1020

• Run Multiple Choice Generator at h:\shares\physics
• Run Core Problems by giving the program qzz as your 'course'--enter this when asked for your 3-digit course number.  
• Do Precalculus Quiz 1018

x^(4/23) = 12

• Raise both sides to the 23/4 and we get x = 12^(23/4) = etc.

(7x)^(-5) = 2

• Take the -1/5 power of both sides to get 7x = 2^(-1/5) so that x = 2^(-1/5) / 7 = .12 approx..

(9x)^(-.5) = 19

• Taking the 1/(-.5) power of both sides we get 9x = 19^(1/(-.5)) = 19^-2 so x = 19^-2 / 9 = .00042.

18 x^(3/7) = 42

• Divide first by 18 to get x^(3/7) = 42/18.  If you took the 7/3 power before doing this you'ld end up with a factor 18^(7/3) which makes things messier.
• Then take the 7/3 power of both sides to get x = (42/18)^(7/3) = 7 or so.

5 ( 14 x / 9) ^( 8/11) = 13

• Divide both sides by 5 to get (14x/9)^(8/11) = 13/5
• Take the 11/8 power of both sides to get 14 x / 9 = (13/5)^(8/11).
• Multiply by 9/14 to get x = 9/14 * (13/5)^(8/11) = 2.4 approx..

If a(n+1) = a(n) + 4, a(0) = 2, then what are a(1), a(2), a(3) and a(4) and what is the pattern of these numbers?

First note that a(n) is function notation, not multiplication. 

Substituting 0 for n we get

a(0+1) = a(0) + 4.   Since we know that a(0) = 2 we find that

a(1) = 2 + 4 = 6.

Substituting 1 for n we get

a(1+1) = a(1) + 4.   Since we now know that a(1) = 6 we find that

a(2) = 6 + 4 = 10.

Substituting 2 for n we get

a(2+1) = a(2) + 4.   Since we now know that a(2) = 10 we find that

a(3) = 10 + 4 = 14.

Substituting 3 for n we get

a(3+1) = a(3) + 4.   Since we now know that a(3) = 14 we find that

a(4) = 14 + 4 = 18.

The numbers we get are a(0), a(1), a(2), ..., or

2, 6, 10, 14, 18, ... .

The pattern is clear.  Every number is 4 greater than the preceding.

Note that the differences of the sequence

2, 6, 10, 14, 18, ... are

4, 4, 4, 4, ...

This means that if we graph a(n) vs. n, each graph point is 4 units higher than the preceding so the graph will have a constant slope of 4, making the graph linear.

If the differences of a sequence are constant the graph is linear and the sequence can be modeled by a linear function of the form y = m x + b.

Do the same with a(n+1) = a(n) + 2 * n, a(0) = 2.

Substituting n = 0 we get a(0+1) = a(0) + 2 * 0, or a(1) = a(0) + 2 * 0.  Knowing that a(0) = 2 we get

a(1) = 2 + 2 * 0 = 2.

Substituting n = 1 we get a(1+1) = a(1) + 2 * 0, or a(2) = a(1) + 2 * 1.  Knowing that a(1) = 2 we get

a(2) = 2 + 2 * 1 = 4.

Substituting n = 2 we get a(2+1) = a(2) + 2 * 0, or a(3) = a(2) + 2 * 0.  Knowing that a(2) = 4 we get

a(3) = 4 + 2 * 2 = 8.

Substituting n = 3 we get a(3+1) = a(3) + 2 * 0, or a(4) = a(3) + 2 * 3.  Knowing that a(1) = 8 we get

a(4) = 8 + 2 * 3 = 14.

The sequence of our results is

2, 4, 8, 14, ...

The differences of this sequence are

2, 4, 6, ...

The differences of the difference sequence, called the second difference sequence, are

2, 2, ...

The second differences will be constant. 

On a graph of the original sequence we see from the first differences that the slopes will be 2, 4, 6, ... .  Slopes increase by the same amount with each new value of n.  This is what should be a familiar property of quadratic functions.

This illustrates the following:

If the sequence of second differences is constant and nonzero, then the sequence of first difference can be modeled by a linear function with nonzero slope, and the original sequence can be modeled by a quadratic function.

Note that a linear function is a first-degree polynomial (highest power of x is 1), and a quadratic function is a second-degree polynomial (highest power of x is 2).

It turns out that if the third differences of a sequence are constant and nonzero then the function would be modeled by a polynomial of degree 3, which might have the form a x^3 + b x^2 + c x + d.  This sequence of reasoning can be followed to higher degrees.

We also note that

• The form a(n+1) = a(n) + c, where c is any constant number, leads to a linear sequence.
• The form a(n+1) = a(n) + c * n leads to a quadratic sequence, which can be modeled by a quadratic function.
• The form a(n+1) = a(n) + c * n^2 would lead to a third-degree sequence, modeled by a third-degree polynomial.
• The pattern continues to higher powers of n.