n = 1 ∑ 1 0 i n 1 = a + b i
In the equation above, a and b are positive real numbers . Find the value of ⌊ a ⌋ − ⌈ b ⌉ .
Notations:
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Tapas, this one you need to explain three notations. I have done that for you.
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Oops. I'll really take care of that from now.
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You may not need to introduce too many of this in you problem. For example the answer can be just ⌊ a + b ⌋ . After all we don't want to test members if they know about ceiling function.
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@Chew-Seong Cheong – Okay. I'll take care of this.
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 is algebraic, so their sum is algebraic; and the real and imaginary parts of an algebraic number are algebraic.
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I see. I've changed it from transcendental to algebraic
Maybe you're right. Can you express a and b as an expression containing rational numbers with only simple mathematical operations of addition, subtraction, multiplication, division, exponents and surds?
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Honestly I didn't try.
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I solved those sums via aid of computing software. I found long non-recurring digits in the first 2 0 − 3 0 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.
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n = 1 ∑ 1 0 i n 1 = n = 1 ∑ 1 0 e 2 n π i = n = 1 ∑ 1 0 ( cos 2 n π + i sin 2 n π ) = 8 . 3 4 2 2 + 3 . 9 0 5 3 i
⟹ ⌊ a ⌋ − ⌈ b ⌉ = ⌊ 8 . 3 4 2 2 ⌋ − ⌈ 3 . 9 0 5 3 i ⌉ = 8 − 4 = 4