Tuesday, August 31, 2010

Solving the Quintic

In my previous post about finding the roots of polynomials, I wrote that no general solution exist for polynomials whose degree is higher than 4. However, we still need to be able to find the roots of these polynomials. In this post I want to talk about some of the methods for finding roots, and about their limitations.

Finding the roots of polynomial might sound an abstract goal, but it is actually a very practical one. Look on the following problem for example - Given a line segment of length one AC, divide it in two parts AB and BC in such a way that AC/AB=AB/BC. This is a purely geometrical problem, but its solution is equivalent to solving the following equation:



If we will simplify it a bit, we w! ill get:]

1-x=x^2
x^2+x-1=0

This is a very simple and geometrical example, but the need to solve an equation similar form may arise in different problems in engineering. In this example, it is easy to see that the solution is irrational. This brings us to the first problem with solving polynomials. The answer might be irrational - and if so, even a computer will not be able to solve an equation by trial and error.

Lets look on the following quintic equation (fifth degree polynomial):



This one cannot be solved using a formula. But it is obvious that 1 a root of this polynomial. However, such a polynomial must have five roots (according to the basic theorem of algebra). What can ! we do to find them? A simple way would be to try to divide thi! s polyno mial by x-1. After such division, the degree of the polynomial we need to solve becomes less by one. We will get then that our original polynomial can be written as:



So, we got now a problem which is very easy to solve. All we need to do is to use the formula we know for the forth degree polynomial, and we are done. If we don't remember the formula, we can try to guess another root, lest call it y. and then divide by x-y. For this polynomial, it is again obvious that 1 is a root. So we can divide by x-1 again. We get:



If this is still too complicated, we can divide again.

I wanted the above e! xample to be as simple as possible, so I selected the polynomial (x-1)^5. However, what about more complex examples? It is practical to try to solve some random polynomial in this way? The answer to this is yes. Firstly, we can always divide two polynomials. This result we get from a theorem I don't want to talk about now. Secondly, while it is often impossible to guess the root, this is a rather practical approach. Unless all of the roots are irrational, the chances to guess them are pretty good (especially if you get this question in an exam). The reason for this is that the last monom (the free number) must be multiplication of all the roots, and the second must be minus the sum of all the roots. In the polynomial I solved above, for example the roots are (1,1,1,1,1). There sum is 5 and there multiplication is 1. And in the polynomial this is exactly what we get - in the second place we have (-5x^4) and in the last we have 1.

Lets look on the following polynom! ial:



This particular polynomial cannot be solved by the method specified. The reason for this is very simple - are the roots are complex. For complex roots there is a very interesting property - if, lets say, (2-i) is a root than (2+i) is also a root. This makes the process slightly easier. If you guess one root you get another one as a bonus. Now, I don't know about you, but I have no idea what are the roots of this particular polynomial, and I don't know how to find then.

So why I am saying that the method I presented in which you need to guess a root is practical? It is only practical when you need to solve a polynomial that is "nice and easy". If the polynomial you need to solve cannot be solved by simply guessing, it is better to use the computer to find the solution or at least an approximation to the solution. Unfortunately, this is the only ! thing we can do without a closed formula.

The problem we have with the quintic, and other polynomials is not unique. We have similar problem with integrals. It is often perfectly clear that an integral have a solution, but we have no way to find it. For example:

.

It is proven that this integral cannot be written using elementary functions, but it exists. And there are many others like this one. In modern mathematics there are more than enough things that are proved to be impossible, but we would like them to be possible. And this is (in my opinion) a strong argument for the claim that math is discovered and not invented - we get result which are hardly expected or desired.

In the next post I will write about how simple polynomials were solved in the ancient world, and I will also ! try to find time to write about the formulas for the third and! forth d egree.


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Solve Your Maths Problem with Tutor Vista

Often the case the parents have difficulty in guiding their children at home, especially if the child has difficulty doing math, especially if your child was in K-12. Your child needs help to understand math and algebra concepts, as you know that 90% is to be understood, while the remaining 10% to be memorized. This tutorial now has developed online, allowing your child can follow this tutorial from home via the internet.

In addition, if your child has difficulty with their homework, Tutor vista can also help your child to solve his homework online by using homework help menu. In this tutorial, there are also menus to solve the problems of mathematics, algebra, and trigonometry. Their tutors are always ready and willing to help you anytime on! line 24 hours a day and seven days a week. Tutor vista also provides tutoring for all math problems such as, equation, fractions, simplify and solving equations.
If you still doubt of their service, Tutor Vista also provide demo for free where you can try their services So, What are you waiting for? Just join Tutor Vista and see the result for yourself.


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Word Problems - Hints and Tips to Solving Them

Word problems are one kind of math question that ALWAYS gives students troubles, probably because making a connection between the math and the 'real world' situation isn't necessarily the easiest thing to do. Despite that, there are a few pointers that you can try to which will hopefully lead you to the correct solution.

The first tip is to MAKE A LIST OF ALL OF THE INFORMATION THAT YOU ARE GIVEN.

Example:

Joe wants to save money to buy a new Playstation 3. It costs $400 to buy from the store. He has a job which pays him $15 per hour, and he works 6 hours per day. However, to get to and from his job, he has to take the bus, which costs him $2 each way, and he buys lunch each day he works for $6. He also owes his dad $500 from when he borrowed it to buy a new stereo. Taking all of these costs into account, how many days will it take Joe to save up enough money to buy his new Playstation? How much more money does Joe need to earn if he w! ants to buy it in 5 days?

LIST WHAT YOU KNOW:
PS3 cost = $400
Wages = $15 / hr
Work day = 6 hr / day
Bus fare = 2 x $2 / trip
Lunch = $6 / day
Owes Dad = $500

Next, it is important to IDENTIFY WHAT YOU WANT TO FIND OUT. In this case, we want to find out how many days it will take Joe to save enough to buy his Playstation. Specifically, we want to know how long it will take him to save $400.

The next pointer is to try to CLASSIFY WHAT YOU ARE GIVEN INTO GROUPS. In this question, you can classify things into "Things Related to Making Money" and "Expenses."

Money:
Wages = $15 / hr
Work day = 6 hr / day
Expenses:
Bus fare = 2 x $2 / trip
Lunch = $6 / day
Owes Dad = $500
From here, it might start becoming more apparent what! need to be done. In this case, you would find out how much money Joe makes in a day, and also how much he spends in a day.
Earnings = $15 / hr x 6 hr / day = $90 / day
Expenses = 2 x ($2 / trip) + $6 / day = $10 / day
Overall = 90 - 10 = $80
So, Joe saves $80 per day.
Now, he owes his Dad $500, and wants to buy his Playstation for $400... totalling $900. At $80/day, Joes can pay off his Dad and then be able to afford his Playstation in:
$900 / ($80/day) = 11.25 days.... = 12 days (since he can't stop working at 11.25 days). :)
Of course, now that you have an answer, make sure that you answer the question completely.... there's still an! other part!
If Joe wants to make his purchase, after paying back his dad, in 5 days.....
KNOW:
total cost = $900
time = 5 days
NEED:
savings per day = ?
When you separate what you know from what you don't know, it is very helpful in seeing how to get to the answer. In this case, Joe needs $900 / 5 days = $180. Compared to what he makes right now, he needs to save another $100 per day!
So to summarize, they keys to word problems are to:
IDENTIFY WHAT YOU KNOW (classifying into groups helps here),
DECIDE WHAT YOU NEED TO KNOW,
and MAKE SURE YOU READ AND ANSWER THE ENTIRE QUESTION.
Simple rules, but important none! theless. :)

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Have questions about math problems

Have questions about math problems, get help on this website. You will learn the toughest questions with the easiest way to solve it!!

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Solve x^2 - 10000x - 10000 = 0 without a calculator! A Precalculus Investigation

BACKGROUND/OVERVIEW
No, there's no mistake in the constant term in that equation. Imagine giving this to your Precalculus/Math Analysis/Adv Math students! Actually, 'solving' it with the TI-84 requires some effort using Solver since one needs to make an approximate guess or adjust the lower bound so that the positive root is obtained. Using the graph is no 'walk in the park' either! The TI-89 or Mathematica would have much less difficulty in displaying the exact radical form or a suitable decimal approximation but they may not be within reach. Perhaps an important issue here is that sometimes technology gives us unexpected or even inaccurate results. That's when students need some understanding of theory to recognize the limitations of the technology and adjust accordingly.

Here's the point of all this. The given quadratic is not factorable over the integers, however we can re! place it with a 'nicer' quadratic that is. The roots of the desired quadratic can be shown to be approximately the same as the 'nice' quadratic and we can show that the absolute error is less than two ten-thousanths (and a much much smaller relative or % error)! Does this 'numerical analysis' have any practical value? Why approximate roots when powerful technology can produce exact answers? Do professionals who need to apply mathematics to the solution of 'real' problems ever use such approximation techniques? Could it be that theory actually provides practical application!

THE INVESTIGATION
(1) Show that the roots of the x2-10000x-10001 = 0 are 10001 and -1 by factoring.

! (2) Show that the roots of x2-10000x-10000=0 can be approximated by 10001 and -1 with an error of less than 0.0002.
(a) By direct calculation: Using the quadratic formula and, yes, you may use the calculator!

(b) (Challenging) By comparing, in general,
(*) the roots of x
2-bx-(b+1)=0 and
(**) the roots of
x2-bx-b=0.
Here we are assuming that b > 0.

(i) First show by factoring that the roots of equation (*) are b+1 and -1.
(ii) Then use the quadratic f! ormula to express the roots of (**) in terms of b.
(iii) Compare the positive roots of these equations by subtracting them and (after algebraic manipulation and simplication), show that the absolute value of the difference is less than 2/(b+1).
Note: For b=10000, this error is therefore less than 0.0002.

(c) Explain intuitively why the roots of the original equation and the 'approximating' equation are virtually the 'same' for 'large' values of b. One possibility here is to consider how the graphs of the associated quadratic functions are related. What do they have in common? How are they different?
Note: Subtle point here for students. Even though the difference of the function values (i.e., y-values) is always 1, this is not true of the difference between their zeros! This may be the essence of the numerical analysis in this i! nvestigation.

EXTE! NSION/PR ACTICE
Ok, now "solve" x2 - (googol)x - googol = 0 without a calculator.

RELATED PROBLEM:
Without your calculator show that √(10001) - √(10000) is less than 0.005.
Does this provide us with an effective method of approximating the square root of some large numbers or is it limited and impractical?

For Calculus students: How does this compare to using linearization to approximate the square root?

For more advanced calculus students: Newton's Method? The Binomial Formula (using fractional exponents)? A Taylor Polynomial approximation? All equivalent?


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Using Algebra to Solve Word Problems

Today's lesson notes and the homework - due tomorrow, June 10th (Thursday).

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An Inexpensive ACT Calculator & a Potpourri of Links

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For the revision to our ACT book, Preparing for the ACT: Mathematics and Science Reasoning, we purchased a cheap scientific calculator from Casio, the fx-115 ES, and I was amazed at the power of this pocket calculator. The calculator sells for about $20 (or less depending o! n where you purchase it), and can:

  • work with percents, fractions, and mixed numbers

  • simplify radical expressions

  • express answers in terms of pi

  • do operations with complex numbers

  • find permutations and combinations

  • perform one-variable statistics (including frequency tables)

  • solve one variable equations

  • do operations with matrices

  • calculate integrals

  • do all the usual things that a scientific calculator can do (like radicals, powers, trig functions, etc.)

To top it off, the darn thing runs on solar power! The geek in me can’t help but say, “Wow!” It’s proof that technology can be cheap and powerful (I’m talking to you, Apple.) I called the ACT twice just to confirm that this calculat! or is allowed on the test.

! ACT Tip: If you are planning on taking the ACT test and can’t afford (or don’t want to bother with) a graphing calculator, you can’t go wrong with the fx-115 ES. You are not going to find a TI model with similar capabilities for the same price.

Note/Warning: The ACT is a timed test – you have an average of 1 minute per question. This means two things: (1) Buying a calculator that you are not accustomed to using right before the test is a big mistake. (2) While using an advanced calculator can be helpful in the classroom, you probably won’t have time to use the advanced features on the ACT test. The Casio fx-115 ES’s features that will be useful on the ACT are fractions, mixed numbers, simplifying radical expressions, expressing answers in terms of pi, and operations with complex numb! ers.

To conclude today’s post, here is the promised potpourri of links:


  • Engineers beat math PH.D’s in math contest. The contest: The Netflix Prize. Make Netflix’s movie recommendation system more accurate by 10%. The math: statistics.

  • We’re all probably going to speak Chinese one day. A group of computer science students from China created one of the most awesome pieces of software I’ve seen in a long time: PhotoSketch (see the video below). It takes a hand-drawn sketch tagged with the name of the object and turns it into a real-world photo. It works in one of those “Why didn’t I think of that?” ways – the software does a web search based on the tags ! and chooses pictures that match the sketch. The best matches a! re then combined together and the user chooses the best looking image. The results are pretty amazing – check out the video below.

  • The Making of a Mathlete. PBS is going to air a documentary about the International Math Olympiad. Need I say more? No, really, it actually looks pretty exciting.


PhotoSketch: Internet Image Montage from tao chen on Vimeo.


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