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Calculus Level 4

$p=1+\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 3 } } +........\frac { 1 }{ \sqrt { 2014 } } \\ Find\quad integral\quad value\quad of\quad p\quad (floor\quad function)$

The answer is 88.

3 solutions

Chew-Seong Cheong
Nov 22, 2014

$p = \sum _{n=1} ^{2014} {\frac {1}{\sqrt{n}}} \approx \int _{ 0.5 }^{ 2014.5 }{ \frac { 1 }{ \sqrt { x } } } dx$

Considering the area under the consistently decreasing function $f(x) = \frac {1}{\sqrt{x}}$ , we have:

$\int _{ 0.5 }^{ 2014.5 }{ \frac {1}{\sqrt {x}}} dx < p < \int _{1}^{ 2014.5 }{ \frac {1}{ \sqrt {x}}} dx + \frac{1}{2} \times \frac {1}{\sqrt{1}}$

$\left[ 2 \sqrt{x} \right] _{0.5} ^{2014.5} < p < \left[ 2 \sqrt{x} \right] _{1} ^{2014.5} + 0.5$

$2(\sqrt(2014.5)-\sqrt{0.5}) < p < 2(\sqrt(2014.5)-\sqrt{1}) + 0.5$

$2(44.88318171-0.707106781) < p < 2(44.88318171-1)+0.5$

$88.35214985<p<88.76636341\quad \Rightarrow \lfloor p \rfloor = \boxed {88}$

Mehul, sorry for late in replying. I have redone the solution.

Hi there! The solution is really nice!

I'm having trouble understanding how you arrived at the approximation for the summation. How did you choose 0.5 and 2014.5 as the limits of the integral?

Thanks

Soham Karwa - 6 years, 6 months ago

It is difficult to show this without graphics. Anyway, I will try.

The technique to solve the problem is by comparing the areas of discrete $p$ with that of continuous curve $f(x)$ . The value of $p$ is given by vertical bars of height $\frac {1} {\sqrt{n}}$ and width of $1$ . There are various ways to superimpose these vertical bars of $p$ on the $f(x)$ . Two ways are used setting the lower and upper limits of $p$ to estimate $p$ .

I set the lower limit (LHS of the inequality) starting the first bar $0.5 \le x < 1.5$ and hence the $2014^{th}$ bar is $2013.5 \le x < 2014.5$ . We note that starting from the first bar $0.5 \le x < 1.5$ the area covered by the first bar above the curve is larger than that covered by the curve above the second bar $1.5 \le x < 2.5$ and this continues until the end, because $f(x)$ is a consistent decreasing function. Therefore, overall the curve covers less are than the bars.

For the upper limit, I shifted $0.5$ to the left. That is why I needed to add $\frac {1}{2}\times \frac {1}{\sqrt{1}}$ in the inequality to account for all the $2014$ bars. In this arrangement, the area covered by the curve above the second bar is larger than that covered by the third bar above the curve and so on. Also the first half bar is larger than the area covered by the first bar above the curve. Therefore overall the area covered by the curve plus the first half bar is larger than the bars.

You can choose other limits which involve whole numbers but I found that the two limits do not fall within $88 \le p < 89$ , therefore not providing the solution.

Chew-Seong Cheong - 6 years, 6 months ago

Prakhar Gupta
Feb 4, 2015

Here is my $JAVA$ code for the problem:-

Brock Brown
Dec 28, 2014

from math import sqrt
i = 1
total = 0
while i <= 2014:
    total += 1/float(sqrt(i))
    i += 1
print int(total)

Are you smarter than me?

The answer is 88.

3 solutions

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