# Math Teachers. Show me the Hook!

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My high school math teachers weren't the worst or the best, average that's what they were.

What killed it for me in spite of an insatiable interest in science of any sort was that fact that basic math BORED me, it was pointless to me. Show me why I should care!

Even when I did learn the formulas and the methods I'd hit summer vacation full of activities and excitement. Come back to math class, and all I had learned was Gone.

That was when I ran into my worst math teacher. He knew which students he would work with and I wasn't one of those. Needless to say even as an A level student in

English, A to B level in Sciences (no math skills) ergo inconsistent. I ended languishing as a student who barely graduated. I did have talent in arts which went nowhere for me.

I've read articles which reported that experts in education and psychology have testified that the system of teaching practiced in the western world is a model that guarantees

Failure. Courses of study are broken up into unmanageable bits and short snippets that don't allow students any opportunity to learn, practice, use, let alone master the skills

they need to complete the courses thus; homework. Thrown out into the wilds of life to figure it out for yourself. (You dummy. Oh Yeah! If you teachers don't care then neither do I).

Screw it. I'm doing what I like, homework my sore patoot. As you know that model, unchanged, still holds as education practiced today.

I even bought pure math books, sat down to read and learn. I gained a lot of seated snooze time. No interest, no spark, no got'cha.

So your failing math? Too bad it's time for French, and I give more homework than any math teacher, so just give up. Now Social Studies. Now Phys/Ed. Now fighting off the school

bullies. Physics might have saved me, but that course required high math skills; I was guided away from that hope. Student counsellors. Realists can fail to see the problem.

So tell me, though it's 50 years too late, what got you hooked on math?

No. Don't tell me. Tell the other students, those with talents and intelligence that are nevertheless failing in math. What is the hook?

Richard Feynman, " Mathematics is the language of nature....."

I weep.

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So tell me, though it's 50 years too late, what got you hooked on math?

I had an inspirational maths teacher who had an enthusiasm for 'A'-level maths. He was instrumental in bringing me out of myself and enabling me to progress beyond anybody's expectations. He did not have the same effect on everybody, I think it was just a few who responded to him. He allowed everybody to work at their own speed, solving any problem which baffled anybody. He gave homework to be done at any speed you wanted, and often said "leave number x out, it's too difficult" and grinned at me, so of course I would stay up half the night solving problem x. I was very lucky.

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Welcome Martin, but I completely failed to understand the point of your OP.

No one is forced to take any A level subject, let alone A level Maths.

It is elective and what's more the number of A levels a student can take during 6th form is limited.

So why did you take A level Maths?

Edited by studiot

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fact that basic math BORED me, it was pointless to me. Show me why I should care!

Sure. I think this is true for a lot of math instruction today. A lot of teachers teach math like it is just a set of rules to memorize.

They should be teaching problem solving skills.

And math is a very good environment for problem solving skills. Until you get to some very high levels, math has clearly right and clearly wrong answers. The tools are very well defined. The environment is set up to enable getting to the answer in a straightforward way. It is a perfect practice environment for problem solving.

Compare to the real life problem solving. Often to real life problems, the tools or methods are sloppy, imperfect, or maybe don't even exist. Real world problems may have 1 answer, no answer, or many, many answers. Real world is messy. But problem solving skills are always, always, always useful.

This is what the math teachers need to emphasizing. Not insisting that the kids need to know how to complete the square, know how to find lowest common denominators, or the equation of a circle. In real life, the vast majority of adults do not need that information. But those are tools used to solve problems, and solving those math problems practice problem solving skills, and we all need problem solving skills.

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This is what the math teachers need to emphasizing. Not insisting that the kids need to know how to complete the square, know how to find lowest common denominators, or the equation of a circle. In real life, the vast majority of adults do not need that information. But those are tools used to solve problems, and solving those math problems practice problem solving skills, and we all need problem solving skills.

It'a a problem. Not every kid will be a mathematician, physicist, or an engineer. Or a biologist, or economist. Anything quantitative. But SOME kids will go on to those fields. I don't think you want to separate kids at age 8, "This one goes to math classes, and that one goes to trade school and learns to make change." Or whatever "practical" skills are.

Should we not teach kids Shakespeare because not everyone will teach English lit? Anyway Shakespeare's a dead white European male so nobody needs to read him.

If you wouldn't teach high school math to high school students, what would you teach them? And why would you eliminate the learning path for the future engineers of the world? We all have to drive over the bridges they build. We should educate them, don't you think?

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In 1967 the A level exam was a privilege.

Approximately 15% of pupils in any one year took A levels at that time.

I don't have reliable data on the % of these who took maths, but it was one of about 15 popular subjects, but by no means the most popular so the % would have been around 1%.

About 3% of pupils in those days went on to become students at University.

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Very much so. They still learn math. They still learn quadratic formula, trig, etc.

But you don't tell them that they have to learn because math. Because a syllabus says so.

You tell them that you are teaching them problem solving skills, and they are learning the quadratic formula because it is another tool in the toolbox to solve math problems. Just like you teach someone what a wrench is instead of trying to loosen a nut with a pair of pliers. And eventually, they will learn what a torque wrench is, and what an impact is, etc. Learn the right tools for the right job.

That is what the formulas in math are... tools to get a job done.

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Very much so. They still learn math. They still learn quadratic formula, trig, etc.

Ok, I feel better now. Somebody using your handle wrote:

Not insisting that the kids need to know how to complete the square ...

But now you are saying you WILL teach them the quadratic equation. And if we're going to teach them the quadratic equation then we MUST teach them completing the square, because that's the trick that makes it possible to UNDERSTAND where the magic formula came from.

You can't fault me for reading what you wrote, right?

But if you're putting the quadratic formula back into the curriculum, along with the method for re-deriving it if we happen to forget it, then we're on the same page.

I agree with the OP that math education is terrible. I don't really know what to do about it.

Edited by wtf

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My high school math teachers weren't the worst or the best, average that's what they were.

What killed it for me in spite of an insatiable interest in science of any sort was that fact that basic math BORED me, it was pointless to me. Show me why I should care!

Even when I did learn the formulas and the methods I'd hit summer vacation full of activities and excitement. Come back to math class, and all I had learned was Gone.

That was when I ran into my worst math teacher. He knew which students he would work with and I wasn't one of those. Needless to say even as an A level student in

English, A to B level in Sciences (no math skills) ergo inconsistent. I ended languishing as a student who barely graduated. I did have talent in arts which went nowhere for me.

I've read articles which reported that experts in education and psychology have testified that the system of teaching practiced in the western world is a model that guarantees

Failure. Courses of study are broken up into unmanageable bits and short snippets that don't allow students any opportunity to learn, practice, use, let alone master the skills

they need to complete the courses thus; homework. Thrown out into the wilds of life to figure it out for yourself. (You dummy. Oh Yeah! If you teachers don't care then neither do I).

Screw it. I'm doing what I like, homework my sore patoot. As you know that model, unchanged, still holds as education practiced today.

I even bought pure math books, sat down to read and learn. I gained a lot of seated snooze time. No interest, no spark, no got'cha.

So your failing math? Too bad it's time for French, and I give more homework than any math teacher, so just give up. Now Social Studies. Now Phys/Ed. Now fighting off the school

bullies. Physics might have saved me, but that course required high math skills; I was guided away from that hope. Student counsellors. Realists can fail to see the problem.

So tell me, though it's 50 years too late, what got you hooked on math?

No. Don't tell me. Tell the other students, those with talents and intelligence that are nevertheless failing in math. What is the hook?

Richard Feynman, " Mathematics is the language of nature....."

I weep.

Having an insatiable interest in science is why you should've cared, but that's not why you failed. Simply put, mathematics is boring to you and you didn't even bother investigating the subject on your own. It's unfortunate that your math teachers didn't know how to inspire you but it's not their fault that you didn't have an interest in understanding mathematics. I actually liked working the equations and solving the problems even though I had a tough time at first memorizing the formulas, but I also understood that math is used to solve problems beyond just learning the equations and processes. As such, I was constantly seeing patterns in numbers and questioning my math teachers about equations that I would solve that predicted cool little patterns that I found.

For instance, in the tenth grade, I learned how to expand binomials, $(a+b)^n$, the hard way by multiplying $(a+b)$ by itself $n$ times. The first night we had homework, I figured out the binomial expansion theorem myself by developing a process that predicted the coefficients and powers of the expanded binomial. I didn't even know there was a binomial expansion theorem because we hadn't learned it yet, but I showed it to my math teacher the next day. I thought it was cool because it allowed me to do my homework using one line instead of half a sheet of paper, and coach Ware agreed. So, he let me teach it to the class and everyone got to use it until we learned the actual binomial expansion theorem.

Now, you might say that I had a good math teacher, but coach Ware was just teaching us math as is normally taught. After all, he was the school's basketball coach and not a mathematician. If I didn't have an interest in mathematics, I would've never got to teach the class that day and explain how I figured out how to expand binomials.

Sure. I think this is true for a lot of math instruction today. A lot of teachers teach math like it is just a set of rules to memorize.

They should be teaching problem solving skills.

And math is a very good environment for problem solving skills. Until you get to some very high levels, math has clearly right and clearly wrong answers. The tools are very well defined. The environment is set up to enable getting to the answer in a straightforward way. It is a perfect practice environment for problem solving.

Compare to the real life problem solving. Often to real life problems, the tools or methods are sloppy, imperfect, or maybe don't even exist. Real world problems may have 1 answer, no answer, or many, many answers. Real world is messy. But problem solving skills are always, always, always useful.

This is what the math teachers need to emphasizing. Not insisting that the kids need to know how to complete the square, know how to find lowest common denominators, or the equation of a circle. In real life, the vast majority of adults do not need that information. But those are tools used to solve problems, and solving those math problems practice problem solving skills, and we all need problem solving skills.

As Bignose has stated, teachers should focus on problem solving skills. This is what mathematics is all about; being able to solve problems. Martin or Nota, I understand that you feel kind of cheated because instead of learning how to use mathematics to solve problems you learned equations and formulas, which are the tools we use to solve such problems. You were given an equation and asked to plug in numbers to get a result, but you were never given a set of numbers and asked to produce the equation. You see, the first step in mastering problem solving is being able to derive an equation that produces an observed pattern of numbers. However, you can only do this if you know all those little equations and rules like factoring that allow you to develop such equations. The problem you had is that you wanted to be a rock star but you didn't want to put in all that boring, hard work of learning how to play and write the music. Do you see what I'm saying?

The binomial expansion theorem is just another equation that allows us to generate Pascal's triangle and calculate coefficients that occur when we expand binomials. Pretty boring stuff huh? However, did you know that the binomial expansion theorem can be found in just about every single problem in number theory? It still blows my mind to this day. When I was learning trigonometry, I learned that there are polynomials that predicts the summation of $x^p$. However, my teacher only knew the polynomial for the summation of $x^1$, which is

$\sum_0^n x=\frac{n(n+1)}{2}$

It took me one year and three months, but I solved the equation that predicts the sum of $x^p$, which allowed me to derive and extend Newton's interpolation formula:

I figured out how to do this when I discovered Newton's interpolation formula my 11th grade year in high school. I didn't know about interpolation at the time. I was in Trigonometry and had learned that summations of $x$ had a polynomial that generalized the sum. It took me one year and three months to discover the equation which predicted the summations of $x^p$. I did not have a computer and I did all of the calculation using paper and a TI-88 (I sure do love Mathematica... It has saved a lot of trees):

$F(s,\, n,\, p)=\sum_{j_{1}=1}^n \sum_{j_{2}=1}^{j_{1}} ... \sum_{x=1}^{j_{S}} (x^p)=\sum_{j=0}^p \left((-1)^{j+p} \left(\sum_{k=0}^j \frac{k^p \, \left \langle -j \right \rangle_{j-k}}{(j-k)!}\right)\left(\frac{\left \langle n \right \rangle_{j+s}}{(j+s)!} \right)\right)$

Where $s$ represents the recursion level of the summation (when $s=0$ we get the original sequence and when $s=1$ we get the first summation of the sequence and so on), and we sum the function, $x^p$, from $0$ to $n$ where $p$ is the power.

The process, which involved recursively taking the deltas of the number sequences and applying the rising factorial or Pochhammer function, allowed me to derive the exponential version by recursively dividing instead of doing a subtraction. This is demonstrated below (note: I replaced the Pochhammer function with a product series operator for those that are not familiar with the rising factorial):

Newton's Interpolation:

Recursively take the deltas of each sequence:

$\begin{matrix} & & & y_4 - 3\, y_3 + 3\, y_2 - y_1\\ & & y_3 - 2\, y_2 + y_1 & y_4 - 2\, y_3 + y_2\\ & y_2 - y_1 & y_3 - y_2 & y_4 - y_3 \\ y_1 & y_2 & y_3 & y_4 \end{matrix}$

We are only interested in the results at the top of each column. Also, we can see that each result is an alternating sum of the original sequence with coefficients that are pascal numbers. Next, we use each result with the rising factorial and standard factorial as follows:

$\left(y_1\right)\frac{1}{0!}\ -\ \left(y_2 - y_1\right)\frac{(1-x)}{1!}\ +\ \left(y_3 - 2\, y_2 + y_1\right)\frac{(1-x)(2-x)}{2!}\ -\ \left(y_4 - 3\, y_3 + 3\, y_2 - y_1\right)\frac{(1-x)(2-x)(3-x)}{3!}\ +\ \text{etc...}$

The formula which defines this process is as defined below (note: I have expanded the formula to include recursively summing the sequence such that when $s=0$ we get the original sequence and when $s=1$ we get the first summation of the sequence and so on):

$F(x, \, s,\, n) = \sum_{i=0}^{n-1}\left( \sum_{j=0}^{i}\left(f(j)\frac{(-1)^{j}\ i!}{j!\ (i-j)!}\right) \frac{(-1)^{s}}{(i+s)!} \prod_{k=1}^{i+s}\left(k-s-x\right)\right)$

Where $x$ is the variable, $s$ is the summation as explained above, and we sum from $0$ to $n-1$. We can start from any number by modifying $f(j)$ to include a starting index, $f(j+start)$.

Daedalus' Exponential Interpolation:

We must do the same thing as above except we will divide the numbers of the sequence:

$\begin{matrix} & & & y_4^1 \times y_3^{-3} \times y_2^{3} \times y_1^{-1}\\ & & y_3^1 \times y_2^{-2} \times y_1^1 & y_4^1 \times y_3^{-2} \times y_2^1\\ & y_2^1 \times y_1^{-1} & y_3^1 \times y_2^{-1} & y_4^1 \times y_3^{-1} \\ y_1 & y_2 & y_3 & y_4 \end{matrix}$

The next part is also similar except additions become multiplications, subtractions become division, and multiplication becomes exponents:

$\left(y_1\right)^{\frac{1}{0!}}\ \times\ \left(y_2^1 \times y_1^{-1}\right)^{-\frac{(1-x)}{1!}}\ \times \ \left(y_3^1 \times y_2^{-2} \times y_1^1\right)^{\frac{(1-x)(2-x)}{2!}}\ \times \ \left(y_4^1 \times y_3^{-3} \times y_2^{3} \times y_1^{-1}\right)^{-\frac{(1-x)(2-x)(3-x)}{3!}}\ \times \ \text{etc...}$

The following formula defines the above process such that when $p=0$ we get the original sequence and when $p=1$ we get the first product of the sequence and so on:

$F(x, \, p,\, n) = \prod_{i=0}^{n-1}\left( \prod_{j=0}^{i}\left(f(j)^{\frac{(-1)^{j}\, i!}{j!\, (i-j)!}}\right)^{ \frac{(-1)^{p}}{(i+p)!} \displaystyle \prod_{k=1}^{i+p}\left(k-p-x\right)}\right)$

Where $x$ is the variable, $p$ is the recursion level of the product as explained above, and we multiply the outputs from $0$ to $n-1$. We can start from any number by modifying $f(j)$ to include a starting index, $f(j+start)$.

This method can be applied to any operator that obeys the associative law (which is why it only works for summations and products). Please forgive any grammar errors in this post. I broke a tooth over thanksgiving and it is causing me a tremendous amount of pain. It's 3:00 am and I have a dentist appointment today. So... I kinda rushed this post. I will correct any issues tomorrow as long as the edit timer has not expired.

As you can see in the above quote, the binomial expansion theorem occurs when taking the deltas or products of the set of numbers when interpolating an equation for them. If I didn't know all of those boring formulas, I would never be able to solve such exciting problems. My latest work on the Collatz conjecture, an unsolved problem in mathematics, actually has the binomial expansion theorem embedded in it where the diagonal numbers in Pascal's triangle determines the number of lines that pass through points that occur for each iteration of the Collatz function. For instance, the following graph shows the lines that goes through the points for one iteration of the Collatz function:

Here is the graph for two iterations of the Collatz function:

Here is the graph for three iterations of the Collatz function:

You'll notice that for the first iteration, the points are equally distributed along each line. However, you'll see that the points occur in repeating clusters for the second and third iterations along each line. The pattern for the second iteration is that the top line has its points equally spaced but there are 2 points clustered along the second line. 3 points clustered along the third line and so forth. The pattern again changes for the third iteration. The top line has its points equally spaced with 3 points clustered along the second line. 6 points clustered along the third line and 10 points clustered along the fourth line and so forth. The green line is the function, $f(x)=x$ and is used to determine where iterations of the Collatz function can cycle, which occurs where the blue lines intersect the green line.

Some really cool, mind-blowing facts from applying my problem solving skills to the Collatz conjecture. First, by analyzing where the blue lines intersect the green line, we can conjecture that cycles can only occur when the points are near zero. The slopes of the blue lines decrease exponentially and therefore diverges from $y=x$ as $x$ approaches infinity. This means that there are no hidden cycles greater than 1 for extremely large values of $x$. It doesn't completely rule out the possibility of a large value of $x$ from being a cycle because each point in a cluster defines a line that is parallel to other points in the cluster, and these parallel lines eventually no longer appear to occur on a common line as the number of iterations approach infinity. However, because I have the equation that predicts all possible lines for all point clusters for any iteration of the Collatz function, I have derived the equation that predicts all possible points that cycle by setting the equations that determines the blue lines equal to the equation for the green line and solving for $x$. For my own purposes, I will withhold this formula, but I don't mind showing you the expanded form of the equation for the first few iterations (if you want to beat me to the proof, you'll need to do the work ):

The intersection points for the first iteration where $m=3$ and $n=1$ as defined by the "normal" Collatz function where $\ell$ is the index of the line cluster and the number of iterations are expressed as constants (I extended it to work with any set of odd numbers $m$ and $n$):

$\left(\frac{\mathit{n}}{2^{1+\ell -1}-\mathit{m}^1}\right)$

The intersection points for the second iteration:

$\left(\frac{\mathit{n}\left(2^{\text{\mathit{x}1}}+\mathit{m}\right)}{2^{2+\ell -1}-\mathit{m}^2}\right)$

The intersection points for the third iteration:

$\left(\frac{\mathit{n}\left(2^{\text{\mathit{x}1}+\text{\mathit{x}2}}+\mathit{m}\left(2^{\text{\mathit{x}1}}+\mathit{m}\right)\right)}{2^{3+\ell -1}-\mathit{m}^3}\right)$

The intersection points for the fourth iteration:

$\left(\frac{\mathit{n}\left(2^{\text{\mathit{x}1}+\text{\mathit{x}2}+\text{\mathit{x}3}}+\mathit{m}\left(2^{\text{\mathit{x}1}+\text{\mathit{x}2}}+\mathit{m}\left(2^{\text{\mathit{x}1}}+\mathit{m}\right)\right)\right)}{2^{4+\ell -1}-\mathit{m}^4}\right)$

You'll notice that the above equations have $x_1,x_2,x_3$ etc.. These variables define the number of twos that actually divide the even numbers that occur when iterating an odd number. However, it doesn't include the number of twos for the last odd number that occurs during a given iteration. If you were to apply the Collatz function to the odd numbers or rational numbers that have odd numerators defined by the above equations, you will find that they do indeed cycle in the specified number of iterations:

Applying the Collatz function to the equation for the cyclic integers / rationals of the first iteration:

$\left(\left(\mathit{m}\left(\frac{\mathit{n}}{2^{1+\ell -1}-\mathit{m}^1}\right)+\mathit{n}\right)\div 2^{(1+\ell )-1}\right)=\left(\frac{\mathit{n}}{2^{1+\ell -1}-\mathit{m}^1}\right)$

Applying the Collatz function to the equation for the cyclic integers / rationals of the second iteration:

$\left(\left(\mathit{m}\left(\left(\mathit{m}\left(\frac{\mathit{n}\left(2^{\text{\mathit{x}1}}+\mathit{m}\right)}{2^{2+\ell -1}-\mathit{m}^2}\right)+\mathit{n}\right)\div 2^{\text{\mathit{x}1}}\right)+\mathit{n}\right)\div 2^{(2+\ell )-(\text{\mathit{x}1})-1}\right)=\left(\frac{\mathit{n}\left(2^{\text{\mathit{x}1}}+\mathit{m}\right)}{2^{2+\ell -1}-\mathit{m}^2}\right)$

Applying the Collatz function to the equation for the cyclic integers / rationals of the third iteration:

$\left(\left(\mathit{m}\left(\left(\mathit{m}\left(\left(\mathit{m}\left(\frac{\mathit{n}\left(2^{\text{\mathit{x}1}+\text{\mathit{x}2}}+\mathit{m}\left(2^{\text{\mathit{x}1}}+\mathit{m}\right)\right)}{2^{3+\ell -1}-\mathit{m}^3}\right)+\mathit{n}\right)\div 2^{\text{\mathit{x}1}}\right)+\mathit{n}\right)\div 2^{\text{\mathit{x}2}}\right)+\mathit{n}\right)\div 2^{(3+\ell )-(\text{\mathit{x}1}+\text{\mathit{x}2})-1}\right)=$

$\left(\frac{\mathit{n}\left(2^{\text{\mathit{x}1}+\text{\mathit{x}2}}+\mathit{m}\left(2^{\text{\mathit{x}1}}+\mathit{m}\right)\right)}{2^{3+\ell -1}-\mathit{m}^3}\right)$

As we can tell by applying the Collatz function to the intersection points, the number of twos that divide the last odd integer / odd rational number is equal to $(r+\ell)-(x_1+x_2+\cdots)-1$. If $x_1,x_2,\cdots$ are greater than 0, and $r+\ell-1 > x_1+x_2+\cdots$, then the numbers defined by the equations for the intersection points do indeed cycle in $r$ iterations of the Collatz function and define odd integers or rational number where the numerator is odd. The next step is to show that 1 is the only integer greater than 0 that cycles and all other positive numbers are rational. These intersection points combined with other equations that I have derived that maps out the odd integers that occur while iterating the Collatz functions indefinitely for an initial odd number can be used to construct a proof of the Collatz conjecture and solve an 80 year old, unsolved math problem.

So, you can see how important it is to know all of these boring equations and formulas because they allow us to delve into the unknown and solve problems that have remained unsolved for decades and even centuries! Although your teachers could never teach you how to solve the Collatz conjecture, they did try to teach you the tools that you could've used to do it. Ultimately, it was up to you to take the leap into the world of mathematics, but it was boring to you, which is perfectly ok. Not everyone was meant to play with numbers Just remember, if you want to be a rock star, you have to practice playing and writing music. Even if the work to get there is boring, the reward is definitely worth the effort \,,/

Edited by Daedalus

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Will this be on the test?

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The worst maths teachers are those who are naturally good at maths. They tend to be incapable of understanding student's problems, since, to them, it is obvious. Give me a teacher who has struggled with maths and understands the difficulties.

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Two thoughts about the replies here.

The OP is talking about A levels, which are intended for 16 - 18 year olds.

Many of the replies are about maths for a much younger age group.

Several have offered problem solving.

I like problem solving and thinking for myself, but many do not.

They prefer to be told what to do.

What do you think they would make of being forced to problem solve?

Of course there is the question "Is the mantra maths, maths and more maths is good education for everyone" actually true?

Or are we boring many with subject matter they don't appreciate and will never need?

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The only thing which got me hooked on Physics (applied Maths) at school as a teenager was that it led to a definite answer - a form of truth which some other subjects lacked. I used to avoid homework until the last minute and be on the streets playing football/soccer or other sports until very late - often seeing the ball by the streetlights.

As a teacher now, I see exactly what the OP is saying - our current model fails the least able or the slow learner and we are leading cattle through a fence. It is tragic, but, when there are moments when a class interacts with, and "clicks" with a teacher, both the teacher and the students improve as a result of this social interaction.

I have had my say about the current education system to the Department for Education in this country (the UK) but no rational response has been given to me.

The motive of the teacher matters as well and I have seen, for the most, part, caring and compassionate teachers who work bloody hard for their pupils. A few egotists or people who don't like children enter the profession to earn their pensions and it looks as if the author of the OP met one of these teachers.

In reality, the school system originally, in the 19th century was in three tiers: a) labourer class; b) middle manager class and c) director and owner class. The success of all those hardworking and self sacrificing teachers throughout the years has been in allowing individuals from a) and b) to enter professions for class c). This success continues to this day.

Edited by jimmydasaint

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Math was one of my favorite subjects, second only to physics. I enjoy finding the solution. I still enjoy mathematical riddles. I liked math independent of the teacher (I was the teachers favourite anyway). I like it for the problem solving, but also for the math on its own.

I guess being good at it helped to enjoy it, but enjoying also helps to be good.

I agree with the sentiment that everyone should get it in sufficient amounts to be able to build on it. I may have known I would do something with math/science at age ten, but others only decide after secondary school (age 18). A former colleague of mine only decided to become an engineer after doing something with languages in secondary school. Now he has a PhD in engineering. I doubt that would have been possible if he hadn't had any math among those languages.

Moreover, even psychologists need a significant amount of math.

I don't think focussing only on the problem solving aspect would work, since it is harder and can only be done well after a solid foundation is established. It usually comes at end of the chapter, often in extras, and is horror for many students who struggle with the basics.

What could be done is closer collaboration between math teachers and eg science teachers.

There are, however, some subjects that could be handled better. I feel we spent eg too much time on solving all kinds of complex integrals. All the integrals I encountered as engineer can be divided in three categories:

- polynomials: by far the most frequent and easy to solve

- something that can be found in the list of integrals on Wikipedia

- something else, which is easily solved numerically.

It would be much more useful to spend more time on composing the integral corretly.

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You can't fault me for reading what you wrote, right?

I'm sure my stream-of-consciousness writing there could have been made clearer. But the point has always remained:

The specifics of math are going to be needed by just a tiny amount of students in their future lives, but teaching math is still important because it is teaching problem solving skills. However, I think most teachers, even most math teachers, don't get that and certainly don't emphasize it enough. The modern emphasis seems to be mostly that they need the kids to memorize enough things to pass the standardized tests, although this is true about every subject of modern education in general.

Edited by Bignose

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If math is about memorisation, the teacher is definitely doing something wrong. It should be the subject that needs the least memorisation.

I don't have this experience, but we also don't have standardised tests.

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I always liked this quote from Robert Heinlein:

“Anyone who cannot cope with mathematics is not fully human. At best, he is a tolerable subhuman who has learned to wear his shoes, bathe, and not make messes in the house.”

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I always liked this quote from Robert Heinlein:

“Anyone who cannot cope with mathematics is not fully human. At best, he is a tolerable subhuman who has learned to wear his shoes, bathe, and not make messes in the house.”

I liked this by my grandad: "Maths is a language, not a problem."

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Having an insatiable interest in science is why you should've cared, but that's not why you failed. Simply put, mathematics is boring to you and you didn't even bother investigating the subject on your own. It's unfortunate that your math teachers didn't know how to inspire you but it's not their fault that you didn't have an interest in understanding mathematics. I actually liked working the equations and solving the problems even though I had a tough time at first memorizing the formulas, but I also understood that math is used to solve problems beyond just learning the equations and processes. As such, I was constantly seeing patterns in numbers and questioning my math teachers about equations that I would solve that predicted cool little patterns that I found.

For instance, in the tenth grade, I learned how to expand binomials, $(a+b)^n$, the hard way by multiplying $(a+b)$ by itself $n$ times. The first night we had homework, I figured out the binomial expansion theorem myself by developing a process that predicted the coefficients and powers of the expanded binomial. I didn't even know there was a binomial expansion theorem because we hadn't learned it yet, but I showed it to my math teacher the next day. I thought it was cool because it allowed me to do my homework using one line instead of half a sheet of paper, and coach Ware agreed. So, he let me teach it to the class and everyone got to use it until we learned the actual binomial expansion theorem.

Now, you might say that I had a good math teacher, but coach Ware was just teaching us math as is normally taught. After all, he was the school's basketball coach and not a mathematician. If I didn't have an interest in mathematics, I would've never got to teach the class that day and explain how I figured out how to expand binomials.

As Bignose has stated, teachers should focus on problem solving skills. This is what mathematics is all about; being able to solve problems. Martin or Nota, I understand that you feel kind of cheated because instead of learning how to use mathematics to solve problems you learned equations and formulas, which are the tools we use to solve such problems. You were given an equation and asked to plug in numbers to get a result, but you were never given a set of numbers and asked to produce the equation. You see, the first step in mastering problem solving is being able to derive an equation that produces an observed pattern of numbers. However, you can only do this if you know all those little equations and rules like factoring that allow you to develop such equations. The problem you had is that you wanted to be a rock star but you didn't want to put in all that boring, hard work of learning how to play and write the music. Do you see what I'm saying?

The binomial expansion theorem is just another equation that allows us to generate <a data-ipb="nomediaparse" data-cke-saved-href="https://en.wikipedia.org/wiki/Pascal"href="https://en.wikipedia.org/wiki/Pascal" s_triangle"="">Pascal's triangle and calculate coefficients that occur when we expand binomials. Pretty boring stuff huh? However, did you know that the binomial expansion theorem can be found in just about every single problem in number theory? It still blows my mind to this day. When I was learning trigonometry, I learned that there are polynomials that predicts the summation of $x^p$. However, my teacher only knew the polynomial for the summation of $x^1$, which is

$\sum_0^n x=\frac{n(n+1)}{2}$

It took me one year and three months, but I solved the equation that predicts the sum of $x^p$, which allowed me to derive and extend Newton's interpolation formula:

As you can see in the above quote, the binomial expansion theorem occurs when taking the deltas or products of the set of numbers when interpolating an equation for them. If I didn't know all of those boring formulas, I would never be able to solve such exciting problems. My latest work on the Collatz conjecture, an unsolved problem in mathematics, actually has the binomial expansion theorem embedded in it where the diagonal numbers in Pascal's triangle determines the number of lines that pass through points that occur for each iteration of the Collatz function. For instance, the following graph shows the lines that goes through the points for one iteration of the Collatz function:

Here is the graph for two iterations of the Collatz function:

Here is the graph for three iterations of the Collatz function:

You'll notice that for the first iteration, the points are equally distributed along each line. However, you'll see that the points occur in repeating clusters for the second and third iterations along each line. The pattern for the second iteration is that the top line has its points equally spaced but there are 2 points clustered along the second line. 3 points clustered along the third line and so forth. The pattern again changes for the third iteration. The top line has its points equally spaced with 3 points clustered along the second line. 6 points clustered along the third line and 10 points clustered along the fourth line and so forth. The green line is the function, $f(x)=x$ and is used to determine where iterations of the Collatz function can cycle, which occurs where the blue lines intersect the green line.

Some really cool, mind-blowing facts from applying my problem solving skills to the Collatz conjecture. First, by analyzing where the blue lines intersect the green line, we can conjecture that cycles can only occur when the points are near zero. The slopes of the blue lines decrease exponentially and therefore diverges from $y=x$ as $x$ approaches infinity. This means that there are no hidden cycles greater than 1 for extremely large values of $x$. It doesn't completely rule out the possibility of a large value of $x$ from being a cycle because each point in a cluster defines a line that is parallel to other points in the cluster, and these parallel lines eventually no longer appear to occur on a common line as the number of iterations approach infinity. However, because I have the equation that predicts all possible lines for all point clusters for any iteration of the Collatz function, I have derived the equation that predicts all possible points that cycle by setting the equations that determines the blue lines equal to the equation for the green line and solving for $x$. For my own purposes, I will withhold this formula, but I don't mind showing you the expanded form of the equation for the first few iterations (if you want to beat me to the proof, you'll need to do the work ):

The intersection points for the first iteration where $m=3$ and $n=1$ as defined by the "normal" Collatz function where $\ell$ is the index of the line cluster and the number of iterations are expressed as constants (I extended it to work with any set of odd numbers $m$ and $n$):

$\left(\frac{\mathit{n}}{2^{1+\ell -1}-\mathit{m}^1}\right)$

The intersection points for the second iteration:

$\left(\frac{\mathit{n}\left(2^{\text{\mathit{x}1}}+\mathit{m}\right)}{2^{2+\ell -1}-\mathit{m}^2}\right)$

The intersection points for the third iteration:

$\left(\frac{\mathit{n}\left(2^{\text{\mathit{x}1}+\text{\mathit{x}2}}+\mathit{m}\left(2^{\text{\mathit{x}1}}+\mathit{m}\right)\right)}{2^{3+\ell -1}-\mathit{m}^3}\right)$

The intersection points for the fourth iteration:

$\left(\frac{\mathit{n}\left(2^{\text{\mathit{x}1}+\text{\mathit{x}2}+\text{\mathit{x}3}}+\mathit{m}\left(2^{\text{\mathit{x}1}+\text{\mathit{x}2}}+\mathit{m}\left(2^{\text{\mathit{x}1}}+\mathit{m}\right)\right)\right)}{2^{4+\ell -1}-\mathit{m}^4}\right)$

You'll notice that the above equations have $x_1,x_2,x_3$ etc.. These variables define the number of twos that actually divide the even numbers that occur when iterating an odd number. However, it doesn't include the number of twos for the last odd number that occurs during a given iteration. If you were to apply the Collatz function to the odd numbers or rational numbers that have odd numerators defined by the above equations, you will find that they do indeed cycle in the specified number of iterations:

Applying the Collatz function to the equation for the cyclic integers / rationals of the first iteration:

$\left(\left(\mathit{m}\left(\frac{\mathit{n}}{2^{1+\ell -1}-\mathit{m}^1}\right)+\mathit{n}\right)\div 2^{(1+\ell )-1}\right)=\left(\frac{\mathit{n}}{2^{1+\ell -1}-\mathit{m}^1}\right)$

Applying the Collatz function to the equation for the cyclic integers / rationals of the second iteration:

$\left(\left(\mathit{m}\left(\left(\mathit{m}\left(\frac{\mathit{n}\left(2^{\text{\mathit{x}1}}+\mathit{m}\right)}{2^{2+\ell -1}-\mathit{m}^2}\right)+\mathit{n}\right)\div 2^{\text{\mathit{x}1}}\right)+\mathit{n}\right)\div 2^{(2+\ell )-(\text{\mathit{x}1})-1}\right)=\left(\frac{\mathit{n}\left(2^{\text{\mathit{x}1}}+\mathit{m}\right)}{2^{2+\ell -1}-\mathit{m}^2}\right)$

Applying the Collatz function to the equation for the cyclic integers / rationals of the third iteration:

$\left(\left(\mathit{m}\left(\left(\mathit{m}\left(\left(\mathit{m}\left(\frac{\mathit{n}\left(2^{\text{\mathit{x}1}+\text{\mathit{x}2}}+\mathit{m}\left(2^{\text{\mathit{x}1}}+\mathit{m}\right)\right)}{2^{3+\ell -1}-\mathit{m}^3}\right)+\mathit{n}\right)\div 2^{\text{\mathit{x}1}}\right)+\mathit{n}\right)\div 2^{\text{\mathit{x}2}}\right)+\mathit{n}\right)\div 2^{(3+\ell )-(\text{\mathit{x}1}+\text{\mathit{x}2})-1}\right)=$

$\left(\frac{\mathit{n}\left(2^{\text{\mathit{x}1}+\text{\mathit{x}2}}+\mathit{m}\left(2^{\text{\mathit{x}1}}+\mathit{m}\right)\right)}{2^{3+\ell -1}-\mathit{m}^3}\right)$

As we can tell by applying the Collatz function to the intersection points, the number of twos that divide the last odd integer / odd rational number is equal to $(r+\ell)-(x_1+x_2+\cdots)-1$. If $x_1,x_2,\cdots$ are greater than 0, and $r+\ell-1 > x_1+x_2+\cdots$, then the numbers defined by the equations for the intersection points do indeed cycle in $r$ iterations of the Collatz function and define odd integers or rational number where the numerator is odd. The next step is to show that 1 is the only integer greater than 0 that cycles and all other positive numbers are rational. These intersection points combined with other equations that I have derived that maps out the odd integers that occur while iterating the Collatz functions indefinitely for an initial odd number can be used to construct a proof of the Collatz conjecture and solve an 80 year old, unsolved math problem.

So, you can see how important it is to know all of these boring equations and formulas because they allow us to delve into the unknown and solve problems that have remained unsolved for decades and even centuries! Although your teachers could never teach you how to solve the Collatz conjecture, they did try to teach you the tools that you could've used to do it. Ultimately, it was up to you to take the leap into the world of mathematics, but it was boring to you, which is perfectly ok. Not everyone was meant to play with numbers Just remember, if you want to be a rock star, you have to practice playing and writing music. Even if the work to get there is boring, the reward is definitely worth the effort \,,/

Do you want to be my professor?

Did you actually wrote these lines of code for this post?

This is very motivational I think

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Do you want to be my professor?

If you truly want to understand mathematics as I understand it, then I'd be more than happy to instruct you in the ways of the force. However, I'm not a professor, but I do tutor people for free and am more than happy to help. A few years ago, before I found a job as the Sr. Software Engineer for Factor, I tutored several people taking Calculus 1 - 3 at Oklahoma City Community College. I was able to help several students understand calculus better and improve their grade because I absolutely love math and have a knack for identifying where someone is struggling, relate the mathematics in way they are familiar with, and help them overcome the problems they are having. So, you can always send me a PM and I'll do my best to help, or you can post your questions here in the math forum because there are plenty of people who are also happy to help and are more knowledgeable than I.

Did you actually wrote these lines of code for this post?

Of course I did. The mathematics that I included in the post are based on real events and actual problems that I solved for myself or in the process of solving. In fact, I often post my own discoveries

http://www.scienceforums.net/topic/61237-discoveries-by-daedalus/

and create math challenges that I post in the Brain Teasers and Puzzles forum for others to solve:

http://www.scienceforums.net/topic/60158-daedalus-first-challenge/

http://www.scienceforums.net/topic/60202-daedalus-second-challenge/

http://www.scienceforums.net/topic/60236-daedalus-third-challenge/

http://www.scienceforums.net/topic/61209-daedalus-fourth-challenge/

http://www.scienceforums.net/topic/67268-daedalus-fifth-challenge/

http://www.scienceforums.net/topic/77503-daedalus-sixth-challenge/

http://www.scienceforums.net/topic/77776-daedalus-seventh-challenge/

http://www.scienceforums.net/topic/78797-daedalus-eighth-challenge/

http://www.scienceforums.net/topic/79134-daedalus-ninth-challenge/

http://www.scienceforums.net/topic/80134-daedalus-tenth-challenge/

http://www.scienceforums.net/topic/87801-daedalus-eleventh-challenge/

http://www.scienceforums.net/topic/93095-daedalus-twelfth-challenge/

http://www.scienceforums.net/topic/95080-daedalus-thirteenth-challenge/

ScienceForums.net allows us to write mathematics in our posts using LaTeX, which contains syntax that can display just about any mathematical symbol or operator. Basically, there are tags that you use in your post that tells the forum software that you are using LaTeX. As seen in the following image, if you hover the mouse over any of the formulas you see here and left click on the equations, you should get a dialog that shows you the tags and the LaTeX used to produce the mathematical output:

This is very motivational I think

That was the intent of my post. Martin or Nota wanted to know what got us mathematicians hooked on math or, at least, tell others why we like it as to inspire them or give them hope. It seems like this person has a grudge because they wanted to be able to understand math but couldn't because their teachers didn't make it exciting or only had them learn equations and formulas without being able to explain how to use them to solve any problem they would like to know the answer. So, my reply was based on my experiences and I wanted to make it clear that my success with understanding math is due to my own curiosity and love for the subject.

If you are truly interested in something, you cannot rely on someone else to teach you everything about it. Sure, teachers are there to teach us various topics that we are required to learn or want to learn so that we can function in society and figure out what we want to do with our lives, but it's ultimately up to us to determine how far we want to take our education. If you really like something, then you will ask questions and find someone who can answer them. I find that we are all in some way experts at things we love to do because we will invest time and effort in learning these things. If something is boring to you, then you are not going to invest the time and effort it takes to master the subject. For instance, I love playing guitar, piano, and writing musical compositions. So, I like to use the rock star analogy because a lot of people would love to be a rock star or famous musician because the idea of having talent, fame, money, and getting the girls or guys is intriguing to them. However, learning to play a musical instrument can be very boring in the beginning because you are learning things like scales, chords, and music theory, which may not pertain to the actual reasons of why you want to be a musician in the first place. Plus, it's really hard to make your fingers hit all of those notes correctly. So, it's extremely frustrating when you start out because you want to be able to play guitar like Eddie Van Halen so you can impress all those girls at school, but you find that you are struggling trying to play "Mary Had a Little Lamb". You really have to put in a lot of time and effort practicing these things and learning all of those boring concepts if you want to be able to play music and even more so if you want to be a rock star.

So, when it comes to why I love math, I gave examples based on the binomial expansion theorem because most people that have taken high school level math should be familiar with that theorem and it still amazes me to this day how often it comes up in problems that I have solved and are currently working on solving such as the Collatz conjecture. I don't expect most people to understand the examples I gave because they are complex. However, I do hope that people who want to learn mathematics and be able to reach a level of understanding that allows them to tackle these type of complex mathematical problems will see that it takes more then just relying on someone else to explain it to them to get there. You have to have a curiosity and love for the subject that enables you to push through all the boring stuff and pursue a mastery of the subject yourself. This applies to everything you want to do in life. If you are truly interested in something, then you will ask the questions and pursue the answers yourself. If your current teachers can't help you, then your motivation and curiosity will force you to find someone who can. Most of my teachers in high school could not answer a lot of the questions I had about the mathematics pertaining to problems that I was solving for myself. However, I didn't let that stop me from finding the answers. Either my teachers would point me in the right direction, or they referred me to someone who could help. All it took from me was the time and effort to pursue the subject.

Edited by Daedalus

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