Hard Induction Proofs (1 Viewer)

phoenix159 · Jan 19, 2014

see attachment:

QZP · Jan 19, 2014

Did you need help?

iJimmy · Jan 19, 2014

probably lol

timeflies · Jan 19, 2014

yea i think thats why he posted lol. or maybe just wanted to share some hard questions?

iJimmy · Jan 19, 2014

phoenix159 · Jan 19, 2014

yes i do need help

QZP · Jan 19, 2014

What have you tried? What are you stuck at?

For the first one, expand the double angle and you'll get it.
The second one is a lot more tricky but I'm trying to find a cleaner method.

phoenix159 · Jan 20, 2014

thanks

phoenix159 · Jan 20, 2014

i got the first one ok but i still need help with the second one

phoenix159 · Jan 20, 2014

QZP said:
What have you tried? What are you stuck at?

For the first one, expand the double angle and you'll get it.
The second one is a lot more tricky but I'm trying to find a cleaner method.

i got the first one ok but i still need help with the second one

mr.habibbi · Apr 13, 2021

phoenix159 said:
i got the first one ok but i still need help with the second one

yo i gotchu bro, sorry im 7 years late

liner · Apr 13, 2021

mr.habibbi · Apr 13, 2021

phoenix159 said:
i got the first one ok but i still need help with the second one

bit messy but here u go, hope u find it helpful

liner · Apr 13, 2021

i don't think he needs it anymore m8

CM_Tutor · Apr 13, 2021

The second one is easier (or at least different) by MX2 methods, by starting:

$\sin{x} + \sin{2x} + \sin{3x} + ... + \sin{nx}$
$= \Im{\left[(\cos{x} + i\sin{x}) + (\cos{2x} + i\sin{2x}) + (\cos{3x} + i\sin{3x}) + ... + (\cos{nx} + i\sin{nx})\right]}$
$= \Im{\left[(\cos{x} + i\sin{x}) + (\cos{x} + i\sin{x})^2 + (\cos{x} + i\sin{x})^3 + ... + (\cos{x} + i\sin{x})^n\right]} \qquad \text{using De Moivre's Theorem}$
$= \Im{\left(z + z^2 + z^3 + ... + z^n\right)} \qquad \text{taking $z = \cos{x} + i\sin{x}$}$
$= \Im{\left(\frac{z\left(z^n - 1\right)}{z - 1}\right)} \qquad \text{on summing the GP}$

$\text{Thus,} \qquad \sin{x} + \sin{2x} + \sin{3x} + ... + \sin{nx} = \Im{\left(\frac{z\left(z^n - 1\right)}{z - 1}\right)} \qquad \qquad \text{. . . . . . . . . (1)}$

We now need only simplify the complex number on the RHS of (1) and find its imaginary part to find the required sum.

$\begin{align*} \text{Now,} \qquad \frac{1}{z-1} &= \frac{\bar{z} - 1}{(z-1)(\bar{z} - 1)} = \frac{\bar{z} - 1}{z\bar{z} - z - \bar{z} + 1} = \frac{\bar{z} - 1}{|z|^2 - 2\Re{(z)} + 1} \qquad \text{as $z+\bar{z} = 2\Re{(z)}$ and as $z\bar{z} = |z|^2$} \\ &= \frac{\cos{x} - i\sin{x} - 1}{1 - 2\cos{x} + 1} \qquad \text{as $z=\cos{x}+i\sin{x}$ and hence $\Re{(z)} = \cos{x}$ and $|z| = 1$} \\ &= \frac{(\cos{x} - 1) - i\sin{x}}{2(1 - \cos{x})} = \frac{\cos{x} - 1}{2(1 - \cos{x})} - i\frac{\sin{x}}{2(1 - \cos{x})} = -\frac{1}{2} - i\frac{\sin{x}}{2(1 - \cos{x})} \\ &= -\frac{1}{2} - i\frac{2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}{2 \times 2\sin^2{\frac{x}{2}}} \qquad \text{recalling that $\sin{2\alpha}=2\sin{\alpha}\cos{\alpha}$ and that $1-\cos{2\alpha}=2\sin^2{\alpha}$ and taking $\alpha=\frac{x}{2}$} \\ \text{So,} \qquad \frac{1}{z-1} &= -\frac{1}{2} - i\frac{\cos{\frac{x}{2}}}{2\sin{\frac{x}{2}}} \qquad \qquad \text{. . . . . . . . . (*)} \end{align*}$

$\begin{align*} \text{And,} \qquad z(z^n-1) &= z^{n+1} - z \\ &= \cos{(n+1)x}+i\sin{(n+1)x} - (\cos{x}+i\sin{x}) \qquad \text{by applying De Moivre's Theorem with as $z=\cos{x} + i\sin{x}$} \\ &= \cos{(n+1)x} - \cos{x} + i\left[\sin{(n+1)x} - \sin{x}\right] \\ &= -2\sin{\frac{(n+2)x}{2}}\sin{\frac{nx}{2}} + 2i\cos{\frac{(n+2)x}{2}}\sin{\frac{nx}{2}} \qquad \text{using the sums-to-products formulae} \\ z(z^n-1) &= -2\sin{\frac{nx}{2}}\left(\sin{\frac{(n+2)x}{2}} - i\cos{\frac{(n+2)x}{2}}\right) \qquad \qquad \text{. . . . . . . . . (**)} \end{align*}$

$\begin{align*} \text{Thus,} \qquad \frac{z(z^n-1)}{z-1} &= \frac{1}{z-1} \times z(z^n - 1) \\ &= -2\sin{\frac{nx}{2}}\left(\sin{\frac{(n+2)x}{2}} - i\cos{\frac{(n+2)x}{2}}\right) \times \left(-\frac{1}{2} - i\frac{\cos{\frac{x}{2}}}{2\sin{\frac{x}{2}}}\right) \qquad \text{using (*) and (**)} \\ \text{And so} \qquad \Im{\left(\frac{z(z^n-1)}{z-1}\right)} &= -2\sin{\frac{nx}{2}}\left(\sin{\frac{(n+2)x}{2}} \times \frac{-\cos{\frac{x}{2}}}{2\sin{\frac{x}{2}}} - \cos{\frac{(n+2)x}{2}} \times -\frac{1}{2}\right) \\ &= \frac{2\sin{\frac{nx}{2}}}{2\sin{\frac{x}{2}}}\left(\sin{\frac{(n+2)x}{2}} \times \cos{\frac{x}{2}} - \cos{\frac{(n+2)x}{2}}\sin{\frac{x}{2}}\right) \\ &= \frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}} \times \sin{\left(\frac{(n+2)x}{2} - \frac{x}{2}\right)} \\ \text{And so,} \qquad \sin{x}+\sin{2x}+\sin{3x}+...+\sin{nx} &= \cfrac{\sin{\cfrac{nx}{2}}\sin{\cfrac{(n+1)x}{2}}}{\sin{\cfrac{x}{2}}} \qquad \text{by applying (1), producing the required result} \end{align*}$

CM_Tutor · Apr 13, 2021

mr.habibbi said:
bit messy but here u go, hope u find it helpful View attachment 30516

In the prove true for $n=1$ , it is not wise to work on both sides at once. This should be set as LHS = and RHS = ... = LHS.

braintic · Apr 17, 2021

Pointing out that questions of this type (even if they involved no complex numbers) cannot be asked in the Extension 1 course.

CM_Tutor · Apr 17, 2021

braintic said:
Pointing out that questions of this type (even if they involved no complex numbers) cannot be asked in the Extension 1 course.

The following question was asked on the 1992 Sydney Grammar 4u trial:

Consider the sequence defined by

$u_1 = 12$
$u_2 = 30$
$u_n = 5u_{n-1}-6u_{n-2} \text{, for $n \geqslant 3$}$

(i) Determine the values of $u_3$ and $u_4$ .

(ii) Show that $u_n = 2 \times 3^n + 3 \times 2^n$ for $n = 1$ and $n = 2$ .

(iii) If $u_k = 2 \times 3^k + 3 \times 2^k$ and $u_{k+1} = 2 \times 3^{k+1} + 3 \times 2^{k+1}$ , where $k$ is a positive integer, prove that $u_{k+2} = 2 \times 3^{k+2} + 3 \times 2^{k+2}$ .

(iv) What conclusion may be reached as a result of parts (ii) and (iii)?

A question like this requires use of algebra and index laws. The conclusion in part (iv) uses the logic of induction but the response is simply that the general formula $u_n = 2 \times 3^n + 3 \times 2^n$ is true for all positive integers $n$ , and thus allows any member of the sequence to be determined. I can argue that the question is not an induction question, especially if part (iv) is excluded. The question could be extended to summing the series $u_1 + u_2 + u_3 + ... + u_n$ by splitting it into two GPs, making the topic sequences and series.

My point is that the syllabus may not allow a particular question in a particular form, but that doesn't mean the same content can't be examined by restructuring a question. A determined examiner can certainly bend the syllabus in places - not so far as breaking it (which using complex numbers would do for MX1) - but enough that I think paying a lot of attention to what is and is not "allowed" is unwise, especially for students who may be unaware of some subtleties. Predicting boundaries is also problematic for trials and internal papers where the scrutiny given in preparation is less stringent.

Another example: many MX1 students were convinced that binomial theorem questions wouldn't appear in the HSC as preliminary materials, but the 2020 HSC proved that to be incorrect. Any preliminary materials can be covered if they are part of a question on HSC course content.

Hard Induction Proofs (1 Viewer)

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