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MCA 202: Discrete Mathematics Instructor Neelima Gupta [email protected] Summation Series under construction Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14) MCA 2012 Summation Series n ak = k 1 lim n→∞ ak k 1 Linearity of Summation n n n k 1 k 1 k 1 (cak +bk) = c ak + bk Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14) , Naveen Kumar(51) MCA 2012 Arithmetic Series ak+1 – ak = d ; where d is a constant n 1 ak = a0 + a1 + a2 +. . . . +an-1 k 0 n 1 n (2a + (n-1)d) a = k 0 2 k 0 Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14) , Naveen Kumar(51) MCA 2012 n Prove : ak = (2a0 + (n-1)d) 2 n 1 k 0 Proof: By mathematical induction, For n=1, LHS = ak = a0 RHS = ½(2a0)=a0 Assume that it is true for n,we will prove it for n+1. For n+1, n n 1 ak = ak + an 0 k 0 k 0 k 0 = n [2a0+(n-1)d]+(a0+nd) 2 Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14) , Naveen Kumar(51) MCA 2012 Contd… n =(n+1)a0 + (n-1+2)d 2 =(n+1)a0 + n (n+1)d 2 = [2a0 + nd] (n 1) 2 Hence proved. Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14) , Naveen Kumar(51) MCA 2012 Excercise: n n(n 1) Prove that k = by induction 2 k 1 Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14) , Naveen Kumar(51) MCA 2012 Alternate Method Sn= 1 + 2 + 3 + . . . . + n Sn= n + (n-1) + (n-2) + . . . . + 1 Adding (i) and (ii), we have, 2Sn = (n+1)+(n+1)+(n+1)+ . .+(n+1) Sn = n(n 1) 2 Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14) , Naveen Kumar(51) MCA 2012 -(i) -(ii) Geometric Series a a k1 k =r r≠1 {finite terms} n a1(1 - r ) k 1 Sum = ak= 1 - r |r|<1 {infinite terms} a1 i 21+x n x = k 1+x+x 1 - r 3+……+xn Sum = n ak= i 0 xi = 1 |x|<1 1-x Differentiating (i) w.r.t x, we get, kxk= x |x|<1 2 i 0 i 0 (1Thanks: - x)Imam Hussain (12), Joseph Vanlalchanchinmawia (14) , Naveen Kumar(51) MCA 2012 …(i) Harmonic Series Hn = 1 + 1 + 1 + 1 + 1 + . . . . . + 1 2 3 4 5 n n = 1 k 1 k Exercise: Prove that, Hn= ln(n) + O(1) Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14) , Naveen Kumar(51) MCA 2012 Telescoping Series n (ak-ak-1) = an - a0 (ak-ak+1) = a0 - an Example: k 1 n 1 k 1 1 k 1 k(k 1) 1 1 = k k(k 1) n 1 1 – (k 1) Therefore, n 1 k 1 n 1 1 1 1 1 =k 1 k – (k 1) = 1 n k(k 1) Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14) , Naveen Kumar(51) MCA 2012 Bounding Summations Example Proven : 3k ≤ c.3n for all n≥no k 0 Proof: By using mathematical induction, we have, Base case: Putting n=0, LHS: 30 = 1 RHS: c.30 =c So, it holds for all c≥1 Thanks: Gunjan Sawhney Roll no. 11 (MCA 2012) Now, Suppose itn holds for all n, i.e. 3k ≤ c.3n k 0 Then, n n 1 k + 3n+1 3k = 3 k 0 k 0 ≤ c.3n + 3n+1 = 3n+1 (c/3 + 1) ≤ c.3n+1 whenever, c/3+1 ≤ c => 1 ≤ 2c/3 => c ≥ 3/2 n k ≤ c.3n is true for all c=3/2 and n≥0 Thus, 3 k 0 Thanks: Gunjan Sawhney Roll no. 11 (MCA 2012) Wrong Application Of Induction Example Prove: n k = O(n) k 0 Proof: By using mathematical induction, we have, Base case: Putting n=1, LHS: 1=O(1) RHS: O(n)=O(1) So , it holds for n=1 Thanks: Gunjan Sawhney Roll no. 11 (MCA 2012) Now, Suppose itn holds for all n, i.e. k = O(n) k 0 Then, n 1 k = O(n) + (n+1) = O(n) k 0 which is incorrect. Thus, By induction we couldn’t prove it appropriately. REASON: We ignored the constant whose value can’t change. Thanks: Gunjan Sawhney Roll no. 11 (MCA 2012) Bounding each term of Series ak 1 r.ak , for r 1 Few Examples : n ( k ) k 1 we have , k n, for all k, hence n n k 1 k 1 2 ( k ) ( n ) n n n k 1 k 1 log(n!) (logk ) (logn) n log n In general : n a k 1 k n n.amax , where amax max{ai } i 1 THANKS Sudhanshu Kumar Singh Roll No. 41 and Sukriti Sinha, Roll No. 42 (MCA 2012) Bounding a series with Another Series b ak bk , where k 1 r bk Example : 2 3 x x x ex 1 .......... 1! 2! 3! r 0 xr r! ( Here, amax 1. So, using previous method will give us 1 . Hence, for a tighter bound,we use another series to bound the given series.) THANKS Sudhanshu Kumar Singh Roll No. 41 and Sukriti Sinha, Roll No. 42 (MCA 2012) Using n! 2 n 1 to bound each term as follows : x3 x3 2, 3! 2 x4 x4 3 ,......... 4! 2 k k x x In general , k 1 k! 2 This gives , e 1 x x 3 (sin ce, | x | 1, amax 1) x 2 THANKS Sudhanshu Kumar Singh Roll No. 41 and Sukriti Sinha, Roll No. 42 (MCA 2012) Tip : For, e x r! x also have r 0 r while bounding each term we k k x x k! (k 1)! Even though this is a tighter bound than our xk xk k 1 , we don’t use it since it does bound k! 2 not lead us anywhere. Hence, we must always try to bound the given series with a series, whose sum is known. THANKS Sudhanshu Kumar Singh Roll No. 41 and Sukriti Sinha, Roll No. 42 (MCA 2012) Exercise: k k Bound the series using bounding the 3 n0 individual terms by another series. ak 1 r , for some r<1 ) ( HINT: show that ak Give some time to students to think!! THANKS Sudhanshu Kumar Singh Roll No. 41 and Sukriti Sinha, Roll No. 42 (MCA 2012) Splitting Summation By Kamal Kishore Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14) MCA 2012 Solution: ak 1 k 1 3 ak k 1 3 k k k 1 3k 1 k 1 3 k 1 2, for k1 k 2 3 i.e. 2k k 1 i.e. k 1 So, ak 1 ak 2/3r THANKS Sudhanshu Kumar Singh Roll No. 41 and Sukriti Sinha, Roll No. 42 (MCA 2012) Now, ak 1 ak r, where r < 1 So, ak 1 rak k 1 And hence, we can say that ak 1 r a1 Thus, we have, k a a r n1 k 1 n1 a1 1r 3a1 (since, r = 2/3) = 1 (since, a1 = 1/3) THANKS Sudhanshu Kumar Singh Roll No. 41 and Sukriti Sinha, Roll No. 42 (MCA 2012) QUES:- Show that 2 k O(1) k 2 k 0 by Splitting the Summation. SOLUTION: ak 1 k 1 2k k 1 . 2 ak 2 k 2 1 k 1 2 k Thanks Kanika Choudhry Roll no.-17 (MCA 2012) 2 For k=1, ak 1 = 2 > 1 ak For k=2, For k=3, For k=4, ak 1 9 = >1 ak 8 8 ak 1 = < 1 9 ak ak 1 ak 25 1 32 ak 1 8 k 3 ak 9 8 ak 9 Thanks Kanika Choudhry Roll no.-17 (MCA 2012) k 3 a3,k 4 So, by Splitting the Summation 2 k2 k2 k2 k k k k 0 2 k 3 2 k 0 2 2 8 8 1 1 a3 1 ... 9 9 2 8 3 k a3 r where r = 9 2 k 0 3 9 93 .9 8 2 8 Hence proved. O1 Thanks Kanika Choudhry Roll no.-17 (MCA 2012) Splitting the Summation contd.. • Bounding the sum of Harmonic Series (not done in class 2013) Thanks Kanika Choudhry Roll no.-17 (MCA 2012) Monotonically Increasing Functions • A function f is said to be monotonically increasing, if for all x and y such that x ≤ y one has f(x) < f(y), so f maintains the order. Thanks Swati Mittal Roll no. 45 (MCA 2012) Monotonically Non- Decreasing Functions • A function f is said to be monotonically nondecreasing, if for all x and y such that x ≤ y one has f(x) ≤ f(y). Thanks Swati Mittal Roll no. 45 (MCA 2012) Approximating Summation By Integrals Let f(k) be a monotonically increasing function, we can approximate it by integrals as follows: n m1 n f ( x)dx f (k ) k m n1 m f ( x)dx Lets prove this first! Thanks Swati Mittal Roll no. 45 (MCA 2012) Y n k m F(n) F(n-1) To Prove : F(m+1) F(m) The integral is the region under the curve and the total (blue) rectangle area represents the value of the summation. F(x) n-2 n-1 n n+1 m-1 m m+1 f (k ) n 1 m f ( x)dx m+2 Thanks Swati Mittal Roll no. 45 (MCA 2012) X Proof: We can see from the figure that, Area under the curve between “m” and “n +1 “ = sum of area of (blue)rectangles + something more(area of small triangles). Sum of areas of rectangles. Hence: n f (k ) k m n 1 m f ( x)dx Hence Proved! Thanks Swati Mittal Roll no. 45 (MCA 2012) Y n m1 n f ( x)dx f (k ) k m n1 m F(x) f ( x)dx F(n) F(n-1) This can be proved by shifting the rectangles in figure one left. F(m+1) F(m) Lets prove this now! m-1 m m+1 n-2 n-1 n n+1 m+2 Sum of area of red rectangles = area under the curve from m-1 to n + area of triangle above it area under the curve from m-1 to n Hence: n n f ( x)d x f (k ) m 1 k m Hence Proved! Thanks Swati Mittal Roll no. 45 (MCA 2012) X EXERCISE Let f(k) be a monotonically decreasing function prove: n 1 m f ( x)dx n f (k ) k m n m1 f ( x)dx Thanks Swati Mittal Roll no. 45 (MCA 2012) Monotonically Decreasing Functions • A function f is said to be monotonically decreasing, if for all x and y such that x ≤ y one has f(x) > f(y) , so f reverses the order. Thanks Swati Mittal Roll no. 45 (MCA 2012) Monotonically Non- Increasing Functions • A function f is said to be monotonically nonincreasing, if for all x and y such that x ≤ y one has f(x) ≥ f(y). Thanks Swati Mittal Roll no. 45 (MCA 2012) Approximating Summation Of Monotonically Decreasing Functions by Integrals • When a summation can be expressed as, f(x), where f(x) is a monotonically decreasing Function, we can approximate it by Integrals as follows: n xm n 1 x m n n f(x) f(x) x m f(x) x m 1 Thanks to Swatantra Kumar Verma Roll no. 43 MCA 2012 PROOF n 1 x m n n f(x) f(x) x m x m 1 Lets Prove this inequality first Thanks to Swatantra Kumar Verma Roll no. 43 MCA 2012 f(x) F(m+1) F(m) F(n) F(n-1) m-1 m m+1 n-1 n n+1 Monotonically Decreasing Function Thanks to Swatantra Kumar Verma Roll no. 43 MCA 2012 To Prove : n n x m x m 1 f(x) f(x) Proof : As we can see from the figure that: n n x m 1 x m f(x) f(x) + Sum of the area of Lower triangle from m to n Therefore :n f(x) x m n f(x) x m 1 Thanks to Swatantra Kumar Verma Roll no. 43 MCA 2012 n 1 x m n f(x) f(x) x m Lets Prove this inequality Now Thanks to Swatantra Kumar Verma Roll no. 43 MCA 2012 n f(x) x m 1 F(m+1) F(m) F(n) F(n-1) m-1 m m+1 n-1 n n+1 Monotonically Decreasing Function Thanks to Swatantra Kumar Verma Roll no. 43 MCA 2012 To prove: n 1 x m n f(x) f(x) x m Proof : As we can see from the figure that: n n 1 x m x m f(x) f(x)+ Sum of the area of upper triangle from m to n Therefore :n 1 n x m x m f(x) f(x) Thanks to Swatantra Kumar Verma Roll no. 43 MCA 2012