I apologies bothering you with a trivial thing, but I’ve been thinking about saving some money in a bank and one of them offers 2.89% nominal benefits on yearly bases, but they award you each and every day a fraction of this yearly benefit. Each subsequent day has the base + previous day’s benefits calculated as a new base for the %. They cite that If you deposit 10k you get 0.79 every day. What are the actual effective yearly benefits in %? Is it 3%? More? Less? Thank you Please Register or Log in to view the hidden image!

\( \left( \frac{1000079}{1000000} \right)^{365} \) is between 2.906% and 2.944% APR or between 2.865% and 2.901% compounded continuously. So this figure looks compatible with continuous compounding at 2.89%. So if the bank quotes 2.89% compounded continuously, then the expected result of investing 10k for a full year is \(10000 \times e^{0.0289} \approx 10293.22 \) which is a APR of 2.9322% or an average of 0.80 / day over 365 days. But the first day you get \(10000 \times \left( e^{\frac{0.0289}{365}} - 1 \right) \approx 0.79 \). In contrast to continuous compounding, simple interest of 0.79 per day on a 1 year term gives you \(10000 + 0.79 \times 365 = 10288.50\) which is a bit less.

2.89% compounded annually gives 2.890000% APR 2.89% compounded semiannually gives 2.910880% APR 2.89% compounded quarterly gives 2.921472% APR 2.89% compounded monthly gives 2.928589% APR 2.89% compounded daily gives between 2.932046% and 2.932048% APR depending on the technical definition of what constitutes a year in banking terms. This may be what your bank was referring to. If r is the rate as a percentage, then the APR for compounding n times in a year is \(100 \times \left( \left( 1 + \frac{r}{100 \times n} \right)^n - 1 \right)\) (the 100's go away if both the rate and the APR are expressed as a decimal like 0.0289). Approximation for when r < 3: APR \( \approx r + \frac{n-1}{200 n} r^2 + \frac{(n - 2) (n - 1)}{60000 n^2} r^3 \) Which is a truncation of \( \sum_{k=1}^{n} \frac{(n-1)!}{k! \, (n-k)! \, 100^{k-1} \, n^{k-1} } r^k \) 2.89% compounded continuously gives 2.932166% APR If r is the rate as a percentage, then the APR for compounding continuously is \(100 \times \left( \textrm{exp}\left( \frac{r}{100} \right) - 1 \right)\). Approximation for when r < 3: APR \( \approx r + \frac{1}{200} r^2 + \frac{1}{60000} r^3 \) Which is a truncation of \( \sum_{k=1}^{\infty} \frac{1}{k! \, 100^{k-1} } r^k \)

That depends. For how long? Forever?! Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image! There is actually an equation answering your question. Simply enter the day i.e. day seven, and the equation tells you how much money you have. Unfortunately I do not know the equation but should it be found, then you may alter it to also tell you which day it is, depending how much money you have. Please Register or Log in to view the hidden image!