What are the best rules, formulas and tricks in math?
February 24, 2007 9:01 AM Subscribe
MeFites: Share with me your favorite math tricks, truisms and formulas that clarified and simplified math problems and made life easier in high school and college.
It can be about algebra, fractions, ratios and proportions, percentages - what have you. If it relates to business math, all the better, since that's what I'm currently taking. Thank you!
It can be about algebra, fractions, ratios and proportions, percentages - what have you. If it relates to business math, all the better, since that's what I'm currently taking. Thank you!
A mnemonic, which I guess is a trick:
Some Officers Had Curly Auburn Hair Till Old Age.
(sin=Opposite/Hypotenuse; cos=Adjacent/Hypotenuse; tan=Opposite/Adjacent)
posted by handee at 9:09 AM on February 24, 2007
Some Officers Had Curly Auburn Hair Till Old Age.
(sin=Opposite/Hypotenuse; cos=Adjacent/Hypotenuse; tan=Opposite/Adjacent)
posted by handee at 9:09 AM on February 24, 2007
The nonsense word "Sohcahtoa" was our go-to for remembering the trig ratios. It also makes a pretty sweet battle cry when the nerds take over.
posted by Schlimmbesserung at 9:16 AM on February 24, 2007 [11 favorites]
posted by Schlimmbesserung at 9:16 AM on February 24, 2007 [11 favorites]
Response by poster: corpse: spell it out brother. I want this to be beneficial for anyone reading this, regardless of how good or bad they are at math.
posted by dropkick at 9:30 AM on February 24, 2007
posted by dropkick at 9:30 AM on February 24, 2007
Response by poster: corpse: It seems I wrongly assumed you to be a male, because the website link on your profile leads to the page of a female author.
If I have made this careless mistake, I apologize in advance.
posted by dropkick at 9:47 AM on February 24, 2007
If I have made this careless mistake, I apologize in advance.
posted by dropkick at 9:47 AM on February 24, 2007
I could go on for a while since one of the tricks at being good at math exams and contests is not only innate ability but also knowing "tricks" as to how to arrive at a solution. But I have to run in 5 minutes....
For calculation, I'd say one of the big problems people have is knowing whether their answer is right. Is there quick checks to make that the answer is sane?
One thing is helpful is Divisibility Rules. This is a good way to cross-check.
Also, use the concept of estimates and Order of Magnitude calculations. Dont churn through things blindly: What is 0.54 * 38.29? Well, I can tell you the answer will be close to 0.5 * 38 which is 19. Its amazing how people will get an answer close to ,say 190 and not blink. So, guess at what the answer should be before you embark on calculation.
Factorization is really useful in re-arranging a problem. What's 5*28? Well, thats the same as 5*2*14. And thats just 10*14 which is 140. This is how I do math in my head. Move the numbers around before calculating.
Same with Order of operations: Whats 14 * (5/2)? Get rid of the 2 by halving the 14. Now you have 7*5=35.
This is actually similar to how high-level mathematical thinking takes place. If a problem looks difficult at first, can it be "translated" into another problem which has the same answer. The actual term is to find an isomorphic problem, then solve that.
posted by vacapinta at 9:48 AM on February 24, 2007 [3 favorites]
For calculation, I'd say one of the big problems people have is knowing whether their answer is right. Is there quick checks to make that the answer is sane?
One thing is helpful is Divisibility Rules. This is a good way to cross-check.
Also, use the concept of estimates and Order of Magnitude calculations. Dont churn through things blindly: What is 0.54 * 38.29? Well, I can tell you the answer will be close to 0.5 * 38 which is 19. Its amazing how people will get an answer close to ,say 190 and not blink. So, guess at what the answer should be before you embark on calculation.
Factorization is really useful in re-arranging a problem. What's 5*28? Well, thats the same as 5*2*14. And thats just 10*14 which is 140. This is how I do math in my head. Move the numbers around before calculating.
Same with Order of operations: Whats 14 * (5/2)? Get rid of the 2 by halving the 14. Now you have 7*5=35.
This is actually similar to how high-level mathematical thinking takes place. If a problem looks difficult at first, can it be "translated" into another problem which has the same answer. The actual term is to find an isomorphic problem, then solve that.
posted by vacapinta at 9:48 AM on February 24, 2007 [3 favorites]
The coolest math trick I ever learned, and the one I use every single day, is called the doomsday rule. This trick alllows you to determine what day of the week any given date is, either in the future or the past. There are several different webpages that try to help you master this trick, but this one is my favorite.
posted by crazyray at 10:00 AM on February 24, 2007 [7 favorites]
posted by crazyray at 10:00 AM on February 24, 2007 [7 favorites]
I recently learned the lattice method of multiplication. It makes multiplying big numbers together relatively foolproof, plus it's a hit at cocktail parties. No, really.
posted by bonheur at 10:20 AM on February 24, 2007
posted by bonheur at 10:20 AM on February 24, 2007
Approximate conversion from miles per hour to metres per second - multiply by 4/9.
The formula for adding counting numbers, eg what is 1+2+3+4+5+6....+n = 1/2 * n * (n+1)
Slightly higher level algebra, but sin(x) / cos (x) = tan (x) can be very useful.
posted by spark at 10:23 AM on February 24, 2007
The formula for adding counting numbers, eg what is 1+2+3+4+5+6....+n = 1/2 * n * (n+1)
Slightly higher level algebra, but sin(x) / cos (x) = tan (x) can be very useful.
posted by spark at 10:23 AM on February 24, 2007
There are lots of fun back-of-the-envelope tricks that can help you create good estimates. These help you quickly double check your answers, as vacapinta pointed out. Off the top of my head:
- 1 meter is pretty close to 1 yard
- If you can remember that a 5k run is about 3 miles, you'll have a ballpark for converting distances
- a liter is a quart, plus a liter bit more
- need a quick measurement? Fold a sheet of paper over diagonally and you'll have an edge that's just barely over 12 inches.
- to help with F->C temperature conversion, remember the following four common and useful conversions:
-- 0°C -> 32°F (freezing of water)
-- 100°C -> 212°F (boiling of water)
-- 37°C -> 98.6°F (human body temp)
-- 25°C -> 77°F (about 75°F, or room temperature)
posted by chrisamiller at 11:08 AM on February 24, 2007
- 1 meter is pretty close to 1 yard
- If you can remember that a 5k run is about 3 miles, you'll have a ballpark for converting distances
- a liter is a quart, plus a liter bit more
- need a quick measurement? Fold a sheet of paper over diagonally and you'll have an edge that's just barely over 12 inches.
- to help with F->C temperature conversion, remember the following four common and useful conversions:
-- 0°C -> 32°F (freezing of water)
-- 100°C -> 212°F (boiling of water)
-- 37°C -> 98.6°F (human body temp)
-- 25°C -> 77°F (about 75°F, or room temperature)
posted by chrisamiller at 11:08 AM on February 24, 2007
Response by poster: This is turning out to be a great thread. Keep'm coming!
posted by dropkick at 11:09 AM on February 24, 2007
posted by dropkick at 11:09 AM on February 24, 2007
Everyone from my high school (even over 10yrs later) can still recite the quadratic formula because one of the algebra teachers put it to the tune of "Pop Goes the Weasel".....
x equals negative b,
plus or minus square root,
b squared minus 4 a c ,
all over 2 a!
posted by dicaxpuella at 11:14 AM on February 24, 2007 [1 favorite]
x equals negative b,
plus or minus square root,
b squared minus 4 a c ,
all over 2 a!
posted by dicaxpuella at 11:14 AM on February 24, 2007 [1 favorite]
A couple things that stood out for me. I hated the traditional way that the forumula for a unit circle is presented as x2 + y2 = 1. I hated having to deal with square roots and so on to get practical coordinates for plotting or otherwise visualizing it, so I taught myself how to do it with two equations instead:
x = sin(θ)
y = cos(θ)
The reason why I like this is that I learned how to quickly parameterize it into easy variations.
x = r × sin(θ) + cx
y = r × cos(θ) + cy
gives the formula for a circle with radius r centered at (cx, cy)
Then I popped in another change:
x = rx × sin(θ) + cx
y = ry × cos(θ) + cy
which now turns this into a formula for an ellipse instead of a circle, but using two different values for rx and ry.
This was eye-opening for me and I discovered ultimately that I really love parametric equations because they were so transparent in comparison to the traditional ones.
Lesson - don't get stuck on one particular formula, look for others - there are usually several, especially with linear equations.
As far practical applications, I took Newtonian physics as a junior and then learned differential calculus as a senior. Holy crap, that should have been the other way around. Most of the physics work became trivial with calculus. I can't say that I would've been able to fully grok calculus as a junior, though.
I agree on rules of divisibility. My 7th grade teacher had a list posted on a bulletin board. I still use those rules. She had a blank for 7 with an invitation for you to try to figure one out. 20 years later I derived one (not necessarily useful, but it works): convert the number to base 8, sum the digits (in base 8) and if they add up to a multiple of 7, the original number is divisible by 7.
posted by plinth at 11:19 AM on February 24, 2007
x = sin(θ)
y = cos(θ)
The reason why I like this is that I learned how to quickly parameterize it into easy variations.
x = r × sin(θ) + cx
y = r × cos(θ) + cy
gives the formula for a circle with radius r centered at (cx, cy)
Then I popped in another change:
x = rx × sin(θ) + cx
y = ry × cos(θ) + cy
which now turns this into a formula for an ellipse instead of a circle, but using two different values for rx and ry.
This was eye-opening for me and I discovered ultimately that I really love parametric equations because they were so transparent in comparison to the traditional ones.
Lesson - don't get stuck on one particular formula, look for others - there are usually several, especially with linear equations.
As far practical applications, I took Newtonian physics as a junior and then learned differential calculus as a senior. Holy crap, that should have been the other way around. Most of the physics work became trivial with calculus. I can't say that I would've been able to fully grok calculus as a junior, though.
I agree on rules of divisibility. My 7th grade teacher had a list posted on a bulletin board. I still use those rules. She had a blank for 7 with an invitation for you to try to figure one out. 20 years later I derived one (not necessarily useful, but it works): convert the number to base 8, sum the digits (in base 8) and if they add up to a multiple of 7, the original number is divisible by 7.
posted by plinth at 11:19 AM on February 24, 2007
I think one thing that often annoys students of math is that all the really useful interesting stuff is... the stuff they teach you. There aren't a bunch of useful methods that the math club kids keep secret.
For calculus there is differentiating under the integral. (I'm not quite at the level to really apply this, but it strikes me as really neat).
Did you know there exists an algorithm to extract arbitrary digits of Pi in certain bases without knowledge of the preceding digits? Not useful, but interesting.
If you've taken calculus you were probably taught Newton's method, which can be used (even without knowledge of calc) to find roots.
posted by phrontist at 11:21 AM on February 24, 2007
For calculus there is differentiating under the integral. (I'm not quite at the level to really apply this, but it strikes me as really neat).
Did you know there exists an algorithm to extract arbitrary digits of Pi in certain bases without knowledge of the preceding digits? Not useful, but interesting.
If you've taken calculus you were probably taught Newton's method, which can be used (even without knowledge of calc) to find roots.
posted by phrontist at 11:21 AM on February 24, 2007
Heron's formula for calculating the area of a triangle given the length of the three sides is fast, easy to remember, and will make you a stand-out star in high school mathletics. You can short-circuit many a trigonometry problem if you know this formula.
posted by IvyMike at 11:26 AM on February 24, 2007
posted by IvyMike at 11:26 AM on February 24, 2007
3-4-5 triangle, oversimplified: if one non-hypotenuse side of a right triangle is 3 units long, and the other non-hypotenuse side is 4 units long, the hypotenuse will be 5 units long. So if you have a right triangle with sides of 6 units and 8 units, you know the hypotenuse is 10 units.
(And yes, since you asked, I'm female, but I'm not offended. A math thread, on Metafilter, on the Interweb -- chances are pretty high any given poster is going to be male.)
posted by The corpse in the library at 11:33 AM on February 24, 2007
(And yes, since you asked, I'm female, but I'm not offended. A math thread, on Metafilter, on the Interweb -- chances are pretty high any given poster is going to be male.)
posted by The corpse in the library at 11:33 AM on February 24, 2007
To determine if a number is divisible by 3, add the digits, repeating until you get a number below 10. If the result is 3, 6 or 9, the original number is divisible by three:
27 -> 2+7 = 9; True
99 -> 9+9 = 18 -> 1=8 = 9; True
15,411 -> 1+5+4+1+1 = 12 -> 1+2 = 3; True
15,412 -> 1+5+4+1+2 = 13 -> 1+3 = 4; False
There's a similar trick for numbers divisible by 7, but I can't remember it right now.
posted by grateful at 11:50 AM on February 24, 2007
27 -> 2+7 = 9; True
99 -> 9+9 = 18 -> 1=8 = 9; True
15,411 -> 1+5+4+1+1 = 12 -> 1+2 = 3; True
15,412 -> 1+5+4+1+2 = 13 -> 1+3 = 4; False
There's a similar trick for numbers divisible by 7, but I can't remember it right now.
posted by grateful at 11:50 AM on February 24, 2007
online book on How to do algebra in your head
Trachtenberg speed math
Vedic math
posted by Zed_Lopez at 11:52 AM on February 24, 2007
Trachtenberg speed math
Vedic math
posted by Zed_Lopez at 11:52 AM on February 24, 2007
My algebra students came up with a mnemonic for y=mx+b (slope-intercept formula of a line): your momma's x-boyfriend.
And since no-one has an Aunt Polly any more, the order of operations can be memorized as Please Excuse My Dumb-Ass Son (something my mother likely had to say more than once tothe principal).
posted by notsnot at 11:54 AM on February 24, 2007 [1 favorite]
And since no-one has an Aunt Polly any more, the order of operations can be memorized as Please Excuse My Dumb-Ass Son (something my mother likely had to say more than once tothe principal).
posted by notsnot at 11:54 AM on February 24, 2007 [1 favorite]
Master simultaneous linear equations (google it). It is a simple technique if you can do basic algebra, and I actually find it occasionally useful in everyday life, unlike, say, calculus, which I haven't used since I left college.
posted by nanojath at 12:06 PM on February 24, 2007
posted by nanojath at 12:06 PM on February 24, 2007
Response by poster: Somewhere, someone with tape on their glasses is reading this thread and having a nerdgasm.
posted by dropkick at 12:31 PM on February 24, 2007 [1 favorite]
posted by dropkick at 12:31 PM on February 24, 2007 [1 favorite]
All those digit divisibility criteria are simple modular arithmetic. As an example: modulo 9, 10 is just 1, 100 = 10*10 = 1*1, so it's also 1, and so on. If we represent a number as a*1 + b*10 + c*100 + d*1000 + ..., that's the same (mod 9) as a + b + c + d + .... Which is another way of saying that a number is divisible by 9 exactly when the sum of its digits is divisible by 9.
In case you encounter sines and cosines in some weird arrangement, it's sometimes a nice trick to use Euler's formula to convert the entire thing to a bunch of (complex) powers of e. Useful for integration or for simplifying trig expressions.
posted by parudox at 1:21 PM on February 24, 2007
In case you encounter sines and cosines in some weird arrangement, it's sometimes a nice trick to use Euler's formula to convert the entire thing to a bunch of (complex) powers of e. Useful for integration or for simplifying trig expressions.
posted by parudox at 1:21 PM on February 24, 2007
My favourite number trick when I was a kid was the number 142,857.
Multiply it by 2, 3, 4, 5 or 6 and it becomes an "anagram" of itself.
I have no idea what it means though.
posted by AmbroseChapel at 1:25 PM on February 24, 2007
Multiply it by 2, 3, 4, 5 or 6 and it becomes an "anagram" of itself.
I have no idea what it means though.
posted by AmbroseChapel at 1:25 PM on February 24, 2007
For Quadratic Formula, I learned it as:
A negative Bee couldn't decide if he wanted to go or not to a radical party where Bees square danced opposite 4 awesome chicks that was all over by 2 AM.
Whenever I need to think of the formula I picture the bees and chickens dancing and it all comes back.
I remembered Sohcahtoa as an Indian princess. My teacher told some story about her but I only remember her name.
posted by avagoyle at 2:16 PM on February 24, 2007
A negative Bee couldn't decide if he wanted to go or not to a radical party where Bees square danced opposite 4 awesome chicks that was all over by 2 AM.
Whenever I need to think of the formula I picture the bees and chickens dancing and it all comes back.
I remembered Sohcahtoa as an Indian princess. My teacher told some story about her but I only remember her name.
posted by avagoyle at 2:16 PM on February 24, 2007
Multiply 142857 by 2, 3, 4, 5 or 6 and it becomes an "anagram" of itself.That's the interesting thing about fractions of 7. Note that all the values are a rotation of 142857.
1/7 = 0.142857 142857 142857 ...
2/7 = 0.285714 285714 285714 ...
3/7 = 0.428571 ...
4/7 = 0.571428 ...
5/7 = 0.714285 ...
6/7 = 0.857142 ...
This comes in very handy to divide by 7 in your head. The string "142857" has been indelibly associated in my head with 7 ever since I saw that pattern.
posted by phliar at 2:40 PM on February 24, 2007 [6 favorites]
Percent change?!?!!! NOO!
percent change=(new-old)/old
posted by croutonsupafreak at 3:16 PM on February 24, 2007 [1 favorite]
percent change=(new-old)/old
posted by croutonsupafreak at 3:16 PM on February 24, 2007 [1 favorite]
For young kids, I always liked the Stand and Deliver method for multiplying any number under ten by nine.
For example, 5 * 9:
Take both your hands and hold them up in front of you. Starting with the pinky of one hand, count until you've reached the number you're multiplying. So, one, two three, four, five is pinky, ring finger, middle, index and thumb. Hold the last finger down and what you'll see is two groups of fingers (separated by the finger that's down). In the case of 5*9, you've got four fingers on one hand, five fingers on the other. Forty-five.
It took way longer to write out the explanation than it would to just do it in front of someone. They'll pick it up right away.
posted by Civil_Disobedient at 3:37 PM on February 24, 2007
For example, 5 * 9:
Take both your hands and hold them up in front of you. Starting with the pinky of one hand, count until you've reached the number you're multiplying. So, one, two three, four, five is pinky, ring finger, middle, index and thumb. Hold the last finger down and what you'll see is two groups of fingers (separated by the finger that's down). In the case of 5*9, you've got four fingers on one hand, five fingers on the other. Forty-five.
It took way longer to write out the explanation than it would to just do it in front of someone. They'll pick it up right away.
posted by Civil_Disobedient at 3:37 PM on February 24, 2007
Also, an easy way to square any number ending in 5: multiply the digits after the five by itself plus one and tack on 25 to the end.
Example: 1052 =
10*11 = 110
+ 25
11025!
Example 2: 652
6 * 7 = 42
+ 25
4225!
posted by Civil_Disobedient at 3:43 PM on February 24, 2007
Example: 1052 =
10*11 = 110
+ 25
11025!
Example 2: 652
6 * 7 = 42
+ 25
4225!
posted by Civil_Disobedient at 3:43 PM on February 24, 2007
Curious Math is basically dedicated to answering your question. I particularly like their "decimal precision" trick.
posted by phrontist at 3:52 PM on February 24, 2007
posted by phrontist at 3:52 PM on February 24, 2007
Susie can tell Oscar has a hard on always
S=o/h
C=a/h
T=o/a
posted by travis08 at 5:05 PM on February 24, 2007
S=o/h
C=a/h
T=o/a
posted by travis08 at 5:05 PM on February 24, 2007
D Hi/Ho = (Ho D Hi - Hi D Ho) / Ho Ho
In words, the derivative of Hi over Ho is
Hodey Hi, minus Hidey Ho, over HoHo.
posted by Wet Spot at 5:25 PM on February 24, 2007
In words, the derivative of Hi over Ho is
Hodey Hi, minus Hidey Ho, over HoHo.
posted by Wet Spot at 5:25 PM on February 24, 2007
I had the same math teacher for the last 4 years of high school, and he was always teaching us number tricks and random rules. Some of the favorites:
I've always loved SohCahToa, and the finger trick for nines.
Another useful one was the ASTC ( All - Sine - Cosine - Tangent) rule fron Trigonometry, to indicate which trig function is positive in that quadrant
S | A
------
T | C
Our math teachers generally taught it as an anagram of "All Stations To Central", but we came up with many, many variations of it over time, to the point where it is permanently fused in our brains.
Another cool rule he taught us was somebody's conundrum, which involved working out how you could work your way over a pattern of lines without going over the same line twice. If a shape has more than three t-intersections, regardless of the size, it will be impossible.
These things just keep coming to mind. I'll post some of the better ones later.
posted by cholly at 5:30 PM on February 24, 2007
I've always loved SohCahToa, and the finger trick for nines.
Another useful one was the ASTC ( All - Sine - Cosine - Tangent) rule fron Trigonometry, to indicate which trig function is positive in that quadrant
S | A
------
T | C
Our math teachers generally taught it as an anagram of "All Stations To Central", but we came up with many, many variations of it over time, to the point where it is permanently fused in our brains.
Another cool rule he taught us was somebody's conundrum, which involved working out how you could work your way over a pattern of lines without going over the same line twice. If a shape has more than three t-intersections, regardless of the size, it will be impossible.
These things just keep coming to mind. I'll post some of the better ones later.
posted by cholly at 5:30 PM on February 24, 2007
somebody being a name which I can't remember right now
posted by cholly at 5:31 PM on February 24, 2007
posted by cholly at 5:31 PM on February 24, 2007
My favourite trick is that the Gelfand transform is just the Fourier transform in the case when the Banach algebra is L^1(G).
posted by number9dream at 5:36 PM on February 24, 2007 [1 favorite]
posted by number9dream at 5:36 PM on February 24, 2007 [1 favorite]
chrisamiller writes "-- 0°C -> 32°F (freezing of water)
"-- 100°C -> 212°F (boiling of water)
"-- 37°C -> 98.6°F (human body temp)
"-- 25°C -> 77°F (about 75°F, or room temperature)"
And -40=-40
posted by Mitheral at 6:08 PM on February 24, 2007
"-- 100°C -> 212°F (boiling of water)
"-- 37°C -> 98.6°F (human body temp)
"-- 25°C -> 77°F (about 75°F, or room temperature)"
And -40=-40
posted by Mitheral at 6:08 PM on February 24, 2007
cholly: I don't know exactly whose conundrum you're referring to, but the question of whether you can "work your way over a pattern of lines without going over the same line twice" is most famously associated with the seven bridges of Königsberg (Wikipedia link). Such a path, i.e., way of drawing the lines in a figure without going over any lines twice, is called Eulerian (since Leonhard Euler solved the general case).
The upshot is, if there are more than two points of odd "degree" -- the degree is the number of lines emitting from that point in the figure -- then it is impossible to draw the figure without going over some lines twice.
posted by sappidus at 7:09 PM on February 24, 2007
The upshot is, if there are more than two points of odd "degree" -- the degree is the number of lines emitting from that point in the figure -- then it is impossible to draw the figure without going over some lines twice.
posted by sappidus at 7:09 PM on February 24, 2007
Oh, not to be too derail-y, but to those mentioning the curiosity of the repeating "142857" and its "anagrams" (actually, cyclic rotational shifts or something) in various fractions with denominator 7... I can't resist but mention that this is not unique to the number 7.
Example:
1/19 = 0.052631578947368421 052631578947368421 ...
2/19 = 0.105263157894736842 105263157894736842 ...
3/19 = 0.210526315789473684 210526315789473684 ...
...and so on. Note the "052631578947368421" and how it rotates as you change the numerator. Not very helpful for quickly dividing by 19 or anything, but there ya go. ;-)
The particular prime numbers that have this kind of property are called "full period primes." Here's another Wikipedia link with some bare-bones information; the Kalman reference therein is a good article as well.
posted by sappidus at 7:24 PM on February 24, 2007
Example:
1/19 = 0.052631578947368421 052631578947368421 ...
2/19 = 0.105263157894736842 105263157894736842 ...
3/19 = 0.210526315789473684 210526315789473684 ...
...and so on. Note the "052631578947368421" and how it rotates as you change the numerator. Not very helpful for quickly dividing by 19 or anything, but there ya go. ;-)
The particular prime numbers that have this kind of property are called "full period primes." Here's another Wikipedia link with some bare-bones information; the Kalman reference therein is a good article as well.
posted by sappidus at 7:24 PM on February 24, 2007
Ai! I was too quick pasting in my arithmetic. My above post should say,
3/19 = 0.157894736842105263 157894736842105263 ...
4/19 = 0.210526315789473684 210526315789473684 ...
The rest stands.
posted by sappidus at 7:26 PM on February 24, 2007
3/19 = 0.157894736842105263 157894736842105263 ...
4/19 = 0.210526315789473684 210526315789473684 ...
The rest stands.
posted by sappidus at 7:26 PM on February 24, 2007
A big collection of great ones at Keith Enevoldsen's Think Zone math fun facts.
posted by LobsterMitten at 11:42 PM on February 24, 2007
posted by LobsterMitten at 11:42 PM on February 24, 2007
And this mental arithmetic shortcut pamphlet written by Enevoldsen's grandfather.
posted by LobsterMitten at 11:48 PM on February 24, 2007
posted by LobsterMitten at 11:48 PM on February 24, 2007
Another point on the subject of cross-checking in geometry. There's this concept called the Degenerate Case or extreme case of a geometrical object or construct.
In general any rule you derive or prove or use should also apply to the degenerate case and so the degenerate case makes a good check.
What is a degenerate case? Consider a triangle for example where the length of two sides adds up to the length of the third. Basically this is a straight line, but its also mathematically a triangle! For example, if you plug in the lengths of the sides into Heron's rule which was linked above, the Area is indeed 0.
Working with and thinking about degenerate cases can make visualization and abstraction easier.
posted by vacapinta at 1:34 AM on February 25, 2007
In general any rule you derive or prove or use should also apply to the degenerate case and so the degenerate case makes a good check.
What is a degenerate case? Consider a triangle for example where the length of two sides adds up to the length of the third. Basically this is a straight line, but its also mathematically a triangle! For example, if you plug in the lengths of the sides into Heron's rule which was linked above, the Area is indeed 0.
Working with and thinking about degenerate cases can make visualization and abstraction easier.
posted by vacapinta at 1:34 AM on February 25, 2007
The lattice method for long multiplication is even neater than the online explanations usually demonstrate. As well as being an extremely straightforward way to do long integer multiplications, it also makes it easy to get the decimal point in the right place when multiplying non-integers.
As this example shows, all you do is follow the grid lines inward from the decimal points in the multiplier and multiplicand; the diagonal passing through the point where they meet shows where the decimal point goes in the product.
posted by flabdablet at 3:46 AM on February 25, 2007
As this example shows, all you do is follow the grid lines inward from the decimal points in the multiplier and multiplicand; the diagonal passing through the point where they meet shows where the decimal point goes in the product.
posted by flabdablet at 3:46 AM on February 25, 2007
My favorite formula generates all the Pythagorean triples, given arbitrary integers p, q, r:
Let
a = r(p2 + q2)
b = r(p2 - q2)
c = 2pqr
Then a2 = b2 + c2.
For example, p = 2, q = 1, r = 1 generates a the classic 3:4:5 triple; p = 3, q=2, r=1 yields 5:12:13 and so on.
I remember being exceedingly pleased with myself in Form 3 after working out a derivation for this while sitting through a boring geography class.
posted by flabdablet at 4:08 AM on February 25, 2007
Let
a = r(p2 + q2)
b = r(p2 - q2)
c = 2pqr
Then a2 = b2 + c2.
For example, p = 2, q = 1, r = 1 generates a the classic 3:4:5 triple; p = 3, q=2, r=1 yields 5:12:13 and so on.
I remember being exceedingly pleased with myself in Form 3 after working out a derivation for this while sitting through a boring geography class.
posted by flabdablet at 4:08 AM on February 25, 2007
vacapinta writes, "What is a degenerate case? Consider a triangle for example where the length of two sides adds up to the length of the third. Basically this is a straight line, but its also mathematically a triangle! For example, if you plug in the lengths of the sides into Heron's rule which was linked above, the Area is indeed 0."
As referenced on the Wikipedia page, Heron's formula itself can be considered the degenerate case of Brahmagupta's formula for the area of a quadrilateral whose four vertices are on a circle (just set d=0 and you have a triangle).
When I found this out, I remember thinking, "Gosh, and this must be the degenerate case of another formula!!" It is, of Bretschneider's formula -- but at this point, I was getting scared of how difficult it was going to be to spell the next formula's name, so I stopped looking for further generalizations.
posted by sappidus at 8:03 AM on February 25, 2007
As referenced on the Wikipedia page, Heron's formula itself can be considered the degenerate case of Brahmagupta's formula for the area of a quadrilateral whose four vertices are on a circle (just set d=0 and you have a triangle).
When I found this out, I remember thinking, "Gosh, and this must be the degenerate case of another formula!!" It is, of Bretschneider's formula -- but at this point, I was getting scared of how difficult it was going to be to spell the next formula's name, so I stopped looking for further generalizations.
posted by sappidus at 8:03 AM on February 25, 2007
Ambrosechapel: if you notice that particular number is one repetition of the decimal version of 1/7. Indeed, the digits are still in the same order after the multiplication, just rotated around a bit.
posted by notsnot at 9:27 AM on February 25, 2007
posted by notsnot at 9:27 AM on February 25, 2007
upon reading, I see that Philar explained more fully. (if I recall, it has something to do with modular arithmetic.)
posted by notsnot at 9:29 AM on February 25, 2007
posted by notsnot at 9:29 AM on February 25, 2007
For multiplication, always multiply by the nearest number, then subtract.
For example, 37 * 7 = (40 * 7) - (7 * 3) .
i.e 37 * 7 = (37 + 3) * 7 - (3 * 7)
It's simple once you get it. You take one factor up to the nearest easy number to multiply. In the process, you have added the number of up-bumps, multiplied by whatever you wanted to multply it with. So you now simply subtract this number away.
Another example:
59 * 11 = (60 * 11) - (1 * 11)
posted by markesh at 1:01 PM on February 25, 2007
For example, 37 * 7 = (40 * 7) - (7 * 3) .
i.e 37 * 7 = (37 + 3) * 7 - (3 * 7)
It's simple once you get it. You take one factor up to the nearest easy number to multiply. In the process, you have added the number of up-bumps, multiplied by whatever you wanted to multply it with. So you now simply subtract this number away.
Another example:
59 * 11 = (60 * 11) - (1 * 11)
posted by markesh at 1:01 PM on February 25, 2007
I've always liked the rule of 72 for doing back-of-the-hand exponential calculations.
If you have a portfolio that is earning 7% per year, how long until it doubles in value?
What if it earns 10%?
As a rough estimate, dividing seven into 72 gives you 10.3, which is close to the doubling time.
At 10%, 10 into 72 = 7.2 years before it doubles.
Once you can easily calculate doubling times with the rule of 72, you can start to do more complicated calculations. Say your retirement portfolio has a certain amount of money in it, and you are about 40 years from retirement. At 10% return, 40/7.2 = doubling 5.5 times
At 7% return, 40/10.3 = doubling 3.9 times.
From that 3% difference in rate of return, you wind up more than tripling the amount of money you wind up with at the end.
posted by Maxwell_Smart at 3:22 PM on February 25, 2007
If you have a portfolio that is earning 7% per year, how long until it doubles in value?
What if it earns 10%?
As a rough estimate, dividing seven into 72 gives you 10.3, which is close to the doubling time.
At 10%, 10 into 72 = 7.2 years before it doubles.
Once you can easily calculate doubling times with the rule of 72, you can start to do more complicated calculations. Say your retirement portfolio has a certain amount of money in it, and you are about 40 years from retirement. At 10% return, 40/7.2 = doubling 5.5 times
At 7% return, 40/10.3 = doubling 3.9 times.
From that 3% difference in rate of return, you wind up more than tripling the amount of money you wind up with at the end.
posted by Maxwell_Smart at 3:22 PM on February 25, 2007
Mitheralwrites: And -40=-40
Also 28°C -> 82°F
and 16°C -> 61°F
(I use those all the time for reference points)
posted by shokod at 2:59 AM on February 26, 2007
Also 28°C -> 82°F
and 16°C -> 61°F
(I use those all the time for reference points)
posted by shokod at 2:59 AM on February 26, 2007
I like the solution to finding the sum of all the numbers between 1 and 100 which is attributed to Gauss when he was a schoolboy.
Rather than add 1+2+3, etc he realised that the sequence involved 50 pairs of numbers that each totalled 101 when added together. So the solution is 50*101 or 5050.
This pattern is generalisable for finding the sum of any arithmetic series via the formula S=na+nd(n-1)/2 where S is the final sum, n is the number of elements in the sequence, a is the value of the first element and d is the difference between consecutive values.
posted by rongorongo at 3:37 AM on February 26, 2007
Rather than add 1+2+3, etc he realised that the sequence involved 50 pairs of numbers that each totalled 101 when added together. So the solution is 50*101 or 5050.
This pattern is generalisable for finding the sum of any arithmetic series via the formula S=na+nd(n-1)/2 where S is the final sum, n is the number of elements in the sequence, a is the value of the first element and d is the difference between consecutive values.
posted by rongorongo at 3:37 AM on February 26, 2007
This is perhaps a bit closer to physics than to math, but here goes:
When you know that a trigonometric function needs to go into a formula, but you're not sure which one, see if you can work out from common sense what the function has to be at 0°, and what it has to be at 90°.
The six standard trig functions have values of 0, 1, or ∞ at 0° and 90°, and each possible combination thereof (as long as the two are unequal) correspond to one of the six functions.
0 at 0°, 1 at 90° = sin
0 at 0°, ∞ at 90° = tan
1 at 0°, 0 at 90° = cos
1 at 0°, ∞ at 90° = sec
∞ at 0°, 0 at 90° = cot
∞ at 0°, 1 at 90° = csc
posted by DevilsAdvocate at 9:12 AM on February 26, 2007
When you know that a trigonometric function needs to go into a formula, but you're not sure which one, see if you can work out from common sense what the function has to be at 0°, and what it has to be at 90°.
The six standard trig functions have values of 0, 1, or ∞ at 0° and 90°, and each possible combination thereof (as long as the two are unequal) correspond to one of the six functions.
0 at 0°, 1 at 90° = sin
0 at 0°, ∞ at 90° = tan
1 at 0°, 0 at 90° = cos
1 at 0°, ∞ at 90° = sec
∞ at 0°, 0 at 90° = cot
∞ at 0°, 1 at 90° = csc
posted by DevilsAdvocate at 9:12 AM on February 26, 2007
not to derail or anything, but I find the lattice method utterly confusing - i'd rather just stick to working it out long hand
1 9 2 x 2 7 =
---------------
(this was going to be all fancy with ascii art intermediate steps, but man, that's just too involved to make a point, so just pretend these text descriptions are visual)
then working right to left...
7x2=14=4 carry 1 (write down 4), 7x9=63+1=64=4 carry 6 (write down 4), 7+6=13=3 carry 1 (write down 3), 1=1 -->1344
same for next line: write down a 0 to account for the tens place, 2x2=4, 2x9=8 carry 1, 2x1=2+1=3 --> 3840
then add the lines
so you get
192 x 27
-----------
1344
3840
-----------
5184
I guess this is not necessarily simpler, but it doesn't require me to add numbers along a diagonal or color-code or anything... and this was good enough for my grandpappy, so it by nabbit is good enough for me... none of this newfangled math
posted by yggdrasil at 9:17 AM on February 26, 2007
1 9 2 x 2 7 =
---------------
(this was going to be all fancy with ascii art intermediate steps, but man, that's just too involved to make a point, so just pretend these text descriptions are visual)
then working right to left...
7x2=14=4 carry 1 (write down 4), 7x9=63+1=64=4 carry 6 (write down 4), 7+6=13=3 carry 1 (write down 3), 1=1 -->1344
same for next line: write down a 0 to account for the tens place, 2x2=4, 2x9=8 carry 1, 2x1=2+1=3 --> 3840
then add the lines
so you get
192 x 27
-----------
1344
3840
-----------
5184
I guess this is not necessarily simpler, but it doesn't require me to add numbers along a diagonal or color-code or anything... and this was good enough for my grandpappy, so it by nabbit is good enough for me... none of this newfangled math
posted by yggdrasil at 9:17 AM on February 26, 2007
I always memorized math formulas by relating them to things that make absolutely no sense. No really.
The opposite of beeswax plus or minus the square ginger root of b two (bingo!) minus 4 air conditioners all over two apples.
Logarithms I got by using WTF.
WT = F (what the fuck)
is equivalent to
logwF = T (what fucking thing)
posted by sephira at 6:28 PM on February 26, 2007
The opposite of beeswax plus or minus the square ginger root of b two (bingo!) minus 4 air conditioners all over two apples.
Logarithms I got by using WTF.
WT = F (what the fuck)
is equivalent to
logwF = T (what fucking thing)
posted by sephira at 6:28 PM on February 26, 2007
The Golden Ratio described as a continued fraction:
1+1/(1+1/(1+1/1 ... to infinity
or
phi = 1 + 1/phi
posted by dragonsi55 at 10:54 AM on March 1, 2007
1+1/(1+1/(1+1/1 ... to infinity
or
phi = 1 + 1/phi
posted by dragonsi55 at 10:54 AM on March 1, 2007
ok, I'm way dumber than anyone in this thread, totally math challenged. Can't even wrap my synapse around any of the comments on this thread. So my tips are for people who still count on their fingers, slowly.
Best tip I learned as a kid to do simple addition in my head was to think in tens and round up or down. (I can only imagine the eyes rolling as I write that *cringe* but there must be other math dumbkopfs out there like me, so I'm writing this for them).
"If you are multiplying 5 times an even number: halve the number you are multiplying by and place a zero after the number. Example: 5 × 6, half of 6 is 3, add a zero for an answer of 30. Another example: 5 × 8, half of 8 is 4, add a zero for an answer of 40."
"If you are multiplying 5 times an odd number: subtract one from the number you are multiplying, then halve that number and place a 5 after the resulting number. Example: 5 × 7: -1 from 7 is 6, half of 6 is 3, place a 5 at the end of the resulting number to produce the number 35. Another example: 5 × 3: -1 from 3 is 2, half of 2 is 1, place a 5 at the end of this number to produce 15."
"To multiply by nine on your fingers, hold up ten fingers - if the problem is 9 × 8 you just put down your 8 finger and there's your answer: 72. (If the problem is 9 × 7 just put down your 7 finger: 63.)"
"Laurie Stryker explains it this way: When you are multiplying by 9, on your fingers (starting with your thumb) count the number you are multiplying by and hold down that finger. The number of fingers before the finger held down is the first digit of the answer and the number of finger after the finger held down is the second digit of the answer.
Example: 2 × 9. your index finder is held down, your thumb is before, representing 1, and there are eight fingers after your index finger, representing 18."
(told you I counted on my fingers, slowly)
Another way of multiplying by 9:
multiply by 10 and subtract the number multiplied.
ex 9 x 8.
First multiply 10 x 8= 80, subtract 9 =71
Fairly easy Math Tips and Tricks.
posted by nickyskye at 12:04 PM on May 28, 2007
Best tip I learned as a kid to do simple addition in my head was to think in tens and round up or down. (I can only imagine the eyes rolling as I write that *cringe* but there must be other math dumbkopfs out there like me, so I'm writing this for them).
"If you are multiplying 5 times an even number: halve the number you are multiplying by and place a zero after the number. Example: 5 × 6, half of 6 is 3, add a zero for an answer of 30. Another example: 5 × 8, half of 8 is 4, add a zero for an answer of 40."
"If you are multiplying 5 times an odd number: subtract one from the number you are multiplying, then halve that number and place a 5 after the resulting number. Example: 5 × 7: -1 from 7 is 6, half of 6 is 3, place a 5 at the end of the resulting number to produce the number 35. Another example: 5 × 3: -1 from 3 is 2, half of 2 is 1, place a 5 at the end of this number to produce 15."
"To multiply by nine on your fingers, hold up ten fingers - if the problem is 9 × 8 you just put down your 8 finger and there's your answer: 72. (If the problem is 9 × 7 just put down your 7 finger: 63.)"
"Laurie Stryker explains it this way: When you are multiplying by 9, on your fingers (starting with your thumb) count the number you are multiplying by and hold down that finger. The number of fingers before the finger held down is the first digit of the answer and the number of finger after the finger held down is the second digit of the answer.
Example: 2 × 9. your index finder is held down, your thumb is before, representing 1, and there are eight fingers after your index finger, representing 18."
(told you I counted on my fingers, slowly)
Another way of multiplying by 9:
multiply by 10 and subtract the number multiplied.
ex 9 x 8.
First multiply 10 x 8= 80, subtract 9 =71
Fairly easy Math Tips and Tricks.
posted by nickyskye at 12:04 PM on May 28, 2007
One that I remember: to multiply any 2-digit number by 11, add the 2 digits of the number together, and stick the result in between the 2 digits. If they add up to 10 or more, add 1 to the first digit and just stick the second digit of the result in the middle. Examples:
11 * 63 = 6 (6+3) 3 = 693
11 * 76 = 7 (7+6) 6 = 7+1 | 3 | 6 = 836
posted by booksherpa at 2:45 PM on May 28, 2007 [1 favorite]
11 * 63 = 6 (6+3) 3 = 693
11 * 76 = 7 (7+6) 6 = 7+1 | 3 | 6 = 836
posted by booksherpa at 2:45 PM on May 28, 2007 [1 favorite]
booksherpa beat me to it ... the 11* 2 digit is really easy and very usefil
posted by radsqd at 9:03 AM on May 29, 2007
posted by radsqd at 9:03 AM on May 29, 2007
This thread is closed to new comments.
posted by The corpse in the library at 9:08 AM on February 24, 2007 [2 favorites]