# 微积分网课代修|微分学代写Differential calculus代考|MATH272 compounded interest

• 单变量微积分
• 多变量微积分
• 傅里叶级数
• 黎曼积分
• ODE
• 微分学

Consider what happens to a $\$ 1000$deposit with$10 \%$annual interest, compounded yearly. It’s a geometric progression; after$x$years we have: $$f(x)=1000 \cdot 1.1^{x} .$$ where$x$is a positive int eger. But what if I want to withdraw my money in the middle of the year? It would be fair to ask the bank for the interest to be compounded now. It would also be fair for the bank to do it in such a way that the annual return remains the same even if we compound twice. What should be the semi-annual interest rate? Suppose the amount is to grow by a proportion,$r$. Then, if applied again, it will give me the same ten percent growth! In other words, we have: $$f(.5)=1000 \cdot r \text { and } r \cdot r=1.1$$ Therefore, according to the definition of the square root, we have $$r=\sqrt{1.1} \approx 1.0488$$ or about$4.9$percent. ## 微积分网课代修|微分学代写Differential calculus代考|bacteria multiplying The radioactive carbon loses half of its mass over a certain period of time called the half-life of the element. It’s a geometric progression again: $$a_{n+1}=a_{n} \cdot \frac{1}{2} .$$ Unfortunately,$n$is not the number of years but the number of half-lives! For example, the percentage of this element.${ }^{14} \mathrm{C}$, left is plotted below against time: However, we only know two points on the graph! Suppose the half-life is 5730 years (i.e, the time it takes to go from$100 \%$to$50 \%$. The model measures time in multiples of the half-life, 5730 years, and any period shorter than that will require a new insight. How much is left after$5730 / 2=2865$years? The answer is below$75 \%$: $$\sqrt{\frac{1}{2}} \approx .707 \text {. }$$ ## 微积分网课代修|微分学代写Differential calculus代考|compounded interest 考虑一下$\$1000$ 存款 $10 \%$ 年度利息，每年复利。这是一个几何级数的进展洉 $x$ 年我们 有:
$$f(x)=1000 \cdot 1.1^{x} .$$

$$f(.5)=1000 \cdot r \text { and } r \cdot r=1.1$$

$$r=\sqrt{1.1} \approx 1.0488$$

## 微积分网课代修|微分学代写Differential calculus代考|bacteria multiplying

$$a_{n+1}=a_{n} \cdot \frac{1}{2} .$$

$$\sqrt{\frac{1}{2}} \approx .707$$