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3 years ago

If I was given the explicit formula how to I find the recursive formula?

Mathematics

1 answer:

maxonik [38]3 years ago

8 0

9514 1404 393

Answer:

solve for f(n) in terms of f(n-1)

Step-by-step explanation:

In general, you solve for f(n) in terms of f(n-k) for k = 1, 2, 3, ....

Usually, such questions arise in the context of arithmetic or geometric sequences.

Arithmetic sequence

The explicit formula for an arithmetic sequence has the general form ...

a(n) = a(1) +d(n -1) . . . . . . . first term a(1); common difference d

The recursive formula for the same arithmetic sequence will look like ...

a(1) = a(1) . . . . . . . the first term is the first term

a(n) = a(n-1) +d . . . the successive terms are found by adding the common difference to the term before

Note: The explicit formula may be given as the linear equation a(n) = dn +b. Then the first term is a(1) = d+b.

Geometric sequence

The explicit formula of a geometric sequence has the general form ...

a(n) = a(1)·r^(n -1) . . . . . . first term a(1); common ratio r

The recursive formula for the same geometric sequence will be ...

a(1) = a(1) . . . . . . the first term is the first term

a(n) = a(n-1)·r . . . the successive terms are found by multiplying the term before by the common ratio

Note: The explicit formula may be given as the exponential equation a(n) = k·r^n. Then the first term is a(1) = kr.

Other sequences

Suppose you're given the quadratic sequence ...

a(n) = pn^2 +qn +r

Since the sequence is known to be quadratic (polynomial degree 2), we expect that we will only need the two previous terms a(n-1) and a(n-2). Effectively, we want to solve ...

a(n) = c·a(n-1) +d·a(n-2) +e

for the values c, d, and e. Doing that, we find ...

(c, d, e) = (2, -1, 2p)

So, the recursive relation is ...

a(1) = p +q +r

a(2) = 4p +2q +r

a(n) = 2a(n-1) -a(n-2) +2p

Additional comment

The basic idea is to write the expression for a(n) in terms of terms a(n-1), a(n-2) and so on. That will be easier for polynomial sequences than for sequences of arbitrary form.

There are some known translations between explicit and recursive formulas for different kinds of sequences, as we have shown above. If you recognize the sequence you have as being of a form with a known translation, then you would make use of that known translation. (For example, Fibonacci-like sequences are originally defined as recursive, but have explicit formulas of a somewhat complicated nature. If you recognize the form, translation from the explicit formula may be easy. If you must derive the recursive relation from the explicit formula, you may be in for a lot of work.)

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k = ½

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Step-by-step explanation:

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Answer:

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3 0

3 years ago

Simplify: [5×(25)^n+1 - 25 × (5)^2n]/[5×(5)^2n+3 - (25)^n+1]

blagie [28]

$\green{\large\underline{\sf{Solution-}}}$

Given expression is 

$\rm :\longmapsto\:\dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} }$

can be rewritten as

$\rm \: = \: \dfrac{5 \times { {(5}^{2} )}^{n + 1} - {5}^{2} \times {5}^{2n} }{5 \times {5}^{2n + 3} - {( {5}^{2} )}^{n + 1} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {( {x}^{m} )}^{n} \: = \: {x}^{mn}}}} \\$

And

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}} \\$

So, using this identity, we

$\rm \: = \: \dfrac{5 \times {5}^{2n + 2} - {5}^{2n + 2} }{{5}^{2n + 3 + 1} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 4} - {5}^{2n + 2} }$

can be further rewritten as

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 2 + 2} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2} - 1)}$

$\rm \: = \: \dfrac{4}{25 - 1}$

$\rm \: = \: \dfrac{4}{24}$

$\rm \: = \: \dfrac{1}{6}$

Hence, 

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} } = \frac{1}{6} }}$

3 0

3 years ago

Help pleaseeeeeee, I need help

Arlecino [84]

Answer:

15, 60, and 61

Step-by-step explanation:

First we have to find the 3rd side length from 2 side length chosen. So let’s pick 8 meters and 11 meters, to find the third side length we use the Pythagorean theorem: c^2 = a^2 + b^2

c = 11^2 + 8^2 = 121 + 64 = 185 = 13.6

13.6 isn’t one of the answers so we can’t pick 11 & 8. Let’s try 11 & 15:

c = 15^2 + 11^2 = 225 + 121 = 346 = 18.6

18.6 isn’t one of the answers either. Let’s try 15 & 60:

c = 60^2 + 15^2 = 3600 + 225 = 3825 = 61.85

61.85 might be it but let’s try one more just in case:

c = 60^2 + 61^2 = 3600 + 3721 = 7321 = 85.6

Nope! That means 15, 60, and 61 are the answer. I hope this is correct!

5 0

3 years ago