January 17

1. I'm trying to find the limit as theta approaches 0 of (1 - cos theta)/(2 sin^2 theta). Direct substitution yields an indeterminate form, so I did the following:

= 1/2 lim as theta approaches 0 of (1 - cos theta)/(1 - cos^2 theta)

Then I did something pretty sketchy involving taking the reciprocal of what I was finding the limit of so that I could get it into a form of the derivative, and then I tried doing something different a second time, but it's hardly worth transcribing all my erroneous work—can anyone point me in the right direction? The right answer is given as 1/4, but I'm having a hell of a time getting there.

2. If f(x) = sin x/2, then there exists a number c in the interval π/2 < x < 3π/2 that satisfies the conclusion for the Mean Value Theorem. I have to find a possible value for c. I tried plugging into the formula f'(c) = (f(b) - f(a))/(b - a), but that just gave me 0, since the numerator = sin 3π/4 - sin π/4 = 0. But apparently the answer is π. I can see that π = 2π/2, the mean of π/2 and 3π/2, but I thought the Mean Value Theorem had more to do with the formula I gave ... again, any help is much appreciated ... thanks, anon. —Preceding unsigned comment added by 70.19.22.49 (talk) 00:19, 17 January 2008 (UTC)[reply]

• 1 - note that ${\displaystyle \sin ^{2}\theta =1-\cos ^{2}\theta =(1-\cos \theta )(1+\cos \theta )}$
• 2 - I'm not entirely sure what exactly it is you want to show. - mattbuck 00:31, 17 January 2008 (UTC)[reply]

I think you have to show that there exists a value c between a and b such that f'(c) = (f(b) − f(a))/(b − a).  --Lambiam 01:11, 17 January 2008 (UTC)[reply]

That's what we have from the MVT. It appears that here we need to find such a c. As the OP has correctly observed, this must be a c such that f'(c)=0. All that remains is to differentiate sin(x/2) and solve. Algebraist 01:37, 17 January 2008 (UTC)[reply]

My belated gratitude. Owing to your hints, I was able to solve both problems and put them up on the board in my calc class. Thanks again! —anon —Preceding unsigned comment added by 70.18.17.227 (talk) 05:01, 20 January 2008 (UTC)[reply]

Dear Wikipedia, I am faced with a math problem and you help would be much appreciated, thanks.

y=2^(-x+3) y=(1/8)(1/(x+2))+9

Intersection, therefore y=y, solve for x.

2^(-x+3)=(1/8)(1/(x+2))+9, Note: there ARE two real solutions.

LS: 2^(-x+3) RS:(1/8)(1/(x+2))+9

   8*2^(-x)             1/(8x+16)+9
                        1/(8x+16)+(9(8x+16)/(8x+16)
                        1/(8x+16)+(72x+144)/(8x+16)
                        (1+72x+144)/(8x+16)
                        (72x+145)/(8x+16)

Combine 8*2^(-x)=(72x+145)/(8x+16) 8*2^(-x)=(72x+145)/(8(x+2) multiply both sides by 8 2^6*2^(-x)=(72x+145)/(x+2) simplify 2^(6-x)=(72x+145)/(x+2) multiply both sides by (x+2) (x+2)(2^(6-x))=72x+145 2x^(-x+6) + 2^(-x+7)-72x=145 and I'm stuck... please help, what is next step? did i mess up along the way? thanks —Preceding unsigned comment added by 99.241.96.136 (talk) 02:55, 17 January 2008 (UTC)[reply]

The equation cannot be solved by using algebra. You can instead solve it numerically by using a root-finding algorithm. The two functions of x are approximately equal when x = −0.1809; the value of y is then about 9.0687.  --Lambiam 08:44, 17 January 2008 (UTC)[reply]

Also when ${\displaystyle x\approx -1.99454,\ y\approx 31.879}$. -- Meni Rosenfeld (talk) 09:15, 17 January 2008 (UTC)[reply]

Put the equation on the form f(x)=0, rather than on the form LS(x)=RS(x). Your result 8·2^−x=(72·x+145)/(8·x+16) leads to the equation f(x)=0 where f is defined by f(x)=(72·x+145)·(2^x)−64·x−128. In addition to the two solutions given above by Lambiam and Meni Rosenfeld, there are also an infinite number of non-real complex solutions. Substitute 2^x=e^x·ln(2) and use the power series for the exponential function to obtain polynomial approximations to f(x). Solve the corresponding equations using the Durand-Kerner method. Bo Jacoby (talk) 22:08, 18 January 2008 (UTC).[reply]

What is 4.8 miles/second in metric? —Preceding unsigned comment added by 67.58.207.35 (talk) 04:37, 17 January 2008 (UTC)[reply]

Metric what? 4.8 (miles per second) = 7.7248512 kilometers per second. You can do this with google. 70.162.25.53 (talk) 04:45, 17 January 2008 (UTC)[reply]

The metric system? There really is only one way to represent that metrically (ignoring prefixes and forgetting about metric time), the question is very unambiguous. Anyway, see this for an example of how to do it with google. risk (talk) 10:33, 17 January 2008 (UTC)[reply]

I guess 70.162.25.53 just wanted to note that in the metric system you can have kilometers/second or meters/second among many others. --Taraborn (talk) 13:00, 17 January 2008 (UTC)[reply]

My [Sharp EL-506V], when calculating 3.5 x 7.6x10^-14 7.6x10^-10 displays the answer 0.000000002 BUT the actual answer is 2.66x10^-9 so it should say exactly that and at the very least, it should round properly and give 0.000000003 (still not acceptable in my opinion). Please check the output of your favourite calculator and enlighten me. :) ----Seans Potato Business 17:27, 17 January 2008 (UTC)[reply]

The result of this calculation is actually ${\displaystyle 2.66\cdot 10^{-13))$. For the value ${\displaystyle 2.66\cdot 10^{-9))$, this depends on the display mode of the calculator. The ones I know have two "normal" modes, where one of them only uses scientific notation when there is no other choice. Try changing it to the other option (I don't know how with this particular calculator). -- Meni Rosenfeld (talk) 17:38, 17 January 2008 (UTC)[reply]

Edit conflict. I was just about to come back and change 14 to 0 (typo). My calculator is in 'degree' mode but I don't suppose you're talking about that. Alobng the top it gives four modes: normal, CPLX, 3-VLE and STAT ----Seans Potato Business 17:43, 17 January 2008 (UTC)[reply]

I've looked at the manual, and it seems you need to switch to scientific notation mode by pressing "2ndF" "FSE" a few times, and then perhaps set the number of digits by pressing "2ndF" "Tab" 9. The relevant information is at the bottom-left corner of the PDF file. -- Meni Rosenfeld (talk) 17:50, 17 January 2008 (UTC)[reply]

Okay, thank you very much. I've changed it and will bear that in mind (and keep a copy of the manual). I do think that the calculator is wrong to do this, even when not in scientific mode - don't you? Does your calculator do the same thing when not in scientific mode? The answer it gives is basically incorrect in my opinion. ----Seans Potato Business 19:52, 17 January 2008 (UTC)[reply]

Yes, my nearest calculator displays the same when in "Normal 2" mode. Which part are you disagreeing with? If it's using the fixed-point notation (thus hiding some digits) then, well, that's exactly what not being in scientific mode means, and you need scientific mode for calculations where this is significant. If it's about rounding down rather than up in this case, I don't have a good answer why that is so. Note that ${\displaystyle 1+2.66\cdot 10^{-9))$ is rounded up on my calculator, perhaps in yours as well. The reason is probably technical, but again, when using fixed-point mode it is assumed that values of ${\displaystyle 10^{-9))$ or less are insignificant for you. -- Meni Rosenfeld (talk) 20:03, 17 January 2008 (UTC)[reply]

Well, I think that a well designed scientific calculator would switch to scientific display mode as soon as it meant that it could display more useful digits of the answer. Even if you disagree with this, I can see that you don't disagree regarding the incorrect rounding. Mine also rounds correctly when I add one. Having scientific mode activated all the time looks weird and makes me uncomfortable... --Seans Potato Business 20:36, 17 January 2008 (UTC)[reply]

You are basically saying the calculator should decide what is important to you and what is not. This is not entirely unplausible, but at the very least it would introduce additional levels of complexity while calculators are meant to be simple. I'd suggest you use fixed-point normally, and in those rare circumstances when it is unsatisfactory switch to scientific mode. -- Meni Rosenfeld (talk) 20:58, 17 January 2008 (UTC)[reply]

I definitely prefer my calculator telling me that 1/7 is 0.142857143 rather than 1.428571429 × 10⁻¹. —Bkell (talk) 21:53, 17 January 2008 (UTC)[reply]

Would you also prefer 0.000000002 instead of 2.66x10^-9? That would have lost me points in an exam. --Seans Potato Business 10:20, 18 January 2008 (UTC)[reply]

I don't use a calculator, but, like Meni, I would prefer mine to do what it's told and not think for itself excessively. Algebraist 11:21, 18 January 2008 (UTC)[reply]

But if you see 0.000000002, then you know there are hidden significant digits and it's "battlescientific mode" time. -- Meni Rosenfeld (talk) 12:29, 18 January 2008 (UTC)[reply]

I would of course prefer 2.66 × 10⁻⁹ to 0.000000002; I was simply making a counterargument to your claim that "a well designed scientific calculator would switch to scientific display mode as soon as it meant that it could display more useful digits of the answer." Somewhere there's a point at which scientific notation will be more useful than standard decimal notation, and that point is somewhere between the two extremes of 0.142857143 and 0.000000002. Exactly where that point should be set, though, is a matter of opinion. Like many others here, I would prefer that point to be in accordance with my own opinion, not the calculator manufacturer's opinion, so I have no problem explicitly switching the calculator to scientific mode if I want scientific notation. —Bkell (talk) 12:51, 18 January 2008 (UTC)[reply]