Another Math qeuestion
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Basic said:
That I got that the both go into 399 but I don thing that is right as I am using a homework program which tells me it is wrong.
My calc gives me 2pi but the program says its wrong
Edit: I give up I need to go to bed to catch my 3 1/2 hours of sleep.
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OK, I'm back.
A prime number is an integer that is evenly divisable by 1 and itself (of course), but not by any other integers.
A prime factorization is to write an integer as a product of prime numbers.
Every integer has a unique prime factorization.
A prime factorization is usually done the brute force way, with trial and error.
Divide the number with prime numbers (2, 3, 5, 7, 11, 13, 17,...), and then keep doing that to the result until you've get a prime number.
Ie: prime factorization of 1638
1638 / 2 = 819
819 / 2 = 409.5 // Not an int, try next prime.
819 / 3 = 273
273 / 3 = 91
91 / 3 = 30.33...
91 / 5 = 18.2
91 / 7 = 13
13 / 7 = 1.86... // Less than the number we divided with (7), so we stop now.
So 1638 = 2*3*3*7*13
Your kind of problem (different figures):
Find smallest integer solution to: 1638a = 2541b
1638a = 2541b
// Prime factorization
2*3*3*7*13*a = 3*7*11*11*b
// Divide by common factors
2*3*13*a = 11*11*b = c
// Since all integers has a unique pime factorization, c must have all the factors on the left.
c = 2*3*11*11*13*d
// We want the smallest a and b => smallest c => d = 1
// So a and b are "the missing factors":
a = 11*11 = 121
b = 2*3*13 = 78
A prime number is an integer that is evenly divisable by 1 and itself (of course), but not by any other integers.
A prime factorization is to write an integer as a product of prime numbers.
Every integer has a unique prime factorization.
A prime factorization is usually done the brute force way, with trial and error.
Divide the number with prime numbers (2, 3, 5, 7, 11, 13, 17,...), and then keep doing that to the result until you've get a prime number.
Ie: prime factorization of 1638
1638 / 2 = 819
819 / 2 = 409.5 // Not an int, try next prime.
819 / 3 = 273
273 / 3 = 91
91 / 3 = 30.33...
91 / 5 = 18.2
91 / 7 = 13
13 / 7 = 1.86... // Less than the number we divided with (7), so we stop now.
So 1638 = 2*3*3*7*13
Your kind of problem (different figures):
Find smallest integer solution to: 1638a = 2541b
1638a = 2541b
// Prime factorization
2*3*3*7*13*a = 3*7*11*11*b
// Divide by common factors
2*3*13*a = 11*11*b = c
// Since all integers has a unique pime factorization, c must have all the factors on the left.
c = 2*3*11*11*13*d
// We want the smallest a and b => smallest c => d = 1
// So a and b are "the missing factors":
a = 11*11 = 121
b = 2*3*13 = 78
No, pi/133 = pi/(7*19)
If I compensate for such an error in the first row, I guess you 're saying:
2pi/126 = pi/(3*3*6*7)
But that's not true either. It's a good idea to check answers of "difficult" calculations, specially when the checking is easy (like here). 3*3*6*7 = 378.
And 6 is not a prime number, since it equal to 2*3.
Second, when you get the right a and b, you'll need to go back and see what they came from. Why did I write the equation that way? What does the left and right hand side mean?
I'll gve you some more help with finding a and b. (OK, actually I'll solve that part for you, since it seems like you've at least tried it.)
a*pi/133 = b*2*pi/126 // Reduce the 2 on RHS
a*pi/133 = b*pi/63 // Multiply both sides with 133 and 63, divide by pi
a*63 = b*133 // Prime factorization
a*3*3*7 = b*7*19 // Divide by common factors
a*3*3 = b*19
a = 19
b = 3*3 = 9
Mintmaster
Veteran
I think Basic is trying to teach you a bit too much here - giving you hints so you can solve both this problem and anything else like it.
Okay, so you know that when x = 2pi, the function will repeat, i.e. f(x) = f(x + 2pi). The question is whether it will repeat any earlier.
Think about dividing a line that's 2pi long into 266 parts (i.e. 2*133) in blue pen, and also into 126 parts in red pen. Where will the blue and red pen marks overlap? The answer is that it will happen n times over that length, where n is the greatest common factor. Now the distance between overlapping marks (i.e. the period of f) is 2pi/n.
Okay, so you know that when x = 2pi, the function will repeat, i.e. f(x) = f(x + 2pi). The question is whether it will repeat any earlier.
Think about dividing a line that's 2pi long into 266 parts (i.e. 2*133) in blue pen, and also into 126 parts in red pen. Where will the blue and red pen marks overlap? The answer is that it will happen n times over that length, where n is the greatest common factor. Now the distance between overlapping marks (i.e. the period of f) is 2pi/n.
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ShootMyMonkey
Veteran
To simplify, you can take
a * pi/133 = b * 2pi / 126
a * pi/133 = b * pi / 63, pi's cancel out...
a/133 = b/63, or.. a * 63 = b * 133
Now you solve for the minimum integer values for which that is true. Once you've found that, you should be able to plug in the values to either side, and that will give you the period.
If I did this right (it's been a while), you should get, a = 19, b= 9. Which in turn gives you a period of pi/7.
EDIT : My method to get to 19 & 9 was slightly different from Basic's... I used
a = b * 133/63.
133/63 = 2.1111111111111 or 2 1/9
That suggests a factor of 9 is necessary to scale up to integer values, which means b must at least be divisible by 9...
a = 9*k*(2.11111111)
a = 19*k....
Hmmmm... 19 is prime as it is... so we can assume that k is 1.
Which means that b, which was substituted as 9*k = 9*1 = 9.
a * pi/133 = b * 2pi / 126
a * pi/133 = b * pi / 63, pi's cancel out...
a/133 = b/63, or.. a * 63 = b * 133
Now you solve for the minimum integer values for which that is true. Once you've found that, you should be able to plug in the values to either side, and that will give you the period.
If I did this right (it's been a while), you should get, a = 19, b= 9. Which in turn gives you a period of pi/7.
EDIT : My method to get to 19 & 9 was slightly different from Basic's... I used
a = b * 133/63.
133/63 = 2.1111111111111 or 2 1/9
That suggests a factor of 9 is necessary to scale up to integer values, which means b must at least be divisible by 9...
a = 9*k*(2.11111111)
a = 19*k....
Hmmmm... 19 is prime as it is... so we can assume that k is 1.
Which means that b, which was substituted as 9*k = 9*1 = 9.
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Basic said:
I see what I did. I didn't mess up the calc b just became 6 because of bad handwriting.
So It should be 9 and 19. But Shoot how do I know which period to use as that is the period of the first function.
Edit: Nevermind they are both equal now so it means it should be the period.
What is with the prime thing anyway? Once I knew to do it I could but before now I've never heard of it.
Oh and thanks for all your help now I just need to remember trig identites in case I need to simplify functions when doing limits, which are quite easy. My problem with Caculas has always been remembring the trig ident to simplify and knowing when I can use them.
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No, it's not the period, it's the number of periods. You've already (correctly) calculated the period of the two parts.
[Edit]Oh, I think I misunderstood you there, you didn't mean that 9 or 19 where the period.[/Edit]
OK, I'll give you the rest too.
Look back at the third question in my first post:
"What's the smallest amount of periods of each before they are in sync?"
The two parts are in sync when you take a periods of the first part, and b periods of the second part, and end up in the same spot. And that's the equation:
a*pi/133 = b*2*pi/126 = d
Where d is the spot where the parts are in sync again -- the period of f(x).
19*pi/133 = 9*2*pi/126 = pi/7
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Basic said:
Yep thats what I did to get the period. Didn't you know that was what I meant? I meant exactly that 19pi/133 = 18pi/126 therefore you could use either to calculate the period.
Did you miss the point where I was thanking you guys because the program excepted the answer as correct.
