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UPSC Mathematics

Linear programming problem Using simplex method.

Question :                    Max Z = 6X + 8Y  subject to,                    30X + 20Y ≤ 300                      5X + 10Y ≤110                         X,Y ≥ 0 Solution : Step 1 : First of all,Convert the above equations in standard form.And,then Add S1 ,S2 in the equation (i)  & (ii) respectively in L.H.S Side with coefficient 1. While in the objective function add or subtract (It doesn't affect the equation) S1 & S2 with coefficient 0 in L.H.S. Note 1: If there was greater or equal to sign (≥) in the equation then you have to subtract S1 & S2 respectively in the L.H.S.  Note 2: We had taken here only two variables S1 & S2 because here we have only two equations other than objective function.              Z-6X-8Y + 0S1 + 0S2 = 0     (Objective function)              30X + 20Y + S1 = 300......................(i)                5X + 10Y+ S2  = 110......................(ii) Step 2: Create a table for Iteration 0 or Basic Iteration & Wri

Isomorphism of graph.

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Definition :  Two graphs G and H are isomorphic, denoted by G ≅ H, if there exists a bijection α:Vg➝Vh such that :                                   uv ∈ Eg  ⇔ α(u)α(v) ∈ Eh ; for all u,v ∈ G Hence G and H are isomorphic if the vertices of H are remaining of those of G.Two isomorphic graphs enjoy the same graph-theoretical properties and they are often identified.In particular,all isomorphic graphs have the same plane figures (excepting the identities of the vertices). This shows in the figures, where we tend to replace the vertices by small circles and talk of 'the graph' although there are, in fact, infinitely many such graphs. Example 1 : The following graphs are isomorphic . Indeed, the required isomorphism is given by                                                     v1 ↦1, v2 ↦ 3,v3 ↦ 4,v4 ↦2, v5 ↦5. Isomorphism Problem : Does there exist an efficient algorithm to check whether any two given graphs are isomorphic or not? The following table lis

Graph theory and tree.

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Introduction : Graph theory may be said to have its beginning in 1736 when Euler considered the (general case of the) Konigsberg bridge problem. Does there exist a walk crossing each of the seven bridges of Konigsberg exactly once? It took 200 years before the first book on graph theory was written. This was " Theorie der endlichen und unendlichen Graphen " by koing in 1936 .Since,then graph theory has developed into an extensive and popular branch of mathematics,which has been applied to many problems in mathematics, computer science and other scientific and not-s-scientific areas. There are no standard notations for graph theoretical objects. This is natural because the names one uses for the objects reflect the applications. Thus, for instance, if we consider a communication network (say, for email) as a graph, the computers taking part in this network, are called nodes rather than vertices or points.On the other hand, other names are used for molecular structures

Prove that a simple graph with n vertices and k components can have at most 1/2(n-k)(n-k-1) edges.

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Theorem : Proof of distance between two points.

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The reflection about the line Y=X is equivalent to the reflection along the X-axis,followed by counter clockwise rotation by θ degrees.Find out the value of θ.

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Show for what condition rotation and scaling are commutative.

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Vector in R space.

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T uple  =  A tuple is a finite ordered list of elements sequence of n-elements where n is a non-negative integer. Set of n-tuples of Real Numbers =>   R n Note:  O-tuple = Empty space  Eg. (0,0,0,0...........) A   particular n tuple in  R n  U =  (  a1,a2,a3,..................an  ) is called a point or vector. [   where, (a1,a2,a3,.............an) is called  Scalar  & U is called  Vector. ]                                            or, U = [a¡] ,where i is called as components/entries/co-ordinates/elements. Note:  Elements of  R =  S calar . 1.) Two vectors U & V are equal i.e U=V if they have same number of components and the corresponding components are equal .  e.g.  U =[a1,a2,a3]  V =[b1,b2,b3]  If U =V then    a1=b1 , a2=b2 , a3=b3 2.) U=[0,0,0..........0] This is Zero vector and is denoted as Bold 0 (𝐨). Example : Q ) Identify the space :  a) (2,-5) ->  R 2  b) (3,4,5) -> R^3 Q ) Find x,y,z such that (x-y , x+y , Z-

Why does 0! is equal to 1 ?

When we have to calculate the factorial of a non-negative integer n, then it is denoted by n!, is the product of all positive integer less than or equal to n . so, 𝑛 ! = 𝑛 ∗ ( 𝑛 − 1 ) ∗ ( 𝑛 − 2 ) ∗ ( 𝑛 − 3 ) ∗ . . .3 ∗ 2 ∗ 1 n ! = n ∗ ( n − 1 ) ∗ ( n − 2 ) ∗ ( n − 3 ) ∗ . . .3 ∗ 2 ∗ 1 However, the recursive definition of factorial is of more use in this proof. 𝑛 ! = { 1 𝑛 ∗ ( 𝑛 − 1 ) ! 𝑛 = 0 𝑛 > 0 n ! = { 1 n = 0 n ∗ ( n − 1 ) ! n > 0 The recursive definition of  Factorial  leads to one interesting way of expressing factorial numbers . 𝑛 ! = ( 𝑛 + 1 ) ! ( 𝑛 + 1 ) n ! = ( n + 1 ) ! ( n + 1 ) This is valid since, as we expand  ( 𝑛 + 1 ) ! ( n + 1 ) !  from the recursive definition, we can cancel  ( 𝑛 + 1 ) ( n + 1 )  term from both numerator and denominator to get  𝑛 ! n ! . Or we can even calculate factorial in the numerator and then evaluate the division For example, 5 ! = 6 ! 6 = 720 6 = 4 ! = 5 ! 5 = 120 5 3 ! = 4 ! 4 = 24 4 2 ! = 3 ! 3