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

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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

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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

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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)

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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

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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

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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

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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

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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

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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

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