IF you would like help with any math problems, feel free to post them

sulexk

IF you would like help with any math problems, feel free to post them

:smile:

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Reply 1

Notnek

I'm trying to prove that every even integer greater than 2 can be expressed as the sum of two primes. Can you help?

Reply 2

sulexk

Original post by notnek

I'm trying to prove that every even integer greater than 2 can be expressed as the sum of two primes. Can you help?

Oh thats a lovely problem!
Although I did mean A level problems, as that is why I put this in that thread, but nonetheless I will give it a try! :smile:

Will take me some time though, but will keep you updated. (Actually this is unproven, it is termed the Goldbach Conjecture, but the above form was formulated by Euler-apparently it has been tested 400,000,000,000,000). With that said, I suppose we must solve it algebraically. May take a while........(will think about it though)

Thank you :smile:

(edited 12 years ago)

Reply 3

Notnek

Original post by sulexk

Oh thats a lovely problem!
Although I did mean A level problems, as that is why I put this in that thread, but nonetheless I will give it a try! :smile:

It's OK, you don't have to prove it!

I'm not sure if this thread is needed since the maths subforum is designed for people to post their questions.

I'm sure you'll be welcome in many threads in the maths subforum to answer questions :smile:

Reply 4

sulexk

Original post by notnek

Yes, you are right.

I just thought another one at this time maybe helpful :smile:

But thank you

I wonder if the Goldbach conjecture can be proved? What do you think?

Reply 5

the bear

i was given a puzzle book for Christmas and am still stuck on this one:

Show that the real part of any non-trivial zero of the Riemann zeta function is 1/2.

i have tried proof by exhaustion but it is taking ages...

Reply 6

sulexk

Original post by the bear

Hi,

That is to do with functions of complex variables. I will try and work on it, although have not yet covered Functions of Complex variables. Will try soon though! :smile:

I would love to work on these in the summer, once all exams are done.

Reply 7

Master Bob

Hey its Bob again,

Noticed you made a thread =) I've been doing some exam style paper today and got myself stuck on this question.

1) A geometric series has a common ratio r (<0). The sum of the first and third terms equals 4 times the sum of the third and fifth terms. Find the value of r.

So i did it like this:

a + ar^2 = 4(ar^2 + ar^4)
a + ar^2 = 4ar^2 + 4ar^4

0 = \frac{4ar^2 + 4ar^4}{a + ar^2}

Which simplifies to 4r^2 + 4r^2 = 0
8r^2 = 0

As you can see, this doesn't make any sense and im not sure what im doing wrong. :s-smilie:

Any help will be very appreciated. Thanks

(edited 12 years ago)

Reply 8

raheem94

Original post by Master Bob

0 = \frac{4ar^2 + 4ar^4}{a + ar^2}

Which simplifies to 4r^2 + 4r^2 = 0
8r^2 = 0

As you can see, this doesn't make any sense and im not sure what im doing wrong. :s-smilie:

Any help will be very appreciated. Thanks

\displaystyle a + ar^2 = 4(ar^2 + ar^4) [br]\displaystyle a(1+r^2) = 4a(r^2+r^4)[br]\displaystyle 1+r^2 = 4(r^2+r^4)[br]\displaystyle 1+r^2 = 4r^2 + 4r^4[br]\displaystyle 4r^4 + 3r^2 -1 =0

Lets take,

\displaystyle r^2 = x

So the equation becomes,

\displaystyle 4r^4 + 3r^2 -1 =0 [br]\displaystyle 4x^2 + 3x - 1 =0[br]

This is a quadratic, solve it, you will get two solution.

One of the solution is a negative value, which you can ignore because it won't agree,

\displaystyle r^2 = x

.

The other will be a positive value, lets say it is 'b'.
So you get,

\displaystyle r^2 = b [br]\displaystyle r = \pm \sqrt{b}[br]

We need the negative solution because the question states that r is less than zero.
Hence,

\displaystyle r = -\sqrt{b}

Reply 9

Master Bob

@raheem94 Thank you. Got the final answer as -0.5 :smile:

Reply 10

raheem94

Original post by Master Bob

@raheem94 Thank you. Got the final answer as -0.5 :smile:

You are welcome.

r=-0.5 is the correct answer.

You can check the answer yourself by substituting it in the initial equation.

\displaystyle a + ar^2 = 4(ar^2 + ar^4) [br]\displaystyle a(1+r^2) = 4a(r^2+r^4)[br]\displaystyle a(1+(-0.5)^2) = 4a((-0.5)^2 + (-0.5)^4)[br]\displaystyle a(1+0.25) = 4a(0.25+0.0625)[br]\displaystyle 1.25a = 4a(0.3125)[br]\displaystyle 1.25a = 1.25a[br]

Reply 11

Brit_Miller

If a mum gives Dave 6 sweets, Sandra 7 sweets, Muhammad 3 sweets and Aailya 2 sweets, what was she thinking when she named her children?

Reply 12

sulexk

Original post by Brit_Miller

If a mum gives Dave 6 sweets, Sandra 7 sweets, Muhammad 3 sweets and Aailya 2 sweets, what was she thinking when she named her children?

Hmm...

Reply 13

sulexk

Original post by Brit_Miller

If a mum gives Dave 6 sweets, Sandra 7 sweets, Muhammad 3 sweets and Aailya 2 sweets, what was she thinking when she named her children?

Just note this is a maths thread! :smile:

Nevertheless, through your pretty clever words, you intended to mean something.
I would think you were saying that a particular way of life, puts one entity in higher status than the other..

Tell me what were you thinking?

Thanks :smile:

Reply 14

starfish232

Hi I was wondering if you could help me with a C2 differentiation Question and it's attached to this post.

Thanks!

Differentiation Question.docx138.0KB

Reply 15

raheem94

Original post by starfish232

Hi I was wondering if you could help me with a C2 differentiation Question and it's attached to this post.

Thanks!

On which part do you need help?

Reply 16

f1mad

Original post by starfish232

Hi I was wondering if you could help me with a C2 differentiation Question and it's attached to this post.

Thanks!

For the first one.

You're told that the length of the wire is 2 i.e the Perimeter= 2.

-> 2= The sum of all the sides on the outside. Note that the perimeter of a semi-circle is 1/2*pi*diameter. Hence express y in terms of x (i.e get it in the form y= something).

Secondly: The area is given by: area of rectangle +area of semi-circle.

*area of semi-circle= 1/2* pi *radius^2

Make sure you replace y here, with what you worked out in part A.

Finally, using the R expression, differentiate R with respect to x twice.