Function

Can someone proove btw that every symmetrical number (although, that's what I suspect) is dividable by 11? (1881, 3649229463, 635536, ...)

That's a lot of text to read What would you do with the tan and cos and the discriminant?

Yes, but acceleration isn't used in the 'trajectory' formula..? The only speed-related variable it uses is velocity, and that velocity should be 38 ms-1, which is the car's velocity at the very edge of the cliff..

Hello everyone I made myself a question about a car flying of a cliff.. The problem is: the answer seems pretty unrealistic to me: A car drives, when he's at 20 m of the edge of a cliff with a height of 147.63 m, at a velocity of 65 km/h and at a constant acceleration of 10 ms-2. Determine how far the car will land of the edge of the cliff. To do this, I'll use the formula [math]y=y_0+x\cdot\tan{\theta}-\frac{g\cdot x^2}{2v^2\cos^2{\theta}}[/math]. I will thus need the car's velocity when its on the edge of the cliff: [math]v = v_0 + a\cdot t[/math] [math] x = \frac{at^2}{2}[/math] [math]t^2 = \frac{20\cdot 2}{10}[/math] [math]t = 4s[/math] [math]v=18 ms^{-1} + 4s\cdot 10 ms^{-2}= 38 ms^{-1}[/math] [math]0 = 147.63 + x\cdot\tan{0}-\frac{g\cdot x^2}{2\cdot 38^2\cdot \cos^2{0}}[/math] [math]\frac{-g}{2\cdot 38^2}x^2+147.63=0[/math] [math]x^2=43,461.31 m[/math] [math]x=208.47 m[/math] Is this realistic? (Of course, these formulas don't include air resistance..) Thanks. -Function

Depends.. Just say what you have to say

Well, it's very simple: I want to know if trigonometric numbers & discriminants influence the signifucant figures in exercises in which the physical quantities are given (exactly). Here's a document in which I made 2 exercises myself. In that document, I assumed that discriminant & trigonometric numbers do influence the number of significant figures. Does someone feel like looking into it? (Don't mind the Dutch title. Note: "Stel" = "let ... be ..." ) Thanks! The document: http://www.mijnbestand.nl/Bestand-7R3UUY7Q8T3P.docx

Thanks! One last thing: are trigonometric numbers (sin, cos, tan) (as in the example) considered mathematical numbers, like the 2 in message #2? Or do they have an influence on the sig digs? I'd say they have an influence, for they are based on a physical quantity (angle)? And how about the discriminant of a quadratic equation? Also here, I'd say it has an influence?

Seems like the formula I have found is correct - well, according to Wolfram Alpha; for both left and right part, it gives the alternative formula: [math]\frac{1}{81}\left(10^{n-1}-1\right)^2[/math] (I secretly wished this wasn't known yet However, I'm glad 'my' formula is right.)

Now that's unsettling. Nevertheless; is 'my' formula in #5 correct? Following that formula, and keeping in mind that my calculator (TI-84 Plus) may have rounding inaccuracies, it gives for both left part and right part of the formula in message #5: 1.234567901E20

(Note that I edited the formula - it contained a small mistake) ...Trachten-what? I don't know who/what that is, so I'd say no.. I just wanted to make a general formula for squaring terms of the series 1, 11, 111, 1111, 11111, ... in function of n, with n = (the amount of ones in the value - 1). ...So I did. However I think this formula I found, i.e. [math]\left[\sum_{i=0}^{n}{10^i}\right]^2=\sum_{i=0}^{n}{\left[10^{2n-i}\left(i+1\right)+i\cdot 10^{i-1}\right]}[/math] was probably already known (or perhaps another formula if this one's incorrect), I'd like to know if it's correct Matter of testing my mathematical capacities (read: insight)

Lol.. Well, my main question now is if my last given formula is right.. (I edited my post a 'bit')?

I don't know how to ask this question, so I'll just give something; my question: is there a formula which can put this 'phenomenon' in a general form? [math]1^2 = 1[/math] [math]11^2 = 121 \rightarrow 1+2+1=4=2^2[/math] [math]111^2=12321\rightarrow 1+2+3+2+1=9=3^2[/math] [math]1,111^2=1234321\rightarrow 1+2+3+4+3+2+1=16=4^2[/math] [math]\cdots[/math] How can this be put in a general, mathematically more plausible expression? I think that this is a 'crucial' part of it: (I found it about a few minutes ago; sorry if it's wrong, haven't seen summations in school yet..) (To be honest, I'm pretty happy with the result that I found this ^^ ) [math]\left[\sum_{i=0}^{n}{10^i}\right]^2=\sum_{i=0}^{n}{\left[i\cdot 10^{i-1}\right]}+\sum_{i=0}^{n}{\left[\left(i+1\right)\cdot 10^{2n-i}\right]}[/math] And I think the sum of the numbers of the square of this is: [math]2\left[\sum_{i=0}^{n}{i}\right]+(n+1)[/math] So, my conclusion would be: [math]\forall n\in\mathbb{N}: 2\left[\sum_{i=0}^{n}{i}\right]+(n+1)=n^2[/math] So, actually, the whole first step wasn't necessary after all.. But, I personally think it's a nice formule to display the series 1, 11, 111, 1111, 11111, ... I'm pretty sure the last thing is correct; my main question (sorry that I have changed it) is now: is this formula correct? [math]\left[\sum_{i=0}^{n}{10^i}\right]^2=\sum_{i=0}^{n}{\left[i\cdot 10^{i-1}\right]}+\sum_{i=0}^{n}{\left[\left(i+1\right)\cdot 10^{2n-i}\right]}[/math] [math]\left[\sum_{i=0}^{n}{10^i}\right]^2=\sum_{i=0}^{n}{\left[10^{2n-i}\left(i+1\right)+i\cdot 10^{i-1}\right]}[/math]

Well, yes, I know... But this topic was mainly about significant figures... About uncertainty; I found out that the 'uncertain' result of a measurement is equal to the answer with no uncertainties plus or minus the most extreme uncertainties: e.g. the measurement of the surface of an A4-sheet of paper without using significant figures: I measure a width of [math]21\pm 0.1 cm[/math] and a length of [math]29.6\pm 0.1cm[/math] The surface is (if I'm correct) i.c.: [math](621.6\pm 5.07)cm^2\left(=21\cdot 29.6\pm (2.1+2.96+0.01)\right)[/math]

Hello everyone I have a question about something I really can't stand in physics: approximates (the rules which say that if you have a sum or a difference, the result may not be more accurate than the less accurate term; if you have a product or a quotient (ratio?), the number of significant figures may not be higher than the number of significant figures of the factor with the less significant figures) (I can't stand these approximates for their inaccuracy ) To exercise this, I've used a self-made example: A bullet is being shot at a height of [math]276.457 m[/math], under an angle [math]\theta[/math] of [math]0.524 rad[/math] and with an initial velocity of [math]87.78\frac{m}{s}[/math]. How much has the bullet dropped after it has flown [math]74.5m[/math] in a one-dimensional way? Using 'the formula' of the trajectory of a projectile: [math]y=y_0+x\cdot \tan{\theta}-\frac{gx^2}{2v^2\cos^2{\theta}}[/math] [math]y = 276.457m + 74.5m\cdot 0.578 - \frac{9,81\frac{m}{s^2}\cdot 74.5\cdot 74.5}{2\cdot 87.78\cdot 87.78 \cdot 0.866\cdot 0.866}[/math] Lowest accuracy is 3 significant figures in numerator and 1 significant figure in denominator [math]=276.457 + 43.1 - \frac{5.44\cdot 10^4}{1\cdot 10^4}[/math] Lowest accuracy is 1 significant figure [math]= 276.457 + 43.1 - 5[/math] Lowest accuracy is one unit [math]= 3\cdot 10^2 m[/math] Is this correct? Thanks. -Function

Thanks

Hello everyone I've seen 2 definitions for [math]e[/math] (one on the internet, one I have proven myself): [math]e=\lim_{n\to\infty}{\left[1+\frac{1}{n}\right]^n}[/math] and [math]e=\lim_{n\to 0}{\left[n+1\right]^{\frac{1}{n}}}[/math] Now, is there a proof that these two are equal? (I'd first like to know if I can change the first expression to something in the form of [math]n\to 0[/math].) Thanks! Function

Not anymore, I edited it.. Sorry for waisting your time!

Hello everyone My book gives us a proof for the rule of de l'Hôpital and to do this, its first statement is: [math]\frac{f(x)}{g(x)}=\frac{f(x)-f(a)}{g(x)-g(a)}[/math] However, I don't agree with this... Or I have no idea where that comes from... e.g.: [math]f(x)=6x^3+2x^2+4[/math] and [math]g(x)=8x^2+3x+1[/math] [math]\frac{f(3)}{g(3)}=2,243...\neq\frac{f(3)-f(7)}{g(3)-g(7)}=5,951...[/math] What's wrong with this? Thanks. Function Nope, wait.. I oversaw that [math]f(a)=g(a)=0[/math] Sorry (...again..)

Hello everyone I love to see Derren Brown (mentalist, illusionist) at work; he manipulates one's brain into believing or thinking what he wants one to believe or think; a very simplistic example I used on a friend of mine (without him knowing of it): I asked this friend to pick a random number from 0 to 10. I would guess that number. I guessed correctly: 8. Well, ok.. There's a chance of 1/10, which is relatively big. How I made him pick 8? Just say a lot of phrases in which the word/pronunciation 'eight' (~ 'ait') is hidden subtle. So, this is a very simple form of what I think is called (hypnotic) suggestion, and I was wondering if anyone knows if this is hypnotic suggestion, and if someone could give me more fun experiments I could practice Thanks! Function

Now that is what I wanted Thanks.

Because I wanted to prove it another way. What you give me is exactly that what I didn't want to see, for we had already proven it in class

Ok, new reasoning: you may assume that it is true that P is the middle of [QR], and the given coördinates for P, Q and R are right. Here's my new reasoning; I only want to know if it's plausible as proof ------- For the expression: [math]|x_q-x_p|=|x_p-x_r|[/math] [math]\Leftrightarrow \left| \frac{a\cdot \cos{t}}{1-\sin{t}} - \frac{a}{\cos{t}} \right| = \left| \frac{a}{\cos{t}} - \frac{a\cdot\cos{t}}{1+\sin{t}} \right|[/math] There are 2 solutions: 1) [math]\frac{a\cos{t}}{1-sin{t}} - \frac{a}{\cos{t}}=\frac{a}{\cos{t}}-\frac{a\cos{t}}{1+\sin{t}}[/math] [math]\Leftrightarrow \frac{\cos^2{t}}{(1-\sin{t})\cos{t}} - \frac{1-\sin{t}}{\cos{t}(1-\sin{t}} = \frac{1+\sin{t}-cos^2{t}}{\cos{t}(1+\sin{t})}[/math] [math]\Leftrightarrow \frac{1-\sin{t}+\cos^2{t}}{\cos{t}(1-\sin{t})} = \frac{1+\sin{t}-\cos^2{t}}{\cos{t}(1+\sin{t})}[/math] [math]\Leftrightarrow 1-\sin{t}+\cos^2{t}+\sin{t}-\sin^2{t}+\sin{t}\cos^2{t}=1+\sin{t}-\cos^2{t}-\sin{t}-\sin^2{t}+\sin{t}\cos^2{t}[/math] [math]\Leftrightarrow \cos^2{t}=-\cos^2{t}[/math] [math]\Leftrightarrow \cos{t}=0[/math] [math]\Leftrightarrow t=\frac{\pi}{2}+k\cdot\pi[/math] 2) [math]\frac{a\cdot \cos{t}}{1-\sin{t}} - \frac{a}{\cos{t}} = \frac{-a}{\cos{t}} + \frac{a\cdot\cos{t}}{1+\sin{t}}[/math] [math]\Leftrightarrow 1-\sin{t}=1+\sin{t}[/math] [math]\Leftrightarrow \sin{t}=0[/math] [math]\Leftrightarrow t=k\cdot\pi[/math] For the expression: [math]|y_r+y_p|=|y_q-y_p|[/math] [math]\Leftrightarrow \left| \frac{-b\cos{t}}{1+\sin{t}} + b\tan{t} \right| = \left| \frac{b\cos{t}}{1-\sin{t}}-b\tan{t} \right|[/math] There should've been 2 solutions, were it not that [math]\frac{\pi}{2}+k\cdot\pi[/math] is also a solution for [math]t[/math]. [math]\Rightarrow \frac{\pi}{2}+k\cdot\pi[/math] is not a solution for [math]t[/math]. [math]\Rightarrow[/math] if the expression [math]|y_r+y_p|=|y_q-y_p|[/math] has [math]k\cdot\pi[/math] as a solution for [math]t[/math], what had to be proven, is proven. [math]\Leftrightarrow \left| \frac{-b\cos{t}}{1+\sin{t}} + b\tan{t} \right| = \left| \frac{b\cos{t}}{1-\sin{t}}-b\tan{t} \right|[/math] [math]\frac{-b\cos{t}}{1+\sin{t}} + b\tan{t}=\frac{-b\cos{t}}{1-\sin{t}}+b\tan{t}[/math] is a solution for the equality. [math]\Leftrightarrow 1+\sin{t}=1-\sin{t}[/math] [math]\Leftrightarrow \sin{t}=0[/math] [math]\Leftrightarrow t=k\cdot\pi[/math] Both the expressions [math]|y_r+y_p|=|y_q-y_p|[/math] and [math]|x_q-x_p|=|x_p-x_r|[/math] both have [math]k\cdot\pi[/math] as a solution for [math]t[/math]. The expressions are true. I'm actually afraid that isn't a proof that the expressions are true... They only express that [math]t[/math] must be [math]k\cdot\pi[/math], which isn't the case. I'm sure my reasoning is true; where's my miscalculation?

Auto-answered: my science (chemistry and physics) teacher (is this a correct term in English? I speak Dutch..) has some test sets to get the information I required.

Now I'm embarrassed.

Thanks, Sensei. Your proof is very clear Another problem: elastic potential energy: [math]E_{pe}=\frac{k\cdot\Delta l}{2}[/math] [math]J=\frac{\frac{N}{m}\cdot m}{2}=\frac{N}{2}[/math] [math]E_{pe}=\frac{F}{2}[/math] [math]E_{pe}=\frac{m\cdot a}{2}[/math] [math]J=\frac{1}{2}\cdot kg\cdot \frac{m}{s^2}[/math] [math]J=\frac{kg\cdot m}{2s^2}\neq \frac{kg\cdot m^2}{2s^2}[/math] Where did the extra m go?