You can use PARI/GP fromagentredlum said:Do you know of any online CAS that does not require the user to download onto the hard drive?

https://cloud.sagemath.com/

You can use PARI/GP fromagentredlum said:Do you know of any online CAS that does not require the user to download onto the hard drive?

https://cloud.sagemath.com/

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
```

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
```

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.

Maybe we should start over again?

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

eddybob123 said:Maybe we should start over again?

mathbalarka said:Yes, I think so.

CRGreathouse said:Go for it.

agentredlum said:Yep , i agree.

That's a whole lot of competitors. I think we should begin?Hoempa said:Sounds good

Eddy, you're in the lead. Start a new thread.

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