Show that two primitives/ antiderivatives are related via a constant
$begingroup$
Let $I subset R$ be an interval. A differentiable function $F: I rightarrow R$ is called a
primitive for the function $f : I rightarrow mathbb{R}$ if
$$F'
(x) = f(x)$$ for all $x in I$.
Show: If $F_ 1$ and $F_2$ are two primitives for $f$ on $I$ then there is a constant
$C in mathbb{R}$ such that $F_2 equiv F_1 + C$, i.e. $F_2(x) = F_1(x) + C$ for all $x in I$.
What we have covered so far is the formal definition of the derivative in terms of the limit:
$$ lim_{h rightarrow 0}frac{F_1(x+h)-F_1(x)}{h}= f(x)$$
$$ lim_{h rightarrow 0}frac{F_2(x+h)-F_2(x)}{h}= f(x)$$
I do not see how I can prove firsthand that these two functions differ by a constant. I know it to be true from my pre-calculus and calculus experience, but how would one make the argument from an analysis point of view, can someone give me a hint as to where this constant 'appears'.
We normally just define $F_2=F_1 +C$ and then it would follow immediately that these two functions have the same derivative.
real-analysis
asked Dec 11 '18 at 10:37
Wesley StrikWesley Strik
2,057423
$endgroup$
add a comment |
$begingroup$
Let $I subset R$ be an interval. A differentiable function $F: I rightarrow R$ is called a
primitive for the function $f : I rightarrow mathbb{R}$ if
$$F'
(x) = f(x)$$ for all $x in I$.
Show: If $F_ 1$ and $F_2$ are two primitives for $f$ on $I$ then there is a constant
$C in mathbb{R}$ such that $F_2 equiv F_1 + C$, i.e. $F_2(x) = F_1(x) + C$ for all $x in I$.
What we have covered so far is the formal definition of the derivative in terms of the limit:
$$ lim_{h rightarrow 0}frac{F_1(x+h)-F_1(x)}{h}= f(x)$$
$$ lim_{h rightarrow 0}frac{F_2(x+h)-F_2(x)}{h}= f(x)$$
I do not see how I can prove firsthand that these two functions differ by a constant. I know it to be true from my pre-calculus and calculus experience, but how would one make the argument from an analysis point of view, can someone give me a hint as to where this constant 'appears'.
We normally just define $F_2=F_1 +C$ and then it would follow immediately that these two functions have the same derivative.
real-analysis
asked Dec 11 '18 at 10:37
Wesley StrikWesley Strik
2,057423
$endgroup$
$begingroup$
math.stackexchange.com/questions/1862231/…
$endgroup$
– Hans Lundmark
Dec 11 '18 at 10:44
1
$begingroup$
It is essential that the domain is an intervall.
$endgroup$
– gammatester
Dec 11 '18 at 10:45
$begingroup$
AH, the glorious $MVT$, cool!
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:46
add a comment |
$begingroup$
Let $I subset R$ be an interval. A differentiable function $F: I rightarrow R$ is called a
primitive for the function $f : I rightarrow mathbb{R}$ if
$$F'
(x) = f(x)$$ for all $x in I$.
Show: If $F_ 1$ and $F_2$ are two primitives for $f$ on $I$ then there is a constant
$C in mathbb{R}$ such that $F_2 equiv F_1 + C$, i.e. $F_2(x) = F_1(x) + C$ for all $x in I$.
What we have covered so far is the formal definition of the derivative in terms of the limit:
$$ lim_{h rightarrow 0}frac{F_1(x+h)-F_1(x)}{h}= f(x)$$
$$ lim_{h rightarrow 0}frac{F_2(x+h)-F_2(x)}{h}= f(x)$$
I do not see how I can prove firsthand that these two functions differ by a constant. I know it to be true from my pre-calculus and calculus experience, but how would one make the argument from an analysis point of view, can someone give me a hint as to where this constant 'appears'.
We normally just define $F_2=F_1 +C$ and then it would follow immediately that these two functions have the same derivative.
real-analysis
asked Dec 11 '18 at 10:37
Wesley StrikWesley Strik
2,057423
$endgroup$
Let $I subset R$ be an interval. A differentiable function $F: I rightarrow R$ is called a
primitive for the function $f : I rightarrow mathbb{R}$ if
$$F'
(x) = f(x)$$ for all $x in I$.
Show: If $F_ 1$ and $F_2$ are two primitives for $f$ on $I$ then there is a constant
$C in mathbb{R}$ such that $F_2 equiv F_1 + C$, i.e. $F_2(x) = F_1(x) + C$ for all $x in I$.
What we have covered so far is the formal definition of the derivative in terms of the limit:
$$ lim_{h rightarrow 0}frac{F_1(x+h)-F_1(x)}{h}= f(x)$$
$$ lim_{h rightarrow 0}frac{F_2(x+h)-F_2(x)}{h}= f(x)$$
I do not see how I can prove firsthand that these two functions differ by a constant. I know it to be true from my pre-calculus and calculus experience, but how would one make the argument from an analysis point of view, can someone give me a hint as to where this constant 'appears'.
We normally just define $F_2=F_1 +C$ and then it would follow immediately that these two functions have the same derivative.
real-analysis
real-analysis
asked Dec 11 '18 at 10:37
Wesley StrikWesley Strik
2,057423
asked Dec 11 '18 at 10:37
Wesley StrikWesley Strik
2,057423
asked Dec 11 '18 at 10:37
Wesley StrikWesley Strik
2,057423
asked Dec 11 '18 at 10:37
Wesley StrikWesley Strik
2,057423
asked Dec 11 '18 at 10:37
Wesley StrikWesley Strik
2,057423
2,057423
$begingroup$
math.stackexchange.com/questions/1862231/…
$endgroup$
– Hans Lundmark
Dec 11 '18 at 10:44
1
$begingroup$
It is essential that the domain is an intervall.
$endgroup$
– gammatester
Dec 11 '18 at 10:45
$begingroup$
AH, the glorious $MVT$, cool!
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:46
add a comment |
$begingroup$
math.stackexchange.com/questions/1862231/…
$endgroup$
– Hans Lundmark
Dec 11 '18 at 10:44
1
$begingroup$
It is essential that the domain is an intervall.
$endgroup$
– gammatester
Dec 11 '18 at 10:45
$begingroup$
AH, the glorious $MVT$, cool!
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:46
$begingroup$
math.stackexchange.com/questions/1862231/…
$endgroup$
– Hans Lundmark
Dec 11 '18 at 10:44
$begingroup$
math.stackexchange.com/questions/1862231/…
$endgroup$
– Hans Lundmark
Dec 11 '18 at 10:44
1
1
$begingroup$
It is essential that the domain is an intervall.
$endgroup$
– gammatester
Dec 11 '18 at 10:45
$begingroup$
It is essential that the domain is an intervall.
$endgroup$
– gammatester
Dec 11 '18 at 10:45
$begingroup$
AH, the glorious $MVT$, cool!
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:46
$begingroup$
AH, the glorious $MVT$, cool!
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:46
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: What is the derivative of $F_1-F_2$?
answered Dec 11 '18 at 10:41
ArthurArthur
116k7116198
$endgroup$
$begingroup$
Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:49
$begingroup$
And then apply the MVT to the interval.
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:49
1
$begingroup$
@WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$?
$endgroup$
– Arthur
Dec 11 '18 at 10:52
$begingroup$
It means the function $F_1 - F_2$ is constant, see my proof below using the MVT.
$endgroup$
– Wesley Strik
Dec 11 '18 at 12:20
add a comment |
$begingroup$
Hint:
If $F_1,F_2$ are primitives of $f$, by linearity of differentiation,
$$(F_1(x)-F_2(x))'=F'_1(x)-F'_2(x)=f(x)-f(x)=0.$$
Remains to prove that the antiderivative of $0$ is a constant function.
answered Dec 11 '18 at 11:02
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
Let $G(x)= F_1 - F_2.$
We observe that $G'(x)=0$. We claim this means that $G(x)$ is constant.
Suppose on the contrary the function is non-constant this means there must exist $a, b in I$ and where $a<b$ such that $G(a) neq G(b)$.
We consider $[a,b]$. Now We apply the mean value theorem to this interval and the function $G(x)$. We get that there must exist a $c$ such that:
Which is a contradiction.
$$G'(c)= frac{G(b)-G(a)}{b-a} land G'(c)=0,$$
$$ frac{G(b)-G(a)}{b-a}=0 implies G(a)=G(b). $$
We conclude that $G(x)$ is constant. Thus, $F_2$ and $F_1$ differ by a constant $square$.
answered Dec 11 '18 at 11:58
Wesley StrikWesley Strik
2,057423
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: What is the derivative of $F_1-F_2$?
answered Dec 11 '18 at 10:41
ArthurArthur
116k7116198
$endgroup$
$begingroup$
Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:49
$begingroup$
And then apply the MVT to the interval.
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:49
1
$begingroup$
@WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$?
$endgroup$
– Arthur
Dec 11 '18 at 10:52
$begingroup$
It means the function $F_1 - F_2$ is constant, see my proof below using the MVT.
$endgroup$
– Wesley Strik
Dec 11 '18 at 12:20
add a comment |
$begingroup$
Hint: What is the derivative of $F_1-F_2$?
answered Dec 11 '18 at 10:41
ArthurArthur
116k7116198
$endgroup$
$begingroup$
Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:49
$begingroup$
And then apply the MVT to the interval.
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:49
1
$begingroup$
@WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$?
$endgroup$
– Arthur
Dec 11 '18 at 10:52
$begingroup$
It means the function $F_1 - F_2$ is constant, see my proof below using the MVT.
$endgroup$
– Wesley Strik
Dec 11 '18 at 12:20
add a comment |
$begingroup$
Hint: What is the derivative of $F_1-F_2$?
answered Dec 11 '18 at 10:41
ArthurArthur
116k7116198
$endgroup$
Hint: What is the derivative of $F_1-F_2$?
answered Dec 11 '18 at 10:41
ArthurArthur
116k7116198
answered Dec 11 '18 at 10:41
ArthurArthur
116k7116198
answered Dec 11 '18 at 10:41
ArthurArthur
116k7116198
answered Dec 11 '18 at 10:41
ArthurArthur
116k7116198
116k7116198
$begingroup$
Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:49
$begingroup$
And then apply the MVT to the interval.
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:49
1
$begingroup$
@WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$?
$endgroup$
– Arthur
Dec 11 '18 at 10:52
$begingroup$
It means the function $F_1 - F_2$ is constant, see my proof below using the MVT.
$endgroup$
– Wesley Strik
Dec 11 '18 at 12:20
add a comment |
$begingroup$
Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:49
$begingroup$
And then apply the MVT to the interval.
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:49
1
$begingroup$
@WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$?
$endgroup$
– Arthur
Dec 11 '18 at 10:52
$begingroup$
It means the function $F_1 - F_2$ is constant, see my proof below using the MVT.
$endgroup$
– Wesley Strik
Dec 11 '18 at 12:20
$begingroup$
Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:49
$begingroup$
Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:49
$begingroup$
And then apply the MVT to the interval.
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:49
$begingroup$
And then apply the MVT to the interval.
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:49
1
1
$begingroup$
@WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$?
$endgroup$
– Arthur
Dec 11 '18 at 10:52
$begingroup$
@WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$?
$endgroup$
– Arthur
Dec 11 '18 at 10:52
$begingroup$
It means the function $F_1 - F_2$ is constant, see my proof below using the MVT.
$endgroup$
– Wesley Strik
Dec 11 '18 at 12:20
$begingroup$
It means the function $F_1 - F_2$ is constant, see my proof below using the MVT.
$endgroup$
– Wesley Strik
Dec 11 '18 at 12:20
add a comment |
$begingroup$
Hint:
If $F_1,F_2$ are primitives of $f$, by linearity of differentiation,
$$(F_1(x)-F_2(x))'=F'_1(x)-F'_2(x)=f(x)-f(x)=0.$$
Remains to prove that the antiderivative of $0$ is a constant function.
answered Dec 11 '18 at 11:02
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
Hint:
If $F_1,F_2$ are primitives of $f$, by linearity of differentiation,
$$(F_1(x)-F_2(x))'=F'_1(x)-F'_2(x)=f(x)-f(x)=0.$$
Remains to prove that the antiderivative of $0$ is a constant function.
answered Dec 11 '18 at 11:02
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
Hint:
If $F_1,F_2$ are primitives of $f$, by linearity of differentiation,
$$(F_1(x)-F_2(x))'=F'_1(x)-F'_2(x)=f(x)-f(x)=0.$$
Remains to prove that the antiderivative of $0$ is a constant function.
answered Dec 11 '18 at 11:02
Yves DaoustYves Daoust
128k675227
$endgroup$
Hint:
If $F_1,F_2$ are primitives of $f$, by linearity of differentiation,
$$(F_1(x)-F_2(x))'=F'_1(x)-F'_2(x)=f(x)-f(x)=0.$$
Remains to prove that the antiderivative of $0$ is a constant function.
answered Dec 11 '18 at 11:02
Yves DaoustYves Daoust
128k675227
answered Dec 11 '18 at 11:02
Yves DaoustYves Daoust
128k675227
answered Dec 11 '18 at 11:02
Yves DaoustYves Daoust
128k675227
answered Dec 11 '18 at 11:02
Yves DaoustYves Daoust
128k675227
128k675227
add a comment |
add a comment |
$begingroup$
Let $G(x)= F_1 - F_2.$
We observe that $G'(x)=0$. We claim this means that $G(x)$ is constant.
Suppose on the contrary the function is non-constant this means there must exist $a, b in I$ and where $a<b$ such that $G(a) neq G(b)$.
We consider $[a,b]$. Now We apply the mean value theorem to this interval and the function $G(x)$. We get that there must exist a $c$ such that:
Which is a contradiction.
$$G'(c)= frac{G(b)-G(a)}{b-a} land G'(c)=0,$$
$$ frac{G(b)-G(a)}{b-a}=0 implies G(a)=G(b). $$
We conclude that $G(x)$ is constant. Thus, $F_2$ and $F_1$ differ by a constant $square$.
answered Dec 11 '18 at 11:58
Wesley StrikWesley Strik
2,057423
$endgroup$
add a comment |
$begingroup$
Let $G(x)= F_1 - F_2.$
We observe that $G'(x)=0$. We claim this means that $G(x)$ is constant.
Suppose on the contrary the function is non-constant this means there must exist $a, b in I$ and where $a<b$ such that $G(a) neq G(b)$.
We consider $[a,b]$. Now We apply the mean value theorem to this interval and the function $G(x)$. We get that there must exist a $c$ such that:
Which is a contradiction.
$$G'(c)= frac{G(b)-G(a)}{b-a} land G'(c)=0,$$
$$ frac{G(b)-G(a)}{b-a}=0 implies G(a)=G(b). $$
We conclude that $G(x)$ is constant. Thus, $F_2$ and $F_1$ differ by a constant $square$.
answered Dec 11 '18 at 11:58
Wesley StrikWesley Strik
2,057423
$endgroup$
add a comment |
$begingroup$
Let $G(x)= F_1 - F_2.$
We observe that $G'(x)=0$. We claim this means that $G(x)$ is constant.
Suppose on the contrary the function is non-constant this means there must exist $a, b in I$ and where $a<b$ such that $G(a) neq G(b)$.
We consider $[a,b]$. Now We apply the mean value theorem to this interval and the function $G(x)$. We get that there must exist a $c$ such that:
Which is a contradiction.
$$G'(c)= frac{G(b)-G(a)}{b-a} land G'(c)=0,$$
$$ frac{G(b)-G(a)}{b-a}=0 implies G(a)=G(b). $$
We conclude that $G(x)$ is constant. Thus, $F_2$ and $F_1$ differ by a constant $square$.
answered Dec 11 '18 at 11:58
Wesley StrikWesley Strik
2,057423
$endgroup$
Let $G(x)= F_1 - F_2.$
We observe that $G'(x)=0$. We claim this means that $G(x)$ is constant.
Suppose on the contrary the function is non-constant this means there must exist $a, b in I$ and where $a<b$ such that $G(a) neq G(b)$.
We consider $[a,b]$. Now We apply the mean value theorem to this interval and the function $G(x)$. We get that there must exist a $c$ such that:
Which is a contradiction.
$$G'(c)= frac{G(b)-G(a)}{b-a} land G'(c)=0,$$
$$ frac{G(b)-G(a)}{b-a}=0 implies G(a)=G(b). $$
We conclude that $G(x)$ is constant. Thus, $F_2$ and $F_1$ differ by a constant $square$.
answered Dec 11 '18 at 11:58
Wesley StrikWesley Strik
2,057423
edited Dec 11 '18 at 12:04
answered Dec 11 '18 at 11:58
Wesley StrikWesley Strik
2,057423
answered Dec 11 '18 at 11:58
Wesley StrikWesley Strik
2,057423
answered Dec 11 '18 at 11:58
Wesley StrikWesley Strik
2,057423
2,057423
add a comment |
add a comment |
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$begingroup$
math.stackexchange.com/questions/1862231/…
$endgroup$
– Hans Lundmark
Dec 11 '18 at 10:44
1
$begingroup$
It is essential that the domain is an intervall.
$endgroup$
– gammatester
Dec 11 '18 at 10:45
$begingroup$
AH, the glorious $MVT$, cool!
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:46