Browsing category binomial

Ask Uncle Colin: Why is it not 4?

Dear Uncle Colin, I have a binomial expansion of $(1+x)^\frac{1}{2}$ and need to approximate $\sqrt{5}$. Apparently you need to substitute in $x=\frac{1}{4}$, but I'd have thought $x=4$ was a more obvious choice. What gives? -- Roots Are Dangerous If Understood Sloppily Hi RADIUS, and thanks for your message! That does

The Mathematical Ninja and the Nineteenths

"Look," said the student, "we all know how this goes down. A nasty-looking fraction comes out of the sum, I reach for the calculator, you commit some act of exaggerated violence and tell me how you, o wondrous one, can do it in your head." "You're not as dumb as

Why $\phi^n$ is nearly an integer

This article is one of those 'half-finished thoughts' put together late at night. Details are missing, and -- in a spirit of collaboration -- I'd be glad if you wanted to fill them in for me. The estimable @onthisdayinmath (Pat in real life) recently posted about nearly-integers, and remarked that

A STEP expansion

A STEP question (1999 STEP II, Q4) asks: By considering the expansions in powers of $x$ of both sides of the identity $(1+x)^n (1+x)^n \equiv (1+x)^{2n}$ show that: $\sum_{s=0}^{n} \left( \nCr{n}{s} \right)^2 = \left( \nCr{2n}{n} \right)$, where $\nCr{n}{s} = \frac{n!}{s!(n-s)!}$. By considering similar identities, or otherwise, show also that: (i)

Ask Uncle Colin: Is my friend crazy?

Dear Uncle Colin, A friend of mine told me that $1 + 2 + 4 + 8 + ... = -1$. Is he crazy, or is there something going on here? -- Somehow Enumerating Ridiculous Infinitely Extended Sum Dear SERIES, There are a couple of 'proofs' of this non-fact that

Ask Uncle Colin: A logarithmic coincidence?

Dear Uncle Colin, I noticed that $2^{\frac{1}{1,000,000}} = 1.000 000 693 147 2$ or so, pretty much exactly $\left(1 + \frac{1}{1,000,000} \ln(2)\right)$. Is that a coincidence? Nice Interesting Numbers; Jarring Acronym Dear NINJA, The easiest way to see that it's not a coincidence is to check out $3^{\frac{1}{1,000,000}}$, which

Why the Maclaurin series gives you Pascal’s Triangle

The Mathematical Ninja, some time ago, pointed out a curiosity about Pascal's Triangle and the Maclaurin1 (or Taylor2 ) series of a product: $\diffn{n}{(uv)}{x} = uv^{(n)} + n u'v^{(n-1)} + \frac{n(n-1)}{2} u'' v^{(n-2)} + ...$, where $v^{(n)}$ means the $n$th derivative of $v$ - which looks a lot like Pascal's

Blazing through the Binomial Expansion

"Where's the Mathematical Ninja?" asked the student. "He's... unavoidably detained," I said. In fact, he was playing Candy Crush Saga. But sh. "What can I help you with today?" "Well, you know the binomial expansion...?" "Intimately," I said. "Well, I got it pretty well at C2... but now we're doing

Why I don’t buy that $1 + 2 + 3 + … = -\frac{1}{12}$

Thanks to Robert Anderson for the question. I told him that the sum of an infinite number of terms of the series: 1 + 2 + 3 + 4 + · · · = −1/12 under my theory. If I tell you this you will at once point out to

Secrets of the Mathematical Ninja: Pascal’s Triangle

You've seen Pascal's triangle before: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 You get the number in each row by adding its two 'parents' - for instance, each 10 in the row that starts with 1