First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

matchalolz · Apr 14, 2017

Re: MATH1131 help thread

I'm currently slogging through 3 weeks worth of maths so I'm probably going to be here for awhile... >.>

matchalolz · Apr 19, 2017

Re: MATH1131 help thread

Screen Shot 2017-04-19 at 12.27.12 pm.png

Can someone please suggest how I should approach this?

InteGrand · Apr 19, 2017

Re: MATH1131 help thread

matchalolz said:
View attachment 33880

Can someone please suggest how I should approach this?

$\noindent Let $w \equiv z - 6 + i \iff z \equiv w + 6 - i$. Find all solutions $w$ to the equation $w^{3} = -27$. Then the solutions for $z$ in terms of the solutions for $w$ are given by $z = w + 6 - i$.$

matchalolz · Apr 19, 2017

Re: MATH1131 help thread

InteGrand said:
$\noindent Let $w \equiv z - 6 + i \iff z \equiv w + 6 - i$. Find all solutions $w$ to the equation $w^{3} = -27$. Then the solutions for $z$ in terms of the solutions for $w$ are given by $z = w + 6 - i$.$

Thanks!

matchalolz · Apr 19, 2017

Re: MATH1131 help thread

Screen Shot 2017-04-19 at 3.37.47 pm.png

I'm not understanding how the MVT can be used to prove that there is a zero. How would you do this question?

leehuan · Apr 19, 2017

Re: MATH1131 help thread

matchalolz said:
View attachment 33881

I'm not understanding how the MVT can be used to prove that there is a zero. How would you do this question?

$\text{Note that }f\text{ is continuous on }[1,3]\text{ and differentiable on }(1,3)\\ \text{so the MVT may be applied.}$

$\text{By the MVT, there is a value of }c\in (1,3)\text{ such that}\\ f^\prime(c) = \frac{f(3)-f(1)}{3-1}$

$\\\text{But note, by computation, that }\\f(1) = (-1)^4\cos 1 = \cos 1\\ \text{and }f(3) = 1^4\cos 1 = \cos 1$

$\\\text{Hence, the value of }c\text{ specified above must satisfy}\\ f^\prime (c) = \frac{\cos 1 - \cos 1}{2} = 0\\ \text{as required.}$

matchalolz · Apr 19, 2017

Re: MATH1131 help thread

leehuan said:
$\text{Note that }f\text{ is continuous on }[1,3]\text{ and differentiable on }(1,3)\\ \text{so the MVT may be applied.}$

$\text{By the MVT, there is a value of }c\in (1,3)\text{ such that}\\ f^\prime(c) = \frac{f(3)-f(1)}{3-1}$

$\\\text{But note, by computation, that }\\f(1) = (-1)^4\cos 1 = \cos 1\\ \text{and }f(3) = 1^4\cos 1 = \cos 1$

$\\\text{Hence, the value of }c\text{ specified above must satisfy}\\ f^\prime (c) = \frac{\cos 1 - \cos 1}{2} = 0\\ \text{as required.}$

oh i see c: ty!!

matchalolz · Apr 19, 2017

Re: MATH1131 help thread

I'm stuck again...

Screen Shot 2017-04-19 at 4.26.52 pm.png

With the first question, I have zero clue what to do...

And for the second question, I managed to do part a and part b but kind of lost with part c.

InteGrand · Apr 19, 2017

Re: MATH1131 help thread

matchalolz said:
I'm stuck again...

View attachment 33882

With the first question, I have zero clue what to do...

And for the second question, I managed to do part a and part b but kind of lost with part c.

$\noindent All of these Q's would use inequalities using Mean Value Theorem. Recall that for a function $f$ that is continuous on the closed interval $[a,b]$ and diffentiable on $(a,b)$, we have $f(b) = f(a) + f'(c) (b-a)$ for some $c \in (a,b)$. So for the last Q., the absolute error in approximating $f(a)$ by $f(b)$ is $\epsilon := |f(b) - f(a)| = |f'(c)| (b-a)$. Thus if you can find an \emph{upper bound for} $|f'(t)|$ \emph{on} $(a,b)$, say $M$, you can upper-bound the error as $\boxed{ \epsilon \leq M(b-a)}$.$

InteGrand · Apr 19, 2017

Re: MATH1131 help thread

$\noindent So for that last one, what you could do is decide what the function $f(t)$ is, what $a$ and $b$ should be for our purposes, calculate the derivative function $f'(t)$, and take $M$ to be the maximum value of $|f'(t)|$ on $[a,b]$. This would apply to the two parts before that one too.$

leehuan · Apr 19, 2017

Re: MATH1131 help thread

matchalolz said:
I'm stuck again...

View attachment 33882

With the first question, I have zero clue what to do...

$\text{Fix }x> 0\text{ and let }f(t) = \sqrt{1+t}\implies f^\prime(t) = \frac{1}{2\sqrt{1+t}}$

$\\ \text{Since }f\text{ is continuous on }[0,x]\text{ and differentiable on }(0,x)\\ \text{By the MVT there is a }c\in (0,x)\text{ such that}\\ f^\prime(c) = \frac{f(x)-f(0)}{x-0}$

$\text{i.e. }\frac{1}{2\sqrt{1+c}}=\frac{\sqrt{1+x}-1}{x}$

$\text{Now, consider the following manipulation of inequalities that stem from }c\in (0,x):\\ 0 < c < x\\ 1 < 1+c < 1+x\\ 1 < \sqrt{1+c} < \sqrt{1+x}\\ 1 > \frac{1}{\sqrt{1+c}} > \frac1{\sqrt{1+x}}$

$\text{So from }\frac1{\sqrt{1+x}} < \frac1{\sqrt{1+c}} < 1\\ \text{Doing a bit of manipulation, noting that }x>0\text{, we have the desired result (remember to substitute in our earlier statement)}$

matchalolz · Apr 19, 2017

Re: MATH1131 help thread

InteGrand said:
$\noindent So for that last one, what you could do is decide what the function $f(t)$ is, what $a$ and $b$ should be for our purposes, calculate the derivative function $f'(t)$, and take $M$ to be the maximum value of $|f'(t)|$ on $[a,b]$. This would apply to the two parts before that one too.$

Ok I got part c. Trying to comprehend the first question

matchalolz · Apr 19, 2017

Re: MATH1131 help thread

leehuan said:
$\text{Fix }x> 0\text{ and let }f(t) = \sqrt{1+t}\implies f^\prime(t) = \frac{1}{2\sqrt{1+t}}$

$\\ \text{Since }f\text{ is continuous on }[0,x]\text{ and differentiable on }(0,x)\\ \text{By the MVT there is a }c\in (0,x)\text{ such that}\\ f^\prime(c) = \frac{f(x)-f(0)}{x-0}$

$\text{i.e. }\frac{1}{2\sqrt{1+c}}=\frac{\sqrt{1+x}-1}{x}$

$\text{Now, consider the following manipulation of inequalities that stem from }c\in (0,x):\\ 0 < c < x\\ 1 < 1+c < 1+x\\ 1 < \sqrt{1+c} < \sqrt{1+x}\\ 1 > \frac{1}{\sqrt{1+c}} > \frac1{\sqrt{1+x}}$

$\text{So from }\frac1{\sqrt{1+x}} < \frac1{\sqrt{1+c}} < 1\\ \text{Doing a bit of manipulation, noting that }x>0\text{, we have the desired result (remember to substitute in our earlier statement)}$

omg how did you get that out so quickly? I'm amazed

leehuan · Apr 19, 2017

Re: MATH1131 help thread

matchalolz said:
omg how did you get that out so quickly? I'm amazed

Reasons that are similar to how IG gets out all of my questions so quickly

matchalolz · Apr 19, 2017

Re: MATH1131 help thread

leehuan said:
Reasons that are similar to how IG gets out all of my questions so quickly

bos is probably going to carry me through math1131 at this rate

Paradoxica · Apr 19, 2017

Re: MATH1131 help thread

matchalolz said:
I'm stuck again...

View attachment 33882

With the first question, I have zero clue what to do...

And for the second question, I managed to do part a and part b but kind of lost with part c.

For all t>0 (and semi-equality at t=0)

$\frac{1}{2} > \frac{1}{2\sqrt{1+t}} > \frac{t}{4(1+t)\sqrt{1+t}}$

Which is simple enough to prove case-wise

Integrate both sides from 0 to x>0 to obtain:

$\frac{x}{2} > \sqrt{1+x} -1 > \frac{x}{2\sqrt{1+x}}$

Then add one throughout the inequality to obtain what was to be shown.

Paradoxica · Apr 19, 2017

Re: MATH1131 help thread

matchalolz said:
I'm stuck again...

View attachment 33882

With the first question, I have zero clue what to do...

And for the second question, I managed to do part a and part b but kind of lost with part c.

$\frac{1}{1002} = \frac{1}{1000} \frac{1}{1+\frac{1}{500}} = \frac{1}{1000} \left( 1-\frac{1}{500} + \frac{1}{500^2} - \frac{1}{500^3} - \cdots \right)$

Paradoxica · Apr 19, 2017

Re: MATH1131 help thread

leehuan said:
$\text{Fix }x> 0\text{ and let }f(t) = \sqrt{1+t}\implies f^\prime(t) = \frac{1}{2\sqrt{1+t}}$

$\\ \text{Since }f\text{ is continuous on }[0,x]\text{ and differentiable on }(0,x)\\ \text{By the MVT there is a }c\in (0,x)\text{ such that}\\ f^\prime(c) = \frac{f(x)-f(0)}{x-0}$

$\text{i.e. }\frac{1}{2\sqrt{1+c}}=\frac{\sqrt{1+x}-1}{x}$

$\text{Now, consider the following manipulation of inequalities that stem from }c\in (0,x):\\ 0 < c < x\\ 1 < 1+c < 1+x\\ 1 < \sqrt{1+c} < \sqrt{1+x}\\ 1 > \frac{1}{\sqrt{1+c}} > \frac1{\sqrt{1+x}}$

$\text{So from }\frac1{\sqrt{1+x}} < \frac1{\sqrt{1+c}} < 1\\ \text{Doing a bit of manipulation, noting that }x>0\text{, we have the desired result (remember to substitute in our earlier statement)}$

Day 436: Still haven't figured out the Mean Value Theorem.

matchalolz · Apr 19, 2017

Re: MATH1131 help thread

Paradoxica said:
Day 436: Still haven't figured out the Mean Value Theorem.

Me

Managed to catch up on 2 weeks of algebra and 2 weeks of calculus this week... but just realised that I still have 1.5 ish weeks of algebra and 2 weeks of calculus left to catch up on, why am I like this fml

matchalolz · Apr 19, 2017

Re: MATH1131 help thread

Screen Shot 2017-04-19 at 10.06.23 pm.png

Do I need to find some sort of interval to use MVT or IVT?

First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

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