Programming in Java
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Find the error in the following and win Vissicomp Technology discount coupon up to Rs.1000/-
32 = 3 + 3 + 3 ( Three added three times. )
52 = 5+5+5+5+5 ( Five added five times. )
X2 = x+x+x+x+ …… x times
Find first derivative w.r.t. x on both sides.
2x = 1+1+1+1+…… x times
2x = x cancelling x on both side
Find the error!
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Edited by Dr. Ashwin I Mehta
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One of the fundamental Data Mining Techniques is finding ‘Association Rules’. This allows retailers to identify purchase pattern of customers. Here I am going to demonstrate a simple technique with Python capabilities at ‘Vissicomp Technologies’ to unravel such a pattern.
Let us assume that a store is selling only three items namely ‘bread’,’butter’ and ‘jam’. A customer can purchase all three items i.e. [‘bread’,’butter’,’jam’], a list in Python . He can also buy any combinations of them. Therefore I need to find all subsets of above set. A program to find subset is given here.
#recursion in arrays
if l == :
return x+[[l]+y for y in x]
The function subs() defined above will generate power set of set [‘bread’,’butter’,’jam’]. In a screen shot given below program and its output both are shown.
Output shows all possible subsets of a universal set . For further analysis and finding “large item set “ I will label these sets as l1, l2 l3…… and so on. Therefore , let
l1=[‘bread’], l2= [‘butter’], l3=[‘jam’],l4=[‘bread’, ‘butter’], l5= [‘bread’,’jam’], l6= [‘jam’,’butter’], l7= [‘bread’,’butter’,’jam’]. We leave out an empty set. Let us say on a particular day we have following sales data:
daysale= [‘l1’,’l2’,’l3’, ‘l4’,’l7’,’l4’, ‘l5’,’l5’,’l6’,‘l4’,’l4’,’l5’,‘l7’,’l4’,’l6’]
Data shows we have 15 sale deal on that particular day. To determine large dataset , we need to determine frequency of each of l1,l2,l3,….. . This is done by following program.
Output shows ‘l4’ has occurred 5 times, ‘l5’ has occurred 3 times and so on . so if we set threshold as 3 only ‘l4’ and ‘l5’ will be candidate for large dataset.
I will welcome suggestions from all of you.
For Machine Learning join Vissicomp. Contact us at /9820681349.
Created and edited by Dr. Ashwin I Mehta at Vissicomp Technology.
There is a 150 percent increase in the job postings for cyber security roles between January 2017 and March 2018, according to a survey by online job portal Indeed .
The data showed that Bengaluru accounts for the highest number (36 percent) of jobs in the sector in India.
Between January 2017 and March 2018, the survey said that there has also been a spike in the number of job postings for data protection roles, which have seen an increase of 143 percent. The number of job searches for the same have risen by 188 percent.
As per the survey, implementation of the General Data Protection Regulation (GDPR) law in Europe has stimulated Indian companies to fortify their databases, leading to an upswing in the search for cyber security and privacy professionals.
Further, Indeed said that the unprecedented increase in the number of cybercrimes in the country has also created a number of job opportunities for data protection and cyber security professionals. According to the Cybersecurity Jobs Report 2018-2021, cybercrime will more than triple the number of job openings over the next five years.
After Bengaluru, Mumbai (17 percent), NCR region (12 percent), Pune (9 percent) and Hyderabad (8 percent) are the other locations for work opportunities in the sector in India.
Sashi Kumar, Managing Director, Indeed India, said that companies across the world are gearing up to ensure compliance to General Data Protection Regulation (GDPR) and ePrivacy requirements.
“While the larger technology giants are more or less equipped to comply, it is the mid-size and smaller firms that are seeking professionals to help them cope with the requirements the new laws entail,” he added.
Indeed’s data also highlights a high level of interest in the field among the age group of 26-30 years. Interestingly, people between the age group of 41-45 years also show high levels of interest in these profiles, as compared to those between 31-35 years of age.
Use the 4th order Runge-Kutta method with h = 0.1 to find the
approximate solution for y(1.1), working to 4 decimal places, for the
initial value problem:
dy/dx = 2xy, y(1) = 1
We have dy/dx = f(x,y) = 2xy.
If you require the 4th order approximation the formula will be:
y(x0 + h) = y(x0) + (1/6)[w1 + 2w2 + 2w3 + w4]
w1 = h*f(x0.y0) = 0.1(2)(1)(1) = 0.2
w2 = h*f(x0 + h/2, y0 + w1/2) = 0.1(2)(1.05)(1.1) = 0.231
w3 = h*f(x0 + h/2, y0 + w2/2) = 0.1(2)(1.05)(1.1155) = 0.234255
w4 = h*f(x0 + h, y0 + w3) = 0.1(2)(1.1)(1.234255) = 0.2715361
y(1.1) = 1 + (1/6)[0.2 + 2(0.231) + 2(0.234255) + 0.2715361]
We can compare this with the exact solution to the problem.
dy/y = 2x dx
ln(y) = x^2 + C
and y = 1 when x = 1
0 = 1 + C
and so C = -1. Therefore:
ln(y) = x^2 – 1
y = e^(x^2-1)
When x = 1.1 this gives:
y = e^(1.1^2 -1)
which is same as Runge-Kutta up to the 5th decimal place.
Edited by- : Dr. Ashwin I Mehta
I’m grateful to Ashwin Mehta Sir for helping out my son with his java assignments.His patient and kind temperament adds to his profound knowledge of the subject. Thanks a lot Sir !