Recent content by Popey

Actually, this was a cheat :biggrin:

No that hard at all :approve:

Yes, of course! it's x^2-x-1

x^4+2x^3-x^2-6x-3 = (x^2-x+1)(x^2+3x+3) The first has the solutions above The second has no real solutions

Hi! Set P=x_{1}x_{2}\ldots x_{n} then g\equiv \sqrt[n]{x_{1}x_{2}\ldots x_{n}} =\sqrt[n]{P} For the real numbers w_{i}=x_{i}/g we have w_{1}w_{2}\ldots w_{n}=(x_{1}/g)(x_{2}/g)\ldots (x_{n}/g)=P/g^n=1 So, (1) holds for wi. Then w_{1}+w_{2}+\ldots+w_{n}\geq n (this is an...

Well, I think that the formula a^0=1 appears when you try to divide a^m by itself: (a^m)/(a^m) = a^(m-m) = a^0 Since the first part of this equation equals 1, we have a^0=1 But if a=0 we can't do the division (0^m)/(0^m)

undefined!

Do you mean, how did we find these formulas? If YES, then, they arise from integration

I'll recall another section what is a power of a^0 ? By definition, a^n=a*a*...a (n factors) we can't find out what a^0 means but a^0=1 works (if, for example, we think of (a^7)/(a^7)=1) So we DEFINE a^0=1 The same as 0! (factorial) - we define it although there is not a...

the limit of f(x) is 1 when (x->0+) or (x->0-) By the way f(x) can't have a value when x=0 But if we define - as you said - that f(0)=1, then we made f(x) continuous (left limit = right limit = f(0))

If you look at the graph, when x tends to -oo then e^x tends to zero set x=1/u and you have (1/u)->-oo => e^(1/u)->0

Shouldn't -ax be an integer anyway? Ok, the limit of (1+1/n)^n is e when n ->oo and n is an integer but what happens when we get the value (1+1/x)^x where x is a very large real but not an integer ?? Shouln't we prove this cases ?

I'm not quite sure about it, but I think that you should now solve these systems x == 2 (mod5) x == 3 (mod7) which gives x == 17 (mod35) and y == 1 (mod5) y == 4 (mod7) which gives y == 11 (mod35)

Yeap! ...

Hi! Could you explain pls this game (or anyway give a link here) cause I don't know it