Challenge 75: Limit, Arithmetic Mean, and Trig

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• Published on Jan 10, 2019
• Congratulations to Vampianist3, Beshoy Nabil, Hiren Bavaskar, Aswini Banerjee, Daulian Doge, Rishav Gupta, adandap, and infrustration for successfully solving the last week's math challenge question! Vampianist3 was the first person to solve the question.
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Welcome, everyone! My channel hosts one weekly math challenge question per week (made by either myself, my family, or my friends), which will be posted every Wednesday. Please comment your proposed answer and explanation below! If you are among the first ten people with the correct answer, you will be recognized in the next math challenge video. The solution to this question and new question will be posted next Wednesday.
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• After simplication , expression reduces to-
Lim(n->0) sum(k=1 to n) [sin(ln(k/n))]/n which then can be reduced by " Integration as sum of limits " to
=Integration of sin(lnx) dx from 0 to 1
= -(x(cos(logx)))/2 - sin(logx) from 0 to 1
= -1/2
-Thank you for the question

• UbuntuLinux 4 months ago

My link on imgur was broken so I have to upload it to another page, thus I have to edit it :((

• Johannes H 5 months ago +1

Maybe there should be a hidden way to submit one's solution (where you can also use mathy (LaTeX) expressions)
After posting a wrong answer and learning by the correct answers there is no point to post another same correct solution.
Probably that's the reason there are not more than 10 people posting their solution...

• Massimo Aster 5 months ago

Ok, so I started by saying that a(n)bar=(1/n)×{sin[ln(1)]+sin[ln(2)]+...sin[ln(n)]} and b(n)bar=(1/n)×{cos[ln(1)]+cos[ln(2)]+...+ln cos[ln(n)]}. Now you have the limit of the first series a(n)bar times cos[ln(n)] minus the second series b(n)bar times sin[ln(n)] as n approaches infinity. Or: lim n->infinity of (1/n)×{sin[ln(1)]×cos[ln(n)]+sin[ln(2)]×cos[ln(n)]+... -cos[ln(1)]×sin[ln(n)]-cos[ln(2)]×sin[ln(n)]-...}. Now you can change the order of addition and subtraction and group up corresponding sines and cosines: sin[ln(1)]×cos[ln(n)]-cos[ln(1)]×sin[ln(n)]. You can do this for all other values for n (2,3,4,5,...) as well! Now you may recall the First Theorem of the Addition of sines: sin(a+b)=sin(a)×cos(b)+cos(a)×sin(b). The connection gets a little more obvious by using the fact that sine is an odd and cosine is an even function to end up with sin(a-b)=sin(a)×cos(b)-sin(b)×cos(a). Comparing this with our problem, we see that sin[ln(1)]×cos[ln(n)]-cos[ln(1)]×sin[ln(n)] is really just equal to sin[ln(1)-ln(n)]. We can now simplify this even further due to the fact that the subtraction of two logarithms with the same base is in fact equal to the quotient of the numeri(the numbers inside of the logarithms). So we end up with sin[ln(1/n)]. We have now finally reduced the limit to the lim n->infinity of (1/n)×{sin[ln(1/n)]+sin[ln(2/n)]+sin[ln(3/n)]+...}. To continue from here on, you will need to know how to integrate functions over the intervall [0;1]. You start by splitting up the intervall in n pieces, so that each piece has length 1/n. This 1/n is the base of all of our n rectangles. The height of of the rectangles however depends on the y or f(x) value. We start at 0, then go 1/n steps to the right. At 1/n, the y value is f(1/n). You then do the same thing again, but now you have f(2/n) as the height. You will correctly find out the area/express the integral, when n goes to infinity. Therefore we have: Integral from 0 to 1 of f(x)dx is equal to lim n->infinity of (1/n)×[f(1/n)+f(2/n)+f(3/n)+...]. Comparing this limit to our original limit, we can conclude, that our limit is nothing more than the integral from 0 to 1 of sin[ln(x)]dx. You can use a u-substitution, letting ln(x)=u. Therefore we get that x=e^u and dx=e^u×du. The bounds on our integral now change to negative infinity to 0 (e^u=0 only works if u=-infinity and e^u=1 only works if u=0).We now have a new integral from negative infinity to 0 of sin(u)×e^u×du. You can now use the DI-method for example to get that the integral is equal to e^u×[sin(u)-cos(u)]/2. Finally you can plug in the important values/the lower and upper bound: e^0×[sin(0)-cos(0)]/2-e^-infinity×[sin(-infinity)-cos(-infinity)]/2. 1×(0-1)/2-0×something=-1/2. (sine and cosine of negative/positive infinity are not defined, however, we know that they must result in a number between -1 and 1. Therefore we can in this case savely say that 0×something is in fact 0!). We now know: lim n-> infinity of a(n)bar×cos[ln(n)]-b(n)bar×sin[ln(n)]= (-1/2)!!!!!

• Massimo Aster 5 months ago

Also, didn't mean to write 0!, it's of course 0.

• Massimo Aster 5 months ago +1

And, no, I'm not taking five consecutive factorials of -1/2 at the end. 😂😂

• mark erena 5 months ago

The limit simplifies to:
Limit as n->infinity of 1/n*sigma of k=1 to n of sin(ln k - ln n). When n aproaches infinity the sum aproaches 0 and 1/n aproaches also 0 and multiplied toghether it gets you 0

• Laurent Eriksen 5 months ago

I used MATLAB to plot the first 123456 elements of that series and see that it's -1/2. Did I win? :)
Plot command if you're curious:
plot(1:123456,cumsum(sin(log(1:123456)))./(1:123456).*cos(log(1:123456)) - cumsum(cos(log(1:123456)))./(1:123456).*sin(log(1:123456)));
Isn't that just beautiful?

• Nathan Müller 5 months ago +4

We have (sin(ln(1))+sin(ln(2))+...+sin(ln(n)))cos(ln(n))/n - (cos(ln(1))+cos(ln(2))+...+cos(ln(n)))sin(ln(n))/n
Multiplying we have:
(Sin(ln(1))cos(ln(n))+...+sin(ln(n))cos(ln(n)) - (sin(ln(n)cos(ln(1)+...+sin(ln(n))cos(ln(n-1))+sin(ln(n))cos(ln(n)))/n
Then rearranging the terms and using the identity sin(a+b)=sin(a)cos(b)+sin(b)cos(a) we have:
(Sin(ln(1) - ln(n)) + sin(ln(2) - ln(n)) +...+ sin(ln(n-1) - ln(n)))/n
The using that ln(a) - ln(b) = ln(a/b) we have:
(Sin(ln(1/n)) + sin(ln(2/n)) + ... + sin(ln((n-1)/n))))/n
That is a Riemman sum, so we have the solutions is equal to:
Integral from 0 to 1 of sin(ln(x))dx
Doing the indefinite integral first:
Calling ln(x)=u then dx=e^u du
The we have:
Integral(sin(u)e^u du)
Integrating by parts:
Integral(sin(u)e^u du)= sin(u)e^u - integral(cos(u)e^u du)
Then doing integral(cos(u)e^u du) by parts:
integral(cos(u)e^u du)= cos(u)e^u + integral(sin(u)e^u du)
Then we have:
Integral(sin(u)e^u du) = sin(u)e^u - cos(u)e^u - integral(sin(u)e^u du)
Rearranging the terms we have:
Integra(sin(u)e^u du) = e^u(sin(u) - cos(u))/2 + C
Then doing the substitution u=ln(x) we have:
Integral(sin(ln(x))dx) = x(sin(ln(x)) - cos(ln(x)))/2 + C
Then doing the definite integral we have:
Integral(sin(ln(x))dx) from 0 to 1 = (sin(ln(1)) - cos(ln(1)))/2 - 0
Then:
Integral(sin(ln(x))dx) from 0 to 1 = (sin(0) - cos(0))/2
Then:
Integral(sin(ln(x))dx) from 0 to 1 = -1/2
🐒

• Isaac YIU Math Studio 5 months ago +1

Hi! I’m the first time to answer the question.
The limit is -0.5
Firstly, using sinacosb-cosasinb=sin(a-b)
We can arrange the question into lim(n to infinity)1/n*(Sum (k=1 to infinity) sin(ln(k/n)))
Then, by definite integral, it is equals to integral from 0 to 1 (sin(lnx)) dx
Finally , by substituting u=lnx, we can find the answer is -1/2( or -0.5)
That is a great challenge!

• Isaac YIU Math Studio 5 months ago

karun mathews
So Congratulations!!!

• karun mathews 5 months ago

Thanks for that first hint, of sinacosb - cosasinb , things
really simplified for me after that and I could solve it- first time I solved one of these problems too!

• The limit is -1/2. We can start by Writing out the AM expression for a_n(bar) and b_n(bar), combine similar terms to form sum of the form: sum_{i=1} ^ {n} (sin(ln(i/n)))/n. This is the Riemann sum of the integral, int_{0}^{1} sin(ln(x)) dx. Which is the same as im(int_{0}^{1} x^i dx). Calculating the integral, yields the value of -1/2.

• Blan Morrison 5 months ago

Holy moly it hasn't even been out for half an hour and there are already over 5 solutions... I'm out.

• Zamir C 5 months ago

• Limit is -1/2
Writing the definition of arithematic mean, we get
{[sin(ln1)+sin(ln2)...sin(ln n)]*cos(ln n) - [cos(ln1)+...+cos(ln n)]*sin(ln n) }/n
Now regroup terms(Guassian pairing) ,ie {[sin(ln1)cos(ln n) -cos(ln1)sin(ln n)]+...+[sin(ln n/2)cos(ln n/2 +1) - cos(ln(n/2)sin(ln n/2+1)]} /n
Since n->infinity, we can approximate n-1,n-2.. to be approximately n and then using formula of sinAcosB -cosAsinB=sin(A-B)..
we get lim n->infinity 1/n [summation from r=1 to n of sin(ln(r/n))] which is definite integral of sin(lnx) from 0 to 1 . This integral can be done by parts (twice) and we get
the integral I= x/2[ sin(lnx) -cos(lnx)] from 0 to 1, which is -1/2 and you are done :)

• Jaleb 5 months ago

Given both a(bar)_n and b(bar)_n are an average of n terms we can factor out 1/n and get:
lim(1/n sum(k=1,n)(a_k cos(ln n) - b_k sin(ln n)).
By looking at a specific case of k we find:
sin(ln k)cos(ln n)-cos(ln k)sin(ln n) = sin(ln k - ln n) = sin(ln (k/n))
Placing it back into the original equation we get
1/n(sin(ln(k/n)))
Taking the sum gives the integral_0^1 sin(ln(x)) dx
By using Euler's formula we can integrate into the form with sin(ln(x)) = im(x^i):
integral_0^1 Im(x^i) dx = Im(integral_0^1 x^i dx) = Im((1-i)/2) = -1/2

• Smokie Bear 🔴🔵 5 months ago

If I take the time to compute the value of the expression in the limit when n = 10, I get -0.527 and when I plug in n=9 I get - 0.526 and when I plug in n=8 I get -0.525 and when I plug in n=11 I also get -0.527 and when I plug in n=13 I also get -0.527 so I’m guessing that the answer will be negative 0.5 because the difference in the values from n=8 to n=13 is very small so as it goes to infinity the value won’t change drastically.

• Nicholas Patel 5 months ago

This is going to take a lot of writing so some parts are sloppy (I apologise) but I think the answer is -1/2
Using sigma notation helps a lot here, so I first wrote the limit as ((sum x=1 to inf sin (lnx)) cos (lnn) )/n - ((sum x=1 to inf cos(lnx)) sin(lnn))/n
Next combine the summations and use the identity 2sinAcosB = sin(A+B) + sin(A-B) to split the first part of the summation to sigma sin(ln(nx))+sin(ln(x/n)) and the second part to -sin(ln(nx)) and -sin(ln(n/x)). Two of these terms in the summation cancel and we are left with lim n->inf sigma((sin(ln(x/n))-sin(ln(n/x)))/2n. Then using the trig identity sinA - sinB = 2cos((A+B))/2)sin((A-B)/2) and log properties lnA+lnB=lnAB, and that cos0=1, the summation nicely simplifies to sigma sin(x/n)/n. And this is just the integral of sin(lnx) dx from 0 to 1, with the function f(x)=sin(lnx) and the ‘gap between bars’ tending to 0 as n-> inf. To solve this we can consider the imaginary part of integral x^i dx from 0 to 1, which evaluates to 1/(1+i), and the imaginary part of this is -1/2

• 123 gogo 5 months ago

limit L
= lim (1/n) * [ (sin(ln 1) + ... + sin(ln n)) * cos(ln n) - (cos(ln 1) + ... + cos(ln n)) * sin(ln n)) * sin(ln n) ]
= lim (1/n) * [ (sin(ln 1)*cos(ln n) - cos(ln 1)*sin(ln n)) + ... + [(sin(ln n)*cos(ln n) - cos(ln n)*sin(ln n)) ]

• Kwekinator117 5 months ago

Rearranging the terms, we obtain (sinln1cosln1-cosln1sinln1 + ... + sinlnncoslnn-coslnnsinlnn)/n. Using the formula for sin(A+B), we can simplify this to sinln(1/n)/n +sinln(2/n)/n +...+sinln(1)/n.
Since n approaches infinity, this sum is simply the integral of sinlnx from x=0 to x=1. The integral of this function is (1/2)*x* (sinln(x)-cosln(x)). At x=1, the function is -1/2. The limit of the function as x approaches 0 is 0. Therefore the limit of the given sum is -1/2.

• PRAKHAR AGARWAL 5 months ago

Write cos(ln(n)) as (e^(iln(n)) +e^(-iln(n))) /2
And write sin(ln n) using the same euler theorem as (e^(iln(n)) - e^(-iln(n))) /2i
Thus obtain a sigma expression for A(n) and B(n)
For the mean divide the sum by n
Now to evaluate tge expression write the means in sigma and once again for the cos(ln n) and sin (ln n) use euler formula
Finally take the sum as n tends to infinity and we can then use the Reimann Summation to approximate it as an integral
Solving the integral leaves us with - 1/2

• Parth Pawar 5 months ago

Let AM of sin(lnx) series and of cos(lnx) series be A and B respectively. Then,
nA=(sin(ln1)+sin(ln2)+...+sin(ln n))cos(ln n)
nB=(cos(ln1)+...+cos(ln n))sin(ln n)
nA-nB=((sin(ln1)cos(ln n)-cos(ln1)sin(ln n)) + ...)
=(sin(ln1-ln(n)) + sin(ln2 -ln(n)) +...)
=∑(r=1 to n) (sin(ln(r/n))
Therefore lim n tending to infinity = (1/n)(∑(r=1 to n) (sin(ln(r/n)))
= integration of sin(lnx) from 0 to 1
Put lnx =t. Then dx=e^t dt
Hence I=(-infinity to 0) of e^xsinx. This can be easily integrated using integration by parts. Answer comes out to be -1/2

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