Defining Trig Funcs: Solving Problem 20-1, Part d

[zhongbeyond]

Homework Statement

From introduction to analysis,by Arthur P. Mattuck,problem 20-1.

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
$$
T(x)=\int_{0}^{x}\frac{dt}{1+t^{2}}
$$

(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?

Homework Equations

The Attempt at a Solution

I am stuck in the sub-problem (d) of this problem,especially the magic number 2.5,please help,thanks.

 

[Ray Vickson]

Homework Statement

From introduction to analysis,by Arthur P. Mattuck,problem 20-1.

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
$$
T(x)=\int_{0}^{x}\frac{dt}{1+t^{2}}
$$

(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?

Homework Equations

The Attempt at a Solution

I am stuck in the sub-problem (d) of this problem,especially the magic number 2.5,please help,thanks.

I don't know where the 2.5 arises; I get instead 1.875, by using 1/(1+t^2) < 1/t^2 for t >= 1 and by choosing an appropriate, simple g(t) such that 1/(1+t^2) <= g(t) on [0,1].

RGV

 

[zhongbeyond]

I don't know where the 2.5 arises; I get instead 1.875, by using 1/(1+t^2) < 1/t^2 for t >= 1 and by choosing an appropriate, simple g(t) such that 1/(1+t^2) <= g(t) on [0,1].

RGV

Thanks for replying,I finally solve it through help of dr.math(http://mathforum.org/dr.math/)

Break the inteval [0,x] to [0,a] + [a,x].
$$
T(x)=\int_{0}^{x}\frac{dt}{1+t^{2}}=
\int_{0}^{a}\frac{dt}{1+t^{2}}+\int_{a}^{x}\frac{dt}{1+t^{2}}
$$

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

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

 

[Ray Vickson]

Thanks for replying,I finally solve it through help of dr.math(http://mathforum.org/dr.math/)

Break the inteval [0,x] to [0,a] + [a,x].
$$
T(x)=\int_{0}^{x}\frac{dt}{1+t^{2}}=
\int_{0}^{a}\frac{dt}{1+t^{2}}+\int_{a}^{x}\frac{dt}{1+t^{2}}
$$

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

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

OK, but the way I got 1.875 was to use 1/(1+t^2) <= g(t) on [0,1], where g(t) = 1 for 0 <= t <= 1/2 and g(t) = 3/2 - t for 1/2 < t <= 1 (and 1/t^2 for t > 1) The integral of g(t) over [0,1] is 7/8.

RGV