How to compare periodically and continuously compounding interest, when time is in-between periods? I am working on a textbook problem, and I think I disagree with the solution. The problem is


  (10.6) Bruce deposits 100 into a bank account. His account is credited interest
  at a nominal rate of interest of 4% convertible semiannually. At the same
  time, Peter deposits 100 into a separate account. Peter’s account is credited
  interest at a force of interest of δ. After 7.25 years, the value of each account
  is the same. Calculate δ.


The solution provided is $\delta = .0396$, which I can get by setting these two accumulation functions equal:
$$
(1+\dfrac{.04}{2})^{2 \times 7.25}=e^{7.25 \delta}
$$
The reason I feel this is incorrect is that the semiannual accumulation function should not permit a time value of $t=7.25$, which is between periods. Instead I feel that a value of $t=7$ should be used, being the end of the most recent period:
$$
(1+\dfrac{.04}{2})^{2 \times 7}=e^{7.25 \delta}
$$
This yields $\delta \approx 0.0382$.
The same principal invested for the same period of time yields the same accumulated value. So over $1$ year we must have the equivalence
$$
\left(1+\frac{i^{(2)}}{2}\right)^2=\mathrm e^\delta
$$
and then
$$
\delta=2\ln\left(1+\frac{i^{(2)}}{2}\right)\approx 0.0396
$$
Usually you use the $\texttt{simple interest}$ for the remaining three months. 

$$\left(1+\frac{0.04}{2}\right)^{2 \cdot  7}\cdot \left(1+\frac{0.04}4\right)=e^{7.25 \cdot \delta}$$

$$7.25 \cdot\delta=\ln\left[1.02^{14}\cdot 1.01 \right]$$

$$7.25 \cdot\delta=0.287187$$

$$\delta=\frac{0.287187}{7.25}=0.039612\approx 3.96\%$$