Mixing
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How is addition and multiplication of step functions defined?
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I'm going through "Calculus" by Tom Apostol. And I'm in this section:
I think the book assumes that from the example I can extrapolate how the graph for any addition of step functions is done; nonetheless, I don't understand that example. So a problem arises now that I have to do the first exercise.
So, when I'm going to do $a)$ I know how to graph $lfloor x rfloor$ and $lfloor 2x rfloor$, I even know how to do the common refinement, but not the graph of $lfloor 2x rfloor$+$lfloor x rfloor$ itself.
So, do you think you can tell me how step functions are added and multiplied? thanks in advance.
calculus algebra-precalculus functions floor-function
asked Jul 19 at 23:24
Daniel Bonilla Jaramillo
38819
add a comment |Â
up vote
2
down vote
favorite
I'm going through "Calculus" by Tom Apostol. And I'm in this section:
I think the book assumes that from the example I can extrapolate how the graph for any addition of step functions is done; nonetheless, I don't understand that example. So a problem arises now that I have to do the first exercise.
So, when I'm going to do $a)$ I know how to graph $lfloor x rfloor$ and $lfloor 2x rfloor$, I even know how to do the common refinement, but not the graph of $lfloor 2x rfloor$+$lfloor x rfloor$ itself.
So, do you think you can tell me how step functions are added and multiplied? thanks in advance.
calculus algebra-precalculus functions floor-function
asked Jul 19 at 23:24
Daniel Bonilla Jaramillo
38819
3
I would think addition and multiplication of step functions are defined as they are for arbitrary functions; namely, point-wise. Do you know how $f+g$ and $fcdot g$ are defined? You have to be careful with the step functions though because they're defined piece-wise.
– Oliver Jones
Jul 19 at 23:35
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm going through "Calculus" by Tom Apostol. And I'm in this section:
I think the book assumes that from the example I can extrapolate how the graph for any addition of step functions is done; nonetheless, I don't understand that example. So a problem arises now that I have to do the first exercise.
So, when I'm going to do $a)$ I know how to graph $lfloor x rfloor$ and $lfloor 2x rfloor$, I even know how to do the common refinement, but not the graph of $lfloor 2x rfloor$+$lfloor x rfloor$ itself.
So, do you think you can tell me how step functions are added and multiplied? thanks in advance.
calculus algebra-precalculus functions floor-function
asked Jul 19 at 23:24
Daniel Bonilla Jaramillo
38819
I'm going through "Calculus" by Tom Apostol. And I'm in this section:
I think the book assumes that from the example I can extrapolate how the graph for any addition of step functions is done; nonetheless, I don't understand that example. So a problem arises now that I have to do the first exercise.
So, when I'm going to do $a)$ I know how to graph $lfloor x rfloor$ and $lfloor 2x rfloor$, I even know how to do the common refinement, but not the graph of $lfloor 2x rfloor$+$lfloor x rfloor$ itself.
So, do you think you can tell me how step functions are added and multiplied? thanks in advance.
calculus algebra-precalculus functions floor-function
asked Jul 19 at 23:24
Daniel Bonilla Jaramillo
38819
asked Jul 19 at 23:24
Daniel Bonilla Jaramillo
38819
asked Jul 19 at 23:24
Daniel Bonilla Jaramillo
38819
asked Jul 19 at 23:24
Daniel Bonilla Jaramillo
38819
38819
3
I would think addition and multiplication of step functions are defined as they are for arbitrary functions; namely, point-wise. Do you know how $f+g$ and $fcdot g$ are defined? You have to be careful with the step functions though because they're defined piece-wise.
– Oliver Jones
Jul 19 at 23:35
add a comment |Â
3
I would think addition and multiplication of step functions are defined as they are for arbitrary functions; namely, point-wise. Do you know how $f+g$ and $fcdot g$ are defined? You have to be careful with the step functions though because they're defined piece-wise.
– Oliver Jones
Jul 19 at 23:35
3
3
I would think addition and multiplication of step functions are defined as they are for arbitrary functions; namely, point-wise. Do you know how $f+g$ and $fcdot g$ are defined? You have to be careful with the step functions though because they're defined piece-wise.
– Oliver Jones
Jul 19 at 23:35
I would think addition and multiplication of step functions are defined as they are for arbitrary functions; namely, point-wise. Do you know how $f+g$ and $fcdot g$ are defined? You have to be careful with the step functions though because they're defined piece-wise.
– Oliver Jones
Jul 19 at 23:35
add a comment |Â
2 Answers
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up vote
3
down vote
accepted
Looks like you have the right idea. You divide the domain into sub-intervals, and evaluate in each sub-interval.
$f(x) + g(x) = begin cases -3-6 = 7 & -3 le x < -2.5\-3-5 = 6 & -2.5 le x < -2\-2-4 = 5 & -2 le x < -1.5\&vdotsendcases$
$f(x)g(x) = begin cases (-3)(-6) = 18 & -3 le x < -2.5\(-3)(-5) = 15 & -2.5 le x < -2\(-2)(-4) = 8 & -2 le x < -1.5\&vdotsendcases$
answered Jul 19 at 23:39
Doug M
39.2k31749
add a comment |Â
up vote
2
down vote
By definition functions have specific (single) values at any point. Given two functions $f$ and $g$ (it is immaterial if they are step functions or not), at any desired point $a$ in the domain get their values $f(a)$ and $g(a)$ (this may need a complicated table look up, lengthy process or could be instantaneous).
Now add these two values to get a well-defined number, and this the value of $f+g$ at $a$. This can be done for every $a$, and so we have the definition of $f+g$. In a similar way can define the product function $fg$. Definitions are conceptually simple and uniform for all kind of functions.
answered Jul 20 at 0:53
P Vanchinathan
13.9k12035
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Looks like you have the right idea. You divide the domain into sub-intervals, and evaluate in each sub-interval.
$f(x) + g(x) = begin cases -3-6 = 7 & -3 le x < -2.5\-3-5 = 6 & -2.5 le x < -2\-2-4 = 5 & -2 le x < -1.5\&vdotsendcases$
$f(x)g(x) = begin cases (-3)(-6) = 18 & -3 le x < -2.5\(-3)(-5) = 15 & -2.5 le x < -2\(-2)(-4) = 8 & -2 le x < -1.5\&vdotsendcases$
answered Jul 19 at 23:39
Doug M
39.2k31749
add a comment |Â
up vote
3
down vote
accepted
Looks like you have the right idea. You divide the domain into sub-intervals, and evaluate in each sub-interval.
$f(x) + g(x) = begin cases -3-6 = 7 & -3 le x < -2.5\-3-5 = 6 & -2.5 le x < -2\-2-4 = 5 & -2 le x < -1.5\&vdotsendcases$
$f(x)g(x) = begin cases (-3)(-6) = 18 & -3 le x < -2.5\(-3)(-5) = 15 & -2.5 le x < -2\(-2)(-4) = 8 & -2 le x < -1.5\&vdotsendcases$
answered Jul 19 at 23:39
Doug M
39.2k31749
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Looks like you have the right idea. You divide the domain into sub-intervals, and evaluate in each sub-interval.
$f(x) + g(x) = begin cases -3-6 = 7 & -3 le x < -2.5\-3-5 = 6 & -2.5 le x < -2\-2-4 = 5 & -2 le x < -1.5\&vdotsendcases$
$f(x)g(x) = begin cases (-3)(-6) = 18 & -3 le x < -2.5\(-3)(-5) = 15 & -2.5 le x < -2\(-2)(-4) = 8 & -2 le x < -1.5\&vdotsendcases$
answered Jul 19 at 23:39
Doug M
39.2k31749
Looks like you have the right idea. You divide the domain into sub-intervals, and evaluate in each sub-interval.
$f(x) + g(x) = begin cases -3-6 = 7 & -3 le x < -2.5\-3-5 = 6 & -2.5 le x < -2\-2-4 = 5 & -2 le x < -1.5\&vdotsendcases$
$f(x)g(x) = begin cases (-3)(-6) = 18 & -3 le x < -2.5\(-3)(-5) = 15 & -2.5 le x < -2\(-2)(-4) = 8 & -2 le x < -1.5\&vdotsendcases$
answered Jul 19 at 23:39
Doug M
39.2k31749
answered Jul 19 at 23:39
Doug M
39.2k31749
answered Jul 19 at 23:39
Doug M
39.2k31749
answered Jul 19 at 23:39
Doug M
39.2k31749
39.2k31749
add a comment |Â
add a comment |Â
up vote
2
down vote
By definition functions have specific (single) values at any point. Given two functions $f$ and $g$ (it is immaterial if they are step functions or not), at any desired point $a$ in the domain get their values $f(a)$ and $g(a)$ (this may need a complicated table look up, lengthy process or could be instantaneous).
Now add these two values to get a well-defined number, and this the value of $f+g$ at $a$. This can be done for every $a$, and so we have the definition of $f+g$. In a similar way can define the product function $fg$. Definitions are conceptually simple and uniform for all kind of functions.
answered Jul 20 at 0:53
P Vanchinathan
13.9k12035
add a comment |Â
up vote
2
down vote
By definition functions have specific (single) values at any point. Given two functions $f$ and $g$ (it is immaterial if they are step functions or not), at any desired point $a$ in the domain get their values $f(a)$ and $g(a)$ (this may need a complicated table look up, lengthy process or could be instantaneous).
Now add these two values to get a well-defined number, and this the value of $f+g$ at $a$. This can be done for every $a$, and so we have the definition of $f+g$. In a similar way can define the product function $fg$. Definitions are conceptually simple and uniform for all kind of functions.
answered Jul 20 at 0:53
P Vanchinathan
13.9k12035
add a comment |Â
up vote
2
down vote
up vote
2
down vote
By definition functions have specific (single) values at any point. Given two functions $f$ and $g$ (it is immaterial if they are step functions or not), at any desired point $a$ in the domain get their values $f(a)$ and $g(a)$ (this may need a complicated table look up, lengthy process or could be instantaneous).
Now add these two values to get a well-defined number, and this the value of $f+g$ at $a$. This can be done for every $a$, and so we have the definition of $f+g$. In a similar way can define the product function $fg$. Definitions are conceptually simple and uniform for all kind of functions.
answered Jul 20 at 0:53
P Vanchinathan
13.9k12035
By definition functions have specific (single) values at any point. Given two functions $f$ and $g$ (it is immaterial if they are step functions or not), at any desired point $a$ in the domain get their values $f(a)$ and $g(a)$ (this may need a complicated table look up, lengthy process or could be instantaneous).
Now add these two values to get a well-defined number, and this the value of $f+g$ at $a$. This can be done for every $a$, and so we have the definition of $f+g$. In a similar way can define the product function $fg$. Definitions are conceptually simple and uniform for all kind of functions.
answered Jul 20 at 0:53
P Vanchinathan
13.9k12035
answered Jul 20 at 0:53
P Vanchinathan
13.9k12035
answered Jul 20 at 0:53
P Vanchinathan
13.9k12035
answered Jul 20 at 0:53
P Vanchinathan
13.9k12035
13.9k12035
add a comment |Â
add a comment |Â
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3
I would think addition and multiplication of step functions are defined as they are for arbitrary functions; namely, point-wise. Do you know how $f+g$ and $fcdot g$ are defined? You have to be careful with the step functions though because they're defined piece-wise.
– Oliver Jones
Jul 19 at 23:35