Sunday, June 28, 2020
THEOREMS ON PROBABILITY OF EVENTS( PART -2)
Saturday, June 27, 2020
A QUESTION FROM IMO 1995 | TITU'S LEMMA | CAUCHY-SCHWARZ INEQUALITY
by the use of Cauchy-Schwarz inequality
Friday, June 26, 2020
BOLZANO-WEIERSTRASS THEOREM FOR SETS | REAL ANALYSIS
Have you remembered Bounded set ,Order completeness of Real numbers.
If not just Click to Read Real Analysis part-1 .
Bounded Set:
A set is said to be bounded if it is bounded above and below.
" I will not explain this just read about it in the above link "
Bolzano-Weierstrass Theorem:
Statement: Every infinite bounded set of real numbers has a limit point.
Proof: Let ' S ' be any infinite bounded set of real numbers.
m ≤ x ≤ M , ∀ x ∈S ---(1)
m = infimum of S M=supremum of S
Let us assume T is a set of Real numbers.
T = { x : x is greater than only for finite number of elements } ------(2)
x ∈ T { x ≥ only a finite number of elements of S } --(3)
From equation (1) and (4)
m ∈ T ⇔ T ≠ Φ
From equation (1)
M ≥ x , ∀ x ∈S
Let us assume another element 'y' from set T.
y ∈ T ⇔ y ≥ only a finite number of elements of S [From equation (3)]
M ≥ y , ∀ y ∈ T ⇔ T is bounded above.
Now we have prove that T is bounded above
So it will follow Order Completeness Property and we know that " There exist a supremum of a set of which set follows the completeness property ".
Let P be the Supremum of T .
Claim: ( P is a limit point of set S )
Let є > 0 be arbitrary number.
Since P + є > P
⇒ P + є ≥ finite number of elements of S -----(5)
[ Because P is the Supremum of T and T is the subset of S]
Since P - є < P
⇒ P - є ≤ infinite number of elements of S ----(6)
From equation (5) and (6) that nbh
] P - є , P + є [ of P contains infinite number of elements of S.
Hence , P is a limit point.
Example:
- Z has no limit point .{it has not bounded}
S = { -1/2 ,2/3 , -3/4 .....}
infinite | Bounded -1 < x < 1
hence lmit point will be 0.
If any doubt , Comment below
THEOREMS ON PROBABILITY OF EVENTS (PART-1)
Before you readout this blog I suggest you to read it's previous session i.e. " Introduction to Probability" , CLICK HERE to read.
THEOREM-1:
PROOF:
REMARK:
It may be noted P( A )=0, does not imply that A is necessarily an empty set. In practice, probability ' 0 ' is assigned to the events which are so rare for example , let us consider the random tossing of a coin. The event that the coin will stand erect on its edge, is assigned the probability 0.
THEOREM -2:
Probability of the complementary event Ā of A is given by P( Ā ) =1-P( A ).
PROOF:
THEOREM - 3:
PROOF:
ADDITION RULE:
★ For any two events A and B
P( A ⋃ B ) = P( A ) + P( B ) - P( A ⋂ B ).
Answer- Since the dice is thrown twice , the sample space ( S ) = | 36 |
Thursday, June 25, 2020
Introduction to Geometric Progression
GEOMETRIC PROGRESSION(G.P.):
⮚A progression is called a G.P. if the ratio of it's each term to it's previous term is always constant. If 'a' be the first term and 'r' be the common ratio then a, ar, ar2, ... ,arn-1 is a sequence of G.P . tn = arn-1.
GENERAL TERM OF A G.P.:
- nth term of G.P. i.e. tn = arn-1. , where the common ratio (r) = t2 / t1 = t3 / t2 = ...
- If G.P. consists of ' n ' terms , then pth term from the end = (n-p+1)th term from the beginning = arn-p. Also , the pth term from the end of the G.P. with the last term ' l ' and common ratio 'r' is = l(1/r)n-1 .
SUM OF FIRST ' n ' TERMS OF A G.P.:
- Sn = a(1-rn )/(1-r) and Sn = (a-lr)/(1-r) , [when | r | < 1. ]
- Sn = a(rn -1)/(r-1) and Sn = (lr-a)/(r-1) , [when | r | > 1. ]
- Sn = na [ when r=1 ]
SUM OF INFINITE TERMS OF G.P.:
⮚ When n→∞ and | r | < 1 ( -1 < r < 1), S∞ = a/(1-r).
⮚ When n→∞ and | r | ≥ 1 , S∞ does not exist.
GEOMETRIC MEAN(G.M.):
⮚ If a , b , c are in G.P. , then b/a = c/b ⇒ b2= ac ⇒ b= √(ac) is the G.M. of a and c.
Similarly G.M. of a , b , c is (abc)1/3 .
G.M. of a1 , a2 , a3 , ... , an is (a1. a2 . a3 ... .an )1/n
⮚ Let n G.M.s are inserted between a & b . Let G1 , G2 , G3 , ... , Gn are n G.M.s ,
then a , G1 , G2 , G3 , ... , Gn , b are in G.P. .
Then common ratio ( r ) = (b/a )1/n+1
So, G1 = ar = a . (b/a )1/n+1
G2 = ar2 = a . (b/a )2/n+1
___________________________
Gn = arn = a . (b/a )n/n+1
PROPERTIES OF G.P. :
✪ If all the terms of a G.P. be multiplied or divided by the same non-zero constant , then it remains a G.P. , with the same common ratio ' r '.
✪ The reciprocal of the terms of a given G.P. form a G.P. with common ratio as reciprocal of the common ratio(r) of the original G.P. i.e. ' 1/r '.
✪ If each term of a G.P. with common ratio ' r ' be raised to the same power k , the resulting sequence also forms a G.P. with common ratio rk .
✪ In a finite G.P. , the product of terms equidistant from the beginning and the end is always the same and is equal to the product of the first and last term i.e. if a1 , a2 , a3 , ... , an be in G.P. . Then a1. an = a2 . an-1 = a3 . an-2 = ... = ar . an-r+1
✪ If a1 , a2 , a3 , ... , an , ... is a G.P. of non-zero , non-negative terms , then loga1 , loga2 ,log a3 , ... , logan , ... is in A.P. and vice-versa.
✪ Three non-zero numbers a , b , c are in G.P. iff b2= ac .
✪ If ax1 , ax2, ax3, ... , axn are in G.P. , then x1 , x2 , x3 , ... , xn are in A.P. .
✪ If the first term of a G.P. of n terms is ' a ' and last term is ' l ' , then the product of all terms of the G.P. is (al )n/2 .
⮚ So, in the below tables , these are terms taken , which should be used in some kinds of questions in G.P. as same as in the A.P. .
TABLE-1 : When the product is given.
No. of terms Terms taken
3 a/r , a , ar4 a/ r3 , a/r , a/r , a/ r3 5 a/ r2 , a/r , a , a/r , a/ r2
3 a , ar , ar2 4a , ar , ar2 , ar3 5a , ar , ar2, ar3, ar4
Newton-Raphson Method ---Solution of Polynomial And Transcendental equation| Engineering Mathematics
≫In the previous session we read about Bisection method which is used to solve the equation for finding root.
if you are not read that blog i suggest you to read that blog first
Click Here to read.
Newton-Raphson Method:
Concept:
1. Given ∱(x)=0 ------(1)
Find initial root x0 such that ∱(x)~0
[ i.e; x0 is near to the root ∱(x)=0]
2. Find ∱(x0) and ∱'(x0)
3. First approximate root by Newton-Raphson method:
Find ∱(x1) and ∱'(x1)
4. Similarly, Second approximate root is
5.Also Third approximate root is
General Formula:
Example
QnS. ∱(x) = x - 2 + lnx has a root near x =1.5 .Use Newton-Raphson Method to obtain a better estimate 2 digit after decimal point.
Ans.
In this question we don't need to figure out the x0, because it has given that the root is lies near x=1.5 .
Here x0 =1.5 and ∱(1.5) = -0.5 + ln(1.5) = -0.0945
∱'(x)= 1+(1/x)
⇒ ∱'(1.5) = 5/3
Hence using the general formula
x1=1.5567
The Newton-Raphson formula can be used again.
so ∱(x1) = ∱(1.5567) = -0.0007
∱'(x1) = ∱'(1.5567) = 1.6424
{ Calculation Part is skipped for Readers}
x2= 1.5571
⇒We will note that the 2digits after decimal point of x1 is same as x2 .So
our answer will be x2.