# How do you manually do a fractional exponent?

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I understand exponents, I eventually figured out how you manually do negative exponents, but how do you do something like: 3^.4? How do you do something like: abs[3]^3? Can you multiply together absolute values? I am only in algebra right now, not even geometry, but nobody else even cares how they do as long as they pass, but i want to understand this stuff!

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Hi.

First of all, I'd like to note that not all results of a fractional power will be rational, meaning you may not be able to solve it by hand, or it would require you to do infinite repetitions. Nonetheless, given that your answer is rational or you don't mind having an irrational component in your solution, then these are the steps.

To do fractional powers manually, it is advisable that you first convert them to proper fractions. For example 3^(0.4) can be expressed as 3^(4/10).

This is easy to do if the number is non repeating. If the number does repeat however, a universal method to do this which I learnt is to multiply the decimal by 100, subtract by itself, and divide by 99. Formula being:

fraction = (100x - x)/99 = 99x/99

Back to the topic. A fractional exponent is basically a root. For example, square root 4 can be expressed as 4^(1/2), cube root 4 as 4^(1/3) and etc. I guess that was the key piece of information you were looking for. In anycase, all the properties used for exponents can be used on fractional exponents as well, so:

4^1.5 = 4^(3/2) = 4^(1 + 1/2) = (4^1)x(4^1/2) = 4*2 = 8

There will be times where the answer is not rational as I mentioned like:

2^(5/2) = 2^(2 + 1/2) = (2^2)x(2^1/2) = 4*(square root 2)

If you have a large number, the general strategy may be to factor that number down into something you know, for example if you have:

64^(1/3) = 8^(2/3) = 2^(3*(2/3)) = 2^2 = 4

You will also find cases where you break down the numbers and you get something ugly which is neither irrational nor integer. In these cases, the only way I know is to resort to iteration, meaning guess the answer, check, adjust, check, adjust until your answer is to an acceptable level of precision. There are formulas that can help you get in the right direction and more quickly, but these are usually used in the realm of programming where computers do the work for us. Conclusion is, fractional exponents are nasty, and even more, fractional exponents of fractional numbers !!!

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You could use the Binomial theorem for non-natural powers. Can't be arsed to quote it for you, do a google.

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On the absolute power thing, just make the number positive and raise it to that power. Like your question, abs[3]^3 is the same as 3^3, which is 27. If it was abs[-3]^3 then you would make it positive and have 3^3 again.

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You could use the Binomial theorem for non-natural powers. Can't be arsed to quote it for you, do a google.

Trouble is, you have to pick the right series. For example, [imath](1+x)^{\frac{1}{2}}[/imath] only converges for [imath]|x|<1[/imath]. So if you want to work out [imath]\sqrt{2}[/imath], you can't just stick in x = 1.

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Thanks, that really helps. All I really needed was to know that x^1/2 is the same as [sqrt x]. But thanks for going into greater detail, now i understand it. As for the absolute value question, I was an idiot. Anyway, how exactly could you do something like this: 3^4!. I thought you just do the factorial than use that as your exponent. But then, I realized that you can't because PEMDAS says you do Exponents first, multiplication second. Would you do 3^24 or 81!? Would it even be either of these? Math gets really confusing sometimes!

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it would be 3^24... instead of just thinking of 4! as just 4*3*2*1, think of it as (4*3*2*1)

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Oh, thanks a bunch

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Originally Posted by □h=-16πT

You could use the Binomial theorem for non-natural powers. Can't be arsed to quote it for you, do a google.

Trouble is, you have to pick the right series. For example, [imath](1+x)^{\frac{1}{2}}[/imath] only converges for [imath]|x|<1[/imath]. So if you want to work out [imath]\sqrt{2}[/imath'], you can't just stick in x = 1.

$(1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k}$

Let x=1, therefore:

$(1+1)^\alpha = 2^\alpha = \sum_{n=0}^{n=\infty} \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k}$

Let alpha =1/2, therefore:

$2^{\frac{1}{2}} = \sqrt{2} = \sum_{n=0}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{1}{2}+1-k}{k}$

Therefore:

$\sqrt{2} = \sum_{n=0}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k}$

Obtain the first five terms of the series.

$\sum_{n=0}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k} = 1+\sum_{n=1}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k}$

$\sum_{n=1}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k} = \prod_{k=0}^{k=1} \frac{\frac{3}{2}-k}{k} + \sum_{n=2}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k}$

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Yeah, OK, that clears it up. Joking. Im an algebra student, I don't even know what the big E means, or what the n=infinity on top and n=1 on the bottom mean, even though I also don't get the big greek sign next to that, as well as why they are underneath or above the big greek signs. I have a couple years to still go over that, I'm only a freshman!

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$\sum$ means sum. like $\sum_{j=1}^{k}x_j=x_1+x_2+x_3+...+x_k$.

$\prod$ is product

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Wait, I kind of get it now. So, the number on top means the last variable that you add to the sum, and the bottom means what the bottom section of the variable equals. Wow, I can see how hard advanced math is going to be already!

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Actually, it is much more complex than I thought it was going to be.

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Actually, it is much more complex than I thought it was going to be.

No it is easy, once you understand your first problem. let me finish the computation of square root of two for you.

Therefore:

$\sqrt{2} = \sum_{n=0}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k}$

Obtain the first five terms of the series.

$\sum_{n=0}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k} = 1+\sum_{n=1}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k}$

$\sum_{n=1}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k} = \prod_{k=0}^{k=1} \frac{\frac{3}{2}-k}{k} + \sum_{n=2}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k}$

Now' date=' focus on the following quantity:

[math'] \prod_{k=0}^{k=1} \frac{\frac{3}{2}-k}{k} [/math]

When k=0, we appear to have division by zero error, however, this is avoided because 0! is defined to be equal to 1. This requires some explanation though.

Look at 1/n! for a moment.

$\frac{1}{n!} = \prod_{k=1}^{k=n} \frac{1}{k}$

In the case where n=3, we have:

$\frac{1}{3!} = \prod_{k=1}^{k=3} \frac{1}{k} = (\frac{1}{1}) \prod_{k=2}^{k=3} \frac{1}{k} = (\frac{1}{1}) (\frac{1}{2}) \prod_{k=3}^{k=3} \frac{1}{k} = (\frac{1}{1}) (\frac{1}{2}) (\frac{1}{3}) = \frac{1}{6}$

Now, consider again the definition of n!

$\frac{1}{n!} = \prod_{k=1}^{k=n} \frac{1}{k}$

Let n=0.

Hence we have:

$\frac{1}{0!} = \prod_{k=1}^{k=0} \frac{1}{k}$

By definition we have:

$0! \equiv 1$

Therefore:

$\frac{1}{1} = 1 = \prod_{k=1}^{k=0} \frac{1}{k}$

Multiplication is commutative therefore:

$\prod_{k=1}^{k=0} \frac{1}{k} = \prod_{k=0}^{k=1} \frac{1}{k}$

Now, if we expand the product, we have:

$\prod_{k=0}^{k=1} \frac{1}{k} = (\frac{1}{0}) (\frac{1}{1})$

By transitivity, it now follows that:

$1 = \frac{1}{0}$

Which is false. This contradiction was caused by the definition 0!=1.

However let us look at our situation...

In the case where the upper indice is equal to n, we will want to evaluate the following iterated product:

$\prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k}$

If we pull out the k from the denominator, we have this:

$\frac{1}{n!} \prod_{k=0}^{k=n} \frac{3}{2}-k$

So here is our series expression for square root of two again:

$\sqrt{2} = \sum_{n=0}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k}$

If we pull the k out of the denominator we have:

$\sqrt{2} = \sum_{n=0}^{n=\infty} \frac{1}{n!}\prod_{k=0}^{k=n} \frac{3}{2}-k$

The first term of the series is:

$\frac{1}{0!}\prod_{k=0}^{k=0} \frac{3}{2}-k = \frac{1}{1} (\frac{3}{2}) = \frac{3}{2}$

But this leads to an error in Newton's binomial formula. Let me explain.

Here is Newton's binomial formula again:

$(1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k}$

Consider the case where alpha = 2, and x is an arbitrary real number.

By direct substitution, we have:

$(1+x)^2 = \sum_{n=0}^{n=\infty} x^n \prod_{k=0}^{k=n} \frac{2+1-k}{k}$

Using only the axioms of the real number system, we can figure out what the left hand side is.

$(1+x)^2 = (1+x)(1+x) = 1+x+x+xx = 1+2x +x^2$

Now figure out what the RHS must equal.

$\sum_{n=0}^{n=\infty} x^n \prod_{k=0}^{k=n} \frac{2+1-k}{k} = \sum_{n=0}^{n=\infty} x^n \prod_{k=0}^{k=n} \frac{3-k}{k} = \sum_{n=0}^{n=\infty} x^n \frac{1}{n!} \prod_{k=0}^{k=n} 3-k$

The first term of the series is:

$x^0 \frac{1}{0!} \prod_{k=0}^{k=0} 3-k = x^0 (3-0) = 3x^0$

Using the axioms of algebra, it is easy to prove that

if not(x=0) then x^0 = 1.

Thus, in the case where x isn't equal to zero, we must have:

$1+2x +x^2 = 3 + \sum_{n=1}^{n=\infty} x^n \frac{1}{n!} \prod_{k=0}^{k=n} 3-k$

Now, lets compute the term of the series, when n=1.

When n=1, we have:

$x^1 \frac{1}{1!} \prod_{k=0}^{k=1} 3-k = x (3-0)(3-1) = 6x$

Thus, we have:

$1+2x +x^2 = 3 + 6x + \sum_{n=2}^{n=\infty} x^n \frac{1}{n!} \prod_{k=0}^{k=n} 3-k$

Now, evaluate the series when n=2. We obtain:

$x^2 \frac{1}{2!} \prod_{k=0}^{k=2} 3-k = \frac{x^2}{2} (3-0)(3-1)(3-2) = 3x^2$

From which it now follows that:

$1+2x +x^2 = 3 + 6x + 3 x^2 + \sum_{n=3}^{n=\infty} x^n \frac{1}{n!} \prod_{k=0}^{k=n} 3-k$

Now, whenever n is greater than or equal to 3, there will be multiplcation by zero since:

$\prod_{k=3}^{k=3} 3-k = (3-3)=0$

Hence there are only three terms in the expansion formula, because:

$0 = \sum_{n=3}^{n=\infty} x^n \frac{1}{n!} \prod_{k=0}^{k=n} 3-k$

Thus, we have:

$1+2x +x^2 = 3 + 6x + 3 x^2$

Which is false. However, it was almost correct. If the RHS is divided by three instead, we would have had:

$1+2x +x^2 = 1 + 2x + x^2$

which is true. Thus, the RHS was off by multiplication of the constant 1/3, but otherwise the series-product expansion formula worked.

Now, look again at the formula for (1+x)^alpha:

$(1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k}$

Suppose instead, that we start off the multiplication from 1, instead of zero, like this:

$(1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k}$

Let alpha=2, so that we have:

$(1+x)^2 = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{2+1-k}{k}$

We already know that the LHS is equal to 1+2x+x^2, so let us write:

$1 + 2x + x^2 = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{2+1-k}{k}$

And 2+1=3, so that we have:

$1 + 2x + x^2 = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{3-k}{k}$

Now, we just have to expand the RHS correctly, so that we obtain equality of the RHS with the LHS.

Let us pull out the k from the denominator of the iterated product, so that we have:

$1 + 2x + x^2 = \sum_{n=0}^{n=\infty} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k$

In the case where x=0, we can readily see that the LHS is equal to 1, from the algebraic field axioms. So in order to have equality, rather than a false statement, the RHS must be equal to 1, in the case where x=0.

We know that the series has at most three terms, since when n is greater than or equal to 3, we all higher terms in the series are multiplied by (3-3)=0, due to the iterated product.

So without loss of generality, we can write:

$1 + 2x + x^2 = \sum_{n=0}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k$

The first term of the series is the only term in the series which isnt multiplied by x, and so must be equal to 1, in order for the RHS to equal the LHS in the case where x=0, so we can write:

$1 + 2x + x^2 = \frac{0^0}{0! }+ \sum_{n=1}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k$

So that, regardless of the value of x, it needs to be true that:

$1 = \frac{0^0}{0!}$

Thus, we can write:

$1 + 2x + x^2 = 1 + \sum_{n=1}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k$

Now, we just have to expand the remaining terms of the series, to see that the RHS equals the LHS. This time however, the product starts off with k=1, instead of k=0. Thus, we have:

$\sum_{n=1}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k = \frac{x^1}{1!} (3-1) + \sum_{n=2}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k$

Thus, we have:

$\sum_{n=1}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k = 2x + \frac{x^2}{2!} \prod_{k=1}^{k=2} 3-k$

Thus, we have:

$\sum_{n=1}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k = 2x + \frac{x^2}{2!} (3-1)(3-2) = 2x+ x^2$

$1 + 2x + x^2 = 1 + \sum_{n=1}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k$

So that by substition, the remaining two terms of the series on the RHS above, are 2x+x^2, so that we have:

$1 + 2x + x^2 = 1 + 2x + x^2$

So it worked, and all is well, provided that we start off the multiplication from k=1, instead of k=0, and that we have:

$1 = \frac{0^0}{0!}$

Thus, we can write Newton's expansion formula as follows:

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k}$

Now, because we have started off the multiplication from 1, instead of zero, we are not faced with division by zero error. The only problem is the formula is not as compact as we would like it to be.

Before going any further, we can try to use the formula above to compute the square root of two, which is the main point of this post. The secondary point, is to show you how to manipulate the series expressions.

Starting off with the formula

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k}$

Let x=1, and let alpha = 1/2, and then write this:

$(1+1)^{\frac{1}{2}}= 1+ \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\frac{1}{2} +1-k}{k}$

However if we stipulate that:

$1 = x^0 \prod_{k=1}^{k=0} \frac{\alpha +1-k}{k}$

Then we can write:

$(1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k}$

And this stipulation had to be made, because of the case where x=0, so that what we really need to stipulate is that:

$1 = 0^0 \prod_{k=1}^{k=0} \frac{\alpha +1-k}{k}$

However, from the top indice, you can see that we have division by zero error. Let us pull out the k, as n!, so that we have this:

$1 = \frac{x^0}{n!} \prod_{k=1}^{k=0} \alpha +1-k$

In fact, lets go back and write this for Newton's formula:

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \prod_{k=1}^{k=n} \alpha +1-k$

Which, provided that not(alpha=-1), can also be written as:

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k$

Now, focus on this part of the formula:

$\sum_{n=1}^{n=\infty} \frac{x^n}{n!} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k$

If it is permissible for n=0, then we have the zeroth term as:

$\frac{x^0}{0!} \frac{1}{\alpha+1} \prod_{k=0}^{k=0} \alpha +1-k$

Which simplifies slightly to:

$\frac{x^0}{0!} \frac{1}{\alpha+1} (\alpha +1) = \frac{x^0}{0!}$

So lets consider carefully where we are:

We are sure that this much is algebraically correct:

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k$

All we want to do is compactify the formula above.

Suppose that we stipulate that the n=0 term of the series is equal to 1, then we can write:

$(1+x)^\alpha = \sum_{n=0}^{n=\infty} \frac{x^n}{n!} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k$

Then, explicitely when n=0 we will have to evaluate:

$\frac{x^0}{0!} \frac{1}{\alpha+1} \prod_{k=0}^{k=0} \alpha +1-k$

And it will have to be the case, that the above formula is equal to 1, so that we must force it to be the case that:

$1 = \frac{x^0}{0!} \frac{1}{\alpha+1} \prod_{k=0}^{k=0} \alpha +1-k$

The iterated product only has one term, that for which k=0, so that we may write:

$1 = \frac{x^0}{0!} \frac{1}{\alpha+1} (\alpha +1-0)$

From which it follows that:

$1 = \frac{x^0}{0!}$

In any case where not(x=0), the numerator of the fraction above is equal to 1, so that in order for Newton's expansion formula to be correct, we must have:

$1 = \frac{1}{0!}$

From which it follows that we must define 0 factorial to be equal to 1.

Let us look at the formula again, and consider the case where x=-1, and alpha =0.

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k$

So that

$(1-1)^0 = 0^0 = 1+ \sum_{n=1}^{n=\infty} \frac{(-1)^n}{n!} \frac{1}{1} \prod_{k=0}^{k=n} 1-k$

Which can be written as follows:

$(1-1)^0 = 0^0 = 1+ \sum_{n=1}^{n=\infty} \frac{(-1)^n}{n!} \frac{1}{1} \prod_{k=0}^{k=n} -(k-1)$

Which can be written as follows:

$(1-1)^0 = 0^0 = 1+ \sum_{n=1}^{n=\infty} \frac{(-1)^n}{n!} \frac{1}{1} (-1)^n \prod_{k=0}^{k=n} (k-1)$

Which can be written as follows:

$(1-1)^0 = 0^0 = 1+ \sum_{n=1}^{n=\infty} \frac{(-1)^{2n}}{n!} \frac{1}{1}\prod_{k=0}^{k=n} (k-1)$

Which can be written as follows:

$(1-1)^0 = 0^0 = 1+ \sum_{n=1}^{n=\infty} \frac{1}{n!} \frac{1}{1}\prod_{k=0}^{k=n} (k-1)$

From which it follows that:

$0^0 = 1+ \frac{1}{1!} \prod_{k=0}^{k=1} (k-1) + \sum_{n=2}^{n=\infty} \frac{1}{n!} \prod_{k=0}^{k=n} (k-1)$

Now, when k=1, (k-1)=0, so that we must have:

$0^0 = 1$

So look at the formula again:

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k$

Now, look at the definition of n factorial for a moment:

$n! = \prod_{k=1}^{k=n} k$

We can perform an indicial operation and write:

$n! = \prod_{k=0}^{k=n-1} k+1$

From which it follows that:

$n! = \frac{1}{n+1}\prod_{k=0}^{k=n} k+1$

Hence, we can write the formula as follows:

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{\frac{1}{n+1}\prod_{k=0}^{k=n} k+1} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k$

Or rather:

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{(n+1) x^n}{\prod_{k=0}^{k=n} k+1} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k$

Or rather:

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} (n+1) x^n \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k+1}$

Now, focus on this part of the formula for a moment:

$\sum_{n=1}^{n=\infty} (n+1) x^n \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k+1}$

If it is permissible for n=0, then the zeroth term of the series above would be given by:

$(0+1) x^0 \frac{1}{\alpha+1} \prod_{k=0}^{k=0} \frac{\alpha +1-k}{k+1}$

Which can be simplified to:

$x^0 \frac{1}{\alpha+1} \frac{\alpha +1-0}{0+1}$

Which can be simplified to:

$x^0 \frac{\alpha +1}{\alpha +1}$

Which can be simplified to:

$x^0$

So let us begin with the following formula we know to be correct, in any case where not(alpha = -1).

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} (n+1) x^n \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k+1}$

If we stipulate that x^0 = 1, for any x, we can write this instead:

$(1+x)^\alpha = \sum_{n=0}^{n=\infty} (n+1) x^n \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k+1}$

As I just showed moments ago.

Which for the sake of reading clarity can be written as follows:

$(1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \frac{n+1}{\alpha+1} \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k+1}$

Now, when n=0, the first term of the iterated product is given by:

$\frac{\alpha +1-0}{0+1} = \alpha +1 = \prod_{k=0}^{k=0} \frac{\alpha +1-k}{k+1}$

Hence we can write Newton's binomial formula as follows:

$(1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \frac{n+1}{\alpha+1} (\alpha+1)\prod_{k=1}^{k=n} \frac{\alpha +1-k}{k+1}$

Which simplifies slightly to:

$(1+x)^\alpha = \sum_{n=0}^{n=\infty} (n+1) x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k+1}$

Thus, I have now removed any possiblity of division by zero error, in the case where alpha=-1. In other words, using an indicial operation, i took out 1/(1+alpha) and opened up the possibility of division by zero error, but now I've put it back into the iterated product, so that we no longer have to worry about whether or not alpha = -1 in our formula. Thus, the following formula is correct:

$(1+x)^\alpha = \sum_{n=0}^{n=\infty} (n+1) x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k+1}$

Which can be written as follows:

$(1+x)^\alpha = \sum_{n=0}^{n=\infty} \frac{d}{dx} (x^{n+1}) \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k+1}$

Which can be written as:

$(1+x)^\alpha = \frac{d}{dx} \sum_{n=0}^{n=\infty} x^{n+1} \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k+1}$

Now, lets perform an indicial operation, and write the above formula as follows:

$(1+x)^\alpha = \frac{d}{dx} \sum_{n=1}^{n=\infty} x^{n} \prod_{k=1}^{k=n-1} \frac{\alpha +1-k}{k+1}$

Now, define U(x) as follows:

$U(x) \equiv \sum_{n=1}^{n=\infty} x^{n} \prod_{k=1}^{k=n-1} \frac{\alpha +1-k}{k+1}$

So that we can now write:

$(1+x)^\alpha = \frac{dU}{dx}$

Now, focus on U(x) for a moment...

$U(x) \equiv \sum_{n=1}^{n=\infty} x^{n} \prod_{k=1}^{k=n-1} \frac{\alpha +1-k}{k+1}$

The idea is that if we differentiate U(x), we have to get (1+x)^alpha.

Now, go back and look at the following formula:

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k$

We can rewrite it as follows:

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \frac{\alpha + 1}{\alpha+1} \prod_{k=1}^{k=n} \alpha +1-k$

Which can be simplified to:

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \prod_{k=1}^{k=n} \alpha +1-k$

Which can be simplified to:

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k}$

Here is U(x) again:

$U(x) \equiv \sum_{n=1}^{n=\infty} x^{n} \prod_{k=1}^{k=n-1} \frac{\alpha +1-k}{k+1}$

We have:

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} x^n \prod_{k=0}^{k=n-1} \frac{\alpha -k}{k+1}$

Or equivalently:

$(1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} x^n \alpha \prod_{k=1}^{k=n-1} \frac{\alpha -k}{k+1}$

Now, let alpha become alpha +1, so that we have:

$(1+x)^{\alpha +1}= 1+ \sum_{n=1}^{n=\infty} x^n (\alpha +1) \prod_{k=1}^{k=n-1} \frac{\alpha +1 - k}{k+1}$

Or equivalently:

$(1+x)^{\alpha +1}= 1+ (\alpha +1)\sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n-1} \frac{\alpha +1 - k}{k+1}$

From which it now follows using the definition of U(x) that:

$(1+x)^{\alpha +1}= 1+ (\alpha +1) U(x)$

Now, we can write the above equation as follows:

$(1+x)(1+x)^\alpha= 1+ (\alpha +1) U(x)$

And recall the earlier formula that:

$(1+x)^\alpha = \frac{dU}{dx}$

Thus, we can write the following differential equation:

$(1+x) \frac{dU}{dx} = 1+ (\alpha +1) U(x)$

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I ran out of time to edit the previous post. The main thing I wanted to show you, was how to use Newton's binomial formula to compute square root of two.

In the previous post you will find the following formula:

$(1+x)^{\alpha}= 1+ \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k}$

Now, let alpha=1/2, and let x=1 in the formula above to obtain:

$(1+1)^{\frac{1}{2}}= 1+ \sum_{n=1}^{n=\infty} 1^n \prod_{k=1}^{k=n} \frac{\frac{1}{2} +1-k}{k}$

From which it follows that:

$\sqrt{2} = 1+ \sum_{n=1}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1 + \prod_{k=1}^{k=1} \frac{\frac{3}{2} - k}{k}+ \sum_{n=2}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1 + (\frac{3}{2} - 1) + \sum_{n=2}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.5 + \sum_{n=2}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.5 + \prod_{k=1}^{k=2} \frac{\frac{3}{2} - k}{k} + \sum_{n=3}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.5 + (\frac{3}{2} - 1)( \frac{\frac{3}{2} - 2}{2}) + \sum_{n=3}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.5 - (.5) (\frac{1}{4}) + \sum_{n=3}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

Now, (.5)(1/4)=(.5)(.25)=(.125) from which it follows that:

$\sqrt{2} = 1.5 - .125 + \sum_{n=3}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.375 + \sum_{n=3}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.375 + \prod_{k=1}^{k=3} \frac{\frac{3}{2} - k}{k}+\sum_{n=4}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.375 + (-1.25)\prod_{k=3}^{k=3} \frac{\frac{3}{2} - k}{k}+\sum_{n=4}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.375 + (-1.25) \frac{\frac{3}{2} - 3}{3}+\sum_{n=4}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.375 + (-1.25)(-.5) +\sum_{n=4}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

Now, (-1.25)(-.5) = .0625.

Hence we have:

$\sqrt{2} = 1.375 + .0625 +\sum_{n=4}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.4375 +\sum_{n=4}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.4375 + \prod_{k=1}^{k=4} \frac{\frac{3}{2} - k}{k}+ \sum_{n=5}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.4375 + (.0625) \prod_{k=4}^{k=4} \frac{\frac{3}{2} - k}{k}+ \sum_{n=5}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.4375 + (.0625)( \frac{\frac{3}{2} - 4}{4}) + \sum_{n=5}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.4375 + (.0625)(-.625) + \sum_{n=5}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.4375 + (-.0390625) + \sum_{n=5}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.3984375 + \sum_{n=5}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.3984375 + \prod_{k=1}^{k=5} \frac{\frac{3}{2} - k}{k} + \sum_{n=6}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.3984375 + (-.0390625)\prod_{k=5}^{k=5} \frac{\frac{3}{2} - k}{k} + \sum_{n=6}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.3984375 + (-.0390625)( \frac{\frac{3}{2} - 5}{5}) + \sum_{n=6}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.3984375 + (-.0390625)(-.7) + \sum_{n=6}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.3984375 + (.02734375) + \sum_{n=6}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.42578125 + \sum_{n=6}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.42578125 + (-.0205078125) + \sum_{n=7}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.4052734375 + \sum_{n=7}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.4052734375 +(-.0205078125)(-.7857142857)+ \sum_{n=8}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.4052734375 +(.01611328125)+ \sum_{n=8}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.42138671875+ \sum_{n=8}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.42193603515625 +(.01611328125)(-.8125)+ \sum_{n=9}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.42193603515625 -(.01309204101562)+ \sum_{n=9}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.40884399414062 + \sum_{n=9}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

And

$\sum_{n=9}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} = (-.01309204101562)(-.8333...)+\sum_{n=10}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.40884399414062 + (.01091003417968)+ \sum_{n=10}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

From which it follows that:

$\sqrt{2} = 1.4197540283203 + \sum_{n=10}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k}$

So as you can see, the infinite series really does converge to what your calculator gives you, namely:

$\sqrt{2} = 1.4142135623731...$

Which was the main point of the exercise.

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Damn, I am impressed. That was one long post! Anyway, thanks a bunch, now I actually am starting to get it. All I have to do is try to understand that a bit more, and maybe I will understand how to do this before summer is over. Thanks for showing me this stuff.

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Please be aware that Johnny's method is very limited in scope (it cannot work out the square root of any number bigger than 2, and doesn't really allow you to do much by hand does it? Are you going to evaluate all those things?

First, there is no reason to evaluate something's decimal expansion. It is for the most part mathematically unimportant.

Second if you really have tot there is a general iterativem method for finding the square root of any number. It is based upon the Newton-Raphson Iterative Method.

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Please be aware that Johnny's method is very limited in scope (it cannot work out the square root of any number bigger than 2)

When working with series expressions' date=' there is something called the radius of convergence.

Consider again the following:

[math']

(1+x)^{\alpha}= 1+ \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k}

[/math]

Now, focus on just the series formula:

$\sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k}$

Now, look at the function of k, inside the iterated product:

$f(k) \equiv \frac{\alpha +1-k}{k}$

Whether or not any given series converges, can be determined by finding its radius of convergence. Define the radius of convergence as follows:

$\rho \equiv |\lim_{k \to \infty} f(k)|$

Now, in the case of Newtons Binomial expansion we have:

$\rho \equiv |\lim_{k \to \infty} \frac{\alpha +1-k}{k}| = |-1| = 1$

So, the radius of convergence is one, just as Matt said.

Now, it turns out that by taking the limit above, you are actually carrying out the very well known "ratio test" for convergence.

Here is a link ratio test.

Consider example 14.1.7 in the link above.

You are faced with the following series:

$\sum_{n=1}^{n = \infty} \frac{1}{n^2}$

We have:

$\sum_{n=1}^{n = \infty} \frac{1}{n^2} = 1 + \sum_{n=2}^{n = \infty} \frac{1}{n^2} = 1 + \sum_{n=2}^{n = \infty} \frac{(n-1)!(n-1)!}{n^2(n-1)!(n-1)!}$

Now,

$\sum_{n=2}^{n = \infty} \frac{(n-1)!(n-1)!}{n^2(n-1)!(n-1)!} = \sum_{n=2}^{n = \infty} \frac{(n-1)!(n-1)!}{n! n!}$

And

$\sum_{n=2}^{n = \infty} \frac{(n-1)!(n-1)!}{n! n!} = \sum_{n=2}^{n= \infty} \prod_{k=1}^{k=n} (\frac{k-1}{k})^2$

Thus, we have:

$\sum_{n=1}^{n = \infty} \frac{1}{n^2} = 1+ \sum_{n=2}^{n= \infty} \prod_{k=1}^{k=n} (\frac{k-1}{k})^2$

Now consider the limit of f(k), as k approaches infinity.

$\lim_{k \to \infty} (\frac{k-1}{k})^2 = 1$

Now, look at excercise 7.8 here.

Let us consider their first example:

$\sum_{n=1}^{n=\infty} \frac{x^n}{2^n}$

For the series above, we have:

$\sum_{n=1}^{n=\infty} \frac{x^n}{2^n} = \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{1}{2}$

Now, investigate the limit as k approaches infinity of f(k).

$\lim_{k \to \infty} \frac{1}{2} = \frac{1}{2}$

The limit above, is one over the radius of convergence, so that:

R =2.

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And the radius of convergence of your series is 1. Please don't hijack a thread with stupidly long and unnecessary posts, Johnny. Not everyone is as ignorant as you and most people understand this, and if they didnt' they would ask about it if they were interested. Your posting is tantamount to vandalism; please desist.

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Damn, I am impressed. That was one long post! Anyway, thanks a bunch, now I actually am starting to get it. All I have to do is try to understand that a bit more, and maybe I will understand how to do this before summer is over. Thanks for showing me this stuff.

You're welcome. To understand more, google for Newton's binomial theorem, power series, convergence of series.

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And the radius of convergence of your series is 1. Please don't hijack a thread with stupidly long and unnecessary posts, Johnny. Not everyone is as ignorant as you and most people understand this, and if they didnt' they would ask about it if they were interested. Your posting is tantamount to vandalism; please desist.

Surely you're joking. I thought it was quite appropriate to the thread, and I quite enjoyed it myself. Anyways, i showed what I wanted to show, which was a computation of square root two.

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It was, in my opinion, a very long post containing nothing of practical interest that could not have been done in 3 lines. You don't compute the square root of 2 and that very power series you cite does not conclude that the expansion you care about is valid in finding the square root of 2, nor does it show it is invalid either before you complain that you have used it. You realize I am talking about the oist on radius of convergence that explained the ratio test? The previous post was just far too long, though well within the scope of this thread.

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Please be aware that Johnny's method is very limited in scope (it cannot work out the square root of any number bigger than 2' date='

[/quote']

$\sqrt{3} = (1+2)^{\frac{1}{2}} = [2(\frac{1}{2}+1)]^{\frac{1}{2}} = \sqrt{2} (1+ \frac{1}{2})^{\frac{1}{2}}$

Given that you know the square root of two, you can now compute the square root of three by using Newton's formula on:

$(1+ \frac{1}{2})^{\frac{1}{2}}$

Just let X=1/2, and alpha = 1/2. Then multiply by root 2.

Regards

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Yes, but that wasn't what you stated (it can easily be extended to get round all the problems I've raised). But then what you do intend to state is often lost in the size of the post.

Your original pair of posts, by the way, contains many misunderstandings, mistatements, and mistakes. Quite what that waffle about 0! was intended to do is something that defies description. And please don't start a debate in this thread about 0!, 0^0 and other things whose standard usage you simply do not accept. They are just formal conventions, you are over analysing.

And you've still not proven that the power series converges in finding the square root of 2.

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Yes' date=' but that wasn't what you stated (it can easily be extended to get round all the problems I've raised). But then what you do intend to state is often lost in the size of the post.

Your original pair of posts, by the way, contains many misunderstandings, mistatements, and mistakes. Quite what that waffle about 0! was intended to do is something that defies description. And please don't start a debate in this thread about 0!, 0^0 and other things whose standard usage you simply do not accept. They are just formal conventions, you are over analysing.

And you've still not proven that the power series converges in finding the square root of 2.[/quote']

Well, point out the mistakes, and I'll discuss them in another thread.

Regards

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