Calculating 3^0.2 without using a calculator?

[bigyan]

How can i calculate 3^0.2 without using a calculator?

We can write 3^4 as 3*3*3*3. Can 3^4.2 be written in the same way (without exponent)?

 

[matt grime]

no. well, not in any meangfully similar fashion. You're asking for the fifth root of 3. This is an irrational number, you cannot write it as a product of integers or rational numbers.

 

[bigyan]

matt,
Do you mean to say that an integer with a fractional exponent cannot be written in terms of multiplication? What about in terms of multiplication, addition and subtraction? Are there any proofs concerning this?? Is there anything like 'level' of operators?

 

[matt grime]

In general x^n, if n is not an integer, and x is an integer, will be irrational, and irrational numbers by definition cannot be written in the form a/b where a and b are integers.

of course 8^{1/3}=2, so it doesn't always hold.

If we assume that by "in terms of multiplication, addition and subtraction" you mean a finite number of operations on integers, then any such expression is a rational number, and so we have answered the question as "no there is no way of wriiting it in the way you ask", and the proof is fairly straightforward - you know how to prove the square root of 2 is irrational?

However, there are other "infinite" operations we could perform it depends on what yoyu want to do with it.

If you know about continued fractions then there are nice representations of lots of irrational numbers.

tell me what you think level of operators means i can think about it.

remember every real number has a decimal expansion, and sometimes we can compute these to as many places as you care to name.

 

[bigyan]

matt,
thanks...i'll do some reading...and post here again in about 15 hours.

 

[Ethereal]

How does the calculator do it?

 

[dextercioby]

How can i calculate 3^0.2 without using a calculator?

We can write 3^4 as 3*3*3*3. Can 3^4.2 be written in the same way (without exponent)?

I don't know what a calculator does,but surely i'd like to tell what i'd do.
Okay,let's call the number u want to compute by 'N'.Then
N=3^{0.2}
Take natural logarithm of both sides.U'll get
\ln N=0.2\ln 3
Now search in a logarithm table 'ln 3',multiply the number by 0.2 (which mean division by 5),call the number by 'a' and then u have this equation:
\ln N=a
Search in the same table of logarithms the N for which it's logarithm equals 'a'.
In the past,when computers were SciFi,logarithm tables were very handy and used.

Daniel.

 

[Ethereal]

Where did the log tables come from?

 

[rachmaninoff]

In general x^n, if n is not an integer, and x is an integer, will be irrational, and irrational numbers by definition cannot be written in the form a/b where a and b are integers.
of course 8^{1/3}=2, so it doesn't always hold.

A more elegant statement is: If a=m/n is rational and non-integral, and x is any integer, then if x^a is not integral it must be irrational.

Proof:

Let x^a=x^{m/n}, x,m,n\in\mathbb{N} be non-integral; assume for contradiction that it is rational. Then x^{m/n}=\frac{p}{q} for some relatively prime p,q. Thus (x^{m/n})^n=x^m=(\frac{p}{q})^n; and since p^n,q^n are relatively prime, (\frac{p}{q})^n must be non-integral; therefore x^m is non-integral, a contradiction.
Therefore x^{m/n} is irrational.

 

[dextercioby]

Where did the log tables come from?

I'll explain why the fundamental transcendental equation is
e^{x} =10 (1)
,with the solution x=\ln 10 [/tex].<br /> This is the fundamental number for the construction of natural logarithm tables.<br /> <br /> Using \int_{1}^{10} \frac{dx}{x} =\ln 10<br /> and computing the integral using Riemann sums and very fine divisions,people were able to compute this fundamental number with great accuracy.<br /> <br /> The fundamemental 2 formulas which help build natural logarithm tables are<br /> \ln a^{b}=b\ln a<br /> and<br /> \ln a+\ln b =\ln (ab)<br /> ,and these two need to be combined with the formula (series) that gives the natural logarithm for every real number between 1 and 2<br /> \ln(1+x)=\sum_{k=1}^{+\infty} (-)^{n+1}\frac{x^{n}}{n}<br /> ,with x\in [0,1].<br /> <br /> <br /> 1)Let&#039;s compute \ln 237 <br /> \ln 237=\ln 200+\ln 1.185=\ln 2+\ln 100+\ln 1.185=\ln 2+2\ln 10+\ln 1.185<br /> So,u make use of the series to find the natural logarithms of 2 and 1.185 and use the fundamental number \ln 10.<br /> <br /> 2)Let&#039;s compute \ln \sqrt[3.15]{\frac{\pi}{8.26}}.Take \pi=3.14<br /> \ln \sqrt[3.15]{\frac{\pi}{8.26}} =\frac{1}{3.15}(\ln \pi-\ln 8.26)=\frac{1}{3.15}(\ln 2+\ln 1.57-3\ln 2-\ln 1.0325)<br /> <br /> And so on and so forth.<br /> <br /> Daniel.

 

[Sirus]

How does the calculator do it?

http://www.du.edu/~jcalvert/math/sqrt.htm is an interesting method I once new, but I'm not sure if it works for fifth roots. Of course, since most roots are irrational, you must stop at some point and round, but it is technically possible to find the square root of any number by hand.

 

[dextercioby]

I've been taught this method when i was in my 6-th grade.10 years ago,to be more exact.For the third order,it's rather tricky,it's not that simple,u have to find cubes and that's a bit difficult if u have large numbers,of 12,20,50 digits.So save it only for the second order and use pencil and logarithm tables for other roots,if u don't have a calculator.

Daniel.

 

[Ethereal]

I see, thanks. I also found the following:

http://www.humboldt.edu/~mef2/Presentations/HSU Colloquia/NEWTLN1.html

 

[krab]

bigyan, yr 15 hrs r up.
A good way to do it is with Newton's method. You need to find the roots of x^5-3=0. First you try a few fractions and find the solution is close to x_1=5/4. Then from Newton's method,
x_{n+1}=x_n-{x_n^5-3\over 5x_n^4}
This gives a next approx: x_2=3893/3125, which is closer than a part in 30,000.
Apply it again and you get 9 decimal places, but that's tedious w/o calculator.

 

[mathwonk]

what about expanding the binomial theorem with fractionl ,exponents, also due to Newton?

 

[Gamish]

If you want to find 3^.2, just go to google, hehe. I would like to see the answer to this also, as earlier, I wanted to find out how to do 25^.5 without a calculator, and just to let you know, raising any number to the power if .5 gives you the square root. root=Number^1/root_number

 

[The Bob]

If you want to find 3^.2, just go to google, hehe. I would like to see the answer to this also, as earlier, I wanted to find out how to do 25^.5 without a calculator, and just to let you know, raising any number to the power if .5 gives you the square root. root=Number^1/root_number

25^0^.^5 = 25^\frac{1}{2}

This is because 0.5 = \frac{1}{2}

x to the power of \frac{1}{2} = \sqrt{x}

Therefore 25^\frac{1}{2} = \sqrt{25} = 5

The Bob (2004 ©)

 

[HallsofIvy]

3^.2= 3^1/5 so if x= 3^.2, x⁵= 3.

Probably the best way to solve that (without a calculator but still using the basic arithmetic operations) is to use "Newton's method". Newton's method for solving the equation f(x)= 0 is to set up a sequence of numbers x_n defined by taking x₀ as some initial "guess" and then x_n+1= x_n- f(x_n/f '(x_n) where f '(x) is the derivative of f.,

Let f(x)= x⁵- 3. Then f '(x)= 5x⁴. The formula becomes
x_n+1= x_n- (x_n⁵-3)/(5x_n⁴)= (4/5)x_n+(1/5)(3/x_n⁴).

If we start with x₀= 2, say, then x₁= 1.6375. Now repeat.

One way of thinking about this is that if x⁵= 3, then x= 3/x⁴.
If, on the other hand, x is less than the actual fifth root of 3, 3/x⁴ will be larger and if x is larger, 3/x⁴ is less than the actual root: in either case, the actual root is between x and 3/x⁴ and any "weighted" average of the two will move close to the true root. Newton's method gives x a weight of 4/5 while 3/x⁴ has a weight of 1/5 in the average.

 

[BobG]

The Babylonians used to have a method that was as accurate as Newton's method, even if less efficient. Actually, if you try Newton's method and the Babylonian method for square roots, you'll notice Newton's method winds up skipping every other guess. I haven't actually tried the Babylonian method for higher roots - Newton's method is more than enough tedium when done manually. (The point being that some kind of effective method for solving such problems has been in existence for a long time).

http://www.rowan.edu/mars/depts/math/osler/Babylon.pdf

 

[DeadWolfe]

Actually, I think that is Newton's Method.

https://www.physicsforums.com/showthread.php?t=58590

 

[jdavel]

Why not a Taylor/MacLaurin series for f(x) = x^0.2?

I'm sure that's how calculators do it.

 

[krab]

That would make sense if you are expanding about a known solution like 1^.2. IOW, Taylor expand (1+x)^.2. But this expansion does not converge for x=2.

Why wouldn't a calculator use Newton's method? It requires very few iterations: each one doubles the number of sig figs. That's a much faster convergence than power series.

 

[arildno]

what about expanding the binomial theorem with fractionl ,exponents, also due to Newton?

Do you happen to know the convergence rate in this case?

From what I know, Newton-Rapson,as krab espouses, is just about the simplest scheme with highest convergence rate.

(I don't doubt more complicated schemes exist which are faster than N-R, but I simply adore N-R for its simplicity, elegance, swiftness and visualizibility)

 

[Integral]

The reason a calculator cannot use Newtons method is that you need the derivative. Numerical derivitives are not very reliable, not nearly as good as a numerical integral. IIRC there are methods similar to a regula falsi method that require a number rather then a interval as a first guess, that also converge well , though certianly not quadradically as Newtons method does.

The origial problem was how to evlaute this funcition WITHOUT a caluclator. Newtons method changes the problem, but you will still find a calculator very handy,if not necessary.

How to do it without a caluclator?... Knowledge of logs and a slide rule.

 

[DeadWolfe]

Slide rules are calculators.

Sort of.

 

[HallsofIvy]

Why not a Taylor/MacLaurin series for f(x) = x^0.2?

I'm sure that's how calculators do it.

I used to think that calculators would use Taylor's series for functions such as exponentials and trig functions but apparently they use a "Cordic" method.
Here's a website:
http://www.dspguru.com/info/faqs/cordic.htm

 

[krab]

In early 70s we used to have a Wang calculator with pixie tubes for the digits. It did trig and exp functions by Taylor expansion. It was fun to watch because the digits tubes were left operating while it added the series. You would see them stablilize from left to right. When the rightmost digit no longer changed, it would stop the calculation and present the result. Took about 5 seconds.

But for square roots, it would have used Newton's method. 3 iterations give 10 digits, and indeed, square roots were much faster than trig functions.

My calculator has a y^x button. I've no idea how it does that. Maybe the Cordic method, as Halls says.

 

[BobG]

Slide rules are calculators.

Sort of.

Sort of?

Slide rules are great calculators. You solve quadratic equations, trig problems, and other things that a cheap calculator can't. If you have a good feel for using a slide rule, there's even some problems where a slide rule is faster and more efficient than a good graphing calculator, especially if you're using a special purpose slide rule for its intended purpose (a Hemmi 257 Chemical Engineering slide rule for Chemistry problems, for example).

Of course, your accuracy is only to three significant digits, but there's really not very many everyday things that require more than that.

 

[dextercioby]

Since side rules have been built using logarithm tables,it makes no difference whether working with logaritm tables,a pencil and a sheet of paper,or with a side rule.It's actually an exercise for the brain to do simple calculations.I find working with a slide rule (in the absence of electronic help) as a proof of laziness,unless u lack logarithm tables,too.In that case,it would be the only viable option.

Daniel.

 

[BobG]

Since side rules have been built using logarithm tables,it makes no difference whether working with logaritm tables,a pencil and a sheet of paper,or with a side rule.It's actually an exercise for the brain to do simple calculations.I find working with a slide rule (in the absence of electronic help) as a proof of laziness,unless u lack logarithm tables,too.In that case,it would be the only viable option.

Daniel.

You're killing me, here.

I swear. People born after the invention of the electronic calculator have no appreciation for art.