# triangle sequence

• Nov 12th 2006, 09:49 AM
Mr_Green
triangle sequence
Given the sequence of isosceles right triangles P to A sub n to A sub n+1
with P to A sub n+1 = A sub n to A sub n+1 and with P to A sub 1 = 1

Attachment 1221

Find the smallest value of n such that A sub n to A sub n+1 < 0.001

one of responces i got was this on a different forum:

"""""""""""""""""""""""""""""""""""""""""""""""""" """"""""""""""""""""""""""
To try to use formal algebraic symbolism, the process and symbolic form is tough to create; but if you have access to a BASIC language then you might find your answer through this, extremely unrefined program:

dim S(50)
i=0
input "Initial Length Hypotenuse? ";So
LET S(1)=1
LET i=1
FOR i=1 to 40
'i=i+1
S(i) = So/(sqr(2))
So=S(i)
print i;" loopwhiletest ";S(i)
i=i+1
NEXT i
END
"""""""""""""""""""""""""""""""""""""""""""""""""" """"""""""""

im only a sophmore in pre calc. i have no idea wat the above responce wanted me to do...???
• Nov 12th 2006, 10:57 AM
CaptainBlack
Quote:

Originally Posted by Mr_Green
Given the sequence of isosceles right triangles P to A sub n to A sub n+1
with P to A sub n+1 = A sub n to A sub n+1 and with P to A sub 1 = 1

Attachment 1221

Find the smallest value of n such that A sub n to A sub n+1 < 0.001

one of responces i got was this on a different forum:

"""""""""""""""""""""""""""""""""""""""""""""""""" """"""""""""""""""""""""""
To try to use formal algebraic symbolism, the process and symbolic form is tough to create; but if you have access to a BASIC language then you might find your answer through this, extremely unrefined program:

dim S(50)
i=0
input "Initial Length Hypotenuse? ";So
LET S(1)=1
LET i=1
FOR i=1 to 40
'i=i+1
S(i) = So/(sqr(2))
So=S(i)
print i;" loopwhiletest ";S(i)
i=i+1
NEXT i
END
"""""""""""""""""""""""""""""""""""""""""""""""""" """"""""""""

im only a sophmore in pre calc. i have no idea wat the above responce wanted me to do...???

$A_{n+1}=A_n/\sqrt{2}$

and $A_1=1$, so:

$A_n=\frac{1}{2^{(n-1)/2}}$

Now set:

$
A_n=\frac{1}{2^{(n-1)/2}}=0.001
$

Taking logs:

$
-\frac{n-1}{2}\log_{10}(2) = \log_{10}(0.001)=-3
$

so:

$
n=-\frac{3 \times 2}{\log_{10}(2)}+1 \approx 20.93
$

To get the smallest integer so that $A_n<0.001$ we need to
round this up to $n=21$.

RonL