Question

Fun Calculus quesiton:
Suppose f(1)=f'(1)=0, f'' is continuous on [0,1] and |f''(x)|<=3 for all x. W.T.S \[ | \int_0^1f(x) dx| \le \frac{1}{2} \]
This is a really fun question. I could give hints if needed to those who have interest.

Answer 1

no @ganeshie8 this is too easy for you
same for you sith

Answer 2

But there are no other contenders I guess

Answer 3

Better get to tagging :P

Answer 4

Hint: Mr. Taylor
Or that is the way I went about it

Answer 5

I will show my way. If anyone else wants to show there way they are welcome to it.
\[f(x) \approx f(1)+f'(1)(x-1)+\frac{f''(1)}{2}(x-1)^2 =\frac{f''(1)}{2}(x-1)^2 \]

\[\int\limits_{0}^{1} f(x) dx \approx \int\limits_{0}^{1}\frac{f''(1)}{2}(x-1)^2 dx =\frac{f''(1)}{2} \frac{(x-1)^3}{3}|_0^1 \\ =\frac{f''(1) (x-1)^3}{6}|_0^1 =\frac{f''(1)}{6}[(1-1)^3-((0-1)^3] \\ \frac{f''(1)}{6}\]
But we have -3<=f''(1)<=3
so this means -3/6 <=f''(1)/6<=3/6
which means |f''(1)/6|<=1/2

Answer 6

And so \[|\int\limits_{0}^{1}f(x) dx | <\frac{1}{2} \]

Answer 7

that is suppose to be a less than or equal to

Answer 8

@ganeshie8 If you want to look at a harder 1 that I haven't attempted yet that looks spicy.
Here it is:
\[\text{ If } 0<a<b ,\text{ find } \lim_{t \rightarrow 0} [\int\limits_{0}^{1}(bx+a(1-x))^t dx ]^\frac{1}{t}\]

Answer 9

well i say it is hard because I'm still thinking where to start

Answer 10

I was thinking about since we have that one exponent we might try rewriting the limit like this:
\[e^{ \lim_{t \rightarrow 0} \frac{1}{t} \int\limits_{0}^{1}(bx+a(1-x))^t dx }\]

Answer 11

on no forgot to write ln( )

Answer 12

\[e^{ \lim_{t \rightarrow 0} \frac{1}{t} \ln( \int\limits\limits_{0}^{1}(bx+a(1-x))^t dx ) }\]

Answer 13

actually it may not be that bad

Answer 14

I am thinking of L'Hospital's rule if the exponent can be shown to be undeterminate form

Answer 15

\[\int\limits_{0}^{1}(bx+a(1-x))^t dx = \frac{(bx+a(1-x))^{t+1}}{t+1}|_0^1=\frac{b^{t+1}}{t+1} -\frac{a^{t+1}}{t+1}=\frac{b^{t+1}-a^{t+1}}{t+1}\]

So we have \[\lim_{t \rightarrow 0}(\frac{b^{t+1}-a^{t+1}}{t+1})^\frac{1}{t}=\]
no wait still thinking

Answer 16

I might have actually missed a constant multiple in my antideviative

Answer 17

*yeah 1/(b-a)

Answer 18

\[\int\limits\limits_{0}^{1}(bx+a(1-x))^t dx = \frac{1}{b-a}\frac{(bx+a(1-x))^{t+1}}{t+1}|_0^1= \frac{1}{b-a} ( \frac{b^{t+1}}{t+1} -\frac{a^{t+1}}{t+1} )= \\ \frac{1}{b-a}\frac{b^{t+1}-a^{t+1}}{t+1}\]

Answer 19

\[\lim_{t \rightarrow 0}( \frac{1}{b-a}\frac{b^{t+1}-a^{t+1}}{t+1})^\frac{1}{t}=\]

Answer 20

there all should be fixed now

Answer 21

f(x)=x^(t+1)

\[\lim_{t \rightarrow 0}(\frac{1}{t+1} \cdot \frac{f(b)-f(a)}{b-a})^\frac{1}{t}\]
Trying to see what use is that given 0<a<b

Answer 22

ln is missing

Answer 23

well that is we write it like e^ln(y)

Answer 24

Ahh I see now :) looks perfect !!!

Answer 25

Since Mr.Taylor gave beautiful solution in previous problem we may try the same trick here and hope something gets simplified :)

f(b) = f(0) + f'(0)b + f''(0)*b^2/2 + ...

Answer 26

wait are you using what I called f(x)?

Answer 27

f(t) = (b^t+1 ) - (a^t+1)

f(t) = (b + blogb*t + blog^2b*t^2/2 + ....) - (a+ aloga*t + alog^2a*t^2/2+...)

Answer 28

\[\begin{align} \\ \lim_{t \rightarrow 0}\left( \frac{1}{b-a}\frac{b^{t+1}-a^{t+1}}{t+1}\right)^\frac{1}{t} &= \lim_{t \rightarrow 0}\left( \frac{1}{b-a}\frac{b-a+(b\ln b - a\ln a)t + \mathcal{O}(t^2)}{t+1}\right)^\frac{1}{t} \end{align}\]

Answer 29

I observe a pattern using wolfram
somehow that is eventually suppose to give us b^b/(a^a*e)

Answer 30

\[\frac{b^b}{a^a e} \]

Answer 31

\[\large \begin{align} \\ &= \dfrac{\lim\limits_{t \rightarrow 0}\left( 1+\frac{+(b\ln b - a\ln a)}{b-a}t\right)^{1/t}}{\lim\limits_{t \rightarrow 0}(1+t)^{1/t}} \end{align}\]

Answer 32

\[\large \begin{align} \\ &= \dfrac{e^{\left( \frac{b\ln b - a\ln a}{b-a}\right)}}{e} \end{align}\]

Answer 33

I found another much better solution, will post shortly :)

Answer 34

coolness
you have more math magic than me

Answer 35

no lol, i cheated... googled ;)

Answer 36

omg you cheater! :p

Answer 37

i haven't reviewed all these cool limit formulas in a while so i don't remember a thing except for computing them using taylor/lhops haha!

the other solution i am having is really nice, wil post after breakfast...

Answer 38

take your time
i look forward to seeing it

Answer 39

I hope you like doing these problems as much a me

Answer 40

\[e^{\ln(\lim_{t \rightarrow 0}(\frac{1}{b-a} \cdot \frac{b^{t+1}-a^{t+1}}{t+1})^{\frac{1}{t}} )}\]
We will work with the exponent for now:
\[\lim_{t \rightarrow 0}\frac{\ln(\frac{b^{t+1}-a^{t+1}}{(b-a)(t+1)})}{t}\]
we have 0/0
differentiating:
\[\lim_{t \rightarrow 0} \frac{\frac{(\ln(b)b^{t+1}-\ln(a)a^{t+1})(b-a)(t+1)-(b^{t+1}-a^{t+1})(b-a)}{(b-a)^2(t+1)^2} }{\frac{b^{t+1}-a^{t+1}}{(b-a)(t+1)}}\]

\[\lim_{t \rightarrow 0} \frac{\ln(b)b^{t+1}-\ln(a)a^{t+1}-\frac{b^{t+1}-a^{t+1}}{t+1}}{b^{t+1}-a^{t+1}}\]

\[\lim_{t \rightarrow 0}(\frac{\ln(b)b^{t+1}-\ln(a)a^{t+1}}{b^{t+1}-a^{t+1}}-\frac{1}{t+1})\]

Function is continuous at t=0 so we have
\[\frac{\ln(b)b-\ln(a)a}{b-a}-\frac{1}{1}=\frac{\ln(b^b)-\ln(a^a)}{b-a}-1\]

\[\text{ so we have } e^{\ln(\frac{b^b}{a^a}) \frac{1}{b-a}\cdot -1 }\]
\[=e^{\ln(\frac{b^b}{a^a}) \cdot \frac{1}{b-a}-\ln(e)}\]
\[=e^{\ln(\frac{b^b}{a^a}) \frac{1}{b-a} } \cdot e^{\ln(e^{-1})} \\ =e^{\ln((\frac{b^b}{a^a})^{^{\frac{1}{b-a}}})}e^{-1}\]


\[(\frac{b^b}{a^a})^{^\frac{1}{b-a}}e^{-1}\]

Answer 41

Wow! L'Hopital's wasn't that scary ! it looks great xD

here is another method that i was studying in the morning :

\[\begin{align}e^{ \lim\limits_{t \rightarrow 0} \frac{1}{t} \ln( \int\limits\limits_{0}^{1}(bx+a(1-x))^t dx ) }& =\lim\limits_{t \rightarrow 0} \left(\int\limits\limits_{0}^{1}(bx+a(1-x))^t dx\right )^{1/t} \\~\\&=\lim\limits_{t \rightarrow 0} \left(1+t\int\limits\limits_{0}^{1}\dfrac{(bx+a(1-x))^t -1}{t}dx\right )^{1/t} \color{Red}{*} \\~\\ &=e^{\int\limits_0^1 \ln\left(bx+a(1-x)\right)dx} \\~\\&=e^{\frac{1}{b-a}\left(b\ln b-a\ln a\right)-1}\end{align}\]

Answer 42

\(\color{Red}{*}\) :- Since \(\large b\gt a\), the expression \(bx+a(1-x)\gt 0\) in the interval \((0,1)\)
using the standard limits, (1+xt)^t = e^x and (a^t-1)/t = lna leads to next step

Answer 43

It looks good I think. I'm just trying to soak up that red step, like how we got from previous step to the red step.

Answer 44

+1 - 1


then multiply by t/t

Answer 45

and another hidden thing : \(\large \int_0^11 dx= 1 \)

Answer 46

Ok I think I see that now
That is kinda cool and way shorter than l'hopital

Answer 47

And l'hopital was scary but I decided to accept my fears and move on

Answer 48

L'hopital's worked nicely too, just a bit of extra algebra !
My first plan was something like using L'hopital+FTC to avoid computing the integral,
but that didn't materialize because the bounds of integral are numbers and not a function of "t".
So had to appeal to taylor :D I liked the taylor/lhops solutions more as they don't require us to memorize a lot many standard limits

Answer 49

Your first was my first thingy
But then I was like well we still may be able to use l'hosptal but I won't like it :p