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January 16

Could someone check my work please?

"Find the equation of a tangent line to the graph of f(x) = cos x that can be used to approximate the value of cos(π/6 + 0.1). Then, find an approximation of cos(π/6 + 0.1)."

Tangent-line equation: f'(x) = -sin x

f(x₀) ≈ f'(c)(x₀ - c) + f(c)

where x₀ = the number whose f( ) value we're trying to approximate = π/6 + 0.1, and
c = a convenient number close to x₀ on the tangent line = π/6

cos (π/6 + 0.1) ≈ -(sin(π/6))(π/6)(π/6 + 0.1 - π/6) + cos(π/6)
cos (π/6 + 0.1) ≈ -(π/6)(1/2)(0.1) + (√3)/2
cos (π/6 + 0.1) ≈ -π/120 + (√3)/2
cos (π/6 + 0.1) ≈ -(π + 60)/120

But somehow it seems a little weird to have π, though it's a cornerstone of radian notation, in a y-value ... Am I on the right track? Thanks, anon.

cos (π/6 + 0.1) ≈ -(sin(π/6))(π/6)(π/6 + 0.1 - π/6) + cos(π/6)

Where did this extra factor of π/6 come from? Also, how did you manage to get rid of the √3 in the last step? —Keenan Pepper 02:14, 16 January 2008 (UTC)[reply]

Think of it this way: the slope of y = cos(x) at x = π/6 is -sin(π/6), which is -1/2. So the tangent line to y = cos(x) at x = π/6 is the line with slope -1/2 that passes through (π/6, cos(π/6)). The point-slope formula tells us that the equation of this line is

y-\cos \left({\frac {\pi }{6}}\right)=-{\frac {1}{2}}\left(x-{\frac {\pi }{6}}\right)

\Rightarrow y=\cos \left({\frac {\pi }{6}}\right)-{\frac {1}{2}}\left(x-{\frac {\pi }{6}}\right)

Now find the value of y when x = π/6 + 0.1 and you have an approximation to cos(π/6 + 0.1). Gandalf61 (talk) 07:42, 16 January 2008 (UTC)[reply]

Thanks Keenan for catching my mistakes and Gandalf for confirming my correction! —anon —Preceding unsigned comment added by 165.155.200.146 (talk) 16:11, 16 January 2008 (UTC)[reply]

About Pi

Where does the expression pi origionate from? —Preceding unsigned comment added by 207.224.29.240 (talk) 03:12, 16 January 2008 (UTC)[reply]

What do you mean, "the expression pi"? If you mean

\pi

, in its context as the ratio of a circle's circumference to its diameter, check out Pi - where the reason for its use is explained in the lead section. If you mean

\Pi

, in the context of an operator to multiply several things together, see Multiplication#Capital pi notation, and while it doesn't explain why Pi was chosen, it would be to stand for "product" just like the sigma of Sigma notation stands for "sum". For other uses of pi, see Pi (disambiguation). Confusing Manifestation(Say hi!) 03:30, 16 January 2008 (UTC)[reply]

Contraction Mapping exam question

In my Metric Spaces examination this morning, the following question was worth 4 marks out of 60:

Suppose that f and (S,p) [are a function and metric space fulfilling the requirements of] Banach's Contraction Mapping Principle and suppose that g:S->S is a function with the property that f(g(x)) = g(f(x)) for all x in S. Show that g has a unique fixed point.

Now, I was easily able to show that g has at least one fixed point (it being the same fixed point as the one f has, as guaranteed by B's CMP). However, I got stuck on the uniqueness part until I realised this:

If g is the identity on S, f(g(x)) = f(x) = g(f(x)) as required.
The identity has as many fixed points as there are points in S, because g(x) = i(x) = x by definition.

Unless I'm missing something... was the question wrong? If so, what would ensure g had a unique fixed point?

Rawling4851 12:06, 16 January 2008 (UTC)[reply]

To further complicate things, if we choose f to be

f(x)=x_{0}

and g to be any function such that

g(x_{0})=x_{0}

(possibly with many other fixed points), then the premises hold but not the conclusion. So it looks like the question is pretty much wrong. However, I think you may be able to prove something if you assume that f is injective and that g is not the identity. -- Meni Rosenfeld (talk) 13:38, 16 January 2008 (UTC)[reply]

Cheers. I ran it past our genius postgrad too and he agreed, so we'll see what the lecturer has to say... Rawling4851 00:25, 17 January 2008 (UTC)[reply]

Crossing Non-Vertical Asymptotes

In the graph of a rational function, I understand that the vertical asymptotes can never be crossed, and the one possible non-vertical asymptote can be crossed "near" the y-axis.

Is there a limit to number of times the non-vertical asymptote can be crossed?

I'm hoping that there is such a limit and that it is related to the degree of the denominator...or at least the degree of the denominator once all common factors have been eliminated.

Thanks, Stableyr —Preceding unsigned comment added by 66.100.0.42 (talk) 13:12, 16 January 2008 (UTC)[reply]

Indeed, assuming the function is not identically equal to its asymptote, the number of crossing points cannot be greater than the degree of the denominator minus 1. -- Meni Rosenfeld (talk) 13:22, 16 January 2008 (UTC)[reply]

I believe the cutoff would be either the degree of the numberator minus 1 or the degree of the denominator minus 1, whichever is greater. Take (x^2-3x+2)/x = (x-1)(x-2)/x, for instance. Its asymptote is x. The difference between the curve and the asymptote is (-3x+2)/x, which is zero precisely when x=2/3. Meni's rule would predict no intersection. Black Carrot (talk) 19:31, 16 January 2008 (UTC)[reply]

Take the function to be (ax^n + P(x))/(bx^m + Q(x)), where P(x) is a polynomial of degree less than n and Q(x) is a polynomial of degree less than m. Its asymptote will be (a/b)x^(n-m). The difference between the two will be one minus the other, (P(x) - ((a/b)x^(n-m))Q(x))/(bx^m + Q(x)). This will be zero no more often than the numerator is zero, and examples can be produced where the numerator is zero only when the denominator isn't, so we can take that as the upper bound. P(x) - ((a/b)x^(n-m))Q(x) = 0, or bP(x)x^m - aQ(x)x^n = 0, with some fiddly details if x = 0. Breaking it into cases, where m<n, m=n, and m>n, there are no more zeros than respectively n-1, n-1=m-1, and m-1. Black Carrot (talk) 19:40, 16 January 2008 (UTC)[reply]

No. The asymptote for your first example is

x-3

, not x. More generally, an asymptote is a curve whose distance from the function tends to 0. In our case we are interested in a polynomial asymptote, so when we calculate the difference all terms from

x^{m}

and up cancel, and we are left with a polynomial of degree at most

m-1

. -- Meni Rosenfeld (talk) 20:16, 16 January 2008 (UTC)[reply]

You're right, I don't know what I was thinking. Black Carrot (talk) 20:22, 16 January 2008 (UTC)[reply]

I do. You were thinking about the notion of asymptoticity that only requires the ratio of the functions to tend to 1. Unfortunately, this doesn't fit the context of the question. -- Meni Rosenfeld (talk) 20:33, 16 January 2008 (UTC)[reply]