and their mean is calculated if the number of data points in a dataset is even.
• RangeIt refers to the value that is calculated by finding the difference between the largest and the smallest value in a dataset.
• Quartile: As the name suggests, quartiles are values that divide the data points in a dataset into quarters. It is calculated by sorting the elements in order and then dividing the dataset into 4 equal parts.
• Three quartiles are identified: The first quartile that is the 25th percentile, the second quartile which is the 50th percentile and the third quartile that is the 75th percentile. Each of these quartiles tells about the percentage of data that is smaller or larger in comparison to other percentiles of data.

Example: 25th percentile suggests that 25 percent of the data set is smaller than the remaining 75 percent of the data set.

Quartile helps understand how the data is distributed around the median (which is the 50th percentile/second quartile).

There are other distributions as well, and it depends on the type of data we have and the insights we need from that data, but Gaussian is considered as one of the basic distributions.

• Variance: The average of the difference between every value and the mean of that specific distribution.
• Standard deviation: It can be understood as the measure that indicates the dispersion that occurs in the data points of the input data.

Conclusion

In this post, we understood why and how statistics is important to understand and work with data science. We saw a few terminologies of statistics that are essential in understanding the insights which statistics would end up giving to data scientist. We also saw a few basic algorithms that every data scientist needs to know, in order to learn other advanced algorithms.

### Amit Diwan

Author

Amit Diwan is an E-Learning Entrepreneur, who has taught more than a million professionals with Text & Video Courses on the following technologies: Data Science, AI, ML, C#, Java, Python, Android, WordPress, Drupal, Magento, Bootstrap 4, etc.

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It is for those candidates who have a master’s degree in a field related to data analytics. It is one of the few vendor-neutral certifications on the list and must be converted to CAP within 6 years, so it offers a good opportunity for those with a long term path in a Data Analytics career. It also needs to be renewed every three years, like the CAP certification. Like its professional counterpart, aCAP helps a candidate step out in a vendor-neutral manner and drastically increases their professional credibility. Prerequisite: Master’s degree in any discipline related to data Analytics. Exam Fee: The base price is$300. For individuals who are members of INFORMS the price is $200. (Source). There is an extensive syllabus which covers: i. Business Problem Framing, ii. Analytics Problem Framing, iii. Data, iv. Methodology Selection, v. Model Building, vi. Deployment, vii. Lifecycle Management of the Analytics process, problem-solving, data science and visualisation and much more.4. 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Exam Fee: The cost for each exam is$180. (Source) An exception is Predictive Modelling using the SAS Enterprise Miner, costing $250, This exam can be taken in the English language. One can join the SAS Academy for Data Science and also take a practice exam beforehand. Salary: You can get a job as a SAS Data Analyst that pays up to$90,000 per year! (Source) 5. IBM Data Science Professional CertificateWhenever someone studies the history of a computer, IBM (International Business Machines) is the first brand that comes up. IBM is still alive and kicking, now having forayed into and becoming a major player in the Big Data segment. The IBM Data Science Professional certificate is one of the beginner-level certificates if you want to sink your hands into the world of data analysis. It shows a candidate's skills in various topics pertaining to data sciences, including various open-source tools, Python databases, SWL, data visualisation, and data methodologies.  One needs to complete nine courses to earn the certificate. It takes around three months if one works twelve hours per week. It also involves the completion of various hands-on assignments and building a portfolio. A candidate earns the Professional certificate from Coursera and a badge from IBM that recognises a candidate's proficiency in the area. Prerequisites: It is the optimal course for freshers since it requires no requisite programming knowledge or proficiency in Analytics. Exam Fee: It costs $39 per month (Source) to access the course materials and the certificate. The course is handled by the Coursera organisation. Expected Salary: This certification can earn you the title of IBM Data Scientist and help you earn a salary of$134,846 per annum. (Source) 6. Microsoft Certified Azure Data Scientist AssociateIt's one of the most well-known certifications for newcomers to step into the field of Big Data and Data analytics. This credential is offered by the leader in the industry, Microsoft Azure. This credential validates a candidate's ability to work with Microsoft Azure developing environment and proficiency in analysing big data, preparing data for the modelling process, and then progressing to designing models. One advantage of this credential is that it has no expiry date and does not need renewal; it also authorises the candidate’s extensive knowledge in predictive Analytics. Prerequisites: knowledge and experience in data science and using Azure Machine Learning and Azure Databricks. Exam Fee: It costs $165 to (Source) register for the exam. One advantage is that there is no need to attend proxy institutions to prepare for this exam, as Microsoft offers free training materials as well as an instructor-led course that is paid. 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It is also behind the SAS Institute for Data Science. Hence, SAS is the organisation you would want to go to if you're aiming for a long-term career in data science. Conclusion To conclude, big data and data Analytics are a field of endless opportunities. By investing in the right credential, one can pave the way to a viable and lucrative career path. Beware though, there are lots of companies that provide certifications, but only recognised and reputed credentials will give you the opportunities you are seeking. Hiring companies look for these certifications as a mark of authenticity of your hands-on experience and the amount of work you can handle effectively. Therefore, the credential you choose for yourself plays a vital role in the career you can have in the field of Data analytics. Happy learning! 5631 Top Data Analytics Certifications What is data analytics?In the world of IT, every s... Read More ## Why Should You Start a Career in Machine Learning? If you are even remotely interested in technology you would have heard of machine learning. In fact machine learning is now a buzzword and there are dozens of articles and research papers dedicated to it. Machine learning is a technique which makes the machine learn from past experiences. Complex domain problems can be resolved quickly and efficiently using Machine Learning techniques. We are living in an age where huge amounts of data are produced every second. This explosion of data has led to creation of machine learning models which can be used to analyse data and to benefit businesses. This article tries to answer a few important concepts related to Machine Learning and informs you about the career path in this prestigious and important domain.What is Machine Learning?So, here’s your introduction to Machine Learning. This term was coined in the year 1997. “A computer program is said to learn from experience E with respect to some class of tasks T and performance measure P, if its performance at the tasks improves with the experiences.”, as defined in the book on ML written by Mitchell in 1997. The difference between a traditional programming and programming using Machine Learning is depicted here, the first Approach (a) is a traditional approach, and second approach (b) is a Machine Learning based approach.Machine Learning encompasses the techniques in AI which allow the system to learn automatically looking at the data available. While learning, the system tries to improve the experience without making any explicit efforts in programming. Any machine learning application follows the following steps broadlySelecting the training datasetAs the definition indicates, machine learning algorithms require past experience, that is data, for learning. So, selection of appropriate data is the key for any machine learning application.Preparing the dataset by preprocessing the dataOnce the decision about the data is made, it needs to be prepared for use. Machine learning algorithms are very susceptible to the small changes in data. To get the right insights, data must be preprocessed which includes data cleaning and data transformation. Exploring the basic statistics and properties of dataTo understand what the data wishes to convey, the data engineer or Machine Learning engineer needs to understand the properties of data in detail. These details are understood by studying the statistical properties of data. Visualization is an important process to understand the data in detail.Selecting the appropriate algorithm to apply on the datasetOnce the data is ready and understood in detail, then appropriate Machine Learning algorithms or models are selected. The choice of algorithm depends on characteristics of data as well as type of task to be performed on the data. The choice also depends on what kind of output is required from the data.Checking the performance and fine-tuning the parameters of the algorithmThe model or algorithm chosen is fine-tuned to get improved performance. If multiple models are applied, then they are weighed against the performance. The final algorithm is again fine-tuned to get appropriate output and performance.Why Pursue a Career in Machine Learning in 2021?A recent survey has estimated that the jobs in AI and ML have grown by more than 300%. Even before the pandemic struck, Machine Learning skills were in high demand and the demand is expected to increase two-fold in the near future.A career in machine learning gives you the opportunity to make significant contributions in AI, the future of technology. All the big and small businesses are adopting Machine Learning models to improve their bottom-line margins and return on investment. The use of Machine Learning has gone beyond just technology and it is now used in diverse industries including healthcare, automobile, manufacturing, government and more. This has greatly enhanced the value of Machine Learning experts who can earn an average salary of$112,000.  Huge numbers of jobs are expected to be created in the coming years.  Here are a few reasons why one should pursue a career in Machine Learning:The global machine learning market is expected to touch \$20.83B in 2024, according to Forbes.  We are living in a digital age and this explosion of data has made the use of machine learning models a necessity. Machine Learning is the only way to extract meaning out of data and businesses need Machine Learning engineers to analyze huge data and gain insights from them to improve their businesses.If you like numbers, if you like research, if you like to read and test and if you have a passion to analyse, then machine learning is the career for you. Learning the right tools and programming languages will help you use machine learning to provide appropriate solutions to complex problems, overcome challenges and grow the business.Machine Learning is a great career option for those interested in computer science and mathematics. They can come up with new Machine Learning algorithms and techniques to cater to the needs of various business domains.As explained above, a career in machine learning is both rewarding and lucrative. There are huge number of opportunities available if you have the right expertise and knowledge. On an average, Machine Learning engineers get higher salaries, than other software developers.Years of experience in the Machine Learning domain, helps you break into data scientist roles, which is not just among the hottest careers of our generation but also a highly respected and lucrative career. Right skills in the right business domain helps you progress and make a mark for yourself in your organization. For example, if you have expertise in pharmaceutical industries and experience working in Machine learning, then you may land job roles as a data scientist consultant in big pharmaceutical companies.Statistics on Machine learning growth and the industries that use MLAccording to a research paper in AI Multiple (https://research.aimultiple.com/ml-stats/), the Machine Learning market will grow to 9 Billion USD by the end of 2022. There are various areas where Machine Learning models and solutions are getting deployed, and businesses see an overall increase of 44% investments in this area. North America is one of the leading regions in the adoption of Machine Learning followed by Asia.The Global Machine Learning market will grow by 42% which is evident from the following graph. Image sourceThere is a huge demand for Machine Learning modelling because of the large use of Cloud Based Applications and Services. The pandemic has changed the face of businesses, making them heavily dependent on Cloud and AI based services. Google, IBM, and Amazon are just some of the companies that have invested heavily in AI and Machine Learning based application development, to provide robust solutions for problems faced by small to large scale businesses. Machine Learning and Cloud based solutions are scalable and secure for all types of business.ML analyses and interprets data patterns, computing and developing algorithms for various business purposes.Advantages of Machine Learning courseNow that we have established the advantages of perusing a career in Machine Learning, let’s understand from where to start our machine learning journey. The best option would be to start with a Machine Learning course. There are various platforms which offer popular Machine Learning courses. One can always start with an online course which is both effective and safe in these COVID times.These courses start with an introduction to Machine Learning and then slowly help you to build your skills in the domain. Many courses even start with the basics of programming languages such as Python, which are important for building Machine Learning models. Courses from reputed institutions will hand hold you through the basics. Once the basics are clear, you may switch to an offline course and get the required certification.Online certifications have the same value as offline classes. They are a great way to clear your doubts and get personalized help to grow your knowledge. These courses can be completed along with your normal job or education, as most are self-paced and can be taken at a time of your convenience. There are plenty of online blogs and articles to aid you in completion of your certification.Machine Learning courses include many real time case studies which help you in understanding the basics and application aspects. Learning and applying are both important and are covered in good Machine Learning Courses. So, do your research and pick an online tutorial that is from a reputable institute.What Does the Career Path in Machine Learning Look Like?One can start their career in Machine Learning domain as a developer or application programmer. But the acquisition of the right skills and experience can lead you to various career paths. Following are some of the career options in Machine Learning (not an exhaustive list):Data ScientistA data scientist is a person with rich experience in a particular business field. A person who has a knowledge of domain, as well as machine learning modelling, is a data scientist. Data Scientists’ job is to study the data carefully and suggest accurate models to improve the business.AI and Machine Learning EngineerAn AI engineer is responsible for choosing the proper Machine Learning Algorithm based on natural language processing and neural network. They are responsible for applying it in AI applications like personalized advertising.  A Machine Learning Engineer is responsible for creating the appropriate models for improvement of the businessData EngineerA Data Engineer, as the name suggests, is responsible to collect data and make it ready for the application of Machine Learning models. Identification of the right data and making it ready for extraction of further insights is the main work of a data engineer.Business AnalystA person who studies the business and analyzes the data to get insights from it is a Business Analyst. He or she is responsible for extracting the insights from the data at hand.Business Intelligence (BI) DeveloperA BI developer uses Machine Learning and Data Analytics techniques to work on a large amount of data. Proper representation of data to suit business decisions, using the latest tools for creation of intuitive dashboards is the role of a BI developer.  Human Machine Interface learning engineerCreating tools using machine learning techniques to ease the human machine interaction or automate decisions, is the role of a Human Machine Interface learning engineer. This person helps in generating choices for users to ease their work.Natural Language Processing (NLP) engineer or developerAs the name suggests, this person develops various techniques to process Natural Language constructs. Building applications or systems using machine learning techniques to build Natural Language based applications is their main task. They create multilingual Chatbots for use in websites and other applications.Why are Machine Learning Roles so popular?As mentioned above, the market growth of AI and ML has increased tremendously over the past years. The Machine Learning Techniques are applied in every domain including marketing, sales, product recommendations, brand retention, creating advertising, understanding the sentiments of customer, security, banking and more. Machine learning algorithms are also used in emails to ease the users work. This says a lot, and proves that a career in Machine Learning is in high demand as all businesses are incorporating various machine learning techniques and are improving their business.One can harness this popularity by skilling up with Machine Learning skills. Machine Learning models are now being used by every company, irrespective of their size--small or big, to get insights on their data and use these insights to improve the business. As every company wishes to grow faster, they are deploying more machine learning engineers to get their work done on time. Also, the migration of businesses to Cloud services for better security and scalability, has increased their requirement for more Machine Learning algorithms and models to cater to their needs.Introducing the Machine learning techniques and solutions has brought huge returns for businesses.  Machine Learning solution providers like Google, IBM, Microsoft etc. are investing in human resources for development of Machine Learning models and algorithms. The tools developed by them are popularly used by businesses to get early returns. It has been observed that there is significant increase in patents in Machine Learning domains since the past few years, indicating the quantum of work happening in this domain.Machine Learning SkillsLet’s visit a few important skills one must acquire to work in the domain of Machine Learning.Programming languagesKnowledge of programming is very important for a career in Machine Learning. Languages like Python and R are popularly used to develop applications using Machine Learning models and algorithms. Python, being the simplest and most flexible language, is very popular for AI and Machine Learning applications. These languages provide rich support of libraries for implementation of Machine Learning Algorithms. A person who is good in programming can work very efficiently in this domain.Mathematics and StatisticsThe base for Machine Learning is mathematics and statistics. Statistics applied to data help in understanding it in micro detail. Many machine learning models are based on the probability theory and require knowledge of linear algebra, transformations etc. A good understanding of statistics and probability increases the early adoption to Machine Learning domain.Analytical toolsA plethora of analytical tools are available where machine learning models are already implemented and made available for use. Also, these tools are very good for visualization purposes. Tools like IBM Cognos, PowerBI, Tableue etc are important to pursue a career as a  Machine Learning engineer.Machine Learning Algorithms and librariesTo become a master in this domain, one must master the libraries which are provided with various programming languages. The basic understanding of how machine learning algorithms work and are implemented is crucial.Data Modelling for Machine Learning based systemsData lies at the core of any Machine Learning application. So, modelling the data to suit the application of Machine Learning algorithms is an important task. Data modelling experts are the heart of development teams that develop machine learning based systems. SQL based solutions like Oracle, SQL Server, and NoSQL solutions are important for modelling data required for Machine Learning applications. MongoDB, DynamoDB, Riak are some important NOSQL based solutions available to process unstructured data for Machine Learning applications.Other than these skills, there are two other skills that may prove to be beneficial for those planning on a career in the Machine Learning domain:Natural Language processing techniquesFor E-commerce sites, customer feedback is very important and crucial in determining the roadmap of future products. Many customers give reviews for the products that they have used or give suggestions for improvement. These feedbacks and opinions are analyzed to gain more insights about the customers buying habits as well as about the products. This is part of natural language processing using Machine Learning. The likes of Google, Facebook, Twitter are developing machine learning algorithms for Natural Language Processing and are constantly working on improving their solutions. Knowledge of basics of Natural Language Processing techniques and libraries is must in the domain of Machine Learning.Image ProcessingKnowledge of Image and Video processing is very crucial when a solution is required to be developed in the area of security, weather forecasting, crop prediction etc. Machine Learning based solutions are very effective in these domains. Tools like Matlab, Octave, OpenCV are some important tools available to develop Machine Learning based solutions which require image or video processing.ConclusionMachine Learning is a technique to automate the tasks based on past experiences. This is among the most lucrative career choices right now and will continue to remain so in the future. Job opportunities are increasing day by day in this domain. Acquiring the right skills by opting for a proper Machine Learning course is important to grow in this domain. You can have an impressive career trajectory as a machine learning expert, provided you have the right skills and expertise.
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text:
# Explanation of a proper time (an add)

I have read on web that:

Proper time $\tau$ is a time interval measured by a clock which at rest relative to the observer.

But lets consider this problem:

The plane is flying with a speed $u=300m/s$. How much time $\Delta t'$ would have to pass for a passanger on that plane for time difference $1s$ to occur between the time $\Delta t'$ on the plane and the time $\Delta t$ on the Earth. We postulate that speed of light equals to the speed of sound $c=343m/s$.

Ok so lets say i set myself in coordinate system $x'y'$ and check the whole situation from a perspective of a passenger on the plane.

I first calculate the Lorenz factor:

$$\gamma=\frac{1}{\sqrt{1-\frac{300^2}{343^2}}} = 2.06$$

If everything is checked from the perspective of the observer on the plane, he is not moving according to his clock so his time is the proper time and i set $\boxed{\Delta t'\equiv\tau}$. Because the proper time is the shortest it follows that:

\begin{align} \Delta t = \gamma \Delta t' = \Delta t' + 1s \xrightarrow{\text{i use the RHS}}\gamma\Delta t' &= \Delta t' + 1s\\ \Delta t' \left( \gamma - 1 \right) &=1s\\ \Delta t' = \frac{1s}{\gamma - 1}\\ \Delta t' = 0.94s \end{align}

So now i reverse the situation and set myself in the coordinate system $xy$ on Earth.

Here the Lorentz factor is the same as before even if i had to use negative speed $u=-300m/s$ (squaring does its trick hehe). So it again holds $\gamma=2,06$.

If now everything is checked from the perspective of an observer on Earth, he is not moving accordint to his clock which means now his time is the proper time and i can set $\boxed{\Delta t=\tau}$. Because the proper time is the shortest it follows that:

\begin{align} \Delta t' = \gamma \Delta t = \Delta t + 1s \xrightarrow{\text{i use the RHS}}\gamma\Delta t &= \Delta t + 1s\\ \Delta t \left( \gamma - 1 \right) &=1s\\ \Delta t = \frac{1s}{\gamma - 1}\\ \Delta t = 0.94s \end{align}

Well this means that the proper time definition needs an add which is:

"In the situation where we have more clocks the proper time is measured by the clock which is situated in the coordinate system from which we observe."

and

"we are free to choose to observe from whichever coordinate system we want."

-
And what is your question? –  Calmarius Jun 22 '13 at 10:06
You have just noticed the apparent paradox known as "the twin paradox". There are many explanations (of varying quality and correctness) of the resolution on the internet. Your suggested additional texts express ideas which are built into the structure of relativity from the get go; you should have been assuming them all along. –  dmckee Jun 22 '13 at 16:53
I'm tempted to close this, because like @Calmarius I can't find an indication of an actual question here. –  dmckee Jun 22 '13 at 16:55

The proper time is a fundamental invarient. By this I mean that all observers, regardless of their co-ordinate system, will measure the same value for the proper time. This applies to General Relativity as well as Special Relativity.

In SR the proper time is defined by:

$$c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2$$

This immediately explains why the proper time is the time measured in the rest frame, because in the rest frame $dx$, $dy$ and $dz$ are all zero so the equation reduces to:

$$c^2d\tau^2 = c^2dt^2$$

or:

$$d\tau = dt$$

The point is that I wouldn't start from the definition that the proper time is the time measured in the rest frame because this is a consequence of a much deeper principle. The equation I gave for $d\tau^2$ may seem a bit arbitrary, but it's a special case of the much deeper statement that:

$$d\tau^2 = \eta_{\alpha\beta} dx^\alpha dx^\beta$$

where the object $\eta$ is the metric, which describes the structure of space time (strictly speaking we use the above equation to calculate the line element not the proper time, but this is a complication we can gloss over for now).

In the example you give, you can calculate the elapsed time on the plane without using the Lorentz factor. Indeed, you can derive the Lorentz transformations from the expression for the proper time.

-


text:
# Which displacement reaction is possible? calcium + sodium carbonate -> sodium + calcium carbonate sodium + calcium carbonate →

###### Question:

Which displacement reaction is possible? calcium + sodium carbonate -> sodium + calcium carbonate sodium + calcium carbonate → calcium + sodium carbonate

### Solve for the variable: -11t + 23 = 56

Solve for the variable: -11t + 23 = 56...

### 5. Describe the type of bonding that exists in the Cl2(g) molecule.How does this type of bonding differ from that found in theHCl(g)

5. Describe the type of bonding that exists in the Cl2(g) molecule. How does this type of bonding differ from that found in the HCl(g) molecule? How is it similar?...

### Help is appreicated! (10 points!)

Help is appreicated! (10 points!) $Help is appreicated! (10 points!)$...

### El año pasado fui a bogotá, colombia para asistir a una conferencia profesional. me

El año pasado fui a bogotá, colombia para asistir a una conferencia profesional. me gusta viajar, (1) particularmente por avión. salí de caracas por la mañana y el vuelo llegó (2) puntualmente a las dos de la tarde. después de pasar por la aduana, fui (3) immediatamente al hotel porque tuve u...

### 6. does blood always carry a lot of oxygen to various parts of your body?

6. does blood always carry a lot of oxygen to various parts of your body?...

### Based on the table, how many official camps are located in areas under Israeli control?A. 8B. 12C. 19D. 27

Based on the table, how many official camps are located in areas under Israeli control? A. 8 B. 12 C. 19 D. 27 $Based on the table, how many official camps are located in areas under Israeli control? A. 8 B. 1$...

### What were junks, and where did they ply the sea road trade?

What were junks, and where did they ply the sea road trade?...

### What are the different options in a page layout feature? Select three options.create multiple columnsadd

What are the different options in a page layout feature? Select three options. create multiple columns add images and shapes add section breaks create table of contents o add line numbers Need help ASAP...

### 2,5,4,1,3,12,4,3,3,4. calculate the following from the data set above. ( show work) a. mean (round to

2,5,4,1,3,12,4,3,3,4. calculate the following from the data set above. ( show work) a. mean (round to the nearest tenths place) b. mode c. median d. range...


text:
# Trigonometric Pattern Design

Triangles are my favorite shape, three points where two lines meet (Tessellate, Alt-J)

Inspired by recurrence plots and by the Gauss error function, I have done the following plots. The first one represents the recurrence plot of $f\left ( x \right )= sec\left ( x \right )$ where distance between points is measured by Gauss error function:

This one is the same for $f\left ( x \right )= tag\left ( x \right )$

And this one represents $latex f\left ( x \right )= sin\left ( x \right )$

I like them: they are elegant, attractive and easy to make. Try your own functions. One final though: the more I use magrittr package, the more I like it. This is the code for the first plot.

library("magrittr")
library("ggplot2")
library("pracma")
RecurrencePlot = function(from, to, col1, col2) {
opt = theme(legend.position  = "none",
panel.background = element_blank(),
axis.ticks       = element_blank(),
panel.grid       = element_blank(),
axis.title       = element_blank(),
axis.text        = element_blank())
seq(from, to, by = .1) %>% expand.grid(x=., y=.) %>%
ggplot( ., aes(x=x, y=y, fill=erf(sec(x)-sec(y)))) + geom_tile() +
RecurrencePlot(from = -5*pi, to = 5*pi, col1 = "black", col2= "white")

# The Zebra Of Riemann

Mathematics is the art of giving the same name to different things (Henri Poincare)

Many surveys among experts point that demonstration of the Riemann Hypothesis is the most important pending mathematical issue in this world. This hypothesis is related to Riemann zeta function, which is supossed to be zero only for those complex whose real part is equal to 1/2 (this is the conjecture of Riemann itself). Confirming the conjecture would imply deep consequences in prime numbers teory and also in our knowledge of their properties. Next plot represents the argument of zeta function over complex with real part between -75 and 5 (x axis) and imaginary part between -40 and 40 (y axis). An explanation of this kind of graph can be found here. Does not it remind you of something?

Solving the hypothesis of Riemann is one of the seven Clay Mathematics Institute Millennium Prize Problems. Maybe zebras keep the secret to solve it in their skins.

This is the code to plot the graph:

require(pracma)
z=outer(seq(-75, 5, by =.1),1i*seq(-40, 40, by =.1),'+')
z=apply(z, c(1,2), function(x) Arg(zeta(x)))
opt=theme(legend.position="none",
panel.background = element_blank(),
panel.grid = element_blank(),
axis.ticks=element_blank(),
axis.title=element_blank(),
axis.text =element_blank())
z=data.frame(expand.grid(x=seq(ncol(z)), y=seq(nrow(z))), z=as.vector(z))
require(ggplot2)
ggplot(z, aes(x=x, y=y, color=z)) + geom_tile() + scale_colour_gradientn(colours=c("black","white", "gray80", "gray40")) + opt


text:
help-gsl
[Top][All Lists]

## Re: [Help-gsl] Integration involving Bessel Function

 From: Brian Gough Subject: Re: [Help-gsl] Integration involving Bessel Function Date: Fri, 04 Jun 2010 11:40:47 -0400 User-agent: Wanderlust/2.15.6 (Almost Unreal) Emacs/23.2 Mule/6.0 (HANACHIRUSATO)

At Wed, 2 Jun 2010 06:33:16 -0400,
> I am a beginer with gsl and I am trying to do an integration of the form:
>
> \int_0^\infty [ x J0(x) F(x) ].
>
> J0(x) being oscillatory makes the integrtal +ve and -ve within its
> consecutive zero's. Form of F(x) is such that the overall integrand is a
> decaying function of x.
>
> How to handle this type of integration using gsl.
>
> I tried using "gsl_integration_qag", but its not giving the correct results.
>

Did you try gsl_integration_qagiu (infinite upper limit)?


text:
## Banach Journal of Mathematical Analysis

### Pictures of $KK$-theory for real $C^{*}$-algebras and almost commuting matrices

#### Abstract

We give a systematic account of the various pictures of $KK$-theory for real $C^{*}$-algebras, proving natural isomorphisms between the groups that arise from each picture. As part of this project, we develop the universal properties of $KK$-theory, and we use CRT-structures to prove that a natural transformation $F(A)\rightarrow G(A)$ between homotopy equivalent, stable, half-exact functors defined on real $C^{*}$-algebras is an isomorphism, provided it is an isomorphism on the smaller class of $C^{*}$-algebras. Finally, we develop $E$-theory for real $C^{*}$-algebras and use that to obtain new negative results regarding the problem of approximating almost commuting real matrices by exactly commuting real matrices.

#### Article information

Source
Banach J. Math. Anal., Volume 10, Number 1 (2016), 27-47.

Dates
Accepted: 6 April 2015
First available in Project Euclid: 15 October 2015

https://projecteuclid.org/euclid.bjma/1444913861

Digital Object Identifier
doi:10.1215/17358787-3163312

Mathematical Reviews number (MathSciNet)
MR3453522

Zentralblatt MATH identifier
1352.46064

#### Citation

Boersema, Jeffrey L.; Loring, Terry A.; Ruiz, Efren. Pictures of $KK$ -theory for real $C^{*}$ -algebras and almost commuting matrices. Banach J. Math. Anal. 10 (2016), no. 1, 27--47. doi:10.1215/17358787-3163312. https://projecteuclid.org/euclid.bjma/1444913861

#### References

• [1] P. Baul and M. Karoubi, On the Baum-Connes conjecture in the real case, Q. J. Math. 55 (2004), no. 3, 231–235.
• [2] B. Blackadar, $K$-Theory for Operator Algebras, 2nd ed., Math. Sci. Res. Inst. Publ. 5, Cambridge Univ. Press, Cambridge, 1998.
• [3] J. L. Boersema, Real $C^{*}$-algebras, united $K$-theory, and the Künneth formula, $K$-Theory 26 (2002), no. 4, 345–402.
• [4] J. L. Boersema, Real $C^{*}$-algebras, united $KK$-theory, and the universal coefficient theorem, $K$-Theory 33 (2004), no. 2, 107–149.
• [5] J. L. Boersema, E. Ruiz, and P. J. Stacey, The classification of real purely infinite simple $C^{*}$-algebras, Doc. Math. 16 (2011), 619–55.
• [6] A. K. Bousfield, A classification of $K$-local spectra, J. Pure Appl. Algebra 66 (1990), no. 2, 121–163.
• [7] A. Connes, Noncommutative Geometry, Academic Press, San Diego, Calif., 1994.
• [8] A. Connes and N. Higson, Deformations, morphismes asymptotiques et K-theorie bivariante, C. R. Acad. Sci. Paris Ser. I Math. 311 (1990), no. 2, 101–106.
• [9] J. Cuntz, A new look at $KK$-theory, $K$-Theory 1 (1987), no. 1, 31–51.
• [10] M. Dadarlat and S. Eilers, On the classification of nuclear $C^{*}$-algebras, Proc. London Math. Soc. (3) 85 (2002), no. 1, 168–210.
• [11] M. Dadarlat and T. Loring, $K$-homology, asymptotic representations, and unsuspended $E$-theory, J. Funct. Anal. 126 (1994), no. 2, 367–383.
• [12] K. R. Davidson, “Essentially normal operators” in A Glimpse at Hilbert Space Operators, Oper. Theory Adv. Appl. 207, Birkhäuser, Basel, 2010, 209–222.
• [13] M. B. Hastings and T. A. Loring, Topological insulators and $C^{*}$-algebras: Theory and numerical practice, Ann. Physics 326 (2011), no. 7, 1699–1759.
• [14] N. Higson, A characterization of $KK$-theory, Pacific J. Math. 126 (1987), no. 2, 253–276.
• [15] N. Higson, Categories of fractions and excision in KK-theory, J. Pure Appl. Algebra 65 (1990), no. 2, 119–138.
• [16] K. K. Jensen and K. Thomsen, Elements of $KK$-theory, Math. Theory Appl., Birkhäuser, Boston, 1991.
• [17] G. G. Kasparov, The operator $K$-functor and extensions of $C^{*}$-algebras, Izv. Akad. Nauk SSSR Ser. Mat. 44 (1980), no. 3, 571–636, 719.
• [18] H. Lin, “Almost commuting selfadjoint matrices and applications” in Operator Algebras and Their Applications (Waterloo, Ont., 1994/1995), Fields Inst. Commun. 13, Amer. Math. Soc., Providence, 1997, 193–233.
• [19] T. A. Loring, Lifting Solutions to Perturbing Problems in $C^{*}$-Algebras, Fields Inst. Monogr. 8, Amer. Math. Soc., Providence, 1997.
• [20] T. A. Loring and A. P. W. Sørensen, Almost commuting unitary matrices related to time reversal, Comm. Math. Phys. 323 (2013), no. 3, 859–887.
• [21] T. A. Loring and A. P. W. Sørensen, Almost commuting self-adjoint matrices—the real and self-dual cases, preprint, arXiv:1012.3494 [math.OA].
• [22] V. Manuilov and K. Thomsen, $E$-theory is a special case of $KK$-theory, Proc. London Math. Soc. (3) 88 (2004), no. 2, 455–478.
• [23] H. Schröder, $K$-Theory for Real $C^{*}$-Algebras and Applications, Pitman Res. Notes Math. Ser. 290, Wiley, New York, 1993.
• [24] P. J. Stacey, Antisymmetries of the CAR algebra, with an appendix by J. L. Boersema and N. C. Phillips, Trans. Amer. Math. Soc. 363 (2011), no. 12, 6439–6452.
• [25] K. Thomsen, Homotopy classes of $*$-homomorphisms between stable $C^{*}$-algebras and their multiplier algebras, Duke Math. J. 61 (1990), no. 1, 67–104.
• [26] D. Voiculescu, Remarks on the singular extension in the $C^{*}$-algebra of the Heisenberg group, J. Operator Theory 5 (1981), no. 2, 147–170.


text:
# How to align different parts of an equation to left or right?

For my homework I am required to replicate (among other things) the following equation:

I was able to achieve this using align:

\begin{align*}
some stuff with & \tag{15} \\
\end{align*}

but I had to use 4 \qquad's because the part on the right would not be on the right! This is ugly, but what bothers me more is the fact that align had worked before for this equation:

I used the following code here:

\begin{align*}
foo &= bar \tag{15} \\
&baz \\
foo &= bar \\
& baz & 0 < x < 1,
\end{align*}

You can see that the part on the right sits exactly where it should without using an embarrassing amount of \quads.

How do make those this equation align properly? How do I make a part of a line flush to the right?

Just for reference, this is how my PDF looks with the code above (this is from Bateman Manuscript Project, Higher Transcendental Functions vol. I, p. 23):

Here is my full code:

\documentclass[12pt,leqno]{article} %leqno is to align equation labels to left

\usepackage[bottom=0in, top=0.4in, left=1.2in, right=1.2in]{geometry} % set margins to look like the original
\usepackage{xfrac, relsize,fancyhdr} % for slanted fractions, bigger integral signs, and some other things
\usepackage[fleqn]{amsmath} % for equations flushed to left, among other things
\usepackage[nice]{nicefrac} % for nice slanted fractions

\DeclareMathOperator{\csch}{csch}

\pagestyle{fancy}
\fancyhf{} % removes default page number
\renewcommand{\headrulewidth}{0pt} % removes the top line

\begin{document}

\noindent \textbf{1.9.1. \; Kummer's series for } \textbf{$\log \Gamma (z)$} \\

The function $\log \Gamma (x)$, $0 < x < 1$, can be expanded in a Fourier series. We shall use the known Fourier expansions (Bromwich, 1947, pp. 356, 393, and 370 respectively):
\begin{align*}
& \log(\sin\pi x) =  -  \log 2  -  \sum_{n=1}^{\infty} \, (1/n) \cos (2\pi n x), \\
& \csch (\nicefrac{1}{2} \, t) \, \sinh (\nicefrac{1}{2}-x)t = 8 \pi \, \sum_{n=1}^{\infty} [n \; \sin(2\pi n x)]/(t^2 + 4{\pi}^2 n^2), \\
& \pi (1 - 2x) = 2 \sum_{n=1}^{\infty} (1/n) \sin(2 \pi nx).
\end{align*}

If these are substituted in (7) with $z=x$, we have to evaluate the integral

$\mathlarger{\int_0^{\infty}} \left(\frac{2\pi n}{t^2 + 4{\pi}^2 \, n^2)} - \frac{e^{-t}}{2\pi n} \right)\frac{dt}{t} = \frac{1}{2\pi n} \mathlarger{\int_0^{\infty}} \left(\frac{1}{1+t^2} - e^{-2 \pi n t}\right)\frac{dt}{t}$

$= \frac{1}{2 \pi n} \left[ \mathlarger{\int_0^\infty} \left( \frac{1}{1+t^2}-\cos t \right) \frac{dt}{t} + \mathlarger{\int_0^\infty} \frac{e^{-t}-e^{-2 \pi n t}}{t}dt + \mathlarger{\int_0^\infty}(\cos t - e^{-t})\frac{dt}{t}\right]$ \\

\noindent and by means of 1.7(21) and 1.7(18) this is $(2 \pi n)^{-1}[\gamma + \log(2 \pi n)]$ since we have for the third integral: \\

$\lim \limits_{\delta \, \to \, 0} \, \int_\delta^{\, \infty} \, (\cos t - e^{-t}) \, t^{-1} dt = \lim \limits_{\delta \to 0} \; [\, \text{Ei}(- \delta) - \text{Ci} (\delta) \,] = 0.$ \\

\noindent Thus we have
\begin{align*}
\log \, \Gamma(x) &= \nicefrac{1}{2} \log(2 \pi) \tag{14}\\
&+ \sum_{n=1}^\infty [(2 \pi)^{-1} \, \cos(2 \pi n x) + (\gamma + \log 2 \pi n) \, (\pi n)^{-1} \, \sin(2 \pi n x)], \\
\log \, \Gamma(x) &= (\nicefrac{1}{2}-x)(\gamma + \log 2) + (1-x) \, \log \pi - \nicefrac{1}{2} \log(\sin \pi x)\\
&+ \sum_{n=1}^\infty (\pi n)^{-1} \log n \, \sin(2 \pi n x) & 0 < x < 1,
\end{align*}

\noindent which is Kummer's series.

A similar representation for $\psi (x)$ is due to Lerch (Nielsen, 1906, p. 204),
\begin{align*}
\psi (x)&  \, \sin(\pi x) = - \nicefrac{1}{2} \pi \cos(\pi x) - (\gamma + \log 2 \pi) \sin \pi x \tag{15} \\
&+ \sum_{n=1}^\infty \log \left( \frac{n}{n+1} \right) \sin(2 \pi + 1) \pi x & \, & \; 0 < x < 1.
\end{align*}

\end{document}

• Note that 2 alignment groups require 3 &. That said, what you want to achieve is not very clear. Mar 11 '17 at 15:30
• Like what? I just tried inserting another & but the thing is still not aligned properly. Mar 11 '17 at 15:32
• Could you post a full compilable code, and explain a little more? It seems you want equation numbering on the left? Mar 11 '17 at 15:34
• Yes, I achieved that with leqno. I can post the code, but it's 64 lines - wouldn't it be too much a mess? Mar 11 '17 at 15:35
• Not really a problem, we'll extract a relevant fraction if necessary. Mar 11 '17 at 15:36

Here's another attempt. Note that I've tried to make the code more "LaTeX-y" by using instructions such as \subsubsection and letting LaTeX place left-aligned equation numbers. I've gotten rid of the \mathlarger instructions (they're not needed) as well as of most instances of \, (thinspace) -- let TeX take care of the appropriate spacing before and after math operators. I've replaced the two instances of \lim\limits with just \lim (the \limits directives aren't needed). I don't think \nicefrac{1}{2} looks good, at least not with Computer Modern; I recommend using \tfrac instead. I've replaced all instances of \left and \right with \biggl and \biggr, resp. I've gotten rid of the super-kludgy \voffset=0.15in directive and, instead, added the option includehead while loading geometry. Eliminating almost all blank lines also permits omitting most \noindent directives. Finally, I've used \mkern directives to position the 0<x<1 terms.

\documentclass[12pt,leqno]{article}
\usepackage[bottom=0in,top=0.4in,hmargin=1.2in,
\usepackage[fleqn]{amsmath}
\DeclareMathOperator{\csch}{csch}
\usepackage{fancyhdr,indentfirst}
\pagestyle{fancy}
\fancyhf{}

\setcounter{section}{1}
\setcounter{subsection}{9}
\setcounter{page}{23}
\setcounter{equation}{13}

\begin{document}
% insert \boldmath directive in arg. of '\subsubsection if '\log\Gamma' should be bolded
\subsubsection{Kummer's series for $\log \Gamma (z)$}

The function $\log \Gamma (x)$, $0 < x < 1$, can be expanded in
a Fourier series. We shall use the known Fourier expansions
(Bromwich, 1947, pp.\ 356, 393, and 370 respectively):
\begin{align*}
& \log(\sin\pi x) =  -\log2 -\sum_{n=1}^{\infty}(1/n)\cos(2\pi nx), \\
& \csch(\tfrac{1}{2}t)\sinh(\tfrac{1}{2}-x)t
= 8\pi\sum_{n=1}^{\infty} [n\sin(2\pi nx)]/(t^2 + 4\pi^2 n^2), \\
& \pi(1-2x) = 2\sum_{n=1}^{\infty}(1/n)\sin(2\pi nx).
\end{align*}

If these are substituted in (7) with $z=x$, we have to evaluate
the integral
\begin{multline*}
\int_0^{\infty} \biggl( \frac{2\pi n}{t^2 + 4\pi^2 n^2} - \frac{e^{-t}}{2\pi n} \biggr)\frac{dt}{t}
= \frac{1}{2\pi n} \int_0^{\infty} \biggl(\frac{1}{1+t^2} - e^{-2 \pi nt}\biggr)\frac{dt}{t}\\
= \frac{1}{2 \pi n} \Biggl[ \int_0^\infty \biggl( \frac{1}{1+t^2}-\cos t \biggr) \frac{dt}{t}
+ \int_0^\infty \frac{e^{-t}-e^{-2\pi nt}}{t}\,dt
+ \int_0^\infty(\cos t - e^{-t})\frac{dt}{t}\Biggr]
\end{multline*}
and by means of 1.7(21) and 1.7(18) this is
$(2\pi n)^{-1}[\gamma+\log(2\pi n)]$
since we have for the third integral:
$\lim_{\delta\to 0} \int_\delta^{\infty} (\cos t-e^{-t}) t^{-1}\, dt = \lim_{\delta\to 0} \,[ \text{Ei}(-\delta) - \text{Ci}(\delta)] = 0\,.$
Thus we have
\begin{align}
\log\Gamma(x) &= \tfrac{1}{2} \log(2 \pi) \\
\notag    &+ \sum_{n=1}^\infty [(2\pi)^{-1}\cos(2\pi n x)
+ (\gamma+\log2\pi n)(\pi n)^{-1}\sin(2\pi nx)], \\
\log\Gamma(x) &= (\tfrac{1}{2}-x)(\gamma+\log2)+(1-x)
\log\pi-\tfrac{1}{2}\log(\sin\pi x)\notag\\
\notag   &+ \sum_{n=1}^\infty(\pi n)^{-1}\log n  \sin(2\pi nx) &  \mkern16mu 0 < x < 1,
\end{align}
which is Kummer's series.

A similar representation for $\psi(x)$
is due to Lerch (Nielsen, 1906, p.\ 204),
\begin{align}
\psi (x)& \sin(\pi x) = -\tfrac{1}{2} \pi\cos(\pi x)
- (\gamma+\log2\pi)\sin\pi x \\
\notag  &+ \sum_{n=1}^\infty \log \biggl( \frac{n}{n+1} \biggr)
\sin(2\pi+1) \pi x & \mkern145mu 0 < x < 1.
\end{align}
\end{document}

• The original clearly doesn't use \boldmath in the section title. The section/subsection number can be added at the top left by suitably redefining \sectionmark and \subsectionmark. Mar 11 '17 at 20:47
• @egreg - Thanks. I was looking at the original more closely just now and had also noticed some of these issues. I'll provide suitable updates shortly.
– Mico
Mar 11 '17 at 20:51
• @egreg no no no no, this is exactly how this stuff appears in the book! I used \textbf but it didn't work on formulae Mar 11 '17 at 20:54
• @Gallifreyan - In the (hopefully not entirely vain) hope that you'll learn a few things about LaTeX and its programmability from this exercise, I deliberately provided a bit of material that shows how to make use of LaTeX's machinery. In the short run (i.e., for your assignment), this may well seem like more work than simply engaging in visual formatting. In the longer run, though, I think (hope?) you'll appreciate knowing a few LaTeX-related programming tricks.
– Mico
Mar 11 '17 at 21:17
• Given I am a Physics major, I don't think I have a choice but to learn LaTeX in full :D Mar 11 '17 at 21:30

Here is my (honest) proposal. I'm not sure I've understood what you want to obtain in full detail, but I did what seemed sensible. In particular, I removed most manual spacings, and defined a macro for the larger integrals, which not include the bounds, but require a manual correction for the placement of the lower bound. Note I gave different alignments for the second lines in eqs 14 and 15, to show the possibilities.

 \documentclass[12pt, leqno]{article} %leqno is to allgn equation labels to left

\usepackage[bottom=0in, top=0.55in,hmargin=1.2in, headheight=15pt, showframe]{geometry} % set margins to look like the original
\usepackage{xfrac, relsize,fancyhdr} % for slanted fractions, bigger integral signs, and some other things
\usepackage[fleqn]{mathtools, nccmath} % for equations flushed to left, among other things
\usepackage[nice]{nicefrac} % for nice slanted fractions

\DeclareMathOperator{\csch}{csch}

\pagestyle{fancy} % I'm not really sure what this does, but the internet guys said to use it ¯ \ _ (ツ) _ / ¯
\fancyhf{} % removes default page number
\renewcommand{\headrulewidth}{0pt} % removes the top line

\newcommand\lint{\mathop{\mathlarger{\int}}\nolimits}

\begin{document}

\noindent \textbf{1.9.1. \; Kummer's series for \boldmath$\log \Gamma (z)$}\bigskip

The function $\log \Gamma (x)$, $0 < x < 1$, can be expanded in a Fourier series. We shall use the known Fourier expansions (Bromwich, 1947, pp. 356, 393, and 370 respectively):
\begin{align*}
& \log(\sin\pi x) = - \log 2 - \sum_{n=1}^{\infty} \, (1/n) \cos (2\pi n x), \\
& \csch (\mfrac{1}{2} \, t) \, \sinh (\mfrac{1}{2}-x)t = 8 \pi \, \sum_{n=1}^{\infty} [n \; \sin(2\pi n x)]/(t^2 + 4{\pi }^2 n^2), \\
& \pi (1 - 2x) = 2 \sum_{n=1}^{\infty} (1/n) \sin(2 \pi nx).
\end{align*}

If these are substituted in (7) with $z=x$, we have to evaluate the integral
%
\begin{multline*}
\lint_{\mkern-15mu 0}^{\infty} \biggl(\frac{2\pi n}{t^2 + 4{\pi }^2 \, n^2)} - \frac{e^{-t}}{2\pi n} \biggr)\frac{dt}{t} = \frac{1}{2\pi n} \lint_{\mkern-15mu 0}^{\infty}\biggl(\frac{1}{1+t^2} - e^{-2 \pi n t}\biggr)\frac{dt}{t} \\
= \frac{1}{2 \pi n} \left[ \lint_{\mkern-15mu 0}^\infty \biggl( \frac{1}{1+t^2}-\cos t \biggr) \frac{dt}{t}
+ \lint_{\mkern-15mu 0}^\infty \frac{e^{-t}-e^{-2 \pi n t}}{t}dt %
+ \lint_{\mkern-15mu 0}^\infty (\cos t - e^{-t})\frac{dt}{t}\right]
\end{multline*}%
and by means of 1.7(21) and 1.7(18) this is $(2 \pi n)^{-1}[\gamma + \log(2 \pi n)]$ since we have for the third integral:
%
$\lim_{\delta \to 0} \int_\delta^{\infty}(\cos t - e^{-t}) \, t^{-1} dt = \lim_{\delta \to 0} \bigl [\text{Ei}(- \delta ) - \text{Ci} (\delta ) \bigr] = 0.$
%
Thus we have
\begin{flalign} \tag{14}
\log \Gamma (x) &= \begin{aligned}[t] & \nicefrac{1}{2} \log(2 \pi ) & \\
& \mathrlap{+ \sum_{n=1}^\infty [(2 \pi )^{-1} \, \cos(2 \pi n x) + (\gamma + \log 2 \pi n) \, (\pi n)^{-1} \, \mathrlap{\sin(2 \pi n x)]},}
\end{aligned} &
\\
\notag
\log \Gamma (x) &= \begin{aligned}[t] & (\nicefrac{1}{2}-x)(\gamma + \log 2) + { (1-x) \, \log \pi -\nicefrac{1}{2} \log(\sin \pi x)}\\
&+ \sum_{n=1}^\infty (\pi n)^{-1} \log n \, \sin(2 \pi n x)
\end{aligned} %
& & \begin{aligned}[t]
& \mbox{}\\[2ex] 0 < x < 1, \end{aligned}
\end{flalign}
%
which is Kummer's series.

A similar representation for $\psi (x)$ is due to Lerch (Nielsen, 1906, p. 204),%
\begin{flalign}\tag{15}%
\psi (x)\sin(\pi x)&= - \nicefrac{1}{2} \pi \cos(\pi x) - (\gamma + \log 2 \pi ) \sin \pi x \\
& + \sum_{n=1}^\infty \log \left( \frac{n}{n+1} \right) \sin(2 \pi + 1) \pi x % This is ugly, but there's nothing I could do
& & 0 < x < 1.\notag
\end{flalign}

\end{document}

• This is weird - it is still not left enough when I compile it Mar 11 '17 at 17:27
• What should be more to the left? The equation numbers are at the left margin. I don't see what you mean. Equation numbers in the left margin, and body of the equation beginning at the left margin? Mar 11 '17 at 17:33
• Sorry :D I meant right. 0<x<1. should be on the right, directly under 0<x<1, in (14). It is correct on your render, but not on mine. Mar 11 '17 at 17:49
• @Gallifreyan: I have a solution with the flalign environment. That's the first I thought of, but it seemed to be conflicting with the leqno option – probably some bug of amsmath. However, loading mathtools (which fixes some of these bugs) in the place of amsmath solves the problem. Please see my updated answer. Is it more like you want? Mar 11 '17 at 20:43
• @Mico: Problem solved. Just add a second ampersand on the first line of flalign`. Mar 12 '17 at 12:58


text:
Conferences

Division:
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- The Division of Applied Geochemistry and Environmental Engineering The Division of Geotechnology The Division of Geodynamics and Environmental Engineering Division of Renewable Energy Sources The Division of Environmental Studies and Waste Management Division of Biogenic Raw Materials The Division of Minerals and Energy Sustainable Development The Division of Mineral Resources Acquisition The Division of Fossil Fuels and Energy Market Research The Division of Mineral Policy The Division of Economic Geology The Division of Strategic Research The Division of Energy Economics
Okres realizacji: -