AAPL Elliott Wave Technical Analysis – 11th April, 2013

Last analysis of AAPL expected price to rise which is what has happened. With a little structure within this new upwards movement to analyse I can now calculate short term targets. I expect this upwards movement to last about another two weeks.

Click on the charts below to enlarge.

AAPL Elliott Wave Chart daily 2013

This main wave count expects a five wave impulse for a cycle degree wave a is unfolding to the downside. Within the impulse primary waves 1 and 2 are complete. Primary wave 3 may be extending. Within primary wave 3 intermediate wave (1) is a complete five wave structure and intermediate wave (2) is unfolding.

Within intermediate wave (2) minor waves A and B are now most likely complete. At 502 minor wave C would reach 1.618 the length of minor wave A and intermediate wave (2) would end close to the 50% mark of intermediate wave (1).

Intermediate wave (2) may not move beyond the start of intermediate wave (1). This main daily wave count is invalidated with movement above 594.59. If price moves above this point then we may use the alternate wave count below.

At 272 primary wave 3 would reach 1.618 the length of primary wave 1. This long term target is months away.

We can now draw a parallel channel about intermediate wave (2). Draw the first trend line from the start of minor wave A to the end of minor wave B, then place a parallel copy upon the high of minor wave A. At 471 minor wave C would reach equality with minor wave A and price would find resistance at the upper end of this channel. At 502 minor wave C would reach 1.618 the length of minor wave A and price would overshoot the channel.

When the zigzag for intermediate wave (2) may be seen as complete then subsequent movement below the channel would confirm that the next wave, intermediate wave (3), would be underway.

AAPL Elliott Wave Chart hourly 2013

Price has moved higher exactly as expected. Minor wave C may subdivide into one of two structures: either an ending diagonal or an impulse. The structure so far looks like an unfolding impulse.

So far we do not have a five wave structure upwards complete and it looks like a triangle has been completing over the last day. This indicates a sharp upwards thrust should occur when the triangle has completed.

Within minute wave i minuette wave (iii) has no Fibonacci ratio to minuette wave (i). This makes it more likely that we shall see a Fibonacci ratio between minuette wave (v) and either of (iii) or (i). At 441 minuette wave (v) would reach equality in length with minuette wave (i). As this is the most common relationship between first and fifth waves this target has the highest probability.

Within minuette wave (iii) there are no Fibonacci ratios between subminuette waves i, iii and v. This gives me a small cause for concern; normally AAPL does exhibit Fibonacci ratios between actionary waves of an impulse. However, the strength of the triangle structure indicates the fourth wave placement is labeled correctly.

Within the triangle for minuette wave (iv) it looks like subminuette wave e is itself completing as a triangle when viewed on the five minute chart. This structure should be completed within a couple of hours, or even sooner.

Movement above 436.35 would confirm the triangle is over and the fifth wave is underway.

When minuette wave (v) completes minute wave i then I would expect a longer lasting downwards correction for minute wave ii which may not move beyond the start of minute wave i. This wave count is invalidated with movement below 419.68.

Draw a channel about minute wave i. Draw the first trend line from the highs of minuette waves (i) to (iii), then place a parallel copy upon the low of subminuette wave ii within minuette wave (iii). Expect minuette wave (v) to end either midway within the channel, or to find resistance about the upper edge. When the channel is clearly breached by subsequent downwards movement that would provide confirmation that minute wave i is complete and minute wave ii is underway.

Alternate Daily Wave Count.

AAPL Elliott Wave Chart daily alternate 2013

If the main daily wave count is invalidated with upwards movement above 594.59 then this alternate would be confirmed.

If cycle wave a is subdividing into a three wave structure then within it primary wave A may be a complete zigzag. Cycle wave a may be unfolding as a flat correction.

Within primary wave A intermediate waves (A) and (C) have no Fibonacci ratio to each other.

The parallel channel drawn here using Elliott’s technique about a correction is clearly breached by upwards movement. The downwards zigzag is over and an upwards trend has begun.

Within primary wave B minor wave 2 may not move beyond the start of minor wave 1. This wave count is firmly invalidated with movement below 419.

Primary wave B must subdivide into a three wave structure, most likely a zigzag.

If cycle wave a is a flat correction then primary wave B must reach at least 90% the length of primary wave A at 676.46.

Primary wave B may make a new high above 705.07. Primary wave B should last about six months.