# Calculating $\frac{1.01^5}{1.01^5-1}$ without calculator with good accuracy?

A computer is sold either for $$19200$$ cash or for $$4800$$ cash down payment together with five equal monthly installments. If the rate of interest charged is $$12\%$$ per annum, then the amount of each installment (nearest to a rupee) is?
Options:
A)$$2965\qquad$$ B)$$2990\qquad$$ C)$$3016\qquad$$ D)$$2896\qquad$$ E)$$2880$$

Boiling down to calculating:

$$\frac{144(1.01)^5}{1.01^5-1}$$ Where if I were to approximate $$1.01^5\sim1+0.01(5)=1.05$$, returns $$3024$$ as answer thus tempting one to select option 'C' which is wrong!

The answer option is Option A (which is itself a bit too off from the more accurate value of $$2967$$ but makes sense anyways as other options are too far apart from this value).

So, without using a calculator, how to calculate the above expression with greater accuracy?

Note:
Just acknowledging the fact that I received a lot of great answers but sadly could accept only $$1$$ which turned out to be a very difficult task.
Finally, unable to select on my own, I went with the one that the community selected (most upvoted).

• use also the degree 2 term in $(1+x)^5=1 + 5x + 10x^2+\dots$, especially in the denominator Dec 24, 2022 at 11:27
• agree with the previous comment, approximating $1.01^5$ as $1.051$ to 3 decimal places in the denominator will give the right answer Dec 24, 2022 at 11:29
• @user8268 thank you. Dec 24, 2022 at 11:39

You just need to note that $$(1+x)^5$$ has a full binomial expansion given by $$(1+x)^5= 1+5x+10x^2+10x^3+5x^4+x^5$$ Therefore, if the approximation $$(1+x)^5 \approx 1+5x$$ does not work, then try $$1 + 5x + 10x^2$$ instead. This leads to $$1.051$$ as an approximation. If necessary, going one further down the expansion $$1.05101$$, which upon substitution will not change the answer much. This is the stipulation of some comments in this thread as well : the answer will stabilize near $$2967$$ by the second order approximation.

##### In the absence of division

Suppose, however, you wanted to completely avoid division : your calculator hates division (or you hate it). Then, you need to focus on the function $$\frac{x^5}{x^5-1}$$ and how it behaves near $$1$$, because $$1.01$$ is close to $$1$$.

We write it as $$\frac{x^5}{x^5-1} = 1 + \frac{1}{x^5-1}$$ so that we only need to focus on $$\frac{1}{x^5-1}$$. However, there's a problem : we cannot "Taylor expand" $$\frac{1}{x^5-1}$$ around the point $$1$$, because it goes to $$+\infty$$ as $$x \to 1$$. However, we can still isolate the "bad" part by finding the rate at which $$\frac{1}{x^5-1}$$ goes to $$+\infty$$ as $$x$$ goes to $$1$$. Then, the remaining part will admit a Taylor expansion.

To do that, observe that $$\frac{1}{x^5-1} = 1+\frac{1}{(x-1)(1+x+x^2+x^3+x^4)} \approx \frac{1}{5(x-1)}$$ Basically, $$\frac{1}{x^5-1}$$ behaves "like" $$\frac{1}{5(x-1)}$$ as $$x$$ is closed to $$1$$. We are led to expect that removing the "bad" part $$\frac{1}{5(x-1)}$$ from $$\frac{1}{x^5-1}$$ should lead to something that is Taylor expandable around $$1$$.

You will observe that the approximation $$\frac{1}{x^5-1} \approx \frac{1}{5(x-1)}$$ is not good enough for the question you're solving (because you'll get $$144 \times 21 = 3024$$ which is not good enough).

As I said, we need to see if removing the "bad" part $$\frac{1}{5(x-1)}$$ from $$\frac{1}{x^5-1}$$ leads to something that is finite around $$x \to 1$$, so that it can be Taylor expanded if necessary. That's why we consider

$$\lim_{x \to 1} \frac{1}{x^5-1} - \frac{1}{5(x-1)} = \frac{5x-x^5-4}{5(x-1)(x^5-1)}$$ A couple of L'Hospitalizations (quite easy ones, because the product below is easy to expand) later, you land at the quantity $$-\frac{2}{5}$$. This will tell you that the function $$g(x) =\frac{1}{x^5-1} - \frac{1}{5(x-1)}$$ satisfies $$\lim_{x \to 1} g(x) = -\frac{2}{5}$$. Now, Taylor expanding $$g$$ around the point $$1$$ (you can expect $$g$$ to have a Taylor expansion : that's not a worry) $$\frac{1}{x^5-1} - \frac{1}{5(x-1)} \approx - \frac{2}{5} + O((x-1))$$ near the point $$x=1$$ (leaving out everything except the constant term in the Taylor expansion of $$g$$).

Now, trying $$x=1.01$$ out gives $$20.6$$, which leads to $$144 \times 2.06 = 2964.4$$, which is also good enough : and avoids any kind of decimal by-hand division.

Note that we used the following heuristic in the second part : even when a function $$h(x)$$ is not differentiable at a point $$x_0$$ , it may happen that we can find $$N$$ so that $$h(x)(x-x_0)^N$$ is differentiable at $$x_0$$. In that case, we can still write down an asymptotic analysis for $$h(x)$$ near $$x_0$$. That's basically what we did here for the function $$h(x) = \frac{1}{x^5-1}$$ at $$x_0=1$$ with $$N=1$$, to find the nature of $$h$$ accurately.

That's an extension of the usual derivative approximation $$h(x+a) \approx h(x) + ah'(x)+\ldots$$ that we're considering. In complex analysis (where it is frequently used e.g. in the theory of generating functions), this kind of analysis appears when we are working with Laurent series , poles and meromorphic functions, for instance.

• +1 solely for coining "L'hospitalizations"! Dec 24, 2022 at 11:49
• @insipidintegrator Thank you, I've been using it for around $9$ years now. Dec 24, 2022 at 11:49
• Thank you for your answer. +1. Dec 24, 2022 at 11:56
• @SarveshRavichandranIyer in your expression with limits, how/why we got the 1st term (from which the 2nd term is being subtracted)? Dec 24, 2022 at 12:04
• @SarveshRavichandranIyer thank you for the addendum. Dec 24, 2022 at 12:23

By using $$a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$$, I would do \begin{align} \frac{144\cdot1.01^5}{1.01^5-1} &= \frac{144\cdot1.01^5}{(1.01-1)(1.01^4+1.01^3+1.01^2+1.01+1)} = \\ &= \frac{144\cdot1.01^5}{0.01(1.01^4+1.01^3+1.01^2+1.01+1)} = \\ &= \frac{14400\cdot1.01^5}{1.01^4+1.01^3+1.01^2+1.01+1} \approx \\ &\approx \frac{14400\cdot1.05}{1.04+1.03+1.02+1.01+1} = \\ &= \frac{14400\cdot1.05}{5.1} \approx 2964.71 \\ \end{align}

• Innovative. Thank you for your answer. +1. Dec 24, 2022 at 11:42
• Just a small comment: The OP does 1.05/0.05, my answer does 1.051/0.051, while this answer does 1.05/0.051. Dec 24, 2022 at 11:43
• @insipidintegrator thank you for pointing it out. Dec 24, 2022 at 11:54
• This can also be rounded to 2965, which is option A. :) Dec 24, 2022 at 16:40

The thing is, the multiplication of the given expression by $$144$$ only serves to amplify the error in your computation $$144$$ times, which makes the answer deviate greatly from the correct answer.

As a general rule of thumb, you should take the Taylor polynomial upto the second degree. If larger numbers, say 2345785, were multiplied to the expression, then I would take the polynomial upto much higher degrees.

So, $$1.01^5=1+(0.01)5+\dfrac12(0.01)^2\cdot5\cdot4=1+0.05+10\times0.0001=1.051,$$ which gives $$2967.53$$.

• It might be worth adding how to divide by 0.051 without a calculator. Dec 24, 2022 at 11:33
• $\dfrac{1.051}{0.051}$ is same as $\dfrac{1051}{51}$, which I think can be done manually easily enough, to the desired accuracy. Dec 24, 2022 at 11:35
• @insipidintegrator thank you for your answer. +1. Dec 24, 2022 at 11:39

Let's compute $$\frac{1.01^5}{1.01^5 - 1}$$. Knowing $$1.01 = 1 + 0.01$$, and using the formula of $$(a+b)^5$$, you get:

$$1.01^5= 1 + 5.1.(0.01) + 10.1.(0.0001) + 10.1.(0.000001)+5.1.(0.00000001)+0.0000000001$$

$$= 1 + 0.05 + 0.001 + 0.00001 + 0.00000005 + 0.0000000001$$

$$= 1.0510100501$$

So you fraction is:

$$\frac{1.0510100501}{0.0510100501} = \frac{1}{0.0510100501} +\frac{0.0510100501}{0.0510100501} = \frac{1}{0.0510100501} + 1$$

So far, I used no approximations. The last step is tricky, here is my advice: as $$0.0510100501$$ is close to $$0.05 = \frac{1}{20}$$, use a Taylor expansion near $$0.05$$:

$$\frac{1}{x} \approx \frac{1}{0.05} - \frac{1}{0.05^2}(x-0.05)$$

$$20 -400.(0.0010100501) + 1 = 21 - 0.40402004 = \color{red}{20.59397996}$$

My calculator gives: $$20.60398$$

So my answer has a relative precision of $$99.951$$% approximately.

• Thank you for your answer. +1. Though a bit too tedious. Dec 24, 2022 at 11:44
• I read the post a little bit too fast, thought you wanted a very high accuracy. Surely, I could have ignored several terms past the comma.
– user1107523
Dec 24, 2022 at 11:50
• Could you tell me a link to getting and solving that taylor expansion (because isn't (x-0.05)=0)? Dec 24, 2022 at 12:07
• If you choose to approximate $0.051...$ to $0.05$, you wouldn't need a Taylor expansion, since your know $0.05 = 1/20$, but it wouldn't be precise enough to choose between (D) and (E). In general, if the approximation around $0.05$ does not fit, you can use any other number $x_0$ for the Taylor expansion. See more information about Taylor here: en.wikipedia.org/wiki/Taylor_series
– user1107523
Dec 24, 2022 at 13:10

For any $$\;x\in\big]0,1\big[\,,\;$$ it results that

$$5x\left(1+2x+2x^2\right)<\big(1+x\big)^5-1<5x\left(1+2x\sum\limits_{n=0}^\infty x^n\right)\;\;,$$

$$5x\left(1+2x+2x^2\right)<\big(1+x\big)^5-1<\dfrac{5x\left(1+x\right)}{1-x}\;.$$

In particular, for $$\;x=\dfrac1{100}\;,\;$$ we get that

$$\dfrac{5101}{100000}<1.01^5-1<\dfrac{101}{1980}\;\;,$$

$$\dfrac{1980}{101}<\dfrac1{1.01^5-1}<\dfrac{100000}{5101}\;\;,$$

$$\dfrac{2081}{101}<1+\dfrac1{1.01^5-1}<\dfrac{105101}{5101}\;\;,$$

$$20.60396<\dfrac{2081}{101}<\dfrac{1.01^5}{1.01^5-1}<\dfrac{105101}{5101}<20.604\;\;,$$

$$\underbrace{144\cdot20.60396}_{>2966.97}<\dfrac{144\cdot1.01^5}{1.01^5-1}<\underbrace{144\cdot20.604}_{=2966.976}\;\;.$$

Hence, it results that

$$2966.97<\dfrac{144\cdot1.01^5}{1.01^5-1}<2966.976\;.$$

• Thank you for your answer. +1. Dec 25, 2022 at 4:46
• You are welcome ! Dec 25, 2022 at 6:37

$$\frac{1.01^5}{1.01^5-1}=1+\frac{1}{1.01^5-1}=1+\frac{1}{(1+\epsilon)^5-1}$$ $$(1+\epsilon)^5-1=\epsilon ^5+5 \epsilon ^4+10 \epsilon ^3+10 \epsilon ^2+5 \epsilon$$ Long division $$\frac{1}{(1+\epsilon)^5-1}=\frac{1}{5 \epsilon }-\frac{2}{5}+\frac{2 \epsilon }{5}-\frac{\epsilon ^2}{5}-\frac{\epsilon ^3}{25}+O\left(\epsilon ^4\right)$$ $$\frac{(1+\epsilon)^5}{(1+\epsilon)^5-1}=\frac{1}{5 \epsilon }+\frac{3}{5}+\frac{2 \epsilon }{5}-\frac{\epsilon ^2}{5}-\frac{\epsilon ^3}{25}+O\left(\epsilon ^4\right)$$ $$\frac{1.01^5}{1.01^5-1}\sim\frac {100}5+\frac 3 5+\frac 2 {500}=\frac{5151}{250}=\frac{20604}{1000}=\color{red}{20.604}$$ instead of $$20.6039799616\cdots$$

• Thank you for your answer. +1. Dec 25, 2022 at 4:47