Euler's Formula for the Quaternions

Transcript

what we're going to do in this video is derive a really interesting extension of oilage formula this time for the querian which is going to allow us to calculate e to a quan power Oilers formula is among the most famous formulas in math and what it allows us to do is calculate the exponentiation of a complex number and what I mean by that is that if we have e as a base and we raise e to some complex number power Oiler formula allows us to give the complex number that results from the computation and it says that e to the IX power is going to be cosine X as the new real part plus I * sinx as the new imaginary part and if you plug in the special case where X is equal to Pi we find that e to the I Pi power is equal to minus one and if you move that to the left hand side you say that e to the I Pi + 1 you find that that's equal to zero which is what we call Oilers identity and to give another special case if we plug in xals 1 into this formula the formula tells us that e to the I power is equal to cosine one is the new real part which is 0 54 plus I * sin 1 as the new imaginary part so 084 I and if we take get another example if we take numbers such as 2 plus I which are some real number plus an imaginary part as opposed to just a purely imaginary number here if we take e to 2 plus I we can use our basic rules of exponentiation here we can break this apart as e^2 * e to the I just like we would normally do if we have e to the a plus b power we can break that apart as e to the a * e to the B power and we find that e to 2 + I is going to be given by E2 times precisely what we found up above for e to the I so we just take what we found up here and multiply by e^2 which will give us 3.99 + 6.22 I and what we're going to do in this formula is derive an extension of Oilers formula for the quorans and what that's going to allow us to do is calculate things like e to the I + J plus K and what we're also going to demonstrate actually what I'm going to do is give you the tools to allow you to demonstrate an extension of Oilers identity for the querian and that's going to say that e to the I Pi e to the J Pi e to the K Pi are all equal to minus one along with a couple others e to the pi overk 2 I plus pi overun 2 J that's also equal to minus one and lastly e to the pi over < tk3 I plus piun over < tk3 J plus pi over < tk3 K is also equal to minus one let me first review the querian a little bit in this video we're going to think of a querian as a four-dimensional object can which can be written as a scalar part plus a three-dimensional Vector part that is if I have a querian such as 1 + I + 2 J plus 3 K I'm going to interpret this as one being the Scala part and the vector 1 2 3 as the three Vector part and this is normally how quion are interpreted anyway so this is really nothing new in terms of notation I I could write it like this one being inserted in that first place there one being the scalar part and this three Vector 1 2 3 being placed in the second component of this order pair in general I could write a querian like this a querian q can be written as the order pair where W is the scalar part and the vector v is going to be the vector part in the above case we had W is equal to 1 and the vector part was 1 two three which I wrote up here and what we're going to need to know in this video is how to multiply pans and if you don't already know multiplying querian follows a very special rule which is that if I have some querian let's say q1 which is given by W1 V1 and some other Quan Q2 which is given by W2 V2 the way in which I multiply q1 Q2 is in the following way the new scaler part is going to be given by W1 * W2 minus the do product between V1 and V2 and the dot product if you don't know that already I encourage you to look that up and the new Vector part is going to be given by W1 * vs2 plus W2 * V1 plus the cross product between V1 and V2 in that order because remember the the cross product you can't commute them it's V1 cross V2 if I'm multiplying q1 Q2 so this is the rule we're going to need to know in this video and in general quum multiplication is non-commutative that is to say that q1 Q2 is not the same as Q2 * q1 now quary multiplication is a little bit complicated but the only thing I'm really going to be doing in this video is is I'm going to take powers of querian which have the form z v that is querian which have zero as a scalar part I'm just going to be taking powers of them that is I'm going to calculate things like q^2 Q cubed Q to 4th Q to 5th and so on so let's start calculating some powers of Q now q^2 is going to be given by 0 V multipli by itself z v and now we just follow the rule that I outlined in the last Slide the zero being in the scaler part is going to simplify the formula so I'm going to have 0 * 0 which is going to give me 0 minus the do product of V with itself so the new scal of part is going to be given by minus v.v and now the new Vector part is going to be given by 0 * V and plus 0 time V plus the cross product of V with v and if you don't know already the cross product of a vector with itself is the zero Vector so all in all the new Vector part is going to be the zero Vector so this is a pretty simple formula for q^2 just going to get negative v.v for the new Scala part and the vector par goes is zero and what is negative v.v when I take the cross when I take the do product of vector with itself that's simply the squared length of the vector so negative v. V is negative the length of v^2 and just to give you an example let's say I have q which is the quitan zero as a scalar part and 111 as a vector part which would also be written as Just i+ J plus K and let's square that so we found up up above that q^2 is going to be the negative do product of this Vector of the vector 111 the dot product of 111 with itself is going to be three so I just negate the three to get minus 3 as q^2 so you notice that we have we start with this quern which is purely a vector has zero scalar part when I Square it it's purely a scalar and no Vector part and now that's actually the the toughest part of doing this calculation from knowledge of q^ s we can now calculate the higher Powers Q cubed is just going to be Q * q^2 and since I found that Q square is just a vector I'm just going to take V and multiply it by negative the squar length of v and that's going to be Q cubed then Q to 4th if q^2 is equal to minus V the length of v^2 then I just square that to get Q to 4 is equal to the length of V to the fourth power I just square that value there and in our example up above I know what Q ^2 was that was minus 3 to find Q cubed I'm just going to take that minus three and multiply by V that vector v so I just take minus 3times 111 and that's going to give me this Vector here which is -3 -3 - 3 then Q to the 4th is just going to be the squared value of minus three which is 9 and what we have to observe here is that for even powers like q^2 Q to 4th we get only the scalar part and for odd Powers such as Q to the 1st Q cubed then if you continue it would be Q to the 5th and so on it's some scaler times the original Vector that's really the observation I want to make from these calculations here and that's going to help in deriving Oilers formula for the querian to derive our formula we should first remember what we mean by exponentiation in general what do we mean by E to X that is when I plug this into my calculator how does my calculator know what to do so what we mean by e to the x is the following tailor series really e to X is given by 1 + x + x^2 over 2 factorial plus X 3r over 3 factorial plus x 4 over 4 factorial and so on where you have plus x to the N Over N factorial that's the definition of exponentiation and if you're wondering why we started by calculating some powers of querian this is why in preparation for this calculation so now that we know a few powers of Q let's plug in Q into this formula so now I'm going to have e to the Q equals something over here on the right hand side and what we're going to do first is assume that it has that form Z V where the the scaler part is zero so we find that e to Q is equal to this form here just this formula with X being equal to q and now from our knowledge of Q ^2 q to3 q to the 4th and so on we can now do the substitution now Q we're assuming that it just has a special form where the scalar part is zero Q is just equal to V the vector v we found that Q ^2 was equal to negative the length squar of V so I just make that substitution there for Q to the 3 I find that it's this value the negative uh squared length of V times V all right and then Q to the 4th I find that that's the length of V to the fourth power then if I continue I'm going to find that Q to the 5th is just that thing the length of the fourth power times V now I continue on Q to the 6 is going to be the negative length of V to the 6 power and then hopefully you can see the pattern by this point then in the denominator I just have n factorial here I have zero factorial here I have one factorial here I have two factorial three four and so on just making that substitution right into the power series for exponentiation and now what we're going to do is we're going to make that observation that the even Powers here that is this is like the zeroth power here this is the second power this is the fourth the sixth power over here these even powers are pure scalar parts that is one is a pure scalar there's no Vector part there this thing right here is a pure scalar this is a pure scalar here and for all the even powers and furthermore all the odd Powers this would be the first Power here this would be the third power here the fifth and so on all of these are pure Vector parts that is there just some quantity some coefficient times the vector v for example in the let's say the third power here third power because it's three factori tutorial it's that coefficient netive the squ length of V time V these are the pure Vector parts so what we're going to do next is break these apart into scalar and Vector Parts up above I've Rewritten what we have so far for e to the Q Power and first what we're going to do is have a look at the scalar part of this formula only and just remind you this is going to be the even powers of Q so what I get when I isolate this it's just this term this term this term this one and so on and let's first observe the pattern here so we see that this series is alternating I have the plus here minus plus minus plus and so on and I see that these lengths of V are in every second power I have the zero of power here I have the second the fourth the sixth the eighth and so on so what does this power series remind you of so this is what we're going to have to do now we're going to have to have to find what function has this thing I have written here as its tailor series and hopefully that observation that we made about this thing being alternating and it comes in every second power I have the zero the second fourth that is to say it skips all of the odd ones I don't have a first power I don't have a third or fifth or so on and hopefully this reminds you of an even function namely the cosine function and if you have a look if you remember what the cosine function Taylor series looks like it's going to be 1 - x^2 over 2 factorial plus x 4 4 factorial minus x 6 6 factorial this is a cosine tayor Series in the case where X takes on the value length of V which means that the scalar part is given by cosine of the length of v and this is half the work already so we found how to calculate the scalar part of the exponentiation of a quum already and next we're going to have a look at the vector part of this formula and again up above I've written out the complete series for e to the Q and now let's have a look at the vector part of this equation that is I'm going to have a look at this term which is the first Power of Q this is the third power here have look at this this and this and so on and I've written it in the following way I've just factored out that V so that this is going to be the coefficient that multiplies V and as before I'm going to want to observe this coefficient and find a nice compact way of representing this that is I want want to know what function this Taylor series corresponds to and now let's make a few observations we see that it's again alternating here's plus minus plus minus plus it comes every second degree so this is in the length of V we have the length of V to the zero power here to the second power to the fourth the sixth the eighth the 10th but in terms of powers if we look at the denominators here we'd really have a one factorial three factorial 5 factorial 7 factorial so it's really coming in only the odd Powers so from those two observations we have an alternating tayor series where only the odd powers are present and just from that observation you should be thinking of an odd function just as before we were thinking of an even function and combined with the fact that this is alternating this resembles a sign function if remember what the tayor series for a sign function is going to look like it's going to look something like x - x Cub we normally have an X Cub over 3 factorial plus X 5th over 5 factorial minus X 7th over 7 factorial and so on we'd have that alternate character there so the only adjustment we really have to make here is that if the series that we see here were multiplied by an extra Factor of the length V we would have the exact series for S of the length of V that is to say if I took this coefficient series here and I multiplied the whole thing by the length of V I would have the length of V minus the length of V cubed over 3 factorial plus length to the fifth power over 5 and that would be S of the length of V exactly which means that what we're looking at is not s of the length of V it's actually s of the length of V divided V so that's the value of the coefficient there s of the length of V / V and then times the vector v so this is the the function that we're really looking at here in this tailor Series so here we've just found what the vector part of the Quan exponential is let's now combine our results for the last two examinations of the Taylor series so what we're saying is that e raised to a querian power that looks like a i plus BJ plus CK remember we're only able to deal with quitan which have zero scalar power for now so the Quan is going to have to look like this this is going to be given by cosine of the length of V where V is just this Vector here the entire thing Plus s of the length of V ID V the length of V times V itself and remember V is just this so V instead of writing the vector v there I just plug in what we have up here so this is really the formula here and again the length of V is going to be the square root of the sum of squares of the components of that Vector as it normally is and now we have the formula let's uh compute a few examples so that means that e to the i+ j i plus J is going to be the vector v that's given by cosine of the length of V what's the length of V it's going to be the square root of 1 + 1 which is a square of two so we find that this is equal to cosine of the square of two as the new scalar part and to find a new Vector part we have S of the length of V which is of two divid the length of V which is again the of two times the vector itself which is I + J and then you just distribute this to the I and the J and we find that the new querian that we get out of taking e to the I plus J is .156 plus 698 I and the same coefficient times J and now let's take another example let's calculate e to 3j plus 4K so what's the length this Vector selected the values pretty nicely so we had the 3 four five thing going going on there so the length of this Vector is going to be five so that means that the new scalar part of this Quan exponential is going to be cosine of five then the new Vector part is going to be given by S of 5 over 5 times the vector itself which was 3j + 4K then again you just do the distrib the distribution here and we find that the Quan exponential of 3j + 4K is equal to 284 - 575 J minus 767 K I'd like to now show you something very nice about the formula that we just found and this will serve really as a sanity check for what we've just done and what I claim is that this formula ought to reduce to Oilers formula when we try to calculate things like e to the i e to 2 I or even e to the J or e to the K so let's do precisely that let's tryal calculating e to the I with this formula so I have e here which means the vector in question is just x * i x is that coefficient in Fr of I the vector only has one component just X so the the length of the vector is pretty simple it's just X which means when I plug into the formula we get cosine X Plus sin of x / x times the vector itself which is IX and notice now that we get the canc of those there and the formula does indeed reduce the oilage formula what this means is that what we just discovered the formula that we just discovered for the querian exponential is really a generalization of Oilers formula which is to say oil's formula is a special case when the J and K coefficients go to zero and now to complete our discussion of this formula remember how in our derivation we making that assumption that the scalar part is zero to be able to calculate the exponential of any Quan we have to take care of the case where it has nonzero scalar parts now I claim that when we go to calculate the Quan exponential of something that has a scal of part we can use a very similar trick as though we were trying to calculate something like e to 2 plus I where we wrote that as e^2 * e to the I power that is we split up the scal part with the imaginary part except now we're going to split up the scal part with the vector part now I claim that if we have a quitan that looks like this a plus b i plus CJ plus DK we can split up the Scala part from the vector part in precisely the same way that we normally do with exponentiation now this is the claim uh as an exercise I encourage you to prove this is this is not very difficult to show you just work through the power series and you'll see that you can indeed split it up like this you can split up the scalar part and the vector part but I warn you not to split up the vector part within itself you can't split up the bi from the CJ for example and to take an example of what I'm talking about let's suppose we want to calculate e to the 1 + I plus J I claim that we can split up that one scal of part from the I plus J vector part and write this as e to the first Power time e to the i+ J and since we know how to take care of something that looks like this already e to the I plus J this is simply going to be e times what we are already able to calculate so if this holds which it does hold we now know how to calculate the querian exponential for any querian that is to say if I have some arbitrary querian let's say a plus b i Plus CJ plus DK I can split the scalar from the vector part and the formula is going to be the following it's going to be e to the a times this entire quantity which again is going to be cosine of the length of v and by V I mean bi plus CJ plus DK so it's be cosine of the length of V plus s of the length of V divided by the length of V times the vector itself which again as bi plus CJ plus DK okay and notice I've changed the letters a little bit here but again the length of V is going to be given by the sum of squares of the components which were b c and d it's going to be the square root of the sum of squares of the components and this allows us to calculate any the the Quan exponential of any Quan and notice in a special case when C and D go to zero I have something that looks like e to the a plus b and if you work through through this formula here you'll find that this is exactly what we have for Oilers formula we have e to the a plus b i is given by e to the a power time cosine B+ I sin B that'll wrap up this video just as a few practice problems if you're really interested in this material first I ask you to show precisely what I claimed in the last slide that I could break apart the scalar and the vector part like I normally do for exponentiation secondly I ask you to convince yourself that the following statement is false that is that e to the i+ J is not the same as e to the I * e to the J and make sure you do the calculation to convince yourself of that and lastly convince yourself of that Oiler identity for the Quan that I claimed in the first slide of this video that the five Quan exponentials I have listed here are all equal to minus one