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How can I best explain that the approximation of these two series is valid?

$$ e^{-t} \cos(2t) \approx 1-t $$

The test should be made by multiplying the series.

I looked at the series and am now wondering if the approximation is valid and how best to explain the behaviour of the multiplied curve (in red).

Ideally I would like to make the proof with series expansion.

$$ e^{-t} \cos(2t) (1-t) \approx e^{-t} \cos(2t) $$

enter image description here

The code for anyone interested:

#!/usr/bin/python3

import numpy as np
from numpy import sin, cos, tan, pi, sqrt, exp, linspace
import matplotlib.pyplot as plt

list1 = []
e = exp(1)
calcrange = 20
resolution = np.linspace(0, calcrange, 200)

for x in np.nditer(resolution):
    list1.append([x, e**(-x) * cos(2*x)])

list1np = np.array(list1)

list2 = []

for x in np.nditer(resolution):
    list2.append([x, 1-x])

list2np = np.array(list2)

list3 = []

for x in np.nditer(resolution):
    list3.append([x, (1-x) * e**(-x) * cos(2*x)])

list3np = np.array(list3)

axes = plt.subplot(111)
axes.set_xlim(0, calcrange)
axes.set_ylim(-2, 2)

plot1 = plt.plot(list1np[:,0], list1np[:,1], label="$e^{-x} \cos(2x)$")
plot2 = plt.plot(list2np[:,0], list2np[:,1], label="$1-x$")
plot3 = plt.plot(list3np[:,0], list3np[:,1], label="$(1-x)e^{-x} \cos(2x)$")

plt.legend(prop={'size':24})

plt.show()
tobias47n9e
  • 177
  • What I saw was $e^t * cos(2t)$. There was no minus sign. I changed it to $e^t \cos(2t)$. ${}\qquad{}$ – Michael Hardy Oct 04 '14 at 16:11
  • Note that you can write $3\times5$ or $3\cdot5$. Using an asterisk to represent ordinary multiplication is for occasions when you're restricted to the symbols on the keyboard. – Michael Hardy Oct 04 '14 at 16:32
  • I made the formula in the picture in TeX now too ;) – tobias47n9e Oct 04 '14 at 16:50
  • But you wrote $cos$ instead of $\cos$. That is not correct usage. The backslash not only prevent italicization, but also results in proper spacing in things like $a\cos b$. ${}\qquad{}$ – Michael Hardy Oct 04 '14 at 17:40

1 Answers1

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Sorry but here something is definitely wrong, since $$ e^{−t}\cos(2t)≈1−t $$ means that the functions on both sides show approximately the same behavior for some limit you don't but I assume that it is infinity. But this is wrong!

Please note inclusion of some program code means that anybody who wants to help you needs to have the same package and will run it for verifying your claims. You should be able to write it out what you want in clear way without recursion to some code elsewhere we are near $42$.

I guess, with my crystalball, that you want to say that for $t\to\infty$: $$e^{−t}\cos(2t)(1-t) \approx e^{−t}\cos(2t)$$ This is true you can just use L'Hopital's rule. If you don't know it, wikipedia is your friend.

Karl
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