Question

e, polynomials, and e^x.

Answer 1

i) Use the fact that \[f(x)=1\le(e^x) \]for \[x \ge 0\] (why is this true?)
to show that \[\int\limits_{0}^{x}f(u)du \le \int\limits_{0}^{x}(e^u)du\]

Answer 2

ii) Show that \[1+x \le e^x\] (for x is larger than 0)
iii) use the same process to show \[1+x+\frac{ x^2 }{ 2 }\le e^x\] for x is bigger than 0
iv) repeat twice more

Answer 3

Yeah, lots of parts haha. Starting from the beginning! :)

Answer 4

@jim_thompson5910 @zepdrix I suck at anything with "e^x" o.o

Answer 5

https://www.desmos.com/calculator plot y = e^x and you'll see

Answer 6

mm..well yes. But I need to "show it" which I guess means write out a proof or a statement for it.

Answer 7

1<e implies 1^x<e^x when x>0

Answer 8

It has to do with how e^x is set up as an infinite polynomial

\[\Large e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!}\]

\[\Large e^x = \frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\]

\[\Large e^x = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\ldots\]

Answer 9

If x > 0, then e^x will be larger than 1 because of the added terms (x, x^2/2, x^3/6, etc) being added to 1

if x > 0 then e^x will be larger than 1+x for similar reasons

Answer 10

So in comparison to
the integral of f(u), e^x is larger because..of the added terms? f(u) doesn't have that?

Answer 11

well you have f(x) = e^x, so f(u) = e^u. They look the same to me, just with a different variable

Answer 12

the question is to show that
f(u)du < [e^u]du

Answer 13

(part i)

Answer 14

that doesn't make any sense though because f(u) = e^u

I don't see how you can make an expression larger than itself

Answer 15

hm yeah o_o
also
f(x) = 1< e^x if that helps..

Answer 16

oh so f(x) = 1?

Answer 17

Yep yep. Look at the first comment (mine) under the question.

Answer 18

ok so basically the function f(x) = 1 is a flat line while e^x is always growing

so naturally the area under the f(x) = 1 line will be a rectangle, the area under e^x will be larger because e^x is increasing (so we're adding on more and more area)

Answer 19

that middle thing is not an equal sign

Answer 20

if f(x) = 1 then f(u) = 1

the left hand side will be the integral of 1 from 0 to x

essentially, a rectangle with base from 0 to x (x units long) and height of 1

area = base*height = x*1 = x

Answer 21

the right hand side will be the integral of e^u from 0 to x

e^u is a nice integral because the result is itself

so we will have e^x - e^0 = e^x - 1 as the area under the curve

Answer 22

for the integral e^u?

Answer 23

yes,

\[\Large \int e^u du = e^u + C\]

\[\Large \int_{0}^{x}e^u du = e^x - e^0 = e^x - 1\]

Answer 24

How does that make it larger than f(u)?

Answer 25

\[\Large \int_{0}^{x}f(u) du \le \int_{0}^{x}e^u du\]

\[\Large \int_{0}^{x}1 du \le \int_{0}^{x}e^u du\]

\[\Large x \le e^x - 1\]

\[\Large 1+x \le e^x\]

Answer 26

heeerm. Interesting o.o

Answer 27

\[\Large 1+x \le e^x\]

\[\Large 1+x \le 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\ldots\]

\[\Large 0 \le \frac{x^2}{2}+\frac{x^3}{6}+\ldots\]

Answer 28

where x >= 0

Answer 29

I see, I hadn't realized e^x made such large numbers.

Answer 30

So I guess that answers part i

Answer 31

keep in mind that e = 2.718 approx

Answer 32

Right, I keep forgetting. So by just looking at it it didn't look "correct"

Answer 33

ok so..I guess in a way that answers both parts i and ii. they sort of go hand in hand.

Answer 34

second could be proved using mean value theorem

Answer 35

e^x approx = 1+x from linear approximation

Answer 36

make a function g(x)=e^x-(1+x)
to use MVT

Answer 37

that's if you don't be that fancy with series expansion

Answer 38

hm. Thinking about it. just a moment.

Answer 39

you know MVT correct

Answer 40

just think about it for some time

Answer 41

@jim_thompson5910 how come you made e^x = 1+x up there (about..3 comments of yours ago xD)

Answer 42

Yes, I know MVT :) just not sure how to apply it here o.o

Answer 43

you mean why did I replace e^x with 1+x+x^2/2+... ?

Answer 44

Yeah

Answer 45

where you have
1+x< e^x
1+x<1+x+ (x^2/2)....

Answer 46

I'm using the idea

\[\Large e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!}\]

\[\Large e^x = \frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\]

\[\Large e^x = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\ldots\]

Answer 47

hm and then the 1+x cancel out on both sides which is why you get
0< (x^2/2) ...

Answer 48

yes

Answer 49

So part iii could be solved the same way? -.- haha

Answer 50

yes there's just that extra term

Answer 51

you have already proved e^x>x+1
extra term x^2/2

Answer 52

\[0<\frac{ x^3 }{ 6 }\] would be enough then..

Answer 53

and for the
iv) repeat twice more
\[0\le \frac{ x^5 }{ 24 }\]\[0\le \frac{ x^5 }{ 120 }\]

Answer 54

I think you meant to say (x^4)/24

Answer 55

but yeah it looks good

Answer 56

Lol, yeah :)
k, so also it asks me to make a graph of the polynomial found at the end of part iv
How would I do that?

Answer 57

at that point I would have the x^5/120 but i'm not sure if the prof wants..more than that?

Answer 58

"graph your polynomial (found at the end of step iv) on the same axis as a graph of
e^x for 0<x<1.5"

Answer 59

The idea is that e^x can be approximated by a polynomial

y = 1 is a horrible approximation, but it's decent around x = 0
y = 1+x is a better approximation, still not that great though (we're still around x = 0)
y = 1+x+(x^2)/2 is better than the second
y = 1+x+(x^2)/2+(x^3)/6 is better than the previous polynomial

and so on

if you keep going forever, then you'll get e^x exactly

if you stop anywhere, then you'll still have an approximation (just more accurate for more and more terms)

Answer 60

so what I think your teacher wants is you to graph

y = 1+x+(x^2)/2+(x^3)/6+(x^4)/24 + (x^5)/120

which will serve as an approximate to y = e^x

Answer 61

compare the graphs

https://www.desmos.com/calculator/qw3a9lwary

notice how the purple graph is fairly close to the red curve for x > 0

Answer 62

Yes!! This is a pretty cool problem :D So I guess the more we were to do this the closer it would be to e^x

Answer 63

if you move the window around a bit

https://www.desmos.com/calculator/ulumqu151q

you'll see that the two curves don't match up very well for larger x values. To get them to match better, you'll need more terms

Answer 64

And then if x < 0 it will never match up. Ahhh

Answer 65

well to be technical, it works for some negative values

it's basically a neighborhood around x = 0 is where the approximating polynomial is good to match up with e^x

Answer 66

Right right. Technicalities:P

Answer 67

Thank you so much!!

Answer 68

you're welcome