AWS Lambda with Spring Boot

The typical deployment scenario for a Spring Boot application in AWS involves running the Java application on an EC2 instance 24 hours a day. Of course, the application could be deployed in AWS ECS as a Docker container, but it still runs continuously on an EC2 instance. In each case, the EC2 instances need to be monitored and you pay for compute capacity used by that EC2 instance.

AWS Lambda provides low cost compute with zero maintenance. Lambda runs your code on demand, without provisioned and managed servers. Lambda automatically runs and scales your code. You are charged for every 100ms your code executes and the number of times your code is triggered. If the code isn’t running, you pay nothing. https://goo.gl/uvqpQb #DataIntegration #ML

Two More Math Problems: Continued Fractions, Nested Square Roots, Digits of Pi

These problems are for college undergrads after a first course in calculus. They are provided with solutions, and could be used by college professors as exercises or exam questions.

1. Digits of Pi/4

Prove that in base b, if b is an even integer, n > 3, and x = Pi/4, then the n-th digit of x, denoted as a(n), is given by the formula below. We start with n = 1 after the decimal point, for the first digit. Also show that the formula below is not valid if the base b is an odd integer, or if x is different from Pi/4. 

where the brackets represent the integer part (also called floor) function.

Solution

Regardless of the number x in [0, 1] and the base b, the n-th digit a(n) of x can be computed as follows:

See here for details. Thus we have

Using the angle difference formula for sinus, the fact that n > 3, b is an even integer, and x = Pi/4, it simplifies to

The result for a(n) follows immediately.

2. Continued Fractions and Nested Square Roots

Let us consider the two following expressions, assuming a is a strictly positive real number:

Prove that x is an integer if and only if a is the product of two consecutive integers. Prove that the same is true for y. 

Solution

Let’s focus on the first case. The second case is almost identical. The strictly positive number x must satisfy x^2 = a + x, thus x = (1 + SQRT(1 + 4a)) / 2. In order for x to be an integer, 1 + 4a must be a perfect square, which is possible  only if a is the product of two consecutive integers. For instance,

* If a = 1 * 2, then x = 2
* If a = 2 * 3, then x = 3
* If a = 3 * 4, then x = 4
* If a = 4 * 5, then x = 5
* and so on.

Note that the expansion of the number x = 2 in the nested square root numeration system, when x tends to 2, has all its “digits” equal to a = 1 * 2. See this spreadsheet for details. More on this here. 

For related articles from the same author, click here or visit www.VincentGranville.com. Follow me on on LinkedIn.

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