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#2 Re: Help Me ! » Integration » 2019-01-26 06:28:43

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:

#3 Re: Exercises » Integration » 2019-01-25 11:16:35

Hi Bob,

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

https://youtu.be/IwmYetE29YU

#4 Re: Exercises » Integration » 2019-01-18 03:03:53

Set

then use the trig identity

#5 Re: Help Me ! » Express ∛(7 + 5√2) in the form x + y√2 » 2019-01-10 07:39:45

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?

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

What is
?

How can you use this to determine
?

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

and
is sufficient to generate all three.

#10 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
.)

#11 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).

#13 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
?

#14 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?

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

Hi zahlenspieler,

I have fixed your LaTeX.

#16 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?

#18 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?

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

Hi vipin_sharma,

Welcome to the forum. Here's a hint:

#20 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,

#22 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?

#23 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.

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

Hi Stuti55,

Welcome to the forum.

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