logging in or signing up Identities in mathematics aSGuest33633 Download Post to : URL : Related Presentations : Let's Connect Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Copy embed code: Embed: Flash iPad Dynamic Copy Does not support media & animations Automatically changes to Flash or non-Flash embed WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 1316 Category: Education License: All Rights Reserved Like it (0) Dislike it (0) Added: December 07, 2009 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript Slide 1: Oh ! Do you want to have a short cut way for mathematics ………identities. World of identities Slide 2: Do you have the pre-requisite knowledge for expansion of (a+b)2 a2+2ab+b2 What is the volume of cube of side “a”? a3 What is an Identity ? : What is an Identity ? Consider the equality (a+1)(a+2)= a 2+3a+2 We shall evaluate both sides of this equality for some value of a, say a=10. For a=10, LHS= (a+1)(a+2)= (10+1)(10+2) = 11x12=132 RHS= a 2+3a+2=10 2+3x10+2 =100+30+2=132 We shall find for any value of a, LHS =RHS. Such an equality, true for every value of the variable in it, is called an identity. Thus (a+1)(a+2)= a 2+3a+2 is an identity What are identities ? : What are identities ? These are the mathematical equations which are satisfied by all values of variables. Algebraic proof of identity. : Algebraic proof of identity. Let us derive (a+b)3 We know that, (a+b)2=a2+2ab+b2. But, (a+b)3=(a+b)2(a+b). Therefore, (a+b)3=(a2+2ab+b2)(a+b). = a2(a+b)+2ab(a+b)+b2(a+b) = a3+a2b+2a2b+2ab2+ab2+b3 = a3+3a2b+3ab2+b3 = a3+3ab(a+b)+b3 Geometric proof of identity. : Geometric proof of identity. a a b b b Aa VOLUME V=a3 VOLUME V=b3 Total volume V1=a3+b3 a Slide 7: a a b b a a b V=a2b V=a2b V=a2b Therefore Total volume of the Cuboides V2=a2b+a2b+a2b=3a2b a a Slide 8: a b b a b b a b b V=ab2 V=ab2 V=ab2 Therefore V3=ab2+ab2+ab2=3ab2. Slide 9: A B a b b C b a E b a G F a3 b3 a2b a+b a+b a+b After arranging jumbled cubes & cuboides…………… WE GET…….. a Therefore V=(a+b)3 But from V1 V2 V3 we have V=V1+V2+V3 (a+b)3=a3+3a2b+3ab2+b3 =a3+3ab(a+b)+b3 Therefore proved. USE OF IDENTITIES. : USE OF IDENTITIES. Find the expansion of (x+5)3 Here a=x b=5 (a+b)3=a3+3ab(a+b)+b3 Therefore (x+5)3=x3+3(x)(5)(x+5)+(5)3 =x3+15x(x+5)+125 =x3+15x2+75x+125 USE OF IDENTITIES IN DAILY LIFE. : USE OF IDENTITIES IN DAILY LIFE. Find the cube of 102 1023=(100+2)3 a=100 b=2 (a+b)3= a3+3ab(a+b)+b3 (100+2)3= (100)3+3(100)(2)(100+2)+(2)3 1023 =1000000+600(102)+8 =1000000+61200+8 =1061208 Slide 12: PASCAL'S TRIANGLE a a + b (a + b)2 = a2 + 2ab+b2 (a + b)3 = a3 + 3a2 b+3ab 2 +b3 (a + b)4 = a4 + 4a3b + 6a2 b2 +4ab3+b 4 (a + b)6 = ------------------------- (a + b)5 =-------------------------- It helps us to write expansion form of the identity (a + b)n You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.