Notes - Analysis II HT23, Binomial theorem

Created: February 27, 2023

Flashcards

Can you state the real binomial theorem, about

\[(1 + x)^p\]

?

Let $p \in \mathbb R$. Then for all $ \vert x \vert < 1$

\[(1 + x)^p = \sum_{k=0}^\infty {p \choose k} x^k\]

The real binomial theorem states that for all $p \in \mathbb R$, and for all $ \vert x \vert < 1$,

\[(1 + x)^p = \sum_{k=0}^\infty {p \choose k} x^k\]

What definition of $p \choose k$ is used here?
\[{p \choose k} = \frac{p(p-1)(p-2)\cdots(p-k+1)}{k!}\]

Quickly prove that

\[{p \choose k} = \frac{p}{k} {p-1\choose k-1} = \frac{p-k+1}{k} {p\choose k-1}\]

and that

\[{p \choose k} + {p \choose k-1} = {p+1 \choose k}\]

First one is immediate from taking out a factor of $\frac p k$. Then

\[\begin{aligned} {p \choose k} + {p \choose k-1} &= \frac{p-k+1}{k} {p \choose k-1} + {p \choose k-1} \\\\ &= \frac{p+1}{k} {p \choose k-1} \\\\ &= {p+1 \choose k} \end{aligned}\]

When proving the real binomial theorem, i.e. that

If $p \in \mathbb R$ then for all $ \vert x \vert < 1$

\[(1 + x)^p = \sum _ {k=0}^\infty {p \choose k} x^k\]

what two lemmas do you use about

\[{p \choose k}\]

that are useful in the proof?
\[{p \choose k} = \frac{p}{k} {p-1\choose k-1} = \frac{p-k+1}{k} {p\choose k-1}\]

and

\[{p \choose k} + {p \choose k-1} = {p+1 \choose k}\]

When proving the real binomial theorem, i.e. that

If $p \in \mathbb R$ then for all $ \vert x \vert < 1$

\[(1 + x)^p = \sum _ {k=0}^\infty {p \choose k} x^k\]

You have two seperate functions

\[(1 + x)^p\]

and

\[\sum_{k=0}^\infty {p \choose k} x^k\]

What relationship do you want to show true about both of them in terms of their derivatives, that you can exploit later?
\[f'(x) = \frac{p}{1+x} f(x)\]

When proving the real binomial theorem, i.e. that

If $p \in \mathbb R$ then for all $ \vert x \vert < 1$

\[(1 + x)^p = \sum _ {k=0}^\infty {p \choose k} x^k\]

One of the steps is to justify that

\[g(x) = \sum_{k=0}^\infty {p \choose k} x^k\]

satisfies

\[(1+x)g'(x) = pg(x)\]

Using the fact that

\[{p \choose k} = \frac{p - k+1}{k}{p \choose k-1}\]

can you justify this?
\[\begin{aligned} (1+x)g'(x) &= (1+x)\sum^\infty_{k=1} {p \choose k} kx^{k-1} \\\\ &=\sum^\infty_{k=1} {p \choose k} kx^{k-1} + \sum^\infty_{k=1} {p \choose k} k x^k \\\\ &= \sum^\infty_{k=0} {p \choose k+1} (k+1)x^k + \sum^\infty_{k=0} {p \choose k} kx^k \\\\ &= \sum^\infty_{k=0} {p \choose k} \frac{p-(k+1)+1}{(k+1)}(k+1)x^k + \sum^\infty_{k=0} {p \choose k} kx^k \\\\ &= \sum^\infty_{k=0} (p-k)x^k + \sum^\infty_{k=0} {p \choose k} k x^k \\\\ &= p\sum^\infty_{k=0} {p \choose k} x^k \end{aligned}\]

Proofs

Prove the real binomial theorem:

Let $p \in \mathbb R$. Then for all $ \vert x \vert < 1$,

\[(1+x)^p = \sum^\infty _ {k=0} {p \choose k} x^k\]

Todo.



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