User:NorwegianBlue/refdesk/mathematics

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1 The area of a square on the surface of a sphere.
2 Cubic equations with three real, irrational roots
3 Algebra stumper
4 Integration in Maxima
5 A couple of calculus questions
6 Maths resources
7 Radicals
8 Solving cubic equations
9 Best calculus textbook
10 Bernoulli numbers
11 Learning maths from scratch

[edit] The area of a square on the surface of a sphere.

[edit] Cubic equations with three real, irrational roots

(Copied from the reference desk, maths section).

The equation 8x³ - 6x + 1 = 0 has the solutions sin 10º, sin 50º and -sin 70º, and is solved readily using trigonometric methods. However, attempting to solve the equation using Cardano's method (see Cubic equation) yields some rather nasty expressions, such as

$\frac{1}{2}\sqrt[3]{-\frac{1}{2}+\frac{1}{2} i\sqrt{3}} + \frac{1}{2}\sqrt[3]{-\frac{1}{2}-\frac{1}{2} i\sqrt{3}}$

Question 1: Is it possible to reduce this to an expression involving radicals and non-complex rational numbers only?
Question 2: Can anybody give an example of a cubic equation with three different real irrational roots, where the roots can be expressed using radicals and non-complex rational numbers only? --NorwegianBlue 16:58, 21 May 2006 (UTC)

I don't know about the first question, but my guess is that no. About the second, the equation

$x^3 - (\sqrt{2}+\sqrt{3}+\sqrt{5})x^2 + (\sqrt{6}+\sqrt{10}+\sqrt{15})x - \sqrt{30} = 0$

Satisfies your conditions. If you want the coefficients to be integers as well, I guess this is equivalent to the first question. -- Meni Rosenfeld (talk) 17:50, 21 May 2006 (UTC)

Yes, I did want the coefficients to be integers. And I suspect you are right that the answer to both questions is no. If so, is anybody aware of a proof? --NorwegianBlue 17:56, 21 May 2006 (UTC)

I think that the answer to Question 1 in general is "no", because of the result refered to here:

"One of the great algebraists of the 20th century, B.L. van der Waerden observes in his book Algebra I, that the Casus Irreducibilis is unavoidable. There will never be an algebraic improvement of the cubic formula, which avoids the usage of complex numbers." [1]

However, it may still be possible that for this particular equation, such a formula exist. Perhaps that reference is enough to get you started if you are really interested. You could also try asking User:Gene Ward Smith. -- Jitse Niesen (talk) 03:05, 22 May 2006 (UTC)

Thank you, Jitse, as far as I can see, the reference answers both questions. And I had no idea that it was this very problem that initiated the study of complex numbers! --NorwegianBlue 16:57, 22 May 2006 (UTC)

Solving the Casus Irreducibilis, the case of three real roots, of the cubic equation requires the use of complex numbers if you insist on doing it with radicals; this was a strong factor in the adoption of complex numbers by European mathematicians, which may be the first clear example of modern European mathematics getting somewhere which the rest of the world hadn't gotten to earlier and better. The article cubic equations goes into great deal on how you can solve the cubic equation algebraically, not just in terms of transcendental functions, without using complex numbers. In that case, the algebraic funcion C_1/3(x) can be used. While this can be computed in terms of trig functions, that does not make it a transcedental function, any more than the fact that P_1/3(x) = x^1/3 can be computed in terms of logs and exponentials makes it a transcendental function. In general, solving a solvable algebraic equation of degree not a power of two, all of whose roots are real, requires complex numbers if you use radicals but no complex numbers if you use Chebychev radicals C_1/n. The cubic equation article is long--would an article on the Casus Irreducibilis help, I wonder? Gene Ward Smith 19:38, 24 May 2006 (UTC)

Thank you Gene, for commenting my question. I do realize that sin 10º, sin 50º and -sin 70º are algebraic numbers, being solutions of the equation 8x³ - 6x + 1 = 0, but I did insist on expressing the solution with radicals. I'm grateful for the response that it cannot be done without using complex numbers. As to whether there should be a separate article on the Casus Irreducibilis, I think that would be overkill, but it wouldn't hurt if, at the end of the section on Cardano's method, you mentioned the fact that, when applied to an equation with three real, irrational roots, it will always give a solution which includes a sum of two conjugate complex numbers, such that the imaginary parts cancel out. --NorwegianBlue 21:47, 25 May 2006 (UTC)

Second thought, maybe for some special classes, it can be done by denesting Nested radicals? (See the reference section, through you may have to use an algebra system) --Lemontea 02:37, 24 May 2006 (UTC)

Well, the Cardano solution is expressed in nested radicals involving complex numbers. I did try to construct cubic equations using nested radicals of real numbers only, but always ended up with irrational coefficients. It is quite some time since this problem nearly drove me nuts (33 years, to be exact), and I didn't have access to a computer algebra system. --NorwegianBlue 14:25, 24 May 2006 (UTC)

Anyway, at least now I know that it is impossible for your example above - see Exact trigonometric constants, which said "No finite radical expressions involving real numbers for these triangle edge ratios are possible because of Casus Irreducibilis. 9×2X-sided 70°-20°-90° triangle - enneagon (9-sided) 80°-10°-90° triangle - octakaidecagon (18-sided)" PS:my fault, the reference I cited actually duels with nested square root more, and for general radicals, not all are denestable. PPS:If question 2 doesn't require there to be three real roots, then

x 3 + 6 x + 2 = 0

satisfy the other requirement. --Lemontea 14:50, 24 May 2006 (UTC)

Thanks. The point was that there be three real roots. The real solution to your example ( $\sqrt[3]{2}-\sqrt[3]{4}$ ) would have satisfied the conditions if only there were two more real, irrational roots that could be expressed in a similar manner. However, the example is only a combination of the equations

y 3 = 2

and

z 3 = 4

.
Subtract the equations, i.e.

y 3 - z 3 = - 2

,
factorize the l.h.s., substitute with

x = y - z

, observe that

y z = 2

and that

x 2 + 6 = (y 2 + y z + z 2)

, and you have your equation. --NorwegianBlue 19:10, 25 May 2006 (UTC)

[edit] Algebra stumper

Years ago, a calculus professor gave me this one. The problem is to either solve the following set of equations by giving x, y and z; or to prove no such solution exists (among the complex numbers)

$x + y + z = 1 \,\!$

$x^2 + y^2 + z^2 = 2 \,\!$

$x^3 + y^3 + z^3 = 3 \,\!$

—The preceding unsigned comment was added by Plf515 (talk • contribs) 03:50, 2006 November 24.

Suppose a is one of x, y, and z. Then

(a - x)(a - y)(a - z) = 0

, so

$a^3 - (x+y+z)a^2 + (xy+xz+yz)a - (xyz)a = 0\,\!$

We already know one of the coefficients, and we can find the other two:

$(x + y + z)^2 = 1\,\!$

$x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 1\,\!$

$2 + 2(xy + xz + yz) = 1\,\!$

$xy + xz + yz = -1/2\,\!$

$(x + y + z)(x^2 + y^2 + z^2) = 2\,\!$

$x^3 + y^3 + z^3 + x^2y + x^2z + xy^2 + xz^2 + y^2z + yz^2 = 2\,\!$

$(x + y + z)(xy + xz + yz) = -1/2\,\!$

$3xyz + x^2y + x^2z + xy^2 + xz^2 + y^2z + yz^2 = -1/2\,\!$

$x^3 + y^3 + z^3 - 3xyz = 5/2\,\!$

$3 - 3xyz = 5/2\,\!$

$xyz = 1/6\,\!$

So, x, y, and z are the roots of the cubic polynomial

a 3 - a 2 - (1 / 2) a - 1 / 6

, which are about 1.43, 0.215 + 0.265i, and 0.215 - 0.265i. —Keenan Pepper 05:30, 24 November 2006 (UTC)

We might explicitly point out that all three equations lead to symmetric polynomials in x, y, and z. This has a double relevance. First, as shown, it assists in finding one solution. Second, it tells us that any permutation of one solution is also a solution.

A modern tool, that may not have been available when the problem was first posed, is the computation of a Gröbner basis for the three polynomials using Buchberger's algorithm. One such basis here is

$\begin{align} &6 z^3-6 z^2-3 z-1 ,\\ &2 y^2+2 z y-2 y+2 z^2-2 z-1 ,\\ &x+y+z-1 . \end{align}$

The cubic in z should look familiar. --KSmrq^T 06:52, 24 November 2006 (UTC)

Expressed algebraically, the real root of the 3rd degree polynomial equals:

$\frac{w}{w^2-w-1}, \mbox{ where } w=\sqrt[3]{5 + \sqrt{26}}.$

This was found by first using the substitution

z : = 1 / ζ

and subjecting the result to Cardano's third degree technique of interrogation. --Lambiam ^Talk 09:14, 24 November 2006 (UTC)

Thanks~ Plf515 10:56, 24 November 2006 (UTC)plf515

[edit] Integration in Maxima

I'm experimenting with open-source computer algebra systems, and am trying out Maxima. I've encountered a problem when doing symbolic integration. The value returned from integrate, although displayed as an algebraic expression, behaves differently from the same expression entered by hand. Here is an example of what is going on:

f(x):=45^(-x^2);

$\, f\left( x\right) :={45}^{-{x}^{2}}$

F(x):=integrate(f(x),x);

$\, F(x):=\operatorname{integrate}(f(x),x)$

F(x);

$\frac{\sqrt{\pi }\,\operatorname{erf}\left( \sqrt{\log\left( 5\right) +2\,\log\left( 3\right) }\,x\right) }{2\,\sqrt{\log\left( 5\right) +2\,\log\left( 3\right) }}$

// Seems reasonable...

// Now I'd like to evaluate F(x) for a specific value of x

F(5);

Attempt to integrate wrt a number: 5#0: F(x=5)

-- an error. To debug this try debugmode(true);

// Hmm, why doesn't it simply substitute x=5?

// Ok, lets try defining G(x) = F(x) from scratch, and see what happens...

G(x):=sqrt(%pi)*erf(sqrt(log(5)+2*log(3))*x)/(2*sqrt(log(5)+2*log(3)));

$\frac{\sqrt{\pi }\,\operatorname{erf}\left( \sqrt{\log\left( 5\right) +2\,\log\left( 3\right) }\,x\right) }{2\,\sqrt{\log\left( 5\right) +2\,\log\left( 3\right) }}$

G(5);

$\frac{\sqrt{\pi }\,\operatorname{erf}\left( 5\,\sqrt{\log\left( 5\right) +2\,\log\left( 3\right) }\right) }{2\,\sqrt{\log\left( 5\right) +2\,\log\left( 3\right) }}$

float(G(5));

$\, 0.45422679973746$

F(t)/G(t);

$\, 1$

F(5)/G(5);

Attempt to integrate wrt a number: 5#0: F(x=5)

-- an error. To debug this try debugmode(true);

My question is this: How do I make F(x) behave as though it had been entered by hand. --NorwegianBlue^talk 19:19, 14 January 2007 (UTC)

When you wrote F(x):=integrate(f(x),x) you intended the right side to be evaluated before the function was defined, but that is not what you actually said. So what is happening is that F(5) means integrate(f(5),5) and the error is exactly what you would expect. Try the quote-quote input:

F(x):=''(integrate(f(x),x));

This forces the integral to be evaluated before the function is defined. --KSmrq^T 23:36, 14 January 2007 (UTC)

Yes, that fixed it. Thanks! --NorwegianBlue^talk 20:59, 15 January 2007 (UTC)

[edit] A couple of calculus questions

Hi, all, hope you can help me with a couple of things.

Firstly, how do I go about evaluating this?

$\int_{-\infty}^{\infty} e^{-x^2}\,dx$

I know what the answer is from looking at a page of integrals, but I'm more interested in the process.

Also, how do you differentiate something in the form of, say, $3 x$ ? I tried using the chain rule to give $\frac{d}{dx} 3^x = \frac{d(3^x)}{d3} \times \frac{d3}{dx} = x(3^{x-1}) \times 0$ , but clearly the answer is not 0 for all x, as this implies. Neither of these are homework, but I don't mind working through them myself if you give me some hints. Thanks, 80.169.64.22 19:45, 29 June 2007 (UTC)

The first example has no antiderivative, except one artifically constructed just for the purpose, the error function. But the second example is easy once we rewrite it as exp(x log(3)), then apply the chain rule. Very creative, your derivative with respect to 3; but it has the fatal flaw that 3 is not a variable! --KSmrq^T 20:18, 29 June 2007 (UTC)

Check out the limits on the first question; we want a specific definite integral. The standard technique for this one is to square the integral and then transform from cartesian co-ordinates to plane polars. Algebraist 20:32, 29 June 2007 (UTC)

I suspect the OP may not be familiar with the tools involved; I will therefore sketch a calculation based on Algebraist's suggestion, and invite requests for anything that needs to be explained further:

$\left(\int_{-\infty}^{\infty} e^{-x^2}\ dx\right)^2 = \int_{-\infty}^{\infty} e^{-x^2}\ dx \int_{-\infty}^{\infty} e^{-y^2}\ dy = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}\ dx\ dy =$

Changing to polar coordinates,

x = r cosθ

and

y = r sinθ

, which has Jacobian

r

$=\int_0^{2\pi}\int_0^{\infty}e^{-r^2}r\ dr\ d\theta = \int_{0}^{2\pi}\frac12\ d\theta = \pi$

$\int_{-\infty}^{\infty} e^{-x^2}\ dx = \sqrt{\pi}$

-- Meni Rosenfeld (talk) 21:03, 29 June 2007 (UTC)

KSmrq: you can differentiate 3, or any other constant (you should know this), it can be considered a function f(x) = 3. The problem is that we are talking about exponentials, rather than the differentiability of a constant function.

Differentiating 3 is something else than differentiating with respect to 3.--P.wormer 07:32, 30 June 2007 (UTC)

On the second (differentiation) problem, you might consider implicit differentiation. Consider

ln y = x ln3

. Thus

$\frac{y'}{y} = \ln 3$

Substitute y and solve for y-prime. All the best (and great questions!), --TeaDrinker 21:13, 29 June 2007 (UTC)

$\frac{d}{dx}\ a^x = a^x ln(a)$

for constant and positive a.

For the first question, check Meni Rosenfeld's rather elegant method for evaluating it.

It's not mine. I was in the middle of trying to asymptotically evaluate an integral in the complex plane when Algebraist gave a gentle reminder of the "right" way to do it. -- Meni Rosenfeld (talk) 10:38, 30 June 2007 (UTC)

[edit] Maths resources

I am delighted by your question. It takes uncommon wisdom and humility to acknowledge a need to learn more and to seek assistance. In this case, your need is a shared one, and there are quality sources on the web. One place to start is The Math Forum at Drexel, where you will find a page of resources. For online mathematics texts, the AMS has links, though many are on more advanced topics. Still, you can find jewels like Calculus, by Gilbert Strang, a Professor of Mathematics at MIT. You might also browse topics at The Mathematical Atlas. If I have a chance, I'll see if I can dig up a good elementary discussion of the binomial theorem. Meanwhile, these sources should keep you in good reading. --KSmrq^T 22:07, 29 December 2006 (UTC)

[edit] Radicals

Just a quick question: is it correct to have a radical inside of another radical, or should I attempt to work the second out of the first radical? I guess "correct" isn't the best word. I guess I'm asking is it more formal format to have it out? 75.18.9.71 03:48, 30 January 2007 (UTC)

It depends on the context, but I would certainly say that there is nothing wrong with having square roots inside other square roots. Do you mean something like $\sqrt{2 + \sqrt{2}}$ ? –King Bee ^{(T • C)} 04:05, 30 January 2007 (UTC)

Yes, that's exactly what I mean. Thanks :-) -75.18.9.71 04:11, 30 January 2007 (UTC)

That's perfectly ok. In fact, there are some numbers defined by an infinite amount of nested radicals. Check out that article. --Ķĩřβȳ ♥Ťįɱé Ø 07:13, 30 January 2007 (UTC)

If there is an unnested version, it will be considered simpler, so by the unwritten rule that (in the absence of other rules to the contrary) the simpler expression is preferred, it is the better form. For example, rather than

$\sqrt[3]{-2+\sqrt 5}\,,$

use

$\frac{-1+\sqrt 5}{2}\,.$

--Lambiam ^Talk 14:13, 30 January 2007 (UTC)

How do you go about to simplify such an expression? I tried the expression in Maxima (which I've only just begun to get acquainted with). It confirmed, of course, that both expressions evaluate to 0.6180339887499. But I was unable to make it simplify the expression (tried procedures ratsimp, radcan, trigsimp and trigreduce, i.e. all the buttons that had the words "simplify" or "reduce" on them). It was no more successful at simplifying

$\frac{\sqrt{5}-1}{2\,{\left( \sqrt{5}-2\right) }^{\frac{1}{3}}}$ .

How does one approach the problem "by hand"? --NorwegianBlue^talk 22:08, 30 January 2007 (UTC)

One possible approach, which can have varying degrees of success, it to conjecture that $\sqrt[3]{-2+\sqrt 5}$ will be of the form $a + b \sqrt{5}$ with a and b rational. This gives $(a+b\sqrt{5})^3=-2+\sqrt{5}$ , or $a^3+3a^2b\sqrt{5}+3ab^25+5b^3\sqrt{5}=-2+\sqrt{5}$ which gives

a 3 + 15 a b 2 = - 2

and

3 a 2 b + 5 b 3 = 1

. Solving this (the substitution $t=\tfrac{a}{b}$ helps) should give the result. -- Meni Rosenfeld (talk) 22:26, 30 January 2007 (UTC)

Thanks for the quick reply! --NorwegianBlue^talk 22:45, 30 January 2007 (UTC)

From Exact_trigonometric_constants

Main article: Nested radicals

In general nested radicals cannot be reduced.

But if for $\sqrt{a + b\sqrt c}$ ,

$R = \sqrt{a^2 - b^2 c}$ is rational,

and both $d = \pm \sqrt{ \frac{ a \pm R }{2}}$

and $e = \pm \sqrt{ \frac{ a \pm R }{2c}}$ are rational,

with the appropriate choice of the four $\pm$ signs,

then $\sqrt{a + b\sqrt c} = d + e\sqrt c$

Example:

$4\sin 18^\circ = \sqrt{6 - 2 \sqrt 5} = \sqrt 5 - 1$

[edit] Solving cubic equations

In school, we learnt how to solve cubic equations of the form ax³ + bx² + cx + d.

To factorise it into (x+r)(x+s)(x+t), we have to find a value for r by trial and error (with the factor theorem), then solve a quadratic equation to get the other two roots, s and t.

In the exams, the value of r will usually be 3 or less, so finding r by trial and error will not take a long time.

However, if r is large, finding r by trial and error will take a long time. Are there any faster, more systematic ways of finding r, besides trial and error? —Preceding unsigned comment added by 218.186.9.3 (talk • contribs) 09:18, 2007 February 13

(Please sign your posts with four tildes, "~~~~".) The roots of a cubic polynomial can be expressed in closed form, so in theory no searching is required. However, the technique is more difficult than using the quadratic formula. When the coefficients, a, b, c, and d, are real numbers (with a nonzero), the cubic is guaranteed to have a real root; but it may have only one. For example,

$x^3-3 x^2+2 x-6 \,\!$

has only one real root.

But we can say much more. The coefficients are normalized so that the x³ term has coefficient 1. If we list them all, (1,−3,2,−6), the −6 has the largest absolute value. The Cauchy bound says that all real roots must lie between −m and +m, where m is one plus the maximum absolute value. In this example, the roots will be between −7 and 7.

We can do better. The positive coefficients before the −6 are 1 and 2, summing to 3; and before the −3 we have just 1. For an upper bound we can choose the maximum of ⁶⁄₃ and ³⁄₁, which will be 3, and again add 1. That is, the maximum real root is 4. Negating every other term (so we are evaluating the polynomial with −x), we can similarly find a lower bound, which here will be −1.

There is also a method known as Descartes' rule of signs that may help us restrict the number of positive or negative real roots. In this example it is no help with the positive roots (we will have three or one), but it tells us we cannot have a negative root.

This example polynomial has integer coefficients, and may have integer roots. If r is such a root, then it must divide the constant term, so the only possibilities here are 1, 2, 3. We have used the known bounds; and we know that 0 cannot be a root unless the constant term is 0.

We also have ways to bracket roots. If the polynomial evaluates to positive for x = a and negative for x = b, then there must be at least one root between a and b. Using a Sturm sequence, we can learn even more.

This does not exhaust our inventory of tools, but perhaps this is enough for now. Computer programs for solving cubics typically use closed form solutions. Above degree four, however, a famous theorem states that we have no closed form solutions. --KSmrq^T 11:48, 13 February 2007 (UTC)

You might want to read cubic equation. – b_jonas 14:17, 13 February 2007 (UTC)

[edit] Best calculus textbook

I've heard very often of Tom Apostle's Calculus I and Calculus II written in the 40s and 50s, which are very proof-heavy, well, entirely proof-based. What are your opinions on the best calculus textbooks? [Mαc Δαvιs] X (How's my driving?) ❖ 08:28, 19 February 2007 (UTC)

I can also recomend Leithold's book. Specially for beginners it is a good choice. Mr.K. (talk) 13:48, 19 February 2007 (UTC)

More information about your expectations would help. Do you want a graphic-rich easy introduction? Well-grounded proofs? Higher mathematics insight? Emphasis on engineering/physics applications? A modern revision? One of the non-standard analysis presentations?

Most texts I've seen do not impress me. Richard Courant and Fritz John do. Try a variety of books online, including those listed by Stef and at AMS. --KSmrq^T 04:27, 20 February 2007 (UTC)

[edit] Bernoulli numbers

In the first paragraph of the Bernoulli number page it is written that there is no known elementary description for Bernoulli numbers .What is the meaning of "elementary description"? I thought that it meant that there is no mathematical rule to find the nth Bernoulli number that is not a sum of an infinite series because there was none on the page .but then I found somewhere on the internet a simple definition that is not a sum of an infinite series and also it does not depend on previous values of $B n$ .So, my question is :what does "elementary description" mean ? --George 04:04, 4 November 2007 (UTC)

I suppose the meaning intended here is that there is no definition in the form of an expression using only elementary functions, in other words, a closed-form solution of a defining equation for the Bernoulli numbers. --Lambiam 06:38, 4 November 2007 (UTC)

George - did you say you have somewhere seen a finite and non-recursive expression for

B n

? Do you remember where that was ? The simplest "rule" for finding the value of

B n

that I can think of is

B 0 = 1

$B_n=\left(\frac{-1}{n+1}\right)\sum_{j=0}^{n-1}{n+1\choose{j}}B_j\mbox{ ; }n>0$

but that is recursive, so it probably doesn't qualify as an "elementary description". Gandalf61 09:57, 4 November 2007 (UTC)

Might it have been a generating function you saw? They're easy to mistake for formulas. Black Carrot 06:42, 5 November 2007 (UTC)

The formula is $B_n=\sum_{k=0}^n\sum_{j=0}^k j^n (-1)^j \frac{\binom{j}{k}}{k+1}$

I'm not sure if this can be considered an elementary description ,if it's not one please explain the reason .Thanks --George 10:20, 5 November 2007 (UTC)

Since j ranges from 0 to k, should $\binom{j}{k}$ in the numerator actually be $\binom{k}{j}$ instead ? Gandalf61 10:55, 5 November 2007 (UTC)

Well, the Gamma function is non-elementary, so in some sense the binomial coefficient function is not elementary, either. This might have some relevance to the claim that no elementary formula exists. -- Meni Rosenfeld (talk) 13:11, 5 November 2007 (UTC)

yes, I'm sorry, it is $B_n=\sum_{k=0}^n\sum_{j=0}^k j^n (-1)^j \frac{\binom{k}{j}}{k+1}$ , I just mistyped the formula. --George 14:59, 5 November 2007 (UTC)

[edit] Learning maths from scratch

I always want to learn maths all over again, but I don’t know how to start with. Now, I’m undergraduate, I “parted” with everything science since high school, and I did almost nothing more than very simple statistics (like poisson distribution, but I don’t really remember it) and calculus (I don’t remember how to differentiate). So, if I want to know “advanced maths”, like what is being taught at university, what sorts of books (maybe in English) may I use (as an adult learner)?

My “junior high school” maths is about factorizing, polynomials, simple geometry (calculating angles) and so on; which areas do high school maths (I mean, for learning science subjects) and university maths cover? I’m looking for books for learning more advanced things (compared to my present level) and I don’t just concentrate on calculus or statistics (or something like that). Any suggestions? --Fitzwilliam (talk) 14:10, 29 February 2008 (UTC)

From my experience, university maths tends to recap all you learnt at school in the first year - try just going along to the first year maths lectures. -mattbuck (Talk) 14:39, 29 February 2008 (UTC)

Especially maths lectures targeted at science students, rather than maths students. My Maths department (at a UK uni, I know it's a little different elsewhere) offers a module called "Mathematics for Scientists and Engineers" which covers a wide range of mathematical topics at a fairly basic level (by Uni standards) without assuming much (if any) prior knowledge. If there are similar modules are your uni, those would be the ones to go to. --Tango (talk) 16:11, 29 February 2008 (UTC)

I'd go to a Maths Department. They love to find new students. Imagine Reason (talk) 23:15, 29 February 2008 (UTC)

What is your motivation? Do you want to learn advanced maths primarily because you feel it will be useful, or is it more for fun? If you have forgotten how to differentiate, maybe you should take a course in calculus (actually analysis) anyway, using a text that does not just give the rules and formulas but also precise definitions of concepts like the real numbers, limit, and continuity, and rigorous proofs. Other topics you may study that don't immediately require much prior knowledge are linear algebra and projective geometry. Also consider elementary number theory and combinatorics. A nice book is Concrete Mathematics; although aiming at hopeful computer scientists, it is also quite valuable for mathematicians. I'd also advice you not to go immediately very deep into one field of maths, but to first build up a fairly broad basic knowledge of various fields. Much of the more advanced stuff in maths requires some knowledge of other fields. --Lambiam 00:26, 1 March 2008 (UTC)