The 22nd Root Problem

Calculus Level 2

$\begin{aligned} S_{1}&=&\sqrt[22]{1887}+\sqrt[22]{1888}+\cdots+\sqrt[22]{2013}+\sqrt[22]{2014} \\ \\ \ S_{2}&=&\sqrt[22]{1888}+\sqrt[22]{1889}+\cdots+\sqrt[22]{2014}+\sqrt[22]{2015} \\ \\ I&=&\int_{1887}^{2015}\!\sqrt[22]{x}\,\mathrm{d}x \end{aligned}$

What can you say about the relative values of $S_{1}$ , $S_{2}$ , and $I$ ?

S_{2}<S_{1}<I

I<S_{1}<S_{2}

S_{1}<I<S_{2}

S_{2}<I<S_{1}

I<S_{2}<S_{1}

S_{1}<S_{2}<I

1 solution

Utsav Banerjee
May 20, 2015

For any function $f(x)$ that is positive, increasing & integrable in the interval $[a,b]$ , where $a$ and $b$ are integers such that $a<b$ , let us consider the quantities

$\displaystyle S_{1}=\sum_{a}^{b-1}\!f(x)$ , $\displaystyle S_{2}=\sum_{a+1}^{b}\!f(x)$ and $\displaystyle I=\int_{a}^{b}\!f(x)\,\mathrm{d}x$

The integral $I$ is the area under the curve $y=f(x)$ in the interval $[a,b]$ as shown in Fig. 1 below.

The sum $S_{1}$ is the numerical approximation of the integral $I$ using rectangular strips of unit width as shown in Fig. 2 below.

The sum $S_{2}$ is the numerical approximation of the integral $I$ using rectangular strips of unit width as shown in Fig. 3 below.

Clearly, $S_{1}<I<S_{2}$ .

In this question, $f(x)=\sqrt[22]{x}$ , $a=1887$ and $b=2015$ .

Moderator note:

Yes, this is essentially Trapezoidal Rule. Bonus question: What would the answer be if all the numbers $22$ in $S_1, S_2, I$ are replaced by its additive inverse, $-22$ ?

Answering the Challenge Master's Bonus Question:

Given $\large f(x)={x}^{-\frac{1}{22}}$

Here, $f(x)$ is a decreasing function. So, $S_{2}<I<S_{1}$ .

Utsav Banerjee - 6 years ago

I know two infinitys alichnol and omega and then ?????? which stands for unknown infinity.

Am Kemplin - 1 month, 3 weeks ago

The 22nd Root Problem

1 solution

Moderator note:

1 pending report

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