Question

Find the interval of convergence of the power series and check the endpoints

Answer 1

\[\sum_{1}^{\infty} \frac{(x-4)^{n-1}}{4^{n-1}}\]

Answer 2

If x is greater than or equal to 4 does this series converge?

Answer 3

Hmm Not too sure on the rules.

Answer 4

Okay, well it is a little funky when x=4. Let's start at x=8 then. What do you think happens when x=8 or more?

Answer 5

also couldn't it be rewritten as sum (0 to infinity) (x-4)^n/4^n?

Answer 6

Yes, it could

Answer 7

if x = 8 then it would be >=2

Answer 8

No, if x=8 we have 1*infinity.

Answer 9

:O?

Answer 10

4/4=1

Answer 11

4/4 sorry.

Answer 12

x=8; x-4 = 4; sum((x-4)/4)^n-1

Answer 13

:)

Answer 14

So that is good. When x = 0 we have a similar but stranger situation. What is that?

Answer 15

-4/4 = -1?

Answer 16

so how do you sum (-1)^n?

Answer 17

0?

Answer 18

-1+1-1+1.......

Answer 19

So what is the sum of the series after an infinite number of terms?

Answer 20

0?

Answer 21

0 if you happen to take an even number of terms; -1 if you take an odd number of terms.

Answer 22

but that's if x = 0, but we also used 0 = 8?

Answer 23

which would be 1.....

Answer 24

x=0 and x=8 are just the obvious endpoints. We say that -1+1-1+1-1+1... diverges.

Answer 25

1+1+1+1+1... definitely diverges.

Answer 26

:)

Answer 27

so how did we know those were the endpoints? OR we just used any #'s that would work?

Answer 28

We looked for where the ratio was critical; such as r=1. Also the series diverges if x is less than 0 or greater than 8.

Answer 29

so the interval of convergence = 0 <= x <=8

Answer 30

\[\sum_0^\infty r^n \text{converges for |r|<1} \]

Answer 31

:O?

Answer 32

So 0<x<8; but we should take a special look at x=4. What is?:
\[\sum_{n=1}^\infty\frac{0^{n-1}}{4^{n-1}}\]

Answer 33

0

Answer 34

What about when n=1?

Answer 35

-3/4

Answer 36

oic :).

Answer 37

A very tricky definition of the power function is that a^0 = 1 for all a so
\[\frac{0^0}{4^0} = \frac{1}{1}=1\]

Answer 38

so why couldn't we use 0 as an endpoint?

Answer 39

We can use x=0 as an endpoint. Just wanted to make the point that the series is funny at x=4.

Answer 40

so if we had (x/6)^n we could say x = 0 and x = 6?

Answer 41

When x is between 4 and 8 you have a positive converging series. When x is between 0 and 4 you have an alternating converging series.

In your x/6 example, we are interested in x = +/- 6

Answer 42

okies.

Answer 43

thhen it becomes -1 and 1 which both converge again...

Answer 44

I see so we are trying to just get 1 and -1?

Answer 45

For problems of convergence of this type. There are other convergence thresholds. But this is the deal with power series.