OHM’S THEOREM

 OHM’S THEOREM 2.1


Martin Ohm Martin Ohm (May 6, 1792 - April 1, 1872) was a German mathematician who earned his doctorate in 1811 at Friedrich-Alexander-University, Erlangen-Nuremberg, under his advisor, Karl Christian von Langsdorf [1]. By the early 1800s he became a professor in the gymnasium at Thorn for mathematics and physics. In 1839, he was chosen to be a professor at the University of Berlin, and delivered lectures at the academy of architecture, and at the school of artillery and engineering [12].


2.2 Ohm’s Rearrangement Theorem

2.2.1 Part I 


Theorem 2.2.1 ([11]). 

For p and q positive integers, rearrange ∑_(n≥1)^∞▒〖(-1)〗^(n-1)/n by taking the first positive terms, then the first q negative terms, then the next p positive terms, the next q negative terms, and so on. The rearranged series converges to 

ln⁡〖(2)+1/2  ln⁡(p/q) 〗. (2.1)

Denote by A (p, q) the series resulting from this rearrangement. Let S_k be the partial sums of A (p, q) where specifically S_k^'= S_k^' (p,q) denotes the k th partial sum of A (p, q), and let C_n= C_n (p,q) be the partial sum of A (p, q), which is the result in S from precisely adding n blocks, where each block consists of p positive terms and q negative terms, consecutively. Denote Sn to be the partial sums of the usual alternating series, where n denotes the terms. Call C_nas C_n  =S_(f(n))  +R_n for some f(n) and some “remainder” R_n. 

It suffices to break the proof of Theorem 2.2.1 into the following: 

I Verifying the explicit formula of C_n  (p,q) for each of the three types of rearrangements:

(a) p = q, 

(b) p > q, and 

(c) p < q.

 II Arguing that Sn and Cn converge ln⁡〖(2)+1/2  ln⁡(p/q) 〗.


EXAMPLE:2.2.11.

Consider A(2,2).we want to show thatC_n (2,2)= S_4n.note that

C_1=1+1/3-  1/2-1/4=1-1/2+1/3-1/4= S_4=S_4.1

C_2=1+1/3-  1/2-1/4+⌈1/5+1/7-1/6-1/8⌉

C_2=1+1/3-  1/2-1/4+1/5+1/7-1/6-1/8= S_8=S_4.2

Proof:

We proceed by induction with the base case shown in example 2.2.1.1. so, suppose that 

C_n (2,2)=1-1/2+1/3-1/4+⋯+1/(4n-4)-1/(4n-1)+1/(4n-1)-1/4n=S_4n

Then, by the inductive hypothesis, 

C_(n+1) (2,2)= C_n (2,2)+(1/(4n+1)+1/(4n-1))-(1/(4n+2)+1/(4n+4))

= C_n (2,2)+1/(4n+1)-1/(4n+2)+1/(4n+3)-1/(4n+4)=S_(4n+4),

as wanted, so f(n)=4n and R_n=0.

General case: A(p,p),p=q.

We now proceed with the more general case:


Theorem 2.2.2. For positive integer p, Cn (p, p) = S2pn for all n.

Proof. Consider A (p, p). We argue that

Cn: = Cn (p, p) = S2pn

by induction. Note that


C_1= (1+1/3+...+1/(2p-1)) - (1/2+1/4+...+1/2p)

      = 1 - 1/2+1/3-1/4+...+1/(2p-1)-1/2p  = S_2p

C_2=  (1+1/3+...+1/(2p-1)) - (1/2+1/4+...+1/2p)

      = C 

      = S^2p + 1/(2p+1)-1/(2p+2)+1/(2p+3)-1/(2p+4) +…+ 1/(4p-1)-1/4p  = S_4p

Now suppose that

C_(n+1)(p, p) = 1-1/2+1/3-1/4+...+1/(2pn-1)+1/2pn  = S_2pn.

By the inductive hypothesis,

C_(n+1)(p, p) = C_(n+1)(p, p) +(1/(2pn+1)+1/(2pn+3)+...+1/(2pn+2p-1))-(1/(2pn+2)+1/(2pn+4)+...+1/(2pn+2p))

= C_n(p, p) + 1/(2pn+1)-1/(2pn+2)+1/(2pn+3)-1/(2pn+4)+...+1/(2pn+2pn-1)-1/(2pn+2p)

= S_2pn + 1/(2pn+1)-1/(2pn+2) + 1/(2pn+3)-1/(2pn+4)+...+1/(2pn+2pn-1)-1/(2pn+2p)

= S_(2pn+2p)   = S_(2p(n+1))


Therefore, f(n) = 2p(n+1) and Rn = 0.  


Example 2.2.2.1. Consider A (3,1). We have

C_1 = 1+ 1/3+1/5+(-1/2)

          = 1 - 1/2+1/3-1/4+1/5-1/6+(1/4+1/6)

         = S_6 + 1/2 (1/2+1/3) = S_6.1+1/2 (1/2+1/3)

C_2 = 1 + 1/3+1/5+(-1/2) + [1/7+1/9+1/11+(-1/4)]

          = 1 - 1/2+1/3-1/4+1/5-1/6+1/7-1/8+1/9-1/10+1/11-1/12+(1/6+1/8+1/10+1/12)

          = S_12+1/2  (1/3+1/4+1/5+1/6) = S_6.2 + 1/2  (1/3+1/4+1/5+1/6)

We claim from our analysis of C1 and C2 that

C_n(3,1) = S_6n + 1/2 (1/(n+1)+1/(n+2)+1/(n+3)+...+1/3n)

So, f(n) = 6n and

R_n   = 1/2 (1/(n+1)+1/(n+2)+1/(n+3)+...+1/3n)

We also find that the last block of Cn (3,1) is

1/(6n-5)+1/(6n-3)+1/(6n-1)-1/2n.

We proceed to verify this.

Theorem 2.2.3. (1) The last block of C_n(3,1) is 1/(6n-5)+1/(6n-3)+1/(6n-1)-1/2n  , and (2) C_n (3,1) = S_6n  + 1/2  (1/(n+1)+1/(n+2)+1/(n+3)+...+1/3n).

Proof. Let us proceed by induction on n. We already have seen that (1) and (2) hold when n = 1,2. Now suppose that (1) and (2) hold for n. We need to show that (1) and (2) hold for n+1.

Proof of (1): By the inductive hypothesis, we know that the positive terms added in the last block of Cn (3,1) are

1/(6n-5)+1/(6n-3)+1/(6n-1).

Hence, the three positive terms in the last block of Cn+1(3,1) that are added together are

1/(6n+1)+1/(6n+3)+1/(6n+5).

Similarly, by the inductive hypothesis, we know that the term subtracted in the last block of Cn (3,1) is 1/2n. Hence, then the term subtracted in Cn+1(3,1) must be 1/(2n+2). We continue to show that these terms added and subtracted in the last block of Cn+1(3,1) by noting that 6(n+1) −5 = 6n+1,6(n+1) −3 = 6n+3,6(n+1) −1 = 6n+5,2(n+1) = 2n+2. This proves (1) by induction, and we have

C_(n+1)(3,1) = C_n(3,1) + 1/(6n+1)+1/(6n+3) + 1/(6n+5)-1/(2n+2)

Proof of (2): By the inductive hypothesis,


C_n(3,1) = S_6n + 1/2  (1/(n+1)+1/(n+2)+1/(n+3)+...+1/3n)

We want to show that

C_(n+1)(3,1) = S_(6(n+1)) + 1/2  (1/((n+1)+1)+1/((n+1)+2)+...+1/(3(n+1))).

From (1), we have shown that

C_(n+1)(3,1) = C_n(3,1) + 1/(6n+1)+1/(6n+3)+1/(6n+5)-1/(2n+2) .

Or in other words,

C_(n+1) (3,1)-C_(n+1) (3,1) =  1/(6n+1)+1/(6n+3)+1/(6n+5)-1/(2n+2).

So, to prove

C_(n+1) (3,1) = S_(6(n+1)) + 1/2  (1/((n+1)+1)+1/((n+1)+2)+...+1/(3(n+1))),

it is enough to show that

(S_(6(n+1))+1/2 (1/((n+1)+1)+1/((n+1)+2)+...+1/(3(n+1)))) –

 (S_6n+1/2 (1/(n+1)+1/(n+2)+1/(n+3)+        ...+1/3n))

                   =  1/(6n+1)+1/(6n+3)+1/(6n+5)-1/(2n+2) .

By definition of Sk, we know that

S_(6(n+1))- S_6n = 1/(6n+1)-1/(6n+2)+1/(6n+3)-1/(6n+4) +1/(6n+5)-1/(6n+6),

and since the expression we are considering equals

S_(6(n+1))-〖 S〗_6n = 1/(6n+1)-1/(6n+2) +1/(6n+3)-1/(6n+4)+1/(6n+5)-1/(6n+6),

we need to prove the identity obtained from substituting the expression S_(6(n+1))-S_6n:

1/(6n+1)-1/(6n+2)+1/(6n+3)-1/(6n+4)+1/(6n+5)-1/(6n+6) + 1/2 (1/((n+1)+1)+1/((n+1)+2)+...+1/(3(n+1))-1/(n+1)-1/(n+2)-1/(n+3)-...-1/3n)

= 1/(6n+1)+1/(6n+3)+1/(6n+5)-1/(2n+2).

Notice that the terms 1/(6n+1),1/(6n+3),1/(6n+5) get canceled out and we are left with

- 1/(6n+2)-1/(6n+4)-1/(6n+6)+1/2 (1/((n+1)+1)+1/((n+1)+2)+...+1/(3(n+1))-1/(n+1)-1/(n+2)-1/(n+3)-...-1/3n) 

= -  1/(2n+2).

Expanding and simplifying the left side of the equation, we need to show,

- 1/(6n+2)  -  1/(6n+4)-1/(6n+6)+1/2 (1/(n+2)+1/(n+3)+1/(n+4)+...+1/2n+1/(2n+1)+...+1/3n+1/(3n+1)+1/(3n+2)+1/(3n+3)-1/(n+1)-1/(n+2)-1/(n+3)-1/(n+4)-...-1/2n-1/(2n+1)-1/(2n+2)-1/(2n+3)-1/(2n+4)-...-1/3n) 

= - 1/(2n+2)

or 

- 1/(6n+2)-1/(6n+4)-1/(6n+6)+1/2 (1/(3n+1)+1/(3n+2)+1/(3n+3)-1/(n+1)) = - 1/(2n+2)

= - 1/(6n+2)-1/(6n+4)-1/(6n+6)+1/(6n+2)+1/(6n+4)+1/(6n+6)-1/(2n+2) 

= - 1/(2n+2),

which simplifies to - 1/(2n+2)  = -1/(2n+2), and so (2) follows


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