Broken Calculator
While giving my Olympiad exam, i found out that root sign in my calculator* was not working. Can i still use it to find the root of a number ?
* similar to standard calculator which comes in Windows OS
* similar to standard calculator which comes in Windows OS
Labels: mathemagic, thinktank
posted by Rajesh Lal at
9:04 AM
20 Comments:
We old engineers use sliderules and don't have this problem. Of course you could use the log function, convert the number to a log and then do simple math to find the root
What? Use the root sign? No, it's broken.
You could simply raise the number to the power of 0.5
If your calculator has an x^y button, you can press [number] x^y 0.5
well the sign is broken, so the button still works. ergo - YES you can.
well the sign is broken, so the button still works. ergo - YES you can.
Depending on what buttons the calculator DOES have, I agree with the above people that raising the number to 1/N will give the Nth root of the number.
Additionally, the use of natural logs can be quite helpful here.
We all know that IF
a^b=c, THEN
b=ln(c)/ln(a)
Assuming the calculator calculates natural logs and raises e to the nth power, then the math is quite simple. Since we know a to be some X, and b to be some fraction 1/N, then voila:
e^(ln(a)/N)= C
example:
sqrt(256)=16
e^(ln(256)/2)=16
however, even the Microsoft default scientific calculator lacks an e^x button, so I don't know if anyone would ever use such a roundabout way of determining a root.
THE ANSWER IS: YES.
We all know that there is a way to solve any mathematical equation without a calculator. That said, if you do have one then it wouldn't matter how limited it's functionality may be. It would still make the calculations that can be performed on said calculator that much faster. The question is pretty clearly of the "yes or no" variety. No showing of work is required on this one, sorry Daniel.
"Clarification" about the calculator
The calculator is similar to the standard calculator which comes with the Windows OS (does not have button for power or log)
Why would you need to get the root of a number if you were giving the test not taking it? or is that just an error in the question?
If by root, you only refer to square root. The answer is YES.
square root of a Number can be found using the +,- and x signs.
As a matter a fact, it can be found without using Calculator.
if time isn't a concern, estimate and check is a pretty good way to go about it.
ie, 115 is larger than 100 but less than 121, so it's near the middle of 10 - 11
if time isn't a concern, estimate and check is a pretty good way to go about it.
ie, 115 is larger than 100 but less than 121, so it's near the middle of 10 - 11
E.g: squareroot of 26115:
______________2 61 15______________
(1) x(1)= 1 ()x()=.or< 2
----------------------
______________1 61 15______________
2(6) x(6)= 1 56 1+1=2,
---------------------2()x()=.or<161
______________ 5 15______________
32(1) x(1)= 3 21 26+6=32
---------------------
______________ 1 94 00
322(6) x(6)= 1 93 56
------------------------
______________ 44 00
3232(0)x(0)= 0
----------------------------
______________ 44 00 00
32320(1)x(1) 32 32 01
------------------------------
So, sq.root 26115=161.601_____
Sorry, this look better:
E.g: squareroot of 26115:
______________2 61 15______________
(1) x(1)=___.___1__________()x()=.or.< 2
--------------------------------
______________1 61 15______________
2(6) x(6)=__.___1 56_______1+1=2,
----------------------------_____2()x()=.or<161
________________ 5 15______________
32(1) x(1)=___.___3 21_____26+6=32
---------------------------------
________________ 1 94 00
322(6) x(6)=______1 93 56
---------------------------------
____________________ 44 00
3232(0)x(0)=_____________0
---------------------------------
___________________ 44 00 00
32320(1)x(1)=_______32 32 01
---------------------------------
So, sq.root 26115=161.601_____
say we want to know the square root of Y.
X is the result
let X=1 first.
Redefine X using the formula below
X=(Y/X+X)/2
As you repeat redefining X, X will get closer and closer to the true value.
yes because you can always simplify the root like the sqroot of 8 can be 2 times the sqroot of 2
You could use the square root key to derive the square root of 1.
Use your brain if you have one!
ANSWER
----------------------
Use of Bakhshali Formula here
http://mathforum.org/library/drmath/view/61909.html
ofcourse as suggested by ed.poore and a few others we have to do a guess work , and yes this bakshali formula is actually a version of newtons method and as you will see its uses two iterations in a single step
q = A ^ 2 + b = A * A + b
example 41 = 6 ^ 2 + 5 = 6 * 6 + 5
Bakhshali Formula says
sqrt(q) = sqrt(A^2 + b) = A + b/(2A) - (b/(2A))^2/[2(A + b/(2A))]
which comes down to
Sqrt(q) = sqrt(A^2 + b) = A + b/(2A) - (b/(2*A))*(b/(2*A)) / (2*(A + b/(2*A)))
http://www.google.com/search?hl=en&lr=&rls=GGLJ%2CGGLJ%3A2006-10%2CGGLJ%3Aen&q=sqrt%2841%29
=
http://www.google.com/search?sourceid=navclient&ie=UTF-8&rls=GGLJ,GGLJ:2006-10,GGLJ:en&q=6+%2b+5%2f%282%2a6%29+%2d+%28%285%2f%282%2a6%29%29%2a%285%2f%282%2a6%29%29%29%2f%282%2a%286+%2b+5%2f%282%2a6%29%29%29
Mathematical operations used = /, * , + , -
GOT RIGHT
----------------------------
Ydrameos , kurt
(Partially )
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