Math Is Fun Forum

Hi joyh
Sorry..............Is This a question or a joke?????????????????=)
Bye

Hi Tobal
for the first function :
The left handside is -∞ and the right handside is  ∞ so the limit does not exist.
for the second function :
The left handside is ∞ and the right handside is  - ∞ so the limit does not exist.
for the third function :
Both left and right limits are ∞ since the denominator is squared.
for the forth function :
The left handside is ∞ and the right handside is  - ∞ so the limit does not exist.
to see the forth
the sign of the denominator when x is less and close to 1  is +ve
and the sign of the denominator when x is greater  and close to 1  is -ve so the two limits do not concide and the limit does not exist , and so for the rest functions
Best Wishes
Riad Zaidan

Hi to all,
I am very pleased with our fellows on this site, although I am teaching at a university and I participate my freinds some ideas concerning mathematics.......thanks for all that whom are dealing with site
greatings for all
salam

Hi Daniel123
this is the proof for natural n:
√3+i=2(√3/2  + (1/2)i)=2(cos 30° +i sin(30° ))  r=2 ,  theta=30° then by demoivers theorem

(√3+i)^n=2^n(cos n(30°) +i sin( n (30° ))  and similarly

√3-i=2(√3/2  - (1/2)i)=2(cos 330° +i sin(330° ))  r=2 ,  theta=30° then by demoivers theorem

(√3-i)^n=2^n(cos n(330°) +i sin( n (330° ))   therefore

(√3+i)^n+(√3-i)^n=2^n(cos n(30°) +i sin( n (30° ))+ 2^n(cos n(330°)+i sin( n (330° )) but
sin( n (30°))=- sin( n (330° )) therefore the imaginary part cancells and the result is in R or
(√3+i)^n+(√3-i)^n=2^n(cos n(30°)+2^n(cos n(330°) but cos n(30°)=(cos n(330°) i.e
(√3+i)^n+(√3-i)^n=2(2^n(cos n(30°))=2^(n+1)   (cos n(30°) ∈R
Best Regards
Riad Zaidan

Hi SgtDawkins
I think that you can do the following:
  1+x+x²+... =1/(1-x) infinite geometric siries  or equavelently
1/(1-x)=1+x+x²+...  by substituting  1-x=z so  x=1-z we get
1/z=1+(1-z)+(1-z)²+...   
∫1/zdz=∫(1+(1-z)+(1-z)²+...)dz  by integrating both sides w.r.t  z
⇒ln(z)=z-((1-z)^2)/2-((1-z)^3)/3-...+ c  where c is a constant
put z=1 we have  ln(1)=1-0+0-0+...+c  ⇒ 0=1+c  so c=-1 therefore
ln(z)=z-((1-z)^2)/2-((1-z)^3)/3-...-1
so  substitute z=2 we get
ln(2)=2-1/2+1/3-1/4+1/5-1/6+...-1
        ln(2) ≈ 1-(1/2)+(1/3)-(1/4)+(1/5)-(1/6)+...
or  ln(2)≈0.64563 as Mr juriguen got
Best wishes
Riad Zaidan

Hi MathsIsFun
Thanks very  much for you  for these pages

Hi dear  Onyx:
You asked the following:
Also can someone tell me what was wrong with my method of parameterization?
The answer is :
the problem was of the point  P(1,3,2) does not lie on the given surface ,  so when you change the constant 20 into 21 in order to have the point on the surface and repeat the procedures you have done , every thing will be right. Try this by yourself.
Best Regards
Riad Zaidan

Dear Onyx
Here is the solution

the point  P(1,3,2) lies on the plane

x^2* y +3yz=20  rewriting the equation as follows

x^2* y +3yz-20=0 so let 

f=x^2* y +3yz-20

df/dx=2xy  so df/dx  (p)  =  2(1)(3)=6

df/dy=x^2+3z  so df/dy  (p)  = 7

df/dz=3y  so df/dz  (p)  =9

so the normal   n = 6i+7j+9k

and the  tnagent plane H to the surface is 6(x-1)+7(y-3)+9(z-2)=0 or

6x+7y+9z=45

Best Regards
Riad Zaidan

Hi ganesh

problem no.  6:

First we choose the three neighbouring positions for the girls by 5 ways so

they can be seated by   5 * 4! * 3! =5(4)(3)(2)(1)*(3)(2)(1)=720 ways

Best Regards

Riad Zaidan

Online

Hi ganesh

problem no.  4:

The no. of diagonals= a(a-3)/2
               

The no. of triangles= aC3=a(a-1)(a-2)/3*2*1

Best Regards

Riad Zaidan

Online

Hi ganesh

problem no.  7 :

6(nC3)=7((n-1)C3)

6*n(n-1)(n-2)/(3*2*1) =  7 (n-1)(n-2)(n-3)/(3*2*1) so by cancellation ((Hint :  n ≥3))

6n  =  7(n-3)  or   6n = 7n - 21  so n=21

Best Regards

Riad Zaidan

Hi ganesh;

For the sixth problem:

Assume that

y= x^3-5x^2+5x+8 and differentiate both sides w.r.t  (t) we get:

dy/dt=(3 x^2 - 10 x + 5) (dx/dt)  ...........(1)

but  dy/dt  =2 *  (dx/dt)  so  substitute  in   (1)    we  have the following:

  2  (dx/dt) =(3 x^2 - 10 x + 5) (dx/dt)  so  if dx/dt ≠0 we  get 

2   =  3 x^2 - 10 x + 5  therfore

3 x^2 - 10 x + 3 = 0  so   

(3x-1)(x-3)=0   so   

either  x= 1/3   or    x= 3    Q.E.D 

Best Regards
Riad Zaidan