Recent content by QED-Kasper

  1. Q

    Feynman - Random Walk <D> and coin flipping

    Thanks codelieb. I have to add though that I misread the text. In it Sands only mentions the expected distance. Which is also known as the MAD (mean average deviation) in statistics. The "square-root of N rule" applies to the RMS (root mean square) distance, aka standard deviation. This is what...
  2. Q

    Feynman - Random Walk <D> and coin flipping

    Thanks, I appreciate that. And thank you for being extra kind :).
  3. Q

    Feynman - Random Walk <D> and coin flipping

    Hello, I have read the probability chapter in Feynman's lectures on physics. And got fascinated by the random walk. There is a statement, that in a game where either a vertical distance of +1 or -1 can be walked each move, the expected value of the absolute distance (lets call it <D>) from...
  4. Q

    Why do forces make an object move?

    people who say that when you push on a box and it doesn't change position you arent giving it energy are wrong. An example of the energy you give to the box is in the form of body heat (if the box is colder than your body). The atoms of your hand have more kinetic energy than the atoms of the...
  5. Q

    Unclear situation involving roots.

    i formulated the problem in my original post.. B's sign isn't given there either. ;p It's confusing what all these people are talking about, about x^(1/6) having six values for example. :s or x^(1/2) and √(x) meaning different things. Especially when it says different in this website: "Any...
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    Unclear situation involving roots.

    All we know is that B squared is equal to some positive value. So B could be negative and positive. Thus the final answer could be negative and positive. So how can the book be correct?
  7. Q

    Unclear situation involving roots.

    Both answers are valid because both satisfy all equations so the ± is justified. The reason why I must use the ± sign is exactly because we don't know the value of B, it could be negative or positive. But we do know the value of r, which is positive. I'm not sure what you mean though. I'm just...
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    Unclear situation involving roots.

    @D H So tiny-tim is wrong? From http://mathworld.wolfram.com/SquareRoot.html There appears to be no difference in meaning between both notations according to wolfram. :s
  9. Q

    Unclear situation involving roots.

    hi again :D √ ± i like these signs, thanks :) how do you make them? OK, to be a bit more extreme, is there any "law book" that I could use as a proof, that the "principal value" rule (when writing just √4 it means its equal to 2) , has indeed been established? "(btw, try asking your friend...
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    Unclear situation involving roots.

    hi tiny-tim :) Thank you! So (r/6)1.5 doesn't by definition include both the positive and negative value? I have a friend who claims otherwise. He said that he wouldn't put a +- sign unless he would calculate out this expression for example if he had a value for r. (He also claims that \sqrt{4}...
  11. Q

    Unclear situation involving roots.

    Let's say we have: A=B3 r=6B2 and we want to rewrite A in terms of r. My way to do it is: B2=r/6 => B=+-\sqrt{(r/6)} => B3=+-(r/6)\sqrt{(r/6)} => A=+-(r/6)\sqrt{(r/6)} In a book I found...
  12. Q

    The answer of a 2 sig calculation 0,098 = 0,10 ?

    So what is the convention in these type of situations? Do you leave it to be 0.098 (i.e. +/- 0,0005) as was the exact result on the calculator (with all values used in the calculation having two significant figures), or do you round up to 0.10 (i.e. +/- 0,005)? Both answers are correct to two...
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    The answer of a 2 sig calculation 0,098 = 0,10 ?

    Ok guys, that should do. Thank you for helping me :)
  14. Q

    The answer of a 2 sig calculation 0,098 = 0,10 ?

    You must have raced trough my post very fast, haha. I know what significant figures are without any problem. The problem is that the solution sheet gives the answer to some question as 0,10. But I left my answer as 0,098. Which is what i got on my calculator after doing the calculation...
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