The following is the section from Introduction to Algorithms by Cormen. et. al. in the Dynamic Tables section.

In the following pseudocode, we assume that $T$ is an object representing the table. The field $table(T)$ contains a pointer to the block of storage representing the table. The field $num(T)$ contains the number of items in the table, and the field $size(T)$ is the total number of slots in the table. Initially, the table is empty: $num(T) = size(T) = 0$.

$text{Table-Insert(T,x)}$

$1quad text{if $size(T)=0$}$

$2quadquad text{then allocate $table(T)$ with $1$ slot}$

$3quadquad size(T) leftarrow 1$

$4quadtext{if } num(T) =size(T)$

$5quadquadtext{then allocate $new-table$ with $2 • size(T)$ slots}$

$6quadquadquadtext{insert all items in $table(T)$ into $new-table$}$

$7quadquadquadtext{free $table(T)$}$

$8quadquadquad table(T) leftarrow new-table$

$9quadquadquad size(T) leftarrow 2 • size(T)$

$10quad text{insert $x$ into $table(T)$}$

$11quad num(T) leftarrow num(T) + 1$

For the amortized analysis for the a sequence of $n$ $text{Table-Insert}$ the potential function which they choose is as follows,

$$Phi(T) = 2.num(T)-size(T)$$

To analyze the amortized cost of the $i$th $text{Table-Insert}$ operation, we let $num_i$ denote the number of items stored in the table after the $i$ th operation, $size_i$ denote the total size of the table after the $i$ th operation, and $Phi_i$ denote the potential after the $i$th operation.

Initially, we have $num_0 = 0, size_0 = 0$, and $Phi_0 = 0$.

If the $i$ th Table-Insert operation does not trigger an expansion, then we have $size_i = size_{i-i}$ and $num_i=num_{i-1}+1$, the amortized cost of the operation is $widehat{c_i}$ is the amortized cost and $c_i$ is the total cost.

$$widehat{c_i}=c_i+Phi_i- Phi_{i-1} = 3 text{ (details not shown)}$$

If the $i$ th operation does trigger an expansion, then we have $size_i = 2 . size_{i-1}$ and $size_{i-1} = num_{i-1} = num_i —1$,so again,

$$widehat{c_i}=c_i+Phi_i- Phi_{i-1} = 3 text{ (details not shown)}$$

**Now the problem is that they do not make calculations for $widehat{c_1}$ the situation for the first insertion of an element in the table (line 1,2,3,10,11 of code only gets executed).**

In that situation, the cost $c_1=1$, $Phi_0=0$ and $num_1=size_1=1 implies Phi_1 = 2.1-1 =1$

We see that $Phi_1=1 tag 1$

So, $$widehat{c_1}=c_1+Phi_1-Phi_0=2$$

But the text says that the amortized cost is $3$, (I feel they should have said the amortized cost is at most $3$, from what I can understand)

Moreover in the plot below,

The text represents graphically the $Phi_1=2$ which sort of contracts $(1)$, but as per the graph if we assume $Phi_1=2$ then $widehat{c_i}=3, forall i$

I do not quite get where I am making the fault.