Question

Sequences/Series help. *question attached below* will give medal.

Answer 1

Can someone walk me through this question please? I had miss this class and now I'm trying to catch up

Answer 2

familiar with proofs by induction ?

Answer 3

Yes

Answer 4

1) prove the given statement is true for n=1
2) assume the given statement is true for n=k
3) prove the given statement is true for n=k+1

Answer 5

start with step 1

Answer 6

When n=1, \[x _{1} ^ {2} + \frac{ 1 }{ 4 }\] ?

Answer 7

yes, is it < 1/2 ?

Answer 8

yes

Answer 9

how ?

Answer 10

1) prove the given statement is true for n=1
we're given \(\large x_1 \lt \frac{1}{2}\)
multiplying both sides you get : \(\large x_1^2 \lt \frac{1}{4}\)
adding \(\large \frac{1}{4}\) both sides you get :
\[\large\begin{align} x_1^2 + \frac{1}{4} &\lt \frac{1}{4} + \frac{1}{4}\\~\\&\lt \frac{1}{2}\end{align}\]

so the given statement is true for n=1

Answer 11

you need to show all the work in a proof ok ?

Answer 12

next assume the given statemetn is true for n=k and show that it will be true for n=k+1

Answer 13

Why are we adding 1/4 to both sides?

Answer 14

because we want to prove that \(\large x_1^2 + \frac{1}{4} \lt \frac{1}{2}\) for step 1

Answer 15

note that when n = 1 :
\[\large x_{n+1} = x_n^2 + \frac{1}{4}\]

becomes :
\[\large x_{1+1} = x_1^2 + \frac{1}{4}\]

Answer 16

let me knw once you're happy wid step1

Answer 17

Hm, I think I get step 1

Answer 18

good, next assume \(\large x_{k+1}\lt \frac{1}{2}\) and prove below :
\[\large x_{k+2}\lt \frac{1}{2}\]

Answer 19

\[\large \begin{align}x_{k+2} &= x_{k+1}^2 + \frac{1}{4}\\~\\&\lt \left(\frac{1}{2}\right)^2 + \frac{1}{4} = \frac{1}{2}\end{align}\]

Answer 20

we're done with part i

Answer 21

Okay, I think I get this

Answer 22

what about part ii

Answer 23

Hmm, we apply \[x _{n+1}= x _{n ^{2}}+\frac{ 1 }{ 4 }\] so now we have

Answer 24

\[x _{n ^{2}}+\frac{ 1 }{ 4 } - x_{n} ?\]

Answer 25

it asks us to consdier \(\large x_{n+1} -x_n\)

Answer 26

\[\large\begin{align} x_{n+1} -x_n &= x_n^2 + \frac{1}{4} - x_n\\~\\&=\left(x_n-\frac{1}{2}\right)^2\end{align}\]

Answer 27

which is always > 0 right ?

Answer 28

\[\large\begin{align} x_{n+1} -x_n &= x_n^2 + \frac{1}{4} - x_n\\~\\&=\left(x_n-\frac{1}{2}\right)^2\\~\\&\gt 0 \end{align}\]

Answer 29

\(\large x_{n+1} - x_n \gt 0 \)

that means \(\large x_n \lt x_{n+1}\)

Answer 30

see if that makes more or less sense

Answer 31

Question, how come that we're squaring?

Answer 32

Oh we're completing the square?

Answer 33

exactly !

Answer 34

and a square of something real can never be negative