Place value function

This has probably already been done before, but I had lots of fun discovering it. :)

I wanted to create a function ff, which, given a number nn, would take in the place value of a digit dd of nn (xx, such that the place value of dd is 10x10^{x}), and output dd itself.

For example, if n=1345.829n = 1345.829, than f(0)=5f(0) = 5, f(3)=1f(3) = 1, and f(2)=2f(-2) = 2.

Here's what I came up with:

f(x)=nmod10x+1nmod10x10xf(x) = \dfrac{n \operatorname{mod} 10^{x+1} - n \operatorname{mod} 10^{x}}{10^{x}}

And a Desmos graph.

I came up with it somewhat inductively, noting patterns while calculating the units digit, the tens digit, and so on, but I understand now why it works. The function essentially chops off everything after the desired place value, subtracts from that everything before the desired place value, and then divides by the power of ten associated with that place value. For example:

n=1345.829n = 1345.829, x=2x = 2 (should retrieve the digit at 10210^2, or 33)

nmod102+1=345.829n \operatorname{mod} 10^{2+1} = 345.829

nmod102=45.829n \operatorname{mod} 10^{2} = 45.829

345.82945.829=300.000345.829 - 45.829 = 300.000

300.000102=3\frac{300.000}{10^{2}} = \boxed{3}

#NumberTheory

Note by David Stiff
3 months, 1 week ago

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Comments

Now you can trick people by making it look a little more complicated using the method I had shown earlier using trig and inverse trig, making it non-obvious to the naked eye(the formula you have shown is what I normally use for getting digits in a computer program, but I was just reminded that I had found the modulus function in terms of trig and inverse trig)

Jason Gomez - 3 months, 1 week ago

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How sinister! :)

I don't seem to remember seeing a note about that... could you add the link here?

David Stiff - 3 months, 1 week ago

It was actually in a daily challenge, in a comment as a reply to Siddarth’s comment(one or two fire), I don’t seem to remember the name so I have to go on treasure hunt too

Jason Gomez - 3 months, 1 week ago

Aha found it(joking it was quite recent) First solution, first comment, first reply to that, first reply to the first reply

Jason Gomez - 3 months, 1 week ago

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Rats, I'm not a Premium member. It says the problem has expired. :/

David Stiff - 3 months, 1 week ago

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@David Stiff Oh in that case I will put down here what I did there

Jason Gomez - 3 months, 1 week ago

@David Stiff "A better one is 10πcot1(cot(3π10x))\frac{10}{π}\cot^{-1}(\cot(\frac{3π}{10}x)) "

"Got the perfect one 5+10π(tan1(tan(3π10xπ2)))5+\frac{10}{π}(\tan^{-1}(\tan(\frac{3π}{10}x-\frac{π}{2}))) "

These are both for 3xmod103x \mod {10}, a little modification should get them to xmod10nx \mod {10^n}

Jason Gomez - 3 months, 1 week ago

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@Jason Gomez Very cool! Confusing at first too. :)

David Stiff - 3 months, 1 week ago

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@David Stiff I tried out the graph using this function, it didn’t look good

Jason Gomez - 3 months, 1 week ago

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@Jason Gomez Hm. For the trig one you mean?

David Stiff - 3 months, 1 week ago

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@David Stiff Yeah I tried the trig, but I assume neither will look good

Jason Gomez - 3 months, 1 week ago

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@Jason Gomez Yeah, the simpler one looks the same I think. It only really looks nice if you keep nn constant and then graph f(x)f(x) for all integers between like x=10x = -10 and x=10x = 10 (or something else depending on nn).

David Stiff - 3 months, 1 week ago

@Jason Gomez In fact when you add or subtract inverse trig stuff, it can be simplified to a single inverse trig expression, so after that it will look all the more confusing

Jason Gomez - 3 months, 1 week ago
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