How to Solve Mathematical Induction Problems

In the previous post we have discussed about Composition of Functions and In today’s session we are going to discuss about How to Solve Mathematical Induction Problems. In Mathematics, sometimes there is a need to prove statements that is in the form of theorem or formula. Mathematical induction is the process in which on the basis of hypothesis ,conclusion is generated and there are mainly threes steps for define mathematical induction. In the blog we are defining all the steps of proving mathematical induction that is as follows:

In the first step we prove the hypothesis or statement that is given for the variable having the value equal to 1, means when the statement is true for n = 1 then go to the next step. (know more about Mathematical induction, here)

In the second step prove that given statement for value of variable equals to k means n = k prove statement for an arbitrary value. And when statement is true for the second step then in the last or final step prove the statement or formula for n = K + 1 that is a value increased by one.

At last when for all the three steps given statement is true that means for all the values statement will be true.

If there is a statement such as 1 + 2 + 3… + s = s (s + 1) / 2

For the statement p (s) have some integer s then according to the mathematical induction p (1) is true for the given expression as 1 ( 1+ 1) / 2 = 1 * 2 / 2 = 2 / 2 = 1 .

In next step find this statement true for any integer value as p (4) = 4 ( 4 + 1) / 2 = 4 * 5 / 2 = 10 that is also true as 1 +2 + 3+ 4 = 10.

at last find that statement is true for p (K +1) that is true ,so p (s) is true for all the values n >= 1.

college algebra solver defines the step by step procedure of solving the algebra and also the word problems of algebra. Cbse board provide CBSE Hindi Syllabus to students of their board that define all the topics of Hindi related to their session. 

Trigonometric form of complex numbers

In this blog we are going to learn about the Trigonometric form of complex numbers that contains any complex number and it can be represented in the trigonometric form that is also called as polar coordinates. (Also look Derivatives of Trig Functions). We explain it by an expression as z = r (cos a + i . sin a), here ‘a’ belongs to Arg (z) that is a modulus or absolute value of ‘z’ and it is fine as:

| x + i .y| = √ x ² + y ² is trigonometric form of a complex number (more information on trigonometric functions is here). You can also play trigonometric identities worksheet available online and can improve you math skills.

But the main question is that how we find for solving the expression. So for finding the value of ‘a’ we know that ‘a’ is not unique but it is found by modulo 2 ∏ .Arg (z) ‘s main value belongs to the interval [0 , 2∏]. If z = x + y i then ‘a’ is angle formed on the x- axis by the radius vector of point (x, y) or these are denoted as the points (x, y) where unit circle is intersected by these points . It is denoted as:

z / |z| = cos a + i . sin a.

Value of ‘a’ can be found apparently as tan a = y / x but it should be noted that for x = 0 it will not work. It is correct to great extent but for x = 0 that is an exceptional case it will be not appropriate.

When there are value of a arg (z) is ∏ / 2 or 3 ∏ / 2 then it depends on the value of sin y. (that is for

y = 0 and z = 0, we know 0 is only complex number that is not related with any argument).

In upcoming posts we will discuss about Polar/rectangular coordinates and Absolute Value in Grade XI. Visit our website for information on higher secondary education Karnataka

 

Mathematical induction

In this blog we are going to discuss about the mathematical induction that is used for mathematical proof to establish the given statement true for all natural numbers these numbers are positive integers. Mathematical induction is define in the way that there are some statements A (n) and a (n) is true for all the ‘n’ statements that belongs to ‘N’ that is a set of statements n .In the mathematic induction proofs for statements A (0), A (1), ………., A(s), here s is a number, then induction is used to describe the difference between continuous and discrete mathematics. For more information on math induction visit this

According to ICSE syllabus there are mainly two steps for defining the mathematical induction that are as:

(a) Base Step: The first step defines the proof for a statement A (0) means it proof the initial statement true.

(b) Inductive step: Next step of mathematical induction is Inductive step that make the statement A(s) true and proof next statement A(s + 1) true.

In the induction step A(s) statement is known as the hypothesis and A (s + 1) statement is known as the conclusion step that is define for the hypothesis statement A(s).

It will be expressed as [A (0) > all k (A(s) → A(s + 1))] → all s A(s),

It is also define by the example of addition of positive integers 1 + 2 ….. + s = s(s + 1) / 2 statements

a (1) = 1 = 1 (1 + 1) / 2 = 1 then for,

A(s) = 1 +2 + ….. + (s) = s(s + 1) / 2 is true and then A(s + 1) is also true as,

A(s + 1) = 1 +2 + ….. + (s) + (s + 1) = (s + 1) (s + 2) / 2.

In upcoming posts we will discuss about Justify the Pythagorean identities and Trigonometric Ratios in grade XI. Visit our website for information on 5th grade math