Infinite sums are not the only heroes!

n = 1 10 i 1 n = a + b i \large \displaystyle \sum_{n=1}^{10} {i^{\frac{1}{n}}} = a + bi

In the equation above, a a and b b are positive real numbers . Find the value of a b \left\lfloor a \right\rfloor - \left\lceil b \right\rceil .

Notations:


The answer is 4.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Chew-Seong Cheong
Sep 21, 2016

n = 1 10 i 1 n = n = 1 10 e π 2 n i = n = 1 10 ( cos π 2 n + i sin π 2 n ) = 8.3422 + 3.9053 i \begin{aligned} \sum_{n=1}^{10} i^\frac 1n & = \sum_{n=1}^{10} e^{\frac \pi{2n}i} = \sum_{n=1}^{10} \left(\cos \frac \pi{2n} + i \sin \frac \pi{2n} \right) = 8.3422+3.9053i \end{aligned}

a b = 8.3422 3.9053 i = 8 4 = 4 \implies \lfloor a \rfloor - \lceil b \rceil = \lfloor 8.3422 \rfloor - \lceil 3.9053i \rceil = 8 - 4 = \boxed{4}

Tapas, this one you need to explain three notations. I have done that for you.

Chew-Seong Cheong - 4 years, 8 months ago

Log in to reply

Oops. I'll really take care of that from now.

Tapas Mazumdar - 4 years, 8 months ago

Log in to reply

You may not need to introduce too many of this in you problem. For example the answer can be just a + b \lfloor a+b \rfloor . After all we don't want to test members if they know about ceiling function.

Chew-Seong Cheong - 4 years, 8 months ago

Log in to reply

@Chew-Seong Cheong Okay. I'll take care of this.

Tapas Mazumdar - 4 years, 8 months ago
Spandan Senapati
Feb 12, 2017

The given sum equals sum(cosπ/2n)+isin(π/2n) for n=(1,2,3.....10).now call the cos part as f.so cos being a decreasing function and the first term being zero its easy to conclude that f<9.and f>(cos45+cos30+7cos22.5) now approximation gives f>8.so we conclude 8<f<9.similarly we can prove for the sin part.starting from back and making use of the increasing nature of sin function.

I don't see why a and b are transcendental numbers.

a and b are not transcendental. Each term i 1 / n i^{1/n} is algebraic, so their sum is algebraic; and the real and imaginary parts of an algebraic number are algebraic.

Rob Waters - 4 years, 8 months ago

Log in to reply

I see. I've changed it from transcendental \text{transcendental} to algebraic \text{algebraic}

Tapas Mazumdar - 4 years, 8 months ago

Maybe you're right. Can you express a a and b b as an expression containing rational numbers with only simple mathematical operations of addition, subtraction, multiplication, division, exponents and surds?

Tapas Mazumdar - 4 years, 8 months ago

Log in to reply

Honestly I didn't try.

Filippo Quattrocchi - 4 years, 8 months ago

Log in to reply

I solved those sums via aid of computing software. I found long non-recurring digits in the first 20 30 20-30 digits after the decimals, so I didn't put any thought behind it and presumed, they were transcendental (although they are definitely irrational, so they can be algebraic too) but I don't know.

Tapas Mazumdar - 4 years, 8 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...