The Nth Root of [Starting Point] = The Answer Sought For
[Starting Point]1/N = The Answer Sought For
The Answer Sought For = [Starting Point]1/N
[The Answer Sought For]N = { [Starting Point]1/N }N
[The Answer Sought For]N = [Starting Point]N/N
Simplifying N/N to 1 and not writing it:
[The Answer Sought For]N = Starting Point
[Partial Answer]N = Starting Point
Procedure:
<Square Root Parenthesis>
- 1a) If N is ODD, then:
- Divide both sides by Partial Answer, i.e.:
- (Partial AnswerN)/(Partial Answer) = (Starting Point)/(Partial Answer)
Partial Answer(N-1) = (Starting Point)/(Partial Answer)
- 1b) If N is EVEN, then:
- Take the square root of both sides, i.e.:
- (Partial AnswerN)1/2 = (Starting Point)1/2
Partial Answer(N/2) = <Starting Point>
- 2) Repeat steps, either 1a or 1b, until N=1
- 3) Beginning with the inmost math process performed on Starting Point and working your way outward and to the right, write down a linear sequence of each math process performed on Starting Point.
- 4) Convert step 3) into a series of calculator steps ---- an example shortly follows.
- 5) Perform the series of calculator steps as steps of a cycle using either a best guess integer (any positive integer, but 1) or the integer 2 in place of Starting Point, repeating each cycle by using its end result in place of every occurance of PARTIAL ANSWER in the cycle. You will notice that, beginning with the tenth's decimal place, each decimal place will slowly grow rightward in non-change throughout continued cycling. The result is accurate to all non-changing decimal places. Take the cycles as far as you like! More explicitly:
- 1] Clear, or turn on, the calculator.
- 2] Enter the positive integer whose root is being sought.
- If the first math step is divide, then:
- 3a] Divide by your best guess (something equal to or less than the square root of the positive integer; or use the number 2).
- Or if the first math step is square root, then:
- 3b] Take the square root.
- 4] Repeat step 3x] until all math steps are used up.
- 5] Add to the calculator's memory (usually something like the M+ key).
- 6] Re-enter the positive integer whose root is being sought.
- 7] If the first math step is divide, then:
- 8a] Divide by (Read) memory (usually something like: RM).
- Or if the first math step is square root, then:
- 8b] Take the square root.
- 9] Repeat step 8x] until all math steps are used up.
- 10] Subtract the calculator's memory from its display by performing: subtract Read Memory (usually: '–' then 'RM').
- 11] Add the calculator's display to its memory by using M+.
- 12] Optionally review your partial answer by invoking read memory (RM).
- 13] Repeat steps 6] through 12] 'ad nauseam' (until you reach your intended accuracy).
For example:
The 3rd Root of 5 = ?
Step One:
?3 = 5
?(3-1) = 5/?
?(2/2) = <5/?>
Step Two:
? = <5/?>
Step Three:
? = < 5 /Divide ? >Square Root
Reading from left to right, there is a divide symbol, followed by a right - square - root - bracket.
Figuring this part out is not unlike unraveling the binary representation of root (three) minus one (equals two), except that the resulting series of math steps on Starting Point will be in reverse order. Two, in binary, is 102 (this reads: the number two in base two). Dividing 102 by two is dividing two by two. It also coincides with taking the square root of 5/2. The number two in binary has now become 12. Subtracting one from 12 yields zero and coincides with dividing 5 by ?.
Steps Four and Five:
- First Cycle:
- Enter 5,
Best Guess or 2, <Take the Square Root>, Add to Memory (usually: M+).
- Second Cycle:
- Enter 5, Read Memory (usually: RM), <Take the Square Root>, Subtract Memory from display by performing: subtract Read Memory (usually:
RM), add display to Memory by using: M+, [At this point, you may view your 'partial' answer by hitting the Read Memory button], go back and repeat this second cycle as many times as you prefer!
Demonstration for the 3rd root of 5:
First Cycle:
5 2 = 2.5
<2.5> = 1.5811388...
1.5811388... M+ = 1.5811388...
Second Cycle:
5 RM (1.5811388...) = 3.1622777...
<3.1622777... > = 1.7782794...
1.7782794... RM (1.5811388...) = 0.1971406...
0.1971406... M+ = 0.1971406...
RM (1.7782794...), Optional step.
Third Cycle:
5 RM (1.7782794...) = 2.8117066...
<2.8117066... > = 1.6768144...
1.6768144... RM (1.7782794...) = 0.101465...
0.101465... M+ = 0.101465...
RM (1.6768144...) Optional step.
Fourth Cycle:
5 RM (1.6768144...) = 2.9818446...
<2.9818446... > = 1.7268018...
1.7268018... RM (1.6768144...) = 0.0499874...
0.0499874... M+ = 0.0499874...
RM (1.7268018...) Optional step.
Fifth Cycle:
5 RM (1.7268018...) = 2.8955262...
<2.8955262... > = 1.7016245...
1.7016245... RM (1.7268018...) = 0.0125425...
0.0125425... M+ = –0.0251773
RM (1.7268018...) Optional step.
Sixth Cycle:
5 RM (1....) = 2.9168686...
<2.9168686... > = 1.7078842...
1.7078842... RM (1.714167...) = 0.0062828...
0.0062828... M+ = 0.0062828...
RM (1.7078842...) Optional step.
The answer so far is 1.7
Alternate Display
And now for a live demonstration: