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3 years ago

A sequence is defined by the recursive formula f(n+1)=f(n)-2. If f(1)=18,what is f(5)?

Mathematics

2 answers:

vladimir2022 [97]3 years ago

7 0

The answer is for this problem is 10

Hatshy [7]3 years ago

4 0

To find f(5), you need to know f(4), to know f(4), you need to find f(3), to know f(3), you need to know f(2)
<span>Use the recursive formula f(n+1)=f(n)-2,
</span>f(2)=f(1)-2, and f(1)=18 is given, so f(2)=18-2=16
f(3)=f(2)-2=16-2=14
f(4)=12
f(5)=10

another way to do it is using the formula: the nth term=the first term +d(n-1)
d represents the difference, which is -2 in this case
so f(5)=f(1)-2(5-1)
f(5)=18-8=10

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2 years ago

Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6

san4es73 [151]

Answer:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

Step-by-step explanation:

Given

$f(x)= \frac{9}{3x+ 2}$

$c = 6$

The geometric series centered at c is of the form:

$\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.$

Where:

$a \to$ first term

$r - c \to$ common ratio

We have to write

$f(x)= \frac{9}{3x+ 2}$

In the following form:

$\frac{a}{1 - r}$

So, we have:

$f(x)= \frac{9}{3x+ 2}$

Rewrite as:

$f(x) = \frac{9}{3x - 18 + 18 +2}$

$f(x) = \frac{9}{3x - 18 + 20}$

Factorize

$f(x) = \frac{1}{\frac{1}{9}(3x + 2)}$

Open bracket

$f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}$

Rewrite as:

$f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}$

Collect like terms

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}$

Take LCM

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}$

$f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}$

So, we have:

$f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}$

By comparison with: $\frac{a}{1 - r}$

$a = 1$

$r = -\frac{1}{3}x + \frac{7}{9}$

$r = -\frac{1}{3}(x - \frac{7}{3})$

At c = 6, we have:

$r = -\frac{1}{3}(x - \frac{7}{3}+6-6)$

Take LCM

$r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)$

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}ar^n$

Substitute 1 for a

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}1*r^n$

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}r^n$

Substitute the expression for r

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n$

Expand

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]$

Further expand:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................$

The power series converges when:

$\frac{1}{3}|x - \frac{7}{3}| < 1$

Multiply both sides by 3

$|x - \frac{7}{3}|$

Expand the absolute inequality

$-3 < x - \frac{7}{3}$

Solve for x

$\frac{7}{3} -3 < x$

Take LCM

$\frac{7-9}{3} < x$

$-\frac{2}{3} < x$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

6 0

2 years ago