Graphs of Logarithmic Functions, y = kln(ax+b)

(14/05/2020, PJ): My Teaching Notes for Online Lessons on Graphs of Logarithmic Functions

 

General Formula to Find n-th Term of Any Diagonal/ Oblique Series of Binomial Coefficients in the Pascal/Yang Hui Triangle

Hello Math Enthusiasts,

27.01.2020 (2nd update, PJ Selangor, Malaysia).

This post is for math enthusiasts - not for students preparing to sit for IB math or IGCSE Math 0580, 0606, 0607 or CIE Math 9709 or Math 9231 unless the students would like to know beyond their exam syllabuses.

It is about the general formula /equation,   N = (n...n+(p-1))!/p!, to work-out:

1) the individual formula for the oblique/diagonal number series in Pascal/Yang Hui triangle
2) the value of any n-th term in the oblique/diagonal number series in the triangle.

The Pascal /Yang Hui triangle:

You see, currently, students of binomial theorem know that the horizontal series of numbers /binomial coefficients can be found by using binomial theorem or, to a limited extent, the Pascal /Yang Hui triangle.

What about the terms in the oblique/diagonal series of binomial coefficients?

As you can see, the binomial, say (x + y) or (a + b):

- to the power 1 yields an oblique number /binomial coefficient series called 'natural numbers': 1 + 2 + 3 + 4 + ... + n, where the n-th term = n;

- to the power 2 yields an oblique number /BC series called triangular number series of: 1 + 3 + 6 + 10 + 15 + ... + n, where the n-th term =n ( n+1)/2;

- to the power 3 yields a diagonal number /BC series called tetrahedral number series of: 1 + 4 + 10 + 20 + ... + n, where the n-th term = n(n+1)(n+2)/6;

- to the power 4 yields a diagonal number /BC series of: 1 + 5 + 15 + 35 + ... + n, where the n-th term = n(n+1)(n+2)(n+3)/24.

The n-th term formula for each oblique/diagonal bc series is currently analysed]and work-out individually - but I saw a pattern that led me to a general formula that can be used:

1) to work-out the individual formula for the n-th term of any oblique/diagonal number /BC series; and hence 

2) to find the value of any n-th term in any oblique/diagonal BC series in the Pascal / Yang Hui triangle. 

I have googled /searched far and wide and it seems that no one had found it before!. ..

The General Formula

N = (n...n+(p-1))!/p!

where,

1) n refers to the n-th term of the diagonal bc series generated by the binomial to the power p;

2) (n...n+(p-1))! refers to an 'up factorial" idea introduced n developed by me, where,

- an up factorial that starts with n and ends with n is:
(n...n)! = n

- an up factorial that starts with n and ends with (n+1) is:
(n...n+1)! = n(n+1)

- an up factorial that starts with n and ends with (n+2) is:
(n...n+2)! = n(n+1)(n+2); and, so on.

3) p! refers to the conventional down factorial which starts with p and ends with 1. Thus,

p! = p(p-1)(p-2)x...x3x2x1

5! = 5x4x...x2x1

1! = 1

0! = 1

In the Pascal/Yang Hui triangle (please see above):

When the power p=1, a natural number series (1+2+3+4+5+...+n) is generated. And, using the general formula
N = (n...n+(p-1))!/p! = (n...n+(1-1))!/1! = (n...n)!/1! = n       i.e. N = n

When p=2, a triangular number series (1+3+6+10+15+21+...+n) is generated. And, using the general formula
N = (n...n+(p-1))!/p! = (n...n+1)!/2! = n(n+1)/(2x1) = n(n+1)/2    i.e. N = n(n+1)/2

When p=3, a tetrahedral number series (1+4+10+20+35+...+n) is generated. And, using the general formula
N = (n...n+(p-1))!/p! = (n...n+2)!/3! =n(n+1)(n+2)/6    i.e. N = n(n+1)(n+2)/6 

When p=4, a number series (1+5+15+35+70+...+n) is generated. And, using the general formula
N = (n...n+(p-1))!/p! = (n...n+3)!/4! = n(n+1)(n+2)(n+3)/24   i.e. N = n(n+1)(n+2)(n+3)/24

And, so on...for p = 5, p = 6,...

Hence, the General Formula, N = (n...n+(p-1))!/p!, is capable of finding out:

1) the individual n-th term formula for any oblique/diagonal number or binomial coefficient series s in the Pascal /Yang Hui triangle - by just substituting the p value into the general formula; and,

2) the value of any n-th term in any oblique/diagonal series of binomial coefficients in the Pascal /Yang Hui triangle - by substituting the n-th value into the individual formula obtained by substituting p!

You'll note that to generate the General Formula, I have to introduce the idea and notation of an up factorial. This idea of 'up factorial' can be used to introduce the idea of partial 'down factorial': for example, 9x8x7x6 can be expressed as (9...6)! So, the binomial coefficient (BC) which is usually expressed as: BC = p!/(n!(p-n))! can now be easily expressed as a 'partial down factorial / another partial down factorial'.

Hahaha, critical comments, anyone?

Arithmetic Progression / Geometric Progression

MY NOTES ON: Arithmetic Progression (AP) and Geometric Progression (GP)

Progression, Sequence or Series – What’s the Difference?

·        Numbers or numerical terms in progression or sequence are given as:

T₁, T₂, T₃, …, T_n.

·        Numbers or numerical terms in series are given as:

o   T₁ + T₂ + T₃ + …+ T_n  ; or,

o   T_m + T_(m+1) +…+ T_n

·        Notations for sum of terms in series:

o   S; (or, any letter or symbol defined in the question); or

o   ∑ (sigma) 

·        Notations for sum of first n terms in series:

o   S_n; (or, any letter or symbol defined in the question); or

o  

·        Notations for sum of m^th to n^th  terms in series: = S_n – S_(m-1)

AP: Arithmetic Progression

o   What is an arithmetic progression (AP)?

An AP is a sequence of numerical terms with a common difference between any two consecutive terms. Thus, if the progression:

            T₁, T₂, T₃, …, T_n, is such that

T_n – T_(n-1) = T₃ - T₂ = T₂ - T₁ = d (common difference);

then, the progression is an AP.

o   Notations:

o   1^st term = a

o   Common difference = d (pl. see above)

o   n^th term = T_n

o   Deriving Basic Formula for n^th Term of AP, T_n:

1^st term 2^nd term            3^rd term            4^th term            n^th term

  T₁                     T₂                     T₃                     T₄                     T_n

  a                    a + d                a + 2d              a + 3d              a + (n – 1)d

Thus, the basic formula for n^th Term of AP is:               

T_n        = a + (n - 1)d ………….Basic Equation (BE)

Four other formulae/methods may be used to find T_n :

T_n = dn + (a - d)….................... (BE Expanded)

T_n = dn + c (where c = a - d)…. (BE in Linear Format)

T_n = T_n-1 + d………………….. (Preceding Term + d)

T_n = S_n – S_(n-1)………………..  (Difference of Sums)

o   Examples of APs:

o   1, 5, 9,…, T_n:               a = 1, d = 4, T_n = 1 + 4(n -1) = 4n – 3

o   2, -2, -6,…, T_n:            a = 2, d = -4, T_n = 2 – 4(n – 1) = 6 – 4n

o   2, 5, 8,…, T_n:               a = 2, d = 3, T_n = 2 + 3(n – 1) = 3n -1

o   Sum of Arithmetic Series (Finite Portion of AP):

o   Sum of first n terms of AP:

§   = S_n    = (n/2)(a + T_n); or,

= (n/2)[2a + (n – 1)d]

Deriving Sum of First n Terms Formula:

        Select 2 identical arithmetic series and proceed as shown before…

o   Sum of m^th term to n^th term of AP:

·      = S_n – S_(m-1)  (apply preceding formula)

o  Mean of Arithmetic Series, n:

      n = S_n/n = (1/n)(n/2)(a + T_n) = (a + T_n)/2

or,  n = S_n/n = (1/n)(n/2)[2a + (n – 1)d] = (1/2)[2a + (n – 1)d]

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GP: Geometric Progression or Sequence

A GP is a sequence of numerical terms with a common ratio (or, multiplier) between any two consecutive terms. Thus, if the progression:

            T₁, T₂, T₃, …, T_n, is such that

T_n / T_(n-1) = T₃ / T₂ = T₂ / T₁ = r (common ratio);

then, the progression is an GP.

o   Notations:

o   1^st term = a

o   Common ratio = r (pl. see above)

o   n^th term = T_n

o   Deriving Basic Formula for n^th Term of GP, T_n:

1^st term 2^nd term            3^rd term            4^th term            n^th term

  T₁                     T₂                     T₃                     T₄                     T_n

  a                      ar                     ar²                    ar³                  ar^{(n – 1)}

Thus, the basic formula for n^th Term of GP is:               

T_n = ar^{(n – 1)} ; or, T_n = (a/r)rⁿ

(Many students find it useful to use the 2^nd eqn to find n when T_n is given:

For example:

The geometric progression 6, -12, 24, …, 6144 consists of  n  terms. Find the value of n.

When you attempt this question and you will find logarithm useful and with your solid foundation in logarithm, you can also solve 2013 Jan P2 Q9 (e) easily – please see end of these notes:

o   Examples of GPs:

o   1, 3, 9, 27, …, T_n (= ar^{(n – 1)}  = (a/r)rⁿ, where r = 3, T_nà +∞, all Ts same sign as a);

o   1, -3, 9, -27, …, T_n (= ar^{(n – 1)} = (a/r)rⁿ, where r = -3, T alternates in sign);   

o   3, 3, 3, 3, …, T_n (= ar^{(n – 1)} = (a/r)rⁿ, where r = 1, same value for all Ts);

o   3/10, 3/100, 3/1000, …, T_n [= ar^{(n – 1)} = (a/r)rⁿ, where r = 1/10 or 0.1, T_n à 0 (i.e. T_n decays or decreases exponentially towards zero) - this type of GP where -1 < r < 1, or ׀r׀ < 1,  is known as a convergent GP where the sum of its series will yield a constant value  = S_∞  = a/(1 + r)];

o   -3, 3, -3, 3, …, T_n (= ar^{(n – 1)} = (a/r)rⁿ, where r = -1, constant modulus value for all Ts i.e. ׀T׀ = constant value (= 3 in this GP) but T alternates in sign – this is known as alternating sequence with constant modulus value)

From the above GPs, it can be seen that the behaviour of GP depends on the value of r (the common ratio). Always, r ≠ 0. If:

o   r is +ve: then all terms will be of the same sign as a (the first term);

(see GP: 1, 3, 9, 27, …, T_n (= ar^{(n – 1)}  = (a/r)rⁿ, where r = 3, T_nà +∞, all Ts same sign as a)

o   r is –ve: then terms will alternate in sign;

(see GP: 1, -3, 9, -27, …, T_n (= ar^{(n – 1)} = (a/r)rⁿ, where r = -3, T alternates in sign)

o   r > 1: then Ts à + ∞, if a is +ve; Ts à - ∞, if a is –ve;

(see GP: 1, 3, 9, 27, …, T_n (= ar^{(n – 1)}  = (a/r)rⁿ, where r = 3, T_nà +∞, all Ts same sign as a)

o   r = 1: then all Ts are of the same value – a constant value GP;

(see GP: 3, 3, 3, 3, …, T_n (= ar^{(n – 1)} = (a/r)rⁿ, where r = 1, same value for all Ts)

o   -1 < r < 1; or, ׀r׀ < 1 (and r ≠ 0): then T_n decays exponentially towards zero i.e. as n à +∞, T_n à 0: All such GPs are known as convergent GPs where the sum of each series yields a constant value  = S_∞  = a/(1 + r);

(see GP: 3/10, 3/100, 3/1000, …, T_n [= ar^{(n – 1)} = (a/r)rⁿ, where r = 1/10 or 0.1, T_n à 0)

o   r = -1: then, this is an alternating GP with a constant modulus value of a.

(see GP: -3, 3, -3, 3, …, T_n (= ar^{(n – 1)} = (a/r)rⁿ, where r = -1, constant modulus value for all Ts i.e. ׀T׀ = constant value (= 3 in this GP) but T alternates in sign.)

o   Sum of Finite Portion of a Geometric Series:

o   Sum of first n terms of GP:

§   = S_n    = a[(rⁿ - 1)/(r -1)]; or,

= a[(1 – rⁿ)/(1 – r)]

Steps in Deriving Sum of First n Terms for a GP:

1.        Select 2 identical geometric series S_n

2.        Multiply the 2^nd series with r

3.        Now, subtract as follows:

a.       rS_n – S_n; or

b.      S_n - rS_n

4.        The subtraction will lead you to these results:

a.       rS_n – S_n = arⁿ – a

S_n(r -1) = a(rⁿ – 1)                                       

S_n         = a[(rⁿ - 1)/(r -1)]; (ideal for r > 1); or

b.      S_n - rS_n = a - arⁿ

S_n(1 –r) = a(1 – rⁿ)

S_n         = a[(1 – rⁿ)/(1 – r)]; (ideal for 0 < r < 1)

o   Sum of m^th term to n^th term of GP:

·      = S_n – S_(m-1)  (depending on r, use the above formulae)

o   Sum to Infinity of a Convergent Series, where -1 < r < 1 or ׀r׀ < 1:

Use:       = S_n = a[(1 – rⁿ)/(1 – r)]

           

When n à ∞,   rⁿ à 0, (1 – rⁿ) à 1

S_n à a(1)/(1 – r)

 S_n à a(1)/(1 – r)

S_∞  = a/(1 + r) =

Therefore,

 = S_∞  = a/(1 + r);

o  Geometric Mean of 3 Consecutive GP Terms: a, b and c:

Geometric Mean = b

and, b² = ac; or b =

Q: Prove that b² = ac.

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Try These 2 Basic Questions on AP and GP:

Q1. A finite portion of an arithmetic series is given as: T_m + T_(m+1), …, T_n. There are 22 terms in this finite portion and the difference in value of T_n and T_m is 252.

a)      Find the common difference, d, of this number series.               (12)

b)      If T_n is 260, find the sum of this finite portion.                           (2948)

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Q2. In a geometric series, the 7^th term is 54 and, the 10^th term is 2. Find:

a)      the common ratio, r, of this series.                                            (1/3)

b)      i) the 6^th term                                                                           (162)

ii) the 1^st term                                                                          (39,366)

c)      the sum to infinity of this series.                                     (59,049)

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