MCA 202: Discrete Mathematics Instructor Neelima Gupta

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

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

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

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

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

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

Alternate Method S n = n -(i) S n = n + (n-1) + (n-2) (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

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

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

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

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)

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)

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)

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)

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)

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)

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)

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

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)

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

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

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

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

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

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 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)

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)

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)

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)

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)

EXERCISE Let f(k) be a monotonically decreasing function prove: 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 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)

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:

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

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

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 :-

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

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

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 :-