## The Termwise Solution

The Sum of All Ions can be calculated termwise, to give one abundance after another after another, until all abundances have been calculated. I started writing from right to left instead of left to right for the termwise simulacrum. I started on the right side with the simulacrum for the Third Critical Ratio, (1+1/a_{3}).

We start with an ‘inside-out’ approach, starting with the smallest variable, and making the simulacrum sum based on the smallest one defining the simulacrum unit. In a Case 2.2.n situation, either AB or BC is the smallest One. In Case 2.2.1, BC is One, and AB is larger than One, as in (BC/AB) = (1/2). So, the sum of AB and BC, based on one of them being One (BC) is (1+(1/(BC/AB))). Since (BC/AB) is a_{3}, this solution for (AB+BC) can be written as (1+1/ a_{3}). Operating the observed ratio of (1/2) in the equation (1+1/ a_{3}) gives (1+1/(1/2)) = (1+2) = 3. This means that the simulacrum sum of (AB + BC), with one of them being One (BC) is 3. When BC is the unit, the sum of the Unit plus the Other is 3. You could say, in this Case, that BC is ‘carrying the Unit’, and that the sum of the Unit plus AB is the simulacrum sum 3. Obviously this means that AB is 2, and so is twice the size of BC, the One. BC is the smallest One. All other abundances are proportional to it through the Critical Ratios.

As mentioned before, we start with an ‘inside-out’ approach, starting with the smallest variable, and making the simulacrum sum based on the smallest one defining the simulacrum unit. In a Case 2.2.n situation, either AB or BC is the smallest One. In Case 2.2.1, BC is the smallest One, and AB is larger than One, as in (BC/AB) = (1/2). So, the sum of AB and BC, based on one of them being One (BC) is (1+(1/(BC/AB))). Since (BC/AB) is a_{3}, this solution for (AB+BC) can be written as (1+1/ a_{3}). Operating the observed ratio of (1/2) in the equation (1+1/ a_{3}) gives (1+1/(1/2)) = (1+2) = 3. This means that the simulacrum sum of (AB + BC), with one of them being One (BC) is 3. When BC is the unit, the sum of the Unit plus the Other is 3. You could say, in this Case, that BC is ‘carrying the Unit’, and that the sum of the Unit plus AB is the simulacrum sum 3. Obviously this means that AB is 2, and so is twice the size of BC, the One. BC is the smallest One. All other abundances are proportional to it through the Critical Ratios.

So, the first term in the termwise solution gives the values of (AB+BC) and is 3 times the Unit. The second term has the first term in the denominator, so when the first sum, (AB+BC), from the simulacrum sum of One plus a_{3}, is multiplied times the second critical ratio, a_{2}, the denominator of a_{2} cancels out, and the value of the variable AC is given. The second term, based on a_{2}, gives the abundance of AC. In the Case 2.2.1 example in Figure 37, a_{2} is = (80/(20+40)) = (80/60) = 4/3. The second term is solved as AC = (S_{3} · a_{2}) = ((1+1/ a_{3}) · a_{2}). When the observed a_{2}, 4/3, is substituted in, it gives AC = (S_{3} · a_{2}) = ((1+1/ (1/2)_{*}) · (4/3)_{*}) = ((1+2) · (4/3)) = 4. AC has a simulacrum solution of 4 times the smallest One.

So far, the simulacrum sum is (AB+BC) = (Three times the One) plus (AC) = (Four times the One), so the sum of all DAG, SDAG = (AB+BC+AC) = (Seven times the One). The sum of diacylglycerol fragment ions equals Seven times the smallest One. The sum of all DAG, SDAG, is the denominator of the First Critical Ratio, a_{1}, so when a_{1} is multiplied times the SDAG, the denominator of a_{1} cancels out and gives the value of MH, the numerator of the First Critical Ratio. The SDAG = (AB+BC+AC) is equal to S_{3} = (AB+BC) plus S_{2} = (AC). When (S_{3}+ S_{2}) is multiplied times the First Critical Ratio, a_{1} the answer is the value of MH, the numerator of the First Critical Ratio. This term in the termwise solution gives MH = ((S_{3}+ S_{2}) · a_{1}) = (SDAG · a_{1}).

When the observed value of a_{1} (5/7) is inserted, we get ((4+3) · (5/7)) = 5. The simulacrum value of the MH is Five times the smallest One. This term in the termwise solution gave (MH), so now we have the sum of all ions, SI = 5 + 4 + 3 = 12. The sum of all ions in terms of One plus Three Critical Ratios is 12 times the One. We already know that the SI in terms of percentage, the desired units, is equal to 2.4, or 240%. The percentage value of the Unit is give by the Percentage Simulacrum Sum divided by the Unit Simulacrum Sum.

Twelve times the Unit (20%) equals the percentage sum, 240%. The MH is Five times the Unit (20%), which equals 100% (base peak). The AC is Four times the Unit, for 80%. AB is Two times the Unit, for 40%. S(AB+BC) gives a simulacrum sum of the Unit, BC = 20%, plus AB. The simulacrum sum of (1+1/a_{3}) was Three Units, or 3 x 20% for 60%.