HSC 2015 MX2 Integration Marathon (archive) (4 Viewers)

glittergal96 · Jul 9, 2015

Re: MX2 2015 Integration Marathon

A kind of open-ended question slightly related to my second method above:

Explain why Simpson's rule with equal spacing is "better" as a tool to numerically approximate definite integrals of smooth functions (functions with continuous derivatives of all orders) than the trapezoidal rule with equal spacing.

Try to use calculus and rigorous statements to justify your explanation, rather than just saying vague qualitative things like "quadratics can approximate curves better than lines".

mrpotatoed · Jul 9, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$let x= \frac{1}{u}$

$I=\int_{\frac{1}{t}}^{t}{\frac{\tan^{-1}(\frac{1}{u})}{u}}$

By dummy variables:

$2I = \int{\frac{1}{x}(tan^{-1}x + tan^{-1}\frac{1}{x}})$

$I= {\frac{\pi}{4}}}\int_{\frac{1}{t}}^t{\frac{1}{x}}$

$I=\frac{\pi ln(t)}{2}$

[]

May I ask what made you think of using a substitution for this? Being the noob I am I would probably try to do integration by parts on that...

Drsoccerball · Jul 9, 2015

Re: MX2 2015 Integration Marathon

mrpotatoed said:
May I ask what made you think of using a substitution for this? Being the noob I am I would probably try to do integration by parts on that...

Look at the boundaries

seanieg89 · Jul 10, 2015

Re: MX2 2015 Integration Marathon

Good stuff glittergal, dyadic decomposition is exactly what I wanted someone to come up with for estimating that integral. (And your exponent is the correct one).

.

porcupinetree · Jul 10, 2015

Re: MX2 2015 Integration Marathon

Next Q:

$\int \frac{dx}{\sqrt{x} + \sqrt[3]{x}}$

Ekman · Jul 10, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Next Q:

$\int \frac{dx}{\sqrt{x} + \sqrt[3]{x}}$

This was asked before. Just use the substitution: x=u^6

VBN2470 · Jul 10, 2015

Re: MX2 2015 Integration Marathon

^ Use the substitution $x=u^6$ to simplify your integral to $6\int u^2-u+1-\dfrac{1}{1+u} \text{ d}u$ .

Ekman · Jul 10, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
^ Use the substitution $x=u^6$ to simplify your integral to $6\int u^3-u^2+u-1-\dfrac{1}{1+u} \text{ d}u$ .

Really? I got something different:

$6\int \frac{u^3}{u+1}du = 6\int \frac{u^3 +1 - 1}{u+1} du = 6\int u^2 - u +1 -\frac{1}{u+1}du$

VBN2470 · Jul 10, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Really? I got something different:

$6\int \frac{u^3}{u+1}du = 6\int \frac{u^3 +1 - 1}{u+1} du = 6\int u^2 - u +1 -\frac{1}{u+1}du$

You're right, I had an extra u factor.

leehuan · Jul 11, 2015

Re: MX2 2015 Integration Marathon

$\int { \sin ^{ -1 }{ \sqrt { x } } \cos ^{ -1 }{ \sqrt { x } } dx }$

glittergal96 · Jul 13, 2015

Re: MX2 2015 Integration Marathon

Here is one of the proofs I had in mind for my factorial question:

$First, we look at the error in single trapezoidal approximations.\\ \\Let $f:[a,b]\rightarrow \mathbb{R}$ be twice-differentiable with $|f''(x)|\leq M$. \\ Let $l:[a,b]\rightarrow\mathbb{R}$ be the trapezoidal approximation, given explicitly by $l(x)=f(a)+\frac{f(b)-f(a)}{b-a}(x-a)$. \\ Let $s(x)=a+x(b-a)$.\\ (The point of $s$ is to move everything to the interval $[0,1]$ for the sake of simplicity.) \\ \\ Now define $g(x):=f(s(x))-l(s(x))$.\\ We have:\\ \\$E=\left|\int_a^b f(x)-l(x)\, dx\right|\\ = (b-a)\left|\int_0^1 g(x)\, dx\right| \\ = \frac{b-a}{2}\left|\int_0^1 \frac{d^2}{dx^2}(x(1-x))g(x)\,dx\right|\\ = \frac{b-a}{2}\left|\int_0^1 x(1-x)g''(x)\, dx\right|\\ \leq \frac{b-a}{a}\int_0^1 x(1-x)|g''(x)|\, dx\\ = \frac{b-a}{2}\int_0^1 x(1-x)|g''(x)|\, dx\\ \leq \frac{(b-a)^3M}{2}\int_0^1 x(1-x) \, dx \\=\frac{(b-a)^3M}{12}.$$

$Now let's use this. \\Since $\log$ is concave, trapezoidal approximations $\mathcal{T}_n$ will be under-approximations to the integrals $\int_1^n \log(t)\, dt$. \\ \\So we have:\\ \\ $\log(n!)=\sum_{k=1}^n \log(k)\\ = \frac{\log(n)+\log(1)}{2}+\mathcal{T}_n\\ \leq \frac{\log(n)}{2}+ \int_1^n \log(t)\, dt \\ = \frac{\log(n)}{2}+[t\log(t)-t]^n_1\\ = \frac{\log(n)}{2}+n\log(n)-n+1.$$

$From our earlier error analysis, we also have:\\ $\sum_{k=1}^n \log(k) \geq \frac{\log(n)}{2}+n\log(n)-n+1-\frac{1}{12}\sum_{k=1}^{n-1} \max_{x\in[k,k+1]} |f''(x)|$\\ $=\sum_{k=1}^n \log(k) \geq \frac{\log(n)}{2}+n\log(n)-n+1-\frac{1}{12}\sum_{k=1}^{n-1} \frac{1}{k^2}\\ \geq \sum_{k=1}^n \log(k) \geq \frac{\log(n)}{2}+n\log(n)-n+1-\frac{\pi^2}{72}$\\ \\ Finally exponentiating gives: \\ \\ $e^{1-\pi^2/72}\leq \frac{n!}{e^{-n}n^{n+1/2}} \leq e.$$

(Of course, we could take much cruder upper bounds of the Basel sum if we do not want to assume the well known series evaluation. For example we could use an integral to bound that sum too! The main point is that it is finite.)

mrpotatoed · Jul 14, 2015

Re: MX2 2015 Integration Marathon

http://integrals.wolfram.com/index.jsp?expr=(x*(sqrt(1-(x-1)^2)&random=false

^ can anyone help with that? Not sure how to sure the latex thingy.

that will show the equation in full, but doesn't include the upper limit of 2 and lower of 0. Any help in working it through appreciated.

nvm... got it.. will leave it up though anyway if anyone wants to try themself

Drsoccerball · Jul 14, 2015

Re: MX2 2015 Integration Marathon

mrpotatoed said:
http://integrals.wolfram.com/index.jsp?expr=(x*(sqrt(1-(x-1)^2)&random=false

^ can anyone help with that? Not sure how to sure the latex thingy.

that will show the equation in full, but doesn't include the upper limit of 2 and lower of 0. Any help in working it through appreciated.

let u=x-1

InteGrand · Jul 14, 2015

Re: MX2 2015 Integration Marathon

mrpotatoed said:
http://integrals.wolfram.com/index.jsp?expr=(x*(sqrt(1-(x-1)^2)&random=false

^ can anyone help with that? Not sure how to sure the latex thingy.

that will show the equation in full, but doesn't include the upper limit of 2 and lower of 0. Any help in working it through appreciated.

$Let $I = \int _0 ^2 x\sqrt{1-(x-1)^2}\text{ d}x$.$

$Let $u=x-1$, then $x=1+u$, $\mathrm{d}x=\mathrm{d}u$, $x=0\Rightarrow u = -1$, $x=2\Rightarrow u = 1$, so$

$I=\int _{-1}^{1}(1+u)\sqrt{1-u^2}\text{ d}u$

$$\Rightarrow I = \int_{-1}^{1} \sqrt{1-u^2}\text{ d}u +\int_{-1}^{1}u \sqrt{1-u^2}\text{ d}u$.$

$The second integral above is clearly $0$ since the integrand is an odd function, and the first integral is $2 \int_{0}^{1} \sqrt{1-u^2}\text{ d}u$ since the integrand is even, so we have $I=2\int _0 ^1 \sqrt{1-u^2}\text{ d}u$. This can be evaluated using either integration by parts (integrate 1 and differentiate $\sqrt{1-u^2}$) or a trigonometric substitution (e.g. $u=\sin \theta$, $0\leq \theta \leq \frac{\pi}{2}$).$

Drsoccerball · Jul 14, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$Let $I = \int _0 ^2 x\sqrt{1-(x-1)^2}\text{ d}x$.$

$Let $u=x-1$, then $x=1+u$, $\mathrm{d}x=\mathrm{d}u$, $x=0\Rightarrow u = -1$, $x=2\Rightarrow u = 1$, so$

$I=\int _{-1}^{1}(1+u)\sqrt{1-u^2}\text{ d}u$

$$\Rightarrow I = \int_{-1}^{1} \sqrt{1-u^2}\text{ d}u +\int_{-1}^{1}u \sqrt{1-u^2}\text{ d}u$.$

$The second integral above is clearly $0$ since the integrand is an odd function, and the first integral is $2 \int_{0}^{1} \sqrt{1-u^2}\text{ d}u$ since the integrand is even, so we have $I=2\int _0 ^1 \sqrt{1-u^2}\text{ d}u$. [b]This can be evaluated using either integration by parts (integrate 1 and differentiate $\sqrt{1-u^2}$) or a trigonometric substitution (e.g. $u=\sin \theta$, $0\leq \theta \leq \frac{\pi}{2}$).$

Or using area of a quarter of a circle with radius 1

porcupinetree · Jul 14, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Or using area of a quarter of a circle with radius 1

That's one of the most enjoyable moments, using that trick/shortcut.

Here's a cool reduction formula question:

If $I_{n} = \int sec^{n}x dx$

Find a reduction formula for $I_{n}$ and hence, find $\int_{0}^{\frac{\pi}{4} } sec^{7}x dx$

glittergal96 · Jul 14, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
That's one of the most enjoyable moments, using that trick/shortcut.

Here's a cool reduction formula question:

If $I_{n} = \int sec^{n}x dx$

Find a reduction formula for $I_{n}$ and hence, find $\int_{0}^{\frac{\pi}{4} } sec^{7}x dx$

Too lazy to do the calculation at the end, but the reduction formula for n >= 2:

$\int \sec^n(x)\, dx \\ = \int \sec^{n-2}(x)\frac{d}{dx}(\tan(x))\, dx \\ = \sec^{n-2}(x)\tan(x)-(n-2)\int \tan^2(x)\sec^{n-2}(x)\, dx \\ \\ \Rightarrow I_n=\frac{\sec^{n-2}(x)\tan(x)+(n-2)I_{n-2}}{n-1}.\\ \\ $Keeping in mind that each of the $I_k$ are only defined up to a constant.$$

glittergal96 · Jul 14, 2015

Re: MX2 2015 Integration Marathon

Ps, you should use a "\" before commonly used trig functions in latex to make things look nicer.

glittergal96 · Jul 14, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
I think he wanted an IBP proof ahaha

What do you think my third line is?

InteGrand · Jul 14, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
$\Rightarrow I_n=\frac{\sec^{n-2}(x)\tan(x)-(n-2)I_{n-2}}{n-1}.$

Probably a typo or something, but we should have +(n – 2) there in the numerator.

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