Prove that in a floating point system with truncation the number of significant digits is $n$.
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I was requested to prove that in a floating point system $text{F}(beta, n, m, M)$ with truncation the number of significant digits is $n$. (where $n$ is the number of digits and $m < text{exponent} < M$.
I tried using the definition that the number of significant digits is the maximum natural $d$ for which the relative error of $x$ is $leqbeta^{1-d}$, and so:
$$ x = b_1b_2ldots b_ncdotbeta^e,$$
$$ fl(x) = b_1b_2ldots b_{n-1}cdotbeta^e.$$
Therefore:
$$dfrac{|x-fl(x)|}{|x|} = dfrac{|b_1b_2ldots b_ncdotbeta^e - b_1b_2ldots b_{n-1}cdotbeta^e}{|b_1b_2ldots b_ncdotbeta^e|} = dfrac{b_n}{b_1b_2ldots b_n} leq beta^{1-d}.$$
And I'm stuck here because I need to prove $d = n$ but I have no idea how to advance. Any suggestions?
Thanks.
numerical-methods floating-point significant-figures
edited Nov 13 at 14:45
Snookie
2558
asked Nov 13 at 11:13
Barak B
224
add a comment |
up vote
0
down vote
favorite
I was requested to prove that in a floating point system $text{F}(beta, n, m, M)$ with truncation the number of significant digits is $n$. (where $n$ is the number of digits and $m < text{exponent} < M$.
I tried using the definition that the number of significant digits is the maximum natural $d$ for which the relative error of $x$ is $leqbeta^{1-d}$, and so:
$$ x = b_1b_2ldots b_ncdotbeta^e,$$
$$ fl(x) = b_1b_2ldots b_{n-1}cdotbeta^e.$$
Therefore:
$$dfrac{|x-fl(x)|}{|x|} = dfrac{|b_1b_2ldots b_ncdotbeta^e - b_1b_2ldots b_{n-1}cdotbeta^e}{|b_1b_2ldots b_ncdotbeta^e|} = dfrac{b_n}{b_1b_2ldots b_n} leq beta^{1-d}.$$
And I'm stuck here because I need to prove $d = n$ but I have no idea how to advance. Any suggestions?
Thanks.
numerical-methods floating-point significant-figures
edited Nov 13 at 14:45
Snookie
2558
asked Nov 13 at 11:13
Barak B
224
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was requested to prove that in a floating point system $text{F}(beta, n, m, M)$ with truncation the number of significant digits is $n$. (where $n$ is the number of digits and $m < text{exponent} < M$.
I tried using the definition that the number of significant digits is the maximum natural $d$ for which the relative error of $x$ is $leqbeta^{1-d}$, and so:
$$ x = b_1b_2ldots b_ncdotbeta^e,$$
$$ fl(x) = b_1b_2ldots b_{n-1}cdotbeta^e.$$
Therefore:
$$dfrac{|x-fl(x)|}{|x|} = dfrac{|b_1b_2ldots b_ncdotbeta^e - b_1b_2ldots b_{n-1}cdotbeta^e}{|b_1b_2ldots b_ncdotbeta^e|} = dfrac{b_n}{b_1b_2ldots b_n} leq beta^{1-d}.$$
And I'm stuck here because I need to prove $d = n$ but I have no idea how to advance. Any suggestions?
Thanks.
numerical-methods floating-point significant-figures
edited Nov 13 at 14:45
Snookie
2558
asked Nov 13 at 11:13
Barak B
224
I was requested to prove that in a floating point system $text{F}(beta, n, m, M)$ with truncation the number of significant digits is $n$. (where $n$ is the number of digits and $m < text{exponent} < M$.
I tried using the definition that the number of significant digits is the maximum natural $d$ for which the relative error of $x$ is $leqbeta^{1-d}$, and so:
$$ x = b_1b_2ldots b_ncdotbeta^e,$$
$$ fl(x) = b_1b_2ldots b_{n-1}cdotbeta^e.$$
Therefore:
$$dfrac{|x-fl(x)|}{|x|} = dfrac{|b_1b_2ldots b_ncdotbeta^e - b_1b_2ldots b_{n-1}cdotbeta^e}{|b_1b_2ldots b_ncdotbeta^e|} = dfrac{b_n}{b_1b_2ldots b_n} leq beta^{1-d}.$$
And I'm stuck here because I need to prove $d = n$ but I have no idea how to advance. Any suggestions?
Thanks.
numerical-methods floating-point significant-figures
numerical-methods floating-point significant-figures
edited Nov 13 at 14:45
Snookie
2558
asked Nov 13 at 11:13
Barak B
224
edited Nov 13 at 14:45
Snookie
2558
asked Nov 13 at 11:13
Barak B
224
edited Nov 13 at 14:45
Snookie
2558
edited Nov 13 at 14:45
Snookie
2558
edited Nov 13 at 14:45
Snookie
2558
2558
asked Nov 13 at 11:13
Barak B
224
asked Nov 13 at 11:13
Barak B
224
asked Nov 13 at 11:13
Barak B
224
224
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Figured out the solution:
$b_n <= beta^{1-n}$ and $b_1b_2…b_n$ is a normalized number therefore bounded by $beta$
so we get: $dfrac{b_n}{b_1b_2…b_n} <= beta^{1-n} =>$ the number of significant digits is $n$.
answered Nov 13 at 19:02
Barak B
224
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Figured out the solution:
$b_n <= beta^{1-n}$ and $b_1b_2…b_n$ is a normalized number therefore bounded by $beta$
so we get: $dfrac{b_n}{b_1b_2…b_n} <= beta^{1-n} =>$ the number of significant digits is $n$.
answered Nov 13 at 19:02
Barak B
224
add a comment |
up vote
0
down vote
Figured out the solution:
$b_n <= beta^{1-n}$ and $b_1b_2…b_n$ is a normalized number therefore bounded by $beta$
so we get: $dfrac{b_n}{b_1b_2…b_n} <= beta^{1-n} =>$ the number of significant digits is $n$.
answered Nov 13 at 19:02
Barak B
224
add a comment |
up vote
0
down vote
up vote
0
down vote
Figured out the solution:
$b_n <= beta^{1-n}$ and $b_1b_2…b_n$ is a normalized number therefore bounded by $beta$
so we get: $dfrac{b_n}{b_1b_2…b_n} <= beta^{1-n} =>$ the number of significant digits is $n$.
answered Nov 13 at 19:02
Barak B
224
Figured out the solution:
$b_n <= beta^{1-n}$ and $b_1b_2…b_n$ is a normalized number therefore bounded by $beta$
so we get: $dfrac{b_n}{b_1b_2…b_n} <= beta^{1-n} =>$ the number of significant digits is $n$.
answered Nov 13 at 19:02
Barak B
224
answered Nov 13 at 19:02
Barak B
224
answered Nov 13 at 19:02
Barak B
224
answered Nov 13 at 19:02
Barak B
224
224
add a comment |
add a comment |
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