Yes my bad, your right it should be x^3+5

nope, that function is not injective

for every odd number in Z, [n/2] = [n+1/2]

if y =/= z then y^3 =/= z^3 so the function f(x) = x^3 +5 is injective, the constant doesn't play a role

&



f^-1 is the inverse function, ie it maps the result of f(x) back to it's original (ie x), so that f^-1(f(x)) = x [it does exist and is unique for any injective function]

here the function g(y) = cube-root(y-5) is f^-1

This post was edited by brmv on Jul 8 2013 01:38am

Thanks alot brmv and do you know how to do these?

2a)simplify
2 3
Σ Σ(i+3j)
i=0 j=1


2b) expand and simplify

n
Σ(a subj - a subj-1 )
j=1

not really sure if i understand what is meant with them but let me try to rewrite 2a) how i understand it:

sum[i=0,2] {sum[j=1,3] (i + 3j)}

even less understand what is meant with 2b) - does 'a subj' mean a with subscript j?

This post was edited by brmv on Jul 8 2013 02:02am

yes correct, sorry wasnt sure how to write it

wasn't written that bad, just some space between the sigmas or even better some parantheses would have made it clearer

not really sure what your teacher understands as simplification but it might be:

sum[i=0,2] {sum[j=1,3] (i + 3j)} = sum[i=0,2] {sum[j=1,3] (i } + sum[i=0,2] {sum[j=1,3] (3j)} = sum[j=1,3] {sum[i=0,2] (i)} + sum[i=0,2] 3 {sum[j=1,3](j)} --- [the 3 can be extracted if wanted]

and going further because there is no j factor in the first sum and 1,2,3 add up to 3 as well there is no i factor in the second sum and 0,1,2 add up to 3 as well, next simplification would be

3 . {sum[i=0,2](i)} + 3 . 3 {sum[j=1,3](j)} --- does that make sense to you in this context and what your teacher wants?

yes, thats great, thank you

let me assume that my suspicion is correct and subj means subscript j and subj-1 means subscript j-1

now i will use <j> for subscript j and assume that you have a sequence of a<0>, a<1>, ... , a<n>

then sum[j=1,n] (a<j> - a<j-1>) = (sum[j=1,n] a<j>) - (sum[j=1,n] a<j-1>) = (sum[j=1,n-1] a<j>) + a<n> - a<0> - (sum[j=2,n] a<j-1>) = a<n> - a<0> since sum[j=1,n-1] a<j> = sum[j=2,n] a<j-1>

all clear?