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Mathematics 22 Online
OpenStudy (lovelyanna):

Fill in the missing information in the problem to find ^115 (square root of 115) by using the Babylonian method. Let 10.5 be your initial guess.

OpenStudy (lovelyanna):

OpenStudy (lovelyanna):

A. 115 B. 2 C. 10.5 D. 10.8 C?

OpenStudy (lovelyanna):

I mean D.. my bad

OpenStudy (whpalmer4):

the Babylonian method for taking square roots is done by taking an initial guess (here, 10.5) and then repeatedly averaging the latest guess with the quotient of the number whose square root you are taking and the current guess. when successive guesses are close enough together, you stop. did my putting it in words confuse you? I'm trying not to just give you the answer :-)

OpenStudy (lovelyanna):

Ya.. it kinda did, can you but it in numbers instead? :P

OpenStudy (lovelyanna):

put*

OpenStudy (whpalmer4):

let's say we are taking the square root of 16, and our initial guess is 10. kind of a poor guess, but that's okay, this method converges to the answer quickly. if our first guess is 10, our next guess is the average of 10 and 16/10. 16/10 = 1.6, 10 + 1.6 = 11.6, 11.6/2 = 5.8. 5.8 is our next guess. now we repeat the process. our third guess is the average of 5.8 and 16/5.8. 16/5.8 = 2.76, approximately. 5.8 + 2.76 = 8.56. divide that by 2 and we get 4.28 as our fourth guess. average of 4.28 and 16/4.28 = (1/2) * (4.28 + 16/4.28) = (1/2)(4.28+3.74) = 1/2*8.02 = 4.01 well, you know that the square of 16 is 4, and hopefully it's obvious that this is zeroing in 4... so there was an example of the process, but with different numbers. have a look at it, and tell me the correct answer for your problem :-)

OpenStudy (whpalmer4):

that should have been "the square ROOT of 16 is 4"

OpenStudy (lovelyanna):

Was D correct? @whpalmer4

OpenStudy (whpalmer4):

:-( to make the next guess, you average the current guess with the quotient of the number and the current guess. what is the current guess in the problem diagram?

OpenStudy (whpalmer4):

alternatively, you could just write out the problem diagram with an X instead of a ? and solve for x! 10.5 + (115/x) ------------- = 10.7262 2

OpenStudy (lovelyanna):

About 10.73..?

OpenStudy (whpalmer4):

10.5 + (115/x) ------------- = 10.7262 2 10.5 + (115/x) = 2*10.7262 115/x = 2*10.7262 - 10.5 115 = x(2*10.7262 - 10.5) x = 115/ (2*10.7262 - 10.5) x = 115/10.9524 x =

OpenStudy (lovelyanna):

about 10.5

OpenStudy (lovelyanna):

Oh, I get it!

OpenStudy (whpalmer4):

but really, it's better that you understand what is going on (which solving that equation doesn't provide). Let me try one more explanation. If you are taking the square root of N, and your current guess is Sn, the next guess is (Sn + N/Sn) ----------- 2 so yes, 10.5 is the answer we were looking for :-)

OpenStudy (lovelyanna):

So this would be C

OpenStudy (whpalmer4):

yes! your original answer :-)

OpenStudy (lovelyanna):

I dont get the Sn stuff.. :/

OpenStudy (whpalmer4):

If we call the first guess S0 (I'd like that to be a subscript 0, but OpenStudy is broken right now, so I can't format nicely), then the next guess is S1, S2, etc, etc. If we want to describe the rule for find "the next guess" or S<n+1> where n is some number, and our current guess is S<n> S<n+1> = S<n> + N/S<n> --------------- 2 which is just a formula for "to find the next guess, average the current guess, S<n> and the quotient of N and the current guess, S<n>"

OpenStudy (lovelyanna):

Oh, ok.. I get it a bit better, thanks :) Can you check my next question?

OpenStudy (whpalmer4):

OpenStudy (whpalmer4):

If you compare that with the original problem, does it make sense?

OpenStudy (lovelyanna):

Use the Babylonian method for your next estimate of ^115 by using 10.7262 as your guess. What is your result? A. 16.0869 B. 10.6107 C. 10.7238 D. 10.8393 Is C correct? Yes or no? @whpalmer4

OpenStudy (whpalmer4):

Here we have our current guess (S<n>) is 10.7262. so our next guess (S<n+1) is: S<n+1> = (10.7262 + (115/10.7262))/2 And if you plug that into the calculator, you get 10.7238. Good job!

OpenStudy (whpalmer4):

What happens if you do another round, using 10.7238 as your guess?

OpenStudy (lovelyanna):

what do you mean?

OpenStudy (whpalmer4):

do the process again. S<n+2> = (S<n+1> + 115/S<n+1>)/2 S<n+2> = (10.7238 + 115/10.7238)/2 S<n+2> = ?

OpenStudy (lovelyanna):

My head hurts.. can you explain more?

OpenStudy (whpalmer4):

think of this as a machine. you put in one guess, turn the crank, and it gives you a better guess. You take the better guess, put it back in the machine, turn the crank, and you get an even better guess. Eventually, the guess will be as close as you want to the exact answer.

OpenStudy (lovelyanna):

Oh, I see

OpenStudy (whpalmer4):

our first guess was 10.5. the machine took that, and gave us 10.7262. we gave the machine 10.7262, and it gave us 10.7238 now we give the machine 10.7238, and what does it give us? by the way, 10.5*10.5 = 110.25 10.7262*10.7262 = 115.051 we're getting closer to the exact square root of 115!

OpenStudy (whpalmer4):

as your head hurts, I'll do the arithmetic for you: (10.7238 + 115/10.7238)/2 = (10.7238 + 10.7238)/2 = 10.7238 !!!!

OpenStudy (whpalmer4):

when our guesses are the same, or close enough for our needs, we stop. here we've gotten the same number back, so we stop, and the square root of 115 is 10.7238, at least we are only looking at 4 decimal places. Really, it's closer to 10.7238052947636083048141596721542847050902118555502875564499023281566\ 5531270700103860671618718519263 but what are a few extra digits between friends? :-)

OpenStudy (lovelyanna):

Oh, sorry, I gtg now! Thanks alot for your help!!

OpenStudy (whpalmer4):

okay, good night!

OpenStudy (tester97):

good job explaining @whpalmer4 ;)

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