Question

HELP!!
http://prntscr.com/4fqwyf

Answer 1

for first one use
\(\sigma (2^{k-1})=\frac{2^{k+1-1}-1}{2-1}=2^k-1=2n-1\)

note that
n=2^{k-1}
\(2n=2^k\)

Answer 2

\(\large \sigma(2^{k-1}) = 1 + 2 + 2^2 + \cdots + 2^{k-1} = 2^{k} - 1 = 2*2^{k-1}-1 = 2n-1\)

Answer 3

or that xD

Answer 4

but why not simply use sigma notation :3

Answer 5

ok the second one ?

Answer 6

\[\large \begin{align}\\

\sigma(2^{k-1}(2^k-1)) &= \sigma(2^{k-1}) * \sigma(2^{k}-1)

\end{align}\]

Answer 7

yeah , so
\(\sigma (2^{k-1})=2^k-1 (2)\)

Answer 8

hmmm

Answer 9

\[\large \begin{align}\\

\sigma(2^{k-1}(2^k-1)) &= \sigma(2^{k-1}) * \sigma(2^{k}-1) \\~\\
&=(2^k-1)(2)
\end{align}\]

Answer 10

hey no wait

Answer 11

solved :)

Answer 12

\[\large \begin{align}\\

\sigma(2^{k-1}(2^k-1)) &= \sigma(2^{k-1}) * \sigma(2^{k}-1) \\~\\
&=(2^k-1)(2^k)
\end{align}\]

Answer 13

its 2 not 2 ^k

Answer 14

since \(2^k-1\) is a prime, the only divisors are \(1\) and \(2^k-1\)
adding gives you \(2^k\)

Answer 15

oh lol ok xD
i thought 2 like tau :3
mixing

Answer 16

\[\large \begin{align}\\

\sigma(2^{k-1}(2^k-1)) &= \sigma(2^{k-1}) * \sigma(2^{k}-1) \\~\\
&=(2^k-1)(2^k) \\~\\
&=(2^k-1)(2^{k-1})*2 \\~\\
&= 2n
\end{align}\]

Answer 17

the edition of this book , is the same one i have xD
but i couldnt find it online before , so thx

Answer 18

lets prove part c also

Answer 19

oh , im invisible :o

Answer 20

and it seems there is a unproven conjecture existing on this

Answer 21

\[\large \begin{align}\\

\sigma(2^{k-1}(2^k-3)) &= \sigma(2^{k-1}) * \sigma(2^{k}-3)

\end{align}\]

Answer 22

well , yeah :o
there exist , but never dicoverd

Answer 23

nvm 2n+1 :o

Answer 24

c sounds like b

Answer 25

i want to work everything for completeness sake :P

Answer 26

\[\large \begin{align}\\

\sigma(2^{k-1}(2^k-3)) &= \sigma(2^{k-1}) * \sigma(2^{k}-3) \\~\\
&= (2^k-1)* (2^{k}-2) \\~\\

\end{align}\]

Answer 27

we might join them in manual

Answer 28

if you want to , making manual solution for burton

Answer 29

six edition :3

Answer 30

not interested, how to conclude part c ?

Answer 31

oh

Answer 32

FOIL

Answer 33

\[\large \begin{align}\\

\sigma(2^{k-1}(2^k-3)) &= \sigma(2^{k-1}) * \sigma(2^{k}-3) \\~\\
&= (2^k-1)* (2^{k}-2) \\~\\
&=2^k* (2^{k}-2) - 1* (2^{k}-2) \\~\\
\end{align}\]

Answer 34

or that :3

Answer 35

\[\large \begin{align}\\

\sigma(2^{k-1}(2^k-3)) &= \sigma(2^{k-1}) * \sigma(2^{k}-3) \\~\\
&= (2^k-1)* (2^{k}-2) \\~\\
&=2^k* (2^{k}-2) - 1* (2^{k}-2) \\~\\
&=2^k* (2^{k}-3+1) - 1* (2^{k}-2) \\~\\
&= 2^k* (2^{k}-3) - 1* (2^{k}-2) + 2^k \\~\\
\end{align}\]

Answer 36

is that part of ur completeness

Answer 37

\[\large \begin{align}\\

\sigma(2^{k-1}(2^k-3)) &= \sigma(2^{k-1}) * \sigma(2^{k}-3) \\~\\
&= (2^k-1)* (2^{k}-2) \\~\\
&=2^k* (2^{k}-2) - 1* (2^{k}-2) \\~\\
&=2^k* (2^{k}-3+1) - 1* (2^{k}-2) \\~\\
&= 2^k* (2^{k}-3) - 1* (2^{k}-2) + 2^k \\~\\
&= 2^k* (2^{k}-3) + 2
\end{align}\]

Answer 38

yup :)
i fear that my arithmetic skills will get more worse if i leave them without seeing the final answer :/

Answer 39

i like foil >.<

2^2k- 3 2^k+2 = 2^k(2^k -3 ) +2

n=2^k-1
2n= 2^k

Answer 40

and this latex >.<
\( \large
(\overbrace{\overbrace{\color{orange}a
+\underbrace{\color{cornflowerblue}b)(\color{seagreen}c}_{\text{Inside}}}^{\text{First}}
+\color{brown}d}^{\text{Outside}})
=\overbrace{\color{orange}a\color{seagreen}c}^\text F
+\overbrace{\color{orange}a\color{brown}d}^\text O
+\underbrace{\color{cornflowerblue}b\color{seagreen}c}_\text I
+\underbrace{\color{cornflowerblue}b\color{brown}d}_\text L\\
\qquad\quad \underbrace{\qquad\qquad}_{\text{Last}}\)

Answer 41

:)

Answer 42

one line better than 3

Answer 43

what about the last statement :

\(\large \sigma(n) = 2n+1\)

god knows is it

Answer 44

ask him

Answer 45

to puruse this is same as pursuing a direct formula for nth prime number

Answer 46

both problems are equivalent

Answer 47

why you think so ?

Answer 48

if we know nth prime number, we can use it to answer \(\large \sigma(n) = 2n+1\)

Answer 49

if you know a direct formula for \(\large \sigma(n)\) or \(\large \tau (n)\) we can use them to find the nth prime

Answer 50

but we cannot find these w/o doing the prime factorization hmm

Answer 51

brilliant !

Answer 52

but still perfect idea !
OMG !

Answer 53

basically if you solve one conjecture, it helps in solving many other related conjectures

Answer 54

i still rather Goldbach

Answer 55

what about this
http://prntscr.com/4frvrq

Answer 56

twin primes means like 5,7 ?

Answer 57

yes primes separated by 2 numbers, another example :
41 and 43

Answer 58

will ovs ,
\(\sigma (n+2)=n+2+1=(n+1)+2=\sigma (n) +2 \)

Answer 59

ahh i see

Answer 60

for 434 :O
and 8575
we can check to make sure , but wait :O
those numbers again >.<

Answer 61

we can prime factorize and check, they don't look special to me

Answer 62

last question of the day
http://prntscr.com/4frxb1

Answer 63

earluer we have seen that \(\large \sigma(n) = 10\) has no solutions, remember ?

Answer 64

yes

Answer 65

now we need to prove that :

\(\large \sigma(n_1) + \sigma(n_2) = n\) always has a solution

Answer 66

2n

Answer 67

Oh wait its wrong,
we need to show : \(\large \tau(n_1) + \tau(n_2) = n\)

Answer 68

i thought u solving the second one

Answer 69

no, i got confusing with first part itself >.<

Answer 70

well first one there is infinitly for each tau ( n ) =k
so
tau(n1)+tau(n2)=k1+k2

there exist n s,t n=k1+k2

Answer 71

cus \(\large \tau(n_1) = k\) always has a solution : \(\large n_1 = p^{k-1}\)

Answer 72

\(\large \tau(p^{n-2}) + \tau(p) = n\)

does this work ?

Answer 73

yeah also that :3

Answer 74

i think this should work :

\(\large \tau(p^{n-3}) + \tau(p) = n\)

Answer 75

n-3+1+2 =n

ohh ok

Answer 76

cool >.<

Answer 77

the solution is :
\(\large n_1 = p^{n-3}\) and \(\large n_2 = p\)

Answer 78

lets see part b

Answer 79

need to review goldbach conjecture,
what does it say ?

Answer 80

wait someone is hunting close .

Answer 81

what

Answer 82

>.<
someone is hunting (birds ) close to the house lol , its rude

ok Goldbach set if there is a prime number then there is infinitly many primes that have same form , right ?

Answer 83

for p>2

Answer 84

still there ?

http://prntscr.com/4fs8sl

Answer 85

yo :o
have any clue /?

Answer 86

no clue yet

Answer 87

goldbach conjecture says every even number can be expressed as sum of two primes :

\[\large p_1 + p_2 = 2n\]

Answer 88

so any prime can =\(\sigma(n)\)

Answer 89

gtg now , gn

Answer 90

gn

Answer 91

what next ?

Answer 92

i couldn't get what ppl were talking here :
http://math.stackexchange.com/questions/907252/prove-that-the-goldbach-conjecture-implies-that-for-each-even-integer-2n-there

Answer 93

hehe i added that thing here >.<
but thought it stupid , so deleted it hehe

Answer 94

well , http://prntscr.com/4fwfrg

2n= sum of two odds = 2r+2=p1+p2+2=(P1+1)+(p2+1)=\(\sigma (p_1)+\sigma(p_2)\)

Answer 95

got it now :)