Math Is Fun Forum

5          x
f(x)= x     •   3

                4       x            5
ƒ¹(x)= 5x    •  3      +   x     •  xln3

And for this:
               0.7x
y=(cosx)
I dunno if u have to calculate the derivative, but u can rewrite it in this way
         
           (7/10)x • ln(cosx)
f(x)= e

O H    M Y   G O D.
I think tat Santa Claus has stolen my brain!!!!!!! For SURE!

ok ok ok ok... ONLY THE LAST NUMBER OF THE CHAIN OR OF THE SEQUENCE MUST BE SUM WITH THE FIRST NUMBER AND THEM HAVE TO BE A PERFECT SQUARE!!!  [sorry again..]

so any lists it's ok,       

LAST RIGHT EXAMPLE:
Let a..z  be numbers:
a  b  c d  e f g h i

a+b=perfect square      b+c=perfect square    c+d=perfect square  d+e=perfect square  e+f=perfect square
f+g=perfect square       g+h=perfect square   h+i perfect square         ANDD      i+a= perfect square!!

A THING LIKE THIS. SORRY FOR ALL MY MISTAKES!!
So it's important tat the sum of 2 consecutive numbers it's always a perfect square and the last with the first have to be a perfect square yet! [to complete the circular sequence]
-And it's not important tat all numbers are present.
THE ONLY 2 THINGS ARE:
- Which is this n which u can arrange  a..b..z..k...to n in a circular sequence
-The longest list that u can make with this n.
TIP:
This minimum n is less than 50.  I think we need to make a circular sequence with all the combinations from 1 since 50  [except 0...in fact n>1]

Thank u again.
Forgive me!

ricky I tried to write the correct version of the question. I think that krass is doing well.
This n is the minimum integer tat allow u to arrange all the numbers  [1.2.4..till this n] in a circular sequence and the sum of 2 consecutive numbers is always a perfect square , but please read my last post.
And the tip was tat this minimum integer n is less than 50. so we need to look for this n as krassi is doing.
I think u wrote n<>9 coz u trying to arrange all the numbers till 9 right? so u wrote <> coz 9 isn't the minimum numbeR? good!
There is only n.

Sorry mathsyperson! Yesterday it was too late and I was so confused
About the problem..I making some mistakes ...You started well
The only 2 things to do are:
1)Choose a n (entire). Try to arrange in a circulare sequence all the numbers from 1 to n. (1,2,3,4,...,n)
  Check if the sum of 2 consecutive numbers is always a perfect square. Which is the smallest n in which u can do  that?
the tip is:  this n is less than 50..so  n<50.

2) Write a sequence (like the example)

Sorry for my  mistakes. I think now it's ok.

sorry!! I wrong in the first post
a)"FOR WHICH ENTIRES  N>1 is possible ARRANGE IN A CERTAIN WAY ALL THE ENTIRE NUMBERS 1,2,3...,n AROUND A CIRCLE (CIRCULAR SEQUENCE),THAT THE SUM OF 2 CONSECUTIVE NUMBERS IS ALWAYS A PERFECT SQUARE."?
b) and then WHICH IS THE SMALLEST ENTIRE NUMBER  N>1 FOR WHICH IS POSSIBLE TO FIND THIS CIRCULAR SEQUENCE?

But your last example was ok!
EXAMPLE: 1 24 12 4 5 11 14 2 7 9 27 22 3 .........

Yes! Right!! That way!
This problem is very nice but it's not easy

However thanx to all!!!!

&& stands for the logic AND

For example a right circular sequence could be this:

1 15 10 6 3          1+15=16 perfect square    15+10=25 perfect square ...till 3 +1 (the first number) is a perfect square
I was also lookin for a program in whatever language..

Hi.
I tried a thing like this  (but I think that it's the same thing)
Let ln log base e
2^x=2x
ln(2^x)=ln2+lnx
xln2=ln2+lnx
x=1+ln(x)/ln(2)
and so

x=1+Log    (x)
               2