Question

g(1) = 4
g(n+1)=4[g(n-1)]

Find a closed form formula.

Answer 1

have you tried to plug in other values of n and see if there is a pattern
this is one way to go about the problem

Answer 2

Yes, but the thing is when you plug it in, you have to know different y-values that are not given, and I do not know how to get those y-values.

Answer 3

so we know g(1)=4
n-1=1 when n=2
so choose n=2
we have
g(2+1)=4g(2-1)
g(3)=4g(1)=4*4=16

so right now we have
g(1)=4
g(3)=16
...we can find g(5) next by inserting n=4
...we can find g(7) by then inserting n=6

Answer 4

now we got to figure out how to evaluate g(even integer)

Answer 5

oh dang that makes sense.

Answer 6

I'm having trouble with g(even integer)

Answer 7

Same.

Answer 8

@zepdrix what do you think?
how to find g(2)?

Answer 9

can i ask what other techniques you have learned for finding closed forms?

Answer 10

if n-1 is odd then n+1 is odd and we are given g(1) which is g(odd number)
if n-1 is even then n+1 is even but we are not given a condition for g(even number)=something so I feel like we only have enough information for a closed form for g(odd integer)

Answer 11

Our teacher really did not say anything about finding them other than using the trend of the output

Answer 12

I can get to:
\[g(2n+1)=4^{n+1} \text{ this is a closed from for the odd integers }\]
hmm... we can try to play with this maybe...

Answer 13

we can assume this will hold for the even integers (this is may not be true though)
x=2n+1
then n=(x-1)/2
So we have
\[g(x)=4^{\frac{x-1}{2}+1}=4^{\frac{x-1+2}{2}}=4^{\frac{x+1}{2}} \\ g(1)=4^{\frac{1+1}{2}}=4^{1}=4 \\ g(2)=4^{\frac{2+1}{2}}=4^{\frac{3}{2}} \\ g(3)=4^\frac{3+1}{2}=4^\frac{4}{2}=4^2=16 \\ g(4)=4^{\frac{4+1}{2}}=4^\frac{5}{2}\]
...

Answer 14

which does make the conditions g(n+1)=4g(n-1) and g(1)=4 hold

Answer 15

you can verify thing by showing g(1)=4 using g(x)=4^((x+1)/2)
and you can also show g(n+1)=4g(n-1)
\[g(n+1)=4^\frac{(n+1)+1}{2} \\ g(n-1)=4^\frac{(n-1)+1}{2} \\ \text{ now inputting into } \\ g(n+1)=4g(n-1) \\ 4^\frac{(n+1)+1}{2}=4 \cdot 4^\frac{(n-1)+1}{2} \\ 4^\frac{n+2}{2}=4 \cdot 4^\frac{n}{2} \\ 4^{\frac{n+2}{2}}=4^{\frac{2}{2} } 4^{\frac{n}{2}} \\ 4^\frac{n+2}{2}=4^{\frac{2+n}{2}} \\ 4^{\frac{n+2}{2}}=4^{\frac{n+2}{2}} \text{ this is a true equation }\]
...so this form does work...
but it is not the only form that works for what was given since we were are given any conditions on the even numbers

Answer 16

we were not given
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