How is this equal

2^a + 2^a = 2(2^a)=2^(a+1)

>>8955076
C'mon anon
[math]
2^w = 2 \cdot (2^{w-1}) = 2^{w-1} +2^{w-1}
[/math]

>>8955076

[math]2^w - 2^{w-1}[/math]

[math]= 2^1*2^{w-1} - 2^{w-1}[/math]

[math]= 2^{w-1}(2 - 1)[/math]

[math]= 2^{w-1}[/math]

w=0
2^0 = 1
2^(-1) = 1/2
1 - 1/2 = 1/2
i saw this immediately so idk if it's what you're looking for

>>8955094
>>8955090
>>8955103
>>8955098

ty didn't know why i didn't see that before, now this property proves

[math]= -x_{w-1}2^w+\sum_{i=0}^{w-2}x_i2^i

= -x_{w-1}2^w + -x_{w-1}2^{w-1}w+\sum_{i=0}^{w-2}x_i2^i

= -x_{w-1}(2^w - 2^{w-1})+\sum_{i=0}^{w-2}x_i2^i

= -x_{w-1}2^{w-1}+\sum_{i=0}^{w-2}x_i2^i

= (x_{w-1}, x_{w-2},...,x_0)[/math]

>>8955152
and i fucked up the latex. saging until archive

Why does this only work for powers of 2, though?

>>8955165
n^k+n^k+...+n^k (n of them)
= n*(n^k)
= n^(k+1)

>>8955165
[eqn]n^w-(n-1)n^{w-1}=n^{w-1}[/eqn]

>>8955172
ohhhhhh, I was missing the (n-1) term, that makes more sense

no bump

[math]= -x_{w-1}2^w+\sum_{i=0}^{w-1}x_i2^i \\

= -x_{w-1}2^w + -x_{w-1}2^{w-1}w+\sum_{i=0}^{w-1}x_i2^i \\

= -x_{w-1}(2^w - 2^{w-1})+\sum_{i=0}^{w-1}x_i2^i \\

= -x_{w-1}2^{w-1}+\sum_{i=0}^{w-1}x_i2^i \\

= ([x_{w-1}, x_{w-2},...,x_0])[/math]

>>8955165
because 2 halves is 1
i suppose if you did 3^0 - 3^(-1) - 3^(-1) = 3^(-1) it'd work

>>8955165
because 2^w=2*2^(w-1)=2^(w-1)+2^(w-1)
if you had 3^w it would be equal to 3^(w-1)+3^(w-1)+3^(w-1)
etc