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Question (may SKIP reading this):

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).

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    $\begingroup$ use also the degree 2 term in $(1+x)^5=1 + 5x + 10x^2+\dots$, especially in the denominator $\endgroup$
    – user8268
    Dec 24, 2022 at 11:27
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    $\begingroup$ agree with the previous comment, approximating $1.01^5$ as $1.051$ to 3 decimal places in the denominator will give the right answer $\endgroup$
    – a1402
    Dec 24, 2022 at 11:29
  • $\begingroup$ @user8268 thank you. $\endgroup$ Dec 24, 2022 at 11:39

6 Answers 6

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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.

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

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    $\begingroup$ Innovative. Thank you for your answer. +1. $\endgroup$ Dec 24, 2022 at 11:42
  • $\begingroup$ 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. $\endgroup$ Dec 24, 2022 at 11:43
  • $\begingroup$ @insipidintegrator thank you for pointing it out. $\endgroup$ Dec 24, 2022 at 11:54
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    $\begingroup$ This can also be rounded to 2965, which is option A. :) $\endgroup$ Dec 24, 2022 at 16:40
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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 $2967.53$.

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  • $\begingroup$ It might be worth adding how to divide by 0.051 without a calculator. $\endgroup$ Dec 24, 2022 at 11:33
  • $\begingroup$ $\dfrac{1.051}{0.051}$ is same as $\dfrac{1051}{51}$, which I think can be done manually easily enough, to the desired accuracy. $\endgroup$ Dec 24, 2022 at 11:35
  • $\begingroup$ @insipidintegrator thank you for your answer. +1. $\endgroup$ Dec 24, 2022 at 11:39
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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 $0.05$:

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

The approximative final answer is:

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

My calculator gives: $20.60398$

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

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  • $\begingroup$ Thank you for your answer. +1. Though a bit too tedious. $\endgroup$ Dec 24, 2022 at 11:44
  • $\begingroup$ 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. $\endgroup$
    – user1107523
    Dec 24, 2022 at 11:50
  • $\begingroup$ Could you tell me a link to getting and solving that taylor expansion (because isn't (x-0.05)=0)? $\endgroup$ Dec 24, 2022 at 12:07
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    $\begingroup$ 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 $\endgroup$
    – user1107523
    Dec 24, 2022 at 13:10
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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

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

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  • $\begingroup$ Thank you for your answer. +1. $\endgroup$ Dec 25, 2022 at 4:46
  • $\begingroup$ You are welcome ! $\endgroup$
    – Angelo
    Dec 25, 2022 at 6:37
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$$\frac{1.01^5}{1.01^5-1}=1+\frac{1}{1.01^5-1}=1+\frac{1}{(1+\epsilon)^5-1}$$ $$(1+\epsilon)^5-1=\epsilon ^5+5 \epsilon ^4+10 \epsilon ^3+10 \epsilon ^2+5 \epsilon$$ 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 $20.6039799616\cdots$

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

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