First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (3 Viewers)

InteGrand · Oct 30, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Could someone help me do this thanks, I'm keen to see the working out so I can get the general idea of things (I'm getting confused with what happens when you get down to the n limit).

$\noindent For any $n\geq 2$, we have$

$$\begin{align*}I_{n} &= \int _{0}^{\frac{\pi}{4}} \tan^{n}x \, \mathrm{d}x \\ &=\int _{0}^{\frac{\pi}{4}} \tan^{n-2}x \tan^2 x\, \mathrm{d}x \\ &= \int _{0}^{\frac{\pi}{4}} \tan^{n-2}x \left(\sec^2x -1\right)\, \mathrm{d}x \\ &= \int _{0}^{\frac{\pi}{4}} \tan^{n-2}x \sec^2 x \, \mathrm{d}x - \int _{0}^{\frac{\pi}{4}} \tan^{n-2}x \, \mathrm{d}x \\ &= \frac{1}{n-1} + I_{n-2} \quad (\text{use a substitution } u = \tan x\text{ for first integral if you don't want to do it by inspection}).\end{align*}$

InteGrand · Oct 30, 2016

Re: MATH1231/1241/1251 SOS Thread

$\noindent Now that we have this formula, just keep using it to find $I_7$ and $I_8$ (noting the initial values of $I_0 = \frac{\pi}{4}$ and $I_1 = \frac{1}{2}\ln 2 $, which we need to find by actually computing those integrals for $n=0,1$). So we have$

$\begin{align*}I_8 &= \frac{1}{7} - I_6 \\ &= \frac{1}{7} - \left(\frac{1}{5} - I_4 \right) \\ &= \frac{1}{7} - \frac{1}{5} + I_4 \\ &\vdots \quad(\text{keep using formula to reduce }I_{4}\text{ to in terms of }I_2,\text{ then } I_{0}) \\ &= \frac{1}{7} -\frac{1}{5} + \frac{1}{3} - \frac{1}{1} + I_{0},\end{align*}$

$\noindent which we can simplify since we know $I_{0} = \frac{\pi}{4}$. We can find $I_{7}$ in a similar way.$

Flop21 · Oct 30, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent Now that we have this formula, just keep using it to find $I_7$ and $I_8$ (noting the initial values of $I_0 = \frac{\pi}{4}$ and $I_1 = \frac{1}{2}\ln 2 $, which we need to find by actually computing those integrals for $n=0,1$). So we have$

$\begin{align*}I_8 &= \frac{1}{7} - I_6 \\ &= \frac{1}{7} - \left(\frac{1}{5} - I_4 \right) \\ &= \frac{1}{7} - \frac{1}{5} + I_4 \\ &\vdots \quad(\text{keep using formula to reduce }I_{4}\text{ to in terms of }I_2,\text{ then } I_{0}) \\ &= \frac{1}{7} -\frac{1}{5} + \frac{1}{3} - \frac{1}{1} + I_{0},\end{align*}$

$\noindent which we can simplify since we know $I_{0} = \frac{\pi}{4}$. We can find $I_{7}$ in a similar way.$

Thanks, what does I0 mean and how does it compare to I1?

InteGrand · Oct 30, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Thanks, what does I0 mean and how does it compare to I1?

$\noindent Note that $I_n$ was defined as $\int_{0}^{\frac{\pi}{4}} \tan ^{n}x \, \mathrm{d}x$ (for $n=0,1,2,\ldots$). So by definition (putting $n=0$), we have $I_{0} =\int_{0}^{\frac{\pi}{4}} \tan ^{0}x \, \mathrm{d}x = \int_{0}^{\frac{\pi}{4}} 1 \, \mathrm{d}x = \frac{\pi}{4}$. Similarly, $I_{1} = \int_{0}^{\frac{\pi}{4}} \tan x \, \mathrm{d}x = \frac{1}{2}\ln 2$.$

Flop21 · Oct 30, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent Note that $I_n$ was defined as $\int_{0}^{\frac{\pi}{4}} \tan ^{n}x \, \mathrm{d}x$ (for $n=0,1,2,\ldots$). So by definition (putting $n=0$), we have $I_{0} =\int_{0}^{\frac{\pi}{4}} \tan ^{0}x \, \mathrm{d}x = \int_{0}^{\frac{\pi}{4}} 1 \, \mathrm{d}x = \frac{\pi}{4}$. Similarly, $I_{1} = \int_{0}^{\frac{\pi}{4}} \tan x \, \mathrm{d}x = \frac{1}{2}\ln 2$.$

OH right, yeah you just sub in n=0!

Okay thanks I understand now!

Flop21 · Oct 31, 2016

Re: MATH1231/1241/1251 SOS Thread

Can someone help me understand what's going on here, what's this Ab stuff?

InteGrand · Oct 31, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Can someone help me understand what's going on here, what's this Ab stuff?

$\noindent Write $\vec{b}$ as a linear combination of $\vec{v}_1$ and $\vec{v}_2$ and then use the given eigenvalue-eigenvector relations to work out the answer. I.e. find (using row reduction or inspection or whatever method you like) the values of $\alpha,\beta$ such that $\vec{b} = \alpha \vec{v}_1 + \beta \vec{v}_2$, and then our answer will be$

$\begin{align*}A \vec{b} &= A\left(\alpha \vec{v}_1 + \beta \vec{v}_{2}\right) \\ &= \alpha A\vec{v}_1 + \beta A \vec{v}_{2} \\ &= \alpha \lambda_{1} \vec{v}_1 + \beta \lambda_{2}\vec{v}_{2},\end{align*}$

$\noindent and we'll be able to end up with the final answer by subbing in the given eigenvectors and eigenvalues.$

InteGrand · Oct 31, 2016

Re: MATH1231/1241/1251 SOS Thread

(The "Ab stuff" just refers to multiplying the matrix A by the vector b.)

Flop21 · Oct 31, 2016

Re: MATH1231/1241/1251 SOS Thread

Can someone please do this integration

int(sin^3[theta] cos^5[theta] d[theta])

I expanded the sin^3 to sin^2 * sin, then made u = cos [theta], getting int( (1-u^2)(u^4)(-du) ). But the answers do it a completely different way, using u = sin, and get a completely different answer.

Drsoccerball · Oct 31, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent Write $\vec{b}$ as a linear combination of $\vec{v}_1$ and $\vec{v}_2$ and then use the given eigenvalue-eigenvector relations to work out the answer. I.e. find (using row reduction or inspection or whatever method you like) the values of $\alpha,\beta$ such that $\vec{b} = \alpha \vec{v}_1 + \beta \vec{v}_2$, and then our answer will be$

$\begin{align*}A \vec{b} &= A\left(\alpha \vec{v}_1 + \beta \vec{v}_{2}\right) \\ &= \alpha A\vec{v}_1 + \beta A \vec{v}_{2} \\ &= \alpha \lambda_{1} \vec{v}_1 + \beta \lambda_{2}\vec{v}_{2},\end{align*}$

$\noindent and we'll be able to end up with the final answer by subbing in the given eigenvectors and eigenvalues.$

This was the question I asked about, trying to figure out the matrix first then multiplying.

InteGrand · Oct 31, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Can someone please do this integration

int(sin^3[theta] cos^5[theta] d[theta])

I expanded the sin^3 to sin^2 * sin, then made u = cos [theta], getting int( (1-u^2)(u^4)(-du) ). But the answers do it a completely different way, using u = sin, and get a completely different answer.

$\noindent We can show the equivalence between two apparently different answers by using the fact that $1-\sin^2 \theta = \cos^2 \theta$. Using your method, we have$

$\begin{align*}\int \sin^3 \theta \cos^5 \theta \, \mathrm{d}\theta &= \int \sin \theta \left(1- \cos^2 \theta\right) \cos ^{5}\theta \, \mathrm{d}\theta \\ &= - \int \left(u^5 - u^7\right)\, \mathrm{d}u \quad (\text{using }u = \cos \theta) \\ &= \frac{\cos^8 \theta }{8} -\frac{\cos^{6} \theta }{6} + c.\end{align*}$

InteGrand · Oct 31, 2016

Re: MATH1231/1241/1251 SOS Thread

Drsoccerball said:
This was the question I asked about, trying to figure out the matrix first then multiplying.

Well you didn't give the full question haha. Here we just needed Ab, not A itself. When you asked, you asked about finding the actual matrix iirc.

Flop21 · Oct 31, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent We can show the equivalence between two apparently different answers by using the fact that $1-\sin^2 \theta = \cos^2 \theta$. Using your method, we have$

$\begin{align*}\int \sin^3 \theta \cos^5 \theta \, \mathrm{d}\theta &= \int \sin \theta \left(1- \cos^2 \theta\right) \cos ^{5}\theta \, \mathrm{d}\theta \\ &= - \int \left(u^5 - u^7\right)\, \mathrm{d}u \quad (\text{using }u = \cos \theta) \\ &= \frac{\cos^8 \theta }{8} -\frac{\cos^{6} \theta }{6} + c.\end{align*}$

I got that exact answer. Thanks!

The book answer is:-1/4sin^4[theta] - 1/3sin^6[theta] + 1/8 sin^8[theta] + C

Drsoccerball · Oct 31, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
I got that exact answer. Thanks!

The book answer is:-1/4sin^4[theta] - 1/3sin^6[theta] + 1/8 sin^8[theta] + C

Wolfram the hell out of integration

InteGrand · Oct 31, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
I got that exact answer. Thanks!

The book answer is:-1/4sin^4[theta] - 1/3sin^6[theta] + 1/8 sin^8[theta] + C

Replace all sin^2 (theta) there with (1 – cos^2 (theta)) and expand and simplify and it should come down to your one. (But it'll be pretty tedious by hand so if you want to confirm it, better to use a computer to do it).

Flop21 · Oct 31, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
Replace all sin^2 (theta) there with (1 – cos^2 (theta)) and expand and simplify and it should come down to your one. (But it'll be pretty tedious by hand so if you want to confirm it, better to use a computer to do it).

sweet!

Now I need help with how to do this sort of integration

Here's the answer (why are there now 3 fractions)

InteGrand · Oct 31, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
sweet!

Now I need help with how to do this sort of integration

Here's the answer (why are there now 3 fractions)

$\noindent They just typoed the integral that they wrote in the answer, but their final answer is correct (so just a pure typo for the first fraction in their integral, which should have instead been $\frac{-x}{x^{2}+2}$).$

$\noindent So they were meant to just split up the $\frac{-x-5}{x^{2}+2}$ part as $\frac{-x}{x^{2}+2} - \frac{5}{x^{2}+2}$.$

Flop21 · Nov 2, 2016

Re: MATH1231/1241/1251 SOS Thread

For some reason I can't seem to do exact ODEs now when I thought I was doing them fine.................

What on earth am I doing wrong??

The equation given is: 2xy + (x^2 + y^2)dy/dx = 0

At first I went, INT[2xy dx] = x^2 y + h(y)

INT[x^2 + y^2 dy] = x^2 y + y^3/3 + g(x)

by comparison, h(y) = y^3/3

therefore the solution is: x^2 y + y^3/3 = C

-

However the answers say the solution is 3x^2 y + y^3 = A.

I tried doing it the way the lecturers showed to do these...

fx(x,y) = 2xy [1]

fy(x,y) = x^2 + y^2

f(x,y) (integration fx with respect to x) = xy^2 + h(y)

fy(x,y) (derivative with respect to y ^^) = 2xy + h'(y) .... this should euqla fx(x,y) / first equation [1], so h(y) is some constant.

therefore the solution is xy^2 = C.

-

Again that's wrong. What am I doing here??

InteGrand · Nov 2, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
For some reason I can't seem to do exact ODEs now when I thought I was doing them fine.................

What on earth am I doing wrong??

The equation given is: 2xy + (x^2 + y^2)dy/dx = 0

At first I went, INT[2xy dx] = x^2 y + h(y)

INT[x^2 + y^2 dy] = x^2 y + y^3/3 + g(x)

by comparison, h(y) = y^3/3

therefore the solution is: x^2 y + y^3/3 = C

-

However the answers say the solution is 3x^2 y + y^3 = A.

I tried doing it the way the lecturers showed to do these...

fx(x,y) = 2xy [1]

fy(x,y) = x^2 + y^2

f(x,y) (integration fx with respect to x) = xy^2 + h(y)

fy(x,y) (derivative with respect to y ^^) = 2xy + h'(y) .... this should euqla fx(x,y) / first equation [1], so h(y) is some constant.

therefore the solution is xy^2 = C.

-

Again that's wrong. What am I doing here??

Your answer of "x^2 y + y^3/3 = C" is the same as the given answer (just multiply yours through by 3 and you get theirs, noting that 3C can just be written as an arbitrary constant itself, like A).

InteGrand · Nov 2, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
For some reason I can't seem to do exact ODEs now when I thought I was doing them fine.................

What on earth am I doing wrong??

The equation given is: 2xy + (x^2 + y^2)dy/dx = 0

At first I went, INT[2xy dx] = x^2 y + h(y)

INT[x^2 + y^2 dy] = x^2 y + y^3/3 + g(x)

by comparison, h(y) = y^3/3

therefore the solution is: x^2 y + y^3/3 = C

-

However the answers say the solution is 3x^2 y + y^3 = A.

I tried doing it the way the lecturers showed to do these...

fx(x,y) = 2xy [1]

fy(x,y) = x^2 + y^2

f(x,y) (integration fx with respect to x) = xy^2 + h(y)

fy(x,y) (derivative with respect to y ^^) = 2xy + h'(y) .... this should euqla fx(x,y) / first equation [1], so h(y) is some constant.

therefore the solution is xy^2 = C.

-

Again that's wrong. What am I doing here??

$\noindent For your second attempt at the question, you equated the $f_{y} (x,y) = 2xy + h'(y)$ to $f_{x}(x,y)$ instead of to $f_{y}(x,y)$, which is why it yielded the wrong answer.$

First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (3 Viewers)

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