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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.

Hi Zeeshan,

I've moved your thread to the Help Me ! forum.

Have you checked out **Maths Is Fun's page on long division**?

Hi Anduin,

Welcome to the forum. What features does the graph y = sin(x) have? How do they compare to the graphs y = 2sin(x) and y = 3sin(x)?

Hello,

I noticed that you've posted several times here as a guest. Why not register an account with us?

What have you tried? What sorts of restrictions do you think you can place on x and y?

Grantingriver wrote:

Hi Zetafunc, if there are duplicate roots of a polymolial you should counts them as distinct roots.

Why? Precisely the opposite is true: if a root is repeated, then we cannot call those repeated roots 'distinct', and whether or not we count them in the same way we would distinct roots depends on the context.

Grantingriver wrote:

For example, if you say that a give polynomial has “exactly” the roots -3, 4 and 8 that means there is no duplicate roots.

I am not convinced that the word 'exactly' necessarily implies distinctness -- I would have thought that the multiplicities were worth consideration, otherwise the problem appears to be a little simplistic.

Grantingriver wrote:

This is not quite true: the multiplicities of the zeroes are important, e.g. and .
The answer for part 1 is, by all means, yes. This is so because the condition that f(x) has the zeros -3,4 and 8 entails that:

f(x)=(x+3)×(x-4)×(x-8)

And since the zeros of g(x) are -5,-3, 2, 4 and 8 we also have:

g(x)=(x+5)×(x+3)×(x-2)×(x-4)×(x-8)

and hence:

g(x)/f(x)=[(x+5)×(x+3)×(x-2)×(x-4)×(x-8)]/[(x+3)×(x-4)×(x-8)]=(x+5)×(x-2)

and since the result of the division (x+5)×(x-2) is a polynomial without a remainer, therefore g(x) is divisible by f(x). The answer of part 2 is very clear from the answer of part 1. For a polynomial g(x) to be divisible by a polynomial f(x) at least all the roots (zeros) of f(x) should be also roots (zeros) of g(x). And the final statement completes the answer.