Question

A function v is defined by
@ganeshie8

Answer 1

\[v:x \rightarrow \frac{ x-1 }{ x+1 },x \neq-1\]

Answer 2

a)Find in similar form

Answer 3

\[i)~v^{18}\]

Answer 4

isn't it just \(\left(\dfrac{x-1}{x+1}\right)^{18}\) ?


are you given answer ?

Answer 5

does similar mean in the form
\[\frac{ax-b}{ax+c}\]

Answer 6

That means that @ganeshie8 is right\[(\frac{ x-1 }{ x+1 })^{18}\]
Thnx @ganeshie8 @myininaya

Answer 7

oh ok

Answer 8

I don't know need a medal
I didn't do anything
I just asked a question

Answer 9

I wasn't saying what @ganeshie8 had was right
I didn't know if there was some magical meaning behind similar or not :p

Answer 10

but yeah normally v=(x-1)/(x+1) implies v^n=[(x-1)/(x+1)]^n
I don't know what else it could mean

Answer 11

I think by similar form, they want you expand numerator and denominator

Answer 12

eww

Answer 13

you could divide the inside but you will end up with (18+1) terms

Answer 14

\[v^{18} : \left(\dfrac{x-1}{x+1}\right)^{18} = \dfrac{(1-x)^{18}}{(1+x)^{18}} = \text{you dont want to expand :/}\]

Answer 15

this question seems weird
i don't know find v^(18) given v
similar form?
it all sounds odd

Answer 16

like I feel like I'm crazy for thinking that

Answer 17

I seem some of his previous questions and they look harder than this

Answer 18

seen*

Answer 19

https://books.google.co.in/books?id=ZZoxLiJBwOUC&pg=PA51&lpg=PA51&dq=%22find+in+similar+form%22&source=bl&ots=Tfb-XbwNxb&sig=_e8Cw-yJO151cqIcjGztE32JshA&hl=en&sa=X&ei=5er3VPfJMJSpuQT1v4LoBQ&ved=0CB0Q6AEwAA#v=onepage&q=%22find%20in%20similar%20form%22&f=false

Answer 20

15th question

Answer 21

wait not to weird but do you think for example in this case v^2 means v composed with v

Answer 22

yeah it seems the exponent is refering to the number of times we need to compose..

Answer 23

that would mean v^2=-1/x

Answer 24

Wow! thats it i guess @MARC_

Answer 25

like @marc did have the answer as -1/x and he deleted it
like I'm not saying that is the answer but that definitely looks more similar to the given form we have above

Answer 26

v^18 = v(v(v(v(... 18 times )))) ?

Answer 27

lol
I think so
but I think we can try to find a pattern maybe for smaller numbers and make a guess for v^(18) based on that pattern

Answer 28

\[v^1=\frac{x-1}{x+1} \\ v^2=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=\frac{x-1-(x+1)}{x-1+(x+1)}=\frac{-2}{2x}=\frac{-1}{x} \\ v^3=v(v(v(x))=v(\frac{-1}{x})=?\]
@marc if you are still here do you want to try to do v(-1/x)?

Answer 29

I think it actually starts over again at a certain number

Answer 30

sorry for the late reply,\[=\frac{ -\frac{ 1 }{ x }-1 }{ -\frac{ 1 }{ x }+1 }\]

Answer 31

\[=\frac{ \frac{ -1-x }{ x } }{ \frac{ x-1 }{ x } }\]

Answer 32

\[=\frac{ -1-x }{ x }\div \frac{ x-1 }{ x }\]

Answer 33

\[=\frac{ -1-x }{ x-1 }\]

Answer 34

@myininaya

Answer 35

yes that is v^3

Answer 36

now to find v^4 plug v^3 result into the v function given

Answer 37

\[v^4=v(v^3)=v(\frac{-1-x}{x-1})=?\]

Answer 38

believe it or not this is going to look pretty very soon

Answer 39

\[v^4=x\] @myininaya

Answer 40

then you see you start over with v^5=v^1
and so therefore v^6=v^2
and v^7=v^3
and v^8=v^4
and v^9=v^5=v^1
and so on...
do you see the pattern?

Answer 41

i got it @myininaya
Assume that every \[v^4\]is \[x\]

Answer 42

\[v^4=x\]\[v^8=x\]\[v^{12}=x\]\[v^{16}=x\]\[v^{17}=\frac{ x-1 }{ x+1 }\]\[v^{18}=-\frac{ 1 }{ x }\]

Answer 43

Thnx @myininaya