Easiest way to draw function with Euler's Number
What is the easiest way to draw a function like this one?$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I calculated the roots $x=-1/2$, $x=1/2$ and limit which is equal to $0$
I can't just substitute some values like $f(0)approx 0,15$ into the formula and draw it, can I?
graphing-functions
add a comment |
What is the easiest way to draw a function like this one?$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I calculated the roots $x=-1/2$, $x=1/2$ and limit which is equal to $0$
I can't just substitute some values like $f(0)approx 0,15$ into the formula and draw it, can I?
graphing-functions
2
Take it's derivative, that often gives you a picture of how the function changes.
– Don Thousand
2 days ago
1
The easiest way is to ask Wolfram Alpha, but that's probably not what you really want.
– Robert Israel
2 days ago
add a comment |
What is the easiest way to draw a function like this one?$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I calculated the roots $x=-1/2$, $x=1/2$ and limit which is equal to $0$
I can't just substitute some values like $f(0)approx 0,15$ into the formula and draw it, can I?
graphing-functions
What is the easiest way to draw a function like this one?$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I calculated the roots $x=-1/2$, $x=1/2$ and limit which is equal to $0$
I can't just substitute some values like $f(0)approx 0,15$ into the formula and draw it, can I?
graphing-functions
graphing-functions
asked 2 days ago
B. Czostek
566
566
2
Take it's derivative, that often gives you a picture of how the function changes.
– Don Thousand
2 days ago
1
The easiest way is to ask Wolfram Alpha, but that's probably not what you really want.
– Robert Israel
2 days ago
add a comment |
2
Take it's derivative, that often gives you a picture of how the function changes.
– Don Thousand
2 days ago
1
The easiest way is to ask Wolfram Alpha, but that's probably not what you really want.
– Robert Israel
2 days ago
2
2
Take it's derivative, that often gives you a picture of how the function changes.
– Don Thousand
2 days ago
Take it's derivative, that often gives you a picture of how the function changes.
– Don Thousand
2 days ago
1
1
The easiest way is to ask Wolfram Alpha, but that's probably not what you really want.
– Robert Israel
2 days ago
The easiest way is to ask Wolfram Alpha, but that's probably not what you really want.
– Robert Israel
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
A good approach is to see what each factor does. The second factor $1 - 4x^2$ is a parabola opening down, and at $x = 0$ reaches its maximum $1$. The first factor takes everything and makes it smaller, unless $x$ is very close to $0$, in which case it just leaves it mostly unchanged.
So in you put these ideas together: you expect a parabola close to the origin, going to zero as $x$ moves away from $0$
A minimal version of the code to produce the graph above in Python
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-2, 2, num = 200)
y1 = np.exp(-2 * x**2)
y2 = 1 - 4 * x**2
plt.plot(x, y1)
plt.plot(x, y2)
plt.plot(x, y1 * y2)
plt.show()
What tools did you use to make that graph?
– lhf
2 days ago
@lhf Python, I just included the code to create the graph
– caverac
2 days ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060829%2feasiest-way-to-draw-function-with-eulers-number%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
A good approach is to see what each factor does. The second factor $1 - 4x^2$ is a parabola opening down, and at $x = 0$ reaches its maximum $1$. The first factor takes everything and makes it smaller, unless $x$ is very close to $0$, in which case it just leaves it mostly unchanged.
So in you put these ideas together: you expect a parabola close to the origin, going to zero as $x$ moves away from $0$
A minimal version of the code to produce the graph above in Python
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-2, 2, num = 200)
y1 = np.exp(-2 * x**2)
y2 = 1 - 4 * x**2
plt.plot(x, y1)
plt.plot(x, y2)
plt.plot(x, y1 * y2)
plt.show()
What tools did you use to make that graph?
– lhf
2 days ago
@lhf Python, I just included the code to create the graph
– caverac
2 days ago
add a comment |
A good approach is to see what each factor does. The second factor $1 - 4x^2$ is a parabola opening down, and at $x = 0$ reaches its maximum $1$. The first factor takes everything and makes it smaller, unless $x$ is very close to $0$, in which case it just leaves it mostly unchanged.
So in you put these ideas together: you expect a parabola close to the origin, going to zero as $x$ moves away from $0$
A minimal version of the code to produce the graph above in Python
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-2, 2, num = 200)
y1 = np.exp(-2 * x**2)
y2 = 1 - 4 * x**2
plt.plot(x, y1)
plt.plot(x, y2)
plt.plot(x, y1 * y2)
plt.show()
What tools did you use to make that graph?
– lhf
2 days ago
@lhf Python, I just included the code to create the graph
– caverac
2 days ago
add a comment |
A good approach is to see what each factor does. The second factor $1 - 4x^2$ is a parabola opening down, and at $x = 0$ reaches its maximum $1$. The first factor takes everything and makes it smaller, unless $x$ is very close to $0$, in which case it just leaves it mostly unchanged.
So in you put these ideas together: you expect a parabola close to the origin, going to zero as $x$ moves away from $0$
A minimal version of the code to produce the graph above in Python
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-2, 2, num = 200)
y1 = np.exp(-2 * x**2)
y2 = 1 - 4 * x**2
plt.plot(x, y1)
plt.plot(x, y2)
plt.plot(x, y1 * y2)
plt.show()
A good approach is to see what each factor does. The second factor $1 - 4x^2$ is a parabola opening down, and at $x = 0$ reaches its maximum $1$. The first factor takes everything and makes it smaller, unless $x$ is very close to $0$, in which case it just leaves it mostly unchanged.
So in you put these ideas together: you expect a parabola close to the origin, going to zero as $x$ moves away from $0$
A minimal version of the code to produce the graph above in Python
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-2, 2, num = 200)
y1 = np.exp(-2 * x**2)
y2 = 1 - 4 * x**2
plt.plot(x, y1)
plt.plot(x, y2)
plt.plot(x, y1 * y2)
plt.show()
edited 2 days ago
answered 2 days ago
caverac
13.8k21130
13.8k21130
What tools did you use to make that graph?
– lhf
2 days ago
@lhf Python, I just included the code to create the graph
– caverac
2 days ago
add a comment |
What tools did you use to make that graph?
– lhf
2 days ago
@lhf Python, I just included the code to create the graph
– caverac
2 days ago
What tools did you use to make that graph?
– lhf
2 days ago
What tools did you use to make that graph?
– lhf
2 days ago
@lhf Python, I just included the code to create the graph
– caverac
2 days ago
@lhf Python, I just included the code to create the graph
– caverac
2 days ago
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060829%2feasiest-way-to-draw-function-with-eulers-number%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
Take it's derivative, that often gives you a picture of how the function changes.
– Don Thousand
2 days ago
1
The easiest way is to ask Wolfram Alpha, but that's probably not what you really want.
– Robert Israel
2 days ago