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Simpson's One _ Third Rule

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Simpson's One_Third Rule :              Let us suppose that interval (a,b) be divided into n equal sub intervals            such that  a = x₀,x₁,.........,xₙ₋₁,xₙ = b  We know from Newton's Forward Difference Interpolation formula that  y = y₀+ sΔy₀+s(s-1)/2!  Δ²y₀+s(s-1)(s-2)/3!  Δ³y₀                        + ................                     .........(1) Where  xₙ = x₀+nh  and x= x₀+sh .......(2) Integrating both sides of equation (1) , between x₀ and xₙ we get   xₙ          xₙ ∫ y dx = ∫ (y₀+sΔy₀+s(s-1)/2!  Δ²y₀ x₀           x₀                               + s(s-1)(s-2)/3!  Δ³y₀+........

Trapezoidal Rule

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Trapezoidal Rule :                   Let us suppose that the interval (a,b) be divided into n equal subintervals such that a = x₀,x₁.......xₙ₋₁,xₙ = b .      We know from Newton's Forward Difference Interpolation Formula that    y = y₀ + sΔy₀ + s(s-1)/2!  Δ²y₀                      +  s(s-1)(s-2)/3!  Δ³y₀.....                                                           ..............(1) Where xₙ = x₀ + nh and  x= x₀ + sh ......(2) Integrating both sides of equation(1) between x₀ and xₙ , we get         xₙ             xₙ         ∫ y dx = ∫ (y₀+sΔy₀+s(s-1)/2!  Δ²y₀       x₀...

Tests For Uniform Convergence

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Tests For Uniform Convergence : Theorem 1 (Mₙ _ Test) :                  Let <fₙ> be a sequence of function defined on a metric space X .      Let  lim  fₙ(x) = f(x)   ∀ x ∈ X and let            n-->∞          Mₙ = Sup {|fₙ(x) - f(x)| : x∈ X}   Then <fₙ> converges uniformly to f iff     Mₙ-->0 as n-->∞. Proof Of Necessary Part :          Let us suppose the sequence <fₙ> of functions converges uniformly to f on X . Then by definition , for a given ε > 0 ∃ a positive integer m (independent of x)     such that  n≥ m ⇒|fₙ(x) - f(x)| < ε ∀ x∈X    Also , Mₙ is the supremum of |fₙ(x) - f(x)|. Therefore |fₙ(x) - f(x)| < ε ∀ n≥m  ∀ x∈X         ⇒ Mₙ = Sup |fₙ(x) - f(x)| < ε...

What Does Uniform Convergence Mean With it's application

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Uniform Convergence :        Uniform Convergence is a part of real analysis which is discussed in detail below. Definition :             Let (X,d) be a metric space and f be a function from X to R . Also for each n ∈ ℕ let fₙ : X ---> R . Then , the sequence of functions <fₙ> converges pointwise to the function f , if for each x ∈ X , the sequence of real numbers < fₙ(x) > converges to the real number f(x) .                   Therefore <fₙ(x) > converges pointwise to f if  lim   fₙ(x) = f(x)    ∀ x ∈ X .                              n-->∞ For Example :            Let < fₙ > be the sequence defined by fₙ  : R --> R such that fₙ(x) = x / n  ∀ x∈R , n∈N  . Show that the sequence conver...

Examples Related To Riemann Integral

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Examples Related To Riemann Integral :  Example 1 :                                                                                                1                  Show that ∫ x⁴ dx = 1 / 5                                     0  Proof :                     Let us consider the partition P in the interval [0,1] as { 0,1/n, 2/n , 3/n .....n/n }.  Since  f(x) = x⁴ , so the supremum and infimum of the function in the interval is (i/n)⁴ and (i-1 / n) ⁴  .    Length of the interval = Δxᵢ = 1-0...

Mean Value Theorem Of Integrability

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Mean Value Theorem Of Integrability :   First Mean Value Theorem :          If a function f is continuous on [a,b] then ∃ a number ξ in [a,b] such that          b         ∫ f dx = f(ξ) (b - a)         a f is continuous , therefore f ∈ R on [ a,b ] . Proof :          Given that function f is continuous on  [ a, b] . Let m , M be the infimum and supremum of f in [ a,b ] . Then clearly       we have                                      b                 m( b - a ) ≤ ∫ f dx ≤ M( b - a )                                     a So , ∃ a number μ ∈ [ m, M ] such that    ...

Fundamental Theorems Of Integral Calculus

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Fundamental Theorems Of Integral Calculus : First Fundamental Theorem Of Integral Calculus : Theorem 1 :                 If a function f is bounded and integrable on [ a,b ] , then the function F defined as                    x                            F(x) = ∫ f(t) dt , a≤x≤b                                         0 is continuous on [ a,b ] and further more , if f is continuous at a point of [ a,b ] , then F is derivable at c and F'(c) = f(c) . Proof :           It is given that the function f is bounded . Then by definition ∃ a number k such that |f(x)|≤ k for x ∈ [ a,b ] ......(1)       Let x₁ , x₂ ∈[ a,b ] such that a≤x₁≤x₂≤b .    ...