Wednesday, March 11, 2015
Tuesday, September 9, 2014
More stuff I made
I've always liked this quote (not clear on who to attribute it to, the internet seems confused on the issue):
I managed to sneak in a little Declaration of Independence to drive home the point.
Available at redbubble on all kinds of stuff. Enjoy!
While I like the undermining of the princess-in-distress-meme, I think we could and should go further and undermine the whole idea of monarchy, especially with neo-monarchists getting a tiny bit of traction (WTF's with that?). My new designed does just that:I'm not a princess, I don't need saving.I'm a queen, I got this shit handled.
I managed to sneak in a little Declaration of Independence to drive home the point.
Available at redbubble on all kinds of stuff. Enjoy!
Tuesday, July 29, 2014
More on 0.9999... = 1!
When the internet talks about 0.9999 equals 1, how many nines do they use? Here's a graph of exactly that.
While there is a bump at 9, people seem to think that 9 nines is nice and symmetric, there isn't a bump at 10. There is also a bump at 32, a nice power of two.
Of course, this plot should remind you of this:
\[\sum_{i = 1}^{\infty} \frac{1}{2^i}\] also converges to 1. It's an interesting fact that in base two, 0.1111... = 1. In fact, for every base b, you can represent one as an infinite sequence of \((b-1)\)s.
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Number of nines in "0.999999", etc, as reported by Google.
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Of course, this plot should remind you of this:
One more thing.
At the end of the last post, I asserted that:\[\sum_{i = 1}^{\infty} \frac{1}{2^i}\] also converges to 1. It's an interesting fact that in base two, 0.1111... = 1. In fact, for every base b, you can represent one as an infinite sequence of \((b-1)\)s.
Sunday, July 27, 2014
It never ends: 0.9999... = 1!
While a mini-vacation/wine-soaked-trip around Napa, an old high-school friend was asking me about the whole 0.9999.... = 1 thing. Sometimes this is written as \(0.\bar{9} = 1\). So I thought I'd dip my toe into and write about that here. This will also server as a test run of putting math equations into the blog. Also, it ties in nicely with my \(\epsilon\) tee-shirt design.
I'm usually careful to not use exclamation points in math equations because they could represent a factorial and not just the normal excitement inherent in any discussion of mathematics but in this case it is ok since \(1! = 1\).
First off, the most popular "proof" that 0.9999... = 1 goes like this. Suppose that
\[
n = 0.9999\ldots
\]
then multiple both sides by 10, which perseveres the equality:
\[
10n = 9.9999\ldots.
\]
Then you can subtract the first equation from the second and get:
\[
9n = 9.0000\ldots
\]
which gives that n = 1. The problem with this is that it's a bit of mathematical slight of hand and hides what's really going on. The problem is what do we even mean by 0.9999...?
0.9999... is usually thought of as an infinity/never-ending sequence of nines following the decimal point. You can write this as a sum:
\[
9/10 + 9/100 + 9/1000 + \cdots = \frac{0}{10^0} + \frac{9}{10^1} + \frac{9}{10^2} + \frac{9}{10^3} + \cdots.
\]
This can be succinctly written as:
\[
\sum_{i = 1}^{\infty} \frac{9}{10^i}.
\]
So 0.9999... isn't number. It's a never ending sum. There is no point where the process stops and spits out the final number. Since it's not a number, what does it mean to ask what number does 0.9999... equal? How can an infinite process equal a number, one is a process and the other is a number? There is a mismatching of concepts.
There is a way (several in fact) where one can assign a value to the above infinite sum. First we start by looking at the partial sum, i.e. the sum where we stop after n terms:
\[
S(n) = \sum_{i = 1}^{n} \frac{9}{10^i}
\]
Two observations:
\[
\forall \epsilon >0 \ \ \exists n \ \ \ \text{s. t.}\ \ \ \forall m > n \left| S(m) - 1\right| < \epsilon.
\]
I think that morass of symbols obscures more than it revivals. The ideas are the thing. In this case, the difference between S(n) and 1 has a simple closed form (left to the reader to derive) and it's clear that we can pick n such that difference is less than \(\epsilon\) for any positive \(\epsilon\). When this is the case, we say that S(n) converges to 1 as n goes to infinity. It's important to note that n never reaches infinity (whatever that might mean).
Thus \(0.9999... = \sum_{i = 1}^{\infty} \frac{9}{10^i} = 1\) is really shorthand for \(\sum_{i = 1}^{n} \frac{9}{10^i}\) converges to 1.
There are lots of infinite sums that converge to 1. Here's another famous one:
\[\sum_{i = 1}^{\infty} \frac{1}{2^i}.\]
Once you wrap your head around that one, what about this:
\[\sum_{i = 1}^{\infty} (-1)^n.\]
Does that converge? If so, to what?
I'm usually careful to not use exclamation points in math equations because they could represent a factorial and not just the normal excitement inherent in any discussion of mathematics but in this case it is ok since \(1! = 1\).
First off, the most popular "proof" that 0.9999... = 1 goes like this. Suppose that
\[
n = 0.9999\ldots
\]
then multiple both sides by 10, which perseveres the equality:
\[
10n = 9.9999\ldots.
\]
Then you can subtract the first equation from the second and get:
\[
9n = 9.0000\ldots
\]
which gives that n = 1. The problem with this is that it's a bit of mathematical slight of hand and hides what's really going on. The problem is what do we even mean by 0.9999...?
Wherefore art thou 0.9999...?
\[
9/10 + 9/100 + 9/1000 + \cdots = \frac{0}{10^0} + \frac{9}{10^1} + \frac{9}{10^2} + \frac{9}{10^3} + \cdots.
\]
This can be succinctly written as:
\[
\sum_{i = 1}^{\infty} \frac{9}{10^i}.
\]
So 0.9999... isn't number. It's a never ending sum. There is no point where the process stops and spits out the final number. Since it's not a number, what does it mean to ask what number does 0.9999... equal? How can an infinite process equal a number, one is a process and the other is a number? There is a mismatching of concepts.
There is a way (several in fact) where one can assign a value to the above infinite sum. First we start by looking at the partial sum, i.e. the sum where we stop after n terms:
\[
S(n) = \sum_{i = 1}^{n} \frac{9}{10^i}
\]
Two observations:
- for a given value of n, the sum S(n) is a finite and has a definite value.
- for larger and larger values of n, S(n) does get closer and closer to 1.
\[
\forall \epsilon >0 \ \ \exists n \ \ \ \text{s. t.}\ \ \ \forall m > n \left| S(m) - 1\right| < \epsilon.
\]
I think that morass of symbols obscures more than it revivals. The ideas are the thing. In this case, the difference between S(n) and 1 has a simple closed form (left to the reader to derive) and it's clear that we can pick n such that difference is less than \(\epsilon\) for any positive \(\epsilon\). When this is the case, we say that S(n) converges to 1 as n goes to infinity. It's important to note that n never reaches infinity (whatever that might mean).
Thus \(0.9999... = \sum_{i = 1}^{\infty} \frac{9}{10^i} = 1\) is really shorthand for \(\sum_{i = 1}^{n} \frac{9}{10^i}\) converges to 1.
There are lots of infinite sums that converge to 1. Here's another famous one:
\[\sum_{i = 1}^{\infty} \frac{1}{2^i}.\]
Once you wrap your head around that one, what about this:
\[\sum_{i = 1}^{\infty} (-1)^n.\]
Does that converge? If so, to what?
Friday, July 11, 2014
More things I made.
The greek letter ε or epsilon is used in mathematics to represent a small real (or real small) number. To be more precise, an arbitrarily small positive real number, most famously used in the δ-ε definition of a limit, which you probably hated in calculus class. This designe commemorates the diversity of ways that ε has been written throughout history. I wanted the last line to be printed on the back but RedBubble doesn't support that (yet?):
I think the tote bag also looks good.
I think the tote bag also looks good.
Sunday, June 29, 2014
Something is rotten in the state of Facebook?
Facebook must have decided that people don't find their social media siteadvertising platform creepy enough, I know I didn't. Then Adam Kramer of Facebook's Core Data Science Team along with collaborators from UCSF and Cornell published the results of a social experiment performed on their users. The authors modified the contents of ~680,000 user's New Feed to see if they were able to manipulate the emotional state of Facebook users.
Federally funded research experiments, such as this, are required to be overseen by an institutional review board (IRB). The main job of an IRB is to protect subjects from physical or psychological harm. The events leading to the formation of IRBs are dark, very dark. A key principle used is that of informed consent. The editor at the journal is quoted as saying that the study was IRB approved (it is not clear which IRB, UCSF or Cornell) based on Facebook's terms of service agreement. The Facebook data use policy does mention research:
I would have felt better if the study had somehow dodged getting IRB approval. That a university IRB and Facebook's own "strong" internal review process would rubber stamp this experiment is depressing. Maybe this is just a brief ethical lapse and not the sign of a troubling disregard of ethical issues in tech, but it is worrying.
Federally funded research experiments, such as this, are required to be overseen by an institutional review board (IRB). The main job of an IRB is to protect subjects from physical or psychological harm. The events leading to the formation of IRBs are dark, very dark. A key principle used is that of informed consent. The editor at the journal is quoted as saying that the study was IRB approved (it is not clear which IRB, UCSF or Cornell) based on Facebook's terms of service agreement. The Facebook data use policy does mention research:
we may use the information we receive about you: ... for internal operations, including troubleshooting, data analysis, testing, research and service improvement.The data use policy is about the privacy, the URL even contains the word. Facebook's response to the storm of criticism unleashed by this paper has focused on the privacy issue. That would be the issue if the paper was an observational study but it is not. I'm not a lawyer but the only information I could find on how Facebook manipulates their user's relates to the pairing of advertisements with social context. There is nothing within a stone's throw of informed consent covering active experimentation on their users.
I would have felt better if the study had somehow dodged getting IRB approval. That a university IRB and Facebook's own "strong" internal review process would rubber stamp this experiment is depressing. Maybe this is just a brief ethical lapse and not the sign of a troubling disregard of ethical issues in tech, but it is worrying.
Wednesday, June 11, 2014
Recipe: Kale Salad
This is the first of several recipes that I hope to share here. Enjoy!
Kale Salad
Ingredients
Salad
1 bushel of kale
½ bushel of cilantro
½ jalapeño pepper (optional)
½ yellow pepper (or 1 small yellow pepper)
1 Belgian endive
⅓ cup slivered or muddled roasted almonds
½ cup shredded romano cheese
Dress
¼ cup olive oil
1 tablespoon sesame oil
1 tablespoon soy sauce
2 tablespoon balsamic vinegar
¼ teaspoon mustard powder (optional)
¼ teaspoon black pepper
Preparation
Finely chop the kale, cilantro, pepper(s), and endive into a salad bowl with the almonds. In another bowl mix all the dressing ingredients. Add the mustard powder last (only used as an emulsifier) and beat with a fork until it has an even, foamy texture. Add the dressing to the salad and toss. Serve at room temperature with the cheese and freshly ground black pepper.
Other Ideas
Use peanut oil instead of sesame oil. Use peanuts or walnuts instead of almonds. Add golden beats, white beans, tofu, anchovies, or quinoa. Add finely green garlic or green onions. Use fresh lime juice instead of balsamic vinegar.
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