Mixing
How do I from variables to growth rates of variables, approximately?
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I'm given a function of the form: $$ y = x + 2 $$
I'm asked to derive the relationship between the growth rate of $ y $ to the growth rate of $ x $, approximately. I'm also given this example without derivation: If $ z = x^a $, then $ g_z = a g_x $, where $ g_z $ and $ g_x $ are the growth rates of $ z $ and $ x $ respectively.
How do I derive the relationship? I can see that $ y = x + 2 $ is very similar to $ z = x^a $, if we have $ a = 1 $ and add in a constant term. Without the constant term, the growth rate of $ x $ and $ y $ would be equal to each other. How do I handle the constant term?
algebra-precalculus
asked Feb 1 at 8:05
WorldGovWorldGov
345211
$endgroup$
add a comment |
$begingroup$
I'm given a function of the form: $$ y = x + 2 $$
I'm asked to derive the relationship between the growth rate of $ y $ to the growth rate of $ x $, approximately. I'm also given this example without derivation: If $ z = x^a $, then $ g_z = a g_x $, where $ g_z $ and $ g_x $ are the growth rates of $ z $ and $ x $ respectively.
How do I derive the relationship? I can see that $ y = x + 2 $ is very similar to $ z = x^a $, if we have $ a = 1 $ and add in a constant term. Without the constant term, the growth rate of $ x $ and $ y $ would be equal to each other. How do I handle the constant term?
algebra-precalculus
asked Feb 1 at 8:05
WorldGovWorldGov
345211
$endgroup$
$begingroup$
Let $2=2x^0$ and continue with your formula
$endgroup$
– Faiq Irfan
Feb 1 at 8:15
$begingroup$
I have some doubts, regarding the relations $z=x^a rightarrow g_z = a g_x$. Shouldn't it be $g_z = a x^{a-1} g_x$?
$endgroup$
– Matti P.
Feb 1 at 8:18
1
$begingroup$
It is not quite clear what you mean. Usually the growth rate (for $xrightarrow infty$) is described with the so-called Landau-symbol $O(f(x))$ In this case we have $x+2=O(x)$ because of $$lim_{xrightarrow infty} frac{x+2}{x}=1$$ This means that the growth rates of $x$ and $x+2$ coincide. It is not necessary that the values are actually equal at some point.
$endgroup$
– Peter
Feb 1 at 8:50
add a comment |
$begingroup$
I'm given a function of the form: $$ y = x + 2 $$
I'm asked to derive the relationship between the growth rate of $ y $ to the growth rate of $ x $, approximately. I'm also given this example without derivation: If $ z = x^a $, then $ g_z = a g_x $, where $ g_z $ and $ g_x $ are the growth rates of $ z $ and $ x $ respectively.
How do I derive the relationship? I can see that $ y = x + 2 $ is very similar to $ z = x^a $, if we have $ a = 1 $ and add in a constant term. Without the constant term, the growth rate of $ x $ and $ y $ would be equal to each other. How do I handle the constant term?
algebra-precalculus
asked Feb 1 at 8:05
WorldGovWorldGov
345211
$endgroup$
I'm given a function of the form: $$ y = x + 2 $$
I'm asked to derive the relationship between the growth rate of $ y $ to the growth rate of $ x $, approximately. I'm also given this example without derivation: If $ z = x^a $, then $ g_z = a g_x $, where $ g_z $ and $ g_x $ are the growth rates of $ z $ and $ x $ respectively.
How do I derive the relationship? I can see that $ y = x + 2 $ is very similar to $ z = x^a $, if we have $ a = 1 $ and add in a constant term. Without the constant term, the growth rate of $ x $ and $ y $ would be equal to each other. How do I handle the constant term?
algebra-precalculus
algebra-precalculus
asked Feb 1 at 8:05
WorldGovWorldGov
345211
asked Feb 1 at 8:05
WorldGovWorldGov
345211
asked Feb 1 at 8:05
WorldGovWorldGov
345211
asked Feb 1 at 8:05
WorldGovWorldGov
345211
asked Feb 1 at 8:05
WorldGovWorldGov
345211
345211
$begingroup$
Let $2=2x^0$ and continue with your formula
$endgroup$
– Faiq Irfan
Feb 1 at 8:15
$begingroup$
I have some doubts, regarding the relations $z=x^a rightarrow g_z = a g_x$. Shouldn't it be $g_z = a x^{a-1} g_x$?
$endgroup$
– Matti P.
Feb 1 at 8:18
1
$begingroup$
It is not quite clear what you mean. Usually the growth rate (for $xrightarrow infty$) is described with the so-called Landau-symbol $O(f(x))$ In this case we have $x+2=O(x)$ because of $$lim_{xrightarrow infty} frac{x+2}{x}=1$$ This means that the growth rates of $x$ and $x+2$ coincide. It is not necessary that the values are actually equal at some point.
$endgroup$
– Peter
Feb 1 at 8:50
add a comment |
$begingroup$
Let $2=2x^0$ and continue with your formula
$endgroup$
– Faiq Irfan
Feb 1 at 8:15
$begingroup$
I have some doubts, regarding the relations $z=x^a rightarrow g_z = a g_x$. Shouldn't it be $g_z = a x^{a-1} g_x$?
$endgroup$
– Matti P.
Feb 1 at 8:18
1
$begingroup$
It is not quite clear what you mean. Usually the growth rate (for $xrightarrow infty$) is described with the so-called Landau-symbol $O(f(x))$ In this case we have $x+2=O(x)$ because of $$lim_{xrightarrow infty} frac{x+2}{x}=1$$ This means that the growth rates of $x$ and $x+2$ coincide. It is not necessary that the values are actually equal at some point.
$endgroup$
– Peter
Feb 1 at 8:50
$begingroup$
Let $2=2x^0$ and continue with your formula
$endgroup$
– Faiq Irfan
Feb 1 at 8:15
$begingroup$
Let $2=2x^0$ and continue with your formula
$endgroup$
– Faiq Irfan
Feb 1 at 8:15
$begingroup$
I have some doubts, regarding the relations $z=x^a rightarrow g_z = a g_x$. Shouldn't it be $g_z = a x^{a-1} g_x$?
$endgroup$
– Matti P.
Feb 1 at 8:18
$begingroup$
I have some doubts, regarding the relations $z=x^a rightarrow g_z = a g_x$. Shouldn't it be $g_z = a x^{a-1} g_x$?
$endgroup$
– Matti P.
Feb 1 at 8:18
1
1
$begingroup$
It is not quite clear what you mean. Usually the growth rate (for $xrightarrow infty$) is described with the so-called Landau-symbol $O(f(x))$ In this case we have $x+2=O(x)$ because of $$lim_{xrightarrow infty} frac{x+2}{x}=1$$ This means that the growth rates of $x$ and $x+2$ coincide. It is not necessary that the values are actually equal at some point.
$endgroup$
– Peter
Feb 1 at 8:50
$begingroup$
It is not quite clear what you mean. Usually the growth rate (for $xrightarrow infty$) is described with the so-called Landau-symbol $O(f(x))$ In this case we have $x+2=O(x)$ because of $$lim_{xrightarrow infty} frac{x+2}{x}=1$$ This means that the growth rates of $x$ and $x+2$ coincide. It is not necessary that the values are actually equal at some point.
$endgroup$
– Peter
Feb 1 at 8:50
add a comment |
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$begingroup$
Let $2=2x^0$ and continue with your formula
$endgroup$
– Faiq Irfan
Feb 1 at 8:15
$begingroup$
I have some doubts, regarding the relations $z=x^a rightarrow g_z = a g_x$. Shouldn't it be $g_z = a x^{a-1} g_x$?
$endgroup$
– Matti P.
Feb 1 at 8:18
1
$begingroup$
It is not quite clear what you mean. Usually the growth rate (for $xrightarrow infty$) is described with the so-called Landau-symbol $O(f(x))$ In this case we have $x+2=O(x)$ because of $$lim_{xrightarrow infty} frac{x+2}{x}=1$$ This means that the growth rates of $x$ and $x+2$ coincide. It is not necessary that the values are actually equal at some point.
$endgroup$
– Peter
Feb 1 at 8:50