# Math Q&A challenge (Part 11)

Forum Staff

#### Hoempa

Math Team
Alternatively you could use a (graphical) calculator with an answer-function; you'd want to use the previous answer.

#### Hoempa

Math Team
For if doing computations are a little hard, here are some more values for the example I gave earlier
x=9; m=2
Code:
u_0=10000
u_1~=5000.000450000000000000000001
u_2~=2500.001124999919000007290000
u_3~=1250.002362499149500426464778
u_4~=625.0047812427707655221975974
u_5~=312.5095905663058873993824080
u_6~=156.2691948412332107580019439
u_7~=78.16339388305804767275154499
u_8~=39.13926864839135056440005116
u_9~=19.68460836830533579654296921
u_10~=10.07090918943413231783420935
u_11~=5.482286148388621759111595382
u_12~=3.561968525146494444660614593
and x=10, m = 2
Code:
u_0=10000
u_1~=5000.000500000000000000000000
u_2~=2500.001249999900000009999999
u_3~=1250.002624998950000584999662
u_4~=625.0053124910750212924468029
u_5~=312.5106561775382028790958509
u_6~=156.2713275431914155561418132
u_7~=78.16765940431098027955658689
u_8~=39.14779477464331043945998763
u_9~=19.70161850236230585337467093
u_10~=10.10459550734079175237914559
u_11~=5.547122113177671295859038962
u_12~=3.674929351352360845093880567

#### Hoempa

Math Team
This problem was open for a while, not sure if you'd like to do more problems but for the case you'd like, I'll give it away,
the sum is $$\displaystyle \sqrt{x}$$, it's merely based on that

$$\displaystyle u_{n+1}=\frac{u_n^2+x}{2 \cdot u_n} = \sqrt{x}$$ with $$\displaystyle u_0 = m$$ and $$\displaystyle m \in \mathbb{R}\backslash \{0\}$$ if n goes to infinity. Computing u1, u2, u3, ... could give the conjecture.

#### eddybob123

Hmmm... no one posted in this thread for so long, I actually forgot it even existed. :lol:

Maybe we should start over again?

#### mathbalarka

Math Team
Yes, I think so. But I'd wait for the other's approval to see how many wants to join.

#### CRGreathouse

Forum Staff
eddybob123 said:
Maybe we should start over again?
Go for it. I doubt I'll have time but that shouldn't stop you!

#### agentredlum

Math Team
Yep , i agree. #### Hoempa

Math Team
Sounds good, who'd like to start?

#### mathbalarka

Math Team
Similar Math Discussions Math Forum Date
General Math
General Math
General Math
General Math