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I'm bored. Give me a challenging maths problem to work on.

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Reply 40
Avatar for Aph
Aph
22
Original post by multiratiunculae
hyperbolic arctan? :biggrin:

I said A-levelish

That is A-level. FP2
Reply 41
Avatar for Andy98
Andy98
20
Original post by multiratiunculae
hyperbolic arctan? :biggrin:

I said A-levelish


I'm doing it now at A-level.....Further maths :colondollar:
Original post by 1 8 13 20 42

If so, note that the derivative of x^2 + 6x is 2x + 6 = 2(x + 3)


I was thinking of using u-substitution: why is the derivative important?
X and Y go shopping. X buys 1 apple and Y buys 2 apples. How many apples all together?
Original post by multiratiunculae
What can I deduce about f(sin(x))? If the original function was f(x)=1 for example, then f(sin(x)) has no effect. Obviously, the original function wasn't so, but what can be deduced?


The info gives away that the first part can be done by IBP. When the first part is done, the second deduction is just an application using the same logic.
(edited 8 years ago)
Reply 45
Avatar for math42
math42
20
Original post by multiratiunculae
I was thinking of using u-substitution: why is the derivative important?


Well this is quite a standard integrating technique; you note that by the chain rule differentiating something of the form (7 - 6x - x^2)^n will give you -2n(x +3)(7 - 6x - x^2)^(n-1); all you need to do is find the right power and the right constant
Original post by 1 8 13 20 42
Idk what is challenging for you

Here's one that's probably too easy, prove that n^3 - n is divisible by 6 for n a natural number


You can write it as n(n^2-1) which is the same as (n-1)n(n+1).
(n-1)(n)(n+1) is 3 consecutive numbers multiplied together.

Therefore, one of your numbers must be divisible by 2 and one must be divisible by 3.
As we know any multiple of 2 multiplied by any multiple of 3 equals a multiple of 6; because 2x3=6
You then have a third number to multiply, and whatever that number is your solution to the equation must still be divisible by 6, since a multiple of 6 times any number is still a multiple of 6.

So you will have (multiple of 2)(multiple of 3)(multiple of 1).
Which is the same as (multiple of 6)(multiple of 1).
Which is the same as (multiple of 6).

With this method you're only really proving that 3 consecutive numbers multiply into a multiple of 6.

I hope there's a more mathematical way to do it :smile:
Reply 47
Avatar for Dabo_26
Dabo_26
9
Original post by multiratiunculae
A-levelish.


here you go:
http://www.damtp.cam.ac.uk/user/stcs/STEP/1999paperIII.pdf
what is 9 + 10
Original post by MathsAstronomy12
X and Y go shopping. X buys 1 apple and Y buys 2 apples. How many apples all together?


that moment when you realise the answer is too obvious lol

3?
I could have sworn you posted that originally and I responded to the "9+10" not "0/0" wtf
Ur a funny ol' scallywag
Reply 52
Avatar for math42
math42
20
Original post by Jamdroid
You can write it as n(n^2-1) which is the same as (n-1)n(n+1).
(n-1)(n)(n+1) is 3 consecutive numbers multiplied together.

Therefore, one of your numbers must be divisible by 2 and one must be divisible by 3.
As we know any multiple of 2 multiplied by any multiple of 3 equals a multiple of 6; because 2x3=6
You then have a third number to multiply, and whatever that number is your solution to the equation must still be divisible by 6, since a multiple of 6 times any number is still a multiple of 6.

So you will have (multiple of 2)(multiple of 3)(multiple of 1).
Which is the same as (multiple of 6)(multiple of 1).
Which is the same as (multiple of 6).

With this method you're only really proving that 3 consecutive numbers multiply into a multiple of 6.

I hope there's a more mathematical way to do it :smile:


Yeah that's the way. I suppose asserting that the number is divisible by 2 and 3 isn't completely rigorous but it is obvious and in terms of this question seems a bit unnecessary to prove, although it can be done quite easily also.
Original post by Awesome Genius
Do they even do Maths at Norwich?


In the University of East Anglia? Yes, yes they do.
On a more serious note did u actually manage to solve
IMAG0058.jpghow do you do 1f?
What about 9b, I keep on getting 1 as k but the answer is 1/2 but how? 14457852692151509479030.jpg
Original post by coconut64
IMAG0058.jpghow do you do 1f?


Use F=ma
The key word is maximum, so you can use values 3ms-2 (acceleration) and 12000N

Let X equal the number of people
Using f=ma
12000N-(300g+75g[x])=(300+75x)3

And find X by making it the subject.

(12000-300g-75gx)/3 = 300+75x
(12000-300g-75gx)/3 - 300 - 75x = 0
(12000-300g)/3 - (75gx)/3 - 300 - 75x = 0

Multiply by 3, and switch sides around.

225x + 75gx = 12000 - 300g - 900
X(225+75g) = " "
X = (12000 - 300g - 900) / (225+75g)

And when rounding, round downwards to the nearest whole number.
So I got 8 people, I hope that is correct! :smile:
(edited 8 years ago)
Original post by Jamdroid
Use F=ma
The key word is maximum, so you can use values 3ms-2 (acceleration) and 12000N

Let X equal the number of people
Using f=ma
12000N-(300g+75g[x])=(300+75x)3

And find X by making it the subject.

(12000-300g-75gx)/3 = 300+75x
(12000-300g-75gx)/3 - 300 - 75x = 0
(12000-300g)/3 - (75gx)/3 - 300 - 75x = 0

Multiply by 3, and switch sides around.

225x + 75gx = 12000 - 300g - 900
X(225+75g) = " "
X = (12000 - 300g - 900) / (225+75g)

And when rounding, round downwards to the nearest whole number.
So I got 8 people, I hope that is correct! :smile:

I did something similar to yours and got 8.5 but the answer is 11 which doesn't ring any bells. also why didnt you round it to 9? Could you please do the other question I posted? Thanks
(edited 8 years ago)
Original post by coconut64
I did something similar to yours and got 8.5 but the answer is 11 which doesn't ring any bells. also why didnt you round it to 9? Could you please do the other question I posted? Thanks


I hate lift questions haha

You have to round downwards because you are using the maximum values to work out X. So the number you get, such as 8.5, is the absolute maximum number of people you can have.

Since you can't have 8.5 people, the maximum, you can only have 8 people.
If you have 9 people the lift will break because it can't hold that force, it can only hold 8.5.

Assuming if 8.5 was the correct answer, of course.

+ the answer for your second question should be 1, I can't see why not.
(edited 8 years ago)

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