Binomial theorem for non integers ? O_o ??

Question

Could we use the binomial theorem for non integers? This comes from:

$$\sqrt{(a+b)}$$ which I can write as $$(a+b)^{1/2}$$

Could I then use the binomial theorem to figure out the value of this expression?

Answer 1

The generalized binomial theorem is actually a special case of Taylor's theorem, which states that

$$f(x)=\sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}(x-a)^k$$

Where $f^{(k)}(a)$ is the $k$th derivative centered at $a$. In particular if we have $f(x)=x^t$, note that

$$f'(x)=tx^{t-1}\\f''(x)=t(t-1)x^{t-2}\\\vdots\\f^{(k)}(x)=t(t-1)(t-2)\dots(t-k+1)x^{t-k}$$

Let $x=a+b$ and we find that

$$f(x)=(a+b)^t=\sum_{k=0}^\infty\frac{t(t-1)\dots(t-k+1)}{k!}a^{t-k}b^k$$

which is the generalized binomial expansion theorem! Special case of $t=-1,-2$ are interesting, as it gives

$$(1-x)^{-1}=1+x+x^2+x^3+\dots\\(1-x)^{-2}=1+2x+3x^2+4x^3+\dots$$

Answer 2

Yes: $$ (a+b)^{\frac{1}{2}}=a^{\frac{1}{2}}\sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}\bigg(\frac{b}{a}\bigg)^k $$ EDIT: fractional binomial coefficients are calculated as the regulae ones. Just replace the fraction with, say, $\alpha$ and use the definition: $$ \binom{\alpha}{k}=\alpha \cdot (\alpha-1) \cdots (\alpha-k+1) \cdot \frac{1}{k!} $$ and then substitute the fraction. What do you get?

Answer 3

You can see the following link, that explanes what occurs when the exponent is a complex number: http://en.wikipedia.org/wiki/Binomial_series

Answer 4

The Binomial Theorem says $$ \begin{align} (1+x)^{1/2} &=\sum_{k=0}^\infty\binom{1/2}{k}x^k\\ &=\sum_{k=0}^\infty\frac{(-1)^{k-1}}{2k-1}\binom{2k}{k}\left(\frac{x}4\right)^k\\ \end{align} $$ Therefore, if $|a|\le b$, $$ (a+b)^{1/2}=b^{1/2}\sum_{k=0}^\infty\frac{(-1)^{k-1}}{2k-1}\binom{2k}{k}\left(\frac{a}{4b}\right)^k $$