Question

Find the finite number which (sqrt2/2)^n-1 approaches. Please include explanation.

Answer 1

is this the question?\[\lim_{n\rightarrow\infty}{(\frac{\sqrt{2}}{2})^n-1}\]

Answer 2

I don't know what that is...

Answer 3

what is bigger,
\(\sqrt{2}\) or \(2\)?

Answer 4

2 is bigger.

Answer 5

why?

Answer 6

because if you have a faction that is less than one and you raise it to higher and higher powers it gets smaller and smaller

Answer 7

yes

Answer 8

so since \(\frac{\sqrt{2}}{2}<1\) we know
\[\lim_{x\to \infty}(\frac{\sqrt{2}}{2})^n=0\]

Answer 9

guess i meant
\[\lim_{n\to \infty}(\frac{\sqrt{2}}{2})^n=0\]

Answer 10

wats the lim thing?

Answer 11

and why does that equal zero?

Answer 12

If I'm taking powers of a, that is, a^n there are basically three things that can happen. The simplest case is when a=1
if a = 1, then
1^1 = 1
1^2 = 1
1^3 = 1 and so on. Any power of a will just give me 1.

The second case is that a is bigger than 1. For example, let's see what happens when a is 2
2^1 =2
2^2 = 4
2^3 = 8
2^4 = 16
So, in general if a is bigger than 1, then powers of a will get bigger and bigger, and they'll grow faster and faster as we take higher powers of n.

The third case is that a is less than 1. For example, let's see what happens when a is (1/2)
(1/2)^1 = (1/2)
(1/2)^2 = (1/4)
(1/2)^3 = (1/8)
(1/2)^4 = (1/16)
So, in general, if a is less than 1, then powers of a will get smaller and smaller as we take higher powers. And they get really small really quickly, giving us tiny fractions that are almost 0.

Answer 13

you are asking what it approaches as n gets large (i assume) the succinct way to write that in math is \(\lim_{n\to \infty}\)

Answer 14

Here is for example a the sequence
\[
\left\{n,\left(\frac{2}{3}\right)^n\right\}
\]
for n =1 to 20
\[\left(
\begin{array}{cc}
1 & 0.666667 \\
2 & 0.444444 \\
3 & 0.296296 \\
4 & 0.197531 \\
5 & 0.131687 \\
6 & 0.0877915 \\
7 & 0.0585277 \\
8 & 0.0390184 \\
9 & 0.0260123 \\
10 & 0.0173415 \\
11 & 0.011561 \\
12 & 0.00770735 \\
13 & 0.00513823 \\
14 & 0.00342549 \\
15 & 0.00228366 \\
16 & 0.00152244 \\
17 & 0.00101496 \\
18 & 0.000676639 \\
19 & 0.000451093 \\
20 & 0.000300729 \\
\end{array}
\right)
What do you think that limit is hwn n goes to infinity?

\]

Answer 15

What do you think that limit is when n goes to infinity?

Answer 16

\[
\left(\frac{2}{3}\right)^{28
}=0.000011734
\]

Answer 17

oh... But the real questoin is does the sum of the perimeters approach a finite number? If so, what number? And the equation that I found for the perimeters is (sqrt 2/2)^n-1. So the first square that I am taking the perimeter of is 4 in perimeter.

Answer 18

This happens as it is explained above by @SmoothMath because
2/3 < 1

Answer 19

The first square has a length of 1 and there is another square inside of it with its corners on the midpoint of each side of the first square. and it goes on and on.

Answer 20

Your formula is not accurate since the limit of your formula is negtive and the perimeter is positive.

Answer 21

How do you know that? How do you find the limit?

Answer 22

the actual equation I should've gotten is 4x(sqrt 2/2)^n-1), isn't it?

Answer 23

Let us call the perimeters
\[p_0 , p_1, .... p_n...

\]
It is easy to show that the second perimeter
\[
p_1= \frac {p_0}{\sqrt 2}
\]
and in general
\[
p_n= \frac {p_{n-1}}{\sqrt 2}
\]
Do we agree on that?

Answer 24

what??? I have no idea what you typed... uh

Answer 25

So
\[
p_n = \frac {p_0}{(\sqrt 2)^n}= \frac {1}{(\sqrt 2)^n} 4= 4 r^n\\
\text { where } \\
r= \frac 1 {\sqrt 2}
\]
The sum is
\[
4\sum_{n=0}^\infty r^n
\]
This is a geometric series. Do you know how to find its sum?
Notice r<1

Answer 26

Every new perimeter is obtained from the previous one by
dividing the previous perimeter by \[ \sqrt 2\]

Answer 27

how do you translate this? do you put it into the equation thing with the sigma right under the comment box?

Answer 28

Have you heard about geometric series?

Answer 29

Read this
http://en.wikipedia.org/wiki/Geometric_series

Answer 30

yes

Answer 31

what is
a + ar + ar^2 + .... + a r^n ?

Answer 32

a(r)^n-1

Answer 33

I just don't know how to find the limit

Answer 34

It is
\[
\frac{a
\left(1-r^{n}\right)}{1-r
}
\]

Answer 35

I dont know what\[ \frac{a \left(1-r^{n}\right)}{1-r } \] is because I can't see what it actually klooks like and I can't make sense of it... I'm sorry

Answer 36

If r < 1, then the limit when n goes to infinity of
\[

r^n
\]
is zero

Answer 37

Read the link that I gave you before to understand what is going on?

Answer 38

But it says here that the answer is 8 +4 sqrt2

Answer 39

Yes it is true the answer is
\[
8+4 \sqrt{2}
\]

Answer 40

how...

Answer 41

?

Answer 42

Because the limit when n goes to infinity of
\[
\frac{a
\left(1-r^{n}\right)}{1-r
}
\]
is
\[
\frac{a
\left(1-0\right)}{1-r
}= \frac 4{ 1 - \frac 1 {\sqrt 2}}=8+4 \sqrt{2}


\]

Answer 43

After simplifying

Answer 44

My friend just told me to use conjugates. Could you explain conjugates to me?

Answer 45

wow this turned out to be a totally different question didn't it?

Answer 46

\[
\frac{4}{1-\frac{1}{\sqrt{2}}}
=\frac{4}{1-\frac{\sqrt{2}}
{2}}=\frac{4}{\frac{1}{2}
\left(2-\sqrt{2}\right)}=\frac{8}{2-\sqrt{2}}
\]

Answer 47

Multiply up and down by the conjugate
\[ 2 + \sqrt 2
\]
You obtain
\[
\frac{8}{2-\sqrt{2}} \frac { 2+ \sqrt 2}{2+ \sqrt 2}=
\frac {8( 2 + \sqrt 2)}
{4-2}= 4 (2 +\sqrt 2)= 8 + 4 \sqrt 2
\]

Answer 48

Thank you for all of your time,. I think I get how to do it now!!!