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solve for ?

To solve this use,

multiply from right of equation both side.

We have:

we use the

So, we have

then multiply frm the left both side of equation above:

\\

changing j to i and i to j, we get:

RH: all nontrivial roots are on real critical line x = 1/2.

Since we know all nontrivial roots are symmetric about the real line x = 1/2, the nontrivial roots are of the form . Therefore, RH is nothing but , which is equivalent to the statement: .

RH, in this interpretation, is to prove . It is no more fancier than that.

We know that both and are roots of , so .

If we can show implies , we will have RH settled.

To show the above, we only need to demonstrate that the equality in the range of is preserved to its domain which is nothing but (C, +), the group of complex addition.

We observe that equality preservation happens in group isomorphism. Therefore, our proof boils down to showing the isomorphism.

The follow-up post is the full proof.

]]>We observe (), () and () are (commutative) groups, where + is the normal complex number addition, and and are respectively the term-wise (Hadamard) product and the inner product of infinite dimension vectors of the form where for complex variable .

Let be defined as , then is homomorphism (where is normal complex number multiplication):

Similarly, let be defined as , then is homomorphism:

and note:

We observe that , for integer set , is kernel for , therefore is isomorphism. Furthermore, since is on principal value, if we define , then .

Suppose and are two roots of , then:

(Q.E.D.)

]]>$a^n+b^n=c^n$ ]]>