Calculus & Analysis Marathon & Questions (2 Viewers)

seanieg89

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Re: First Year Uni Calculus Marathon

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.
 

seanieg89

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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.
 

seanieg89

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Re: First Year Uni Calculus Marathon

(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).

This contradiction completes the proof.


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.

What about for higher k?
 

seanieg89

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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?)
 

Paradoxica

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Re: First Year Uni Calculus Marathon

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).

This contradiction completes the proof.


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.

What about for higher k?
I suppose the problem splits up on the parity of k?
 

Paradoxica

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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.
 

seanieg89

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Re: First Year Uni Calculus Marathon

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



 
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seanieg89

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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.
 

Paradoxica

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Re: First Year Uni Calculus Marathon

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

 

seanieg89

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Re: First Year Uni Calculus Marathon

Related:

 

leehuan

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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.

 

leehuan

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Re: First Year Uni Calculus Marathon

Good exercise question from a past paper.



 

InteGrand

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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?
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).
 

leehuan

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Re: First Year Uni Calculus Marathon

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
 

leehuan

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Re: First Year Uni Calculus Marathon

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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