# Thread: Calculus & Analysis Marathon & Questions

1. ## Re: First Year Uni Calculus Marathon

Originally Posted by porcupinetree
Yup. I'm guessing you guys haven't?
shhh....

We need to give UNSW students the impression that they're ahead....

2. ## Re: First Year Uni Calculus Marathon

Originally Posted by porcupinetree
Yup. I'm guessing you guys haven't?
Taylor series are second semester for us.

3. ## Re: First Year Uni Calculus Marathon

Originally Posted by porcupinetree
$\noindent Find the n-th order Taylor polynomial for f(x) = \frac{1}{1+x} . Hence find the n-th order Taylor polynomial for g(x) = \tan^{-1}x .$
$\frac{1}{1+x} = \frac{1}{1 - (-x)} = \sum_{n=0}^{\infty} (-x)^n\\\frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-x^2)^n\\ \int_{0}^{x} \frac{1}{1+t^2}\,dt = \int_{0}^{x} \sum_{n=0}^{\infty} (-1)^nt^{2n}\,dt\\\arctan{x} = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}$

The nth order Taylor polynomials are the just the infinite series truncated at the nth degree summand.

4. ## Re: First Year Uni Calculus Marathon

Originally Posted by KingOfActing
$\frac{1}{1+x} = \frac{1}{1 - (-x)} = \sum_{n=0}^{\infty} (-x)^n\\\frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-x^2)^n\\ \int_{0}^{x} \frac{1}{1+t^2}\,dt = \int_{0}^{x} \sum_{n=0}^{\infty} (-1)^nt^{2n}\,dt\\\arctan{x} = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}$

The nth order Taylor polynomials are the just the infinite series truncated at the nth degree summand.
Don't forget to state the radius of convergence.

Although in this case, it is fairly obvious.

5. ## Re: First Year Uni Calculus Marathon

$\noindent i) Show that the n-th order Taylor polynomial for \mathrm{e}^{x} about 0 is P_{n} (x) = 1+x+\frac{x^2}{2!}+\cdots + \frac{x^{n}}{n!}.$

$\noindent ii) Using the Lagrange form for the error term (or otherwise), show that the Taylor series 1+x+\frac{x^2}{2!}+\cdots converges to \mathrm{e}^{x} for all real x.$

$\noindent iii) For x<0, show using induction or otherwise that$

$\mathrm{e}^{x} < 1+x+\frac{x^2}{2!}+\cdots + \frac{x^{n}}{n!} if n \geq 0 is even$

$\noindent and$

$\mathrm{e}^{x} > 1+x+\frac{x^2}{2!}+\cdots + \frac{x^{n}}{n!} if n \geq 0 is odd.$

6. ## Re: First Year Uni Calculus Marathon

Originally Posted by InteGrand
$\noindent i) Show that the n-th order Taylor polynomial for \mathrm{e}^{x} about 0 is P_{n} (x) = 1+x+\frac{x^2}{2!}+\cdots + \frac{x^{n}}{n!}.$

$\noindent ii) Using the Lagrange form for the error term (or otherwise), show that the Taylor series 1+x+\frac{x^2}{2!}+\cdots converges to \mathrm{e}^{x} for all real x.$

$\noindent iii) For x<0, show using induction or otherwise that$

$\mathrm{e}^{x} < 1+x+\frac{x^2}{2!}+\cdots + \frac{x^{n}}{n!} if n \geq 0 is even$

$\noindent and$

$\mathrm{e}^{x} > 1+x+\frac{x^2}{2!}+\cdots + \frac{x^{n}}{n!} if n \geq 0 is odd.$
Nice Q's. I think I can do them all in my head but I'll have a proper go tomorrow

7. ## Re: First Year Uni Calculus Marathon

Originally Posted by seanieg89
And some much easier ones for students who don't want to do the above questions:

E1. Prove that if a function f: (a,b) -> R is differentiable and f'(x) is non-negative in this interval, then f is non-decreasing in this interval.

E2. Prove that if a function f: (a,b) -> R is differentiable and f'(x) = 0 in this interval, then f is constant.
Unsolved.

8. ## Re: First Year Uni Calculus Marathon

Originally Posted by InteGrand
$\noindent i) Show that the n-th order Taylor polynomial for \mathrm{e}^{x} about 0 is P_{n} (x) = 1+x+\frac{x^2}{2!}+\cdots + \frac{x^{n}}{n!}.$

$\noindent ii) Using the Lagrange form for the error term (or otherwise), show that the Taylor series 1+x+\frac{x^2}{2!}+\cdots converges to \mathrm{e}^{x} for all real x.$

$\noindent iii) For x<0, show using induction or otherwise that$

$\mathrm{e}^{x} < 1+x+\frac{x^2}{2!}+\cdots + \frac{x^{n}}{n!} if n \geq 0 is even$

$\noindent and$

$\mathrm{e}^{x} > 1+x+\frac{x^2}{2!}+\cdots + \frac{x^{n}}{n!} if n \geq 0 is odd.$
$Let f(x) = e^{x}.$
$Given that f^{(n)} (x) = e^{x}, f^{(n)} (0) = e^{0} = 1.$
$Hence the n-th order Taylor polynomial around x=0 is given by:$
$P_{n} (x) = \frac{1}{0!} x^{0} + \frac{1}{1!} x^{1} + \frac{1}{2!} x^{2} + \frac{1}{3!} x^{3} + \cdots + \frac{1}{n!} x^{n}$
$= 1+x+\frac{x^2}{2!}+\cdots + \frac{x^{n}}{n!} as required.$
$\noindent Using Taylor's theorem, R_{n} (x) = f(x) - P_{n} (x) = \frac{e^{c}}{(n+1)!} x^{n+1} for some c \in (0,x) (assuming that x \geq 0 for now.)$
$\noindent Since the exp function is strictly increasing on the interval (0,x), the maximum value of e^{c} is e^{x}. Hence,$
$R_{n} (x) \leq \frac{e^{x} x^{n+1}}{(n+1)!},$
$\noindent which can be made a small as desired for a given x by choosing n sufficiently large (the proof of which is omitted, given that the result is fairly obvious).$
$\noindent In the case for x < 0, \left| R_{n} (x) \right| \leq \left| \frac{x^{n+1}}{(n+1)!} \right|, which obviously approaches zero for a given x as n approaches infinity.$
$\noindent In other words, for any given x, the error of the Taylor series 1+x+\frac{x^2}{2!}+\cdots converges to 0 as n approaches infinity. Hence, as f(x) = P_{n} (x) + R_{n} (x), the required result follows.$

Don't have time to attempt part (iii) right now, but I might later

9. ## Re: First Year Uni Calculus Marathon

Whilst it is fairly obvious that r^n/n! converges to zero if r is positive (in fact regardless of sign but we can wlog take it positive), I don't think it is a fact that should be glossed over by first year students.

In any case, the methods used to bound such a term can be useful in analysing far more subtle "remainder" terms.

10. ## Re: First Year Uni Calculus Marathon

Originally Posted by seanieg89
(A bit tricky)

For which positive integers k is it possible to find a continuous function f:R->R that such that

f(x)=y

has exactly k solutions x for every real number y?
Also unanswered. I will get people started:

For k=2 it is not possible.

Proof:
Assume f is a cts function taking every value exactly twice.
There must exist two zeros a < b of the function f.
Between a and b f must have fixed sign (otherwise the intermediate value theorem would provide the existence of a third zero).
Without loss of generality, we assume f is positive in (a,b).
Then f must attain a positive maximum m at some p in (a,b) (Extreme value theorem applied to [a,b]).

Now this means that f(x)=2m must have its solutions OUTSIDE the interval [a,b], without loss of generality we may assume that one such solution q is greater than b. Then by the intermediate value theorem, the equation f(x)=m/2 must have:
-at least one solution in (a,p)
-at least one solution in (p,b)
-at least one solution in (b,q).

As for k=3, it IS actually possible.
Picture the square [0,1]x[0,1] and draw line segments between O, (1/3,1), (2/3,0), (1,1). (Looks like a zigzag.)
Now replicate this curve in each square [k,k+1]x[k,k+1].
The resulting curve is the graph of a continuous function with the sought property.

11. ## Re: First Year Uni Calculus Marathon

One can also consider the same problem dropping the condition of surjectivity.

(For which positive integers k is it possible to find a continuous function f defined on R that takes every value in its range exactly k times?)

12. ## Re: First Year Uni Calculus Marathon

Originally Posted by seanieg89
Also unanswered. I will get people started:

For k=2 it is not possible.

Proof:
Assume f is a cts function taking every value exactly twice.
There must exist two zeros a < b of the function f.
Between a and b f must have fixed sign (otherwise the intermediate value theorem would provide the existence of a third zero).
Without loss of generality, we assume f is positive in (a,b).
Then f must attain a positive maximum m at some p in (a,b) (Extreme value theorem applied to [a,b]).

Now this means that f(x)=2m must have its solutions OUTSIDE the interval [a,b], without loss of generality we may assume that one such solution q is greater than b. Then by the intermediate value theorem, the equation f(x)=m/2 must have:
-at least one solution in (a,p)
-at least one solution in (p,b)
-at least one solution in (b,q).

As for k=3, it IS actually possible.
Picture the square [0,1]x[0,1] and draw line segments between O, (1/3,1), (2/3,0), (1,1). (Looks like a zigzag.)
Now replicate this curve in each square [k,k+1]x[k,k+1].
The resulting curve is the graph of a continuous function with the sought property.

I suppose the problem splits up on the parity of k?

13. ## Re: First Year Uni Calculus Marathon

Use the squeeze theorem to find the derivative of ex at x=0

Note you must differentiate from first principles.

14. ## Re: First Year Uni Calculus Marathon

Here is a question related to a basic tool in the analysis I do.

$For positive \lambda let I(\lambda):=\int_\mathbb{R} a(x)e^{i\lambda \phi(x)}\, dx\\ \\where a,\phi are smooth and a is zero outside of a bounded interval [-M,M].\\ \\ Expressions like this come up a lot in analysis, and we often want to understand how this expression behaves as \lambda\rightarrow\infty. The general intuition is that it gets small because increasing \lambda makes the integrand more oscillatory, so it is small at more values of x.\\ \\ 1. Suppose \phi'(x)\neq 0 on [-M,M]. Show that for each n> 0, we can find a positive constant c_n such that |I(\lambda)|\leq c_n\lambda^{-n}. (i.e I(\lambda) decays faster than any polynomial in \lambda.)\\ \\ 2. Let \phi(x)=x^2. Find an asymptotic expansion for I(\lambda).\\ \\ 3. Suppose \phi(x) has exactly one stationary point in [-M,M] and has nonzero second derivative there. Find an asymptotic expransion for I(\lambda). (What if it has a finite number of such stationary points?)$

$Note: An asympotic expansion in this sense is an increasing real sequence \rho_j\rightarrow \infty and a sequence of reals b_j such that \lambda^{\rho_{j+1}}(I(\lambda)-\sum_{j=1}^N b_j \lambda^{-\rho_j}) is bounded for each positive integer N. Think of Taylor series but exponents in such an expansion need not be integers, and the formal series \sum_{j=1}^\infty b_j \lambda^{-\rho_j} need not converge to be a valid asymptotic expansion.$

15. ## Re: First Year Uni Calculus Marathon

Use the squeeze theorem to find the derivative of ex at x=0

Note you must differentiate from first principles.
You kinda need to include the definition of e / exponentiation that you would like students to work from to make a first principles question like this well-defined.

16. ## Re: First Year Uni Calculus Marathon

Originally Posted by seanieg89
You kinda need to include the definition of e / exponentiation that you would like students to work from to make a first principles question like this well-defined.
Alright.

Use the limit definition of ex

$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n \equiv e^x$

17. ## Re: First Year Uni Calculus Marathon

Related:

$Prove that \lim_{n\rightarrow \infty }\left(1+\frac{x}{n}\right)^n and \lim_{n\rightarrow \infty }\left(\sum_{k=0}^n \frac{x^k}{k!}\right) both exist and coincide for all real numbers x.$

18. ## Re: First Year Uni Calculus Marathon

I didn't even get up to the last part of my quadrature question but I feel this last part is a good marathon question so I'll drop it here... I could not get it out though.

$e) Check that e=\lim_{k\rightarrow \infty}{{\left(\frac{2+\sqrt{3+9k^2}}{3k-1}\right)}^k}$

19. ## Re: First Year Uni Calculus Marathon

Good exercise question from a past paper.

$Let f be the function defined by f(x)\int_2^{x^2+x}{\sqrt{1+t^4}dt}$

$\\ a) Find f^\prime (x)\\ b) Find the stationary point of f and determine its nature. \\ c) Find the largest interval I, containing 0, such that f, as a function defined on I, is invertible.\\ d) Find \left(f^{-1}\right)^\prime (0)$

20. ## Re: First Year Uni Calculus Marathon

By the way, would you say that the fact each monotone interval on a function attains at most one zero is a corollary of Rolle's theorem?

Originally Posted by porcupinetree

$Use Rolle's Theorem and the Intermediate Value Theorem to show that the equation \ln x + \frac{1}{x} = 2 has exactly 2 real solutions.$

21. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
By the way, would you say that the fact each monotone interval on a function attains at most one zero is a corollary of Rolle's theorem?
What do you mean? Do you mean if a function is monotone on an interval, it has at most one zero on that interval? If so, that follows from the definition of monotone (if it's monotone, it can't take the same value at two different places).

22. ## Re: First Year Uni Calculus Marathon

Originally Posted by InteGrand
What do you mean? Do you mean if a function is monotone on an interval, it has at most one zero on that interval? If so, that follows from the definition of monotone (if it's monotone, it can't take the same value at two different places).
Oh right. Ok yep basically what I needed to see - just quote one-to-one

23. ## Re: First Year Uni Calculus Marathon

$A series S_n:=\sum_{k=1}^n x_k converges as n\rightarrow \infty iff for every \epsilon >0 there exists a N>0 such that for every m\geq n\geq N we have |S_m-S_n|<\epsilon. (You can prove this if you like, but you don't have to. It just comes from the completeness of the reals.)\\ \\ Using this criterion or otherwise, show that for any \theta, the series \sum_{k=1}^n e^{ik\theta}g(k) converges, where g(k) is a sequence of positive real numbers monotonically decreasing to zero.\\ \\ (The case \theta=\pi is the hopefully familiar alternating series test.)$

24. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
Good exercise question from a past paper.

$Let f be the function defined by f(x)\int_2^{x^2+x}{\sqrt{1+t^4}dt}$

$\\ a) Find f^\prime (x)\\ b) Find the stationary point of f and determine its nature. \\ c) Find the largest interval I, containing 0, such that f, as a function defined on I, is invertible.\\ d) Find \left(f^{-1}\right)^\prime (0)$
$a) Let g(t) = \sqrt{1+t^4} \\ \\ \therefore f(x) = G(x^2+x) - G(2) \\ \\ f'(x) = (2x+1)g(x^2+x) - 0 = (2x+1) \sqrt{1+(x^2+x)^4}$

$b) at \ \ f'(x) = 0 \ \ x = -\frac{1}{2} it's a minimum point since the slope function goes from negative to positive$

$c) \ \ f'(x) > 0 \ \ for \ \ x > -\frac{1}{2} \ \ so f is invertible at that interval$

$d) \ \ \left(f^{-1}\right)^\prime (0) = \frac{1}{f'(f^{-1}(0))} \\ \\ f^{-1}(0) = x \Rightarrow x^2+x = 2 \\ \\ \Rightarrow x = -2,1 \\ \\ But only \ \ x=1 \ \ exists for the given inverse, thus the final answer should be \frac{1}{f'(1)} = \frac{1}{3\sqrt{5}}$

25. ## Re: First Year Uni Calculus Marathon

Originally Posted by integral95
$a) Let g(t) = \sqrt{1+t^4} \\ \\ \therefore f(x) = G(x^2+x) - G(2) \\ \\ f'(x) = (2x+1)g(x^2+x) - 0 = (2x+1) \sqrt{1+(x^2+x)^4}$

$b) at \ \ f'(x) = 0 \ \ x = -\frac{1}{2} it's a minimum point since the slope function goes from negative to positive$

$c) \ \ f'(x) > 0 \ \ for \ \ x > -\frac{1}{2} \ \ so f is invertible at that interval$

$d) \ \ \left(f^{-1}\right)^\prime (0) = \frac{1}{f'(f^{-1}(0))} \\ \\ f^{-1}(0) = x \Rightarrow x^2+x = 2 \\ \\ \Rightarrow x = -2,1 \\ \\ But only \ \ x=1 \ \ exists for the given inverse, thus the final answer should be \frac{1}{f'(1)} = \frac{1}{3\sqrt{5}}$
Yeah basically, but part a) could've been smacked straight away with the first FTC and the chain rule

Hiccup on the final answer for d) though

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