Need help, URGENT maths question: (2 Viewers)

leehuan · Apr 25, 2016

1008 said:
Thanks! Why didn't I think of that?!

Because L'Hopitals comes after that stuff in the MATH1151 book

InteGrand · Apr 25, 2016

1008 said:
Thanks! Why didn't I think of that?!

If we use L'Hospital's rule, we'll end up having to assume that f' is continuous at a after the differentiation step in order to take the limit as h -> 0 of the resulting expression. But we aren't given this fact, so we can't assume it, so can't really use L'Hospital like this.

1008 · Apr 25, 2016

InteGrand said:
If we use L'Hospital's rule, we'll end up having to assume that f' is continuous at a after the differentiation step in order to take the limit as h -> 0 of the resulting expression. But we aren't given this fact, so we can't assume it, so can't really use L'Hospital like this.

$\\\lim_{h \rightarrow 0}\frac{f(a+ph)-f(a-ph)}{h}\\\lim_{h\rightarrow 0}\frac{pf'(a+ph)+pf'(a-ph)}{1}=2pf'(a)\\\text{I don't understand where we have to make that assumption...}$

InteGrand · Apr 25, 2016

1008 said:
$\\\lim_{h \rightarrow 0}\frac{f(a+ph)-f(a-ph)}{h}\\\lim_{h\rightarrow 0}\frac{pf'(a+ph)+pf'(a-ph)}{1}=2pf'(a)\\\text{I don't understand where we have to make that assumption...}$

We used that assumption in the last equality. I.e. assuming that f'(a + ph) will tend to f'(a) as h -> 0. This essentially is using the assumption that f' is continuous at a (by definition of continuity).

(In general it could happen that f'(a + t) has no limit even as t -> 0.)

1008 · Apr 25, 2016

InteGrand said:
We used that assumption in the last equality. I.e. assuming that f'(a + ph) will tend to f'(a) as h -> 0. This essentially is using the assumption that f' is continuous at a (by definition of continuity).

(In general it could happen that f'(a + t) has no limit even as t -> 0.)

But if the question states that f is differentiable at x=a, can't we assume that f' is continuous at a?

InteGrand · Apr 25, 2016

1008 said:
But if the question states that f is differentiable at x=a, can't we assume that f' is continuous at a?

No. There exist functions that are differentiable somewhere but not continuously differentiable there. An example of such a function is found here: https://en.wikipedia.org/wiki/Differentiable_function#Differentiability_classes .

1008 · Apr 25, 2016

1008 said:
Just another question:

$\\\text{Let f be continuous on [a,b]. Prove there exists c}\in(a,b)\text{ such that}\\\int_{a}^{b}f(t)\text{ }dt = f(c)(b-a).\\\\\text{The way I think I can do it:}\\\text{Consider the function }F(t)\text{ where }F'(t) = f(t)\\\text{As }f(t)\text{ is continuous on [a,b], }F(t)\text{ is differentiable on [a,b] and hence is also continuous.}\\\text{Hence we apply the Mean Value Theorem (MVT) over [a,b]}\\F'(c) =\frac{F(b) - F(a)}{b-a}, c\in(a,b)\\\text{As }F'(t) = f(t),\\f(c)(b-a) = F(b) - F(a).\text{ Now }F(b) - F(a) = \int_{a}^{b}f(t)\text{ }dt\\\therefore\int_{a}^{b}f(t)\text{ }dt=f(c)(b-a)$

Is this right, or am I missing something

In continuation to this question, I have another:
$\\\text{Let }f\text{ be continuous on [a,b]. Prove, provided }g(x)\neq0\text{ for }x\in[a,b],\\\text{that there exists c}\in(a,b)\text{ such that:}\\\int_{a}^{b}f(t)g(t)dt=f(c)\int_{a}^{b}g(t)dt\\\text{Hint: Generalise the previous proof.}$

InteGrand · Apr 25, 2016

1008 said:
In continuation to this question, I have another:
$\\\text{Let }f\text{ be continuous on [a,b]. Prove, provided }g(x)\neq0\text{ for }x\in[a,b],\\\text{that there exists c}\in(a,b)\text{ such that:}\\\int_{a}^{b}f(t)g(t)dt=f(c)\int_{a}^{b}g(t)dt\\\text{Hint: Generalise the previous proof.}$

In order for this to be true, we need g to be of fixed sign (and integrable of course), rather than simply non-zero. (If g is given to be continuous though, then g(x) =/= 0 suffices.)

A proof of that result is given here: https://en.wikipedia.org/wiki/Mean_...rst_Mean_Value_Theorem_for_Definite_Integrals .

leehuan · Apr 25, 2016

I love how I'm seeing questions that I already asked InteGrand.

1008 · Apr 26, 2016

InteGrand said:
In order for this to be true, we need g to be of fixed sign (and integrable of course), rather than simply non-zero. (If g is given to be continuous though, then g(x) =/= 0 suffices.)

A proof of that result is given here: https://en.wikipedia.org/wiki/Mean_...rst_Mean_Value_Theorem_for_Definite_Integrals .

Thanks InteGrand, I get the proof on Wikipedia. But does that connect to my previous proof at all?

I also have another question:
$\\\\\text{Find all the values of }a\text{ and }x,\text{ both in [0,2}\pi],\text{ where}\\f(x) = \cos a+2\cos(2x)+\cos(4x-a)\\\text{has a horizontal point of inflexion.}$

InteGrand · Apr 26, 2016

That last Q. is essentially just based on using the higher-order derivative test ( https://en.wikipedia.org/wiki/First_derivative_test#Higher-order_derivative_test ). So you look for places where the first and second derivatives are simultaneously 0 for a horizontal point of inflexion, and out of these points, the points where the first non-zero derivative at the point is an odd derivative (e.g. third or fifth derivative) will be a horizontal point of inflexion.

1008 · Apr 26, 2016

InteGrand said:
That last Q. is essentially just based on using the higher-order derivative test ( https://en.wikipedia.org/wiki/First_derivative_test#Higher-order_derivative_test ). So you look for places where the first and second derivatives are simultaneously 0 for a horizontal point of inflexion, and out of these points, the points where the first non-zero derivative at the point is an odd derivative (e.g. third or fifth derivative) will be a horizontal point of inflexion.

Sorry, but I still don't get the question. Because when I differentiate it, I get something very complex, and it is very hard to solve it. Is there a simpler way to do this?

InteGrand said:
In order for this to be true, we need g to be of fixed sign (and integrable of course), rather than simply non-zero. (If g is given to be continuous though, then g(x) =/= 0 suffices.)

A proof of that result is given here: https://en.wikipedia.org/wiki/Mean_...rst_Mean_Value_Theorem_for_Definite_Integrals .

Is there any way this proof connects to this question I answered before it, because it says if I generalise the proof below, I will be able to answer the question.

1008 said:
Just another question:

$\\\text{Let f be continuous on [a,b]. Prove there exists c}\in(a,b)\text{ such that}\\\int_{a}^{b}f(t)\text{ }dt = f(c)(b-a).\\\\\text{The way I think I can do it:}\\\text{Consider the function }F(t)\text{ where }F'(t) = f(t)\\\text{As }f(t)\text{ is continuous on [a,b], }F(t)\text{ is differentiable on [a,b] and hence is also continuous.}\\\text{Hence we apply the Mean Value Theorem (MVT) over [a,b]}\\F'(c) =\frac{F(b) - F(a)}{b-a}, c\in(a,b)\\\text{As }F'(t) = f(t),\\f(c)(b-a) = F(b) - F(a).\text{ Now }F(b) - F(a) = \int_{a}^{b}f(t)\text{ }dt\\\therefore\int_{a}^{b}f(t)\text{ }dt=f(c)(b-a)$

Is this right, or am I missing something

InteGrand · Apr 26, 2016

1008 said:
Sorry, but I still don't get the question. Because when I differentiate it, I get something very complex, and it is very hard to solve it. Is there a simpler way to do this?

Is there any way this proof connects to this question I answered before it, because it says if I generalise the proof below, I will be able to answer the question.

You could also have answered the first Q. in a similar way to the Wikipedia proof linked. Then the Wikipedia proof becomes a generalisation of the first proof.

1008 · Apr 26, 2016

InteGrand said:
You could also have answered the first Q. in a similar way to the Wikipedia proof linked. Then the Wikipedia proof becomes a generalisation of the first proof.

Right, thanks. Do you mind (I know this sounds stupid, but) also giving a proof of the question below, because I get the derivative test, but when I differentiate it I get something very complex, so its hard to determine the points of inflexion.

1008 said:
Thanks InteGrand, I get the proof on Wikipedia. But does that connect to my previous proof at all?

I also have another question:
$\\\\\text{Find all the values of }a\text{ and }x,\text{ both in [0,2}\pi],\text{ where}\\f(x) = \cos a+2\cos(2x)+\cos(4x-a)\\\text{has a horizontal point of inflexion.}$

leehuan · Apr 26, 2016

I asked my calculus tutor that one

$f(x)=\cos{\alpha}+2\cos{2\alpha}+cos{(4x-\alpha)}\\f^\prime(x)=-4\sin{2x}-4\sin{(4x-\alpha)}$

$$For the S.P. component, let $f^\prime(x)=0\\ \Rightarrow \sin{2x}=\sin{(4x-\alpha)}\\-\sin{(\pi+2x)}=-\sin{(4x-\alpha)}$

$\pi+2x=4x-\alpha\pm 2n\pi\\ \Leftrightarrow -2x=4x-\alpha\pm(2n+1)\pi $ (from supplementary angles - general solution for sine in both 3rd and 4th quad)$$

$f^{\prime\prime}(x)=-8\cos{2x}-16\cos{(4x-\alpha)}\\$For the P.O.I. component, let $f^{\prime\prime}(x)=0\\ \Rightarrow \cos{(4x-\alpha)}=\frac{1}{2}\cos{2x}$

$$Using the above results, by substituting:$ \\ \cos{\left[2x+(2n+1)\pi\right]}=\frac{1}{2}\cos{2x} \Rightarrow $crumbles down to $\cos{2x}=0$

As does the other case

$$So if $\cos{2x}=0$ we have $x=\frac{\pi}{4}, \frac{3\pi}{4}, \dots$

And then sub them back into the above equations.

Note: One of them does yield the answers at the back of the textbook

And he was way too lazy to test for P.O.I.

1008 · Apr 26, 2016

leehuan said:
I asked my calculus tutor that one

$f(x)=\cos{\alpha}+2\cos{2\alpha}+cos{(4x-\alpha)}\\f^\prime(x)=-4\sin{2x}-4\sin{(4x-\alpha)}$

$$For the S.P. component, let $f^\prime(x)=0\\ \Rightarrow \sin{2x}=\sin{(4x-\alpha)}\\-\sin{(\pi+2x)}=-\sin{(4x-\alpha)}$

$\pi+2x=4x-\alpha\pm 2n\pi\\ \Leftrightarrow -2x=4x-\alpha\pm(2n+1)\pi $ (from supplementary angles - general solution for sine in both 3rd and 4th quad)$$

$f^{\prime\prime}(x)=-8\cos{2x}-16\cos{(4x-\alpha)}\\$For the P.O.I. component, let $f^{\prime\prime}(x)=0\\ \Rightarrow \cos{(4x-\alpha)}=\frac{1}{2}\cos{2x}$

$$Using the above results, by substituting:$ \\ \cos{\left[2x+(2n+1)\pi\right]}=\frac{1}{2}\cos{2x} \Rightarrow $crumbles down to $\cos{2x}=0$

As does the other case

$$So if $\cos{2x}=0$ we have $x=\frac{\pi}{4}, \frac{3\pi}{4}, \dots$

And then sub them back into the above equations.

Note: One of them does yield the answers at the back of the textbook

And he was way too lazy to test for P.O.I.

Thanks

I have another question, though:

$\\\text{Suppose that }f\text{ is a continuous function on }\mathbb{R}\text{ and that }\lim_{x \rightarrow \infty}f(x) =\lim_{x \rightarrow -\infty}f(x)=0 \\\text{(c) Show that if there is a real number }\xi\text{ such that }f(\xi)>0\text{ then }f\text{ attains}\\\text{ a maximum value on }\mathbb{R}.{\\\text{Note that the maximum-minimum theory only applies to }finite\text{ closed intervals [a,b]}}$

1008 · Apr 26, 2016

1008 said:
Thanks I have another question, though:

$\\\text{Suppose that }f\text{ is a continuous function on }\mathbb{R}\text{ and that }\lim_{x \rightarrow \infty}f(x) =\lim_{x \rightarrow -\infty}f(x) =0\\\text{(c) Show that if there is a real number }\xi\text{ such that }f(\xi)>0\text{ then }f\text{ attains}\\\text{ a maximum value on }\mathbb{R}.{\\\text{Note that the maximum-minimum theory only applies to }finite\text{ closed intervals [a,b]}}$

$\\{\text{My approach:}}\\\text{We know f is continsous on }\mathbb{R}\text{ and}\lim_{x \rightarrow \infty}f(x) =\lim_{x \rightarrow -\infty}f(x) =0\\\text{Hence, regardless of smaller values of f(x), as x gets bigger or smaller, it approaches zero.}\\\text{So if }\exists\text{ }\xi\text{ such that }f(\xi)>0\text{ then to remain continuous, f must be close to zero near }-\infty\\\text{and then raise to }\xi\text{ or higher point }\theta\text{ such that }f(\theta)>f(\xi)\text{ and all other positive values of }f(x),\\\text{ and then turn back to 0 as }x\rightarrow\infty.\text{Hence the point }\theta\\\text{ (regardless of if }\theta=\xi)\text{, would be the maximum of }f.$

InteGrand · Apr 26, 2016

1008 said:
Thanks I have another question, though:

$\\\text{Suppose that }f\text{ is a continuous function on }\mathbb{R}\text{ and that }\lim_{x \rightarrow \infty}f(x) =\lim_{x \rightarrow -\infty}f(x)=0 \\\text{(c) Show that if there is a real number }\xi\text{ such that }f(\xi)>0\text{ then }f\text{ attains}\\\text{ a maximum value on }\mathbb{R}.{\\\text{Note that the maximum-minimum theory only applies to }finite\text{ closed intervals [a,b]}}$

I think the same Q got asked and answered here: http://community.boredofstudies.org...271/laters-maths-help-thread.html#post7133125 .

1008 · May 2, 2016

InteGrand said:
I think the same Q got asked and answered here: http://community.boredofstudies.org...271/laters-maths-help-thread.html#post7133125 .

Thanks InteGrand! I had this another question:
$\\\text{Suppose that }f\text{ is a continuous function on the interval [0,1] and that }f(0) = f(1)\\\text{(a) Show that }f(a) = f(a+\frac{1}{2})\text{ for some }a\in[0,\frac{1}{2}].\\\text{Hint: Let }g(x)=f(x+\frac{1}{2})-f(x)\text{ and apply the Intermediate Value Theorem.}\\\text{(b) If }n\text{ is an integer greater than 2, show that: }\\f(a) = f(a+\frac{1}{n})\text{ for some }a\in[0,1-\frac{1}{n}].$

I get the first part, but can't do the second one...

InteGrand · May 2, 2016

1008 said:
Thanks InteGrand! I had this another question:
$\\\text{Suppose that }f\text{ is a continuous function on the interval [0,1] and that }f(0) = f(1)\\\text{(a) Show that }f(a) = f(a+\frac{1}{2})\text{ for some }a\in[0,\frac{1}{2}].\\\text{Hint: Let }g(x)=f(x+\frac{1}{2})-f(x)\text{ and apply the Intermediate Value Theorem.}\\\text{(b) If }n\text{ is an integer greater than 2, show that: }\\f(a) = f(a+\frac{1}{n})\text{ for some }a\in[0,1-\frac{1}{n}].$

I get the first part, but can't do the second one...

I believe this Q. was asked here before, and answered by Sy123: http://community.boredofstudies.org...1437/intermediate-value-theorem-question.html .

Need help, URGENT maths question: (2 Viewers)

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