Cambridge 2F Q13 and Q 21 (1 Viewer)

Michael12901 · Dec 23, 2023

Pls help with 13d) and e) and 21 b)

kendricklamarlover101 · Dec 23, 2023

Michael12901 said:
Pls help with 13d) and e) and 21 b)
View attachment 41930

View attachment 41931

13b):
$t <= x \implies t^{2N+2} \le x^{2N+2}$
$\text{Since } |t| < 1, 1-t^2 >0$
$\therefore \frac{ t^{2N+2}}{1-t^2} \le \frac{ x^{2N+2}}{1-t^2}$
$\implies \int_{0}^{x} \frac{ t^{2N+2}}{1-t^2} dt \le \int_{0}^{x} \frac{ x^{2N+2}}{1-t^2} dt$
13e):
From a) and b),
$1+t^2 + t^4 +... +t^{2n} + \frac{t^{2N+2}}{1-t^2} = \frac{1}{1-t^2}$
Integrating both sides from 0 to x with respect to t,
$\int_{0}^{x}1+t^2 + t^4 +... +t^{2n} + \frac{t^{2N+2}}{1-t^2} dt = \int_{0}^{x} \frac{1}{1-t^2} dt$
$\int_{0}^{x} \frac{1}{1-t^2} dt =\int_{0}^{x}1+t^2 + t^4 +... +t^{2n} dt + \int_{0}^{x} \frac{t^{2N+2}}{1-t^2} dt$
$\int_{0}^{x} \frac{1}{1-t^2} dt \le \int_{0}^{x}1+t^2 + t^4 +... +t^{2n} dt + \int_{0}^{x} \frac{ x^{2N+2}}{1-t^2} dt$
$\int_{0}^{x} \frac{1}{1-t^2} dt \le x +\frac{x^3}{3} + \frac{x^5}{5} + ... + \frac{x^{2N+1}}{2N+1} +\int_{0}^{x} \frac{ x^{2N+2}}{1-t^2} dt$
$\text{Solving } \int_{0}^{x} \frac{1}{1-t^2} dt, \text{ we get:}$
$\int_{0}^{x} \frac{1}{1-t^2} dt = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$
$\implies \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \le x +\frac{x^3}{3} + \frac{x^5}{5} + ... + \frac{x^{2N+1}}{2N+1} +\int_{0}^{x} \frac{ x^{2N+2}}{1-t^2} dt$
$\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \le x +\frac{x^3}{3} + \frac{x^5}{5} + ... + \frac{x^{2N+1}}{2N+1} +\frac{x^{2N+2}}{2}\ln\left(\frac{1+x}{1-x}\right)$
$\text{From c), } x +\frac{x^3}{3} + \frac{x^5}{5} + ... + \frac{x^{2N+1}}{2N+1} < \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$
$\therefore x +\frac{x^3}{3} + ... + \frac{x^{2N+1}}{2N+1} < \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \le x +\frac{x^3}{3} + ... + \frac{x^{2N+1}}{2N+1} + \frac{x^{2N+2}}{2}\ln\left(\frac{1+x}{1-x}\right)$
$\text{As } x \rightarrow \infty, x^{2N+2} \rightarrow 0$
$\therefore x +\frac{x^3}{3} + ... + \frac{x^{2N+1}}{2N+1} < \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \le x +\frac{x^3}{3} + ... + \frac{x^{2N+1}}{2N+1}$
$\implies x +\frac{x^3}{3} + ... + \frac{x^{2N+1}}{2N+1} = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$
but thats just my attempt im sure there are better ways to do it lol

kendricklamarlover101 · Dec 23, 2023

Michael12901 said:
Pls help with 13d) and e) and 21 b)
View attachment 41930

View attachment 41931

For 21 b) use the fact that e^(-x) is decreasing on [0,1] to get the inequality and the integrand is non negative in that same domain to get,
$0< e\int_{0}^{1} \frac{x^n}{n!}e^{-x} dx < \int_{0}^{1} \frac{x^n}{n!}dx$
the rest is trivial

Michael12901 · Dec 23, 2023

kendricklamarlover101 said:
For 21 b) use the fact that e^(-x) is decreasing on [0,1] to get the inequality and the integrand is non negative in that same domain to get,
$0< e\int_{0}^{1} \frac{x^n}{n!}e^{-x} dx < \int_{0}^{1} \frac{x^n}{n!}dx$
the rest is trivial

kendricklamarlover101 said:
13b):
$t <= x \implies t^{2N+2} \le x^{2N+2}$
$\text{Since } |t| < 1, 1-t^2 >0$
$\therefore \frac{ t^{2N+2}}{1-t^2} \le \frac{ x^{2N+2}}{1-t^2}$
$\implies \int_{0}^{x} \frac{ t^{2N+2}}{1-t^2} dt \le \int_{0}^{x} \frac{ x^{2N+2}}{1-t^2} dt$
13e):
From a) and b),
$1+t^2 + t^4 +... +t^{2n} + \frac{t^{2N+2}}{1-t^2} = \frac{1}{1-t^2}$
Integrating both sides from 0 to x with respect to t,
$\int_{0}^{x}1+t^2 + t^4 +... +t^{2n} + \frac{t^{2N+2}}{1-t^2} dt = \int_{0}^{x} \frac{1}{1-t^2} dt$
$\int_{0}^{x} \frac{1}{1-t^2} dt =\int_{0}^{x}1+t^2 + t^4 +... +t^{2n} dt + \int_{0}^{x} \frac{t^{2N+2}}{1-t^2} dt$
$\int_{0}^{x} \frac{1}{1-t^2} dt \le \int_{0}^{x}1+t^2 + t^4 +... +t^{2n} dt + \int_{0}^{x} \frac{ x^{2N+2}}{1-t^2} dt$
$\int_{0}^{x} \frac{1}{1-t^2} dt \le x +\frac{x^3}{3} + \frac{x^5}{5} + ... + \frac{x^{2N+1}}{2N+1} +\int_{0}^{x} \frac{ x^{2N+2}}{1-t^2} dt$
$\text{Solving } \int_{0}^{x} \frac{1}{1-t^2} dt, \text{ we get:}$
$\int_{0}^{x} \frac{1}{1-t^2} dt = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$
$\implies \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \le x +\frac{x^3}{3} + \frac{x^5}{5} + ... + \frac{x^{2N+1}}{2N+1} +\int_{0}^{x} \frac{ x^{2N+2}}{1-t^2} dt$
$\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \le x +\frac{x^3}{3} + \frac{x^5}{5} + ... + \frac{x^{2N+1}}{2N+1} +\frac{x^{2N+2}}{2}\ln\left(\frac{1+x}{1-x}\right)$
$\text{From c), } x +\frac{x^3}{3} + \frac{x^5}{5} + ... + \frac{x^{2N+1}}{2N+1} < \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$
$\therefore x +\frac{x^3}{3} + ... + \frac{x^{2N+1}}{2N+1} < \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \le x +\frac{x^3}{3} + ... + \frac{x^{2N+1}}{2N+1} + \frac{x^{2N+2}}{2}\ln\left(\frac{1+x}{1-x}\right)$
$\text{As } x \rightarrow \infty, x^{2N+2} \rightarrow 0$
$\therefore x +\frac{x^3}{3} + ... + \frac{x^{2N+1}}{2N+1} < \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \le x +\frac{x^3}{3} + ... + \frac{x^{2N+1}}{2N+1}$
$\implies x +\frac{x^3}{3} + ... + \frac{x^{2N+1}}{2N+1} = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$
but thats just my attempt im sure there are better ways to do it lol

How do u get the first line? x>=t. I understand the rest

kendricklamarlover101 · Dec 23, 2023

Michael12901 said:
How do u get the first line? x>=t. I understand the rest

its an assumption we have to make since otherwise we would be integrating over the asymptote, which will lead to a divergent integral.

kendricklamarlover101 · Dec 23, 2023

kendricklamarlover101 said:
its an assumption we have to make since otherwise we would be integrating over the asymptote, which will lead to a divergent integral.

wait no lemme think this over this is wrong

Michael12901 · Dec 23, 2023

kendricklamarlover101 said:
For 21 b) use the fact that e^(-x) is decreasing on [0,1] to get the inequality and the integrand is non negative in that same domain to get,
$0< e\int_{0}^{1} \frac{x^n}{n!}e^{-x} dx < \int_{0}^{1} \frac{x^n}{n!}dx$
the rest is trivial

Also how do u deal with the e at the front of the expression in the middle? Because presumably u need to multiply the whole inequality by e as well.

Thanks btw ur help v much appreciated.

kendricklamarlover101 · Dec 23, 2023

Michael12901 said:
Also how do u deal with the e at the front of the expression in the middle? Because presumably u need to multiply the whole inequality by e as well.

Thanks btw ur help v much appreciated.

that middle term is simply just e - S_n

for the other question im not too sure theres probably some deeper geometric interpretation behind it but i havent encountered enough functions defined as integrals to grasp it fully.

Michael12901 · Dec 23, 2023

kendricklamarlover101 said:
that middle term is simply just e - S_n

for the other question im not too sure theres probably some deeper geometric interpretation behind it but i havent encountered enough functions defined as integrals to grasp it fully.

Yeah but the integral in the middle (the one with e^(-x) has an e in front of the whole integral which u haven't accounted for in the rightmost integral (in the inequality). Cos presumably that e makes the middle integral larger and I don't understand how it can be smaller than the rightmost one which doesn't have an e multiplied to it

kendricklamarlover101 · Dec 23, 2023

Michael12901 said:
Yeah but the integral in the middle (the one with e^(-x) has an e in front of the whole integral which u haven't accounted for in the rightmost integral (in the inequality). Cos presumably that e makes the middle integral larger and I don't understand how it can be smaller than the rightmost one which doesn't have an e multiplied to it

in the full working out i would have,
$e \int_{0}^{1} \frac{x^n}{n!} e^{-x} dx < e \int_{0}^{1} \frac{x^n}{n!} dx < 3\int_{0}^{1} \frac{x^n}{n!} dx$
as e < 3
sorry for any confusion i didnt actually do this question i kinda just glanced at it lol

Michael12901 · Dec 23, 2023

kendricklamarlover101 said:
in the full working out i would have,
$e \int_{0}^{1} \frac{x^n}{n!} e^{-x} dx < e \int_{0}^{1} \frac{x^n}{n!} dx < 3\int_{0}^{1} \frac{x^n}{n!} dx$
as e < 3
sorry for any confusion i didnt actually do this question i kinda just glanced at it lol

Ah yes ty so much

Also for the x>= t matter, could it be because in part c) we're integrating a function in terms of t over the bounds 0 to x. Meaning that 0<= t <= x?? I've seen that logic used in some questions.

Luukas.2 · Dec 24, 2023

Michael12901 said:
Ah yes ty so much
Also for the x>= t matter, could it be because in part c) we're integrating a function in terms of t over the bounds 0 to x. Meaning that 0<= t <= x?? I've seen that logic used in some questions.

Yes. If I integrate from (say) $x = 0$ to $x = 1$ then I know that $0 \le x \le 1$ .

The same logic applies to integrating from $t = 0$ to $t = x$ , it follows that $0 \le t \le x$ or $t \in [0,\,x]$ .

Cambridge 2F Q13 and Q 21 (1 Viewer)

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