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Post by Admin on Jul 9, 2020 11:27:52 GMT
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Jul 9, 2020 14:46:14 GMT
Vasco,
I included my answer to this exercise as part of Exercise 10, which I posted in the wee hours this morning. I'll take a look at yours.
Gary
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jul 9, 2020 14:59:08 GMT
Vasco,
I included my answer to this exercise as part of Exercise 10, which I posted in the wee hours this morning. I'll take a look at yours.
Gary
Vasco,
Nice solution. I found a repetition in mine. I will move the [exercise] to my notes section.
Gary
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Post by mondo on Aug 31, 2022 17:10:59 GMT
I have a simple question to the setup for this exercise:
$|i - k| = \sqrt{1 + k^2}$ - this is because of vector subtraction right? A subtraction of vertical vector $i$ and a horizontal vector $k$ (90 degrees angle between them). So the difference is the connecting vector |i -k| which we can get from a Pythagoras formula?
Thank you.
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Post by Admin on Aug 31, 2022 18:09:18 GMT
I have a simple question to the setup for this exercise: $|i - k| = \sqrt{1 + k^2}$ - this is because of vector subtraction right? A subtraction of vertical vector $i$ and a horizontal vector $k$ (90 degrees angle between them). So the difference is the connecting vector |i -k| which we can get from a Pythagoras formula? Thank you. Mondo That's one way of looking at it and it is correct. However, $|i-k|$ is not a vector. $i-k$ is a complex number(vector). $|i-k|$ is defined as the modulus(length) of the complex number $i-k$ (see page 6) and this is defined as $\sqrt{[\text{Re}(i-k)]^2+[\text{Im}(i-k)]^2}=\sqrt{k^2+1}$. I think you need to read and study chapter 1. Vasco |
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