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 $\delta \approx 0.0382$.
 A: 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
$$
\delta=2\ln\left(1+\frac{i^{(2)}}{2}\right)\approx 0.0396
$$
A: I'm a little late to this one, but if you google the FM exam problem it shows up, and I think that the fundamental problem here still hasn't been addressed, so it's worth talking about.
The question says that after 7.25 years, the value of each account is the same. They very carefully did not mention the balance of the accounts. If they had, this would be a different problem, since one account credits interest every six months and the other does so constantly. The value of the account credited semiannually still increases constantly, since you are always getting closer to the next payout.
Here is a slightly different way to approach the problem:
The value of both accounts is governed by exponential equations that start at the same place and grow at the same rate, so they must be equivalent. Since we're looking for an annual force of interest let's make everything annual. The effective annual rate is $$i = \left(1 + \frac{i^{(m)}}{m}\right)^{m} - 1 $$ so $i = .0404$. Then the force of interest given $i$ is $$\delta = \ln(1+i)$$ so $\delta = .0396$
A: Usually you use the $\texttt{simple interest}$ for the remaining three months. 
$$\left(1+\frac{0.04}{2}\right)^{2 \cdot  7}\cdot \left(1+\frac{0.04}4\right)=e^{7.25 \cdot \delta}$$
$$7.25 \cdot\delta=\ln\left[1.02^{14}\cdot 1.01 \right]$$
$$7.25 \cdot\delta=0.287187$$
$$\delta=\frac{0.287187}{7.25}=0.039612\approx 3.96\%$$
