- #1

bogdan

- 191

- 0

Any ideas ?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter bogdan
- Start date

- #1

bogdan

- 191

- 0

Any ideas ?

- #2

plus

- 178

- 1

The sum from k to infinity of (x/(n*(n+1))) is x/k, so you could sum up to a certain point and then use this to approximate the truncation error.

- #3

rgrig

- 34

- 1

I think the answer is tan(x).

sin(x/(n(n+1))) = sin(x/n)cos(x/(n+1)) - sin(x/(n+1))cos(x/n)

After some simplifications you get:

sum tan(x/n)-tan(x/(n+1))

That is:

tan x - tan x/2 +

+ tan x/2 - tan x/3

...

which is

tan x - tan 0 = tan x

- #4

bogdan

- 191

- 0

(eram doar curios sa vad cine stie sa-l rezolve...e dintr-o carte de exercitii de analiza...cu tot cu raspunsuri)

Share:

- Last Post

- Replies
- 7

- Views
- 234

- Last Post

- Replies
- 10

- Views
- 350

- Last Post

- Replies
- 4

- Views
- 661

- Last Post

- Replies
- 3

- Views
- 760

MHB
Sum of digits

- Last Post

- Replies
- 7

- Views
- 571

- Replies
- 26

- Views
- 558

MHB
Finite sum

- Last Post

- Replies
- 1

- Views
- 498

- Last Post

- Replies
- 1

- Views
- 436

- Last Post

- Replies
- 8

- Views
- 698

- Replies
- 4

- Views
- 498