From 70a582bd111a83e13c0a4b1547252e687269a988 Mon Sep 17 00:00:00 2001 From: Noah Smith Date: Sat, 17 Aug 2024 07:51:10 +0200 Subject: [PATCH] backwards to i --- _drafts/backwards-to-i.md | 22 ++++++++++++++++++++++ _drafts/henry-briggs.md | 22 ---------------------- _posts/2024-08-17-backwards-to-i.md | 22 ++++++++++++++++++++++ 3 files changed, 44 insertions(+), 22 deletions(-) create mode 100644 _drafts/backwards-to-i.md delete mode 100644 _drafts/henry-briggs.md create mode 100644 _posts/2024-08-17-backwards-to-i.md diff --git a/_drafts/backwards-to-i.md b/_drafts/backwards-to-i.md new file mode 100644 index 0000000..6aea463 --- /dev/null +++ b/_drafts/backwards-to-i.md @@ -0,0 +1,22 @@ +--- +layout: post +title: "square root of negative one" +--- + +a wheel is turning around its axle at a steady rate. the wheel's radius is one and a spot on its edge has coordinates x = cos(t) and y = sin(t). the distance of the spot from the axle is x^2 + y^2 = 1. + +rather than represent x and y separately, can they be at least somewhat unified? for example, a tuple (x, y). or a vector. or, even better, a unit-vector. or an arithmetic expression. one possibility is x + k * y. the factor 'k' can be plus or minus itself and x - k * y is in some sense equivalent, and leads onward to writing down and contemplating some algebra. + + (x + k * y)(x - k * y) = x^2 - k^2 * y^2 + +if this expression equals the wheel's radius of one, then 'k' is the square root of negative one, and the symbol 'i' has appeared as a scaling factor that makes everything work within the geometry of a spot on a steadily turning wheel. + +this is a 'backwards' approach to what feynman is discussing towards the [end of this chapter](https://www.feynmanlectures.caltech.edu/I_22.html). basically, starting at the end of the chapter and 'reverse engineering' back towards the beginning. + +to start at the beginning and go forwards, instead of backwards, what feynman did was to gradually and step-by-step get to an imaginary exponent. in particular, ten to the 'i' times 's', 10^is. he then noted that 10^is = x + i * y, and that 10^-is = x - i * y, and that their product is one. + + (x + i * y)(x - i * y) = x^2 + y^2 = 1 + +in the end, after changing the base from ten to 'e', he arrived at the final destination. + + e^is = cos(s) + i * sin(s) diff --git a/_drafts/henry-briggs.md b/_drafts/henry-briggs.md deleted file mode 100644 index 15e0810..0000000 --- a/_drafts/henry-briggs.md +++ /dev/null @@ -1,22 +0,0 @@ ---- -layout: post -title: "henry briggs" ---- - -currently working towards tops-10 assembly and fortran for some of what feynman is discussing towards the [end of this chapter](https://www.feynmanlectures.caltech.edu/I_22.html). earlier in the chapter, he mentions briggs of halifax, 1620. this conjures a picture of a hermit living in a rock hut on the nova scotia coast. the true history turns out to be surprisingly pleasant. - -[henry briggs](https://en.wikipedia.org/wiki/Henry_Briggs_(mathematician)) 1561-1630 was a contemporary of shakespeare 1564-1616, cervantes 1547-1616, and [robert burton](https://en.wikipedia.org/wiki/Robert_Burton) 1577-1640. - -briggs was born near halifax yorkshire england and remained in england. one of his two sons did go to the new world. he was associated with cambridge, [gresham college london](https://en.wikipedia.org/wiki/Gresham_College), and oxford, where he surely knew burton. gresham college was effectively the precursor of the royal society, which incorporated in 1660. - -he worked on navigation and astronomy, strategic fields for england. two hundred years later, his work was a cornerstone of the reality shaping moderns like naval captain, navigator, and astronomer [jack aubrey](https://en.wikipedia.org/wiki/Jack_Aubrey). - -he quickly adopted and then improved the logarithm, invented by his scottish contemporary john napier 1550-1617. he visited napier to coordinate on advancing common base 10 'briggsian logarithms', which are more practical than napier's originals and became a standard. he invented the long division algorithm taught in elementary schools. and he used his own advances to publish logarithm, sine, tangent, and secant tables. - -here's feynman's comments. - -_This is how logarithms were originally computed by Mr. Briggs of Halifax, in 1620. He said, “I computed successively 54 square roots of 10.” ... all logarithm tables for three hundred years were borrowed from Mr. Briggs’ tables by reducing the number of decimal places. Only in modern times, with the WPA and computing machines, have new tables been independently computed._ - -and a comment from a colleague at oxford. - -_a man of great probity, a condemner of riches, and contented with his own station, preferring a studious retirement to all the splendid circumstances of life_ diff --git a/_posts/2024-08-17-backwards-to-i.md b/_posts/2024-08-17-backwards-to-i.md new file mode 100644 index 0000000..ca73fe8 --- /dev/null +++ b/_posts/2024-08-17-backwards-to-i.md @@ -0,0 +1,22 @@ +--- +layout: post +title: "square root of negative one" +--- + +a wheel is turning around its axle at a steady rate. the wheel's radius is one and a spot on its edge has coordinates x = cos(t) and y = sin(t). the distance of the spot from the axle is x^2 + y^2 = 1. + +rather than represent x and y separately, can they be somewhat unified? for example, a tuple (x, y). or a vector. or, even better, a unit-vector. or an arithmetic expression. one possibility is x + k * y. the factor 'k' can be plus or minus itself and x - k * y is in some sense equivalent, and leads onward to writing down and contemplating some algebra. + + (x + k * y)(x - k * y) = x^2 - k^2 * y^2 + +if this expression equals the wheel's radius of one, then 'k' is the square root of negative one, and the symbol 'i' has appeared as a scaling factor that 'makes the geometry work'. + +this is a 'backwards' approach to what feynman is discussing towards the [end of this chapter](https://www.feynmanlectures.caltech.edu/I_22.html). basically, starting at the end of the chapter and 'reverse engineering' back towards the beginning. + +to start at the beginning and go forwards instead of backwards, what feynman did was to gradually and step-by-step get to an imaginary exponent. in particular, ten to the 'i' times 's', 10^is. he then noted that 10^is = x + i * y, and that 10^-is = x - i * y. he was then able to write some algebra. + + (x + i * y)(x - i * y) = x^2 + y^2 = 1 + +from there, he soon arrived at the final destination, after changing the base from ten to 'e'. + + e^is = cos(s) + i * sin(s)