Algebra discussion

1. Arithmetic sequence

Arithmetic sequence is a sequence of real numbers. with a constant common difference of adjacent terms.

ex. 5, 8, 11, 14, 17, ... is arithmetic sequence with first term is 5 and common difference is 3.

Let aₙ is the n-term of the arithmetic sequence. / d is common difference. / Sₙ is a sum of n first term of an arithmetic sequence. Then

aₙ = a₁+(n-1)d = a₂+(n-2)d = ...

Sₙ = n(a₁+aₙ)/2 = n(2a₁+(n-1)d)/2

2. Geometic sequence

Geometic sequence is a sequence of real numbers. with a constant common ratio of adjacent terms.

ex. 4, 20, 100, 500, 2500, ... is geometic sequence with first term is 4 and common ratio is 5.

Let aₙ is the n-term of the geometic sequence. / d is ratio difference. / Sₙ is a sum of n first term of aₙ geometic sequence. Then

aₙ = a₁r^(n-1) = a₂r^(n-2) = ...

Sₙ = a₁(r^n -1)/(r-1)

3. Harmonic sequence

Harmonic sequence is a sequence of real numbers. with a constant common difference of reciprocal of adjacent terms.

ex. 1/2, 1/6, 1/10, 1/14, 1/18, ... is harmonic sequence.

Note: If a₁, a₂, a₃, ... is an arithmetic sequence, then 1/a₁, 1/a₂, 1/a₃, ... is a harmonic sequence.

Positive.

I'm not sure where the line A+C came from. However C² = A²+2(2Ax+x²) = A²+4Ax+2x²

A+C = 4A+2x is therefore wrong because c ≠ 3A+2x

SOLUTION

A², B², C² are arithmetic sequence. So B²-A² = C²-B² -> (B-A)(B+A) = (C-B)(C+B)

Multiply A+C both sides; (B-A)(B+A)(A+C) = (C-B)(C+B)(A+C) -> (B-A)/[(C+B)(A+C)] = (C-B)/[(B+A)(A+C)]

(C+B-(A+C))/[(C+B)(A+C)] = (C+A-(B+A))/[(B+A)(A+C)]

1/(A+C) - 1/(C+B) = 1/(B+A) - 1/(C+A)

1/(B+C) - 1/(A+C) = 1/(A+C) - 1/(A+B)

So 1/(A+B), 1/(A+C), 1/(B+C) are arithmetic sequence.

A+B, A+C, B+C are harmonic sequence.

Problem 8:

Find all the solutions of the equation.

3(15ˣ) - 75(3ˣ) - 5ˣ + 25 = 0

HINT: let a = 3ˣ and b = 5ˣ

Then you will get the equation 3ab - 75a - b + 25 = 0

Problem 9:

Let p(x) be a second degree polynomial such that p(0)=1, p(1)=10 and p(2)=100.

Find the value of p(3).

Oh this one is a softball compared to the others. You could find a simpler version of this in a Highschool class.

Anyways, the equation for a second degree polynomial can be expressed as a(x)^2+bx+c.

When x is set to 0, it is evident that c=1.

When x is set to 1, the equation boils down to a+b=9

When x is set to 2, the equation boils down to

4a+2b=98

This turns into an even easier problem, a system of equations.

a=9-b

4(9-b)+2b=98

36-2b=98

b=-31

a-31=9

a=40

p(3)=40(3)^2-31(3)+1=266

GREAT SOLUTION!

A slight miscalculation: when x is set to 2, the equation boils down to 4a+2b=99

From a=9-b ; 4(9-b)+2b=99 -> 36-2b=99 -> b=-31.5

and a= 9-(-31.5) = 40.5

So p(3)= 40.5(3)²-31.5(3)+1= 364.5-94.5+1 = 271

SOLUTION: problem 8

Find all the solutions of the equation.

3(15ˣ) - 75(3ˣ) - 5ˣ + 25 = 0

3(3ˣ)(5ˣ) - 75(3ˣ) - 5ˣ + 25 = 0

let a = 3ˣ and b = 5ˣ

Then we will get the equation 3ab -75a -b +25 =0

(3ab-75a)-(b-25)=0 -> 3a(b-25)-(b-25)=0 -> (3a-1)(b-25)=0

Thus 3a-1=0 or b-25=0 -> a=1/3 or b=25

Therefore 3ˣ=1/3 or 5ˣ=25

x = -1, 2

Problem 10:

Arrange a, b, c and d from least to greatest when

a = 25⁷⁰ , b = 32⁵⁰ , c = 125⁴⁵ , d = 128³⁵

Note: You may or may not use a calculator and/or the following estimates.

log(2) = 0.301 , log(5) = 0.699

Let the compound interest rate be equal to P times per year.

A principal of $10,000 will yield a total of $10,000(1+P)³ after 3 years.

10,000(1+P)³ = 13,310

(1+P)³ = 1.331

1+P = 1.1

P = 0.1

Therefore, the interest rate is equal to 0.1 times or 10% per year.

Problem 12:

Find the distance from the origin to the vertex of the parabola x² - 6x - 4y - 7 = 0

Can you find the vertex of the parabola?

What is fuction p ?

Tough challenge.

Do you have a solution? I give up.

I've tried searching it. This is a theory that has been widely studied. This is the research I found. The method proved beyond my ability to understand.

https://ir.library.oregonstate.edu/downloads/7m01bm43v

Solution:

a = (5²)⁷⁰ = 5¹⁴⁰ , c = (5³)⁴⁵ = 5¹³⁵ ; a > c

b = (2⁵)⁵⁰ = 2²⁵⁰ , d = (2⁷)³⁵ = 2²⁴⁵ ; b > d

c = 5¹³⁵ > 4¹²⁵ = 2²⁵⁰ = (2⁵)⁵⁰ = b ; c > b

Therefore, a > c > b > d

Problem 14:

Let f(x) = x²-5 , g(x) = 3x

Find the value of (f○g⁻¹)(15).

Problem 15:

From the graph, find the intersection point of line L1 and line L2.

Hint: find the equation of the two lines.