Why are you able to ignore this factor when looking for local maximum?
$begingroup$
This is from a math gre subject test practice.
We have a function $$f(x) = int_{x}^{2x} frac{sin(t)}{t} dt $$
We want to find a point $x$ where it takes a local maximum on the interval $(0, frac{3pi}{2})$
The argument for finding it in the book goes as follows:
Let:
$$ h(x) = int frac{sin(t)}{t} dt $$
Then:
$$ f(x) = h(2x) - h(x) $$
Local maximum is a point where $f'(x) = 0$ . So we have (by differentiating the "h" expression that)
$$ 2 frac{sin(2x)}{2x} - frac{sin(x)}{x} = 0$$
$$ sin(2x) - sin(x) = 0$$
$$ sin(x)(2 cos(x) - 1) = 0 $$
And now the book states that "trivially it can't be the case that $sin(x)=0$ since $0 < x < frac{3pi}{2}$" and just goes on to explore the $2 cos(x) - 1 = 0$ case to find the desired $x$.
I absolutely don't understand what's trivial here, $pi < frac{3 pi}{2}$ is a well known zero of $sin(x)$. And checking that is in fact NOT the local maxima is extremely expensive (since you have to draw graphs and intuitively reason about it which is risky when you only want to spend 1 - 1.5 minutes on the question). I don't understand how they were able to ignore it.
(Not mentioned in the book)
One could try to use convexity, i.e. by differentiating
$$ frac{sin(2x)}{x} - frac{sin(x)}{x} $$
We get
$$ f'' = 2frac{cos(2x)}{x} - frac{sin(2x)}{x^2} +frac{sin(x)}{x^2} - frac{cos(x)}{x} $$
Evaluating at $x = pi$
$$ f'' = frac{2}{pi} - 0 + 0 - frac{1}{pi} = frac{1}{pi} $$
So therefore this must be a local "minima".
This is effort and time-wise less expensive than graphing the thing and staring at it, but still pretty costly, i'm hoping for an even simpler technique than this.
calculus
asked Jan 15 at 23:51
frogeyedpeasfrogeyedpeas
7,71372055
$endgroup$
add a comment |
$begingroup$
This is from a math gre subject test practice.
We have a function $$f(x) = int_{x}^{2x} frac{sin(t)}{t} dt $$
We want to find a point $x$ where it takes a local maximum on the interval $(0, frac{3pi}{2})$
The argument for finding it in the book goes as follows:
Let:
$$ h(x) = int frac{sin(t)}{t} dt $$
Then:
$$ f(x) = h(2x) - h(x) $$
Local maximum is a point where $f'(x) = 0$ . So we have (by differentiating the "h" expression that)
$$ 2 frac{sin(2x)}{2x} - frac{sin(x)}{x} = 0$$
$$ sin(2x) - sin(x) = 0$$
$$ sin(x)(2 cos(x) - 1) = 0 $$
And now the book states that "trivially it can't be the case that $sin(x)=0$ since $0 < x < frac{3pi}{2}$" and just goes on to explore the $2 cos(x) - 1 = 0$ case to find the desired $x$.
I absolutely don't understand what's trivial here, $pi < frac{3 pi}{2}$ is a well known zero of $sin(x)$. And checking that is in fact NOT the local maxima is extremely expensive (since you have to draw graphs and intuitively reason about it which is risky when you only want to spend 1 - 1.5 minutes on the question). I don't understand how they were able to ignore it.
(Not mentioned in the book)
One could try to use convexity, i.e. by differentiating
$$ frac{sin(2x)}{x} - frac{sin(x)}{x} $$
We get
$$ f'' = 2frac{cos(2x)}{x} - frac{sin(2x)}{x^2} +frac{sin(x)}{x^2} - frac{cos(x)}{x} $$
Evaluating at $x = pi$
$$ f'' = frac{2}{pi} - 0 + 0 - frac{1}{pi} = frac{1}{pi} $$
So therefore this must be a local "minima".
This is effort and time-wise less expensive than graphing the thing and staring at it, but still pretty costly, i'm hoping for an even simpler technique than this.
calculus
asked Jan 15 at 23:51
frogeyedpeasfrogeyedpeas
7,71372055
$endgroup$
add a comment |
$begingroup$
This is from a math gre subject test practice.
We have a function $$f(x) = int_{x}^{2x} frac{sin(t)}{t} dt $$
We want to find a point $x$ where it takes a local maximum on the interval $(0, frac{3pi}{2})$
The argument for finding it in the book goes as follows:
Let:
$$ h(x) = int frac{sin(t)}{t} dt $$
Then:
$$ f(x) = h(2x) - h(x) $$
Local maximum is a point where $f'(x) = 0$ . So we have (by differentiating the "h" expression that)
$$ 2 frac{sin(2x)}{2x} - frac{sin(x)}{x} = 0$$
$$ sin(2x) - sin(x) = 0$$
$$ sin(x)(2 cos(x) - 1) = 0 $$
And now the book states that "trivially it can't be the case that $sin(x)=0$ since $0 < x < frac{3pi}{2}$" and just goes on to explore the $2 cos(x) - 1 = 0$ case to find the desired $x$.
I absolutely don't understand what's trivial here, $pi < frac{3 pi}{2}$ is a well known zero of $sin(x)$. And checking that is in fact NOT the local maxima is extremely expensive (since you have to draw graphs and intuitively reason about it which is risky when you only want to spend 1 - 1.5 minutes on the question). I don't understand how they were able to ignore it.
(Not mentioned in the book)
One could try to use convexity, i.e. by differentiating
$$ frac{sin(2x)}{x} - frac{sin(x)}{x} $$
We get
$$ f'' = 2frac{cos(2x)}{x} - frac{sin(2x)}{x^2} +frac{sin(x)}{x^2} - frac{cos(x)}{x} $$
Evaluating at $x = pi$
$$ f'' = frac{2}{pi} - 0 + 0 - frac{1}{pi} = frac{1}{pi} $$
So therefore this must be a local "minima".
This is effort and time-wise less expensive than graphing the thing and staring at it, but still pretty costly, i'm hoping for an even simpler technique than this.
calculus
asked Jan 15 at 23:51
frogeyedpeasfrogeyedpeas
7,71372055
$endgroup$
This is from a math gre subject test practice.
We have a function $$f(x) = int_{x}^{2x} frac{sin(t)}{t} dt $$
We want to find a point $x$ where it takes a local maximum on the interval $(0, frac{3pi}{2})$
The argument for finding it in the book goes as follows:
Let:
$$ h(x) = int frac{sin(t)}{t} dt $$
Then:
$$ f(x) = h(2x) - h(x) $$
Local maximum is a point where $f'(x) = 0$ . So we have (by differentiating the "h" expression that)
$$ 2 frac{sin(2x)}{2x} - frac{sin(x)}{x} = 0$$
$$ sin(2x) - sin(x) = 0$$
$$ sin(x)(2 cos(x) - 1) = 0 $$
And now the book states that "trivially it can't be the case that $sin(x)=0$ since $0 < x < frac{3pi}{2}$" and just goes on to explore the $2 cos(x) - 1 = 0$ case to find the desired $x$.
I absolutely don't understand what's trivial here, $pi < frac{3 pi}{2}$ is a well known zero of $sin(x)$. And checking that is in fact NOT the local maxima is extremely expensive (since you have to draw graphs and intuitively reason about it which is risky when you only want to spend 1 - 1.5 minutes on the question). I don't understand how they were able to ignore it.
(Not mentioned in the book)
One could try to use convexity, i.e. by differentiating
$$ frac{sin(2x)}{x} - frac{sin(x)}{x} $$
We get
$$ f'' = 2frac{cos(2x)}{x} - frac{sin(2x)}{x^2} +frac{sin(x)}{x^2} - frac{cos(x)}{x} $$
Evaluating at $x = pi$
$$ f'' = frac{2}{pi} - 0 + 0 - frac{1}{pi} = frac{1}{pi} $$
So therefore this must be a local "minima".
This is effort and time-wise less expensive than graphing the thing and staring at it, but still pretty costly, i'm hoping for an even simpler technique than this.
calculus
calculus
asked Jan 15 at 23:51
frogeyedpeasfrogeyedpeas
7,71372055
asked Jan 15 at 23:51
frogeyedpeasfrogeyedpeas
7,71372055
edited Jan 15 at 23:59
frogeyedpeas
asked Jan 15 at 23:51
frogeyedpeasfrogeyedpeas
7,71372055
asked Jan 15 at 23:51
frogeyedpeasfrogeyedpeas
7,71372055
asked Jan 15 at 23:51
frogeyedpeasfrogeyedpeas
7,71372055
7,71372055
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1 Answer
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$begingroup$
That is because, near $pi$, $;2cos x-1<0$, and $sin x>0$ if $x<pi$, $,<0$ if $x>pi$, so that the function is decreasing for $x<pi$, increasing for $x>pi$ (in a neighbourhood of $pi$), and we have a local minimum.
answered Jan 16 at 0:02
BernardBernard
125k743119
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$begingroup$
That is because, near $pi$, $;2cos x-1<0$, and $sin x>0$ if $x<pi$, $,<0$ if $x>pi$, so that the function is decreasing for $x<pi$, increasing for $x>pi$ (in a neighbourhood of $pi$), and we have a local minimum.
answered Jan 16 at 0:02
BernardBernard
125k743119
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add a comment |
$begingroup$
That is because, near $pi$, $;2cos x-1<0$, and $sin x>0$ if $x<pi$, $,<0$ if $x>pi$, so that the function is decreasing for $x<pi$, increasing for $x>pi$ (in a neighbourhood of $pi$), and we have a local minimum.
answered Jan 16 at 0:02
BernardBernard
125k743119
$endgroup$
add a comment |
$begingroup$
That is because, near $pi$, $;2cos x-1<0$, and $sin x>0$ if $x<pi$, $,<0$ if $x>pi$, so that the function is decreasing for $x<pi$, increasing for $x>pi$ (in a neighbourhood of $pi$), and we have a local minimum.
answered Jan 16 at 0:02
BernardBernard
125k743119
$endgroup$
That is because, near $pi$, $;2cos x-1<0$, and $sin x>0$ if $x<pi$, $,<0$ if $x>pi$, so that the function is decreasing for $x<pi$, increasing for $x>pi$ (in a neighbourhood of $pi$), and we have a local minimum.
answered Jan 16 at 0:02
BernardBernard
125k743119
answered Jan 16 at 0:02
BernardBernard
125k743119
answered Jan 16 at 0:02
BernardBernard
125k743119
answered Jan 16 at 0:02
BernardBernard
125k743119
125k743119
add a comment |
add a comment |
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