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#1 Re: Exercises » Formula » 2018-11-29 03:12:42

What is
?

How can you use this to determine
?

#2 Re: Help Me ! » Number Properties » 2018-11-15 22:26:31

and
is sufficient to generate all three.

#3 Re: Help Me ! » Quasilinear 2nd order PDEs with inital data » 2018-10-29 22:32:14

Hi Emma22,

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

Emma22 wrote:

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)

What is
here? (You have later called this
.)

#4 Re: This is Cool » Repeated cosine converges! » 2018-10-10 01:34:18

Hi Βεν,

Nice contribution! Yes, the repeated iteration of the cosine function converges: actually, it converges to the fixed point of the cosine function, i.e. the solution to
. (I think the two different answers come from using degrees versus radians rather than real vs imaginary.) There are ways of calculating this in terms of the Lambert W function or some nice infinite sums of Bessel functions I think.

You can prove that a solution to the above equation exists via Brouwer's fixed point theorem (and probably the contraction mapping theorem too).

#6 Re: This is Cool » Something ineteresting » 2018-10-10 00:33:14

Βεν Γ. Κυθισ wrote:

it should be (a^m)n^a.

Do you mean
rather than
?

#7 Re: This is Cool » Something ineteresting » 2018-10-09 23:34:30

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

I know what you mean, but this can be misleading:
is multiplied by itself
times. For example,
, where
gets multiplied by itself one time.

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

The brackets should go around the
here, i.e.
for even
.

a^n÷a^m=a^(m-n) (Makes sense right?)

On one side of the equation, the
and
are the wrong way round. It should read:

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

What did you mean here?

#8 Re: Euler Avenue » Sine and cosine » 2018-10-03 01:13:18

Hi zahlenspieler,

I have fixed your LaTeX.

#9 Re: Help Me ! » Algebra: Polynomial Question » 2018-10-01 01:12:02

Hi math9maniac,

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

#11 Re: Help Me ! » Exponential expression » 2018-09-28 22:32:53

OK, suppose we had
. Then:

In this case, we have
. We want to calculate
, which means we need to multiply the index (
) by
. Can you see how to do this?

#12 Re: Help Me ! » Exponential expression » 2018-09-28 21:14:39

Hi vipin_sharma,

Welcome to the forum. Here's a hint:

#13 Re: Help Me ! » answer for this series: 4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 1 » 2018-09-23 05:52:10

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,

#15 Re: Help Me ! » Conditional Statements and Venn Diagrams » 2018-07-28 04:59:30

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?

#16 Re: Help Me ! » Find all real solutions: Proof help » 2018-07-27 21:23:51

!nval!d_us3rnam3 wrote:

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.

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 you

Since
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.

#18 Re: Introductions » Hi there. » 2018-07-03 03:03:20

Hi Stuti55,

Welcome to the forum.

#19 Re: Help Me ! » Division » 2018-07-02 20:30:53

Hi Zeeshan,

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

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

#20 Re: Help Me ! » Precalculus Help » 2018-06-19 09:28:17

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)?

#21 Re: Help Me ! » algebra » 2018-05-27 22:57:19

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?

#22 Re: Help Me ! » Algebra » 2018-05-07 08:27:45

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.

#23 Re: Help Me ! » Algebra » 2018-05-05 20:41:44

Grantingriver wrote:

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.

This is not quite true: the multiplicities of the zeroes are important, e.g.
  and 
.

#24 Re: Help Me ! » Help Me ! » 2018-04-30 06:42:02

Hello bishop.bell,

Welcome to the forum. This problem has been asked at least twice in the past -- you can find some discussions about it here and here.

#25 Re: Help Me ! » a question of a dice in probabilty » 2018-03-30 21:10:17

The denominator will not be 36. Remember: you want to restrict the number of possible outcomes to include only those that have a '5' in them.

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