# exponent

问题 In Python, scientific notation always gives me 2 digits in exponent: print('%17.8E\n' % 0.0665745511651039) 6.65745512E-02 However, I badly want to have 3 digits like: 6.65745512E-002 Can we do this with a built-in configuration/function in Python? I know my question is basically the same question as: Python - number of digits in exponent, but this question was asked 4 years ago and I don't want to call such a function thousand times. I hope there should be a better solution now. 回答1:

## Infinite Loop During Calculation of Power of Big Integers Java

问题 I have been staring at this code and cannot figure out what is wrong with it, maybe a fresh pair of eyes could help. public static BigInteger powerOfBigInteger (BigInteger base, BigInteger power){ if (power == BigInteger.valueOf(0)){ return BigInteger.valueOf(1); } if (power == BigInteger.valueOf(1)){ return base; } BigInteger x = BigInteger.valueOf(1); while (x != power ){ base.multiply(base); x.add(BigInteger.valueOf(1)); System.out.println(x + " " + power); return base; } return base; I

## Matlab - matrix to the power of 2

问题 In Matlab , I have typed the following commands: >> a = [1 2; 3 4] a = 1 2 3 4 When I tried the command a^2 , I got the following: >> a^2 ans = 7 10 15 22 I was actually expecting to get: ans = 1 4 9 16 In other words, I was expecting to get each element of the matrix to be raised to 2 . Why was the result as shown above? Thanks. 回答1: In MATLAB, all single-character operators are matrix operators. So, you are using the matrix power, e.g., a^2 == a*a if you want to square each element , you'll

## R Exponent Produces NaN

问题 I am running into an issue when I exponentiate floating point data. It seems like it should be an easy fix. Here is my sample code: temp <- c(-0.005220092) temp^1.1 [1] NaN -0.005220092^1.1 [1] -0.003086356 Is there some obvious error I am making with this? It seems like it might be an oversight on my part with regard to exponents. Thanks, Alex 回答1: The reason for the NaN is because the result of the exponentiation is complex, so you have to pass a complex argument: as.complex(temp)^1.1 [1]

## How to make exponents in the SwiftUI

问题 I found the new method in SwiftUI to allow you to create an exponent. Here what I write code is based in SwiftUI on the Swift Playground. import SwiftUI import PlaygroundSupport struct V: View { var body: some View { HStack { Text("8") Text("2\n").font(Font.system(size: 10)) } } } PlaygroundPage.current.setLiveView(V()) The \n allows you to break the line and make a new line in UIKit. However, SwiftUI gives you to make a top while break line and make a new line inside the second Text. You can

## How to make exponents in the SwiftUI

## How do i make my answer print correctly in java

问题 I am having an issue where when my program prints (base)^0= , it doesn't print the answer (1) (I shorend down the out put examples as I'm only having an issue with the first line of the output) expected output: 2^0=1 2^1=2 2^2=2*2=4 2^3=2*2*2=8 2^4=2*2*2*2=16 actual output: > 2^0= > 2^1=2=2 > 2^2=2*2=4 > 2^3=2*2*2=8 > 2^4=2*2*2*2=16 code: else if(option == 2){ base = Input.nextInt(); for(int i = 0; i<10; i+=1){ System.out.print(base+"^"+i+"="); for(int j = 0; j < i; j+=1){ if(j != i -1){

## How to write a loop that calculates power?

问题 I'm trying to write a loop that calculates power without using the pow() function. I'm stuck on how to do that. Doing base *= base works for even powers upto 4, so there is something totally weird that I can't seem to figure out. int Fast_Power(int base, int exp){ int i = 2; int result; if(exp == 0){ result = 1; } if(exp == 1){ result = base; } else{ for(i = 2; i < exp; i++){ base *= base; result = base; } } return result; } 回答1: First off, I agree it was probably a mistake to use base *=

## R: How can I calculate large numbers in n-choose-k? [duplicate]

问题 This question already has answers here : How would you program Pascal's triangle in R? (2 answers) How to work with large numbers in R? (1 answer) Closed 3 years ago . For a class assignment, I need to create a function that calculates n Choose k. I did just that, and it works fine with small numbers (e.g. 6 choose 2), but I'm supposed to get it work with 200 choose 50, where it naturally doesn't. The answer is too large and R outputs NaN or Inf, saying: > q5(200, 50) [1] "NaN" Warning