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May 12

Graphic calculator.

Hey, I've never used a graphic calculator before but I really need one because I'm not understanding maths and need to learn a lot really quickly and so hopefully it'll be able to help learn, and in the exams soon. Any calculator is allowed. Hopefully some of you are students, or use calculators, and can help. I need a calcualtor that I can put equasions in, in forms y=mx+c, y=ax^2+bc+c and that kind of stuff, and use to solve equasions, and normal calculatory stuff. Which is best? There are so many! So which is best TI-84 Plus series, TI-89 series, TI-Nspire, HP-50G, Sharp EL9900, or any other? Also on Wikipedia:Reference_desk/Miscellaneous Thanks for any help you can provide. 86.178.214.25 (talk) 00:20, 12 May 2010 (UTC)[reply]

You are right that a graphing calculator (or any software function plotter) is often a great aid in understanding functions. One interesting option that you did not include in your list is the Casio ClassPad. But if you are at home and need a function plotter, a Windows/Linux/Mac application is far more usable than a pocket calculator. You might be interested in professional (read: "expensive") tools such as Mathematica, Maple, or Matlab or free alternatives such as Maxima or AlgoSim.

When it comes to the calculators in your list, the major difference between the 84 and the 89 series is that only the latter is able to do symbolic calculations, i.e. algebraically manipulate furmulae. For instance, it can solve equations algebraically (find an expression for x as a function of the other quantities in the equation) in addition to numerically (find a numerical approximation to one of the roots), do simplification, factorization, symbolic differentiation/integration etc. --Andreas Rejbrand (talk) 08:46, 12 May 2010 (UTC)[reply]

[ec] The professional tools are available at a great discount for students (I think the cost is around $100-200). You can also use Wolfram Alpha - here's an example. Note that calculators and programs are great if used properly, but if you use them to do the thinking for you instead of learning how to do things yourself, they'll only hurt your understanding. -- Meni Rosenfeld (talk) 08:59, 12 May 2010 (UTC)[reply]

Yes, Wolfram|Alpha is really great. It is free, available from any Internet-connected computer, and you can ask it almost anything (it does not even need to be about math). Try it if you haven't already. --Andreas Rejbrand (talk) 09:11, 12 May 2010 (UTC)[reply]

...but if you use Wolfram|Alpha for function plotting, you need to be careful when you interpret the graphs. Sometimes Wolfram|Alpha chooses peculiar and unintuitive axes. One time, for instance, I was confused for several minutes before I realized that the horizontal axis was not positioned at y = 0, but at y = 1... For a few minutes I thought the function had a zero at the origin, where it in fact was equal to 1. --Andreas Rejbrand (talk) 09:15, 12 May 2010 (UTC)[reply]

Prove that the tangent to the x value which is the average of two zeros of a cubic passed through the third zero.`

Suppose you are given two zeros for a cubic $y=ax^{3}+bx^{2}+cx+d$ . Call these values $m$ and $n$ . The tangent at the point ${\frac {m+n}{2}}$ on the cubic passes through the third zero. My goal is to prove that this is true.
The slope at this point would be $3a({\frac {m+n}{2}})^{2}+2b{\frac {m+n}{2}}+c$ (taking the derivative of the cubic and sub (m+n)/2 for x)
The y value would be $a({\frac {m+n}{2}})^{3}+b({\frac {m+n}{2}})^{2}+c({\frac {m+n}{2}})+d$
Therefore the equation for the tangent would be $y-(a({\frac {m+n}{2}})^{3}+b({\frac {m+n}{2}})^{2}+c({\frac {m+n}{2}})+d)=(3a({\frac {m+n}{2}})^{2}+2b{\frac {m+n}{2}}+c)$
and when y = 0, obviously $-(a({\frac {m+n}{2}})^{3}+b({\frac {m+n}{2}})^{2}+c({\frac {m+n}{2}})+d)=(3a({\frac {m+n}{2}})^{2}+2b{\frac {m+n}{2}}+c)$
If I rearrange that equation to the form "x = " and then sub the right hand side of that equation into x for $y=ax^{3}+bx^{2}+cx+d$ , I expect that that expression will evaluate to y=0 if the conjecture is true. However, I have been unsuccessful in doing so. I know that if I encounter the expression $am^{3}+bm^{2}+cm+d$ or $an^{3}+bn^{2}+cn+d$ then I can replace that with 0.
Could I please have some help with my problem? --Alphador (talk) 09:12, 12 May 2010 (UTC)[reply]

Your slope should be

3a\left({\frac {m+n}{2}}\right)^{2}+2b\left({\frac {m+n}{2}}\right)+c

and the equation for your tangent should be

y-a\left({\frac {m+n}{2}}\right)^{3}-b\left({\frac {m+n}{2}}\right)^{2}-c\left({\frac {m+n}{2}}\right)-d=\left(3a\left({\frac {m+n}{2}}\right)^{2}+2b\left({\frac {m+n}{2}}\right)+c\right)\left(x-\left({\frac {m+n}{2}}\right)\right)

Gandalf61 (talk) 09:35, 12 May 2010 (UTC)[reply]

Oops that was a typing error. I meant to write 3a, not 3a^2--Alphador (talk) 09:53, 12 May 2010 (UTC)[reply]

and I forgot to put in brackets--Alphador (talk) 09:58, 12 May 2010 (UTC)[reply]

Put it this way. Let x₀ be a zero of the third degree polynomial p(x). Then, for any λ, x₀ is still a zero of the polynomial p_λ(x):=p(x)-λ(x-x₀). Moreover, the constant λ does not affect the third and second degree coefficients of p_λ(x). Therefore the sum of the zeros is constant, for all values of λ. Now, assume that λ is such that the other two zeros of p_λ(x) coincide. What should be the double zero z ? What does it mean, geometrically, that the equation p_λ(x)=0, that is, p(x)=λ(x-x₀), has solutions x₀ and z with multiplicity 2, in terms of the intersection of the graph of y=p(x) and the graph of y=λ(x-x₀)? --pm a 10:02, 12 May 2010 (UTC)[reply]

The double zero of the polynomial p_λ(x) would be the average of the other two zeros of p(x) I guess? Hmm... don't really see how that's helpful sorry :P--Alphador (talk) 10:52, 12 May 2010 (UTC)[reply]

well, that's because you haven't answered the second question :) I have reedited and rephrased it. You do not have to compute anything (but to know some basic little things: I do not know what's your background). Your first answer is ok, c'mon just 50% to go!--pm a 11:23, 12 May 2010 (UTC)[reply]

P.S. answer: it means, of course, that the line passes through x₀ and it is the tangent to the graph of p(x) at z=(n+m)/2, QED. No computation. --pm a 21:29, 12 May 2010 (UTC)[reply]

Yet another approach. Let m, n and q be the zeros of your cubic:

y=(x-m)(x-n)(x-q)

y^{\prime }=(x-n)(x-q)+(x-m)(x-q)+(x-m)(x-n)

h=(m+n)/2

Then:

y(h)=(n-m)(m-n)(h-q)/4

{\begin{aligned}y^{\prime }(h)&=(m-n)(h-q)/2+(n-m)(h-q)/2+(n-m)(m-n)/4\\&=(n-m)(m-n)/4\end{aligned}}

Now it is obvious

y(h)=y^{\prime }(h)\cdot (h-q)

which means the tangent passes through (q,0), what you were to prove.
--CiaPan (talk) 12:53, 12 May 2010 (UTC)[reply]

How does

y(h)=y^{\prime }(h)\cdot (h-q)

prove that the tangent passes through (q,0)?--115.178.29.142 (talk) 22:19, 12 May 2010 (UTC)[reply]

See definition I began with: q is one of the cubic's roots, so

y(q)=0

. That means for argument difference

\Delta x=h-q

we have the function value difference

\Delta y=y(h)-y(q)=y(h)-0=y(h)

, so the slope of a secant line, passing through

(h,y(h))

and the 'third zero'

(q,0)

is

\Delta y/\Delta x=y(h)/(h-q)

and that equals the tangent slope

y^{\prime }(h)

. So the secant line and tangent line are the same, which means tangent line passes the third zero.
All of that assuming we can divide by

(h-q)

, that is the third zero does not happen to be the mean of the first and second one. In that case, however, the answer would be immediately 'yes', without any calculations.

CiaPan (talk) 06:13, 13 May 2010 (UTC)[reply]

Actually the answer is immediate without calculation in any case; check my answer above. --pm a 07:22, 13 May 2010 (UTC)[reply]

What mathematical operation to you perform on a number to reverse its digits?

e.g
$f(10428)=82401$
$f(102000)=201$
define f(x)
--Alphador (talk) 10:04, 12 May 2010 (UTC)[reply]

\sum _{n=0}^{\lfloor \log x\rfloor }10^{n}*\left(\left\lfloor {\frac {x}{10^{n}}}\right\rfloor -10*\left\lfloor {\frac {x}{10^{n+1}}}\right\rfloor \right)

Should do the trick. I don't think there is a closed form. Taemyr (talk) 10:20, 12 May 2010 (UTC)[reply]

LOL that's just like f(82401) = 1 + 0 + 400 + 2000 + 80000 haha... pretty useless but thanks anyway :)--Alphador (talk) 10:57, 12 May 2010 (UTC)[reply]

Well, yeah. Reversing the digits is not an operation which is intrinsic to the number, it depends on the way you represent it (e.g., if you used another base the results would be different). So you shouldn't expect some magical formula that gives this result without explicitly decomposing the number to its digits. -- Meni Rosenfeld (talk) 11:04, 12 May 2010 (UTC)[reply]

This gives back x. It should be

\sum _{n=0}^{\lfloor \log x\rfloor }10^{\lfloor \log x\rfloor -n}*\left(\left\lfloor {\frac {x}{10^{n}}}\right\rfloor -10*\left\lfloor {\frac {x}{10^{n+1}}}\right\rfloor \right)

.

-- Meni Rosenfeld (talk) 10:59, 12 May 2010 (UTC)[reply]

Doh! Taemyr (talk) 11:07, 12 May 2010 (UTC)[reply]

Functions help

"Find f(-3) if f(x) = x^2"

I keep getting -9 but the answer should be 9. What am I missing?

"Select the points that lie on the function h(x) = 3x^2"

Available options are (1, 3), (1, 9), (-1, -3), and (-1, 3). I hate to say it but I'm completely lost here.

--Glaesisvellir (talk) 17:20, 12 May 2010 (UTC)[reply]

Calculating

f(x)

means taking the number x and squaring it. I'm guessing that what you've written is

f(-3)=-3^{2}=-9\;\!

. However,

-3^{2}

is not taking the number -3 and squaring it - it is taking the number 3, squaring it, and taking the minus of that. The operation of taking the minus needs to be done before the squaring, and to clarify this we put the -3 in parentheses. Thus you have

f(-3)=(-3)^{2}=(-3)\cdot (-3)=9

.

For the second question - the graph of a function is the collection of all points of the form

(x,y)

where

y=h(x)

. Thus, a point

(x,y)

is on the graph of h if and only if

y=h(x)

. Can you solve the problem now? -- Meni Rosenfeld (talk) 17:35, 12 May 2010 (UTC)[reply]

Nowhere in my schoolwork can I find where it's specified that I have take the minus? I'm not criticising you or anything, it's just I'm still confused. There have been other similar problems that haven't wanted the minus. I should also note that I had no way of knowing that the problem wanted the minus of the answer until I saw the answer, as it was a multiple choice. --Glaesisvellir (talk) 18:11, 12 May 2010 (UTC)[reply]

The question was to find

f(-3)

. -3 is minus 3. The minus is a part of the question. There's no extra minus involved - the function f squares a number, which means multiplying it by itself. If the number is -3, you need to multiply -3 by -3.

Can you specify exactly which part (or parts) of my answer you didn't understand? Can you also give some examples of similar problems that you are able to solve? -- Meni Rosenfeld (talk) 18:28, 12 May 2010 (UTC)[reply]

-3 times -3 is indeed. I think it may be an issue with my calculator. (a Casio fx-115ES). Punching in -3^2 into it gives me -9, but -3(-3) (or any other way to multiply) gives me the correct answer, 9. I don't know if you would be able to help me past here, but thanks in advance if you can. As for the other problem, do you have ideas? I've tried graphing it but I'm still clueless. I know what they're asking but I don't know how to find the answer (and I can't seem to find information on how for some reason). --Glaesisvellir (talk) 22:11, 12 May 2010 (UTC)[reply]

Your second question gives you a choice of (x,y) options as solutions to the expression y=3x². All you do is plug in the x and see if you get the y. The expression works for (1,3), that is, x=1, y=3, because 3=3×1². It also works for (-1,3) but not for any of the others. ~Amatulić (talk) 22:25, 12 May 2010 (UTC)[reply]

"Glaesisve", you're misusing your calculator. You calculator is correctly telling you that

-3^{2}=-9.\,

The calculator is right, but you are wrong. What you should be looking for is

(-3)^{2},{\text{ not }}-3^{2}.\,

Michael Hardy (talk) 00:17, 13 May 2010 (UTC)[reply]

sigh* There I go again forgetting basic rules... I wish I wasn't braindead. Thank you all so much. --Glaesisvellir (talk) 02:07, 13 May 2010 (UTC)[reply]

In fact, it is generally not a good idea to write down your calculator's answer in an exam, if you feel that you can logically reason the truth of another answer. As Michael Hardy mentions, the calculator can only output what you tell it to output. Often in exams, especially in those involving sophisticated calculators, it is extremely easy to input an completely different expression to that you wish to input, and write down a wrong answer. That is why, even with the calculator at hand, you should always trust yourself rather than trust what you see on your calculator, especially if you are confident that you know the theory well. It is only too easy to input −3² instead of (−3)² in your calculator.

With regards to "forgetting basic rules", it is important to note that if, when confronted with a basic rule, you display a real attempt to understand the rule, then you will not forget it as easily. For instance, in "advanced mathematics", when a mathematician proposes a new result, he/she must supplement the result with what is known as a "formal proof"; the logical deduction of the result from certain other true results. Furthermore, when one reads a mathematics textbook (let us assume, a university-level mathematics textbook), he/she will frequently encounter proofs of many results within the textbook. I doubt that there is a mathematician who remembers, word-by-word, all the proofs he/she has studied. Indeed, a professional mathematician could easily "reproduce" the proofs without remembering them word-by-word, simply because they would have actually understood the ideas the first time they learnt them. Even in basic mathematics (like high-school mathematics), the principle remains the same; you should never be content with what you read if you do not feel that you understand it thoroughly. And by "understand", I do not just mean mere memorization. Understanding a textbook means, that in the event a knowledgeable person were to engage in a 1 hour discussion about the textbook material with you, you would be able to explain the material and reply to the person's questions with fluency. When you read anything (whether it is mathematics, literature, science etc.), you should ask yourself whether you can do exactly this. PS T 02:36, 13 May 2010 (UTC)[reply]

It may also be useful to listen to a new piece of music each time you attempt to understand a new set of results. For some people (and I have found, for myself), the music associated to a particular result sticks in your head whenever you attempt to recall the result later on, and this aids your ability to remember. PS T 02:42, 13 May 2010 (UTC)[reply]

This may vary between calculators, but the minus key is really for subtraction. To enter "minus three" on some calculators, you should enter 3 then the "+/-" key to give a negative three in the display. Squaring this should give a positive nine. Dbfirs 18:38, 13 May 2010 (UTC)[reply]

I think this applies mostly to older (or more primitive) calculators. These days scientific calculators usually have a line where you can input the calculation. In the one I have here, there are separate keys for subtraction and negation, but typing

-3^{2}

using either gives -9. They also have parentheses which obviously fix the problem. You can also type -3 and then Ans². -- Meni Rosenfeld (talk) 06:31, 14 May 2010 (UTC)[reply]

Poisson distribution, but where events are not independent of the time of the last event

I'm trying to model some data that I have where events may occur 0 or 1 times on a given day over many years. However, I believe that the occurrence (1) of an event is not independent of the time of the last time it occurred. (Or in other words, it's not memoryless.) What other distributions or processes should I be looking at? Renewal Theory and processes look like a good start. Any suggestions or pointers (especially to code in R or other languages) would be greatly appreciated. Thank you. --Rajah (talk) 18:44, 12 May 2010 (UTC)[reply]

Markov chain? 69.228.170.24 (talk) 23:42, 12 May 2010 (UTC)[reply]

Yeah, makes sense. Thanks! --Rajah (talk) 20:49, 13 May 2010 (UTC)[reply]

y=x^x

For some quirky reason, I've always been fascinated with the equation y=x^x, and the graph it forms, especially as it approaches 0 from either side. Does this equation show up in any mathematical discussions, proofs, practical applications, etc., or is it just a curious expression? — Michael J 22:18, 12 May 2010 (UTC)[reply]

I've not seen it come up myself. I find it interesting also that this function has a minimum at x=1/e. ~Amatulić (talk) 22:30, 12 May 2010 (UTC)[reply]

Tetration perhaps. 76.229.213.156 (talk) 22:56, 12 May 2010 (UTC)[reply]

Certainly it's commonplace in calculus textbooks for this to appear in an exercise challenging the student to find its extreme values. It has a single absolute minimum for x > 0. Michael Hardy (talk) 00:11, 13 May 2010 (UTC)[reply]

You might like to know that the Lambert W function crops up when you invert y=x^x. Enjoy, Robinh (talk) 07:37, 13 May 2010 (UTC)[reply]

You might be interested in Sophomore's dream, and also Exponentiation#Zero to the zero power. -- Meni Rosenfeld (talk) 08:13, 13 May 2010 (UTC)[reply]

I've seen it in thermodynamics (applied science) - specifically statistical thermodynamics - counting numbers of arrangements - such as permuatations and combinations - coming from approximations of n! Stirling's approximation - to calculate large permuations - such as found in theoretical entropy calculations (integers only).77.86.10.27 (talk) 15:56, 18 May 2010 (UTC)[reply]

Integration help

Hey guys. I have a figure which is formed by a square with a side length of 3, and a quadrant of a circle with radius of 3 as well. The quadrant and square overlap, such that the radii forming the sides of the quadrant overlap with two adjacent sides of the square. I want to find the area of the part of the square NOT covered by the quadrant. I need to use integration to do this. I'm did the following steps, but putting it into my calculator gives me something diferent. Any idea where I go wrong? I think I set the bounds wrong, but I'm not sure how to do that.
The equation of the circle is $(x-3)^{2}+(y-3)^{2}=9$
Since I'm only interested in the part that curves up, I can change it to
$y=-{\sqrt {9-(x-3)^{2}}}+3$
The $(x-3)$ will just cause me trouble later on. I can discard -3 since I don't need it; I can just set the bounds different.
$y=-{\sqrt {9-x^{2}}}+3$
Now's the fun part: I need to integrate this function from -3 to 0.
$\int _{-3}^{0}ydx=\int _{-3}^{0}(-{\sqrt {9-x^{2}}}+3)dx$
I can take the -1 out because it is simply multiplication, and separate it into two integrals.
$\int _{-3}^{0}ydx=-\int _{-3}^{0}{\sqrt {9-x^{2}}}dx+\int _{-3}^{0}(3)dx$
$\int _{-3}^{0}(3)dx$ evaluates immediately to 9, I think. I factor 9 out of the first integral.
$\int _{-3}^{0}ydx=-3\int _{-3}^{0}{\sqrt {1-{\frac {x^{2}}{9}}}}dx+9$
I substitute: ${\theta }=arcsin{\frac {x}{3}}$ This is most likely where I rechose my bounds wrong
$\int _{-3}^{0}ydx=-3\int _{\frac {\pi }{2}}^{0}{\sqrt {1-sin^{2}{\theta }}}d{\theta }+9$
And obviously
$\int _{-3}^{0}ydx=-3\int _{\frac {\pi }{2}}^{0}(cos^{2}{\theta })d{\theta }+9$
$\int _{-3}^{0}ydx=-3\int _{\frac {\pi }{2}}^{0}{\frac {1+cos2{\theta }}{2}}d{\theta }+9$
$\int _{-3}^{0}ydx=-3({\frac {\theta }{2}}{\bigg |}_{\frac {\pi }{2}}^{0}+{\frac {sin2{\theta }}{2}}{\bigg |}_{\frac {\pi }{2}}^{0})+9$
${\frac {sin2{\theta }}{2}}{\bigg |}_{\frac {\pi }{2}}^{0}$ clears out so I'm lef with
$\int _{-3}^{0}ydx={\frac {-3{\pi }}{2}}+9$
But this isnt what I get when I do it in my calculator. Where di I go wrong?

You neglected the fact that if

{\frac {x}{3}}=\sin \theta

then

dx=3\cos \theta \,d\theta .

You also changed

{\sqrt {1-\sin ^{2}\theta }}\,

to cos²θ, thereby neglecting the radical. It should have been simply cos θ. Michael Hardy (talk) 00:33, 13 May 2010 (UTC)[reply]

...or maybe you did write

dx=3\cos \theta \,d\theta

but then wrote dx above where you should have had dθ. And forgot the 3. Also, looks like π/2 appears where π/4 should be.

I'll look again later when I'm not as rushed. Michael Hardy (talk) 00:37, 13 May 2010 (UTC)[reply]

OOPS! I completely forgot to put in dθ for dx after I substituted! But Likely this is more of an issue of transcription than what's really wrong. But It's interesting about the dx becoming cos θ dθ. Why does it??

With free manipulation of increments, the reason is that if

\theta =\arcsin {\frac {x}{3}}

, then

\sin(\theta )={\frac {x}{3}}

, so that

3\sin(\theta )=x

and x is a function of

\theta

. Consequently, we can differentiate x with respect to

\theta

, and we obtain that

{\frac {dx}{d\theta }}=3\cos(\theta )

. Therefore,

dx=3\cos(\theta )d\theta

. PS T 02:52, 13 May 2010 (UTC)[reply]

OK, I think I found my mistake (it was forgetting that after substituting, dx becomes 3cos(θ) dθ). I would copy my new work, but i'm sleepy and the TeX is getting messy and confusing ;). Thanks everybody. I would makr this one with the resolved thingy but I'm not sure how to do that (something like ((resolved))?). 76.229.213.156 (talk) 03:06, 13 May 2010 (UTC)[reply]

Tangent to a cubic

Consider a cubic $f(x)=a(x-m)(x-n)(x-q)$
It can be shown that $f({\frac {m+n}{2}})=f^{\prime }({\frac {m+n}{2}})\cdot ({\frac {m+n}{2}}-q)$
Apparently this equation shows that the tangent to the point ${\frac {m+n}{2}}$ of the cubic $f(x)$ passes through (0, q), but how? --Alphador (talk) 23:12, 12 May 2010 (UTC)[reply]

Hint. What is the relation of the coefficient of x^2 of a cubic to the roots of that cubic? Note that equating the tangent to the cubic yields another cubic equation with a double root at the point where it touches the original cubic. Also, the new cubic has the same x^3 and x^2 terms as the old cubic. Count Iblis (talk) 23:44, 12 May 2010 (UTC)[reply]

a(x-m)(x-n)(x-q)=ax^{3}-(am+an+aq)x^{2}+(amn+amq+anq)x-amnq

So the x^2 coefficient is equal to the sum of its roots times –a.--Alphador (talk) 00:10, 13 May 2010 (UTC)[reply]

So

a({\frac {m+n}{2}})^{3}-(am+an+aq)({\frac {m+n}{2}})^{2}+(amn+amq+anq){\frac {m+n}{2}}-amnq=(3a({\frac {m+n}{2}})^{2}-2(am+an+aq)({\frac {m+n}{2}})+(amn+amq+anq))\cdot ({\frac {m+n}{2}}-q)

—Preceding unsigned comment added by Alphador (talk • contribs) 00:18, 13 May 2010 (UTC)[reply]

Ok, so the coefficient of x^2 equals the sum of the rootstimes minus a. Forthe cubic tis equals -a(m + n + q). Now, if you equate your cibic to the equation of the tangent line, that equation will be another cubic, but this has the same valyue for a (the coefficient of x^3) and the coefficient of x2 is also the same. So, the sum of the roots of that equation times minus a will also be equal to -a(m + n + q). But, by cponstrution, this equation is known to have a double root located at x = (m + n)/2. Minus a times the sum of these two roots is thus equal to -a(m+n). So, the third root must be at x = q. Count Iblis (talk) 00:45, 13 May 2010 (UTC)[reply]

Do you mean the point (q, 0)? The statement as written doesn't work. I'm going to write c = (m+n)/2 just for readability. The tangent line has a slope of f'(c) and passes through the point (c, f(c)) so the claim that is also passes through (q, 0) would be equivalent to the statement that f'(c) = Δy/Δx = (f(c) - 0)/(c - q) which is what you've got. Rckrone (talk) 23:39, 14 May 2010 (UTC)[reply]