UNSW 1st year finals Q (1 Viewer)

integral95 · Nov 10, 2014

Hey Ya I just had my final maths exam today, and some questions rekt me hard lol, and ODE (ordinary differential equation) question and a weird limit one

$Suppose that y satisfies the initial value problem$ \\ \frac{dy}{dx} + y^2 = cos(x) \ \ with \ \ y(0) = 0 \\ $find the first non-zero terms of the maclaurin series of y (using implicit differentiation or otherwise)$ \\ \\ \\ \\ $Let c be a positive real number and define the sequence$ \ \ {a_n} \ \ for \ \ n\geq 1 \\ \\ a_n = n(c^{\frac{1}{n}}-1) \\ \\ $find, with reason, ${a_n} $as n goes to infinity$

need a pro guy to help here hehe

integral95 · Nov 15, 2014

BUMP1!!

RenegadeMx · Nov 15, 2014

y=1/Pi

seventhroot · Nov 15, 2014

false

seventhroot · Nov 15, 2014

can't do the first one; probably some syntax error

Q2) ln(c)

2ez

NOTE: utilized the "otherwise"

integral95 · Nov 15, 2014

seventhroot said:
can't do the first one; probably some syntax error

Q2) ln(c)

2ez

NOTE: utilized the "otherwise"

damn i wonder who got ln(c)

seventhroot · Nov 15, 2014

integral95 said:
damn i wonder who got ln(c)

<

well tbh I've done a similar question to that before so I just wrote that down

anomalousdecay · Nov 15, 2014

I got zero for the second one in the test, then my mate convinced me its just c by using calculator trial and error, now I'm convinced its ln (c)

D94 · Nov 15, 2014

Sure that first one is written correctly?

MATLAB couldn't solve it; did not even give a complicated output: http://puu.sh/cRuoW/e47a4dcbd7.png

anomalousdecay · Nov 15, 2014

Oh yeah that question was difficult.....

I got:

$x - \frac{3x^2}{2!}$

Not even sure if correct or even remotely right....

D94 said:
Sure that first one is written correctly?

MATLAB couldn't solve it; did not even give a complicated output: http://puu.sh/cRuoW/e47a4dcbd7.png

Its because you are supposed to only find the first two terms of the Maclaurin series for it.

anomalousdecay · Nov 15, 2014

I found the first derivative as:

$\frac{dy}{dx} + 0 = 1 \\ \\ \\ \therefore \frac{dy}{dx} = 1$

So we get that result from y(0) = 0.

I found the implicit derivative with respect to x as:

$\frac{d^2 y}{dx^2} + 2y \frac{dy}{dx} = -\sin (x)$

Then I got that when we substitute y(0) = 0, we get the second derivative is equal to zero.

Now derive once more....

$\frac{d^3 y}{dx^3} + 2 \left ( \frac{dy}{dx} \right ) ^2 + 2y \frac{d^2 y}{dx^2} = - \cos (x) \\ \\ \\ $Substituting we get: $ \frac{d^3 y}{dx^3} = 1$

Yeah I stuffed up in the exam.... The answer actually is:

$y = x + \frac{x^3}{3!} $ (first two non-zero terms only about x = 0) $$

I did too many sillies that exam.... Gonna cost me HD. Just really wasn't my day... That exam made me feel so sad because I knew everything (except one probability question) but in the exam couldn't do it and write everything down properly. Missed the hyperbolic sine integral for q4 completely too....

seanieg89 · Nov 15, 2014

For the limit one, its just

$\lim_{n\rightarrow \infty} n(c^{1/n}-1)=\lim_{t\rightarrow 0^+} \frac{c^t-1}{t}=c^t\log(c)|_{t=0}=\log(c)$

by L'Hopital.

anomalousdecay · Nov 15, 2014

seanieg89 said:
for the limit one, its just

$\lim_{n\rightarrow \infty} n(c^{1/n}-1)=\lim_{t\rightarrow 0^+} \frac{c^t-1}{t}=c^t\log(c)|_{t=0}=\log(c)$

by l'hopital.

LOOOOOOOOOL

No-one expected it after a whole semester of not having to touch L'Hopital's whatsoever haha.

seanieg89 · Nov 15, 2014

Whenever you see two competing limiting behaviours you should think L'Hopital.

integral95 · Nov 15, 2014

NOOOOOO WHY DID NOT THINK OF L'HOPITAL!!!!! I WILL NEVER FORGET THIS NEXT TIME
Ahh well thanks anyways

anomalousdecay · Nov 15, 2014

Still not sure if I did the other one correctly lol....

Does it look valid?

seanieg89 · Nov 15, 2014

anomalousdecay said:
I found the first derivative as:

$\frac{dy}{dx} + 0 = 1 \\ \\ \\ \therefore \frac{dy}{dx} = 1$

So we get that result from y(0) = 0.

I found the implicit derivative with respect to x as:

$\frac{d^2 y}{dx^2} + 2y \frac{dy}{dx} = -\sin (x)$

Then I got that when we substitute y(0) = 0, we get the second derivative is equal to zero.

Now derive once more....

$\frac{d^3 y}{dx^3} + 2 \left ( \frac{dy}{dx} \right ) ^2 + 2y \frac{d^2 y}{dx^2} = - \cos (x) \\ \\ \\ $Substituting we get: $ \frac{d^3 y}{dx^3} = 1 (*)$

Yeah I stuffed up in the exam.... The answer actually is:

$y = x + \frac{x^3}{3!} $ (first two non-zero terms only about x = 0) $$

I did too many sillies that exam.... Gonna cost me HD. Just really wasn't my day... That exam made me feel so sad because I knew everything (except one probability question) but in the exam couldn't do it and write everything down properly. Missed the hyperbolic sine integral for q4 completely too....

Are you sure about that answer? I think (*) is incorrect, as well as the Maclaurin series.

I think it should be

y(x)=x-x^3/2 to 3rd order (and y'''(0)=-3).

anomalousdecay · Nov 16, 2014

seanieg89 said:
Are you sure about that answer? I think (*) is incorrect, as well as the Maclaurin series.

I think it should be

y(x)=x-x^3/2 to 3rd order (and y'''(0)=-3).

Yep that's a mistake thanks for spotting that. Your final solution fixes up my error.

Ironically, I got closer to that in the exam (just had x^2 instead of x^3 as I have written down on my question paper when I wanted to check later on). I was more worried that my method seemed wrong completely.

We actually had no practice on such a question in our homework questions I believe (integral, seventh, confirm if this is true or not), so I just wanted to know that I had made the most of my understanding of ODE's from the course.

Thanks for the help again

Need to spread more rep.

seventhroot · Nov 16, 2014

anomalousdecay said:
Yep that's a mistake thanks for spotting that. Your final solution fixes up my error.

Ironically, I got closer to that in the exam (just had x^2 instead of x^3 as I have written down on my question paper when I wanted to check later on). I was more worried that my method seemed wrong completely.

We actually had no practice on such a question in our homework questions I believe (integral, seventh, confirm if this is true or not), so I just wanted to know that I had made the most of my understanding of ODE's from the course.

Thanks for the help again

Need to spread more rep.

done for you

FrankXie · Nov 17, 2014

integral95 said:
Hey Ya I just had my final maths exam today, and some questions rekt me hard lol, and ODE (ordinary differential equation) question and a weird limit one

$Suppose that y satisfies the initial value problem$ \\ \frac{dy}{dx} + y^2 = cos(x) \ \ with \ \ y(0) = 0 \\ $find y (using implicit differentiation or otherwise)$ \\ \\ \\ \\ $Let c be a positive real number and define the sequence$ \ \ {a_n} \ \ for \ \ n\geq 1 \\ \\ a_n = n(c^{\frac{1}{n}}-1) \\ \\ $find, with reason, ${a_n} $as n goes to infinity$

need a pro guy to help here hehe

are you sure the first question is really like that? I guess you missed a y on RHS? like $y^\prime+y^2=y\cos x$

the second question can be solved, other than L'Hospital rule, using equivalent substitution like

$c^{\frac1n}-1\sim \ln c^{\frac1n}=\frac1n\ln c$

UNSW 1st year finals Q (1 Viewer)

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