Differential Theorem with fractions and negatives

[Dalmain]

Good day All

I don't have a specific math problem to post but a general question regarding calculations with the binomial theorem that have either a fraction or a negative exponent.

As I understand it if the binomial has a positive integer as the exponent, after expansion there will be that number plus one terms. Example a binomial to the power of 2 has three terms after expansion. Exponent of 6 then seven terms etc. When using n choose k, you stop calculating when n=k.

How many terms are there when the exponent is a fraction or negative?

I realise I have many hole in my maths fundamentals, but I am determined to get this right. Any assistance is always much appreciated.

 

[topsquark]

Good day All

I don't have a specific math problem to post but a general question regarding calculations with the binomial theorem that have either a fraction or a negative exponent.

As I understand it if the binomial has a positive integer as the exponent, after expansion there will be that number plus one terms. Example a binomial to the power of 2 has three terms after expansion. Exponent of 6 then seven terms etc. When using n choose k, you stop calculating when n=k.

How many terms are there when the exponent is a fraction or negative?

I realise I have many hole in my maths fundamentals, but I am determined to get this right. Any assistance is always much appreciated.

Look up Taylor Series. In both cases the series will be infinite.

-Dan

 

[Dr.Peterson]

I don't have a specific math problem to post but a general question regarding calculations with the binomial theorem that have either a fraction or a negative exponent.

As I understand it if the binomial has a positive integer as the exponent, after expansion there will be that number plus one terms. Example a binomial to the power of 2 has three terms after expansion. Exponent of 6 then seven terms etc. When using n choose k, you stop calculating when n=k.

How many terms are there when the exponent is a fraction or negative?

I realise I have many hole in my maths fundamentals, but I am determined to get this right. Any assistance is always much appreciated.

Try that with an example, and see what happens! (You'll find that it never ends ... and you'll see why.)

Often the way to start on a general question is with a specific example; at the least, that gives you some sense of how it works. It's also good when asking questions, as you can point out exactly what you are asking.

 

[Dalmain]

Thank you very much topsquark and Dr.Peterson.
And thank you for the advice Dr.Peterson, I will remember that for next time.

 

[JeffM]

I believe that the OP has lost interest.

Standard statement of the binomial theorem.

[math]n \in \mathbb Z \text { and } n \ge 0 \implies (a + b)^n \equiv \left ( \sum_{j=0}^n \dbinom{n}{j} * a^{(n-j)} * b^j \right )\\ \dbinom{n}{j} \equiv \dfrac{n!}{j!*(n-j)!}.[/math]
Although the factorial function itself, and therefore the binomial coefficient) is defined only for non-negative integers, the definition of

[math]n \in \mathbb Z \text { and } x \ne 0 \implies x^{-n} \equiv \left ( \dfrac{1}{x} \right )^n \equiv \dfrac{1}{x^n}[/math]
lets us take care of negative integers with ease.

[math](a + b)^{-n} \equiv \left ( \dfrac{1}{a + b} \right )^n = \dfrac{1}{(a + b)^n} \implies\\ (a + b)^{-n} \equiv 1 \div \left ( \sum_{j=0}^n \dbinom{n}{j} * a^{(n-j)} * b^j \right )[/math]
Things are not so obvious if n is a rational number.

Let’s start here.

[math]n, \ j \in \mathbb Z \text { and } n \ge j \ge 1.\\ j = 1 \implies \dfrac{n!}{(n - j)!} = n = \left ( \prod_{i=1}^1 n + 1 - i \right).\\ j= 3 \implies \dfrac{n!}{(n - j)!} = n(n - 1)(n - 2) = \left ( \prod_{i=1}^2 n + 1 - i \right ). \\ j = n \implies \dfrac{n!}{(n - j)!} = \dfrac{n!}{0!} = n! = \left (\prod_{i=1}^n n + 1 - i \right ).\\ \text {By induction, } \dfrac{n!}{(n - j)!} = \left ( \prod_{i=1}^j n + 1 - i \right ). [/math]
So if we have integers j and n such that

[math]n \ge j \ge 1 \implies (a + b)^n = a^n + \left \{ \sum_{j=1}^n \dfrac{1}{j!} * a^{(n-j)} * b^j * \left ( \prod_{i=1}^j n + 1 - i \right) \right \} [/math]
But

[math] n + 1 - (n + 1) = 0 \implies \left ( \prod_{i=1}^j n+ 1 - j \right ) = 0 \text { if } j = n +1 \implies \\ (a + b)^n = a^n + \left ( \sum_{j=1}^{\infty} \dfrac{1}{j!} * a^{(n-j)} * b^j * \left ( \prod_{i=1}^j n + 1 - i \right ) \right \}.[/math]
Now we have no factorials that can possibly be based on a non-negative integer. We can extend to rationals.

 

[Dr.Peterson]

Or you can go with Newton's generalized binomial theorem (basically the same thing) which leads to an infinite series for negative numbers or non-integers.

 

[Dalmain]

I believe that the OP has lost interest.

Standard statement of the binomial theorem.

[math]n \in \mathbb Z \text { and } n \ge 0 \implies (a + b)^n \equiv \left ( \sum_{j=0}^n \dbinom{n}{j} * a^{(n-j)} * b^j \right )\\ \dbinom{n}{j} \equiv \dfrac{n!}{j!*(n-j)!}.[/math]
Although the factorial function itself, and therefore the binomial coefficient) is defined only for non-negative integers, the definition of

[math]n \in \mathbb Z \text { and } x \ne 0 \implies x^{-n} \equiv \left ( \dfrac{1}{x} \right )^n \equiv \dfrac{1}{x^n}[/math]
lets us take care of negative integers with ease.

[math](a + b)^{-n} \equiv \left ( \dfrac{1}{a + b} \right )^n = \dfrac{1}{(a + b)^n} \implies\\ (a + b)^{-n} \equiv 1 \div \left ( \sum_{j=0}^n \dbinom{n}{j} * a^{(n-j)} * b^j \right )[/math]
Things are not so obvious if n is a rational number.

Let’s start here.

[math]n, \ j \in \mathbb Z \text { and } n \ge j \ge 1.\\ j = 1 \implies \dfrac{n!}{(n - j)!} = n = \left ( \prod_{i=1}^1 n + 1 - i \right).\\ j= 3 \implies \dfrac{n!}{(n - j)!} = n(n - 1)(n - 2) = \left ( \prod_{i=1}^2 n + 1 - i \right ). \\ j = n \implies \dfrac{n!}{(n - j)!} = \dfrac{n!}{0!} = n! = \left (\prod_{i=1}^n n + 1 - i \right ).\\ \text {By induction, } \dfrac{n!}{(n - j)!} = \left ( \prod_{i=1}^j n + 1 - i \right ). [/math]
So if we have integers j and n such that

[math]n \ge j \ge 1 \implies (a + b)^n = a^n + \left \{ \sum_{j=1}^n \dfrac{1}{j!} * a^{(n-j)} * b^j * \left ( \prod_{i=1}^j n + 1 - i \right) \right \} [/math]
But

[math] n + 1 - (n + 1) = 0 \implies \left ( \prod_{i=1}^j n+ 1 - j \right ) = 0 \text { if } j = n +1 \implies \\ (a + b)^n = a^n + \left ( \sum_{j=1}^{\infty} \dfrac{1}{j!} * a^{(n-j)} * b^j * \left ( \prod_{i=1}^j n + 1 - i \right ) \right \}.[/math]
Now we have no factorials that can possibly be based on a non-negative integer. We can extend to rationals.

Thank you very much for your reply. I have not lost interest, Dr.Peterson and topsquark had answered my question. But again, thank you for your reply.