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|p-p*|/|p|<=5*10^(-t)

said by numerical analysis textbook burden

then

how we define significant digits? do we have to choose any integer t?

example of this text book

p=0.54617, p*=0.5462 how we get significant digits by definition?

this text book say four significant digits but I don't know why

seyunkim
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  • Welcome to MSE. Please use MathJax to format your posts. Your question is not clear at all. Do you mean how do you get that $.54617$ to four significant figures $.5462$, or do you mean why has the book chosen to use four significant figures instead of say, three? – saulspatz Jul 22 '19 at 15:15
  • "this text book say four significant digits but I don't know why" Because that's what the book wanted. If I asked you to find $\pi$ to $2$ significant figures or to $17$ significant figures there wouldn't be any reason why I wanted those figures. ... or maybe there would be a practical reason. But there is no reason in general for why you choose $t$ to be $4$. It's just an example. – fleablood Jul 22 '19 at 16:08

1 Answers1

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You take the largest $t$ that satisfies the requirement. Just plug into the definition $$\frac {|p^*-p|}p=\frac {|0.54617-0.5462|}{0.54617}=\frac{0.00003}{0.54617}\approx 0.000055$$ This is greater than $0.5\cdot 10^{-4}$, so it should only be $3$ significant digits. Most people will just count the number of accurately rounded digits and say this has four, but that is not the definition you have quoted.

Ross Millikan
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  • is this text book not correct? |p*-p|/|p|<=10^(-k)/2 largest k integer ? – seyunkim Jul 22 '19 at 15:33
  • 5,4,6 is significant digits? – seyunkim Jul 22 '19 at 15:34
  • For your first comment, see my calculation. Do you understand it? Why did you change from $t$ to $k$, though that is not important? Yes, the significant digits are $546$ – Ross Millikan Jul 22 '19 at 15:40
  • significant digits is various definition each textbook . can i accept your definition? – seyunkim Jul 22 '19 at 15:48
  • Yes, the definition varies. You can use mine, but you may get marked down in this class. Your book's definition tries to handle the glitch that happens when you change from a number about $99$ to one about $101$. In my definition that adds one significant digit, but it really shouldn't. I wrote this answer a while back. This question has some more information – Ross Millikan Jul 22 '19 at 15:56