You are not logged in.
Hi ganesh;
I won't hide the answer because no one is checking this topic anymore.
Answer to 33:
Generally an expert player (rating around 2000) could easily force mate from that position in under 50 moves. A grandmaster (rating about 2500 - 2600) could force mate in that position in under 35 moves. Most computer chess engines can do it in around 30 moves.
1 count
2 pleasant
3 coherent
Hi random_fruit;
It could be a quiz problem and I agree it appears to have only 1 solution in R.
Hi emadstats
Hi gogeomth
Hi quittyqat;
I did click it. It says "Highlight this message." Is that what is supposed to happen?
1 mean
2 delectable
3 incision
Hi ganesh;
Here is my answer to #10
Now multiply numerator and denominator by
OK. quittyqat now you got me, I just see an empty box. Was that on purpose? Art is even more difficult for me than math or chess.
bobbym
Hi suresh1969
Thats because I am a simpleton. Just ask anybody for confirmation of that fact. You really find out how much you know about something when you try to explain it in simple terms. Its the hardest thing in the world to do. I remember a story about David Hilbert along those lines.
I really do think that identity's post to you
http://www.mathisfunforum.com/viewtopic.php?id=11996
is really a good example of a simple and lucid answer.
bobbym
Hi quittyqat;
I got this one!
My answer
bobbym
Hi Moshe!
Hi ganesh;
The answer to #4 is (3) 555681
Hi quittyqat;
The one you picked looks best but maybe only to me.
bobbym
Hi Mikau;
Typically they haven't used taylor series for a long time. They were replaced by Pade approximants and economized polynomials. These provide better accuracy over a wider range than a Taylor series. A human thinks of the number line as having all the real numbers but to a computer there are lots of missing numbers on it. Simply, because many numbers are not expressible in binary on a finite machine. Like 1/3,1/6,1/9... for instance.
bobbym
Hi ganesh;
My answer for #6
This is a bernoulli trial so:
Hi quittyqat;
The 3rd one looks best to me.
Sorry Jane, I was being nosy. Thanks for the invite. I already do a lot of lurking there.
(Hope you did not see the rest of this post before I got a chance to delete it. Stupidly, put my foot in my mouth with that thoughtless comment.)
Hi Tigeree;
I was being funny, it looks fine here too.
bobbym
Hi smiyc86;
Here is my answer to #1, I hope it is what you require.
As you know the way to sum all the digits of any number is to take that number mod 9.
1999 is congruent to 1 mod 9 so 1999^1999 is also going to be congruent to 1 mod 9.
Therefore D = sum of the digits of 1999^1999 = 1
Also for fun:
B=29656
C=28
bobbym
Hi Jane;
Thanks for providing the solution. Checking out the scienceforum link you provided their is a member that is like you but under another alias. On checking out her profile it leads to the invision forum where you appear as Jane Fairfax.
bobbym
Hi Tony123;
What are you asking for? Are you asking for what value of n that would make that sum equal 400, because I don't think that is possible.
bobbym
Hi smiyc86;
Here is a quick and very ugly way to get the real roots of (sinx)^4-(sinx)^5-(cosx)^5 = 2
First I notice that big 2 on the rhs and guess that one of the extrema of sin(x) and cos(x) are the answer. We have:
Just trying the first one yields (-1)^4 - (-1)^5 - 0 which equals 2 so x = -π/2 is one solution. Since sin is periodic
There are also complex solutions but I could not get them without resorting to an iterative method so I left them out.
bobbym
Hi;
It is apparent to me that Jane needs bigger smilies or I need bigger eyes, I totally missed that little guy.
bobbym
Hi Mikau;
Your counterexample n=5 and x=5.1 should have convinced your teacher that you were right and that there is something wrong with his proof. Unfortunately, he is not wrong about calculators and even frontline computer algebra systems. In some cases they will experience loss of significant digits due to subtractive cancellation and or smearing.
bobbym