First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (1 Viewer)

InteGrand · Oct 24, 2016

Re: MATH1231/1241/1251 SOS Thread

iforgotmyname said:
Hi everyone, someone please help me out with this question. How did they turn the maclaurin series for 1/(x+1) into series 1/(x^2+1) just like that...

Just replace x with x^2. The series will be convergent for |x^2| < 1, i.e. -1 < x < 1.

iforgotmyname · Oct 24, 2016

MATH1231/1241/1251 SOS Thread

InteGrand said:
Just replace x with x^2. The series will be convergent for |x^2| < 1, i.e. -1 < x < 1.

Holy smokes, Thanks for the ultra fast reply.

I tried deriving the series, but didn't really work out very well cause everything is Zero, not sure what went wrong there.

And does it also mean that for 1/(1+x^3) you just need to replace x with x^3?

InteGrand · Oct 24, 2016

Re: MATH1231/1241/1251 SOS Thread

iforgotmyname said:
Holy smokes, Thanks for the ultra fast reply.

I tried deriving the series, but didn't really work out very well cause everything is Zero, not sure what went wrong there.

Basically you should try and avoid having to derive the series unless you absolutely need to (or are specifically asked to). Just use well-known Taylor series to obtain the desired one if you can (like by replacing x with x^2 here).

Regarding your differentiations, I think you've misdifferentiated after some point. Like to get f"(x), f⁽³⁾(x), etc., we should be using the quotient rule.

iforgotmyname · Oct 24, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
Basically you should try and avoid having to derive the series unless you absolutely need to (or are specifically asked to). Just use well-known Taylor series to obtain the desired one if you can (like by replacing x with x^2 here).

Regarding your differentiations, I think you've misdifferentiated after some point. Like to get f⁽³⁾(x), we should be using the quotient rule.

omg I did.... exam revision past 2 is not a good idea

thanks so much mate

InteGrand · Oct 24, 2016

Re: MATH1231/1241/1251 SOS Thread

iforgotmyname said:
Holy smokes, Thanks for the ultra fast reply.

I tried deriving the series, but didn't really work out very well cause everything is Zero, not sure what went wrong there.

And does it also mean that for 1/(1+x^3) you just need to replace x with x^3?

Yep to get 1/(1+x^3), replace x with x^3.

Flop21 · Oct 24, 2016

Re: MATH1231/1241/1251 SOS Thread

Can someone help me wrap my head around this sort of probability:

Question:

Answer (underlined things i don't understand):

So initially I just went 0.8 * (0.3 + 0.6). But that's wrong obviously.

iforgotmyname · Oct 24, 2016

MATH1231/1241/1251 SOS Thread

Flop21 said:
Can someone help me wrap my head around this sort of probability:

Question:

Answer (underlined things i don't understand):

So initially I just went 0.8 * (0.3 + 0.6). But that's wrong obviously.

Probability of a losing AND b OR c losing .3*.6 accounts for b and c losing at the same time

InteGrand · Oct 24, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Can someone help me wrap my head around this sort of probability:

Question:

Answer (underlined things i don't understand):

So initially I just went 0.8 * (0.3 + 0.6). But that's wrong obviously.

$\noindent Remember there is a general probability rule that $\mathbb{P}\left(C\cup D\right) = \mathbb{P}\left(C\right) + \mathbb{P}\left(D\right) - \mathbb{P}\left(C\cap D\right)$ for events $C,D$. The solutions there typo'd, in the second thing you underlined it was meant to be a $-\mathbb{P}\left(C \cap D\right)$, not $C\cup D$. It was just a pure typo I think, because the numbers they subbed in for that part were for $\mathbb{P}\left(C \cap D\right)$.$

He-Mann · Oct 24, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Can someone help me wrap my head around this sort of probability:

Answer (underlined things i don't understand):

So initially I just went 0.8 * (0.3 + 0.6). But that's wrong obviously.

I think there is a type when they used Inclusion-Exclusion Principle.

Should be this P(C U D) = P(C) + P(D) - P(C n D)

Then by independence P(C n D) = P(C)xP(D)

Flop21 · Oct 24, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent Remember there is a general probability rule that $\mathbb{P}\left(C\cup D\right) = \mathbb{P}\left(C\right) + \mathbb{P}\left(D\right) - \mathbb{P}\left(C\cap D\right)$ for events $C,D$. The solutions there typo'd, in the second thing you underlined it was meant to be a $-\mathbb{P}\left(C \cap D\right)$, not $C\cup D$. It was just a pure typo I think, because the numbers they subbed in for that part were for $\mathbb{P}\left(C \cap D\right)$.$

Oh yeah that's right! Thanks

Forgot, but now it's coming back to me.

Flop21 · Oct 24, 2016

Re: MATH1231/1241/1251 SOS Thread

Question

A random variable Y is defined by Y = -2X + 3. Determine E(Y) and Var[(]Y).

I know how to do E() and Var(), but what do I do with the -2X + 3? Do I go, [-2(0) + 3]*0.2 + [-2(1) + 3]*1 + etc... when doing E()

also i was given the probability distribution of the random variable X before this

InteGrand · Oct 24, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Question

A random variable Y is defined by Y = -2X + 3. Determine E(Y) and Var[(]Y).

I know how to do E() and Var(), but what do I do with the -2X + 3? Do I go, [-2(0) + 3]*0.2 + [-2(1) + 3]*1 + etc... when doing E()

E is linear so E(Y) = -2E(X) + 3.

For variance, use the property that Var(aX + b) = a² Var(X), where a and b are constants. So here, Var(Y) = (-2)²Var(X) = 4Var(X).

Mr_Kap · Oct 26, 2016

Re: MATH1231/1241/1251 SOS Thread

HEY. Im currently tilted with this:

So during a question i end up getting this:

What i cant grasp is how this is 4*sqrt(2)...even when i looked up the indefinite integral and tried subbing in i get ZERO...any help would be GREATLY APPRECIATED!

InteGrand · Oct 26, 2016

Re: MATH1231/1241/1251 SOS Thread

Mr_Kap said:
HEY. Im currently tilted with this:

So during a question i end up getting this:

What i cant grasp is how this is 4*sqrt(2)...even when i looked up the indefinite integral and tried subbing in i get ZERO...any help would be GREATLY APPRECIATED!

$\noindent Need to use absolute value signs. Note that $1+\sin x = \cos^2 \frac{x}{2} + 2\sin \frac{x}{2} \cos \frac{x}{2} + \sin^2 \frac{x}{2}= \left(\cos \frac{x}{2} + \sin \frac{x}{2}\right)^2$. So $\sqrt{1+\sin x} = \left | \cos \frac{x}{2}+ \sin \frac{x}{2} \right |$. So this is the function to integrate.$

Mr_Kap · Oct 26, 2016

Re: MATH1231/1241/1251 SOS Thread

Cheers for that. However could someone show me the working out as I'm still getting zero.

InteGrand · Oct 26, 2016

Re: MATH1231/1241/1251 SOS Thread

Mr_Kap said:
Cheers for that. However could someone show me the working out as I'm still getting zero.

$\noindent The integral has become $I = \int _{0}^{2\pi}\left|\cos \frac{x}{2} + \sin \frac{x}{2}\right|\, \mathrm{d} x = 2\int_{0}^{\pi} \left|\cos u + \sin u\right|\, \mathrm{d}u$ (substituting $u = \frac{x}{2}$). Let $f(u)=\cos u + \sin u$ for $u\in [0,\pi]$. Note that $f(u) = 0$ precisely when $\tan u = -1\iff u = \frac{3\pi}{4}$. Note $f(0) = 1>0$ and $f(\pi) = -1 < 0$, so we can see that $f(u)$ is positive when $0<u < \frac{3\pi}{4}$ and negative when $\frac{3\pi}{4} < u < \pi$. So we obtain$

$\begin{align*}I &= 2\left( \int_{0}^{\frac{3\pi}{4}} \left( \cos u + \sin u\right)\, \mathrm{d}u +\int_{\frac{3\pi}{4}} ^{\pi} -\left(\cos u + \sin u\right) \, \mathrm{d}u\right) \\ &= 2\left( \sin \frac{3\pi}{4} + \cos 0 - \cos \frac{3\pi}{4} + \sin \frac{3\pi}{4} + \cos \pi - \cos \frac{3\pi}{4}\right) \\ &= 2\left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right) \\ &= 4\sqrt{2}.\end{align*}$

InteGrand · Oct 26, 2016

Re: MATH1231/1241/1251 SOS Thread

The main thing is that we need to split up the interval into cases based on where the thing in the absolute values is positive/negative in order to deal with the absolute values.

Flop21 · Oct 26, 2016

Re: MATH1231/1241/1251 SOS Thread

What's going on here:

[Integral sign] coshx (1 + sinh^2 x)dx

= sinhx + (sinh^3 x)/3

What did they do to get to that answer (where did the +1 go)?

InteGrand · Oct 26, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
What's going on here:

[Integral sign] coshx (1 + sinh^2 x)dx

= sinhx + (sinh^3 x)/3

What did they do to get to that answer (where did the +1 go)?

Substitute u = sinh(x), so du = cosh(x) dx. Then the integral becomes

int (1+u^2) du = u + (u^3)/3 + c

= sinh(x) + (sinh^3 (x))/3 + c.

iforgotmyname · Oct 26, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
What's going on here:

[Integral sign] coshx (1 + sinh^2 x)dx

= sinhx + (sinh^3 x)/3

What did they do to get to that answer (where did the +1 go)?

=Int(Coshx+coshx*sinh^2x )

Since coshx*sinh^2x is of the form int(y'*y)
We can treat it as x^2 so int(x^2) is (x^3)/3

Thus Int(Coshx+coshx*sinh^2x )=sinhx+sinh^3x/3

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