Question

Help with sequence questions

Answer 1

so the first one is a recursive formula, it basically means take the previous term (an) and the next term becomes an^(an)

so starting with a1 = (1/2)

a2 is (1/2)^(1/2) = sqrt(2)/2

then a3 becomes [ sqrt(2)/2 ] ^ [sqrt(2)/2)]. etc.

Answer 2

similar logic for the other three problems

Answer 3

okay so

A3=s [ sqrt(2)/2 ] ^ [sqrt(2)/2)]. =1

a4=1^1=1

Answer 4

for a

Answer 5

hm, i don't think sqrt(2)/2 ^ [sqrt(2)/2] equals 1, I end up getting something weird on my calculator

Answer 6

do you get 0.78

Answer 7

i get about 0.866 or 2^(1/2 - sqrt(2)/2)

this is a really weird question

Answer 8

for a4?

Answer 9

for a3

Answer 10

then a4 becomes 2^(1/2 - sqrt(2)/2) ^ 2^(1/2 - sqrt(2)/2)

Answer 11

thankfully the other ones aren't so weird with the exponents >>

Answer 12

0.883?

Answer 13

idk, i'd probably just write out the exponents instead of moving to decimals

Answer 14

um 2.5

Answer 15

we can probably get back to this one, I'll try to find the best way to express the terms

Answer 16

okay

Answer 17

anyway for b)

a1 = 4 , as given

so a2 = 5/(6-2) = 5/4, then you'd plug in 5/4 back into the equation to get a3, and so on

Answer 18

wait if a1 equals four where is the four in the a2 equation?

Answer 19

ah good catch

Answer 20

so a2 = 5/(6-4) = 5/2 whoops

Answer 21

1.4?

Answer 22

for a 3

Answer 23

yes, but as a general rule in math try to give the most accurate possible answer

a3 = 5/(6-5/2) = 10/7 which is more precise than 1.4

Answer 24

35/32 for a4

Answer 25

yeah that's what i got too

Answer 26

Im getting 2.5x10^-1 for a2 for the next one

Answer 27

because a1 is 3/2

Answer 28

oh this is actually a tricky one

since a1 = 3/2

then the denominator is 2(3/2) - 3 which is 0

making a2 undefined

i'm actually not sure you can progress from there, making terms a3 and a4 also undefined :S

Answer 29

okay so for d it would be a1=1

Answer 30

a2=1/2

Answer 31

yes

Answer 32

from this point onward i'd just leave it in log form i gues

Answer 33

a3=1/4-ln(1/2)/2

Answer 34

for a4 i'm getting 1-ln(1/2)+2ln(1-2ln(1/2)/4)ln(1/2)

Answer 35

srry was helping somebody else with something

anyway a3 looks good, i'm not sure about a4

Answer 36

okay thats alright... did you figure out the best way to do 1a

Answer 37

i just left it like this

[1/4-ln(1/2)/2 ](1/2 - ln(1/4-ln(1/2)/2) )
without doing any distribution or expansion

Answer 38

for the first one we did ?

Answer 39

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
we can probably get back to this one, I'll try to find the best way to express the terms
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 40

this one

Answer 41

I didn't really get a response so i'd just go with

a1 = 1/2
a2 = (1/2)^(1/2) = sqrt(2)/2
a3 = sqrt(2)/2 ^ sqrt(2)/2
a4 = [sqrt(2)/2 ^ sqrt(2)/2] ^ [sqrt(2)/2 ^ sqrt(2)/2]

Answer 42

is there any way to contact your teacher/the person who gave out the assignment, about the formatting?

Answer 43

Yeah. Im doing the assignment now and just gonna have my professor check it over tomorrow.