# IntMath Newsletter: Derivative graphs, roller door problem, online math in remote India

By Murray Bourne, 05 Dec 2012

5 Dec 2012

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 Calculus can be a bit of a mystery at first. Explore these graphs to get a better idea of what differentiation actually means. Derivative graphs interactive

## 2. Roller Door problem 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.)

## 3. Online math education for Ajab, Gujarat Students in Ajab village, India, enjoy a math lesson via Web conference from the US. Online math education for Ajab, Gujarat

## 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 (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. Friday math movie: How things in nature tend to sync up (b) Science is for everyone, kids included Meet 12-year-old Amy O'Toole presenting a scientific paper that starts, "Once upon a time..." Friday math movie: Science is for everyone, kids included

## 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.

### 33 Comments on “IntMath Newsletter: Derivative graphs, roller door problem, online math in remote India”

1. Christian Mills says:

(50*99)/2 = 2475

2. Jagdish says:

Is it 2525?

3. Michael Dean says:

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

4. Miguel Angel Velo says:

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. John Cherpak says:

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

6. Barneel says:

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?

7. Colin Fraser says:

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!

8. devanie says:

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. 🙂

9. pedram66 says:

1525

10. Thomas A Buckley says:

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

2875

12. Mawanda Ismail says:

= 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.

13. Yeoh Kiat Boon says:

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.

14. Guido says:

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

15. Kaitlyn and Emily says:

n^2+.5*n=50^2+.5*50=2500+25=2525

16. Nicos Mavrommatis says:

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.

17. qi says:

18. Tomas Garza says:

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.

19. jamesfathiaraj says:

Hai,
This site is very useful to improve our maths skils.

20. jamesfathiaraj says:

[1+2+3....99]/2 =(99 * 100)/2 =4950

21. Vasu Jain says:

The answer to the puzzle is 2475.

What I did
(1+2+...+99)/2

22. ARUN KUMAR says:

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

23. Tom Barrett Dublin Ireland says:

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.

24. Rosa says:

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 🙂

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();
}

26. Dineth says:

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

27. hamid says:

2475

28. Gautam says:

1275

29. Lachezar Borisov says:

I have used the formula for arithmetic progression a1=1/2, d=1/2, n=99. The answer is 2475.

30. shshtawi says:

97.5=sum ( 0.5 + ( 1/3+2/3) + ( 1/4 + 2/4 + 3/4 ) ....

31. shshtawi says:

(1/2) + (1/3+2/3) + (1/4 + 2/4 + 3/4) + ........ + ( 1/100 - 99/100) = 2474

32. Lidia says:

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

33. vipul prajapati says:

### Comment Preview

HTML: You can use simple tags like <b>, <a href="...">, etc.

To enter math, you can can either:

1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):
a^2 = sqrt(b^2 + c^2)
(See more on ASCIIMath syntax); or
2. Use simple LaTeX in the following format. Surround your math with $$ and $$.
$$\int g dx = \sqrt{\frac{a}{b}}$$
(This is standard simple LaTeX.)

NOTE: You can mix both types of math entry in your comment.

## Subscribe

* indicates required