. Sorry not to participate. I'm trying to stay away from this thread to get some work done.
Here’s my contribution to this great thread. Not a challenge but Funny and good for changing the mood as we get to the weekend
Likes:
takbetak and spinhead
And in 2025, they will remove the question all together, because calling a quadrilateral a rectangle is culturally insensitive toward all the other quadrilaterals who do not have the "right" angles.
Likes:
spinhead and Payandeh Iran
Here is a problem I came up with during my college years. I certainly hope someone has a better answer than me (assuming mine is even correct) as my answer is very long and convoluted. Here is the question:
You have a bow fixed into the ground and with an adjustable angle. If you shoot a first arrow at angle α1 at what angle (α2) should you shoot a second area t0 seconds later in order to hit the first arrow in mid air?
You have a bow fixed into the ground and with an adjustable angle. If you shoot a first arrow at angle α1 at what angle (α2) should you shoot a second area t0 seconds later in order to hit the first arrow in mid air?
I thought about this problem a lot and finally decided to murder it!
Here it is:
We know x<1, so we take the first two terms of the infinite series for cos(x) and ignore the higher order terms, to obtain:
1 - x^2/2 ~= x
where ~= means "almost equals to"
Solving for x, we obtain our initial estimate:
x(0) = sqrt(3) - 1
Now to find the estimate at iteration 1, x(1), take the first two terms of the Taylor series expansion of cos(x) about x(0):
x(1) ~= cos[x(0)] - sin[x(0)] ( x(1) - x(0) ) and solve for x(1) to obtain:
x(1) = { cos[x(0)] + x(0).sin[x(0)] } / { 1 + sin[x(0)] }
and so on for the subsequent iterations:
x(n+1) = { cos[x( n )] + x( n ).sin[x( n )] } / { 1 + sin[x( n )] }
So here's what I get for the first few iterations:
x(0) = .732050807568877
cos[x(0)] = .743805214977555
x(1) = .739096139913583
cos[x(1)] = .739077718925933
x(2) = .73907912080266
cos[x(2)] = .739089183235187
x(3) = .739085133223142
cos[x(3)] = .739085133209784
As you can see cos(x) is fucking close to x after only 3 iterations.
Here it is:
We know x<1, so we take the first two terms of the infinite series for cos(x) and ignore the higher order terms, to obtain:
1 - x^2/2 ~= x
where ~= means "almost equals to"
Solving for x, we obtain our initial estimate:
x(0) = sqrt(3) - 1
Now to find the estimate at iteration 1, x(1), take the first two terms of the Taylor series expansion of cos(x) about x(0):
x(1) ~= cos[x(0)] - sin[x(0)] ( x(1) - x(0) ) and solve for x(1) to obtain:
x(1) = { cos[x(0)] + x(0).sin[x(0)] } / { 1 + sin[x(0)] }
and so on for the subsequent iterations:
x(n+1) = { cos[x( n )] + x( n ).sin[x( n )] } / { 1 + sin[x( n )] }
So here's what I get for the first few iterations:
x(0) = .732050807568877
cos[x(0)] = .743805214977555
x(1) = .739096139913583
cos[x(1)] = .739077718925933
x(2) = .73907912080266
cos[x(2)] = .739089183235187
x(3) = .739085133223142
cos[x(3)] = .739085133209784
As you can see cos(x) is fucking close to x after only 3 iterations.
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Likes:
Gaines
Very nice. You are right, there is no closed form equation for the solution. So here is the back story behind this problem for me......
In the early-mid 70s in Iran, and my older brother had just bought a very fancy (for back in the early 70s) Canon calculator. And of course, he would not let us touch it. Except, when he was not in his room, we would go and find it (because he would hide it from us) and play with it pushing various buttons and functions to see what happens.
One of the things I discovered back then, was that regardless of what number you entered, if you hit the cos button over and over again, you would get the same number: 0.999847741 every time. Where as if I did the same thing with the sin button, you would get 0. It was later that I understood this activity to be an iterative solution to the x=cos(x) equation and that 0.9998.. was the answer to this problem.
So your answer is correct in Radians. In degrees the answer is what I wrote above.
==========================================
Recently, I googled this problem to see what else has been done. So just for laughs, I quote the most interesting answer from one of the websites:
https://www.tiger-algebra.com/drill/x-cosx=0/
Step by step solution :
STEP1Pulling out like terms
1.1 Pull out like factors :
x - xcos = -x • (cos - 1)
Equation at the end of step1:
STEP2:Theory - Roots of a product
2.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation:
2.2 Solve : -x = 0
Multiply both sides of the equation by (-1) : x = 0
Solving a Single Variable Equation:
2.3 Solve cos-1 = 0
In this type of equations, having more than one variable (unknown), you have to specify for which variable you want the equation solved.
We shall not handle this type of equations at this time.
One solution was found :
x = 0
In the early-mid 70s in Iran, and my older brother had just bought a very fancy (for back in the early 70s) Canon calculator. And of course, he would not let us touch it. Except, when he was not in his room, we would go and find it (because he would hide it from us) and play with it pushing various buttons and functions to see what happens.
One of the things I discovered back then, was that regardless of what number you entered, if you hit the cos button over and over again, you would get the same number: 0.999847741 every time. Where as if I did the same thing with the sin button, you would get 0. It was later that I understood this activity to be an iterative solution to the x=cos(x) equation and that 0.9998.. was the answer to this problem.
So your answer is correct in Radians. In degrees the answer is what I wrote above.
==========================================
Recently, I googled this problem to see what else has been done. So just for laughs, I quote the most interesting answer from one of the websites:
https://www.tiger-algebra.com/drill/x-cosx=0/
Step by step solution :
STEP1Pulling out like terms
1.1 Pull out like factors :
x - xcos = -x • (cos - 1)
Equation at the end of step1:
STEP2:Theory - Roots of a product
2.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation:
2.2 Solve : -x = 0
Multiply both sides of the equation by (-1) : x = 0
Solving a Single Variable Equation:
2.3 Solve cos-1 = 0
In this type of equations, having more than one variable (unknown), you have to specify for which variable you want the equation solved.
We shall not handle this type of equations at this time.
One solution was found :
x = 0
Here's one for your weekend enjoyment:
Javad and Sudabeh go to a party with four other couples. Each person there shakes hand with everyone he or she does not know. Later, Javad asks around and discovers that every one of the other nine shook hands with a different number of people.
How many people did Sudabeh shook hand with?
Javad and Sudabeh go to a party with four other couples. Each person there shakes hand with everyone he or she does not know. Later, Javad asks around and discovers that every one of the other nine shook hands with a different number of people.
How many people did Sudabeh shook hand with?
LOL. This reminds me of another puzzle I heard in school way back. Here it is....
حسن و حسین ۲۰ ریال داشتند. آنها دو پرتقال میخرند. پیدا کنید اسم پرتقال فروش را.
حسن و حسین ۲۰ ریال داشتند. آنها دو پرتقال میخرند. پیدا کنید اسم پرتقال فروش را.
Likes:
Payandeh Iran
For me the important point was that the numbers would range from 0 to 8. The maximum number anyone can shake hands with is 8. If all the numbers are unique, then then 9 numbers from everyone (given to Javad) is 0,1,2,3,4,5,6,7,8.
The next point was that there is a person with 0 handshakes. In that case, the spouse of that person must be the one with 8 handshakes. Next and Similarly, the person with 1 handshake must be married to the one with 7, and so on. Then there is a couple of 4 & 4 between them. That couple must be Javad and Soodabeh, because if it was some other couple, then the numbers given to Javad, would not have been unique.
As you mentioned, there may be an argument based on symmetry as well, but I have not worked it out.
The next point was that there is a person with 0 handshakes. In that case, the spouse of that person must be the one with 8 handshakes. Next and Similarly, the person with 1 handshake must be married to the one with 7, and so on. Then there is a couple of 4 & 4 between them. That couple must be Javad and Soodabeh, because if it was some other couple, then the numbers given to Javad, would not have been unique.
As you mentioned, there may be an argument based on symmetry as well, but I have not worked it out.
Likes:
spinhead