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Fundamental Theorem of Algebra in Complex Analysis

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  In this page, we are going to analysis the important topic or theorem in Complex Analysis.    Fundamental Theorem of Algebra in Complex Analysis is a very important topic to understand that can enhance your Higher Mathematics knowledge. Fundamental Theorem of Algebra in Complex Analysis Statement: If f(z) =  a 0 z n + a 1 z n-1 + ... + a n , a 0 ≠ 0 be a polynomial in z, then f(z) = 0 has at least one root in z-plane. Proof: If possible, let f(z) = 0 has no root in z-plane. Therefore, f(z)  ≠ 0 for any value of z in z-plane. We define, because, f(z) is analytic and f(z)  ≠ 0 therefore, F(z) is analytic for all values of z. Also, |F(z)| --> 0 as |z| -->  ∞  therefore, F(z) is bounded therefore, By Liouville's Theorem, F(z) is constant. ⇒ f(z) is constant, which is contradiction [because, f(z) is not constant polynomial] therefore, f(z) = 0 has at least one root in z-plane. Explanation in YouTube:

The Union of two subgroups of a group G may not be a subgroup of G

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Theorem: The Union of two subgroups of a group G may not be a subgroup of G    Proof:  Let G = Z be the additive group of integers.    ∀ ∈  H 1 ∪H 2 Let H1 = {2n: n = Z} = {....-4, -2, 0, 2, 4, 6, ....} H2 = {3n: n = Z} = {....-6, -3, 0, 3, 6, ....} Since          0 =   2(0)  ∈  H 1 , so  H 1 is a non-empty subset of G Let a, b =  H 1 . Then a = 2n 1 and b = 2n 2 for some n 1 , n 2   ∈  Z.  Now, a-b = 2n 1 - 2n 2 = 2(n 1 -n 2 )  ∈  H 1   a-b  ∈  H 1 a, b  ∈  H1 and so H1 is a subgroup of G.  Similarly, H2 is a subgroup of G. Also,  H 1 ∪H 2  = {..... -6, -4, -3, -2, 0, 2, 3, 4, 6, ...}  Now, 2, 3  ∈   H 1 ∪H 2 but 2-3=-1  ∉   H 1 ∪H 2  Thus,  H 1 ∪H 2  is not a subgroup of G. Youtube Video on this Theorem

Solve the Non-Exact differential equation (x^2 + y^2 + 2x)dx + 2ydy = 0 | Study With Nitin

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Solution of (x² + y² 2xy)dx +2ydy=0 is a non-exact differential EQUATION  In this page, now you have to see that the Question (x^2 + y^2 + 2x)dx + 2ydy = 0 is the non exact differential equation and here is the solution of this question. In our website you have to see that we are updated higher Maths question and videos on it, that why you can not confuse for understand the question. Here, we find the integrating factor of this equation and also we solve the full question. This is the differential equation equation and we will solve it. SOLUTION: (x^2 +y^2 +2x)dx + 2ydy + 0 Comparing this with Mdx + Ndy = 0 Here, M = x^2 + y^2 + 2x and N= 2y dM/dy = 2y and dN/dx = 0 (dm/dy-dn/dx)/N = (2y-0)/2y = 1 or x^0 or f(x) e^integral f(x)dx = e^integral x^0dx = e^integral dx = e^x Now, Multiplying throughout in above equation e^x(x^2 + y^2 + 2x)dx + e^x2ydy = 0 Which is Exact differential equation integral e^x x^2 dx + integral e^xy^2dx + integral e^x 2xdx = c x^2 e^x - integral 2xe^x dx...