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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!

Draw a Venn diagram with 3 overlapping circles, labelling the four bits that overlap as A, B, C, D.

The information given in the question then allows you to form some equations for A, B, C and D. Can you see how?

!nval!d_us3rnam3 wrote:

That depends on what you consider to be a 'rigorous demonstration'. Clearly is a factor, since both the LHS and RHS vanish. Dividing your equation by (taking ) gives youSince vanishes whenever , then the numerator of the term on the RHS also vanishes for these values of . That's three solutions, and you can 'rigorously demonstrate' that there aren't any more solutions over via the method in post #4.
Ok, but I wasn't looking for a proof with the answers in the beginning. I've gotten pretty far, but I need to find all the answers to

EDIT: With a rigorous demonstration that this is the answer.

Hi Stuti55,

Welcome to the forum.