# Documentation

### This is machine translation

Translated by
Mouseover text to see original. Click the button below to return to the English version of the page.

Note: This page has been translated by MathWorks. Please click here
To view all translated materials including this page, select Japan from the country navigator on the bottom of this page.

# besseli

Modified Bessel function of first kind

## Syntax

```I = besseli(nu,Z) I = besseli(nu,Z,1) ```

## Description

`I = besseli(nu,Z)` computes the modified Bessel function of the first kind, Iν(z), for each element of the array `Z`. The order `nu` need not be an integer, but must be real. The argument `Z` can be complex. The result is real where `Z` is positive.

If `nu` and `Z` are arrays of the same size, the result is also that size. If either input is a scalar, it is expanded to the other input's size.

`I = besseli(nu,Z,1)` computes `besseli(nu,Z).*exp(-abs(real(Z)))`.

## Examples

collapse all

Create a column vector of domain values.

`z = (0:0.2:1)';`

Calculate the function values using `besseli` with `nu = 1`.

```format long besseli(1,z)```
```ans = 0 0.100500834028125 0.204026755733571 0.313704025604922 0.432864802620640 0.565159103992485 ```

Define the domain.

`X = 0:0.01:5;`

Calculate the first five modified Bessel functions of the first kind.

```I = zeros(5,501); for i = 0:4 I(i+1,:) = besseli(i,X); end```

Plot the results.

```plot(X,I,'LineWidth',1.5) axis([0 5 0 8]) grid on legend('I_0','I_1','I_2','I_3','I_4','Location','Best') title('Modified Bessel Functions of the First Kind for v = 0,1,2,3,4') xlabel('X') ylabel('I_v(X)')```

## More About

collapse all

### Bessel’s Equation

The differential equation

`${z}^{2}\frac{{d}^{2}y}{d{z}^{2}}+z\frac{dy}{dz}-\left({z}^{2}+{\nu }^{2}\right)y=0,$`

where ν is a real constant, is called the modified Bessel's equation, and its solutions are known as modified Bessel functions.

Iν(z) and Iν(z) form a fundamental set of solutions of the modified Bessel's equation. Iν(z) is defined by

`${I}_{\nu }\left(z\right)={\left(\frac{z}{2}\right)}^{\nu }\sum _{\left(k=0\right)}^{\infty }\frac{{\left(\frac{{z}^{2}}{4}\right)}^{k}}{k!\Gamma \left(\nu +k+1\right)},$`

where Γ(a) is the gamma function.

Kν(z) is a second solution, independent of Iν(z). It can be computed using `besselk`.

## See Also

Was this topic helpful?

#### The Manager's Guide to Solving the Big Data Conundrum

Download white paper