Mixing
find $p$, $q$ such that $f$ is differentiable at zero
$begingroup$
I am trying to understand tasks like that
$$f(x)=begin{cases} q+sin(px)+qx &text{for } xge0 \ frac{-1}{xsin(x)} + frac{cos(x)}{xsin(x)} &text{for } xin (-pi,0) end{cases}$$
On my lecture I had theorem that
$f$ is differentiable at zero if and only if exists finite:
$$ lim_{hrightarrow0} frac{f(a+h)-f(a)}{h}= f'(a) $$
My current way of thinking
Ok, after small discussion in comments, I think that these conditions should be fulfilled:
1. $$ lim_{hrightarrow0^-} frac{f(a+h)-f(a)}{h}=lim_{hrightarrow0^+} frac{f(a+h)-f(a)}{h} $$ what is equal to:
$$lim_{xrightarrow0^-}f'(x) = lim_{xrightarrow0^+} f'(x) $$
- Function should be continuous:
$$lim_{xrightarrow0^-}f(x) = lim_{xrightarrow0^+} f(x) = f(0)$$
My current problem
I am trying to solve this with differential quotient but I stucked
$$ lim_{h to 0^-} frac{f(h) - f(0)}{h} $$
there is a problem:(
real-analysis
asked Jan 24 at 0:07
VirtualUserVirtualUser
1,096117
$endgroup$
|
show 3 more comments
$begingroup$
I am trying to understand tasks like that
$$f(x)=begin{cases} q+sin(px)+qx &text{for } xge0 \ frac{-1}{xsin(x)} + frac{cos(x)}{xsin(x)} &text{for } xin (-pi,0) end{cases}$$
On my lecture I had theorem that
$f$ is differentiable at zero if and only if exists finite:
$$ lim_{hrightarrow0} frac{f(a+h)-f(a)}{h}= f'(a) $$
My current way of thinking
Ok, after small discussion in comments, I think that these conditions should be fulfilled:
1. $$ lim_{hrightarrow0^-} frac{f(a+h)-f(a)}{h}=lim_{hrightarrow0^+} frac{f(a+h)-f(a)}{h} $$ what is equal to:
$$lim_{xrightarrow0^-}f'(x) = lim_{xrightarrow0^+} f'(x) $$
- Function should be continuous:
$$lim_{xrightarrow0^-}f(x) = lim_{xrightarrow0^+} f(x) = f(0)$$
My current problem
I am trying to solve this with differential quotient but I stucked
$$ lim_{h to 0^-} frac{f(h) - f(0)}{h} $$
there is a problem:(
real-analysis
asked Jan 24 at 0:07
VirtualUserVirtualUser
1,096117
$endgroup$
2
$begingroup$
Start by getting continuity at $0$. Take the limits from the left and the right (in terms of $p$ and $q$).
$endgroup$
– Michael Burr
Jan 24 at 0:11
$begingroup$
I wrote that in topic
$endgroup$
– VirtualUser
Jan 24 at 0:13
$begingroup$
The lazy way of doing this is just by considering two things: continuity and $f’_+ (0)=f’_-(0)$, where $f_+$ is $f$ for $x geq 0$ and $f_-$ for $x in (-pi,0)$. And indeed all the things need to be limits.
$endgroup$
– Dani
Jan 24 at 9:23
$begingroup$
Why lazy? What do you mean by that?
$endgroup$
– VirtualUser
Jan 24 at 18:19
$begingroup$
I edited topic - can you look at this again? I don't want solution, only I want to be sure about my way of thinking
$endgroup$
– VirtualUser
Jan 25 at 19:14
|
show 3 more comments
$begingroup$
I am trying to understand tasks like that
$$f(x)=begin{cases} q+sin(px)+qx &text{for } xge0 \ frac{-1}{xsin(x)} + frac{cos(x)}{xsin(x)} &text{for } xin (-pi,0) end{cases}$$
On my lecture I had theorem that
$f$ is differentiable at zero if and only if exists finite:
$$ lim_{hrightarrow0} frac{f(a+h)-f(a)}{h}= f'(a) $$
My current way of thinking
Ok, after small discussion in comments, I think that these conditions should be fulfilled:
1. $$ lim_{hrightarrow0^-} frac{f(a+h)-f(a)}{h}=lim_{hrightarrow0^+} frac{f(a+h)-f(a)}{h} $$ what is equal to:
$$lim_{xrightarrow0^-}f'(x) = lim_{xrightarrow0^+} f'(x) $$
- Function should be continuous:
$$lim_{xrightarrow0^-}f(x) = lim_{xrightarrow0^+} f(x) = f(0)$$
My current problem
I am trying to solve this with differential quotient but I stucked
$$ lim_{h to 0^-} frac{f(h) - f(0)}{h} $$
there is a problem:(
real-analysis
asked Jan 24 at 0:07
VirtualUserVirtualUser
1,096117
$endgroup$
I am trying to understand tasks like that
$$f(x)=begin{cases} q+sin(px)+qx &text{for } xge0 \ frac{-1}{xsin(x)} + frac{cos(x)}{xsin(x)} &text{for } xin (-pi,0) end{cases}$$
On my lecture I had theorem that
$f$ is differentiable at zero if and only if exists finite:
$$ lim_{hrightarrow0} frac{f(a+h)-f(a)}{h}= f'(a) $$
My current way of thinking
Ok, after small discussion in comments, I think that these conditions should be fulfilled:
1. $$ lim_{hrightarrow0^-} frac{f(a+h)-f(a)}{h}=lim_{hrightarrow0^+} frac{f(a+h)-f(a)}{h} $$ what is equal to:
$$lim_{xrightarrow0^-}f'(x) = lim_{xrightarrow0^+} f'(x) $$
- Function should be continuous:
$$lim_{xrightarrow0^-}f(x) = lim_{xrightarrow0^+} f(x) = f(0)$$
My current problem
I am trying to solve this with differential quotient but I stucked
$$ lim_{h to 0^-} frac{f(h) - f(0)}{h} $$
there is a problem:(
real-analysis
real-analysis
asked Jan 24 at 0:07
VirtualUserVirtualUser
1,096117
asked Jan 24 at 0:07
VirtualUserVirtualUser
1,096117
edited Jan 29 at 8:26
VirtualUser
asked Jan 24 at 0:07
VirtualUserVirtualUser
1,096117
asked Jan 24 at 0:07
VirtualUserVirtualUser
1,096117
asked Jan 24 at 0:07
VirtualUserVirtualUser
1,096117
1,096117
2
$begingroup$
Start by getting continuity at $0$. Take the limits from the left and the right (in terms of $p$ and $q$).
$endgroup$
– Michael Burr
Jan 24 at 0:11
$begingroup$
I wrote that in topic
$endgroup$
– VirtualUser
Jan 24 at 0:13
$begingroup$
The lazy way of doing this is just by considering two things: continuity and $f’_+ (0)=f’_-(0)$, where $f_+$ is $f$ for $x geq 0$ and $f_-$ for $x in (-pi,0)$. And indeed all the things need to be limits.
$endgroup$
– Dani
Jan 24 at 9:23
$begingroup$
Why lazy? What do you mean by that?
$endgroup$
– VirtualUser
Jan 24 at 18:19
$begingroup$
I edited topic - can you look at this again? I don't want solution, only I want to be sure about my way of thinking
$endgroup$
– VirtualUser
Jan 25 at 19:14
|
show 3 more comments
2
$begingroup$
Start by getting continuity at $0$. Take the limits from the left and the right (in terms of $p$ and $q$).
$endgroup$
– Michael Burr
Jan 24 at 0:11
$begingroup$
I wrote that in topic
$endgroup$
– VirtualUser
Jan 24 at 0:13
$begingroup$
The lazy way of doing this is just by considering two things: continuity and $f’_+ (0)=f’_-(0)$, where $f_+$ is $f$ for $x geq 0$ and $f_-$ for $x in (-pi,0)$. And indeed all the things need to be limits.
$endgroup$
– Dani
Jan 24 at 9:23
$begingroup$
Why lazy? What do you mean by that?
$endgroup$
– VirtualUser
Jan 24 at 18:19
$begingroup$
I edited topic - can you look at this again? I don't want solution, only I want to be sure about my way of thinking
$endgroup$
– VirtualUser
Jan 25 at 19:14
2
2
$begingroup$
Start by getting continuity at $0$. Take the limits from the left and the right (in terms of $p$ and $q$).
$endgroup$
– Michael Burr
Jan 24 at 0:11
$begingroup$
Start by getting continuity at $0$. Take the limits from the left and the right (in terms of $p$ and $q$).
$endgroup$
– Michael Burr
Jan 24 at 0:11
$begingroup$
I wrote that in topic
$endgroup$
– VirtualUser
Jan 24 at 0:13
$begingroup$
I wrote that in topic
$endgroup$
– VirtualUser
Jan 24 at 0:13
$begingroup$
The lazy way of doing this is just by considering two things: continuity and $f’_+ (0)=f’_-(0)$, where $f_+$ is $f$ for $x geq 0$ and $f_-$ for $x in (-pi,0)$. And indeed all the things need to be limits.
$endgroup$
– Dani
Jan 24 at 9:23
$begingroup$
The lazy way of doing this is just by considering two things: continuity and $f’_+ (0)=f’_-(0)$, where $f_+$ is $f$ for $x geq 0$ and $f_-$ for $x in (-pi,0)$. And indeed all the things need to be limits.
$endgroup$
– Dani
Jan 24 at 9:23
$begingroup$
Why lazy? What do you mean by that?
$endgroup$
– VirtualUser
Jan 24 at 18:19
$begingroup$
Why lazy? What do you mean by that?
$endgroup$
– VirtualUser
Jan 24 at 18:19
$begingroup$
I edited topic - can you look at this again? I don't want solution, only I want to be sure about my way of thinking
$endgroup$
– VirtualUser
Jan 25 at 19:14
$begingroup$
I edited topic - can you look at this again? I don't want solution, only I want to be sure about my way of thinking
$endgroup$
– VirtualUser
Jan 25 at 19:14
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Fix $p,q$ and assume $f'(0)$ exists. Then $f$ is continuous at $0.$ Because $f(0)=q,$ we must have $lim_{xto 0^-}f(x)=q.$ But
$$lim_{xto 0^-}f(x) = lim_{xto 0^-}frac{cos x-1}{xsin x} = -frac{1}{2}.$$
Therefore $q= -1/2.$
Now take $x>0$ and consider
$$frac{f(x)-f(0)}{x}= frac{-1/2 +sin (px) -x/2 +1/2}{x} = frac{sin (px) - x/2}{x}.$$
The limit of this from the right is $p-1/2.$ This is the derivative of $f$ from the right at $0.$
Thus the derivative from the left equals $p-1/2.$ To find out what that implies, we look at the difference quotient from the left:
$$frac{(cos x-1)/(xsin x)-(-1/2)}{x}= frac{cos x-1+(xsin x)/2}{x^2sin x}.$$
Using $cos x = 1-x^2/2 + O(x^4)$ and $sin x = x + O(x^3)$ shows the above equals
$$frac{O(x^4)}{x^2sin x}to 0.$$
Thus the derivative from the left is $0.$ Therefore $p-1/2 = 0.$ So $p=1/2.$
Summary: Fixing $p,q$ and assuming $f'(0)$ exists, we showed $q=-1/2, p=1/2.$ We're not quite done though. Now we have to reason in the other direction: Assuming $q=-1/2, p=1/2,$ it follows that $f'(0)$ exists. This is simple, and I'll leave this to you for now.
answered Jan 28 at 21:03
zhw.zhw.
74.2k43175
$endgroup$
$begingroup$
Ok, so if I assume that $ f'(a)$ exists, (a) will I find all "possible" (but they may be wrong) values? (b) If I want to check if these values are correct, I should put them to $ lim_{h to 0^-} frac{f(h) - f(0)}{h} $ and check if derivative exists, is that right? @zhw.
$endgroup$
– VirtualUser
Jan 28 at 23:59
1
$begingroup$
Right. It's an iff statement. Thus we must show $f'(0)$ exists $implies q=-1/2,p=1/2$ and $q=-1/2,p=1/2implies f'(0)$ exists. I showed the first $implies.$
$endgroup$
– zhw.
Jan 29 at 17:05
$begingroup$
Thanks for explanation!
$endgroup$
– VirtualUser
Jan 29 at 17:19
add a comment |
$begingroup$
First observe that
$$ frac{cos(x) - 1}{x sin(x)} = frac{cos(x) - 1}{x^2} cdot frac{x}{sin(x)} longrightarrow -frac 1 2 $$
for $x to 0$. Moreover, it holds
$q + sin(px) + qx to q$. So since we need continuity of $f$ in the first place, one obtains $q = -frac 1 2$.
Now calculate the left and the right derivate in $0$ for the function
$$f(x)=begin{cases} -frac 1 2+sin(px)- frac 1 2 x &text{for } xgeq 0, \ frac{-1}{xsin(x)} + frac{cos(x)}{xsin(x)} &text{for } xin (-pi,0). end{cases}$$
If you want $f$ to be differentiable in $x = 0$ you need to have $lim_{h to -0} frac{f(h) - f(0)}{h} = lim_{h to +0} frac{f(h) - f(0)}{h}$. This relation yields you then all possible values for $p$.
answered Jan 28 at 12:44
YaddleYaddle
3,106829
$endgroup$
$begingroup$
Ok, but there I have problem: $lim_{h to -0} frac{f(h) - f(0)}{h} $ - There I can't use derivate - I should calculate this from $ frac{f(h) - f(0)}{h} $
$endgroup$
– VirtualUser
Jan 28 at 12:57
$begingroup$
And that makes me trouble, I am not able to finish calculating because I get some weird stuff @Yaddle
$endgroup$
– VirtualUser
Jan 28 at 14:57
$begingroup$
You miscalculated in your first claim, the limit there is $-1/2.$
$endgroup$
– zhw.
Jan 29 at 17:07
$begingroup$
@zhw. You're right. I corrected it.
$endgroup$
– Yaddle
Jan 29 at 18:49
add a comment |
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2 Answers
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$begingroup$
Fix $p,q$ and assume $f'(0)$ exists. Then $f$ is continuous at $0.$ Because $f(0)=q,$ we must have $lim_{xto 0^-}f(x)=q.$ But
$$lim_{xto 0^-}f(x) = lim_{xto 0^-}frac{cos x-1}{xsin x} = -frac{1}{2}.$$
Therefore $q= -1/2.$
Now take $x>0$ and consider
$$frac{f(x)-f(0)}{x}= frac{-1/2 +sin (px) -x/2 +1/2}{x} = frac{sin (px) - x/2}{x}.$$
The limit of this from the right is $p-1/2.$ This is the derivative of $f$ from the right at $0.$
Thus the derivative from the left equals $p-1/2.$ To find out what that implies, we look at the difference quotient from the left:
$$frac{(cos x-1)/(xsin x)-(-1/2)}{x}= frac{cos x-1+(xsin x)/2}{x^2sin x}.$$
Using $cos x = 1-x^2/2 + O(x^4)$ and $sin x = x + O(x^3)$ shows the above equals
$$frac{O(x^4)}{x^2sin x}to 0.$$
Thus the derivative from the left is $0.$ Therefore $p-1/2 = 0.$ So $p=1/2.$
Summary: Fixing $p,q$ and assuming $f'(0)$ exists, we showed $q=-1/2, p=1/2.$ We're not quite done though. Now we have to reason in the other direction: Assuming $q=-1/2, p=1/2,$ it follows that $f'(0)$ exists. This is simple, and I'll leave this to you for now.
answered Jan 28 at 21:03
zhw.zhw.
74.2k43175
$endgroup$
$begingroup$
Ok, so if I assume that $ f'(a)$ exists, (a) will I find all "possible" (but they may be wrong) values? (b) If I want to check if these values are correct, I should put them to $ lim_{h to 0^-} frac{f(h) - f(0)}{h} $ and check if derivative exists, is that right? @zhw.
$endgroup$
– VirtualUser
Jan 28 at 23:59
1
$begingroup$
Right. It's an iff statement. Thus we must show $f'(0)$ exists $implies q=-1/2,p=1/2$ and $q=-1/2,p=1/2implies f'(0)$ exists. I showed the first $implies.$
$endgroup$
– zhw.
Jan 29 at 17:05
$begingroup$
Thanks for explanation!
$endgroup$
– VirtualUser
Jan 29 at 17:19
add a comment |
$begingroup$
Fix $p,q$ and assume $f'(0)$ exists. Then $f$ is continuous at $0.$ Because $f(0)=q,$ we must have $lim_{xto 0^-}f(x)=q.$ But
$$lim_{xto 0^-}f(x) = lim_{xto 0^-}frac{cos x-1}{xsin x} = -frac{1}{2}.$$
Therefore $q= -1/2.$
Now take $x>0$ and consider
$$frac{f(x)-f(0)}{x}= frac{-1/2 +sin (px) -x/2 +1/2}{x} = frac{sin (px) - x/2}{x}.$$
The limit of this from the right is $p-1/2.$ This is the derivative of $f$ from the right at $0.$
Thus the derivative from the left equals $p-1/2.$ To find out what that implies, we look at the difference quotient from the left:
$$frac{(cos x-1)/(xsin x)-(-1/2)}{x}= frac{cos x-1+(xsin x)/2}{x^2sin x}.$$
Using $cos x = 1-x^2/2 + O(x^4)$ and $sin x = x + O(x^3)$ shows the above equals
$$frac{O(x^4)}{x^2sin x}to 0.$$
Thus the derivative from the left is $0.$ Therefore $p-1/2 = 0.$ So $p=1/2.$
Summary: Fixing $p,q$ and assuming $f'(0)$ exists, we showed $q=-1/2, p=1/2.$ We're not quite done though. Now we have to reason in the other direction: Assuming $q=-1/2, p=1/2,$ it follows that $f'(0)$ exists. This is simple, and I'll leave this to you for now.
answered Jan 28 at 21:03
zhw.zhw.
74.2k43175
$endgroup$
$begingroup$
Ok, so if I assume that $ f'(a)$ exists, (a) will I find all "possible" (but they may be wrong) values? (b) If I want to check if these values are correct, I should put them to $ lim_{h to 0^-} frac{f(h) - f(0)}{h} $ and check if derivative exists, is that right? @zhw.
$endgroup$
– VirtualUser
Jan 28 at 23:59
1
$begingroup$
Right. It's an iff statement. Thus we must show $f'(0)$ exists $implies q=-1/2,p=1/2$ and $q=-1/2,p=1/2implies f'(0)$ exists. I showed the first $implies.$
$endgroup$
– zhw.
Jan 29 at 17:05
$begingroup$
Thanks for explanation!
$endgroup$
– VirtualUser
Jan 29 at 17:19
add a comment |
$begingroup$
Fix $p,q$ and assume $f'(0)$ exists. Then $f$ is continuous at $0.$ Because $f(0)=q,$ we must have $lim_{xto 0^-}f(x)=q.$ But
$$lim_{xto 0^-}f(x) = lim_{xto 0^-}frac{cos x-1}{xsin x} = -frac{1}{2}.$$
Therefore $q= -1/2.$
Now take $x>0$ and consider
$$frac{f(x)-f(0)}{x}= frac{-1/2 +sin (px) -x/2 +1/2}{x} = frac{sin (px) - x/2}{x}.$$
The limit of this from the right is $p-1/2.$ This is the derivative of $f$ from the right at $0.$
Thus the derivative from the left equals $p-1/2.$ To find out what that implies, we look at the difference quotient from the left:
$$frac{(cos x-1)/(xsin x)-(-1/2)}{x}= frac{cos x-1+(xsin x)/2}{x^2sin x}.$$
Using $cos x = 1-x^2/2 + O(x^4)$ and $sin x = x + O(x^3)$ shows the above equals
$$frac{O(x^4)}{x^2sin x}to 0.$$
Thus the derivative from the left is $0.$ Therefore $p-1/2 = 0.$ So $p=1/2.$
Summary: Fixing $p,q$ and assuming $f'(0)$ exists, we showed $q=-1/2, p=1/2.$ We're not quite done though. Now we have to reason in the other direction: Assuming $q=-1/2, p=1/2,$ it follows that $f'(0)$ exists. This is simple, and I'll leave this to you for now.
answered Jan 28 at 21:03
zhw.zhw.
74.2k43175
$endgroup$
Fix $p,q$ and assume $f'(0)$ exists. Then $f$ is continuous at $0.$ Because $f(0)=q,$ we must have $lim_{xto 0^-}f(x)=q.$ But
$$lim_{xto 0^-}f(x) = lim_{xto 0^-}frac{cos x-1}{xsin x} = -frac{1}{2}.$$
Therefore $q= -1/2.$
Now take $x>0$ and consider
$$frac{f(x)-f(0)}{x}= frac{-1/2 +sin (px) -x/2 +1/2}{x} = frac{sin (px) - x/2}{x}.$$
The limit of this from the right is $p-1/2.$ This is the derivative of $f$ from the right at $0.$
Thus the derivative from the left equals $p-1/2.$ To find out what that implies, we look at the difference quotient from the left:
$$frac{(cos x-1)/(xsin x)-(-1/2)}{x}= frac{cos x-1+(xsin x)/2}{x^2sin x}.$$
Using $cos x = 1-x^2/2 + O(x^4)$ and $sin x = x + O(x^3)$ shows the above equals
$$frac{O(x^4)}{x^2sin x}to 0.$$
Thus the derivative from the left is $0.$ Therefore $p-1/2 = 0.$ So $p=1/2.$
Summary: Fixing $p,q$ and assuming $f'(0)$ exists, we showed $q=-1/2, p=1/2.$ We're not quite done though. Now we have to reason in the other direction: Assuming $q=-1/2, p=1/2,$ it follows that $f'(0)$ exists. This is simple, and I'll leave this to you for now.
answered Jan 28 at 21:03
zhw.zhw.
74.2k43175
edited Jan 29 at 17:36
answered Jan 28 at 21:03
zhw.zhw.
74.2k43175
answered Jan 28 at 21:03
zhw.zhw.
74.2k43175
answered Jan 28 at 21:03
zhw.zhw.
74.2k43175
74.2k43175
$begingroup$
Ok, so if I assume that $ f'(a)$ exists, (a) will I find all "possible" (but they may be wrong) values? (b) If I want to check if these values are correct, I should put them to $ lim_{h to 0^-} frac{f(h) - f(0)}{h} $ and check if derivative exists, is that right? @zhw.
$endgroup$
– VirtualUser
Jan 28 at 23:59
1
$begingroup$
Right. It's an iff statement. Thus we must show $f'(0)$ exists $implies q=-1/2,p=1/2$ and $q=-1/2,p=1/2implies f'(0)$ exists. I showed the first $implies.$
$endgroup$
– zhw.
Jan 29 at 17:05
$begingroup$
Thanks for explanation!
$endgroup$
– VirtualUser
Jan 29 at 17:19
add a comment |
$begingroup$
Ok, so if I assume that $ f'(a)$ exists, (a) will I find all "possible" (but they may be wrong) values? (b) If I want to check if these values are correct, I should put them to $ lim_{h to 0^-} frac{f(h) - f(0)}{h} $ and check if derivative exists, is that right? @zhw.
$endgroup$
– VirtualUser
Jan 28 at 23:59
1
$begingroup$
Right. It's an iff statement. Thus we must show $f'(0)$ exists $implies q=-1/2,p=1/2$ and $q=-1/2,p=1/2implies f'(0)$ exists. I showed the first $implies.$
$endgroup$
– zhw.
Jan 29 at 17:05
$begingroup$
Thanks for explanation!
$endgroup$
– VirtualUser
Jan 29 at 17:19
$begingroup$
Ok, so if I assume that $ f'(a)$ exists, (a) will I find all "possible" (but they may be wrong) values? (b) If I want to check if these values are correct, I should put them to $ lim_{h to 0^-} frac{f(h) - f(0)}{h} $ and check if derivative exists, is that right? @zhw.
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– VirtualUser
Jan 28 at 23:59
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Ok, so if I assume that $ f'(a)$ exists, (a) will I find all "possible" (but they may be wrong) values? (b) If I want to check if these values are correct, I should put them to $ lim_{h to 0^-} frac{f(h) - f(0)}{h} $ and check if derivative exists, is that right? @zhw.
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– VirtualUser
Jan 28 at 23:59
1
1
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Right. It's an iff statement. Thus we must show $f'(0)$ exists $implies q=-1/2,p=1/2$ and $q=-1/2,p=1/2implies f'(0)$ exists. I showed the first $implies.$
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– zhw.
Jan 29 at 17:05
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Right. It's an iff statement. Thus we must show $f'(0)$ exists $implies q=-1/2,p=1/2$ and $q=-1/2,p=1/2implies f'(0)$ exists. I showed the first $implies.$
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– zhw.
Jan 29 at 17:05
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Thanks for explanation!
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– VirtualUser
Jan 29 at 17:19
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Thanks for explanation!
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– VirtualUser
Jan 29 at 17:19
add a comment |
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First observe that
$$ frac{cos(x) - 1}{x sin(x)} = frac{cos(x) - 1}{x^2} cdot frac{x}{sin(x)} longrightarrow -frac 1 2 $$
for $x to 0$. Moreover, it holds
$q + sin(px) + qx to q$. So since we need continuity of $f$ in the first place, one obtains $q = -frac 1 2$.
Now calculate the left and the right derivate in $0$ for the function
$$f(x)=begin{cases} -frac 1 2+sin(px)- frac 1 2 x &text{for } xgeq 0, \ frac{-1}{xsin(x)} + frac{cos(x)}{xsin(x)} &text{for } xin (-pi,0). end{cases}$$
If you want $f$ to be differentiable in $x = 0$ you need to have $lim_{h to -0} frac{f(h) - f(0)}{h} = lim_{h to +0} frac{f(h) - f(0)}{h}$. This relation yields you then all possible values for $p$.
answered Jan 28 at 12:44
YaddleYaddle
3,106829
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Ok, but there I have problem: $lim_{h to -0} frac{f(h) - f(0)}{h} $ - There I can't use derivate - I should calculate this from $ frac{f(h) - f(0)}{h} $
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– VirtualUser
Jan 28 at 12:57
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And that makes me trouble, I am not able to finish calculating because I get some weird stuff @Yaddle
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– VirtualUser
Jan 28 at 14:57
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You miscalculated in your first claim, the limit there is $-1/2.$
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– zhw.
Jan 29 at 17:07
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@zhw. You're right. I corrected it.
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– Yaddle
Jan 29 at 18:49
add a comment |
$begingroup$
First observe that
$$ frac{cos(x) - 1}{x sin(x)} = frac{cos(x) - 1}{x^2} cdot frac{x}{sin(x)} longrightarrow -frac 1 2 $$
for $x to 0$. Moreover, it holds
$q + sin(px) + qx to q$. So since we need continuity of $f$ in the first place, one obtains $q = -frac 1 2$.
Now calculate the left and the right derivate in $0$ for the function
$$f(x)=begin{cases} -frac 1 2+sin(px)- frac 1 2 x &text{for } xgeq 0, \ frac{-1}{xsin(x)} + frac{cos(x)}{xsin(x)} &text{for } xin (-pi,0). end{cases}$$
If you want $f$ to be differentiable in $x = 0$ you need to have $lim_{h to -0} frac{f(h) - f(0)}{h} = lim_{h to +0} frac{f(h) - f(0)}{h}$. This relation yields you then all possible values for $p$.
answered Jan 28 at 12:44
YaddleYaddle
3,106829
$endgroup$
$begingroup$
Ok, but there I have problem: $lim_{h to -0} frac{f(h) - f(0)}{h} $ - There I can't use derivate - I should calculate this from $ frac{f(h) - f(0)}{h} $
$endgroup$
– VirtualUser
Jan 28 at 12:57
$begingroup$
And that makes me trouble, I am not able to finish calculating because I get some weird stuff @Yaddle
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– VirtualUser
Jan 28 at 14:57
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You miscalculated in your first claim, the limit there is $-1/2.$
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– zhw.
Jan 29 at 17:07
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@zhw. You're right. I corrected it.
$endgroup$
– Yaddle
Jan 29 at 18:49
add a comment |
$begingroup$
First observe that
$$ frac{cos(x) - 1}{x sin(x)} = frac{cos(x) - 1}{x^2} cdot frac{x}{sin(x)} longrightarrow -frac 1 2 $$
for $x to 0$. Moreover, it holds
$q + sin(px) + qx to q$. So since we need continuity of $f$ in the first place, one obtains $q = -frac 1 2$.
Now calculate the left and the right derivate in $0$ for the function
$$f(x)=begin{cases} -frac 1 2+sin(px)- frac 1 2 x &text{for } xgeq 0, \ frac{-1}{xsin(x)} + frac{cos(x)}{xsin(x)} &text{for } xin (-pi,0). end{cases}$$
If you want $f$ to be differentiable in $x = 0$ you need to have $lim_{h to -0} frac{f(h) - f(0)}{h} = lim_{h to +0} frac{f(h) - f(0)}{h}$. This relation yields you then all possible values for $p$.
answered Jan 28 at 12:44
YaddleYaddle
3,106829
$endgroup$
First observe that
$$ frac{cos(x) - 1}{x sin(x)} = frac{cos(x) - 1}{x^2} cdot frac{x}{sin(x)} longrightarrow -frac 1 2 $$
for $x to 0$. Moreover, it holds
$q + sin(px) + qx to q$. So since we need continuity of $f$ in the first place, one obtains $q = -frac 1 2$.
Now calculate the left and the right derivate in $0$ for the function
$$f(x)=begin{cases} -frac 1 2+sin(px)- frac 1 2 x &text{for } xgeq 0, \ frac{-1}{xsin(x)} + frac{cos(x)}{xsin(x)} &text{for } xin (-pi,0). end{cases}$$
If you want $f$ to be differentiable in $x = 0$ you need to have $lim_{h to -0} frac{f(h) - f(0)}{h} = lim_{h to +0} frac{f(h) - f(0)}{h}$. This relation yields you then all possible values for $p$.
answered Jan 28 at 12:44
YaddleYaddle
3,106829
edited Jan 29 at 18:49
answered Jan 28 at 12:44
YaddleYaddle
3,106829
answered Jan 28 at 12:44
YaddleYaddle
3,106829
answered Jan 28 at 12:44
YaddleYaddle
3,106829
3,106829
$begingroup$
Ok, but there I have problem: $lim_{h to -0} frac{f(h) - f(0)}{h} $ - There I can't use derivate - I should calculate this from $ frac{f(h) - f(0)}{h} $
$endgroup$
– VirtualUser
Jan 28 at 12:57
$begingroup$
And that makes me trouble, I am not able to finish calculating because I get some weird stuff @Yaddle
$endgroup$
– VirtualUser
Jan 28 at 14:57
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You miscalculated in your first claim, the limit there is $-1/2.$
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– zhw.
Jan 29 at 17:07
$begingroup$
@zhw. You're right. I corrected it.
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– Yaddle
Jan 29 at 18:49
add a comment |
$begingroup$
Ok, but there I have problem: $lim_{h to -0} frac{f(h) - f(0)}{h} $ - There I can't use derivate - I should calculate this from $ frac{f(h) - f(0)}{h} $
$endgroup$
– VirtualUser
Jan 28 at 12:57
$begingroup$
And that makes me trouble, I am not able to finish calculating because I get some weird stuff @Yaddle
$endgroup$
– VirtualUser
Jan 28 at 14:57
$begingroup$
You miscalculated in your first claim, the limit there is $-1/2.$
$endgroup$
– zhw.
Jan 29 at 17:07
$begingroup$
@zhw. You're right. I corrected it.
$endgroup$
– Yaddle
Jan 29 at 18:49
$begingroup$
Ok, but there I have problem: $lim_{h to -0} frac{f(h) - f(0)}{h} $ - There I can't use derivate - I should calculate this from $ frac{f(h) - f(0)}{h} $
$endgroup$
– VirtualUser
Jan 28 at 12:57
$begingroup$
Ok, but there I have problem: $lim_{h to -0} frac{f(h) - f(0)}{h} $ - There I can't use derivate - I should calculate this from $ frac{f(h) - f(0)}{h} $
$endgroup$
– VirtualUser
Jan 28 at 12:57
$begingroup$
And that makes me trouble, I am not able to finish calculating because I get some weird stuff @Yaddle
$endgroup$
– VirtualUser
Jan 28 at 14:57
$begingroup$
And that makes me trouble, I am not able to finish calculating because I get some weird stuff @Yaddle
$endgroup$
– VirtualUser
Jan 28 at 14:57
$begingroup$
You miscalculated in your first claim, the limit there is $-1/2.$
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– zhw.
Jan 29 at 17:07
$begingroup$
You miscalculated in your first claim, the limit there is $-1/2.$
$endgroup$
– zhw.
Jan 29 at 17:07
$begingroup$
@zhw. You're right. I corrected it.
$endgroup$
– Yaddle
Jan 29 at 18:49
$begingroup$
@zhw. You're right. I corrected it.
$endgroup$
– Yaddle
Jan 29 at 18:49
add a comment |
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Start by getting continuity at $0$. Take the limits from the left and the right (in terms of $p$ and $q$).
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– Michael Burr
Jan 24 at 0:11
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I wrote that in topic
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– VirtualUser
Jan 24 at 0:13
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The lazy way of doing this is just by considering two things: continuity and $f’_+ (0)=f’_-(0)$, where $f_+$ is $f$ for $x geq 0$ and $f_-$ for $x in (-pi,0)$. And indeed all the things need to be limits.
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– Dani
Jan 24 at 9:23
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Why lazy? What do you mean by that?
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– VirtualUser
Jan 24 at 18:19
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I edited topic - can you look at this again? I don't want solution, only I want to be sure about my way of thinking
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– VirtualUser
Jan 25 at 19:14