stevem

It would be easier to say [math]\frac{x^2 + 4x}{2y^2 + y} = 6 \implies x^2+4x=6\left(2y^2+y \right)[/math] Hence [math]2x+4=6\left(4y\frac{dy}{dx}+\frac{dy}{dx}\right)=6\frac{dy}{dx}(4y+1)[/math] So [math]\frac{dy}{dx}=\frac{2x+4}{6(4y+1)}[/math]

And if you want a free equation editor to generate the LaTeX code for you, have a look at TeXAide

I agree with Dave that the most sensible thing to do is install LaTeX on your system. But if you really don't want to do that you'll need a LaTeX emulator like mimeTeX. Assuming that you are using Windows, then download mimetex.exe from there. You can then call mimetex from the DOS command line, but the simplest thing to do is write a batch file: @echo off mimetex -f mimetemp.tex -e mimetemp.gif Save your latex code in mimetemp.tex, run the batch file and the image will appear as mimetemp.gif.

! wasn't worried about that because everyone happily agrees that 0!=1 because, for example, it makes a nice formula for [math]e^x[/math]. The problem that really concerns me is that you have used [math]x^0[/math] for any x. But you need to define [math]x^0[/math] for x=0 before you can use it in the formula for [math]e^x[/math] and that is where you get the circular argument. What you can do is say: I like the formula [math] e^x = \frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+... [/math] so I will define [math]0^0=1[/math] and [math]0!=1[/math] and then I can use this version rather than the one that starts 1 + x. The only snag is that not everyone would agree with this definition.

Unfortunately, your argument is circular because of the way you've defined [math]e^x[/math]. If you define it as [math] e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+... [/math] then your argument no longer works.

How do you get that? I repeat what I said - there is no proof that [math]0^0=1[/math]. If you wish to use that definition then feel free to do so; it's just that not everyone likes that definition, whereas 0!=1 is, as Matt says, used by everyone, though, again you are free to leave it undefined.

There is no proof. You can define [math]0^0[/math] to be whatever proves convenient. I prefer to say it is indeterminate because [math]\lim_{x \to 0^+}x^0=1[/math] but [math]\lim_{x \to 0^+}0^x=0[/math] but others disagree. See http://mathforum.org/dr.math/faq/faq.0.to.0.power.html for a discussion of this topic.

If you want arrows you can have them! [math]\ce{-> ->[\alpha] <- <=> <=>> <-> v ^}[/math] and even [math]\ce{Zn^2+ <=>[\ce{+ 2OH-}][\ce{+ 2H+}] {\underset{\text{amphoteres Hydroxid}}{\ce{Zn(OH)2 v}}} <=>[\ce{+ 2OH-}][\ce{+ 2H+}]{\underset{\text{Hydroxozikat}}{\cf{[Zn(OH)4]^2-}}}}[/math] There's more in the mhchem documentation (mhchem provides the Chemistry [math]\LaTeX[/math] here)

As Matt said and you will find that the entries will be aligned - in this case in the centre because of the c in \begin{array}{cccc} Unfortunately there's a bug in the [math]\LaTeX[/math] implementation on the forum so it shows wrongly in [math]A \times B = \left| \begin{array}{ccc} \hat i & \hat j & \hat k \\ A_x & A_y & A_z \\ B_x & B_y & B_z \\ \end{array} \right| [/math] but in normal [math]\LaTeX[/math] use it will do what you wanted.

It's not possible to solve this algebraically, if by that you mean using the usual 'elementary' functions. You can solve it using a computer package because it reduces to a non-elementary function called the LambertW function. This function is useful precisely because it allows one to solve these sorts of equations. It is a function that is one step beyond logs in the same way as logs are one step beyond polynomials. If we start with [math]3x+lnx=3[/math] then [math]e^{3-3x}=x[/math] which can be rewritten as [math]3e^3=3xe^{3x}[/math] or [math]3e^3=we^w[/math] where [math]w=3x[/math] w is the value of the LambertW function, which, as I've said isn't on a calculator, but is in most computer maths packages and you find [math]x=\frac{1}{3}\text{LambertW}(3e^3)[/math] which gives 1. Don't understand or (hopefully) want to learn more? Then there's an excellent beginners guide to LambertW (including why it's called that) in the American Scientist at Why W?

Not sure which you can't prove - that [math]\lambda[/math] is an injection or that it is a homomorphism or both so here is how to start for both these steps: 1. Let [math]a,b \in G[/math] and [math]\lambda_a=\lambda_b[/math]. Apply these 2 functions [math]\lambda_a,\lambda_b[/math] to the identity of G and what can you deduce? 2. The binary operation in Sym(G) is composition of functions so for any g in G [math]\left(\lambda_a \circ \lambda_b\right)(g)=\lambda_a\left(\lambda_b(g) \right)=\lambda_a\left(bg\right) = \dots[/math] Then use the fact that 2 functions [math]\alpha,\beta[/math] in Sym(G) are equal if and only if [math]\alpha(g)=\beta(g)[/math] for all [math]g \in G[/math] Finally, apply what you have shown to give results about [math]\lambda[/math] using [math]\lambda(g)=\lambda_g[/math] Hope that helps.

If n=2 then [math]\left(T^{-1}AT\right)^2=T^{-1}ATT^{-1}AT=T^{-1}AIAT=T^{-1}A^{2}T[/math] This should help you to construct a proof by induction.

Just realised that you can get an anthology of Manifold in: Seven Years of Manifold: 1968-80 Ian Stewart (Editor), John Jaworski (Editor) Publisher: Shiva Pub, Nantwich ISBN: 0906812070

He did indeed. One must be more careful in stating results; I should have said that Theorem n! is even for all integers n > 1 Oh and, this is for Dave's benefit, the proof was thought up in the old Maths Institute at Warwick before Ian Stewart showed us how to do maths humour properly in a magazine called Manifold. If there are any still lying around somewhere at Warwick they are worth reading. My favourite cartoon in that magazine had a picture like this

Er, which theorem are you referring to?

The theorem that Pulkit is referring to is often known as the Odd Order Theorem and is Feit, W. and Thompson, J. G. "Solvability of Groups of Odd Order." Pacific J. Math. 13, 775-1029, 1963 As its title says, the paper proves that all groups of odd order are soluble (the British version of solvable) and as an immediate consequence you get the fact that a finite simple non-cyclic group has even order. Humorous proofs in maths are hard to come by but this impressive theorem is the source of one that, as students, we thought of many years ago. Theorem n! is even for all positive integers n Proof Suppose, for a contradiction, n! is odd. It follows that the symmetric group [math]S_n[/math], which has order n!, is soluble by the Odd Order Theorem. But for n>4 this is well-known to be false, so n! must be even. This leaves the case where n<5 which can be checked individually

I wrote it out in detail but you should be able to do most of it in your head

© can be done by using the conjugate and without using L'Hopital [math]\frac{\sqrt{x}-2}{\sqrt{2x+1}-3\sqrt{x-3}}[/math] [math]=\frac{(\sqrt{x}-2)(\sqrt{2x+1}+3\sqrt{x-3})}{(\sqrt{2x+1}-3\sqrt{x-3})(\sqrt{2x+1}+3\sqrt{x-3})}[/math] [math]=\frac{(\sqrt{x}-2)(\sqrt{2x+1}+3\sqrt{x-3})}{2x+1-9(x-3)}[/math] [math]=\frac{\sqrt{x}-2}{x-4}.\frac{\sqrt{2x+1}+3\sqrt{x-3}}{-7}=\frac{1}{\sqrt{x}+2}.\frac{\sqrt{2x+1}+3\sqrt{x-3}}{-7}[/math] [math]\to \frac{1}{4}.\frac{6}{-7}=-\frac{3}{14} \text{ as } x \to 4[/math]

One example is to classify quadrics (3 or more dimensions) or conics (2 dimensions) For example, suppose you want to know which type of quadric [math]5x^2+3y^2+3z^2-2xy+2yz-2xz-10x+6y-2x-9=0[/math] is. In geometrical terms you rotate it and translate it so it has one of the standard forms listed in quadrics In algebraic terms, you put the equation into the form [math]\mathbf{x}^T\mathbf{Ax}+\mathbf{J}^T+H=0[/math] where [math]\mathbf{A}[/math] is a [math]3 \times 3[/math] matrix, [math]\mathbf{J}[/math] and [math]\mathbf{x}[/math] are column vectors and [math]H[/math] is a real number. You then diagonalize [math]\mathbf{A}[/math] to get [math]\mathbf{P}^T\mathbf{A}\mathbf{P}=\mathbf{D}[/math] where [math]\mathbf{P}[/math] is an orthogonal matrix. Then you transform the equation using [math]\mathbf{P}[/math] and that effectively gives you new perpendicular axes, which is in effect a rotation. After that completing the square gives you a translation and you end up with [math]\frac{x^2}{3}+\frac{y^2}{6}+\frac{z^2}{9} =1[/math] which is an ellipsoid

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Mix maths and text to give this: Let [math]\{v_1,v_2,...,v_k\}[/math]be the basis set for the vector space V which are linearly independent, then it is possible to construct another basis set [math]\{e_1,e_2,...,e_k\}[/math]from the original basis set, such that the basis vectors in the new basis set are all orthogonal to each other, i.e. [math]<e_i,e_j>=0[/math] for [math]i \neq j[/math] where [math]<,>[/math] is the inner product defined on the Vector Space V That looks fine to me. Don't forget you need to put \ before { and } to tell [math]\LaTeX[/math] that you really mean those. If dave could put a math button to insert [ math] and [ /math] tags then it will be even easier to type.