# Thread: How to prove the definition of arctangent by G. H. Hardy through integral?

1. ## How to prove the definition of arctangent by G. H. Hardy through integral?

From introduction to analysis,by Arthur P. Mattuck,problem 20-1.
I am stuck in the sub-problem (d) of this problem,especially the magic number 2.5,please help,thanks.

Problems 20-1
One way of rigorously defining the trigonometric functions is to start with the definition of the arctangent function. (This is the route used for example in the classic text Pure Mathematics by G. H. Hardy.)
So, assume amnesia has wiped out the trigonometric functions (but the rest of your knowledge of analysis is intact). Define
$\displaystyle T(x)=\int_{0}^{x}\frac{dt}{1+t^{2}}$

(a) Prove T(x) is defined for all x and odd.
(b) Prove T(x) is continuous and differentiable, and find T(x).
(c) Prove T(x) is strictly increasing for all x; find where it is convex, where
concave, and its points of inflection.
(d) Show T(x) is bounded for all x, and |T(x)| < 2.5, using comparison
of integrals. Can you get a better bound?

2. ## Re: How to prove the definition of arctangent by G. H. Hardy through integral?

(a) Since the integrand is continuous for all t, the integral is defined for all x. I don't know what "and odd" means.

(b) Do you know the "fundamental theorem of Calculus"? I presume you mean "T'(x)", not "T(x)".

(c) Once you have found T'(x), T is "strictly increasing for all x", as long as T'(x)> 0. A function is convex if and only if its second derivative is positive, and concave if and only if its second derivative is negative. Its points of inflection are where the seocnd derivative is 0.

(d) The integrand is clearly less than [tex]\frac{1}{t^2}= t^{-2}[/itex]. So what can you say about the integral?

3. ## Re: How to prove the definition of arctangent by G. H. Hardy through integral?

Originally Posted by HallsofIvy
(a) I don't know what "and odd" means.
I take it he means "an odd function".
Show $\displaystyle T(-x)=-T(x)$.

4. ## Re: How to prove the definition of arctangent by G. H. Hardy through integral?

Originally Posted by HallsofIvy
(a) Since the integrand is continuous for all t, the integral is defined for all x. I don't know what "and odd" means.

(b) Do you know the "fundamental theorem of Calculus"? I presume you mean "T'(x)", not "T(x)".

(c) Once you have found T'(x), T is "strictly increasing for all x", as long as T'(x)> 0. A function is convex if and only if its second derivative is positive, and concave if and only if its second derivative is negative. Its points of inflection are where the seocnd derivative is 0.

(d) The integrand is clearly less than [tex]\frac{1}{t^2}= t^{-2}[/itex]. So what can you say about the integral?
Thanks for reply.
I have worked out (a),(b),(c),but the magic value 2.5 in (d) puzzle me,it is a little bit large.

5. ## Re: How to prove the definition of arctangent by G. H. Hardy through integral?

Originally Posted by Plato
I take it he means "an odd function".
Show $\displaystyle T(-x)=-T(x)$.
Thanks for reply.
I have worked out (a),(b),(c),but the magic value 2.5 in (d) puzzle me,it is a little bit large.

6. ## Re: How to prove the definition of arctangent by G. H. Hardy through integral?

Originally Posted by HallsofIvy
(a) Since the integrand is continuous for all t, the integral is defined for all x. I don't know what "and odd" means.

(b) Do you know the "fundamental theorem of Calculus"? I presume you mean "T'(x)", not "T(x)".

(c) Once you have found T'(x), T is "strictly increasing for all x", as long as T'(x)> 0. A function is convex if and only if its second derivative is positive, and concave if and only if its second derivative is negative. Its points of inflection are where the seocnd derivative is 0.

(d) The integrand is clearly less than [tex]\frac{1}{t^2}= t^{-2}[/itex]. So what can you say about the integral?
Thanks for reply,I finally solve it through help of dr.math(The Math Forum - Ask Dr. Math)

Break the inteval [0,x] to [0,a] + [a,x].

$\displaystyle \int_{0}^{x}\frac{dt}{1+t^{2}}=\int_{0}^{a}\frac{d t}{1+t^{2}}+\int_{a}^{x}\frac{dt}{1+t^{2}}$

For the first inteval [0,a], $\displaystyle \frac{1}{1+t^{2}} <= 1$
For the second inteval [a,x], $\displaystyle \frac{1}{1+t^{2}} <= \frac{1}{t^{2}}$

and change the value of a will get the magic value 2.5

7. ## Re: How to prove the definition of arctangent by G. H. Hardy through integral?

Originally Posted by Plato
I take it he means "an odd function".
Show $\displaystyle T(-x)=-T(x)$.
Thanks for reply,I finally solve it through help of dr.math(The Math Forum - Ask Dr. Math)

Break the inteval [0,x] to [0,a] + [a,x].

$\displaystyle \int_{0}^{x}\frac{dt}{1+t^{2}}=\int_{0}^{a}\frac{d t}{1+t^{2}}+\int_{a}^{x}\frac{dt}{1+t^{2}}$

For the first inteval [0,a], $\displaystyle \frac{1}{1+t^{2}} <= 1$
For the second inteval [a,x], $\displaystyle \frac{1}{1+t^{2}} <= \frac{1}{t^{2}}$

and change the value of a will get the magic value 2.5

8. ## Re: How to prove the definition of arctangent by G. H. Hardy through integral?

I approached part d) as follows:

We want to find a positive value for $\displaystyle a$ such that:

$\displaystyle \frac{a}{(t+1)^2}>\frac{1}{t^2+1}$ for all $\displaystyle t$ in the given domain.

This means we want:

$\displaystyle (a-1)t^2-2t+(a-1)>0$

This means the discriminant must be negative:

$\displaystyle 4-4(a-1)^2<0$

$\displaystyle 0<a(a-2)$

which implies:

$\displaystyle 2<a$. Now, since:

$\displaystyle \lim_{x\to\infty}\left[a\int_0^x\frac{1}{(t+1)^2}\,dt \right]=a$ we may state:

$\displaystyle 0<T(x)<2$.

9. ## Re: How to prove the definition of arctangent by G. H. Hardy through integral?

Originally Posted by MarkFL2
I approached part d) as follows:

We want to find a positive value for $\displaystyle a$ such that:

$\displaystyle \frac{a}{(t+1)^2}>\frac{1}{t^2+1}$ for all $\displaystyle t$ in the given domain.

This means we want:

$\displaystyle (a-1)t^2-2t+(a-1)>0$

This means the discriminant must be negative:

$\displaystyle 4-4(a-1)^2<0$

$\displaystyle 0<a(a-2)$

which implies:

$\displaystyle 2<a$. Now, since:

$\displaystyle \lim_{x\to\infty}\left[a\int_0^x\frac{1}{(t+1)^2}\,dt \right]=a$ we may state:

$\displaystyle 0<T(x)<2$.
Excellent!