Author Topic: 4U Maths Question Thread (Read 665358 times) Tweet Share

katherine123 · « **Reply #75 on:** March 15, 2016, 05:38:54 pm »

how do you normally start with these questions involving i(n) I(n-1)

RuiAce · « **Reply #76 on:** March 15, 2016, 07:22:33 pm »

Quote from: katherine123 on March 15, 2016, 05:38:54 pm

how do you normally start with these questions involving i(n) I(n-1)

$\text{By far, with reduction formulae in integration, there are multiple methods that one needs to} \textbf{ know and remember.} \\ \text{However, over 70 percent of these questions will involve an application of integration by parts.}\\ \text{An exception to this case is the integral of }\tan^{n}{x}$
$\text{Also, note that in the exam if they want you to show you that }{I}_{n}+{I}_{n-1}=something \\ \text{then you use that. Anyway, for these problems:}$
$\begin{align*}{I}_{n}&=\int{\sin^{n}{x}\, dx} \\ &= \int{\sin{x}\sin^{n-1}{x}\,dx} \\ &=-\cos{x}\sin^{n-1}{x} + \int{\left(n-1\right)\cos^2{x}\sin^{n-2}{x}\,dx} \, \text{via I.B.P.} \\ &=-\cos{x}\sin^{n-1}{x}+\left(n-1\right) \int{\left(1-\sin^{2}{x}\right)\sin^{n-2}{x}\,dx} \, \text{by recalling the Pythagorean identity} \\ &=-\cos{x}\sin^{n-1}{x}+\left(n-1\right)\int{\sin^{n-2}{x} \, dx}-\left(n-1\right) \int{\sin^{n}{x} \, dx}\\ &=-\cos{x}\sin^{n-1}{x}+\left(n-1\right) {I}_{n-2}-\left(n-1\right) {I}_{n} \\ \Leftrightarrow n{I}_{n}&=\cos{x}\sin^{n-1}{x}+\left(n-1\right) {I}_{n-2}\end{align*}$
$\text{The method is similar for the integral of}\sec^{n}{x}\text{, however you will split out }\sec^{2}{x} \\ \text{as that integrates into }\tan{x} \\ \text{Once again, a Pythagorean identity will be necessary later on.}$
$\text{Q19. requires TWO applications of integration by parts to achieve the desired result.}$
$\text{As for Q20. we use algebraic manipulation in the 'plus one minus one' trick} \\ \begin{align*}{I}_{n}&=\int_{0}^{1}{x{\left(1-{x}^{3}\right)}^{n}\,dx}\\&=\left[\frac{{x}^{2}{\left(1-{x}^{3}\right)}^{n}}{2}\right]_{0}^{1}-\int_{0}^{1}{-3n{x}^{4}{\left(1-{x}^{3}\right)}^{n-1}\,dx}\\ &=0-3n\int_{0}^{1}{x\left(-{x}^{3}\right){\left(1-{x}^{3}\right)}^{n-1}\,dx} \\&=-3n\int_{0}^{1}{x\left(\left(1-{x}^{3}\right)-1\right){\left(1-{x}^{3}\right)}^{n-1}\,dx} \, \text{(via +1 -1 trick)}\\&=-3n\int_{0}^{1}{x\left(1-{x}^{3}\right){\left(1-{x}^{3}\right)}^{n-1}\,dx}+3n\int_{0}^{1}{x{\left(1-{x}^{3}\right)}^{n-1}\,dx} \, \text{(by expanding extremely carefully)} \end{align*}\\ \text{The rest is left as an exercise}$
Edit: Found a mistake

Happy Physics Land · « **Reply #77 on:** March 15, 2016, 07:49:12 pm »

Quote from: RuiAce on March 15, 2016, 07:22:33 pm

$\text{By far, with reduction formulae in integration, there are multiple methods that one needs to} \textbf{ know and remember.} \\ \text{However, over 70 percent of these questions will involve an application of integration by parts.}\\ \text{An exception to this case is the integral of }\tan^{n}{x}$
$\text{Also, note that in the exam if they want you to show you that }{I}_{n}+{I}_{n-1}=something \\ \text{then you use that. Anyway, for these problems:}$
$\begin{align*}{I}_{n}&=\int{\sin^{n}{x}\, dx} \\ &= \int{\sin{x}\sin^{n-1}{x}\,dx} \\ &=-\cos{x}\sin^{n-1}{x} + \int{\left(n-1\right)\cos^2{x}\sin^{n-2}{x}\,dx} \, \text{via I.B.P.} \\ &=-\cos{x}\sin^{n-1}{x}+\left(n-1\right) \int{\left(1-\sin^{2}{x}\right)\sin^{n-2}{x}\,dx} \, \text{by recalling the Pythagorean identity} \\ &=-\cos{x}\sin^{n-1}{x}+\left(n-1\right)\int{\sin^{n-2}{x} \, dx}-\left(n-1\right) \int{\sin^{n}{x} \, dx}\\ &=-\cos{x}\sin^{n-1}{x}+\left(n-1\right) {I}_{n-2}-\left(n-1\right) {I}_{n} \\ \Leftrightarrow n{I}_{n}&=\cos{x}\sin^{n-1}{x}+\left(n-1\right) {I}_{n-2}\end{align*}$
$\text{The method is similar for the integral of}\sec^{n}{x}\text{, however you will split out }\sec^{2}{x} \\ \text{as that integrates into }\tan{x} \\ \text{Once again, a Pythagorean identity will be necessary later on.}$
$\text{Q19. requires TWO applications of integration by parts to achieve the desired result.}$
$\text{As for Q20. we use algebraic manipulation in the 'plus one minus one' trick} \\ \begin{align*}{I}_{n}&=\int_{0}^{1}{x{\left(1-{x}^{3}\right)}^{n}\,dx}\\&=\left[\frac{{x}^{2}{\left(1-{x}^{3}\right)}^{n}}{2}\right]_{0}^{1}-\int_{0}^{1}{-3{x}^{4}{\left(1-{x}^{3}\right)}^{n-1}\,dx}\\ &=0-3\int_{0}^{1}{x\left(-{x}^{3}\right){\left(1-{x}^{3}\right)}^{n-1}\,dx} \\&=-3\int_{0}^{1}{x\left(\left(1-{x}^{3}\right)-1\right){\left(1-{x}^{3}\right)}^{n-1}\,dx} \, \text{(via +1 -1 trick)}\\&=-3\int_{0}^{1}{x\left(1-{x}^{3}\right){\left(1-{x}^{3}\right)}^{n-1}\,dx}+3\int_{0}^{1}{x{\left(1-{x}^{3}\right)}^{n-1}\,dx} \, \text{(by expanding extremely carefully)} \end{align*}\\ \text{The rest is left as an exercise}$

You are an absolute god Rui

katherine123 · « **Reply #78 on:** March 17, 2016, 09:33:11 pm »

How do you do these questions? thanks

RuiAce · « **Reply #79 on:** March 17, 2016, 09:58:58 pm »

Quote from: katherine123 on March 17, 2016, 09:33:11 pm

How do you do these questions? thanks

$\begin{align*}\int{\ln{\left({x}^{2}-1\right)} \,dx}&=x\ln{\left({x}^{2}-1\right)}-\int{x\cdot \frac{2x}{{x}^{2}-1} \,dx} \text{ (via integration by parts)}\\ &=x\ln{\left({x}^{2}-1\right)}-2\int{\frac{{x}^{2}-1+1}{{x}^{2}-1}\,dx} \text{ (using +1 -1 trick)}\\&=x\ln{\left({x}^{2}-1\right)}-2\int{dx}-2\int{\frac{1}{{x}^{2}-1}\,dx}\\&=x\ln{\left({x}^{2}-1\right)}-2x+\int{\left(\frac{1}{x+1}-\frac{1}{x-1}\right)\,dx} \text{ (via partial fractions)} \\ &= x\ln{\left({x}^{2}-1\right)}-2x+\ln{\left|\frac{x+1}{x-1}\right|} +C\end{align*}$

Observe that a way to avoid the partial fractions is to decompose ln(x^2-1) into ln(x+1) + ln(x-1) and perform two applications of integration by parts

$x=\sin{\theta}\Rightarrow dx=\cos{\theta}\,d\theta\\ \begin{align*}\int_{0}^{1}{\frac{{x}^{2}}{\sqrt{1-{x}^{2}}}dx}&=\int_{0}^{\frac{\pi}{2}}{\frac{\sin^{2}{\theta}}{\sqrt{1-\sin^{2}{\theta}}}\cos{\theta}\,d\theta}\\&=\int_{0}^{\frac{\pi}{2}}{\sin^{2}{\theta}\,d\theta}\\&=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\left(1-\cos{2\theta}\right)d\theta}\end{align*} \\ \text{etc.}$

Neutron · « **Reply #80 on:** March 19, 2016, 08:29:32 pm »

Hey guys!

Sort of a dumb question but I was doing some questions and one of the questions was:

"If P(x)=x^4-2x^3-x^2+6x-6 has a zero 1-i, find the zeros of P(x) over C"

Now I know that 1+i is also a root (conjugate root theorem) but how do you know there aren't 4 complex roots? Like how come you can assume it's just two complex roots idk I'm tired and confused so sorry if this is very basic!

Neutron

RuiAce · « **Reply #81 on:** March 20, 2016, 07:48:18 am »

Quote from: Neutron on March 19, 2016, 08:29:32 pm

Hey guys!

Sort of a dumb question but I was doing some questions and one of the questions was:

"If P(x)=x^4-2x^3-x^2+6x-6 has a zero 1-i, find the zeros of P(x) over C"

Now I know that 1+i is also a root (conjugate root theorem) but how do you know there aren't 4 complex roots? Like how come you can assume it's just two complex roots idk I'm tired and confused so sorry if this is very basic!

Neutron

$\text{By the fact that }1\pm i\text{ are both roots, it can be directly inferred that }\\ {x}^{2}-2x+2 \text{ is a factor of }P\left(x\right)$
$\text{Using polynomial long division, it can therefore be easily shown that }\\ P\left(x\right) = \left({x}^{2}-2x+2\right) \left({x}^{2}-3\right)$
$\text{There should be no surprise that the other factor is also a quadratic} \\ \text{ because when you divide a degree 4 polynomial by a degree 2 polynomial (which is a factor), you should get another degree 2 polynomial.} \\ \text{You can always tell if a quadratic has two real roots or not by the quadratic discriminant}$

jamonwindeyer · « **Reply #82 on:** March 20, 2016, 01:44:52 pm »

Quote from: RuiAce on March 20, 2016, 07:48:18 am

$\text{By the fact that }1\pm i\text{ are both roots, it can be directly inferred that }\\ {x}^{2}-2x+2 \text{ is a factor of }P\left(x\right)$
$\text{Using polynomial long division, it can therefore be easily shown that }\\ P\left(x\right) = \left({x}^{2}-2x+2\right) \left({x}^{2}-3\right)$
$\text{There should be no surprise that the other factor is also a quadratic} \\ \text{ because when you divide a degree 4 polynomial by a degree 2 polynomial (which is a factor), you should get another degree 2 polynomial.} \\ \text{You can always tell if a quadratic has two real roots or not by the quadratic discriminant}$

And I'll add that, in case you aren't sure, we can infer that first quadratic root by multiplying the two complex roots together:

$(x-(1-i))(x-(1+i))=(x-1+i)(x-1-i)=(x^2-x-ix-x+1+i+ix-i+1)=(x^2-2x+2)$

RuiAce · « **Reply #83 on:** March 20, 2016, 06:51:53 pm »

Quote from: jamonwindeyer on March 20, 2016, 01:44:52 pm

And I'll add that, in case you aren't sure, we can infer that first quadratic root by multiplying the two complex roots together:

$(x-(1-i))(x-(1+i))=(x-1+i)(x-1-i)=(x^2-x-ix-x+1+i+ix-i+1)=(x^2-2x+2)$

$\text{Very much true, but it can be builded on further as there is a very famous expansion that }\\\left(x-z\right)\left(x-\overline{z}\right)={x}^{2}-2x\,\text{Re}\left(z\right)+{\left|z\right|}^{2}\\\text{This can be easily proved using the method of difference of two squares and can easily be applied here}$
$\text{Also note that actually, difference of two squares could have been used with the above expansion.}\\(x-(1-i))(x-(1+i))=((x-1)-i)((x-1)+i)={\left(x-1\right)}^{2}-{i}^{2}$

jamonwindeyer · « **Reply #84 on:** March 20, 2016, 08:36:10 pm »

Quote from: RuiAce on March 20, 2016, 06:51:53 pm

$\text{Very much true, but it can be builded on further as there is a very famous expansion that }\\\left(x-z\right)\left(x-\overline{z}\right)={x}^{2}-2x\,\text{Re}\left(z\right)+{\left|z\right|}^{2}\\\text{This can be easily proved using the method of difference of two squares and can easily be applied here}$
$\text{Also note that actually, difference of two squares could have been used with the above expansion.}\\(x-(1-i))(x-(1+i))=((x-1)-i)((x-1)+i)={\left(x-1\right)}^{2}-{i}^{2}$

Ahh cool, wasn't sure if that result is taught explicitly or is just derived in practice

xXCandyDXx · « **Reply #85 on:** March 21, 2016, 06:57:25 pm »

Hey guys I need some help here.. How do I do this question? I know that q, r and s are coefficients but ... How would you relate them ?.. I don't know, I'm lost and confused thanks a heap! @.@

RuiAce · « **Reply #86 on:** March 21, 2016, 07:50:43 pm »

Quote from: xXCandyDXx on March 21, 2016, 06:57:25 pm

Hey guys I need some help here.. How do I do this question? I know that q, r and s are coefficients but ... How would you relate them ?.. I don't know, I'm lost and confused thanks a heap! @.@(Image removed from quote.)

$\text{The fact that two roots are the negatives of each other immediately tells me that the sum of one root at a time will be useful.} \\ \begin{align*}\alpha-\alpha+\beta&=-q\\\beta&=-q\end{align*}\\\text{We then consult the product of roots, because we already know what one of the roots is}$
$\begin{align*}\alpha\cdot-\alpha\cdot-q&=-s\\{\alpha}^{2}&=-\frac{s}{q}\end{align*} \\ \text{And this will give us all we need to apply the sum of two roots at a time. Note how two terms do cancel out}$
$\begin{align*}{\alpha}^{2}-q\alpha-q\left(-\alpha\right)&=r\\ -\frac{q}{s}&=r\\qr&=-s\end{align*}$

And I threw the pencil here; Somehow I accidentally got a negative stuck in there so unless the question has a typo, I had a mistake in algebra at some point.

xXCandyDXx · « **Reply #87 on:** March 21, 2016, 08:15:54 pm »

Quote from: RuiAce on March 21, 2016, 07:50:43 pm

$\text{The fact that two roots are the negatives of each other immediately tells me that the sum of one root at a time will be useful.} \\ \begin{align*}\alpha-\alpha+\beta&=-q\\\beta&=-q\end{align*}\\\text{We then consult the product of roots, because we already know what one of the roots is}$
$\begin{align*}\alpha\cdot-\alpha\cdot-q&=-s\\{\alpha}^{2}&=-\frac{s}{q}\end{align*} \\ \text{And this will give us all we need to apply the sum of two roots at a time. Note how two terms do cancel out}$
$\begin{align*}{\alpha}^{2}-q\alpha-q\left(-\alpha\right)&=r\\ -\frac{q}{s}&=r\\qr&=-s\end{align*}$

And I threw the pencil here; Somehow I accidentally got a negative stuck in there so unless the question has a typo, I had a mistake in algebra at some point.

Ohhh I got it!!!

but how about the next part

(((

RuiAce · « **Reply #88 on:** March 21, 2016, 08:22:23 pm »

Quote from: xXCandyDXx on March 21, 2016, 08:15:54 pm

Ohhh I got it!!! (Image removed from quote.) but how about the next part (((

$\text{Inspection of the coefficients (all real) means that our two complex roots are, as you would expect, two complex conjugates} \\ \text{But there is a twist. Not only are these roots conjugates, but they are also the negative of each other.} \\ \text{This results in the fact that the two complex roots must have a real part of 0. A formal proof succeeds this immediate assumption.}$
$\text{Let }\alpha=a+bi\\\text{We have}\\\begin{align*}\overline{a+bi}&=-(a+bi)\\a-bi&=-a-bi\\2a&=0\\a&=0\end{align*}$

RuiAce · « **Reply #89 on:** March 22, 2016, 08:13:07 am »

Quote from: xXCandyDXx on March 21, 2016, 08:15:54 pm

Ohhh I got it!!! (Image removed from quote.) but how about the next part (((

Actually, hang on (in regards to the diagram). Shouldn't the product of roots be negative s?

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