Order 1 Recurrence

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Question statement

Find the closed form of the relations:

$\begin{cases} a_0 &= \var{a0}\\ a_n &= \simplify{{recurrence}} \end{cases}$

Give any introductory information the student needs.

			Name	Type	Generated Value

			k	integer	-7
			r	integer	3

			Name	Type	Generated Value

			v	integer	-28
			closedForm	expression	-28*(-7)^n - 3

			Name	Type	Generated Value

a0	integer	-31
d	integer	-24
recurrence	expression	-7*sub(a,n - 1) + -24

			Name	Type	Generated Value

a1rec	expression	-7*sub(a,0) + -24
a1	integer	193
b0	integer	224
an1rec	expression	-7*sub(a,n) - 24
bnrec	expression	-7*sub(b,n - 1)
anrec	expression	-7*sub(a,n - 1) - 24
bndiff	expression	-7sub(a,n) - 24 - (-7sub(a,n - 1) - 24)

			Name	Type	Generated Value

bnclosed

expression

224*(-7)^n

			Name	Type	Generated Value

bnstep1	expression	-7*sub(a,n) - 24 - sub(a,n)
bnstep2lhs	expression	224*(-7)^n + 24
bnstep2rhs	expression	-8*sub(a,n)

			Name	Type	Generated Value

a2	integer	-1375
a22	number	-1375
test	expression	-28*(-7)^0 - 3
test2	expression	-28*(-7)^3 - 3

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(-10..10 except 0 except 1)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `integer`

-7

This variable doesn't seem to be used anywhere.

Advice

To solve Order 1 recurrences, we need to apply the following steps:

Express the recurrence relation for the first difference (which will be a geometric sequence).
Find the closed form for that geometric sequence.
Find the closed form of the original sequence using its first difference.

Finding the first difference.

We need two things here: the first element, and the recurrence formula for the first difference.

For the first element, we have $b_0 = a_1-a_0$ .

$a_0 = \var{a0}$
$a_1 = \var{a1rec} = \var{a1}$ .

So, $b_0 = \var{a1} - \var{a0} = \var{b0}$ .

For the recurrence formula, we have

$\begin{aligned} b_n &= a_{n+1} - a_{n}\\ &= \var{bndiff}\\ &= \var{bnrec} \end{aligned}$

Closed form of the first difference.

Now we know that the first difference is a geometric sequence, and we know its recurrence form. From Activity 2, we can find the closed form: $b_0\times k^n$ , where $k$ is the multiplier in the recurrence. So

$b_0 = \var{bnclosed}$

Closed form for the sequence $a_n$ :

Finally we can use the closed form of the first difference to find $a_n$ .

$\begin{aligned} b_n &= a_{n+1} - a_{n}\\ \var{bnclosed} &= \var{an1rec} - a_{n}\\ &= \var{bnstep1}\\ \var{bnstep2lhs} &= \var{bnstep2rhs}\\ \var{closedForm} &= a_n &\text{divide both sides by }\simplify{{k}-1} \end{aligned}$

And so we find the closed form for $a_n$ .

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Julien Ugon 2 years, 7 months ago

Julien Ugon 2 years, 7 months ago

Julien Ugon 2 years, 7 months ago

Julien Ugon 5 years, 1 month ago

Error in variable testing condition

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Warning

Generated value: `integer`

Parts

Mathematical expression

Advanced settings

Checking accuracy

Variable value generators

Pattern restriction

Variables

String restrictions

Finding the first difference.

Closed form of the first difference.

Closed form for the sequence $a_n$ :

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Order 1 Recurrence

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Julien Ugon 2 years, 7 months ago

Julien Ugon 2 years, 7 months ago

Julien Ugon 2 years, 7 months ago

Julien Ugon 5 years, 1 month ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Mathematical expression

Advanced settings

Checking accuracy

Variable value generators

Pattern restriction

Variables

String restrictions

Finding the first difference.

Closed form of the first difference.

Closed form for the sequence ana_n:

Generated value: `integer`

Closed form for the sequence $a_n$ :