October 2023
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In the Math Corner of my October 2023 Diary I posted the following brainteaser.
Define SigN to be sin(1) + sin(2) + sin(3) + … + sin(N). The numbers there are all radians, of course.
Prove that −1/7 < SigN < 2 for any positive whole number N.
(Credit here belongs, I think, to Sam Walters.)
The nifty thing about this result — at any rate to me — is that it's not what you'd expect at a first glance. Given the apparently random y-coordinates of whole-number radians along a graph of the π-periodic sine function, you'd expect the lower and upper limits on the sum to be symmetrical about the x-axis.
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• Solution
The keys to a solution are familiar identities from algebra, trig and complex-number arithmetic.
From algebra:
(i) (1 − xN+1) / (1 − x) = 1 + x + x2 + x3 + … + xN
From trig:
(ii) cos² A + sin² A = 1
(iii) cos 2A = 1 − 2 sin² A
(iv) sin (A + B) = sin A cos B + cos A sin B
From complex-number arithmetic:
(v) eiθ = cos(θ) + i sin(θ)
(vi) (a + ib) (a − ib) = a² + b²
Substituting ei for x in (i) and expanding with (v), it easily follows that what I have called SigN is just a way of writing the imaginary component of
(vii) (1 − ei(N+1)) / (1 − ei)
So is there another way of writing the imaginary component of (vii) that will give us the inequalties we seek?
First, let's express the denominator of (vii) as a plain complex number. ei is just cos(1) + i sin(1). The denominator of (vii) is therefore (1 − cos(1)) − i sin(1).
With (vi) in mind, let's now multiply the top and bottom of (vii) — both the numerator and the denominator — by (1 − cos(1)) + i sin(1), to get rid of that pesky i in the denominator. Using (vi), (ii), and (iii), the denominator resolves into 4 sin²(½).
So the denominator is no longer a complex number. It's a real number, 0.91939538826372…
We now have to compute the numerator of (vii) — which, remember, we just multiplied by (1 − cos(1)) + i sin(1) — then, when we've figured out its imaginary component, we'll divide it by that 4 sin²(½).
That numerator is now (1 − cos(1)) + i sin(1) times (1 − ei(N+1)). Moving from the exponential to its trig equivalent via (v), that is
((1 − cos(1)) + i sin(1)) × (1 − (cos(N+1) + i sin(N+1))).
What is the imaginary component of that? Well, we just multiply out and ignore real terms. That, with help from (iv), gets us:
sin(1) − sin(N + 1) + sin(N).
We can further reduce the number of occurrences of N in that expression by rewriting it as
sin(1) − sin((N + ½) + ½) + sin((N + ½) − ½)and then using (iv) again:
sin(1) − (sin(N + ½) cos(½) + cos(N + ½) sin(½)) + (sin(N + ½) cos(½) − cos(N + ½) sin(½))
That cancels down to
sin(1) − 2 cos(N + ½) sin(½)
Since the value of the cosine — of any cosine — may range from −1 to 1, that expression can go from sin(1) − 2 sin(½) to sin(1) + 2 sin(½), which is to say from −0.11738009240051 to 1.80032206201630. Dividing at last by our denominator 4 sin²(½), the limits for SigN are
−0.12767096061052 < SigN < 1.95815868232297
Since −1/7 is −0.14285714285714…, it is certainly the case that −1/7 < SigN < 2 , with room to spare. QED.