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[tallguytutoring]

Cauchy Sequences Problem 1

So I was brushing up on my Cauchy sequences the other day. Man what a fucking blast!

I was having a tough time trying to figure out how Cauchy sequences relate to convergent sequences. My conclusion is that they actually describe two different properties about sequences. The Cauchy criterion establishes a condition on how close terms must get to each other. Convergence establishes how close terms get to a destination. 

My intuition tells me that in general, they are very much related. For example, convergent sequences are all Cauchy sequences. But sometimes you can’t go the other way around (check out  completeness).  

For this short series on Cauchy sequences, we’ll use that we already have two Cauchy sequences of real numbers, \(\{a_n\}_{n \geq 1}\) and  \(\{b_n\}_{n \geq 1}\).

Problem 1

Show that each of \(\{a_n\}\) and  \(\{b_n\}\) are bounded above and below

Solution Intuition

As always, take some time to piece together WHY this is true. The way I like to do this is to start by trying to “break” it. In this case, I’ll look for a Cauchy sequence that isn’t bounded above. 

My first try was the sequence \(0.9,1,1.99,2,2.999,3,...\)

I thought it was pretty clever because the terms, consecutively, get infinitely close together, and there is no upper bound to the sequence.

Learning Opportunity: 

Can you tell why my example isn’t actually a counter example?

Once you realise WHY my example doesn’t work, the solution to this problem becomes quite clear. 

Terms is a Cauchy sequence are eventually ALL arbitrarily close. It’s not enough that some weird consecutive pairing happens to have terms which become arbitrarily near.

Formalish Solution

Let \(\epsilon>0\) be randomly chosen by fairies, and let \(N \in \mathbb{N}\) be chosen such that \( |a_n-a_m|<\epsilon\) whenever \(n,m>N\). 

All the \(a_n\) terms for \(n>N\) fall within an \(\epsilon\) stripe so they are obviously bounded. The terms before and including the \(N\)th term form a finite set, which admits both a maximum and a minimum.

Idea: set the upper bound to \(max\{a_1,a_2,...,a_{N+1}\}+\epsilon\) and the lower bound to  \(min\{a_1,a_2,...,a_{N+1}\}-\epsilon\).

Learning Opportunity 

Why must we go to \(N+1\), and not just \(N\)? Provide an example of a sequence whose upper bound is NOT  \(max\{a_1,a_2,...,a_{N}\}+\epsilon\)

Learning Opportunity Answer

Only when you’ve had time to think about the question, and try it...

Thinking Space. Thinking Space. Thinking Space.

Try the sequence \(1,2,3,3.9,3.99,3.999,...\) and set \(\epsilon=0.1\). Then \(N=3\) works, with \(max\{1,2,3\}+\epsilon=3+0.1\) which is not an upper bound. Daaaaammmmmmnnnnn.

[tallguytutoring]

The Fundamental Theorem of Calculus actually makes sense?

The Fundamental Theorem of Calculus makes SO much sense now

Note that to view the math typography you must open the full blog post.

Warning: This article is NOT for high school students. I assume you'll know some more advanced math syntax that you'd normally pick up in university. I also assume you're familiar with FTC, or that you've been exposed to it, at the very least. I might do a more in depth version later, but not right now.

I know, I know, the FTC tells us that we can integrate continuous functions and that the derivative of an integral gives you back your original function.

That's pretty much the least sexy way to think about it ever. Guys and girls, remember that math is the most sexy thing you can do for your partner, so it's important that you get it right.

Here's the full unsexy version of the FTC:

Suppose that \(f\) is continuous on \( I=\{x:a\leq x \leq b\}\) and that \(F\) is defined by \[F(x)=\int_a^xf(t)dt.\] Then \(F\) is continuous on \(I\) and \(F’=f(x)\) for each \(x \in I_1=\{x:a<x<b\}\).

Let's take a moment to understand what they are saying (I'll keep it short) and then we'll dive into why it makes SO much sense geometrically.

Do you know the geometrical significance of \(F(x)\)? If you don't, take a look at how it's defined. It's just the area under the \(f\) curve, starting at \(t = a\) and ending at \(t = x\). Nothing new here. 

What we claim next is where things get interesting. We claim that the derivative, \(F’\) , gives us back our original function \(f\). 

Actually I kinda lied to you when I said that was the interesting part. I mean, I suppose that was the interesting part of the FTC when you first saw it. By now, however, most students are quite comfortable with the relationship between derivatives and anti-derivatives. Everyone and their dog knows that they "undo" each other. 

As it turns out, there is a super intuitive way to think about the FTC. And this my friends, is why we are gathered here today.

The first thing I want you to see is that if \(F'(x)=f(x)\), that means that \[\frac{dF}{dx}=f(x) \]

or in other words:

Every super small change in the \(x\) variable (we use \(dx\) to represent this change) will produce a change of \(f(x)\) in the \(F\) variable.

Re-read that sentence until you get it get it get it get it. Another way to see it is that for every increment of \(dx\) in the \(x\) variable, we get an increment of \(f(x)\) in the \(F\) variable (which is like the \(y\) variable here).

The next thing I want you to remember is that our function \(F\) represents the area under the \(f\) curve up to whatever \(x\) value you specify.

With that said, we can now re-state the FTC in a super sweet way.  \[\frac{dF}{dx}=f(x) \]

means that:

every super small change in the \(x\) variable (\(dx\)) will produce a change of \(f(x)\) in the area under the \(f\) curve.

OK, now that's pimpin sweet. THAT, is something I can try to wrap my puny brain around and visualize! Awwwww yeah!

So let's do that. We're so close!

Shel, shut up and show us a pimpin sweet picture!

Shut up people! I'll show you a damn picture! And I'll make it pimpin sweet, as commanded!

Pic 1: Notice that when \(x=2\), the added area, or change in area, is a box of height approximately equal to \(f(2)\)

Now, everything should be coming together for you with regards to FTC, and life in general. OF COURSE it makes sense now. Of COURSE the change in area at some \(x\) value must be the area of the little \(f(x)dx\) sliver that you add on to the end of it. 

Thanks for reading you weirdos! I hope this brought some intuition to the FTC, something that has eluded me for some time! 

Now go and think about math! :P Until next time!

[tallguytutoring]

Epsilon delta definition of continuous functions: A soft 101-introduction

Remember my rule: you aren't allowed to read this if it doesn't interest you. Well, you can start reading but then if it bores you you gotta stop! I only do this for fun and because I love it and so should you.

Note that to view the math typography you must open the full blog post.

I'm going to teach you a lesson taught in university math. Even if you’re just starting high school, I’ll bet that if you try hard, you’ll be able to understand this. This lesson usually causes a lot of headaches for students, but keep reading, and I think you’ll be fine.

For everyone who has no idea wtf a continuous function is, it’s a function that is relatively unbroken. Very loosely speaking, here is a continuous function and a non continuous one:

Pic 1: This red graph is a continuous function

Pic 2: This blue graph is NOT a continuous function

See that “break” in the blue curve? That’s the problem spot. That’s the point at which this function is not continuous. At every other \(x\) value, the blue function is continuous. 

The high school definition of a continuous function is essentially, “A function is continuous if you can draw it without taking your pen off the paper”. This definition is OK for intuition, but in university you’re going to need to understand some language and symbols and concepts that are more precise than this. 

That’s what this post is about. It’s about getting a taste for university precision regarding the definition of continuity. 

Here’s the definition (broken into easy language, but still retaining most of its precision) of a continuous function at a point \(x\) - which I want you to READ but you are NOT expected to understand it. Just see the words, and store them in your brain so that you are familiar with them as we go:

A function \(f\) is continuous at a point \(x\) if for any small interval around the image of \(x\) (on the \(y\)-axis), we can find a related interval around \(x\) (on the \(x\)-axis) where the images of any point in that interval are completely within the initial interval we picked around the image of \(x\).

Woah. Our goal in this post is to understand THAT. Fully. Completely. Now get pumped up!

Let’s talk about what a few of those words mean. That’s always the first step into understanding a statement: figure out exactly what all the words mean. Let’s talk about intervals, and image. Those are really the only two terms that might be new.

An interval around a point is just a group of points that are within a certain distance. For example, you might say that a \(5\)km radius from your house is a \(5\)km interval (note that this would then be a \(10\)km diameter). Another example of an interval would be the interval of radius \(1\) on the real line, centred around the point \(x=5\). That would look like this:

Pic 3: Interval on the real line of radius \(1\) around the point \(x=5\).

See how \(5\) is in the centre, and how every point on the red line is at most a distance of \(1\) away from \(5\)? That my friends, is an interval of radius \(1\) around the \(x\) value \(5\) (assuming we are restricted to the real line).The image of a point \(x\) is the point that our function sends \(x\) to. The image then, is a \(y\) value. It’s a point on the \(y\)-axis. If \(x\) is the input, \(f(x)\) is the image, or output. In the usual case of having a function that sends a real number to another real number, the image of a point \(x\) is just the height of the function at that particular \(x\) value. I like to think of it as the length of the shadow of \(x\).

Pic 4: The length of the blue line represents the image of \(x=5\)

Here is the definition broken into several chunks, which we’ll build on until we have the full definition. I’ll explain each number separately so it’ll be easier to understand.

A function \(f\) is continuous at a point \(x\) if for any small interval around the image of \(x\)

A function \(f\) is continuous at a point \(x\) if for any small interval around the image of \(x\), we can find a related interval around \(x\)

A function \(f\) is continuous at a point \(x\) if for any small interval around the image of \(x\), we can find a related interval around \(x\) where the images of every point in that interval fall completely within the initial interval we picked around the image of \(x\).

Let's start from the top.

A function \(f\) is continuous at a point \(x\) if for any small interval around the image of \(x\)

First thing to note is that for now, we are not asking if the whole function is continuous. We are only interested in seeing if a function is continuous at a single point \(x\). So the very first thing we do is pick the \(x\) value we’d like to analyse.

Now, we know what the image of \(x\) is (it’s the length of that blue line in Pic 4, directly above). Now what is an interval around the image? In this case, it’s important to note that an interval around the image is referring to an interval on the \(y\)-axis.

For example,

Pic 5: The green lines form an interval around the image of \(x=5\).

In this graph, the horizontal orange line represents the height of the image of \(x=5\). The two green lines are each a small distance away, creating a small interval around the image.

Important question for later: Can you see the different \(x\) values that produce images that fall within the two green lines?? Think about this until you know the answer or are completely stumped. The answer is below so don't read it until you've tried to figure it out. Invest one or two minutes.

Answer only after you've thought about it: It looks like the image of \(x=3\) would definitely be in the range. Same with \(x=5.1\), or \(4.7\). There are many many many \(x\) values whose images fall within the accepted range. In fact, Here is a picture showing all the \(x\)-values whose images fall within the desired range. Any \(x\)-value that’s part of a purple column will do the trick.

Pic 6:  The purple columns are there to help you visualize which \(x\) values have images that fall within the accepted green lines. Note that all \(x\) values that are not part of a purple column will have an image that is not between the green lines.

It's important to note that for this particular green interval around the image of \(x=5\), we CAN find an interval around \(5\) whose images all fall within the desired range.

What happens if we pick a smaller interval around the image of \(x=5\)? Let’s take a look:

Pic 7: the green lines are tighter due to a smaller interval around the image of \(x\). As a result, we now have fewer accepted \(x\) values whose image falls within the green lines.

With a smaller interval, the green lines hug the orange line a little closer. As a consequence, you’ll notice that there are far fewer \(x\) values whose images fall within the new, smaller interval. since fewer \(x\) values now have images that fall within the green lines, the purple columns will get thinner as the green lines get tighter.

OK so now we know what an interval around the image is, and we know that if you make the size of your interval around the image smaller, your purple column gets smaller as there are now fewer acceptable \(x\) values.

2. For ALL intervals around the image of \(x\), we can find a corresponding interval around \(x\)

If you understood point \(1\), this one will be a cinch. The “interval around \(x\)” are all the \(x\) values in the purple column. The only new concept here is that we’re finding a corresponding interval around \(x\), which sits somewhere on the \(x\)-axis. In other words, our interval around the image of \(x\) is on the \(y\)-axis, while this interval is on the \(x\)-axis.

3. For ALL intervals around the image of \(x\), we can find a corresponding interval around \(x\) where all their images fall within the interval around \(x\)’s image.

Note that every \(x\) value in the purple column has an image that fits between the green lines, since that’s the way we chose the purple column.

The essence of a function to be continuous at a point \(x\), is that we must be able to guarantee that for ANY interval around the image of \(x\) (on the \(y\)-axis), we can find a corresponding interval around \(x\) (on the \(x\)-axis) such that any of the images of the points in the \(x\) interval fall within the interval on the \(y\)-axis.

If our function \(f\) is continuous at \(x\), it means that no matter how tightly you make the green lines hug the orange line, you will always be able to find a corresponding purple column, with accepted \(x\) values on BOTH sides of your chosen \(x\) value (that's part of what an interval AROUND something means). If you can always do this around the point \(x=5\), we say that the function \(f\) (the red line) is continuous at the point \(x=5\).

That’s the language portion of this post! Now onto the logic!

Pic 8: A clearly non-continuous function at \(x=2\) 

Clearly, this function is NOT continuous at \(x=2\). There is a jump there which fucks us over, hard. But that’s not the interesting part. The interesting part is determining WHY this function fails the definition of continuity that we just took the time to understand.

This function fails the test for continuity because there are some intervals around the image of \(2\) that do NOT have a corresponding interval around \(2\) with the required property that those \(x\) values all have images that fall within the interval around the image. Man, that is such a mouthful.

To help understand, go here. Play with the \(a\) value, which in this case is acting like our epsilon (from this point forward, we will use the Greek letter \(\epsilon\) instead of the word epsilon) . Note that for \(\epsilon\) values less than \(1\), we can find no interval around \(2\) on the \(x\) axis that meets the requirements given in the definition. Thus, Not continuous at \(x=2\).

Now, if you understand everything up to here, then you actually already understand what it means for a function to be continuous according to the definition you’d see in university. If you feel like taking it one step further, which I certainly do, then let’s try and understand the language that is typically used in more advanced textbooks.

Let’s start using the following version of the definition, which means the EXACT same thing, except it’s a little more precise, and has way fewer words.

“For any \(\epsilon>0\), we can find a \(\delta>0\) such that \(|f(x)-f(a)|<\epsilon\) whenever \(|x-a|<\delta\).”

If you need to brush up on your absolute values, go ahead and do that. I won’t talk too much about them here. I will say this: If you write \(|x-2|<3\), for example, it represents an interval on the \(x\) axis. Which interval? It’s all the \(x\) values that are a distance of less than \(3\) away from \(2\).

If you don’t believe me, which you shouldn’t unless you actually SEE what’s going on, you should plug a few numbers into \(|x-2|<3\) to see which numbers satisfy the condition and which numbers do not.

Let’s make \(x=1\). Does it satisfy \(|x-2|<3\)? Check: \(|1-2|=|-1|=1<3\), so yes, \(x=1\) is a solution.

Using similar logic, you’ll see that the only \(x\) values that work are the \(x\) values that are less than a distance of \(3\) away from the number \(2\).

If we then replace the \(2\_- with an \(a\), like in \(|x-a|<3\), we’re now talking about all the \(x\) values that are a distance of less than \(3\) away from an unknown value \(a\). So, as \(a\) moves around the \(x\)-axis, our set of solutions will also move around to hug whatever value we assign to \(a\).

Finally, if we write \(|x-a|<\delta\), it still represents all the \(x\) values that are less than a certain distance away from \(a\), it’s just that we haven’t specified our distance, and we leave it as the value \(\delta\) (the greek letter delta).

Now, the other part of the new definition is \(|f(x)-f(a)|<\epsilon\). This will work the same way as our interval on the \(x\)-axis, but now we’re on the \(y\) axis. How can I tell? I can tell because we are using \(f(x)\), and \(f(a)\), which are the images of \(x\) and \(a\), respectively. Images are always represented by values on the \(y\)-axis.

As you might then expect, \(|f(x)-f(a)|\) represents the distance that \(f(x)\) is away from \(f(a)\). In our definition, we want it to be less than an unknown value, \(\epsilon\).

OK so let’s compare the two definitions we’ve used thus far, and try and see if we can see that they are really the same thing

Kinda shitty definition: A function \(f\) is continuous at a point \(x\) if for any small interval around the image of \(x\) (on the \(y\)-axis), we can find a related interval around \(x\) (on the \(x\)-axis) where the images of any point in that interval are completely within the initial interval we picked around the image of \(x\).

Much better def, which is different in language but IDENTICAL in content: For any \(\epsilon>0\), we can find a \(\delta>0\) such that \(|f(x)-f(a)|<\epsilon\) whenever \(|x-a|<\delta\).

Can you believe that’s the whole definition? Crazy right? You can see why mathematicians take the time to introduce complicated symbols. If they didn’t, it would take PAGES to explain concepts that you could otherwise explain in a sentence.

Here's an example to help drive home the concepts.

Pic 9: A continuous function given by the line \(y=x+1\) 

At which \(x\) values is the function continuous?

For any \(\epsilon\) value, which \(\delta\) value will guarantee the continuity condition is satisfied?

Answers:

1: This function is continuous at every \(x\) value.

2: This line has a slope of \(1\). That means that if we take an \(\epsilon\) interval around the image of any point, that top line of the interval will intersect our function exactly \(\epsilon\)  units to the right (slope of \(1/1\) means the line always has the same rise and run), and our bottom line of the interval will intersect our function exactly \(\epsilon\) units to the left. So, why not make \(\delta=\frac{\epsilon}{2}\) (anything less than \(\epsilon\) will do the trick)? That way we guarantee that we’ll always be within the required range.

To help see this, check this graph out. Notice that no matter how small you make your \(a\) value, there is always an interval around \(2\) that produce images within \(a\) units of \(f(2)\).

OK, if you made it this far you totally rock! Like, really, you do! I had a really fun time making this doc. Please send any comments you like, and yes I’ll read every single one. If you weren’t able to follow this whole thing, please tell me where it got confusing!

[tallguytutoring]

Formal definition of convergence: How close is infinitely close?

Do you guys remember why I write these posts? Cause math is awesome, and it's a ton of fun. You should have fun with it too! The shit mathematicians write in books is sometimes super awesome and cool. I challenge every reader to find a math topic/definition/theorem that they actually enjoy piecing together. Don’t do it because you should, do it because it’s fun.

Yes, it's true, there is a certain unparalleled nerdiness that comes with the "mathematician" label. But we like to get close to things too.

In fact, we so much like to get close to things, that our predecessors have come up with a way to precisely define what it means for some value to get infinitely close to another value, but while never touching it.

When a group of values, let’s call them \(x_1, x_2, x_3,\) etc.   "gets close" to another value, which we’ll call \(x\), mathematicians call this convergence. 

An example would be to let  \(x_1=4, x_2=4.9, x_3=4.99\), and so on into infinity. The pattern here is easy; at each ensuing index, we add an additional 9 to the end of the number.

The result is that we have a sequence of values that are converging to 5. Loosely speaking, they converge because they get infinitely close. They don’t have to ever be equal to 5; it’s enough that they get infinitely close!

The point of this post is to talk about the formal definition of convergence (which is the notion of getting infinitely close), the one that you’d see in university. but using only words that a high school student can understand. Maybe even an elementary school student, if they're diligent. My first examples are with sequences, as I figure that's a pretty good starting point. 

Let's start turning the ol' thinkin wheels

So that we can easily talk about different terms in our sequence, let’s let \(n\) represent an arbitrary index value.  \(x_n\) could represent the first term in the sequence, or the 1,000,000,000th, we don’t know. That will depend on what value \(n\) takes on in the end. What we do know, is that \(x_n\) is indeed one of the terms from the sequence, so it looks like a 4 with a bunch of 9′s after the decimal.    

I want you to take a second to think about what it could mean for a sequence of terms to be infinitely close to a value, in this case 5, but never touch it. Specifically, I want you to ask yourself:

As \(n\) gets infinitely large, what’s the smallest distance we can achieve between the terms in \(x_n\) and 5?

In math speak, we write that distance as  \(|x_n-5|\). 

To help us out, I’m going to slowly introduce some language for the beginners. Let’s let  \(\epsilon\) represent a distance goal for \(|x_n-5|\). In other words, if someone picks an  \(\epsilon\) value of 0.1, let’s see if we can pick our \(n\) large enough so that the distance between the terms in  \(x_n\) are now always less than 0.1 away from 5. Once we find the index that guarantees we hit our distance goal, we can set that index as a marker, meaning that every term with an index greater than our marker will also reach the distance goal. We label marker indecies with \(N\).

In math speak: Let \(N\) be any index value such that \(|x_n-5|<0.1\) as long as \(n>N\). Then \(N\) is our marker.

You are not allowed to read further until you have either re-read the preceding 2 paragraphs 5 times, or you understand it completely!

I think we can reach our distance goal of 0.1, if we let  \(n\) be anything larger than 2. Let’s check that claim, and then further check the distance for  \(n=4\), and \(n=10\).

If  \(n=2\), that means  \(x_n=4.9\), and then;

\[|x_2-5|=|4.9-5|=|-0.1|=0.1\]

Oh shit, I was WRONG. I was wrong because \(x_2\) is a distance of exactly 0.1 away from 5, but I wanted it to be less. Luckily, it’s an easy fix. I’m pretty sure  \(x_3\) will work.

It does.

As promised, let’s look at the distances that each of  \(x_4,\) and  \(x_{10}\) are away from 5. \[|x_4-5|=|4.999-5|=|-0.001|=0.001\] \[|x_{10}-5|=|4.999999999-5|=|-0.000000001|=0.000000001\] Everything make sense? If not, don’t be afraid to read it through again.

Something to notice

You know what’s kind of weird? I’m pretty sure that no matter which term I take in my sequence, it will never be equal to 5, as there will always be some small positive number I can add to \(x_n\) to make it equal to 5.

The second thin to notice is that indeed, no matter how small I make \(\\epsilon\), I can always find an  \(n\) value that makes \(|x_n-5|<\epsilon\).

Believe it or not, we’re essentially done.

The Formal Definition

The formal definition of a sequence on numbers  \(x_n\) converging to some value,  \(a\), follows. Know that what you’re about to read is equivalent to the definition you’ll see in university, and it’s using the language that you’re expected to understand. If it confuses you, know that I have fully explained everything in this post barring small notation enhancements. So, if you don’t understand, keep trying. You’ll definitely get there!

A sequence of terms \(x_n\) converges to a value \(a\) if for any and all \(\epsilon>0\), no matter how small, we can find a marker number \(N\) (which is always a natural number) so that \(|x_n-5|<\epsilon\) whenever \(n>N\).

Cool! Sweet job everyone! As usual let me know if there is a part I can explain better! hope you had a rockin’ sweet time!

Cheers,

shel

[bigblueboo]

Exploring square waves

[mindfuckmath-deactivated2016062]

Every math classroom in the world, from elementary schools to elite universities, needs this.  

[chaosophia218]

Interesting Facts about Pi. 1. Pi is the most recognized mathematical constant in the world. Scholars often consider Pi the most important and intriguing number in all of mathematics.

2. Pi has about 6.4 billion known digits which would take a person roughly 133 years to recite without stopping.  The world record holder for the most memorized digits of Pi took nine hours to recite over 44,000 digits of Pi.

3. We can never truly measure the circumference or the area of a circle because we can never truly know the value of Pi. Pi is an irrational number, meaning its digits go on forever in a seemingly random sequence. 4. The “squaring the circle” method of understanding pi has fascinated mathematicians because traditionally the circle represents the infinite, immeasurable, and even spiritual world while the square represents the manifest, measurable, and comprehensive world. 5. Egyptologists and followers of mysticism have been fascinated for centuries by the fact that the Great Pyramid at Giza seems to approximate pi. The vertical height of the pyramid has the same relationship to the perimeter of its base as the radius of a circle has to its circumference. 6. One of the earliest known records of pi was written by an Egyptian scribe named Ahmes (c. 1650 B.C.) on what is now known as the Rhind Papyrus. He was off by less than 1% of the modern approximation of pi (3.141592). 7. In 2002, a Japanese scientist found 1.24 trillion digits of pi using a powerful computer called the Hitachi SR 8000, breaking all previous records. 8. There are people who believe that Pi contains the answer to the universe, or that information is held in the digits. It has even been suggested it contains the Voice of God. In Carl Sagan’s book ‘Contact’ the places of Pi are found to contain a message from the beings that built the universe.

[utrippy]

[fossilporn]

A Whale Inside A Whale, That Was Eaten By A Shark, Discovered In Egypt

t’s like a prehistoric turducken, but on a colossal scale: Scientists in Egypt have discovered the remains of a 40 million-year-old whale with another whale inside it, and researchers believe the two were then eaten by sharks.

The smaller whale found inside the fossils of the 60-foot-long basilosaurus may have been a fetus. However, basilosaurus had some pretty unusual dining habits. Like today’s orcas, this fierce and ancient predator was known to eat other whales, so the whale inside might not have been a fetus so much as a meal.  Read more…

Be sure to follow the Fossil Porn Tumblr Blog for more fossils news, links and photos.

[notyouraveragemedgirl-deactivat]

For the first in my life I am actually understanding how to do math. I wish I had better teachers before just like now to have delivered this feeling before.

[muspeccoll]

We’re wrapping up our Beautiful Math series with Isaac Newton. How could we consider beauty and mathematics without him?  

Isaac Newton first published his Philosophiae naturalis principia mathematica in 1687.  One of the landmarks of scientific thought, the book explains the physics, or “mathematical principles,” behind various phenomena in the natural world.  The Principia was the first explication of differential calculus, as well as the new science of dynamics.  Concepts discussed in the book also include mass, centripetal force, gravity and inertia.

The Principia was originally published in Latin and divided into three books: “The Motion of Bodies” (books 1 and 2) and “The System of the World” (book 3).  In the first two books Newton laid out his laws of motion, concerning the relationships between force, mass, and direction.  The third part applies these laws to the motions of the heavenly bodies, fusing physics and astronomy into a science later called celestial mechanics.

We don’t have a first edition of the Principia, but we do have a first edition of the first translation of Newton’s work into English, made by Andrew Motte in 1729.  Motte’s translation provided the basis for several later English editions, including the first American edition, published in 1846, and was important because it made Newton’s theories accessible to a broader public.

Newton, Isaac, 1642-1727. The mathematical principles of natural philosophy / by Sir Isaac Newton. Tr. into English by Andrew Motte. To which are added, The laws of the moon’s motion according to gravity. By John Machin. London : Printed for B. Motte, 1729.  MU Ellis Special Collections Rare Vault QA803 .A45  1 -2

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11.06.15 // preparing for finals.. 😣

[themathkid]

Probability of Yahtzee outcomes over time. Via. 

[ohstarstuff]

“Man is… related inextricably to all reality, known and unknowable…All things are one thing and that one thing is all things—plankton, a shimmering phosphorescence on the sea and the spinning planets and an expanding universe, all bound together by the elastic string of time. It is advisable to look from the tide pool to the stars and then back to the tide pool again.”  ― John Steinbeck, The Log from the Sea of Cortez

Image & Source Credit: Michael Shainblum

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[graftheory]

Platonic Solids by Moni

[science-junkie]

“In the first place, then, as is evident to all, fire and earth and water and air are bodies. And every sort of body possesses solidity, and every solid must necessarily be contained in planes; and every plane rectilinear figure is composed of triangles…” – Plato. Timaeus circa 360 BC.

Icosahedron –> Water Tetrahedron –> Fire Cube –> Earth Octahedron –> Air Dodecahedron –> The Fifth Element (The Universe).

[scishow]

How to Upload Your Mind

Uploading your mind to a computer might one day let humans cheat death. The technology’s a long way off, but researchers are working on closing that gap. This episode was brought to you and inspired by the movie Self/less.