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MCA 202: Discrete Mathematics Instructor Neelima Gupta ngupta@cs.du.ac.in

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Summation Series under construction Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14) MCA 2012

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Summation Series lim Linearity of Summation (ca k +b k ) = c a k + b k n→∞ akak a k = Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14), Naveen Kumar(51) MCA 2012

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Arithmetic Series a k+1 – a k = d ; where d is a constant a k = a 0 + a 1 + a 2 +.... +a n-1 a k = (2a 0 + (n-1)d) Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14), Naveen Kumar(51) MCA 2012

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Prove : a k = (2a 0 + (n-1)d) Proof: By mathematical induction, For n=1, LHS = a k = a 0 RHS = ½(2a 0 )=a 0 Assume that it is true for n,we will prove it for n+1. For n+1, a k = a k + a n = [2a 0 +(n-1)d]+(a 0 +nd) Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14), Naveen Kumar(51) MCA 2012

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Contd… =(n+1)a 0 + (n-1+2)d =(n+1)a 0 + (n+1)d = [2a 0 + nd] Hence proved. Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14), Naveen Kumar(51) MCA 2012

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Excercise: Prove that k = by induction Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14), Naveen Kumar(51) MCA 2012

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Alternate Method S n = 1 + 2 + 3 +.... + n -(i) S n = n + (n-1) + (n-2) +.... + 1 -(ii) Adding (i) and (ii), we have, 2S n = (n+1)+(n+1)+(n+1)+..+(n+1) S n = Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14), Naveen Kumar(51) MCA 2012

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Geometric Series = r Sum = a k = r≠1 {finite terms} Sum = a k = |r|<1 {infinite terms} x i = 1+x+x 2 +x 3 +……+x n x i = |x|<1…(i) Differentiating (i) w.r.t x, we get, kx k = |x|<1 Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14), Naveen Kumar(51) MCA 2012

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Harmonic Series H n = 1 + + + + +..... + = Exercise: Prove that, H n = ln(n) + O(1) Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14), Naveen Kumar(51) MCA 2012

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Telescoping Series (a k -a k-1 ) = a n - a 0 (a k -a k+1 ) = a 0 - a n Example: = – Therefore, = – = 1 - Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14), Naveen Kumar(51) MCA 2012

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Bounding Summations Example Prove : 3 k ≤ c.3 n for all n≥n o Proof: By using mathematical induction, we have, Base case: Putting n=0, LHS: 3 0 = 1 RHS: c.3 0 =c So, it holds for all c≥1 Thanks: Gunjan Sawhney Roll no. 11 (MCA 2012)

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Now, Suppose it holds for all n, i.e. 3 k ≤ c.3 n Then, 3 k = 3 k + 3 n+1 ≤ c.3 n + 3 n+1 = 3 n+1 (c/3 + 1) ≤ c.3 n+1 whenever, c/3+1 ≤ c => 1 ≤ 2c/3 => c ≥ 3/2 Thus, 3 k ≤ c.3 n is true for all c=3/2 and n≥0 Thanks: Gunjan Sawhney Roll no. 11 (MCA 2012)

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Wrong Application Of Induction Example Prove: k = O(n) 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)

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Now, Suppose it holds for all n, i.e. k = O(n) Then, k = O(n) + (n+1) = O(n) 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)

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Bounding each term of Series Few Examples : we have,for all k, hence In general : THANKS Sudhanshu Kumar Singh Roll No. 41 and Sukriti Sinha, Roll No. 42 (MCA 2012)

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Bounding a series with Another Series Example : ( Here,So, using previous method will give us 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)

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Usingto bound each term as follows : In general, This gives, THANKS Sudhanshu Kumar Singh Roll No. 41 and Sukriti Sinha, Roll No. 42 (MCA 2012)

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Tip : THANKS Sudhanshu Kumar Singh Roll No. 41 and Sukriti Sinha, Roll No. 42 (MCA 2012)

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Exercise: Bound the series using bounding the individual terms by another series. ( HINT: show that, for some r<1 ) Give some time to students to think!! THANKS Sudhanshu Kumar Singh Roll No. 41 and Sukriti Sinha, Roll No. 42 (MCA 2012)

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Splitting Summation By Kamal Kishore Thanks: Imam Hussain (12), Joseph Vanlalchanchinmawia (14) MCA 2012

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Solution: So, THANKS Sudhanshu Kumar Singh Roll No. 41 and Sukriti Sinha, Roll No. 42 (MCA 2012)

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Thanks Kanika Choudhry Roll no.-17 (MCA 2012) QUES:- Show that O(1) by Splitting the Summation. SOLUTION:

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Thanks Kanika Choudhry Roll no.-17 (MCA 2012) For k=1, = 2 > 1 For k=2, = > 1 For k=3, = < 1 For k=4,

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Thanks Kanika Choudhry Roll no.-17 (MCA 2012) So, by Splitting the Summation Hence proved. where r =

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Splitting the Summation contd.. Bounding the sum of Harmonic Series (not done in class 2013) Thanks Kanika Choudhry Roll no.-17 (MCA 2012)

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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)

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Monotonically Non- Decreasing Functions A function f is said to be monotonically non- decreasing, if for all x and y such that x ≤ y one has f(x) ≤ f(y). Thanks Swati Mittal Roll no. 45 (MCA 2012)

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Let f(k) be a monotonically increasing function, we can approximate it by integrals as follows: Approximating Summation By Integrals Lets prove this first! Thanks Swati Mittal Roll no. 45 (MCA 2012)

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F(n-1) F(m) F(m+1) m-1mm+1 m+2 n-1 n n+1 The integral is the region under the curve and the total (blue) rectangle area represents the value of the summation. To Prove : F(n) n-2 F(x)Y X Thanks Swati Mittal Roll no. 45 (MCA 2012)

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Hence Proved! 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: Thanks Swati Mittal Roll no. 45 (MCA 2012)

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F(n-1) F(m) F(m+1) F(n) m-1mm+1 m+2 n-1 n n+1 n-2 F(x)Y X Lets prove this now! This can be proved by shifting the rectangles in figure one left. 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: Hence Proved! Thanks Swati Mittal Roll no. 45 (MCA 2012)

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EXERCISE Let f(k) be a monotonically decreasing function prove: Thanks Swati Mittal Roll no. 45 (MCA 2012)

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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)

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Monotonically Non- Increasing Functions A function f is said to be monotonically non- increasing, if for all x and y such that x ≤ y one has f(x) ≥ f(y). Thanks Swati Mittal Roll no. 45 (MCA 2012)

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Thanks to Swatantra Kumar Verma Roll no. 43 MCA 2012 Approximating Summation Of Monotonically Decreasing Functions by Integrals When a summation can be expressed as,, where f(x) is a monotonically decreasing Function, we can approximate it by Integrals as follows:

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PROOF Thanks to Swatantra Kumar Verma Roll no. 43 MCA 2012 Lets Prove this inequality first

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F(n-1) F(m) F(m+1) F(n) Monotonically Decreasing Function Thanks to Swatantra Kumar Verma Roll no. 43 MCA 2012 m-1 m m+1 n-1 nn+1

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Thanks to Swatantra Kumar Verma Roll no. 43 MCA 2012 To Prove : Proof : As we can see from the figure that: + Sum of the area of Lower triangle from m to n Therefore :-

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Thanks to Swatantra Kumar Verma Roll no. 43 MCA 2012 Lets Prove this inequality Now

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F(n-1) F(m) F(m+1) F(n) Monotonically Decreasing Function Thanks to Swatantra Kumar Verma Roll no. 43 MCA 2012 m-1 m m+1 n-1 nn+1

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Thanks to Swatantra Kumar Verma Roll no. 43 MCA 2012 To prove: Proof : As we can see from the figure that: + Sum of the area of upper triangle from m to n Therefore :-

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