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Fingers crossed!

Monox D. I-Fly wrote:

And here I thought that A' in the context of sets meant the complement of A.

It can indeed -- the notation can be quite varied!

A' is read as 'A dash' in British English and 'A prime' in American English, although at my university 'A prime' was far more common, probably due to American influence.

In *Littlewood's Miscellany*, Littlewood jokes about this notation used in the context of sets (in point-set topology, A' is the set of all limit points of A, so that A' is called the *derived set* of A).

John E. Littlewood wrote:

I have had occasion to read aloud the phrase "where E' is any dashed (i.e. derived) set". It is necessary to place the stress with care.

Hi Amartyanil,

Yes: try looking at that equation modulo 2 and modulo 3. From there you can deduce that *x* is a multiple of 2 and that *y* is a multiple of 3, which allows you to reduce that equation into something much simpler!

Welcome to the forum, Kamov!

Hi Math 1122,

Welcome! Why not register an account with us?

You can start by using the identity

You'll also need (some, if not all of) these facts:

Hi Bob,

I made a video explanation here, if Zeeshan 01 would like to have a look.

Set

then use the trig identity

Hi segfault,

Welcome to the forum!

Suppose that there are some values x and y for which ∛(7 + 5√2) = x + y√2. What happens if you cube both sides of that equation?

Happy New Year!

Merry Christmas!

What is ?How can you use this to determine ?

and is sufficient to generate all three.

Hi Emma22,

Welcome to the forum. Have you considered registering an account with us?

Emma22 wrote:

What is here? (You have later called this .)
The general solution obtained is u(x,t) =F(x^2-t^2*exp(

u)) and the initial condition is u(x,0)=2ln(x)

Hi Βεν,

Nice contribution! Yes, the repeated iteration of the cosine function converges: actually, it converges to theYou can prove that a solution to the above equation exists via Brouwer's fixed point theorem (and probably the contraction mapping theorem too).

Hi Benjamin,

Welcome to the forum!

Βεν Γ. Κυθισ wrote:

Do you mean rather than ?
it should be (a^m)n^a.

Hi Βεν,

Welcome to the forum! Thanks for your contribution. That looks like a nice list. Some comments:

a^n=a multiplied by itself n times

If n is even then (-a^n)=a^n

a^n÷a^m=a^(m-n) (Makes sense right?)

If n>0 then (a^m)na=a^(m+n)

What did you mean here?

Hi zahlenspieler,

I have fixed your LaTeX.

Hi math9maniac,

Can I just check that the equation you wrote down is correct?

What is ?

Hi vipin_sharma,

Welcome to the forum. Here's a hint:

Hi niravashah,

Thanks for posting this. Have you considered creating an account with us?

For reference, the sequence is:

4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 14, 6, 4, 4, 6, 6, 4, 4, 6, 22, 4, 4, 9, 6,

and we're asked to find the next two terms. Perhaps this will give you a hint:

**4**, **4**, 341, 6, **4**, **4**, 6, 6, **4**, **4**, 6, 10, **4**, **4**, 14, 6, **4**, **4**, 6, 6, **4**, **4**, 6, 22, **4**, **4**, 9, 6,

Good to hear -- welcome to the forum!