Author Topic: 3U Maths Question Thread (Read 1466686 times) Tweet Share

legorgo18 · « **Reply #1530 on:** February 22, 2017, 11:15:55 pm »

rui i tried that then idk whats happening after, do i group a part and it becomes a limiting sum or something

RuiAce · « **Reply #1531 on:** February 22, 2017, 11:20:50 pm »

Actually, here's something I thought of now.

After doing the splitting, evaluate the partial sums the way they are (they aren't limiting cause they're not infinite). So use $S_n=\frac{a(1-r^n)}{1-r}$ n times. Or maybe n-1 times since you might not need it for the first term (the 0.2)

And then, look at your terms closely. If you do some clever algebra you'll see that there's (hopefully) just ONE more partial sum left...

RuiAce · « **Reply #1532 on:** February 22, 2017, 11:37:59 pm »

I put my theory to the test. Do NOT look at it if you don't want to.

Answer

Are you sure you want to look at it?

Are you REALLY REALLY SURE you want to look at it?

Ok, here goes.

Last chance...

$\begin{align*}&\quad 0.2 + 0.22 + 0.222 + \dots + 0.\underbrace{22\dots2}_{n\, 2\text{'s}}\\ &= 0.2 + (0.2+0.02) + (0.2+0.02+0.002)+\dots+(0.2+0.02+\dots+\underbrace{0.0002}_{n\, 0\text{'s}})\\ &= \frac{0.2(1-0.1)}{1-0.1}+\frac{0.2(1-0.1^2)}{1-0.1}+\frac{0.2(1-0.1^3)}{1-0.1}+\dots+\frac{0.2(1-0.1^n)}{1-0.1}\\ &= \frac{0.2}{0.9}\left(n - (0.1+0.1^2+\dots+0.1^n)\right)\\ &= \frac29 \left(n-\frac{0.1(1-0.1^n)}{1-0.1}\right)\\ &=\frac29 \left(n-\frac{1-0.1^n}9\right)\end{align*}$ There may be a couple of silly mistakes.

Zainbow · « **Reply #1533 on:** February 24, 2017, 10:47:57 am »

Use the sum of a geometric series to find the sum of the first n powers of 3 (starting with n=0) and prove this result by mathematical induction.

Hey guys, I'm stuck with this can you help me

It doesn't work for n=0 so idk

RuiAce · « **Reply #1534 on:** February 24, 2017, 11:07:41 am »

Quote from: Zainbow on February 24, 2017, 10:47:57 am

Use the sum of a geometric series to find the sum of the first n powers of 3 (starting with n=0) and prove this result by mathematical induction.

Hey guys, I'm stuck with this can you help me
It doesn't work for n=0 so idk

$\text{We're interested in the sequence }1,3,9,27,81,\dots, 3^{n-1}\\ \text{Recall that }3^0=1\text{ which is our first term.}\\ \therefore 3^{n-1}\text{ is our }n\text{th term}$
$\text{Using the formula, we have}\\ S_n=\frac{1(3^n-1)}{3-1}=\frac{1}{2}\left(3^n-1\right)$
$\text{Hence, we wish to prove this proposition by induction:}\\ \boxed{1+3+9+\dots+3^{n-1}=\frac{1}{2}(3^n-1)}$
$\text{The proposition is clearly true when }n-1=0\\ \text{or equivalently }n=1\\ \text{as }RHS=\frac12(3^n-1)=1=LHS$
$\text{Assuming the truth of the proposition when }n=k\text{ gives}\\ 1+3+9+\dots+3^{k-1}=\frac{1}{2}(3^k-1)\\ \text{We wish to use the truth of this to prove the truth of the case }n=k+1\\ \text{i.e. } 1+3+9+\dots+3^{k-1}+3^k=\frac{1}{2}(3^(k+1)-1)$
$\text{By considering the assumption}\\ \begin{align*}LHS&=\frac{1}{2}(3^k-1) + 3^k\\ &= \frac{1}{2}(3^k) - \frac12 + 1(3^k)\\ &= \frac32 (3^k)-1\\ &= \frac12(3^{k+1})-\frac12\\ &= \frac12(3^{k+1}-1)\\ &= RHS\end{align*}\\ \text{Make sure to not get stuck in algebra}$

Zainbow · « **Reply #1535 on:** February 24, 2017, 11:11:10 am »

Quote from: RuiAce on February 24, 2017, 11:07:41 am

$\text{We're interested in the sequence }1,3,9,27,81,\dots, 3^{n-1}\\ \text{Recall that }3^0=1\text{ which is our first term.}\\ \therefore 3^{n-1}\text{ is our }n\text{th term}$
$\text{Using the formula, we have}\\ S_n=\frac{1(3^n-1)}{3-1}=\frac{1}{2}\left(3^n-1\right)$
$\text{Hence, we wish to prove this proposition by induction:}\\ \boxed{1+3+9+\dots+3^{n-1}=\frac{1}{2}(3^n-1)}$
$\text{The proposition is clearly true when }n-1=0\\ \text{or equivalently }n=1\\ \text{as }RHS=\frac12(3^n-1)=1=LHS$
$\text{Assuming the truth of the proposition when }n=k\text{ gives}\\ 1+3+9+\dots+3^{k-1}=\frac{1}{2}(3^k-1)\\ \text{We wish to use the truth of this to prove the truth of the case }n=k+1\\ \text{i.e. } 1+3+9+\dots+3^{k-1}+3^k=\frac{1}{2}(3^(k+1)-1)$
$\text{By considering the assumption}\\ \begin{align*}LHS&=\frac{1}{2}(3^k-1) + 3^k\\ &= \frac{1}{2}(3^k) - \frac12 + 1(3^k)\\ &= \frac32 (3^k)-1\\ &= \frac12(3^{k+1})-\frac12\\ &= \frac12(3^{k+1}-1)\\ &= RHS\end{align*}\\ \text{Make sure to not get stuck in algebra}$

Thanks Rui

kiwiberry · « **Reply #1536 on:** February 24, 2017, 12:32:14 pm »

Prove n! > n² for n>=4 by mathematical induction
This is how far I've gotten:
Assume true for n=k: k! > k²
Prove true for n=k+1, ie. (k+1)! > (k+1)²
LHS = (k+1)! = k!(k+1) > k²(k+1) (by assumption)

To prove it true from here you have to prove that k² >= (k+1) right? But is there any other way of doing this without drawing a graph? I've also tried moving the inequality to one side and proving that it's greater than 0 but I got stuck as well

jakesilove · « **Reply #1537 on:** February 24, 2017, 12:38:59 pm »

Quote from: kiwiberry on February 24, 2017, 12:32:14 pm

Prove n! > n² for n>=4 by mathematical induction
This is how far I've gotten:
Assume true for n=k: k! > k²
Prove true for n=k+1, ie. (k+1)! > (k+1)²
LHS = (k+1)! = k!(k+1) > k²(k+1) (by assumption)

To prove it true from here you have to prove that k² >= (k+1) right? But is there any other way of doing this without drawing a graph? I've also tried moving the inequality to one side and proving that it's greater than 0 but I got stuck as well

To prove that

$k^2 \geq k+1$
$k^2-k-1 \geq 0$
$(k-\frac{1}{2})^2-\frac{5}{4} \geq 0$
$(k-\frac{1}{2})^2 \geq \frac{5}{4}$

So, we know that as k gets bigger, the left hand side will always get bigger. The minimum value for k is 4. If you sub 4 in, that's clearly greater than the right hand side! So, for all further values of k, the left hand side will also be bigger than the right hand side.

This is a round about way of proving something usually done in a more mathematically rigorous way, but that is reserved for university mathematics

Actually, you could also solve for k. If you find that the equation is true for all k>x, and than x<4, then you've proven the thing rigorously!

kiwiberry · « **Reply #1538 on:** February 24, 2017, 12:56:02 pm »

Quote from: jakesilove on February 24, 2017, 12:38:59 pm

To prove that

$k^2 \geq k+1$
$k^2-k-1 \geq 0$
$(k-\frac{1}{2})^2-\frac{5}{4} \geq 0$
$(k-\frac{1}{2})^2 \geq \frac{5}{4}$

So, we know that as k gets bigger, the left hand side will always get bigger. The minimum value for k is 4. If you sub 4 in, that's clearly greater than the right hand side! So, for all further values of k, the left hand side will also be bigger than the right hand side.

This is a round about way of proving something usually done in a more mathematically rigorous way, but that is reserved for university mathematics

Actually, you could also solve for k. If you find that the equation is true for all k>x, and than x<4, then you've proven the thing rigorously!

Ahh I see, thanks Jake

bsdfjnlkasn · « **Reply #1539 on:** February 24, 2017, 08:09:37 pm »

Hey dudes,

Can I please get some help with finding the general solution and all solutions for the following for -180°<x<180° (in radians and the inequality signs are supposed to have the line indicating equality too - sorry i'm not sure how to type these)

THANK YOU!! I would also really appreciate explanations not just the method as my teacher didn't explain this topic at all ...

RuiAce · « **Reply #1540 on:** February 24, 2017, 08:33:57 pm »

Quote from: bsdfjn;lkasn on February 24, 2017, 08:09:37 pm

Hey dudes,

Can I please get some help with finding the general solution and all solutions for the following for -180°<x<180° (in radians and the inequality signs are supposed to have the line indicating equality too - sorry i'm not sure how to type these)

THANK YOU!! I would also really appreciate explanations not just the method as my teacher didn't explain this topic at all ...

You're asking for an answer in degrees but... your question is in radians?
$\text{Using the formula on the reference sheet}\\ \begin{align*}\tan \left(2x-\frac{\pi}6\right)&=-\sqrt3\\ 2x-\frac{\pi}6&=n\pi + \tan^{-1}(-\sqrt3)\qquad (n\in \mathbb{Z})\\2x-\frac{\pi}{6} &= n\pi - \frac{\pi}{3}\\ 2x&={n}\pi-\frac{\pi}{6}\\ x&= \frac{n\pi}{2}-\frac{\pi}{12}\end{align*}$
$\text{Given the general solution, we can find our particular solutions}\\ \text{satisfying }-\pi \le x \le \pi\\ \text{just by reading off the general solution}$
$\text{By inspection the solutions occur when }n=-1,0,1,2\\ \text{So they are }x=-\frac{7\pi}{12}, -\frac{\pi}{12}, \frac{5\pi}{12}, \frac{11\pi}{12}$
Note that when I say by inspection, since we know that n must be an integer, we just guess and check values of n that are between -pi and pi
______________________________
$\text{Or if we don't have the general solution and want a purely 2U approach}\\ \text{Commence by shifting the domain to match the argument}$
$-\pi \le x\le \pi\\ -2\pi \le 2x\le 2\pi\\ \therefore \boxed{-\frac{13\pi}6\le 2x-\frac{\pi}{6} \le \frac{11\pi}{6}}$
The purpose of changing the argument is because we're taking the tan of the WHOLE thing. So we need to compromise.
$\text{For this domain, solving }\tan \left(2x-\frac{\pi}6\right)=-\sqrt3\\ \text{noting that tan is negative in the}\\ \text{2nd quadrant, 4th quadrant, negative 1st quadrant AND negative 3rd quadrant}\\ \text{and that }\tan \frac{\pi}{3}=\sqrt3$
Regarding the quadrants, you're going to have to draw the diagram out. Ensure that you know what the 1st, 2nd, 3rd, 4th quadrants are, AND the negative 1st to 4th quadrants.

The terminology used here depends really on what textbook you used. I know Cambridge calls it the "related angle" that you combine with the rule of ASTC.
$\begin{align*}\tan \left(2x-\frac{\pi}6\right)&=-\sqrt3\\ 2x-\frac{\pi}6&=-\frac{4\pi}{3},-\frac{\pi}{3},\frac{2\pi}3,\frac{5\pi}{3}\\ 2x&=-\frac{7pi}{6},-\frac{\pi}{6},\frac{5\pi}{6},\frac{11\pi}{6}\\x&=-\frac{7pi}{12},-\frac{\pi}{12},\frac{5\pi}{12},\frac{11\pi}{12} \end{align*}$

bsdfjnlkasn · « **Reply #1541 on:** February 24, 2017, 08:42:20 pm »

Quote from: RuiAce on February 24, 2017, 08:33:57 pm

You're asking for an answer in degrees but... your question is in radians?
$\text{Using the formula on the reference sheet}\\ \begin{align*}\tan \left(2x-\frac{\pi}6\right)&=-\sqrt3\\ 2x-\frac{\pi}6&=n\pi + \tan^{-1}(-\sqrt3)\qquad (n\in \mathbb{Z})\\2x-\frac{\pi}{6} &= n\pi - \frac{\pi}{3}\\ 2x&={n}\pi-\frac{\pi}{6}\\ x&= \frac{n\pi}{2}-\frac{\pi}{12}\end{align*}$
$\text{Given the general solution, we can find our particular solutions}\\ \text{satisfying }-\pi \le x \le \pi\\ \text{just by reading off the general solution}$
$\text{By inspection the solutions occur when }n=-1,0,1,2\\ \text{So they are }x=-\frac{7\pi}{12}, -\frac{\pi}{12}, \frac{5\pi}{12}, \frac{11\pi}{12}$
Note that when I say by inspection, since we know that n must be an integer, we just guess and check values of n that are between -pi and pi
______________________________
$\text{Or if we don't have the general solution and want a purely 2U approach}\\ \text{Commence by shifting the domain to match the argument}$
$-\pi \le x\le \pi\\ -2\pi \le 2x\le 2\pi\\ \therefore \boxed{-\frac{13\pi}6\le 2x-\frac{\pi}{6} \le \frac{11\pi}{6}}$
The purpose of changing the argument is because we're taking the tan of the WHOLE thing. So we need to compromise.
$\text{For this domain, solving }\tan \left(2x-\frac{\pi}6\right)=-\sqrt3\\ \text{noting that tan is negative in the}\\ \text{2nd quadrant, 4th quadrant, negative 1st quadrant AND negative 3rd quadrant}\\ \text{and that }\tan \frac{\pi}{3}=\sqrt3$
Regarding the quadrants, you're going to have to draw the diagram out. Ensure that you know what the 1st, 2nd, 3rd, 4th quadrants are, AND the negative 1st to 4th quadrants.

The terminology used here depends really on what textbook you used. I know Cambridge calls it the "related angle" that you combine with the rule of ASTC.
$\begin{align*}\tan \left(2x-\frac{\pi}6\right)&=-\sqrt3\\ 2x-\frac{\pi}6&=-\frac{4\pi}{3},-\frac{\pi}{3},\frac{2\pi}3,\frac{5\pi}{3}\\ 2x&=-\frac{7pi}{6},-\frac{\pi}{6},\frac{5\pi}{6},\frac{11\pi}{6}\\x&=-\frac{7pi}{12},-\frac{\pi}{12},\frac{5\pi}{12},\frac{11\pi}{12} \end{align*}$

I don't think you understand how clear and helpful this has been for me - THANK YOU SO VERY MUCH!!

jamonwindeyer · « **Reply #1542 on:** February 24, 2017, 08:54:14 pm »

Quote from: bsdfjn;lkasn on February 24, 2017, 08:42:20 pm

I don't think you understand how clear and helpful this has been for me - THANK YOU SO VERY MUCH!!

Rui is a superhero

bsdfjnlkasn · « **Reply #1543 on:** February 24, 2017, 09:20:06 pm »

Damn what did I do wrong :/ - I'm still getting the hang of this

bsdfjnlkasn · « **Reply #1544 on:** February 24, 2017, 09:21:31 pm »

Sorry for asking so many questions, but could I please get a proper method for 8d and a hint for 8b? Thank you so much

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