Author Topic: werid polynomial q (Read 4068 times) Tweet Share

massive · « **on:** July 12, 2016, 09:03:06 pm »

how do you do this question guys?

Prove that 1 + x + x^2/2! +...+ x^n/n! has no multiple roots for any n>=1. (greater than or equal to)

thnx

RuiAce · « **Reply #1 on:** July 12, 2016, 09:09:30 pm »

Quote from: massive on July 12, 2016, 09:03:06 pm

how do you do this question guys?

Prove that 1 + x + x^2/2! +...+ x^n/n! has no multiple roots for any n>=1. (greater than or equal to)

thnx

$\text{If a root is a multiple root, it is at least a double root. Therefore it must satisfy }f(x)=f^\prime(x)=0$
$f^\prime(x)=1+x+\frac{x^2}{2!}+\dots+\frac{x^{n-1}}{(n-1)!}$
$\begin{align*}\text{For }f(x)&=f^\prime(x)\\ 1+x+\frac{x^2}{2!}+\dots+\frac{x^{n-1}}{(n-1)!}+\frac{x^n}{n!}&=1+x+\frac{x^2}{2!}+\dots+\frac{x^{n-1}}{(n-1)!}\\ \frac{x^n}{n!}=0\\ x=0\end{align*}$
$\text{Thus }x=0\text{ is the only candidate for a multiple root.}$
$\text{But }f(0)=1+0+\frac{0^2}{2!}+\frac{0^3}{3!}+\dots+\frac{0^n}{n!}=1\neq 0$
$\text{So }x=0\text{ isn't even a root, thus contradicting what we require.}$
$\text{Hence there are no roots.}$

massive · « **Reply #2 on:** July 12, 2016, 09:17:50 pm »

You are the saviour to my 4U dreams and aspirations!

brenden · « **Reply #3 on:** July 12, 2016, 09:45:28 pm »

Quote from: RuiAce on July 12, 2016, 09:09:30 pm

$\text{If a root is a multiple root, it is at least a double root. Therefore it must satisfy }f(x)=f^\prime(x)=0$
$f^\prime(x)=1+x+\frac{x^2}{2!}+\dots+\frac{x^{n-1}}{(n-1)!}$
$\begin{align*}\text{For }f(x)&=f^\prime(x)\\ 1+x+\frac{x^2}{2!}+\dots+\frac{x^{n-1}}{(n-1)!}+\frac{x^n}{n!}&=1+x+\frac{x^2}{2!}+\dots+\frac{x^{n-1}}{(n-1)!}\\ \frac{x^n}{n!}=0\\ x=0\end{align*}$
$\text{Thus }x=0\text{ is the only candidate for a multiple root.}$
$\text{But }f(0)=1+0+\frac{0^2}{2!}+\frac{0^3}{3!}+\dots+\frac{0^n}{n!}=1\neq 0$
$\text{So }x=0\text{ isn't even a root, thus contradicting what we require.}$
$\text{Hence there are no roots.}$

Quote from: massive on July 12, 2016, 09:03:06 pm

how do you do this question guys?

Prove that 1 + x + x^2/2! +...+ x^n/n! has no multiple roots for any n>=1. (greater than or equal to)

thnx

Can I just point out that Rui answered this question 6 minutes after it was asked? Nowhere else can you get such phenomenal help so quickly. This is why I love ATAR Notes.

conic curve · « **Reply #4 on:** July 12, 2016, 09:51:21 pm »

Quote from: brenden on July 12, 2016, 09:45:28 pm

Can I just point out that Rui answered this question 6 minutes after it was asked? Nowhere else can you get such phenomenal help so quickly. This is why I love ATAR Notes.

Of course. Who doesn't love ATAR Notes, best HSC forum out there, especially the free essay marking from all the moderators. Thanks for all your hard work guys

If I ace my HSC, I'll dedicate a speech to you all ahaha

massive · « **Reply #5 on:** July 14, 2016, 03:26:40 am »

Hey everyone howdu do part i and part iii. :S

RuiAce · « **Reply #6 on:** July 14, 2016, 09:29:59 am »

Quote from: massive on July 14, 2016, 03:26:40 am

Hey everyone howdu do part i and part iii. :S

$\text{Direct substitution shows that }P(0)=1\neq 0\text{ so }\alpha \neq 0$
$\text{Assume that }x=1\text{ is a root of }P(x)=0\text{. Then}\\ 1+A+B+A+1=0 \implies 2+B = -2A$
$\text{Similarly, if }x=-1\text{ is a root of }P(x)=0\text{ we have }2+B = 2A$
$\text{But }(2+B)^2\neq 4A^2 \implies 2+B \neq \pm 2A\\ \text{So the above two statements are contradictions and hence the root cannot be either of }x=\pm 1$
_________________
$\text{One way, albeit a very weird way to show that the multiplicity cannot be greater than 2 is to use product of roots:} \\ \text{Assume }x=\alpha\text{ has a multiplicity of 3. Then by the product of roots: } \alpha^3 \times \frac{1}{\alpha}=\frac{1}{1} \implies \alpha^2 = 1\\ \text{But this cannot be possible! So we have a contradiction.}\\ \text{Also, }\alpha\text{ definitely does not have multiplicity 4 otherwise }\frac{1}{\alpha}\text{ cannot be a root.}\\ \text{ (This is corollary of fundamental theorem of algebra - exactly 4 roots in P(x))}\\ \text{Hence, }\alpha\text{ is a root of multiplicity 2.}$
Where was my inspiration to use product of roots and proof by contradiction? Out of thin air. I ran out of things obvious using the second derivative so I thought ahead about this. The fact that 1/alpha is a root helped me greatly.
$\text{Suppose that the other root is }\beta\\ \text{Then: }\alpha^2 \times \frac{1}{\alpha} \times \beta = 1 \implies \beta = \frac{1}{\alpha}\\ \text{Hence the roots are }\alpha,\alpha,\frac{1}{\alpha},\frac{1}{\alpha}$
$\text{Sum of 1 root at a time: }\alpha+\alpha+\frac{1}{\alpha}+\frac{1}{\alpha}=-A \implies \alpha+\frac{1}{\alpha}=-\frac{A}{2} \\ \therefore \left(\alpha+\frac{1}{\alpha}\right)^2=\frac{A^2}{4} \implies \alpha^2+\frac{1}{\alpha^2}+2={A^2}{4}$
$\text{Sum of 2 roots at a time: }\alpha^2+1+1+1+1+\frac{1}{\alpha^2}=B\\ \therefore \alpha^2+\frac{1}{\alpha^2}+2=B-2$
$\therefore \frac{A^2}{4}=B-2 \implies 4B=A^2+8$

jakesilove · « **Reply #7 on:** July 14, 2016, 10:18:25 am »

Quote from: RuiAce on July 12, 2016, 09:09:30 pm

$\text{If a root is a multiple root, it is at least a double root. Therefore it must satisfy }f(x)=f^\prime(x)=0$
$f^\prime(x)=1+x+\frac{x^2}{2!}+\dots+\frac{x^{n-1}}{(n-1)!}$
$\begin{align*}\text{For }f(x)&=f^\prime(x)\\ 1+x+\frac{x^2}{2!}+\dots+\frac{x^{n-1}}{(n-1)!}+\frac{x^n}{n!}&=1+x+\frac{x^2}{2!}+\dots+\frac{x^{n-1}}{(n-1)!}\\ \frac{x^n}{n!}=0\\ x=0\end{align*}$
$\text{Thus }x=0\text{ is the only candidate for a multiple root.}$
$\text{But }f(0)=1+0+\frac{0^2}{2!}+\frac{0^3}{3!}+\dots+\frac{0^n}{n!}=1\neq 0$
$\text{So }x=0\text{ isn't even a root, thus contradicting what we require.}$
$\text{Hence there are no roots.}$

As a side note, I always loved this proof

RuiAce · « **Reply #8 on:** July 14, 2016, 10:41:12 am »

Quote from: jakesilove on July 14, 2016, 10:18:25 am

As a side note, I always loved this proof

Pretty nice classic

massive · « **Reply #9 on:** July 14, 2016, 06:42:48 pm »

Thanks mate, also this one if you dnt mind, i got everything except part iv

RuiAce · « **Reply #10 on:** July 14, 2016, 07:30:54 pm »

Quote from: massive on July 14, 2016, 06:42:48 pm

Thanks mate, also this one if you dnt mind, i got everything except part iv

Doing questions out of the 1994 HSC is a really big risk.
$\text{Using our definition of }x+\frac{1}{x}=u \iff x^2-ux+1=0\\ \text{No solutions for }x\text{ exist if }\Delta=u^2-4 < 0 \\ \therefore -2 < u < 2$
$\text{Now, if }R(u)\text{ has no solutions, working backwards it is impossible for }P(x)\text{ to as well.}\\ \text{This occurs when }A^2-4(B-2) < 0 \\ \implies B > \frac{A^2}{4}+2$
$\text{Matching regions, }B=\frac{A^2}{4}+2\text{ must be the curved portion }c\text{ of the graph.}$
$\text{The next part is tricky. Assume that }u=\beta \text{ is a root of }R(u)\\ \text{Then, using the info earlier we must have }-2 < \beta < 2$
$\text{When this sort of stuff happens, consider the boundary cases.}\\ \textbf{This is because the boundary case must be where }\mathcal{D}\textbf{cuts off from the rest of the plane.}\\ R(2)=4+2A+B-2=0\\ \therefore B=-2A-2\\ R(-2)=4-2A+B-2=0\\ \therefore B=2A-2$
$\text{Note that the region in our diagram is only the first quadrant. Hence, the line }l\text{ must be the line }B=2A-2$
$\textbf{One thing to note: The line and the curve are dashed - this is why we can treat safely the less than but not equal to.}\\ \text{The line }l\text{ and curve }c\text{ do not actually belong to the region }\mathcal{D}$
$\text{Solving simultaneous equations as }T\text{ is the point of intersection:}\\ 2A-2=\frac{A^2}{4}+2 \implies \dots \implies A=4,B=6$
$\therefore T\text{ is }(4,6)$

massive · « **Reply #11 on:** July 14, 2016, 07:41:31 pm »

Thankss but why is it a big risk?

RuiAce · « **Reply #12 on:** July 14, 2016, 07:42:54 pm »

Quote from: massive on July 14, 2016, 07:41:31 pm

Thankss but why is it a big risk?

The difficulty of it isn't worth it (unless you're aiming for state rank or something). And you can easily feel distraught for not having any idea as to how to tackle it.

Time could easily be spent on current level difficulty questions and ensuring you don't make silly mistakes.

massive · « **Reply #13 on:** July 29, 2016, 04:44:10 pm »

How do you do part iii ? :S

RuiAce · « **Reply #14 on:** July 29, 2016, 05:06:09 pm »

Quote from: massive on July 29, 2016, 04:44:10 pm

How do you do part iii ? :S

$\textbf{1995 HSC 4U Additional}$
$\text{We have }f^\prime (t) = 3t^2 + c\\ \text{The roots of }f^\prime(t)=0\text{ will correspond to the points }u\text{ and }v\\ 3t^2+c=0\text{ solves to give }t=\pm \sqrt{-\frac{c}{3}}$
$\text{We wish to somehow make use of the fact that }f(u)f(v)< 0\\ \text{Notice from above how }v=-u\\ \begin{align*}\therefore f(u)f(v)&=f(u)f(-u)\\ &= \left(u^3+3cu+d\right)\left(-u^3-3cu+d\right)\\ &= \left(u(u^2+c)+d\right)\left(-u(u^2+c)+d\right)\end{align*}$
$\text{Recall that }f^\prime(u) = 0 \iff 3u^2+c=0\\ \begin{align*}\therefore f(u)f(v)&= \left(u\left(\frac{c}{3}+c\right)+d\right)\left(-u\left(\frac{c}{3}+c\right)+d\right)\\ &= \left(d+\frac{2uc}{3}\right)\left(d-\frac{2uc}{3}\right)\\ &= d^2 - \frac{4c^2u^2}{9}\\ &=d^2 - \frac{4c^2}{9}\left(-\frac{c}{3}\right) \tag{!}\\ &= d^2+\frac{4c^3}{27}\end{align*}$
$\text{So using the information given: }d^2+\frac{4c^3}{27} < 0 \iff 27d^2 + 4c^3 < 0$

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