Order 1 Recurrence

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Julien Ugon 2 years, 8 months ago

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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	9
			r	integer	2

			Name	Type	Generated Value

			v	integer	27
			closedForm	expression	27*9^n - 2

			Name	Type	Generated Value

a0	integer	25
d	integer	16
recurrence	expression	9*sub(a,n - 1) + 16

			Name	Type	Generated Value

a1rec	expression	9*sub(a,0) + 16
a1	integer	241
b0	integer	216
an1rec	expression	9*sub(a,n) + 16
bnrec	expression	9*sub(b,n - 1)
anrec	expression	9*sub(a,n - 1) + 16
bndiff	expression	9sub(a,n) + 16 - (9sub(a,n - 1) + 16)

			Name	Type	Generated Value

bnclosed

expression

216*9^n

			Name	Type	Generated Value

bnstep1	expression	9*sub(a,n) + 16 - sub(a,n)
bnstep2lhs	expression	216*9^n - 16
bnstep2rhs	expression	8*sub(a,n)

			Name	Type	Generated Value

a2	integer	2185
a22	number	2185
test	expression	27*9^0 - 2
test2	expression	27*9^3 - 2

Automatically regenerate variables?

Name

Data type

Value

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

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Description

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

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Generated value: `integer`

→ Used by:

a0
a1
a1rec
a2
an1rec
anrec
bnclosed
bnrec
bnstep2rhs
closedForm
d
recurrence

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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Taxonomy: mathcentre

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Taxonomy: Context

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Julien Ugon 2 years, 8 months ago

Julien Ugon 2 years, 8 months ago

Julien Ugon 2 years, 8 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 $a_n$ :

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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, 8 months ago

Julien Ugon 2 years, 8 months ago

Julien Ugon 2 years, 8 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 $a_n$ :