IntMath Newsletter: Derivative graphs, roller door problem, online math in remote India
By Murray Bourne, 05 Dec 2012
5 Dec 2012
In this Newsletter:
1. Derivative graphs interactive
2. Roller door problem
3. Online math education for Ajab, Gujurat
4. Math puzzles
5. Friday math movies
6. Final thought - candles
1. Derivative graphs interactive
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Calculus can be a bit of a mystery at first. Explore these graphs to get a better idea of what differentiation actually means. |
2. Roller Door problem
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Here's a real-life application of calculus posed by a roller door engineer. I actually made an error when first writing the article. I should have done an estimation first, something I always preach. (My graph worked fine, so I thought all was good. I'd used the correct formula for the graph, but not for the integration.) See: Roller Door problem |
3. Online math education for Ajab, Gujarat
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Students in Ajab village, India, enjoy a math lesson via Web conference from the US. |
4. Math puzzles
(a) Last puzzle: The puzzle in the last Newsletter asked about the number of messages you can send using a flashlight.
The correct answer was 30 messages. People who got it correct and showed working were: Soutrik, Pradyut De, Guido, Nicos Mavrommatis, Thomas A Buckley and Rosa.
(b) New puzzle: What is the value of the following?
You can respond in the comments.
5. Friday math movies
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(a) How things in nature tend to sync up Well-known mathematician Steven Strogatz talks about how flocking behavior naturally occurs - and some mathematical reasons for it. |
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(b) Science is for everyone, kids included Meet 12-year-old Amy O'Toole presenting a scientific paper that starts, "Once upon a time..." |
6. Final thought - candles
This quote suggests it's up to us to make sure we know what's going on.
If you're not lighting any candles, don't complain about being in the dark. [Anon]
Until next time, enjoy whatever you learn.
See the 33 Comments below.
5 Dec 2012 at 7:49 pm [Comment permalink]
(50*99)/2 = 2475
5 Dec 2012 at 7:51 pm [Comment permalink]
Is it 2525?
5 Dec 2012 at 9:41 pm [Comment permalink]
2525
You can use an arithmetic progression of (n/2) to work out what the sum of the fractions are, but in order to work out what the sum of the sum is we need to use a sigma formula. We know that the sum of N is (1/2)n(n+1) so half of that (since the arithmetic progression is (1/2)n take half of the sum of N:
(1/2)(1/2)n(n+1) --> (1/4)n(n+1) since there are 100 sets of fractions to add up we do (1/4)*100*101 = 2525
5 Dec 2012 at 9:52 pm [Comment permalink]
Response to the puzzle: The arithmetic sum is S = first term pus the last term multiply by the number of terms divided by 2. In this case is 1/2 + 2/2 + 3/2 + ... + 99/2.
Then the common factor is 2 and the total sum is: 100 x 99 / 4 = 99 x 25 = 2475
5 Dec 2012 at 11:14 pm [Comment permalink]
We have an Arithmetic Series with t(1)= 1/2 and t(99)= 99/2
The sum of these terms can be found using the formula :
S(n)= n/2( a+l) ....where a=1/2 and l=99/2 and n=99
The substitution into the formula would produce the sum. : 2475
5 Dec 2012 at 11:53 pm [Comment permalink]
The given sequence can be written in general form as:
1/2+1/3(1+2)+1/4(1+2+3)+....(upto nth term)
=(sigma)1/(n+1)*(1+2+3+..+n)
=(sigma)n/2
=n(n+1)/4
Putting n=100,we've got the sequence to be equal to=2525
I oughta be right,aren't I?
6 Dec 2012 at 1:14 am [Comment permalink]
Probably not what you really want, but ...
Sum asked = (1/2) + (1 + 0/3) + (1 + 2/4) + (1 + 5/5) + (1 + 9/6) + ... + (1 + 4752/99) + (1 + 4850/100)
= (0.5) + (1) + (1.5) + (2) + (2.5) + (3) + ... (49) + (49.5)
= 2(1 + 2 + 3 ... + 49) +50(0.5)
= 2(1225) + 25
= 2475
If I improve on this, I'll re-submit!
6 Dec 2012 at 1:20 am [Comment permalink]
100 (m-1)
= Σ ( Σ n ) / m = 2475
m=2 n=1
OR
u can just add 1->99 and divide by 2 u should get
2475. because all the numbers when u add them give fractions over 2 in ascending order, e.g. 1/2+ 2/2+ 3/2 all the way up to 99/2 =2475. 🙂
6 Dec 2012 at 1:42 am [Comment permalink]
1525
6 Dec 2012 at 10:54 am [Comment permalink]
Arithmetical Series, Common difference d=0.5
n = number of terms, a = value of 1st term.
Sum of terms
= n[2a+d(n-l)]/2
= 99[2x0.5 +0.5(99-1)]/2
= 2475
6 Dec 2012 at 5:48 pm [Comment permalink]
2875
6 Dec 2012 at 6:31 pm [Comment permalink]
= 1/2+2/2+3/2+4/2+5/2+....+100/2
= 1/2(1+2+3+4+5+....+100)
= 1/2*1/2(101+101+101+101+101+...+101)
= 1/4*101*100
= 101*25
= 2525.
6 Dec 2012 at 6:35 pm [Comment permalink]
This is actually an Arithmetic Progression with the first term of half and total number of terms = 99, and the difference is half.
Hence the correct answer is 2,475. Thank you.
6 Dec 2012 at 8:11 pm [Comment permalink]
1/2 + (1+2)/3 + (1+2+3)/4 + .... + (1+2+3+...+99)/100
= 0.5 + 1 + 1,5 + 2 + ... + 49,5
This is an arithmetic progression with a constant difference between the consecutive terms of : v=0.5
So: U{n} = U{1} + (n-1)v = U{99} = 0.5 +98*0.5 = 49.5
And: S{n} = n/2 * (U{1} + U{99}) = 99/2 * 50 = 2475
7 Dec 2012 at 1:10 am [Comment permalink]
n^2+.5*n=50^2+.5*50=2500+25=2525
7 Dec 2012 at 6:30 am [Comment permalink]
It's easy to see that the given sequence is an arithmetic progression P: 0.5, 1, 1.5, 2, 2.5,... where the common difference is d=0.5 and the number of terms is given by the last numerator of each subsequence. So, the total number of terms is n=99. The sum of a progression is S=a*n+n(n-1)d/2 and "a" is the initial term. In that case, the total sum of our progression P is S=0.5*99+99*98*0.5/2=2475. The value of the series above is 2475.
7 Dec 2012 at 9:53 am [Comment permalink]
the answer is: 51*99/2
7 Dec 2012 at 10:05 am [Comment permalink]
The n-th term of the sequence is the sum of the first n integers divided by n + 1, i.e., n (n + 1)/(2 (n + 1)) = n/2. So, the sum of the first 99 terms of the sequence is (99 x 100/2)/2 = 99 x 25 = 2475.
7 Dec 2012 at 11:36 am [Comment permalink]
Hai,
This site is very useful to improve our maths skils.
7 Dec 2012 at 11:39 am [Comment permalink]
Puzzles Answer:
[1+2+3....99]/2 =(99 * 100)/2 =4950
7 Dec 2012 at 3:54 pm [Comment permalink]
The answer to the puzzle is 2475.
What I did
(1+2+...+99)/2
7 Dec 2012 at 6:20 pm [Comment permalink]
THE GIVEN SUM IS 1/2 + 2/2 + 3/2 + 4/2 +...+99/2 WHICH IS NOTHING BUT
1/2(1 + 2 + 3 + 4 +... + 99)
=1/2.99(99+1)/2
=99*100/4
= 2475
7 Dec 2012 at 10:59 pm [Comment permalink]
Each term in this question is itself an Arithmetical Progression.
By evaluating the first 4 terms we obtain ½+ 1 + 1 ½ + 2 + …
It can be inferred by inspection of the original question that there are 99 terms, since the highest numerator value in each term of the original question gives the number of the term.
We thus have an Arithmetic Progression(AP) where a = ½ , d = ½ and n = 99.
Thus S = n/2(2a + (n-1) d) = 99/2(1 + (98) ½ ) = (99 * 50) /2 = 2475.Answer.
The above workings can be proved as follows:
The last term L of the question is an AP where the sum of the numerators can be found by
S = n/2(2a + (n-1) d) where n = 1 , d = 1 and n = 99.= 99/2(2 + (98) 1 ) = 4950.
So the last term is 4950/100 = 49 ½ .
The sum of an AP can also be expressed as S = n/2(a + L) where L is the last term.
Therefore in the original question a = ½ , L = 49 ½ and n = 99. resulting in 99/2( ½ + 49 ½) = (99 * 50) /2 = 2475 as before.
8 Dec 2012 at 12:03 am [Comment permalink]
u(1)=1/2, u(2)= 1, u(3)= 3/2 it is a p.a of racio 1/2.
So, u(99)= 1/2 + 98* 1/2 = 99/2 then the sum will be
S = 1/2*( 99/2+1/2) * 99 =25*99 =2475
Even if it is not correct, thank you for this iniciative 🙂
8 Dec 2012 at 1:05 pm [Comment permalink]
Solution for this series is seems such like that,…
1/2 + (1/3+2/3)+1/4+2/4+3/4+…+(1/100+…+99/100)
Method 1: by mathematically
1/2 + (1/3+2/3)+1/4+2/4+3/4+…+(1/100+…+99/100)
=?(i=2 to 100) ?(j=1 to i-1) [ j/i]
=?(i=2 to 100) [1/i] ?(j=1 to i-1) [j]
= ?(i=2 to 100) [1/i] [(i)(i+1)/2]
= ?(i=2 to 100) [ ((i+1))/2]
= 100(100+1)/2 + 100/2
=2475
Method 2 : by pragmatically (in c)
#include
#include
void main()
{
float sum=0,sumj=0,div;
float i,j;
clrscr();
for(i=2;i<=100;i++)
{
sumj=0;
for(j=1;j<=i-1;j++)
{
sumj=sumj+(j/i);
}
sum=sum+sumj;
}
printf("%f",sum);
getch();
}
8 Dec 2012 at 9:29 pm [Comment permalink]
if we look at the value(Y) of a particular therm (Xth term) the relationship can be expressed by following linear equation
Y = 0.5X
therefore, this is nothing but the simple series with...
First Term (A) = 0.5
Sequential Difference (D) = 0.5
Last Term (L) = 49.5
(Note that, though last term has 100 as the denominator, it's 99th term.)
Therefore, S = (99/2)(0.5+49.5)
= 2475
9 Dec 2012 at 12:27 pm [Comment permalink]
2475
10 Dec 2012 at 12:46 pm [Comment permalink]
1275
11 Dec 2012 at 6:58 pm [Comment permalink]
I have used the formula for arithmetic progression a1=1/2, d=1/2, n=99. The answer is 2475.
14 Dec 2012 at 7:37 am [Comment permalink]
97.5=sum ( 0.5 + ( 1/3+2/3) + ( 1/4 + 2/4 + 3/4 ) ....
14 Dec 2012 at 7:49 am [Comment permalink]
(1/2) + (1/3+2/3) + (1/4 + 2/4 + 3/4) + ........ + ( 1/100 - 99/100) = 2474
20 Dec 2012 at 9:46 pm [Comment permalink]
Arithmetic progression.
First term 0.5 , difference 0.5, number of terms 99.
Therefore Sum = 99/2 (2 x 0.5 + 98 x 0.5)= 99.5
17 Jan 2013 at 2:33 am [Comment permalink]
my answer is 2475