Your calculator is LIAR. (1 Viewer)

Uncle

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Exhibit A:

What you see:

ex

What your calculator really has:

[maths]e^{x} = 1 + x + \frac{1}{2!}x^{2}+\frac{1}{3!}x^{3}[/maths]
[maths]+\frac{1}{4!}x^{4}+\frac{1}{5!}x^{5}+\frac{1}{6!}x^{6}[/maths]
[maths]+\frac{1}{7!}x^{7}+\frac{1}{8!}x^{8}+\frac{1}{9!}x^{9}[/maths]
[maths]+\frac{1}{10!}x^{10} + \frac{1}{11!}x^{11} + \frac{1}{12!}x^{12}[/maths]
[maths]+\frac{1}{13!}x^{13}+\frac{1}{14!}x^{14}+\frac{1}{15!}x^{15}[/maths]
[maths]+\frac{1}{16!}x^{16}+\frac{1}{17!}x^{17}+\frac{1}{18!}x^{18}[/maths]
[maths]+\frac{1}{19!}x^{19}+\frac{1}{20!}x^{20} + \frac{1}{21!}x^{21} [/maths]
[maths]+ \frac{1}{22!}x^{22} + \frac{1}{23!}x^{23} + \frac{1}{24!}x^{24}[/maths]
[maths]+ \frac{1}{25!}x^{25} + \frac{1}{26!}x^{26} + \frac{1}{27!}x^{27}[/maths]
[maths]+ \frac{1}{28!}x^{28} + \frac{1}{29!}x^{29} + \frac{1}{30!}x^{30} + ...[/maths]

The calculator uses a Taylor Polynomial, it requires around a degree of 30 to maintain accuracy to 12 decimal places.


You probably can't enter a 30th degree polynomial into your calculator but why don't you try entering "what your calculator says" and an actual Taylor polynomial representation of degree 6 with some reasonable accuracy:

Enter e2 on your calculator:

Answer:
7.389056099

Enter

[maths]1 + 2 + \frac{1}{2!}(2)^{2}+\frac{1}{3!}(2)^{3}[/maths]
[maths]+\frac{1}{4!}(2)^{4}+\frac{1}{5!}(2)^{5}+\frac{1}{6!}(2)^{6}[/maths]

on your calculator:

Answer:
7.355555556



Sketches of:

ex in red.
[maths]1 + x + \frac{1}{2!}x^{2}+\frac{1}{3!}x^{3}+\frac{1}{4!}x^{4}+\frac{1}{5!}x^{5}+\frac{1}{6!}x^{6}[/maths] in blue.



The exponential function's 6th degree Taylor polynomial representation is almost perfect for a small value of x (at around x = 2).


The moral of the story is, your calculator doesn't know the elementary functions sin x, cos x, log x, it uses Taylor polynomial approximations of very high degree to maintain accuracy on its little 12 digit LCD screen and you probably didn't even know it until you read this thread which is one of the most glorious ones of all time.


EDIT:

For those in first year physics the formula for the length change due to temperature is given by the formula:

L = L0[1 + α(T - T0)]

However it originally came from:

L = L0eα(T - T0)

To make life easier, the approximation ex ≈ 1 + x is only valid if x is much much less than 1.
This is allowed because the coefficient of thermal expansion/contraction (α) for many materials is as small as 10-6.
Also you may encounter problems maybe because, say a bar is moved from your warm bedroom to the cold winter night and the temperature change is very small, compared to shoving, say, simonloo or a metal bar in a high speed furnace from 200C to 55000C like the ones in industrial plants.

Substitute the values and therefore:

L = L0[1 + α(T - T0)]
 
Last edited:

clintmyster

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LOL so this is what you do in your spare time.
 

loller

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not that wow hes just a freak with numerals

hope it puts all you azns and shit in your place
 

Trebla

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This what you learn in uni maths...when you actually cover Taylor polynomials. Anyone who's gone through first year maths would know this anyway. A hand held calculator is sufficient for most general purposes. If you want a more accurate tool, use computer software...
 

cutemouse

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lol, I'm thinking of setting ppl's calculators to round to the nearest integer, and setting the mode to radians for 2U or Physics candidatures before their trial exam.

"Hey mate, can I borrow your calculator just to check something...?"

xD wonder what would happen.
 

annabackwards

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Woah oO

lol, I'm thinking of setting ppl's calculators to round to the nearest integer, and setting the mode to radians for 2U or Physics candidatures before their trial exam.

"Hey mate, can I borrow your calculator just to check something...?"

xD wonder what would happen.
I like the way you think :rofl:
 

Studentleader

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So would typing in

((1+1/n)^n)^X be accurater than e^x on your calculator as

e^1 = lim(x->inf) (1+1/x)^x

(given n is 9 million or something stupidly large)
 

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